B 417800 DUPL ARTES 1837 יד VERITAS LIBRARY SCIENTIA OF THE UNIVERSITY OF MICHIGAN E-PLURIBUS-UNUM TUE BOR 指​11. SI-QUAERIS PENINSULAM AMOENAMU CIRCUMSPICE DEPARTMENT OF ENGINEERING Engineering です ​TJ 275 F31 1900 FOLDER.DIN A LIBRARY OF STEAM ENGINEERING 115852 BY JOHN FEHRENBATCH, M. E. EX-PRESIDENT INTERNATIONAL ASSOCIATION OF MECHANICAL ENGINEERS; PAST PRESIDENT NATIONAL ASSOCIATION OF STATIONARY ENGINEERS; AND UNITED STATES SUPER- VISING INSPECTOR OF STEAM VESSELS FOR THIRTEEN YEARS. CINCINNATI: THE OHIO VALLEY COMPANY. MICH UNIV PARY 1900. PENGECALITA COPYRIGHT, 1894, BY JOHN FEHRENBATCH. PRESS OF His VALLE CINCINNATI INTRODUCTION. In the production of this work the great aim of the author has been to produce a book that would embrace the entire field of the science of steam engineering, and to present that important science, in all of its various branches, in its simplest possible form, so as to bring it within the understanding of engineers of ordinary education, and thus give them an equal chance in competition with those who are more fortunately situated with regard to the possession of knowledge in the higher branches of learning. To accomplish this desirable object, the rules and formulæ appertaining to the mathematics and geometry of steam engineering have been reduced to their utmost sim- plicity, and brought within the easy comprehension of every engineer who understands the simple rules of ordinary arithmetic. It has also been the purpose and aim of the author to embrace, in this same volume, all the information necessary to enable engineers to pass a most successful and rigid examination by inspection officers; to enable engineers to fill positions of inspectors; to become experts in making evaporative and calorimeter boiler tests, and indicating steam engines, adjusting valves and valve-gear, and draughting plans and specifications for the construction of steam boilers; in short, to become proficient in all the arts and sciences of steam engineering-stationary, locomotive and marine. The measure of success that has attended the efforts of the author toward the accomplishment of these objects, must be determined by those who follow the profession of steam engineering. To them, then, this volume is respectfully dedicated. THE AUTHOR. CONTENTS. CHAPTER I. MATHEMATICS OF STEAM ENGINEERING. The basis of the theory and practice of steam engineering-Familiarity with the rules of arithmetic necessary to complete the education of engineers-Famil- iarity with the mathematics of steam engineering necessary to pass examination before inspection officers-Thorough knowledge of mathematics of steam engi- neering necessary to attain a high position in the profession and command the highest standard of wages-Familiarity with the ordinary terms of mathematics necessary to the acquirement of mathematical knowledge-The ordinary terms of mathematics simplified and explained-The use of the constants 3.1416 and .7854 -How the constant .7854 is obtained, explained and illustrated-How to deter- mine the circumference, diameter, and area of a circle-Decimal parts of square inches, explained and illustrated-Multiplication and explanation of decimal frac- tions-The correct diameter of a safety-valve-How to determine the required. diameter of a valve for a given area-How to find the diameter required for a valve of any given circumference-The annexation of ciphers in division-How to reduce common fractions to decimal fractions-How to determine the number of cubic inches in a cylindrical weight-How to determine the number of pounds in a cast-iron weight mathematically-How to determine the number of pounds in a wrought-iron lever of safety-valve mathematically-How to determine the weight of a valve spindle mathematically-How to determine the weight of a valve mathematically-How to determine the weight of a cast-iron ball mathe- matically, Pages 33-47 CHAPTER II. THE SAFETY-VALVE MATHEMATICALLY CONSIDERED. Adjustment of a safety-valve by the aid of a diagram-Names of different parts of the lever-The center of gravity of a lever explained, and how determined -The amount of pressure produced on the valve seat by the weight of the lever, illustrated-How to determine the pressure required to raise a given valve-How to determine the area of the valve to complete adjustment of valve for determin- ing pressure required to raise it-The adjustment completed for determining pressure required to raise it-Rules for determining pressure required to raise the valve-The various operations required to determine the pressure required to raise a valve, illustrated-How to determine the distance on the lever to place a given weight to allow the valve to rise at a given pressure, with various operations for determining the same, illustrated-How to determine the amount of weight to be placed a given distance on the lever to allow the valve to rise at a given press- ure, illustrated and explained-How to determine the required length of the sbort arm of the lever to allow the valve to rise at a given pressure, with various 8 CONTENTS. operations and rules for determining the same-How to determine the diameter of a valve to allow it to rise at a given pressure, explained and illustrated-How to determine the pressure required to raise a valve without weighing any of its parts-How to determine the weight of the lever, weight, valve, and spindle mathematically-Simple method of adjusting safety-valves by the use of a bar of iron of uniform section, illustrated-How to determine the pressure produced on the valve seat by the valve spindle, lever, and weight-How to graduate a safety- valve lever-Proper area of safety-valves-How to determine the proper diameter of a safety-valve, Pages 48-94 CHAPTER III. VALVES GENERALLY. How to determine the area of opening produced by the lift of a flat-seated valve-How to determine the size of a square hole, the area of opening produced by the lift of a flat-seated valve will make-How to determine the size of a round hole, the area of opening produced by the lift of a flat-seated valve will make- Bevel-seated safety valves-How to determine the area of opening produced by the lift of any bevel-seated valve when the lift is not above the seat-How to determine the area of opening produced by the lift of a bevel-seated valve when the lift is above the seat-Sines and cosines, illustrated and explained-How to determine the area of opening produced by the lift of a bevel-seated valve having its seat made at an angle of 45° to the center line of the valve's axis, without the use of sines or cosines, illustrated-Valve proportions-To determine the lift of a required flat-seated valve to produce an area of opening equal to the area of open- ing produced by the given lift of a given flat-seated valve-How to determine the lift of a required valve to produce an area of opening equal to the area of the diameter of a given valve, or area of cross section of the diameter of a given pipe-How to determine the required diameter of a required flat-seated valve to have a given lift to produce an area of opening equal to that produced by the given lift of a given valve-How to determine the lift of a bevel-seated valve when the area of opening required and the diameter of the valve are given-How to determine the diameter of a required bevel-seated valve to have a given lift that will produce an area of opening equal to that produced by the lift of a given valve, Pages 95-111 CHAPTER IV. BOILER PLATE. Quality of material necessary to render boilers safe - The necessity for testing all boiler plate - Plate made with coupon attached for testing, illus- trated-Coupons, how prepared for testing, illustrated-How to make meas- urements of coupons before and after breaking-Tensile strength, how deter- mined — Ductility of boiler plate, how determined-Percentage of ductility re- quired, Pages 112-116 - CHAPTER V. STEAM BOILERS. How to determine safe-working pressure for single-riveted longitudinal seams-How to determine bursting pressure for single riveting-How to deter- CONTENTS. 9 mine required thickness of plate-How to determine the required diameter for single-riveted boilers when the working pressure, thickness, and tensile strength of material are given-Tensile strength of material required for single-riveted boilers-Safe-working pressure for double riveting-Simple method for determin- ing safe-working pressure for double riveting-Bursting pressure for double riveting Simple method for determining bursting pressure-Thickness of plate for double-riveted boilers-Required diameter for given pressure-Required tensile strength of plate for given pressure, given thickness of material, and given diam- eter of a double-riveted boiler-Strain on longitudinal seams-Strain longitudi- nally on each inch in the circumference of the boiler-Simple method for deter- mining longitudinal strain-How to determine the weight of steam discharged through a pipe or orifice of given diameter, Pages 117-131 CHAPTER VI. RIVETED JOINTS. Distance of rivet holes from edge of plate-Shearing strength of rivets-Dis- tance between centers of rows of rivet holes for chain riveting-Distance between centers of rivet holes for zigzag riveting-Diameter of rivet holes for single- riveted lap joints, iron plates and iron rivets-Diameter of rivet holes for double- riveted lap joints, iron plates and iron rivets-Diameter of rivet holes for single- riveted lap joints, steel plates and steel rivets-Diameter of rivet holes for double- riveted lap joints, steel plates and steel rivets-Pitch of rivet holes for single- riveted lap joints, iron plates and iron rivets-Pitch of rivet holes for double-riv- eted lap joints, iron plates and iron rivets-Pitch of rivet holes for triple-riveted lap joints, iron plates and iron rivets-Pitch of rivet holes for single-riveted lap joints, steel plates and steel rivets-Pitch of rivet holes for double-riveted lap joints, steel plates and steel rivets-Pitch of rivet holes for triple-riveted lap joints, steel plates and steel rivets-Strength of single-riveted lap joints, iron plates and iron rivets-Strength of double-riveted lap joints, iron plates and iron rivets Strength of triple-riveted lap joints, iron plates and iron rivets Strength of single-riveted lap joints, steel plates and steel rivets-Strength of double- riveted lap joints, steel plates and steel rivets-Strength of triple-riveted lap joints, steel plates and steel rivets-Butt-strap joints-Pitch of rivet holes for single- riveted double butt-strap joints, steel plates, steel butt-straps and steel rivets- Pitch of rivet holes for double-riveted double butt-strap joints, steel plates, steel butt-straps and steel rivets-Pitch of rivet holes for triple-riveted double butt- strap joints, steel plates, steel butt-straps and steel rivets-Strength of single- riveted double butt-strap joints, steel plates, steel butt-straps and steel rivets— Strength of double-riveted double butt-strap joints, steel plates, steel butt-straps and steel rivets-Strength of triple-riveted double butt-strap joints, steel plates, steel butt-straps and steel rivets-Table of riveted joints, Pages 132-166 CHAPTER VII. BOILER FLUES. Experiments made by Sir WILLIAM FAIRBAIRN, as to the influence of length of flue on the collapsing pressure-Theory in the construction of boiler flues not to be followed-Rules should be based on what has been found safe in practice- Safe-working pressure-Required thickness of material-Required diameter for 10 CONTENTS. given thickness, length and pressure-Required length based on thickness, diam- eter and pressure-Collapsing pressure, CHAPTER VIII. PUMPS. Pages 167-174 Horse power required to elevate water-How to determine the horse power required-Feed pumps Should be of sufficient capacity for all emergencies-Size should be regulated by the evaporative capacity of the boiler under all circum- stances To compute the dimensions of feed pumps-Rule based on consumption of coal and evaporation of water-The rule in detail-Examples, Pages 175-183 CHAPTER IX. SMOKE STACK PROPORTIONS. Area of stack-Diameter of square stack-Diameter of round stack-Height of stack-Percentage of air space required for grate surface-How to determine the required height-Passage of furnace gas from furnace to stack-Construction of breeching Height of stack for bituminous lump coal-Height of stack for bitum- inous nut coal-Height of stack for bituminous slack coal, CHAPTER X. SPECIFICATIONS AND PLANS FOR TUBULAR BOILERS. Pages 184-187 Specifications-Plans-Horizontal tubular boiler (front and back view), illus- trated-Eclipse man hole, illustrated-Safety water column, illustrated-Plans for single setting, with full arch front, with dimensions-Plans for single setting, with half arch front, with dimensions-Plans for a different single setting, with full arch front, with dimensions-Plans for single setting, with half arch front, with dimensions-Different plans for single setting, with half arch front, with specifi- cations and dimensions--Plans (different styles) for single setting, with half arch front, with specifications and dimensions-Plans for full arch front and settings for same boiler, with specifications and dimensions-Plans for double setting, with half arch front, with specifications and dimensions-Plans for double setting, with full arch front, with specifications and dimensions-Plans for triple setting, with half arch front, with specifications and dimensions-Plans for triple setting, with full arch front, with specifiations and dimensions-Table of materials for tubular boilers-Plan and elevation of right-hand engine with tubular boilers-Plan and elevation of left-hand engine with tubular boilers-How to draft plans and speci- fications for boilers-How to determine strain on each longitudinal inch in a boiler-Actual strength of a strip of boiler plate one inch in width of given thick- ness-How to determine required diameter of boiler-Heating surface required per horse power-How to determine required thickness of material-How to deter- mine required tensile strength of plate, Pages 188-222 CHAPTER XI. EVAPORATIVE TESTS OF STEAM BOILERS. Heat units-Temperature and heat units not synonymous terms-The ther- mometer registers the sensible heat, but does not register the amount of work CONTENTS. 11 done in converting water into steam-Evaporation, heat units, and temperature fully explained-The heat unit the standard for measuring work performed-How to determine the heat units in any given weight of steam-How to determine the factor of evaporation-How to determine the equivalent evaporation of water to steam from and at a temperature of 212° Fahrenheit-How to determine the per- centage and moisture of combustible matter in coal-How to reduce evaporative tests to the standard evaporation of from and at 212° Fahrenheit-Table of prop- erties of saturated steam, Pages 223-233- CHAPTER XII. CALORIMETER TESTS. The making of calorimeter tests requiring great care and accuracy-The- rules and formulæ for making calorimeter tests reduced to the utmost simplicity -Calorimeter tests indispensable in determining the real efficiency of steam boilers in making evaporative tests-Object of calorimeter tests is to determine the percentage of water carried off by the steam-Illustration showing construc- tion of simple device for making calorimeter tests, with full instructions-How to determine the percentage of water in steam-Table of properties of water from 32° to 212° Fahrenheit-United States Navy evaporative and calorimeter tests. with coil boilers, illustrated and explained-Tables showing details of tests, with formula employed in determining the results-Evaporative and calorimeter test. of land boilers-Quality of steam determined by a Barrus universal calorimeter -How to determine the weight of steam discharged per hour by the orifice in the heat gauge of the calorimeter-Algebraic formula reduced to common arith- metic Percentage of moisture in steam, how determined-How to compute the moisture in the steam after leaving the drip box, and indicated by the ther- mometer of the heat gauge-Formula and rule for determining the same-Moist- ure in coal, how determined-Calorimeter experiments-Observed data and com- puted results-How to determine capacity results-How to determine economic evaporation-How to determine the rate of evaporation-The Centennial standard of horse power for a boiler-How to determine the commercial horse power-The Centennial standard reduced to equivalent evaporation as from and at 212° Fah- renheit Illustrations of the various types of steam boilers-Illustrations of the Hawley down draft furnace-Temperature of boiler furnaces-How to compute the heat of a boiler furnace, Pages 234-276 CHAPTER XIII. SPEED OF PULLEYS AND GEARING. PULLEYS. How to determine the required diameter-How to determine the speed- How to determine the speed of driven pulley in a combination of driving and. driven pulleys, illustrated-How to determine the speed of any given pulley in a combination of driving and driven pulleys, illustrated-How to determine the required diameter of any intermediate pulley in a combination of driving and driven pulleys, illustrated--How to determine the diameter of an intermediate driven pulley in a combination of driving and driven pulleys, illustrated-General rule for determining the diameter and speed of pulleys in a series-How to deter- mine the respective diameters of intermediate pulleys in a series, illustrated- How to determine the respective diameters of intermediate driving and driven 12 CONTENTS. pulleys in a series, illustrated-Simple rule for determining the required diameter of a given pulley in a series on a given shaft. GEARING. Rules relating to gearing similar to those of pulleys, with one exception— Gears generally required to run a given distance apart-Calculations made with gearing as to speed must be based on diameter at pitch line, or the number of cogs wheels are to contain-How to determine the required diameters of gear wheels required to run a given distance apart from center to center of shafts-A simple method for determining the required diameter of the driving wheel-How to determine the required diameter of the driven wheel, CHAPTER XIV. ELECTRICITY. Pages 227-297 Importance of engineers studying this newly developed science-Definitions of electrical terms-Electrical units-The ohm and the volt-The unit of resist- ance (the ohm) represented by a column of mercury-The current produced by a volt through an ohm, called an ampere-Coulomb the unit of quantity-Farad the unit of capacity-Electrical resistance-Electro motive force-The dynamo, differ- ent parts illustrated and explained-How to build a dynamo to produce a given volt machine-How to wind the armature-How to handle the armature-How to make and connect the commutator-How to connect the armature-How to wind the field magnets-How to insulate brush holders-How to make the brushes-How to operate the machine-How to increase the power of the machine by different winding-How to connect the lamps, multiple method-How to connect the lamps, series method, Pages 298-306 CHAPTER XV. THE STEAM ENGINE INDICATOR. Employment of the indicator as an instrument of communication with the interior of the steam engine while in operation-Deficiencies, inaccuracies and defects communicated to the engineer by means of the indicator diagram-De- scription of construction and use of the indicator-Management and care of the indicator-How to indicate a steam engine-How to attach the indicator, illus- trated-Reducing lever-Description of lazy tongs, with illustrations-Description of the pantagraph-Manner of taking diagrams-Essential features of the indicator diagram, illustrated-Regulating the exhaust-A well-formed diagram-How dis- arrangement of the valve is revealed-Pipes and passages for the admission and exhaust of steam-Examination of valves of engines long in use-Usefulness of the indicator-Tables showing percentage of loss by cylinder condensation-Com- bining diagrams from compound engines-Engravings showing how to perform the operation-Volume of high and low-pressure cylinders--How to compute the horse power of engines from indicator diagrams, with illustrations showing opera- tion-Planimeters and their use, illustrated, Pages 307-334 • CHAPTER XVI. UNITED STATES MARINE BOILER INSPECTION AND ENGINEERS' LICENSE LAWS.. Students of steam engineering advised not to commit examination questions to memory-Such so-called examination questions of no assistance to engineers in CONTENTS. 13 passing examinations-The importance of becoming thoroughly educated in the science of steam engineering to pass required examinations-The enactment of laws for the protection of life and property-Ignorant and incompetent engineers excluded from the management of steam plants-Required qualifications of United States inspectors of boilers-Required qualifications of marine engineers. -Strength of steam boilers-Bursting pressure of single-riveted boilers-Bursting pressure of double-riveted boilers-Safe-working pressure required by the United States laws-Safe-working pressure for single-riveted boilers-Actual strength of boiler plate-Strain on each inch in the length of the boiler tending to sever the material in circular directions-Thickness of material required-Safe-working pressure of double-riveted boilers with holes drilled-Simple rule for determining safe-working pressure for double-riveted boilers-Tensile strength of boiler plate- Ductility of boiler plate-To determine the ductility of boiler plate-Safe-working. pressure based upon construction of any given marine boiler-Form of test piece for determining ductility and tensile strength of iron plate-United States Gov- ernment requirements as to tensile strength, ductility and thickness of plates, and manner of constructing boilers-Percentage of ductility of boiler plate, tensile strength and safe-working pressure-Form of test piece determining elongation and tensile strength of steel plate-Required length of reduced section of test piece-How to determine the percentage of elongation of test piece-Special test. of steel plate under the rules of the British Board of Trade, British Lloyds and Bureau Veritas-How to determine the elongation-Bumped heads for boilers- How to determine the safe-working pressure-How to determine the safe-working pressure by a simple rule-How to determine the thickness of material required for any given radius to which the head may be bumped-How to determine the thickness of material required by a simple rule-How to determine the radius to which the head is required to be bumped-Hydrostatic pressure for marine boilers. -How to determine the hydrostatic pressure for single-riveted boilers-How to determine the hydrostatic pressure for double-riveted boilers-Reduction of work- ing steam pressure on account of deterioration-Thickness of material for marine boilers limited-Thickness of material required for single-riveted boilers-Work- ing steam pressure based on the hydrostatic pressure-Table of pressure allowable on marine boilers-Rules relating to the construction of flat surfaces in steam boilers-Engineers required to calculate strain on stays, stay bolts and braces- How to determine area of surface stayed by a brace or stay bolt, with illustrations -Construction of flat surfaces and steam pressure allowable-How to determine required diameter of stay bolts-How to determine the pitch of stay bolts based on diameter and steam pressure-How-to determine the required thickness of plate based on pitch of bolts and steam pressure-How to determine the steam pressure based on diameter and pitch of bolts-How to determine the thickness of plate and pitch of bolts-Stay bolts exceeding 14 inches and not exceeding 23 inches in diameter allowed a strain of 8000 pounds to the square inch of cross- section-How to determine the diameter of stay bolts-How to determine the steam pressure based on diameter and pitch of bolts-Steel stay bolts exceeding 2 inches in diameter allowed a strain of 9000 pounds to the square inch of cross section-How to determine the diameter of stay bolts-How to determine the steam pressure based on diameter and pitch of bolts-Fire pumps and required. dimensions-How to determine the size of pump required-How to determine the horse power of marine engines-How to determine the mean effective pressure- -Government regulations relating to safety-valves-How to determine the required area of a lever safety-valve-How to determine the required diameter of 14 CONTENTS. a lever safety-valve-Spring-loaded safety-valves and government regulation. relating to the same-How to determine the required area of a spring-loaded safety-valve-How to determine the required diameter of a spring-loaded safety- valve-Practical adjustment of safety-valves-How to determine the pressure per square inch required to raise a valve, with illustrations-How to determine the distance from the fulcrum a given weight is to be placed on the lever to allow the valve to rise at a given pressure, with illustrations-How to determine the size of a cubic cast-iron weight required to be placed a given distance from the fulcrum to allow the valve to rise at a given pressure, with illustrations-How to deter- mine the pressure required to raise a valve having a spherical weight of given diameter attached to a lever a given distance from the fulcrum, with illustrations -How to determine pressure required to raise a valve with cast-iron conic weight of given dimensions placed a given distance on the lever from the fulcrum, with illustrations-How to determine the pressure required to raise a valve with two or more cast-iron weights of given number of pounds in each placed a given dis- tance from the fulcrum, with illustration-United States lever safety-valve, illus- trated-Specifications for United States lever safety-valve-Spring-loaded valve, illustrated and explained-How to determine the required diameter of steam drum legs for externally-fired marine boilers-Boiler flues, with instructions in regard to their construction-How to determine the safe-working pressure for cor- rugated and ribbed flues-How to determine the thickness of material required for corrugated flues required to carry a given pressure-How to determine the safe-working pressure for plain cylindrical furnace flues-Furnace flues with Adamson and angle-iron rings, explained and illustrated-How to dermine the safe-working pressure for plain cylindrical furnace flues-How to determine thick- ness of material required for a plain furnace flue of given diameter, given length, and required to carry given steam pressure per square inch-How to determine the diameter required of a furnace flue of a given length, given thickness of material and given steam pressure per square inch-How to determine required length of a furnace flue, or length between the flanges of such a flue made in sections, when the thickness of material, steam pressure per square inch, and diameter of the flue are given-Half-round iron strengthening rings for flues, illustrated and explained-How to determine the required diameter of a half-round iron strength- ening ring-How to determine the working steam pressure allowable for flues hav- ing half-round iron strengthening rings-How to determine thickness of material required for flues reinforced with half-round iron strengthening rings-How to determine the required distance from center to center of half-round iron strength- ening rings-How to determine the required diameter for a flue supported with half-round iron strengthening rings-Riveted and lap-welded flues made in sections, and having a diameter exceeding 6 inches and not exceeding 40 inches— Table of pressure, thickness of material, diameter and length of sections of riveted and lap-welded flues exceeding 6 inches and not exceeding 40 inches in diameter -How to determine safe-working pressure for any flue over 10 and not over 40 inches in diameter-Thickness of material required for flues and tubes not other- wise provided for, with table ot dimensions-Working steam pressure allowable on lap-welded flues not exceeding 6 inches in diameter-Steam pressure allow- able for lap-welded flues exceeding 6 inches in diameter and not exceeding 16 inches in diameter and not exceeding 18 feet in length-Lap-welded and riveted flues exceeding 6 inches and not exceeding 16 inches in diameter and not ex- ceeding 18 feet in length, required to carry a steam pressure exceeding 60 pounds and not exceeding 120 pounds-Lap-welded and riveted flues exceeding CONTENTS. 15 6 inches and not exceeding 40 inches in diameter, required to carry a working pressure exceeding the pressure prescribed in the table, to be governed by Section 15, of Rule 2-Thickness of material for copper steam pipes, and how to deter- mine the same-United States marine examination questions and answers for engineers and inspectors, Pages 335-470 CHAPTER XVII. A UNITED STATES BATTLE SHIP. General arrangement of machinery of a triple screw-protected cruiser of 7350 tons-Boilers and machinery designed to propel the vessel at a maintained speed of 21 knots per hour-Machinery capable of developing 21,000 indicated horse power-The power of the vessel divided into three parts, the power of each being developed by a triple expansion engine driving its own screw-In case of accident to any engine the vessel would have a reserve power to drive her at a good rate of speed-Almost impossible to conceive a combination of circum- stances that would render the vessel entirely helpless-Clutch coupling leading feature of design-Either propeller can be disconnected from its engine and left free to revolve-Boilers designed for high steam pressure, and capable of furnish- ing power to drive the vessel 15 knots per hour with one engine, and 19 knots per hour with two engines-Reasons for adopting plan of triple screws-Advan- tages and disadvantages of triple compared with two-screw propellers-Three screws placed abreast impair the efficiency of each other-Side screws in this vessel placed fifteen feet forward of the center screw-Arrangement of screws, described and illustrated-Construction of screws and arrangement for setting blades to any pitch found most effective, described and illustrated-Description of main engines, with illustrations showing plan-Sectional elevation and end view-Air pumps and condensers, described and illustrated-Main and auxiliary boilers, described and illustrated-Forced draught employed-Feed water, how supplied to boilers, Pages 471-482 CHAPTER XVIII, WESTERN RIVER STEAMBOAT PRACTICE Specifications for main and auxiliary engines, boilers and attachments for a Western river stern-wheel steamer-Plans of construction for non-condensing engine for a Western river side-wheel steamer-Feed water heaters, illustrated- Fire-box tubular boilers, described and illustrated-The Sweeney marine engine, explained and illustrated - Detailed description and explanation of the valve gear-How to adjust the valve gear and how to set the valves-The Frisbie improved marine engine, with illustrations showing construction-Full descrip- tion of valves and valve gear-Cams in use on Western river engines-How to lay off a full-stroke cam, described and illustrated-How to set a full-stroke cam-How to lay off a cut-off cam, described and illustrated-How to set the cut-off cam-The Rees adjustable cut-off marine engine-The adjustable cut off explained and illustrated-The movements of the full-stroke cam, described and illustrated-How to set the full-stroke cam, and how to put the engine on the dead center-Specifications and plans for Western river steamboat boilers and attachments, . Pages 483-527 16 CONTENTS. CHAPTER XIX. THE CORLISS ENGINE. The first of all high class engines invented by GEORGE H. CORLISS-The skill and genius of the world have approximated but never excelled the Corliss engine -Previous to the advent of the Corliss engine the common D slide invented by Watt was in common use on land engines for the admission and exhaust of steam-The speed of the engine was regulated by wire-drawing steam through a valve, in the supply pipe, operated by a pendulum governor-The difficul- ties overcome by the invention of Corliss-The leading features of the Corliss engine―The employment of four valves for controlling the admission and ex- haust of steam-The employment of a rotating valve operated by mechanism out- side of the steam space-The employment of a single eccentric for operating the four valves-A simple and effective method for detaching the steam valves from the driving mechanism to produce any required expansion of steam in the cylin- der-Direct connection with the steam valve mechanism to produce the required ratio of expansion of steam in the cylinder adjusting the power of the engine to correspond with the amount of work to be done-Adjustments to throw the least possible work on the regulating mechanism-Mechanism for operating the valves, explained and illustrated-The dash pot, how attached to pedestals under the cylinder-Limitation of engine speed-Friction of moderate speed engines com- pared with high speed engines-The Columbian Corliss engine, explained and illustrated-Improvements in the construction of the cylinder, explained and illustrated-Objections to the use of wood lagging for the top of Corliss engines- Steam and exhaust valves, explained and illustrated-Steam and exhaust chests of the Columbian Corliss compared with those of other Corliss engines-Cylinder heads, how fitted-Frames of Corliss engines, compared and illustrated-Various forms of slides, explained and illustrated-Piston packing, explained and illus- trated-How to adjust the cross head, explained and illustrated-Connecting rods, illustrated and explained--Valves and valve rods, explained and illustrated—The governor illustrated, with instructions how to adjust the same for various speeds- The crab claw and its functions, explained and illustrated-The bonnet on the valve-motion side, explained and illustrated-The-construction, operation and adjustment of the dash pot and connections, explained and illustrated-The pil- low blocks, explained and illustrated-Description of what constitutes a com- pound engine-Condensers and their functions, with illustration of air pump and condenser-What constitutes first-class engine performance, with table for high and low pressure engines-Economy in the generation and use of steam-Explan- ation, comparison and illustration of slide valve and Corliss engine diagram-- Feed-water heaters, their use and economy, with table of factors of evaporation and table showing percentage of saving in fuel by the use of heaters-How to put an engine on the dead center by various methods, explained and illustrated-How to set the valves of Corliss engines, with complete instructions for making adjust- ments of various parts of the valve gear, illustrated, Pages 528-569 CHAPTER XX. THE ATLAS AUTOMATIC ENGINE. Novel features, construction and operation of the Atlas engine, explained and illustrated-In engines of moderate size the slide valve almost universally employed as a means to effect the distribution of steam-For large engines and CONTENTS. 17 for high pressures balancing of valve necessary when required to regulate the motion of the engine-The setting of the engine and building of foundations, illustrated-Material required for foundations-Specifications for engines and attachments-Dimensions for foundations-Construction of the valves, illustrated and explained-The shaft governor and its details, explained and illustrated- How to set the valves of the engine and how to adjust the various working parts of the governor-Cycloidal engines, simple and compound, illustrated and described in detail, Pages 570-590 CHAPTER XXI. THE BROWN ENGINE. A distinct type of American steam engines-A most important combination of ingenious mechanism-The importance of engineers and students of steam engineering becoming familiar with the construction and operation of the Brown engine-Full description of the valves, their construction and operation, with illustrations-Details of valve gear explained and illustrated-The governor, its mechanism and operation, illustrated and explained-How to set the valves and adjust the governor of the Brown engine-Performance of the engine, illustrated with indicator diagrams-Steam and exhaust ports, how to determine the proper area and dimensions, Pages 591-605 CHAPTER XXII. THE BUCKEYE ENGINE. Interesting features of the Buckeye automatic high speed engine, explained and illustrated-How to put the engine together-Shop marks-Main eccen- tric-The governor containing wheel-The eccentric rods-How to make the adjustments independent of shop marks--Difference between the main valve and. the common slide valve-How to set the main valve-How to set the eccentric- The amount of compression-How to adjust for the amount of compression- Adjustment for equalizing compression-How to set the governor wheel-Cut off equalization-Adjustment for equalization-Choice of methods of adjustment- How to determine point of cut off of valve closure-Adjustment by indicator cards-Distortion from elasticity of the indicator cord-Test diagrams, showing what adjustments are necessary to make-The governor-Names of details, and explanations as to the functions of each-Close regulation-How to secure close regulation-Frictional disturbance-Centripetal friction-How to change engine speed-How to determine required weight of weighted lever-Minor changes of speed increase-Different methods of adjusting for same-To decrease speed- Reversal of direction of motion of the engine-Valve adjustment, illustrated and explained-Rudiments and analysis of indicator diagrams, from automatic and common slide-valve engines-The steam or initial pressure line-Diagrams from throttling engines-The point of cut off-The expansion curve-The exhaust line-The back pressure line-The point of exhaust closure-The compression curve-The lead or admission line-The atmospheric line-The vacuum line— The clearance line-Theoretical expansion curve-The clearance of Buckeye engines compared with four-port and single valve high speed engines-Mathe- matical method of drawing the isothermal curve-Geometric method for deter- mining the theoretical curve in indicator diagrams-The adiabatic curve-How to trace the adiabatic curve, illustrated by two different methods-Theoretical 18 CONTENTS. clearance-Horse power calculations-The mean effective pressure determined by ordinates in the indicator diagram-Spacing ordinates-A quick method of esti- mating mean effective pressure-Compound engines-How to reduce the mean effective pressure of the low pressure cylinder to an equivalent pressure for the high pressure cylinder-Theoretical water consumption-How to calculate the theoretical consumption of water, explained and illustrated by indicator dia- grams Slide valve performance-Impossibility of equalizing all the functions of the common slide valve-To determine the points of cut off and exhaust clos- ure, explained and illustrated by diagrams-To determine the consequence of other events of setting the same valve for equal cut off, illustrated geometrically- To locate the exhaust closure-To find the valve laps required for given results- Condensed form for determining slide valve functions, illustrated geometric- ally-The Buckeye tandem and cross compound engines-Ratio of cylinders of compound engines, Pages 606-655 CHAPTER XXIII. THE PORTER-ALLEN ENGINE. The Porter-Allen engine peculiar to itself-Differs distinctly from other en- gines-Its central feature a link operated by a single eccentric-Eccentric placed on shaft in same position with the crank-The lead of valve adjusted by other means-The shaft and eccentric one piece as shown by illustrations-The Link- The Link motion, illustrated and explained-Distinguishing features of the valve motion-Exhaust valves-Differential valve movement-Adjustable pressure plates-How to set the valves-How to set the engine―The foundation—The bed- plate-The Shaft-To line the engine with shaft placed at a higher or lower level, with diagram and instructions-How to line the engine with a shaft coupled direct- Action of reciprocating parts-The unit of centrifugal force-Compound condensing engines-Compound expansion-Conversion of heat into work-How to determine the number of heat units expended per horse power-Attainable limit of economy-Steam-jacketed intermediate receiver-Diagrams showing bene- ficial effect of the receiver on work performed-Independent air pump and con- denser-Indicator diagrams from various Porter-Allen engines-Capacity in horse power of various size engines-Table of high pressure engines, CHAPTER XXIV. THE WEIR-HARDEN COMPOUND ENGINE. Pages 656-697 A recent invention and formidable competitor for first place among American steam engines-A separate and distinct type of steam engines and presenting many new and novel features for the careful study of engineers and students of steam engineering-The engine set forth in detail with full instructions in regard to construction and management of valve gear-The valves consisting of a main and a cut off valve, described and illustrated-The compound cylinder illustrated and fully described-Balancing feature of the main valve described and illustra- ted-The operation of the cut-off valve, explained and illustrated, showing its operation in the admission and cutting off of steam-Balancing feature of the cut-off valve, described and illustrated-The shaft governor fully described and illustrated-Instructions in setting the valves-How to set the main valve fully explained-The cut-off valve, and how to set it described-How to adjust the shaft governor for engine to run under or over, explained in detail, Pages 698-719 CONTENTS. 19 CHAPTER XXV. SLIDE-VALVE ENGINES. The valve gear the main feature of all steam engines, and the chief difference in the various types-Importance to engineers to study the anatomy of the various types of steam engines to enable them to acquire a thorough knowledge of the con- struction and operation of their valve gear-Engineers should understand the con- struction and operation of the various distinctly different types of engines in order to enable them to claim proficiency in their profession-The slide-valve engine the pioneer of modern steam engines, and no other type of engine in such gen- eral use-Its construction and operation better understood by engineers than any other type of steam engines-But no work on steam engineering complete without the slide-valve engine-Explanation and illustration of two high class engines of the slide valve type-The Sinker-Davis slide-valve engine, explained and illustra- ted-Valve and crank movement described and illustrated, showing the position of the valve in relation to the crank pin at various points of the stroke-How to set the valves fully described-Throttling slide-valve non-condensing compound engines-The slide valve the simplest instrument for the distribution of steam to the cylinder of a steam engine-The automatic cut-off gear more economical when applied to single cylinder engines-The advantages of the slide valve in giving easy relief to water and in being easily made and kept tight-Limitation of use- fulness of slide valve disappeared with the advent of the compound and triple expansion engine-The number of expansion no longer depend upon the cut off, but chiefly upon the ratio of cylinder area-The principles of expansions, the basis of modern marine engine design where cut off five-eighths and one-half are used, and with triple and quadruple expansion engines, fourteen to twenty expan- sions are obtained-Advantages of non-condensing compound slide-valve engines set forth and fully explained-The Houston, Stanwood & Gamble cross com- pound engines-Advantages of cross compound engines described-How to find the horse power of compound engines-How to set the valves of cross compound engines-General instructions with illustration for setting the valves of slide- valve engines, Pages 720-733 CHAPTER XXVI. LOCOMOTIVES. The science of steam engineering applicable to locomotives as well as to sta- tionary or marine engines-The knowledge of the science of steam engineering, requisite to make a stationary or a marine engineer proficient in his calling, indis- pensable to the completion of the education of locomotive engineers-Things pecu- liar to the profession of locomotive engineering, not met with in other branches of steam engineering, fully treated-Adjustment of link-motion valve gear a chief feature of the science of locomotive engineering-To secure accuracy in setting valves great care must be exercised in putting engines on the dead center-The accuracy of all subsequent movements depend upon the accuracy of putting engines on dead centers - Various methods heretofore employed for putting engines on dead centers lack the accuracy essential to correct valve setting- Full instructions in regard to putting engines on the dead center-Inaccuracies pointed out in putting engines on dead centers under various methods generally employed-Correct method of putting locomotives on dead centers, fully explained 20 CONTENTS. and illustrated, with simple instructions in regard to how to perform the opera- tion-How to adjust the valve gear and how to set the valves described in simple form-How to set the eccentrics of locomotives correctly before driving wheels are placed under the engine, illustrated and explained-How to adjust locomotive valve gear, in case of derangement on the road, simplified-Compound locomotives illustrated and described-The Weir-Harden compound locomotive a great im- provement in locomotive construction-The construction and operation of the compound locomotive fully illustrated and explained-The high and low pressure cylinder a continuous casting combining both high and low pressure engines- Full description of the cylinder, with illustrations-Details of the valve, piston and valve gear-The Weir-Harden compound locomotive capable of attachment to ordinary locomotives without change of valve gear except valve, cylinder and piston-Valve and valve gear adjusted and set the same as those of an ordinary locomotive-The engine capable of being converted into a simple engine from compound, and from a simple engine to a compound, at the pleasure of the engi- neer- Description of the converting valves with illustrations of mechanism for operating them-The main valve and its functions, illustrated and described in detail-Description and illustration of valve stem marks for valve setting-Rules and examples for determining the piston speed of locomotives, Pages 734-758 CHAPTER XXVII. STEAM BOILER INJECTORS. • Discovery of the capacity of a moving jet of steam, or other fluid, to produce a vacuum in properly formed ducts, to raise air, water or other fluids-Perfection of the injector by the eminent French engineer, H. J. GIFFARD, to utilize the kinetic energy of a moving mass of fluid to overcome the static energy of a mass of water under boiler pressure-The active parts of an injector, their construction and operation, illustrated and explained-Automatic injectors, double jet injectors, explained and illustrated-Causes which prevent injectors working-Supply tank for steamboats, illustrated and explained-Speculative theory of the injector— Why an injector works-Velocity of steam flowing into the atmosphere-How to determine the velocity of steam flowing into the atmosphere-Proportion of water to that of steam-How to determine the velocity of water discharged by injectors- How to determine the velocity with which water will flow from a boiler under a given pressure-How to determine the capacity of injectors-Efficiency of in- jectors as boiler feeders-Delivering capacity of injectors, how determined theo- retically, Pages 759-782 ILLUSTRATIONS. FIG. Dimensions of .7854 of a square inch.. 1 Dimensions of .5 of a square inch.... 2 Dimensions of .25 of a square inch... 3 Dimensions of .125 of a square inch.. 4 How the constant .7854 is obtained.. 5 How the constant .7854 is obtained.. 6 How the constant .7854 is obtained.. 7 Adjustment of the safety-valve ...... 8 Pressure of safety-valve lever on the valve seat Adjustment of safety-valve as found in practice.... 9 FIG. First operation for determining the distance of weight from fulcrum..26 Second Fourth do. do. do. 27 do. do. do. 28 do. do. do. 29 Sixth Adjustment of safety-valve for deter- mining weight required to be placed a given distance from ful- crum.. .....30 First operation for determining weight required.. Second ..10 Fourth Operation for determining area of valve. Adjustment of safety-valve complete for determining pressure required. to raise it.. .11 Sixth ...31 do. do. do. 32 do. do. do 33 do. do. do. 34 Adjustment of safety-valve for deter- mining the length of short arm of lever First operation for determining length of short arm of lever. ..35 .36 do. do. 37 do. do. 38 .12 First operation for determining press- ure required to raise the valve.. 13 Second Second Third do. do. do. 14 Fourth do. do. do. 15 Fourth do. do. do. 16 Adjustment of the safety-valve......17 do. do. Operation for determining area of valve..... ....18 Adjustment of safety-valve completed for determining pressure required to raise the valve.... ....19 First operation for determining press- ure required to raise the valve...20 Adjustment of safety-valve for deter- mining diameter of valve for a given pressure First operation for determining diam- ....39 eter of valve.... 40 Second do. do. do. 41 Third do. do. do. 42 Fifth do. do. do. 43 Second Third Fifth do. do. do. 21 do. do. do. 22 do. do. do. 23 Adjustment of safety-valve completed for determining pressure required .....24 to raise it.. Adjustment of safety-valve for deter- mining the distance to place weight from fulcrum.. ...25 Adjustment of safety-valve for deter- mining weight of various parts mathematically 44 First operation showing how weight of lever is determined..... .....45 Second operation showing weight of weight and lever..... ..46 Third operation showing weight of weight, lever and spindle........47 Fourth operation showing weight of weight, lever, spindle and valve..48 21 22 ILLUSTRATIONS FIG. First operation for determining press- ure required to raise the valve...49 Second Third Fifth do. do. do. 50 do. do. do. 51 do. do. do. 52 Adjustment of valve for determining pressure required to raise it by a simple method.... .53 .54 First operation.... Adjustments with a bar of uniform section for determining pressure required to raise a valve..... ...55 Sines and cosines illustrated.........56 Opening produced by the lift of a bev- el-seated valve.... ....57 Sheet of boiler plate with test piece attached.... Coupon prepared for testing. Coupon after testing. FIG. Front elevation of tubular boiler with full front.... ....79 Ground plan for half arch front tubu- lar boiler... ...80 Cross section of tubular boiler and setting. tubular boiler .....81 Longitudinal sectional elevation of ....82 Front elevation with half arch front .83 Longitudinal sectional elevation of tubular boiler, full arch front...84 Cross section of Fig. 84..... ..85 Ground plan of foundation for full arch front.... ...86 Front elevation of tubular boiler, full arch front..... ....87 Longitudinal sectional elevation of tubular boiler, full arch front....88 Cross section of Fig. 88... ...58 .59 ...60 ...89 Section of single-riveted lap joint for steam boiler...... Ground plan ...90 Front elevation.. .....61 .91 Section of zig-zag double-riveted lap Side elevation of plan of tubular joint.. .... ....62 boiler, half arch front. .92 Section of chain riveting, double-riv- Front elevation... .93 eted joint.... ...63 Cross section of Fig. 92... .94 Section of triple-riveted joint........64 Section of single-riveted butt-strap Ground plan. ..95 joint.... ...65 Section of single-riveted, double butt- ....66 strap joint..... Section of staggered double-riveted double butt-strap joint..... ....67 Section of chain double-riveted Side elevation, sectional view of tubu- lar boiler, half arch front........96 Front elevation.... Cross sectional view of Fig. 96.......98 Ground plan..... .97 .99 Sectional side elevation, full arch front, of tubular boiler. .....100 .69 Front elevation... Cross section of Fig. 100.... Ground plan.... ..101 .102 ..103 double butt-strap joint.. .....68 Section of triple-riveted double butt- strap joint.... Horizontal tubular boiler, front view.70 Horizontal tubular boiler, back view.71 Eclipse man hole, back head........72 Eclipse man hole, front head........73 Safety-water column complete with water-gauge and guage-cocks....74 Safety-water column, detailed section- al engraving.. Ground plan of foundation for tubu- lar boiler.. ...76 Longitudinal sectional elevation of Sectional side elevation, half arch front, of double setting, tubular boilers.... Front elevation of battery.. Cross section of battery... Ground plan.... ..104 .105 ...106 ...107 Sectional side elevation, full arch front, of double setting tubular boilers ....75 Front elevations of battery. Cross section of battery.. tubular boiler and setting.......77 Cross section of tubular boiler and setting.. ..78 ..108 .109 .110 .111 Ground plan . Sectional side elevation, half arch front, of triple setting tubular boilers.. ..112 ILLUSTRATIONS. 23 FIG. Front elevation of battery..... .113 Cross section of battery.. .114 Ground plan. ...115 Sectional side elevation full arch front, of triple setting tubular boilers... .116 Front elevation of battery... 117 Cross section of battery.. 118 Ground plan.... ...119 Plan of left-hand engine with tubu- Intermediate driven pulleys in a series Intermediate driving and driven FIG. ...146 ......147 pulleys in a series... Intermediate driving and driven pulleys on intermediate shafts..148 .149 Dynamo Section showing upper part of indi- cator, steam cylinder and mech- anism.... lar boilers.. ..120 Front view of indicator.. Elevation of left-hand engine with tubular boilers.... Sectional view of indicator. ..121 Indicator spring.. Plan of right-hand engine with tubu- lar boilers... 122 tubular boilers.. paper drum .123 ..... ..124 125 Elevation of right-hand engine with Simple device for making calorim- eter tests... Sectional elevation of the Ward coil boiler.... Plan of the Ward coil boiler........126 Side view of general arrangements.. 127 Front elevation, partly in section, of Cowles' water tube boiler......128 Vertical longitudinal section.......129 Perspective view of Hawley down draft furnace grate... Perspective view of boiler with Haw- .....130 ley down draft furnace attached 131 Hawley down draft furnace front...132 Side elevation of tubular boiler.....133 Side elevation of National water tube boiler..... 134 Side elevation of Heine boiler..... 135 Side elevation of Stirling boiler.....136 Side elevation of Munroe water tube safety boiler.... ..150 ..151 .152 ..153 Three-way cock for indicator.......154 Manner of communicating motion to ....155 ..157 Manner of attaching lazy tongs.....156 Lazy tongs.. Manner of attaching the Pantagraph.158 Essential features of an indicator diagram.. 159 Diagram from high pressure cylinder 160 Diagram from low pressure cylinder. 161 Combination of diagrams from com- ....162 pound engine... Diagram showing how to compute mean effective pressure........163 Expansion line of non-condensing engine with light load... Planimeter.. Polar Planimeter... .....164. ...165 ..166 Form of iron test-piece... 167 Form of steel test-piece.. ....168 Form of steel test-piece for special tests.... ...169 ..137 Side elevation of Babcock & Wilcox boiler.... .138 Side elevation of Peck boiler... .139 Side elevation of Caldwell standard boiler.... 140 Side elevation of Campbell & Zell water tube boiler.... Sectional side elevation of Galloway 141 boiler...... .142 Series of driving and driven pulleys. 143 Series of driving and driven pulleys in 143 reversed....... .....144 Intermediate driving pulleys in a series... .145 Section of flat surface of a boiler with stay-bolts.... ...170 Section of flat surface showing area stayed by a stay-bolt...........171 Section of flat surface showing area stayed by each bolt.... Method for determining average ..172 ..173 .....174 steam pressure in a cylinder dur- ing stroke of engine.... Safety-valve adjusted for determin- ing steam pressure required to raise the valve..... Diagram for determining distance to place weight from fulcrum.....175 Diagram for determining the weight to be placed a given distance from fulcrum.. .176 24 ILLUSTRATIONS. Diagram representing safety-valve with cubic weight attached to lever....... .... FIG. ....177 Diagram representing safety-valve with spherical weight of given diameter attached to lever.....178 Diagram representing safety-valve ...179 with spherical weight of given number of pounds attached to lever.... Diagram representing safety-valve with conic weight of given di- mensions attached to lever.....180 Diagram representing safety-valve with conic weight of given num- ber of pounds attached to lever. 181 Diagram representing safety-valve with several weights attached to lever... Sectional elevation of U. S. Govern- FIG. Section through line 54.... Section through line 37.. Section through line 29..... ...200 ..201 ...202 Section through lines 72 and 82.....203 Section through lines 63 and 72.....204 Plan of main engines.... .205 Vertical section of main engines....206 Elevation of starboard engine, look- ing forward...... .....207 Elevation of starboard engine, look- ....208 Side elevation of air pumps and ing aft... engine. engine. .....209 Vertical section of air pumps and Section on line A. B... ...210 .211 Section on line E. F.... ..212 Section on line C. D.... .213 ....182 Plan of condenser... .214 End view of condenser... ...215 ment safety-valve... ..183 Sectional top view of U. S. Govern- ment safety-valve... End view, showing man hole ar- rangements.. .....216 ...184 Side view of U. S. Government safe- Longitudinal section of condenser..217 Section on line A. B.. .218 ty-valve... ..185 Section on line E. F. .219 Top view of U. S. Government safe- ty-valve...... ..186 Vertical section of spring-loaded Detail of section on X. Y... Detail of section on X. Y.. 220 221 valve..... ..187 Arrangement of link to which lever is attached for lifting the valve from its seat... ..188 Detail of section on X. Y... Detail of section on X. Y... Detail of section on X. Y... Side elevation of main boiler, partly .22:2 ..223 .....224 Top view of valve..... in section... ..189 ..225 Plan showing brass bush and nickel seat.. ....190 Section of Adamson ring for furnace flues... ..... .....191 Section of angle-iron ring for furnace flues.... ..192 Section of half-round iron ring for furnace flues..... Plan of general arrangement of ma- chinery of a United States battle ship... .....193 .194 Vertical view of general arrangement of machinery.... Plan of propeller wheels and shafts.196 Vertical longitudinal section through center propeller shaft..........197 Frame 89, looking aft... View of stern of vessel and propeller ...195 Front view of main boiler, partly in section...... ....226 Side elevation of main boiler, partly in section... ....227 Front view, showing arrangement of tubes, furnaces, etc...... ....228 Sectional side elevation of auxiliary boiler ..229 Front view, showing arrangement of tubes, furnaces, etc.... Section of boiler, showing triple-riv- eted butt-strap joint..... Perspective view of steamer Hudson.232 Side elevation of a Western river .230 .....231 wheels.... stern-wheel steamer. ..233 Plan of cabin deck... ..234 ...198 Plan of main deck... 235 Cabin stanchion... .236 ....199 Western river steamboat doctor.....237 ILLUSTRATIONS. 25 FIG. Western river nigger engines.......238 An old-time Missouri river mountain steamer... ...239 A Western river electric light en- river side-wheel steamer.......241 gine..... ....240 Sectional elevation of a Western Side elevation of vessel draught line...... above .....242 A Western river side-wheel steamer.243 Side elevation of main engine, partly Cross section of cylinder and valves.246 .244 .245 FIG. Side view of the Rees marine engine.273 Side elevation of cut-off gear at- tached to engine.... ...274 Plan of cut-off gear attached to en- gine... Vertical cross section of cylinder and valves.. Plan of Rees marine engine... Side elevation... ....275 .276 .277 .278 Front elevation of a battery of West- ern river steamboat boilers.....279 Rear elevation of a battery of West- ern river steamboat boilers.....280 Plan of a battery of Western river steamboat boilers.... ....281 Side elevation of a battery of West- ern river steamboat boilers.....282 Top view of fire-box marine boiler..283 Longitudinal sectional view of fire- in section.... Plan of main engine.. End view of cylinder and cut off....247 Front elevation of cut off and dash pot. ..248 Plan of cut off and dash pot... Vertical section of cut off.. 249 ..250 Plan of vertical section of cut off...251 ...284 Shell of heater.. ..252 Sectional front elevation of fire-box Shell of heater in longitudinal sec- tion... boiler.... Inside view of back head... .253 254 ...285 Front elevation of fire-box boiler...286 Back elevation of fire-box boiler....287 Sectional back elevation of fire-box boiler ..288 .289 box boiler.... View of tube sheet and tubes.......255 Outside view of back head.........256 Inside view of back head...... ....257 Longitudinal section through center of boiler..... Front elevation... 258 .259 Vertical cross section through boiler and forward steam drum leg...260 Side elevation of the Sweeney ma- rine engine... Plan of the Sweeney engine... Exhaust lifter.. Reclining lifter..... ...261 ....262 ..263 ..264 Center line of engine, showing vari- ous positions of full-stroke cam.265 Side elevation of the Frisbie marine engine. ...266 Vertical cross section of cylinder and valves.... ....267 Side elevation, showing position of cams and valve connections....268 Top view of valve.... ....269 Full-stroke cam, process of laying off the same.. ....270 A cam, process of laying off the same... Ağcam... 271 ..272 Plan of bottom of fire-box... Perspective view of the Corliss en- gine..... ..290 Rear elevation of Corliss engine....291 End view, partly in section.........292 Plan, partly in section... Front elevation... Longitudinal section of cylinder .293 ....294 showing position of the valves..295 Cross section through center of cyl- inder..... ....296 Cross section through steam and exhaust valve..... ....297 Cross section of cylinder of Colum- bian Corliss engine, showing steam and exhaust chest.......298 Cross section of cylinder, showing steam and exhaust chest of or- dinary Corliss engine... .....299 Cross section of frame of ordinary Corliss engine... ....300 Cross section of frame of ordinary ...301 Corliss engine... Cross section of frame of Lane & Bodley Corliss engines.. ...302 26 ILLUSTRATIONS. Cross section of flat slides and cross head...... FIG. ....303 Cross section of V shaped slides and cross head..... .....304 Cross section of exaggerated V shaped slides and cross heads .........305 Cross section of slides, cross head and frame of Columbian Corliss engine.. Piston, partly in section, showing arrangement of rings... Section of packing ring..... FIG. .....336 Longitudinal section through hood and valve.... Perspective view of the valve and seat with hood detached........337 Face view of sleeve of Atlas shaft governor.. ....306 ...307 ..308 .....338 Top view of sleeve of Atlas shaft governor...... .....339 End view of sleeve of Atlas shaft governor... ...340 Face view of eccentric of Atlas shaft governor. ....341 Top view of eccentric of Atlas shaft governor...... ..342 End view of eccentric of Atlas shaft governor... ..343 …..313 Lug piece of eccentric of Atlas shaft 314 governor .344 .... End view of lug piece.. ..345 ..315 Face view of eccentric link. ..346 The Babbitt & Harris packing......309 Cross head, partly in section.......310 Crank end of connecting rod........311 Cross head end or connecting rod...312 Cross section of valve.... Perspective view of valve.. Perspective top and side view of Perspective side and bottom view of valve rod... valve rod... ..316 The Lane & Bodley governor.......317 Crab claw .318 Longitudinal section of valve and .319 .320 ..321 .322 ..323 bonnet..... Dash pot.... Main pillow block.... Pillow block... Air pump and condenser.. Indicator, diagram of ordinary slide valve and of Corliss engine.....324 How to put an engine on the dead center.... Device for putting engine on the dead center... ..326 Method of putting engine on the ..325 ...327 dead center... Reynolds Corliss engine...........328 Position of wrist plate when crank is on the center.... .....329 Position of wrist plate when crank is on the center and eccentric set to give steam valve proper lead....330 Dash pot.. ...331 The Atlas automatic engine, per- .....332 .... Top view of eccentric link..........347 Vertical elevation of dead wheel of governor .....348 ...349 Vertical longitudinal section through bore of governor... Front view of weight of governor..350 Top view of weight of governor....351 Top view of weight link of governor.352 Face view of weight link of gov- ernor.. Details of governor.... .....353 ...354-362 Face view of governor set to run 364 over, greatest travel..... ...363 Top view of dead wheel.. Face view of governor set to run un- der, least travel... ...365 ....366 Top view of dead wheel.... Diagrams of the Atlas automatic en- gine, with full load...... .....367 Diagrams of the Atlas automatic en- gine, with light load..... ..368 Cycloidal engine... ..369 Steam valve bracket of cycloidal engine ....370 Exhaust valve bracket of cycloidal engine. ...371 ....333 Cross section through valves and cylinder.. ..372 Perspective view of compound cy- cloidal engine ...373 spective view. Foundation plan and end eleva- tion. Engine foundation template........334 Perspective view of hood and valve.335 Steam valve of cycloidal engine....374 ILLUSTRATIONS. 27 FIG. Exhaust valve of cycloidal engine..375 Longitudinal section of compound cycloidal engine..... .....376 The Brown engine, perspective view.377 Side view, with covers of steam chest removed. Sectional plan.. Sectional end elevation.. main shaft.. ..378 ...379 ..380 General arrangement from valves to ....381 Sectional view showing application of gauge for locating position of exhaust valve and for verifying length of exhaust valve rod....382 Sectional view showing application of gauge for locating position of steam valve and for verifying length of steam valve rod......383 The governor.. Side view of cross head. End view of cross head.... ..384 ..385 ..386 Main bearing.. ...387 Plan and elevation of engine.......388 Diagram showing engine carrying Diagram from automatic engine for FIG. ....406 analysis for students. Diagram from throttling engine for analysis... .....407 ...409 Diagram from single valve auto- matic engine for analysis.......408. Diagram from automatic engine showing how to locate the clear- ance line... Geometric method for determining the theoretical curve...........410 Diagram showing how to locate the clearance line... Diagram showing how to determine 411 the mean effective pressure.....412 Diagram showing how to calculate the theoretical consumption of water... Geometric method of determining slide valve points of cut off and exhaust closure Geometric method of determining 413 ....414 consequences to other events of setting slide valve for equal cut off.... ...415 Geometric method of locating slide valve exhaust closure..........416 Geometric method of finding slide valve laps for given results......417 Geometric method, condensed form, to determine slide valve func- tions Perspective rear view of Buckeye tandem compound engine......419 Sectional plan of Buckeye tandem compound engine... .418 light load..... ....389 Diagram showing engine carrying full load...... ....390 Perspective back view of the Buck- eye engine.... ...391 Plan view and foundation..........392 Front elevation of engine and foun- dation... ....393 End elevation of engine and founda- tion.... ...394 Diagram representing perfect engine performance. ..395 ...420 Diagram showing slow main eccen- tric.... ..396 Diagram showing fast eccentric.....397 Diagram showing main valve dis- Plan of cross compound engine.....421 End elevation of cross compound engine. .....422 placement.. ...398 Diagram showing unequal cut off..399 Shaft governor. 400 How to find weight necessary for change of speed. ....401 .402 Longitudinal section of cylinder and valves.... Section on line B B (Fig. 402).. ..403 Section on line C C (Fig. 402).......404 Cross section of girder bed and com- pound rocker arm... ...405 The Porter-Allen engine, view from ...423 The Porter-Allen engine, view from ....424 fly wheel side.... Eccentric and link, side elevation...425 Eccentric and link, front elevation. .426 Diagram showing movements of the link in incorrect adjustment....427 Diagram showing movements of the link in correct adjustment......428 Horizontal section of cylinder through steam and exhaust valves......429 crank side... 28 ILLUSTRATIONS. Vertical section through cylinder and valves.... Plan of connections .. FIG. FIG. .430 ..431 Longitudinal section through center of steam chest. ... Longitudinal section through center ...465 Elevation of connections.... .432 Section showing valve commencing to open..... ...433 Section showing valve at the end of of cylinder.... Diagrams from high and low press- ure cylinders 1 cut off.........467 Diagrams from high and low pressure ..466 ...434 cylinders cut off.... Diagrams from high and low pressure ..468 ..469 .435 ....470 its lap.... Section showing pressure plate forced up and away from the valve... Section showing pressure plate in working position..... .....436 Diagram for lining an engine with shaft placed at a higher or lower level .. ....437 Front elevation of Porter-Allen en- gine and foundation.... ..438 Foundation plan of Porter-Allen en- gine..... ....439 Rear elevation of engine and foun- dation.... ....440 View from crank end, pulley partly in section.. ..441 View from cylinder end.... .442 Diagram of force on crank pin......443 Plan of tandem compound condens- ...444 ing engine... Front elevation of tandem com- pound condensing engine...... 445 Diagrams from high pressure cylin- der showing action of steam- jacketed receiver.... .....446 Diagrams from low pressure cylinder showing action of steam-jacketed receiver... .447 Independent air pump and conden- ser.... ..448 Diagrams taken from Porter-Allen engines. ...449-456 Diagrams taken from Porter-Allen tandem compound engine..457-460 Front elevation of Weir-Harden ver- tical compound engine, partly in section.... 461 End elevation of vertical compound engine.... End elevation of pillow block.......463 Cross section of cylinder and steam. 462 chest.... ..464 cylinders cut off... Diagrams from high and low pressure cylinder cut off..... Sectional view of cut-off valve, show- ..... ing valve at point of opening...471 Sectional view of cut-off valve at full opening ..472 Sectional view of cut-off valve, show- ing valve at extreme position of travel... .473 ..... Sectional view of cut-off valve, show- ing valve in center of travel and ports closed.. ..474 Side elevation of Weir-Harden hori- zontal compound engine.......475 Plan of Weir-Harden horizontal compound engine.. ..476 Plan view of governor, ports at full stroke..... ....477 Plan view of governor, ports at re- duced stroke. ...478 Section on line x x (Fig. 477)... .....479 Plan view of oscillating disc..... ..480 Cross section of oscillating disc.....481 Plan view of eccentric ring.. Plan view of pendulus lever........483 Plan view of governor, modification .482 .484 of Fig. 477.... Vertical section on line x x (Fig. 484) 485 Sectional elevation on line y y (Fig. 484)... .... ..486 Longitudinal elevation of slide-valve engine, partly in section.... ..487 Plan of slide-valve engine. ..488 Diagram showing position of valve at beginning of forward stroke....489 Diagram showing position of crank when rear steam port is full open... .490 Diagram showing position of crank when rear steam port is closed..491 ILLUSTRATIONS. 29 Diagram showing position of crank at which valve closes the ex- haust..... FIG. ..492 Diagram showing position of crank when rear steam port is open to exhaust, and forward steam port open to lead.... 493 .494 Diagram showing rear steam port full open to return movement of the crank... Diagram showing various positions of the crank with valve in posi- tions shown in Figs. 489-494....495 Perspective view of cross compound slide-valve engine... Diagrams showing engine perform- ance... ...496 ...497, 498 Diagram for setting slide valves.....499 Perspective view of device for putting locomotives on the dead center.500 Perspective view of modified form of device for putting locomotives on dead center.. ....501 Diagram showing how to put crank pin on center. ....502 Diagram showing how to adjust ec- centrics of locomotives.. .....503 Diagram showing how to set loco- motive eccentrics before wheels are placed under engine........504 Longitudinal view of compound loco- FIG. ..507 Rear elevation of section back of rock shaft and section front of rock shaft.... Front elevation, and front elevation showing pilot removed.........508. Longitudinal section of steam chest, valve and cylinder... ..509 Cross section of cylinder and steam chest, showing converting valves.510 Longitudinal elevation of valve gear with cylinder in longitudinal section... ....511 Longitudinal section of steam chest, valve and upper portion of cylin- der... .....512 Exterior view of Monitor injector...513 Interior view of Monitor injector....514 Longitudinal interior view showing working parts of injector. .....515 Exterior view of automatic injector.516. Interior view of automatic injector..517 Exterior view of the World injec- tor. .....518 Interior view of the World injec- tor... .519 Exterior view of automatic injector. 520- Interior view of automatic injector.521 Steam nozzle of automatic injector..522 Intermediate nozzle of automatic ....523 Condensing nozzle of automatic in- injector... motive..... ...505 jector... ..524 Rear elevation and section showing position of levers... ...506 Perspective view of supply tank for steamboats..... ..525. A LIBRARY OF STEAM ENGINEERING. 31 TIBETKI OF 2LEVA EXCIZEEBIZC CHAPTER I. MATHEMATICS OF STEAM ENGINEERING. The profession of steam engineering is composed of three separate and distinct branches, namely: "The Theory of Steam Engineering, The Practice of Steam Engineering, and The Mathematics of Steam Engineering." The latter is the fundamental principle upon which the whole science of steam engineering is based; and the engineer who is deficient in this indispensable branch of his profession can neither be a first-class, reliable, safe nor an economical engineer. Therefore, to the completion of his education, as an engineer, it is absolutely indis- pensable that he should be familiar with, at least, the ordinary rules of arithmetic, if not with the higher branches of mathematics. There is another reason why every man who intends to follow the profession of steam engineering should thoroughly familiarize himself with the mathematics of steam engineering. The U. S. Government has set the example, and the various states in the union are rapidly following in the passage of laws providing for the examination and licensing of steam engineers. The time therefore is fast approaching when no man who is deficient in this important science will be per- mitted to take charge of any steam plant, or follow the profession of steam engineering anywhere in the United States. To this may be added still another, and by no means unimportant reason. The engineer who wishes to attain a high position in his profession and command the highest standard of wages can never realize his expecta- tions without a thorough knowledge of the mathematics of steam engineering. But the great difficulty which invariably confronts the student in his initial effort to acquire that knowledge is familiar to every engineer. To obviate that difficulty it is essential that a course of instruction for the student should be arranged in its utmost sim- plicity and kept throughout within the comprehension of every engineer of the most ordinary education. To accomplish that. desirable object is the purpose and aim of the course of instruction laid down in this work; and as this work is devoted largely to the mathematics of steam engineering, it is essential that the student should familiarize himself with at least the ordinary terms employed in mathematics; as these terms are employed universally in all works. 33 34 A Library of Steam Engineering. on steam engineering as well as in all others devoted to mathematical science. We will therefore begin with addition, the first branch of mathematics. ADDITION. The Sum.-When two or more numbers are added together, the total is called the sum. Thus: 1890 1891 1892 5673 The Sum. SUBTRACTION. The Remainder.-When a smaller number is taken from a greater number that which remains is called the remainder. Thus: 250 Minuend. 175 Subtrahend. 75 Remainder. MULTIPLICATION. Multiplicand, Multiplier, Product.-The multiplicand is the number to be multiplied; the multiplier is the number with which to multiply the multiplicand, the result is called the product. Thus: 248 Multiplicand. 25 Multiplier. 1240 496 6200 The Product. LONG DIVISION. Divisor, Dividend, Quotient, Remainder.-The divisor is the number by which the dividend is divided; the dividend is the number divided by the divisor; the quotient is the number that represents the number of times the divisor is contained in the dividend; the remainder is that part of the dividend remaining which has not been divided by the divisor. Thus: Dividend. Divisor. 25) 7859 (314 Quotient. 75 35 25 109 9 Remainder. 100 Mathematics of Steam Engineering. 35 DECIMAL FRACTIONS. In addition to the above, decimal fractions, square root and cube root enter largely into the mathematics of steam engineering and par- ticularly decimal fractions, which branch of mathematics is constantly called into play and the necessity of which is met by the student at the very threshold of steam engineering. Therefore, before the student of steam engineering enters upon the study of the science of the pro- fession he must, if he desires to make any progress whatever, familiarize himself with the branch of mathematics called decimal fractions. CONSTANTS 3.1416. Such numbers as 3.1416 and .7854 are called constants and as these particular constants enter more largely into steam engineering calculations than any other constants they will be here explained. TO DETERMINE THE CIRCUMFERENCE OF A CIRCLE. The constant 3.1416 is universally employed in determining the circumference of things circular in form, and the operation is per- formed by multiplying the constant 3.1416 by the diameter of the circle in inches and the product will give the circumference of the circle in inches. Example.-Taking a circle 2 inches in diameter and we have: 3.1416 Constant. 2 Diameter of circle. 6.2832 Circumference of circle in inches and decimals of an inch. TO DETERMINE THE DIAMETER OF A CIRCLE. RULE.-Divide the circumference of the circle in inches by 3.1416 and the quotient will give the diameter in inches. - Example. Taking a circle 6.2832 inches in circumference and we have: 3.1416) 6.2832 (2 Diameter of circle in inches. 6.2832 .7854 As to the constant .7854 a more extended explanation will be re- quired. There is hardly an engineer who has not frequently made use of that constant in making his calculations for determining the area of things circular in form; yet very few of those who constantly use it have any knowledge as to what it represents, or from what or whence it is derived. In the first place, the student is informed that it repre- 36 A Library of Steam Engineering. sents the area of any circle whose diameter is unity. That is, whose diameter is, for example, one inch, one foot, one yard, one mile or one anything in diameter. But as it is almost invariably employed in determining the number of square inches in a given circle, the rules here laid down will be arranged so as to be made applicable to that purpose. When we square the diameter of any circle we obtain for our answer the number of circular inches in the circle; and as each circular inch contains .7854 of a square inch, by multiplying the number of circular inches contained in a circle of any given diameter by .7854 we obtain the number of square inches contained in the circle. The area of a circular inch as expressed in decimal fractions is .7854 of a square inch, which means that the area is one inch by .7854 of an inch in dimensions and not .7854 of an inch square as many engineers erroneously consider it. A square inch has four equal sides, each one inch in length, and all fractions of a square inch when ex- pressed in decimals, contain a length of one inch, and the decimals, whatever they may be, express the width. TO REDUCE THE AREA OF A CIRCLE TO A PERFECT SQUARE. To reduce decimals of a square inch or any number of square inches to a perfect square it is necessary to extract the square root of the decimals, or of the number of square inches. As thus As thus expressed, A 4-inch circle contains 12.5664 square inches. its dimensions are one inch in height and 12.5664 inches in length; and if reduced to a perfect square would contain four equal sides each 3.5449 inches in length. Performing the operation we have: 12.5664 (3.5449+ inches, length of side of square. 9 65) 356 325 704) 3164 2816 7084) 34800 28336 70889) 646400 638001 DECIMAL PARTS OF A SQUARE INCH. The following figures contain an illustration of the dimensions of parts of square inches as expressed in decimal fractions. Fig. 1 contains .7854 of a square inch as expressed in decimal fractions. Mathematics of Steam Engineering. 37 Fig. 2 contains .5 of a square inch as expressed in decimal fractions. Fig. 3 contains .25 of a square inch as expressed in decimal fractions. 125' 25Y --.7854"--- Fig. 1 1" -.5" - k.25" V Fig. 2 K-- 1" Fig. 3 Fig. 4 Fig. 4 contains .125 of a square inch as expressed in decimal fractions. HOW THE CONSTANT .7854 IS OBTAINED. The next step is to show how the constant .7854 is obtained. To illustrate one of the methods, a circle is drawn one inch in diameter, as shown in Fig. 5, which is divided into 24 equal parts, and the parts then arranged in the shape of a parallelogram, as shown in Fig. 6. A circle, one inch in diameter, as shown in Fig. 5, has a circumference of 3.1416 inches, and if arranged as shown in Fig. 6 will have a width of inch, and a length of 1.5708 inches nearly, according to the number of pieces into which the circle is divided; the greater the number of pieces the nearer the length will approach one-half of the circumference of the circle, and the nearer the width will approach one-half of its diameter. One-half of the circle is represented by the wavy line at the top of Fig. 6, and the other half is represented by the wavy line at the bottom of the figure. Although this illustration is often given as being correct, it will be observed that it is not absolutely correct, and Fig. 5 k. 1.5708" Fig. 6 can not be made so, no matter into how many pieces the circle is di- vided; it can, however, be made near enough correct to answer the purpose of illustration, hence the following rule. RULE.-Multiply the length of the parallelogram by its width and the product will give the area. 38 A Library of Steam Engineering. Example. Let 1.5708 inches equal length of Fig. 6, .50 (fifty one- hundredths) of an inch equal width of Fig. 6. Then we have: 1.5708 .50 .785400 Area of a circle one inch in diameter. Now as it is well known that all ciphers to the right of the last decimal neither increase nor diminish the value of the decimals, the ciphers in the answer are dropped for convenience, and we have for our purpose the much-used, and little-known, mysterious constant .7854. It is well for the student to know also that the constant .7854 itself is not strictly correct, but it is near enough for all practical pur- poses, so much so that its adoption by mathematicians has become universal. Another method of showing how the constant .7854 is obtained is by geometrical rule according to the following illustration. A sector of a circle, as shown in Fig. 7, is the part included by the two radii and the intercepted arc A B C. A B C Fig. 7 RULE.-Ascertain the area of a sector by multiplying the distance cir- cumferentially between the two points in the arc, A B, as shown in Fig. 7 by the radius C, and dividing the product by 2. As the circle is one inch in diameter its circumference is 3.1416 inches; and if the circle be divided according to the previous illustra- tion (Fig. 5) into 24 sectors, the distance of each, circumferentially, will be 3.1416-24-.1309 inch; which multiplied by the radius C A or C B, inch, or reduced to a decimal fraction .5 (five-tenths) of an inch, would be .1309X.5=.06545, which divided by 2 would be .06545÷÷2== .032725, which is the area of a sector. As the circle, in this case, is divided into 24 sectors, we multiply the area of one sector by 24 and the product will give the area of the circle, and consequently the constant .7854. Thus: .032725 24 0 130900 0 65450 0.785400 Mathematics of Steam Engineering. 39 As in the former case, the ciphers being dropped for convenience, we have the constant .7854. TO FIND THE AREA OF A CIRCLE. Among the first things the student is required to learn is the rule for determining the area of a circle or a valve, which is as follows: RULE.-Square the diameter in inches or decimals of an inch of the circle or valve, and multiply the product by .7854. What is meant by the term squaring the diameter of a circle or a valve is this: It is multiplying the diameter by the diameter. For example: If a circle or valve is 3.25 inches in diameter, we say 3.25 times 3.25. Thus: 3.25 3.25 1625 650 975 10.5625 The square of the diameter of the valve. MULTIPLICATION OF DECIMAL FRACTIONS. In multiplying decimal fractions always point off as many deci- mals in the product as there are decimals in the multiplicand and multiplier. In the above case there are two in each; we therefore point. off four decimals, counting, always, from right to left, and all figures to the left of the decimal point are whole numbers, and all figures to the right of the decimal point are the fractional part of unity or of one whole number. The next step is to multiply the square of the diameter of the circle or valve by the constant .7854. Thus: 10.5625 .7854 422500 528125 845000 7 39375 8.29578750 Area of circle or valve in square inches. THE CORRECT DIAMETER OF A VALVE. In taking the diameter of a valve, engineers are cautioned not to be governed by the valve opening leading into the valve chamber-a mistake commonly made, but they must be governed by the smallest diameter of the valve's bearing on the seat. A valve may cover an 40 A Library of Steam Engineering. opening, say three inches in diameter, and the valve may have a bear- ing all around the opening a quarter of an inch from the opening, in which case the diameter of the valve would be 3 inches instead of 3 inches. Example.-Taking a valve whose smallest diameter of its bearing on the seat is 4 inches. Then we have: 4X4X.7854-12.5664, area of valve. Thus: 5664 4 times 4 are 16, and 16 times .7854 are 12 and 10,000 square inches. Or, we may put it in the following shape. First, square the diameter of the valve thus: 4 4 16 Then multiply the decimal .7854 by 16-the square of the diameter of the valve. Thus: .7854 16 4 7124 7 854 12.5664 Area of the valve in square inches. TO DETERMINE THE REQUIRED DIAMETER OF A VALVE OF ANY GIVEN AREA. In determining the required diameter of a valve, the first thing to be done is to ascertain the amount of area required to make the valve effective for the purpose for which it is intended. After the amount of area required is ascertained, proceed as follows: RULE.-Divide the number of square inches in the area by .7854, and then extract the square root of the quotient. Example.-Taking a valve required to have an area of 12.5664 square inches, what must be its diameter? Then we have: .7854) 12.5664 (16 Square of the diameter 7 854 of the valve. 4 7124 4 7124 The next step is to extract the square root of the quotient, which in this case represents the square of the diameter of the valve. Thus: 16 (4 Diameter of a valve required to 16 contain an area of 12.5664 square inches. Mathematics of Steam Engineering. 41 Performing the operation in a shorter method we have: 112.5664 ---4 Diameter of the valve in inches. N.7854 TO DETERMINE THE CIRCUMFERENCE OF A VALVE. RULE.-Multiply the smallest diameter of its bearing on the seat by the constant 3.1416, the product will give the circumference of the valve or of a circle. Example. Taking a valve having a diameter of 3.5 inches, and we have: 3.1416 3.5 157080 9 4248 10.99560 Circumference of the valve in inches. TO FIND THE DIAMETER REQUIRED FOR A VALVE OF ANY GIVEN CIRCUMFERENCE. RULE.-Divide the given circumference by the constant 3.1416, the quotient will give the required diameter of a valve or circle. Example. Taking a valve or circle having a circumference of 10.99560 inches, then we have: 3.14160) 10.99560 (3.5 This quotient represents 9 42480 diameter required in inches. 1 570800 1 570800 ANNEXATION OF CIPHERS IN DIVISION. It will be noticed that a cipher has been added to the decimal in the divisor. This is done under the rule to avoid confusion between the whole numbers and the decimal numbers in the quotient. In dividing with decimal fractions, or in other words, in using decimal fractions when performing the operation of division, the divisor and dividend should each contain the same number of decimals. If one contains more than the other, a sufficient number of ciphers should be annexed to the lesser whichever it may be, either the divisor or the dividend, to equal the number of decimals in the other, then, so long as there is any figure or cipher in the dividend which has not been brought down in performing the operation, or in other words, so long as the entire dividend has not been exhausted the quotient produced by dividing the remainder will be a whole number. But as soon as 42 A Library of Steam Engineering. the dividend has become exhausted and a cipher has been borrowed to continue the operation, a decimal point must be put immediately after the last whole number in the quotient, and all the figures which follow the decimal point to the right are decimals. Thus: 20) 343 (17 20 143 140 3 Remainder. According to this we have 3 as a remainder and this represents a 3 vulgar fraction Now if we wish to continue the operation until 20 we have a sufficient number of decimals-four are generally sufficient for all practical purposes-or until there is no remainder, we proceed as follows: 20) 343 (17.15 20 143 140 30 20 100 100 This cipher is borrowed and we therefore put a decimal point in the quotient, and all figures which follow that point are decimals, and we continue to bor- row until there is no remainder. TO REDUCE COMMON FRACTIONS TO DECIMALS. In reducing a vulgar or common fraction to a decimal we divide the numerator by the denominator. The numerator is the number above the line and the denominator is the number below the line. Thus 3 is called three-twentieths. 20 In reducing it to a decimal we say: 20 into 3 goes no times, and we write a cipher in the quotient. Now as we are compelled to borrow a cipher we place a decimal point after the cipher and continue the operation until there is no remainder. First operation: Second operation : 20) 3 (0. 20) 3.0 (0.1 20 10 Third operation : 20) 3.0 (0.15 Fifteen one hundredths. 20 100 1 00 Mathematics of Steam Engineering. 43 15 Therefore is equal to 100 thus: .15. 3 and in decimal fractions is expressed 20 In mathematics of steam engineering all vulgar fractions should first be reduced to decimal fractions. For example: Taking a valve 3 inches in diameter, the first thing to be done is to reduce the to a decimal fraction. Thus: 4) 3.0 (0.75 This represents three-fourths of an inch; and the diameter of the valve will be expressed, 3.75. 28 220 We now square this diameter if we wish to determine the area. Thus: 3.75 3.75 1875 2 625 11 25 14.0625 Square of the diameter of the valve. We now multiply the square of the diameter of the valve by the constant .7854 and the product will give the area of the valve in square inches. Thus: 14.0625 .7854 562500 703125 1 125000 9 84375 11.04468750 Area of valve. It will be seen that we have four decimals in the multiplicand and four decimals in the multiplier, making eight decimals in both. We now count eight from right to left in the product and place a decimal point to the left of the eighth figure. Now all figures to the left of the decimal point are whole numbers and all figures to the right of the decimal point are fractions or decimals. TO DETERMINE THE NUMBER OF CUBIC INCHES IN A CYLINDRICAL WEIGHT. RULE.-Square the diameter and multiply .7854 by the product; then multiply the last product by the length of the weight. Example.-Taking a cylindrical weight 8 inches in diameter and 12 inches in length, and we have: 44 A Library of Steam Engineering. .7854 8X8 64 Square of the diameter of the weight. 3 1416 47 124 50.2656 12 Length of the weight. 100 5312 502 656 603.1872 Number of cubic inches contained in the weight. TO DETERMINE THE NUMBER OF POUNDS CONTAINED IN A CAST-IRON WEIGHT. RULE.-Multiply the number of cubic inches contained in the weight by .2607-the weight of a cubic inch of cast-iron. Example. Taking a cast-iron weight 8 inches in diameter and 12 inches in length and we have: .7854 8X8-64 Square of diameter. 3 1416 47 124 50.2656 12 Length of weight. 100 5312 502 656 603.1872 Number of cubic inches in the weight. .2607 Weight of a cubic inch of cast-iron. 42223104 36 191232 120 63744 157.25090304 Number of pounds contained in the weight. TO DETERMINE THE NUMBER OF POUNDS CONTAINED IN A WROUGHT- IRON LEVER OF A SAFETY-VALVE. RULE.-Multiply the number of cubic inches of iron, in the lever by .2817; which is the weight of one cubic inch of wrought-iron. Example. Taking a lever of uniform section, for example, having a length of 40 inches, a width of 2.5 inches and a thickness of .5 of an inch, we then multiply the length by the width, the product by the thickness, and the last product by .2817. Thus: Mathematics of Steam Engineering. 45 40 Length of lever. 2.5 Width of lever. 200 80 100.0 5 Thickness of lever. 50.00 Cubic inches. Hence there are 50 cubic inches of iron contained in the lever. The next operation is to find the weight of the lever. Thus: .2817 Weight in decimals of a pound of one cubic inch of wrought-iron. 50 Number of cubic inches contained in lever. 14.0850 Number of pounds weight contained in the lever, which it will be observed are 14 and 8 one hundredths of a pound. TO DETERMINE THE NUMBER OF POUNDS WEIGHT CONTAINED IN A WROUGHT-IRON VALVE SPINDLE. RULE.-First square the diameter of the spindle; then multiply the product by .7854, then multiply the product of that operation by the length of the spindle; then multiply the last product by .2817; the answer will give the number of pounds contained in the spindle. Example.-Taking a spindle of an inch in diameter and 8 inches in length, and we have, First: 8) 7.0 (0.875 The diameter of the spindle in decimals of an inch. 64 66 60 56 40 40 40 Then: .875 .875 4375 6125 7000 .765625 Square of the diameter of the spindle. .7854 3062500 3828125 6125000 5359375 Am't carried forward, .6013218750 46 A Library of Steam Engineering. Am't brought forward, .6013218750 8 4.8105750000 Number of cubic inches of iron in the .2817 33674025 4810575 38484600 9621150 spindle. As the ciphers have no value they are dropped, and that leaves but 6 decimals. 1.3551389775 Number of pounds contained in the spindle. It will be observed that it weighs a fraction over one pound and one-third of a pound. TO DETERMINE THE WEIGHT OF THE VALVE WITHOUT WEIGHING IT. RULE.-Immerse the valve in a vessel containing enough water to sub- merge the valve, and just large enough in diameter to contain the valve, so as to show as much displacement of water as possible; note the rise of water in the vessel, and from the measurement of this rise determine the number of cubic inches of water displaced by the submerging of the valve, and that will represent the number of cubic inches of metal in the valve. If the valve is of cast-iron, multiply the number of cubic inches of water displaced by .2607; but if the valve is of brass, multiply the number of cubic inches of water displaced by .3194. Example.-Taking a vessel 6" in diameter, and 8" in height, and containing 6" of water before the valve is submerged therein, and 71" of water after the valve is submerged, we then have a rise of 1" of water in a vessel 6" in diameter. Then 6 times 6 are 36, the square of the diameter of the space occupied by the water. Then: .7854 6x6 36 Square of the diameter of the vessel. 47124 23562 28.2744 Number of square inches in the diameter. 1.5 Rise of water in the vessel. 14 13720 28 2744 42.41160 Number of cubic inches contained in the rise of water by submerging the valve. Now suppose the valve to be of brass, then we have: 42.4116 .3194 1696464 3817044 424116 12 72348 13.54626504 Number of pounds contained in tne valve. Mathematics of Steam Engineering. 47 In order that the student may not be misled, it will be here observed that this is a large valve, say 6" in diameter, with wings for guides extending down into the valve opening below the seat. TO DETERMINE THE WEIGHT OF A CAST-IRON BALL OR SPHERE. RULE.-Multiply the cube of the diameter of the ball by .5236, and then multiply the last product by .2607, the answer will give the weight in pounds. Example.-Let 8" equal diameter of a spherical cast-iron weight. Then we have 8X8X8X.5236X.2607-69.88+lbs. Or 83X.5236X.2607-69.88+lbs. Putting it in the ordinary way, we have: 8 Diameter of sphere. 8 Diameter of sphere. 64 Square of diameter of sphere. 8 Diameter of sphere. 512 Cube of diameter. 5236 1 0472 5 236 261 80 268.0832 Area of sphere in square inches. .2607 Weight of a cubic inch of cast-iron. 18765824 16 084992 53 61664 69.88929024 lbs. Weight of the sphere. The student is now prepared to advance one step and consider the adjustment of the safety-valve mathematically, hence, the next chapter will be devoted to that subject. 48 A Library of Steam Engineering. CHAPTER II. THE SAFETY-VALVE MATHEMATICALLY CONSIDERED. The first step necessary in the elucidation of this subject is to adjust, by the aid of a diagram, a safety-valve as it is found in practice, and then to explain in detail the various points connected therewith. Recourse will be had to skeleton drawings because of their simplicity and because they come more readily within the comprehension of those in need of the information here imparted. The following drawing shows a practical illustration of the adjustment of the safety-valve and it will constitute the foundation upon which the instructions in this chapter are based. F S VS C V W Fig. 8 NAMES OF DIFFERENT PARTS OF THE LEVER. F represents the fulcrum-the point on which the lever turns. A common error committed by many engineers is to call the dis- tance from F to S the fulcrum when it is nothing of the kind. S represents the length of the short arm of the lever; and the short arm of the lever is that portion of the lever between F and S. C represents the center of gravity of the lever, that is the point on which the lever will balance on a knife-edge. In a lever of uniform section the center of gravity will be at or near the middle in the length of the lever. If the lever is larger at the fulcrum and tapers toward the outer end, the center of gravity will be nearer the fulcrum. L represents the long arm of the lever, and therefore the long arm of the lever is that portion of the lever between F and L. Vs represents the valve spindle. V represents the valve. CENTER OF GRAVITY OF THE LEVER. As the center of gravity of the lever plays an important part in the adjustment of all lever safety-valves, mathematically, and without The Safety-valve Mathematically Considered. 49 a consideration of which no lever valve can be accurately adjusted mathematically, it is therefore necessary that the student should thor- oughly understand what use is to be made of the center of gravity of the lever in making his calculations in determining the pressure required to raise the valve, the distance the weight is to be placed from the fulcrum to allow the valve to rise at any given pressure, to deter- mine the number of pounds a weight must contain to be placed a given distance on the lever from the fulcrum; in short, in making all calculations connected with the adjustment of the valve, mathemat- ically, or in determining the weight or dimensions of any and all por- tions of the safety valve. All this may seem complicated to the beginner, and yet the whole matter connected with the center of gravity of the lever can be summed up in the following words: The only use to be made of the center of gravity of the lever is to determine the amount of pressure produced on the valve by the weight of the lever. It is well known that the pressure produced on the valve by the lever depends upon the weight of the lever, the distance of the center of gravity of the lever from the fulcrum, and the length of the short arm of the lever. Hence, to determine the total pressure-not per square inch-the lever produces on the valve, we multiply the distance in inches of the center of gravity of the lever from the fulcrum by the weight of the lever in pounds, and then divide the product by the length of the short arm of the lever in inches, the quotient will give the total pressure on the valve produced by the weight of the lever. To illustrate we will make use of the diagram (Fig. 9). 4" 15" Λ 16 lbs. weight of lever. Fig. 9 4" represents the length of the short arm of the lever in inches or the distance from the fulcrum to the center of the valve. 15" represents the distance in inches of the center of gravity of the lever from the fulcrum. The weight of the lever is 16 pounds, as shown in Fig. 9, and to determine the amount of pressure it produces on the valve we will pro- ceed according to the rule above laid down. Then we have: 15" Distance of center of gravity from the fulcrum.. Weight of the lever. 16 lbs. 90 15 Length of short arm of the lever. 4 4") 240 60 lbs. The total pressure produced on the valve by the weight of the lever. 50 A Library of Steam Engineering. This is the same as though a weight of 60 pounds was placed on the top of the valve. If it is required to determine the amount of pressure per square inch produced on the valve by the weight of the lever, we divide the total pressure produced on the valve by the area of the valve, that is by the number of square inches in the valve, and the quotient will give the pressure per square inch produced on the valve by the weight of the lever. To illustrate this subject a little further, the student is informed that he must treat the center of gravity of the lever the same as he would a weight equal to the weight of the lever, hung on the lever at the center of gravity. In the case above mentioned the distance of the center of gravity from the fulcrum is 15 inches and the weight of the lever is 16 pounds, and the total pressure produced on the valve by the weight of the lever is 60 pounds. Now, in addition to that, if a weight of 16 pounds is placed on the lever-that is the center of the weight-15 inches from the fulcrum that weight would produce a pressure on the valve of 60 pounds also, and the two combined would produce a pressure of twice 60 pounds or 120 pounds. We are now prepared to advance another step, and for the purpose of simple illustration we will continue to employ the diagram method, and adjust a safety-valve complete as found in every day practice. TO DETERMINE THE PRESSURE PER SQUARE INCH REQUIRED TO RAISE THE FOLLOWING VALVE. F 4" 8 lbs. weight of valve and spindle. 4" diam. of valve. 12" 大 ​16 lbs. weight of lever. 24" 150 lbs. Fig. 10 The information required to determine the pressure per square inch necessary to raise a safety-valve as shown in Fig. 10, is shown in the diagram, with the exception of the area of the valve, which should be determined in order to make a diagram complete. In order then to aid the student and to simplify this subject we will construct a diagram for each operation required to be made, and place only such figures in the diagram as may be required to perform each particular operation. The first thing, then, to make the diagram complete is to determine the area of the valve, for which purpose we will employ the following diagram. The Safety-valve Mathematically Considered. 51 4" diam. of valve. Fig. 11 TO DETERMINE THE AREA OF THE VALVE. RULE.-Square the diameter of the valve, that is, multiply the diameter by the diameter; then multiply the product by .7854, the last product will give the area of the valve. Example.-Taking a valve 4" in diameter, we have: 4X4X.7854-12.5664 square inches. Area of the valve. To put it in simple form, we have: 4 4 16 .7854 4 7124 7 854 12.5664 Area of the valve in square inches. This enables us to make the diagram complete as follows: TO DETERMINE THE PRESSURE PER SQUARE INCH REQUIRED TO RAISE THE FOLLOWING VALVE. F 4" 8 lbs. weight of valve and spindle. 4" 12.5664 area of valve. 12" Λ 24" 16 lbs. weight of lever. Fig. 12 150 lbs. F represents the fulcrum. 4" represents the length of the short arm of the lever, or, the dis- tance of the center of the valve from the fulcrum. 12" represents the distance of the center of gravity of the lever from the fulcrum. 24" represents the length of the long arm of the lever, and also the distance of the weight from the fulcrum. 8 lbs. represents the weight of the valve and spindle. 16 lbs. represents the weight of the lever. 4" represents the diameter of the valve. 12.5664 represents the area of the valve in square inches. 150 lbs. represents the number of pounds contained in the weight. 52 A Library of Steam Engineering. The first thing the student is required to do is to familiarize him- self thoroughly with either of the following rules in determining the pressure, per square inch, required to raise the valve. RULES FOR DETERMINING THE PRESSURE REQUIRED TO RAISE THE VALVE. RULE I.-First, divide the distance of the center of gravity of the lever from the fulcrum by the length of the short arm of the lever, and then multiply the quotient by the weight of the lever. Call this product "Product No. 1." Second, divide the distance in inches the center of the weight is hung on the lever from the fulcrum by the distance in inches the center of the valve is from the fulcrum, and then multiply the quotient by the number of pounds contained in the weight, and call this product "Product No. 2." Third, add the weight of the valve and the spindle and Products Nos. 1 and 2 together and call this "The Sum." Fourth, divide "The Sum" by the area of the valve, and the quotient will give the pressure, per square inch, required to raise the valve. Where there are fractions in the weights or distances the following method will be found the simplest : RULE II.-First, multiply the weight of the valve and spindle in pounds by the distance in inches of the center of the valve from the fulcrum, and call this product "Product No. 1." Second, multiply the distance in inches of the center of gravity of the lever from the fulcrum by the weight of the lever in pounds, and call this product "Product No. 2." Third, multiply the distance in inches of the center of the weight from the fulcrum by its weight in pounds, and call this product "Product No. 3." Fourth, add Products Nos. 1, 2 and 3 together and call the answer "The Sum." Fifth, multiply the area of the valve by the distance in inches of its center from the fulcrum, and call this product "Product No. 4." Sixth, divide "The Sum" by "Product No. 4" and the quotient will give the pressure per square inch required to raise the valve. These rules, it will be observed, differ very materially from each other, yet the answer in both cases will be exactly the same. We will now proceed according to the first rule to determine the pressure required to raise the valve as adjusted in Fig. 12, by the aid of the following diagrams. LL F 4" FIRST OPERATION. 12" Center of gravity of lever. A 16 lbs. weight of lever. Fig. 13 The Safety-valve Mathematically Considered. 53 Figure 13, it will be observed, contains only such figures as are required to perform the first operation according to the first paragraph of the first rule, as follows: "First, divide the distance of the center of gravity of the lever from the fulcrum by the length of the short arm of the lever, and then multiply the quotient by the weight of the lever, and call the product 'Product No. 1.'" Example.-Let 4 inches equal length of short arm of the lever. Let 12 inches equal distance of center of gravity of the lever from the fulcrum. Let 16 pounds equal weight of the lever. Then we have: 12 4 X16-48 lbs. Pressure produced on the valve by the weight of the lever. Putting it in different form, we have: 4|12 3 16 48 lbs. This we set aside and call it "Product No. 1." We will perform the second operation by the aid of the following diagram, in determining the pressure produced on the valve by the weight F 4" SECOND OPERATION.olt 24" 150 lbs. Fig. 14 Figure 14 contains only such figures as are required to perform the operation as laid down in the second paragraph of the first rule, which is as follows: "Second, divide the distance in inches of the center of the weight hung on the lever from the fulcrum by the length of the short arm of the lever in inches, and then multiply the quotient by the number of pounds contained in the weight, and call this product 'Product No. 2.'" Example. Let 4 inches equal length of short arm of the lever. Let 24 inches equal distance of the center of the weight from the fulcrum. Let 150 pounds equal the number of pounds contained in the weight. 54 A Library of Steam Engineering. Then we have: 24 4 X150-900 lbs. Pressure produced on the valve by the weight of the weight. Putting it in different form, we have: 4) 24 6 150 900 Which we set aside and call it "Product No. 2" 900 lbs. We will perform the next operation by determining the pressure produced on the valve seat by the weight of the valve and spindle. 8 lbs. weight of valve and spindle. THIRD OPERATION. Fig. 15 It will be observed that there is but one figure connected with the diagram proper, and that represents the weight of the valve and spin- dle, and as that weight is directly over the valve seat, no operation is required to determine the pressure produced on the seat, but simply follow the directions as laid down in the third paragraph of the first rule, which is as follows: 'Third, add the weight of the valve and spindle to Products Nos. 1 and 2 and call this 'The Sum."" Example.-Let 8 pounds equal the weight of the valve and spindle. Let 48 pounds equal "Product No. 1." Let 900 pounds equal "Product No. 2." And we have: 8 48 900 956 Total pressure on the valve seat produced by the weight of valve, spindle, lever and weight, which we set aside and call it "The Sum" 956 lbs. FOURTH OPERATION. "The sum". 956 lbs. 12.5664 area of valve. Fig. 16 Here, also, all the figures required to perform the operation are in The Safety-valve Mathematically Considered. 55 the diagram; so we will proceed according to directions laid down in the fourth paragraph of the first rule as follows: "Fourth divide The Sum' by the area of the valve and the quotient will give the pressure, per square inch, required to raise the valve." Example.-Let 12.5664 square inches equal the area of the valve. Let 956 pounds equal "The Sum." Then we have: 12.5664) 956.0000 (76.07+ Pounds per square inch required 879 648 to raise the valve. 76 3520 75 3984 953600 879648 Putting the example in shorter form, we have: 956.0000 12.5664 =76.07+ lbs. We will now proceed to perform the various operations in accord- ance with directions contained in Rule 2. TO DETERMINE THE PRESSURE, PER SQUARE INCH, REQUIRED TO RAISE THE FOLLOWING VALVE, ACCORDING TO RULE 2. F 4" 8 lbs. weight of valve and spindle. 4" diam. of valve. 12" 16 lbs. weight of lever. 24" 150 lbs. Fig. 17 It will be observed that the area of the valve is not given in the diagram. The first thing then to be done to complete the diagram is to determine the area of the valve, and the method for determining that is the same in Rule 2 as it is in Rule 1, so we will begin with the use of the following diagram: 4" diam. of valve. Fig. 18 TO DETERMINE THE AREA OF THE VALVE. RULE.-Square the diameter of the valve, that is, multiply the diameter by the diameter, then multiply the product by .7854, the last product will give the area of the valve. $56 A Library of Steam Engineering. Example.-Taking a valve 4 inches in diameter, as shown in Fig. 18, we have: 4X4X.7854-12.5664 square inches-area of the valve. Putting it in simple form, we have: 4 4 16 .7854 4 7124 7 854 12.5664 Area of the valve in square inches. This enables us to complete the diagram as follows: TO DETERMINE THE PRESSURE PER SQUARE INCH REQUIRED TO RAISE THE FOLLOWING VALVE. F 4" 8 lbs. weight of valve and spindle. 4" diam. of valve. Area of valve 12.5664. 12" 24" 16 lbs. weight of lever. Fig. 19 150 lbs. F represents the fulcrum. 4" represents the length of the short arm of the lever, or the dis- tance in inches of the center of the valve from the fulcrum. 12″ represents the distance in inches of the center of gravity of the lever from the fulcrum. 24" represents the length in inches of the long arm of the lever, and also the distance of the weight from the fulcrum. 8 lbs. represents the weight of the valve and spindle. 16 lbs. represents the weight of the lever. 150 lbs. represents the weight of the weight. 4" represents the diameter of the valve in inches. 12.5664 represents the area of the valve in square inches. We will now proceed according to Rule 2 to determine the pressure per square inch required to raise the valve as adjusted in Fig. 19. FIRST OPERATION. F 4" 8 lbs. weight of valve and spindle. Fig. 20 The Safety-valve Mathematically Considered. 57 RULE.-Multiply the weight of the valve and spindle by the length of the short arm of the lever, and call this product "Product No. 1." Example.-Let 4 inches equal length of the short arm of the lever. Let 8 pounds equal weight of valve and spindle. Then we have 8X4-32, "Product No. 1." Putting it in simpler form, we have: F 84 32 "Product No. 1." SECOND OPERATION. 12" т 16 lbs. weight of lever. Fig. 21 RULE.-Multiply the distance in inches of the center of gravity of the lever from the fulcrum by the weight of the lever in pounds, and call the pro- duct "Product No. 2." Example. -Let 12 inches equal distance of center of gravity of the lever from the fulcrum. Let 16 pounds equal weight of lever. Then we have 12X16-192, "Product No. 2." Putting it in different form, we have: 12 10 16 32 16 192 "Product No. 2." THIRD OPERATION. 24" 150 lbs. Fig. 22 RULE.-Multiply the number of pounds contained in the weight by the distance in inches of its center from the fulcrum, and call this product "Pro- duct No. 3." 58 A Library of Steam Engineering. Example.-Let 24 inches equal distance of center of weight from the fulcrum. Let 150 pounds equal weight of the weight. Then we have 150X24-3600, "Product No. 3." Putting it in the following form, we have: 150 24 600 300 3600 "Product No. 3." FOURTH OPERATION. As this operation has reference to the result of the operations relat- ing to the preceding diagrams, it needs no diagram for the illustration. So we will proceed according to the following rule: purpose of RULE.-Add Products Nos. 1, 2 and 3 together and call the answer "The Sum." Example.-Let 32 equal "Product No. 1." Let 192 equal "Product No. 2." Let 3600 equal "Product No. 3." Then we have 32+192+3600-3824. "The Sum." Putting it in the ordinary form, we have: F 4" 12.5664 area of valve. 32 192 3600 3824 "The Sum.' FIFTH OPERATION. Fig. 23 RULE.-Multiply the area of the valve in square inches by the distance in inches of its center from the fulcrum, and call this product " Product No. 4." Example.-Let 4 inches equal distance of center of valve from the fulcrum. Let 12.5664 square inches equal area of the valve. Then we have 12.5664X4=50.2656. "Product No. 4." The Safety-valve Mathematically Considered. 59 Putting it in the ordinary form, we have: 12.5664 4 50.2656 "Product No. 4. " SIXTH OPERATION. As the various answers requiring the use of diagrams have been obtained by the aid of such diagrams, this operation does not require any diagram. So we will proceed at once according to the rule. RULE.—Divide " The Sum" by "Product No. 4," and the quotient will give the pressure, per square inch, required to raise the valve as adjusted in Fig. 19. Example.-Let 3824 equal "The Sum. Let 50.2656 equal "Product No. 4," then we have: 3824 50.2656 76.07+ Pounds pressure per square inch required to raise the valve. Performing the operation in the ordinary manner, we have: 50.2656) 3824.0000 (76.07+ Pounds pressure per square 3518592 305 4080 301 5936 3 814400 3 518592 inch required to raise the valve. TO DETERMINE THE DISTANCE ON THE LEVER TO PLACE THE GIVEN WEIGHT. Having determined the pressure required to raise the valve, we will now proceed to determine the distance a given weight will have to be placed on the lever to allow the valve to rise at a given pressure; and as these problems will gradually become more and more complicated, as we proceed toward the end of the chapter, we will employ a valve requiring the use of small numbers, and in this way keep the work down to its utmost simplicity. We will, therefore, construct a small valve, and first determine the pressure per square inch, according to Rule 2, required to raise it, and all calculations made hereafter to the end of this chapter, will have reference to this valve for the purpose of showing the correctness of the rule as well as the correctness of the calculations made. 60 A Library of Steam Engineering. TO DETERMINE THE PRESSURE PER SQUARE INCH REQUIRED TO RAISE THE FOLLOWING VALVE. 4" x F 2 8" 2 lbs. weight of valve and spindle. 4 lbs. weight of lever. -2" diam. of valve. 2x2x.7854 = 3.1416 area of valve. Fig. 24 36.77 lbs. Proceeding according to Rule 2, we have: First, the distance in inches of the center of the valve from the fulcrum multiplied by the weight of the valve and spin- dle in pounds, 2X2="Product No. 1." Second, the distance in inches of the center of gravity of the lever from the fulcrum multiplied by the weight of the lever in pounds, 4X4 " Product No. 2. " Third, the distance in inches of the center of the weight from the fulcrum multiplied by the number of pounds con- tained in the weight, 8×36.77 equal "Product No 3." Fourth, Products 1, 2 and 3 added together " The Sum," Fifth, the area of the valve, multiplied by the distance in inches of its center from the fulcrum equals "Product No. 4" 2X2X.7854-3.1416 area of valve, and 3.1416 X2=6.2832," Product No. 4." Sixth, "The Sum" divided by "Product No. 4" 4 16 294.16 314.16 The Sum" 314.16 -=50 pounds. "Product No. 4" 6.2832 This, it will be observed, gives 50 pounds pressure per square inch required to raise the valve. Putting it in another form, we have: (2x2)+(4X4)+(8x36.77) 50 pounds. 2X2X.7854×2 Performing the operation, we have: 2" Distance in inches of center of valve from fulcrum. 2 Weight of valve and spindle in pounds. 4 "Product No. 1." 4" Distance of center of gravity of lever from fulcrum. 4 Weight of lever in pounds. 16 "Product No. 2. دو The Safety-valve Mathematically Considered. 61 36.77 Distance of weight from fulcrum. Weight of weight in pounds. 8" 294.16 "Product No. 3. " Adding Products Nos. 1, 2 and 3 together, we have: 4 16 294. 16 314. 16 "The Sum." Performing the operation below the line in the example, we have: 2" ลด Diameter of valve. Diameter of valve. 2" 4 Square of the diameter of valve. .7854 3.1416 2" 6.2832 A constant. Area of valve.. Distance of center of valve from fulcrum. "Product No. 4. ' Dividing "The Sum" by "Product No. 4," we have: 6.2832) 314.1600 (50 Pounds pressure per square inch required to raise the valve. 314 160 0 TO DETERMINE THE DISTANCE IN INCHES A GIVEN WEIGHT IS TO BE PLACED ON THE LEVER FROM THE FULCRUM TO ALLOW THE VALVE TO RISE AT A GIVEN PRESSURE PER SQUARE INCH. To determine the distance from the fulcrum the weight in Fig. 25 will be required to be placed on the lever to allow the valve to rise at a pressure of 50 pounds per square inch. F 2 2 lbs. weight of 4" S" 4 lbs. weight of lever. 2" diam. of valve. valve and spindle. 2x2x .7854 = 3.1416 area of valve. 50 lbs. pressure. Fig. 25 36.77 lbs. RULE.-First, multiply the distance in inches the center of valve is from the fulcrum by the weight of the valve and spindle in pounds, and call the product "Product No. 1." 62 A Library of Steam Engineering. Second, multiply the distance in inches the center of gravity of the lever is from the fulcrum by the weight of the lever in pounds, and call the product "Product No. 2." Third, add Products Nos. 1 and 2 together and call the answer "The Sum." Fourth, multiply the distance in inches the center of the valve is from the fulcrum by the area of the valve in square inches and multiply the product by the required pressure per square inch, and call the last product" Product No. 3." Fifth, subtract "The Sum" from "Product No. 3" and call the answer "The Remainder." Sixth, divide "The Remainder" by the number of pounds contained in the given weight and the quotient will give the distance in inches the given weight is to be placed on the lever from the fulcrum to allow the valve to rise at the given pressure. F represents the fulcrum. 2" represents the distance in inches of the center of valve from the fulcrum. 4" represents the distance in inches of center of gravity of the lever from the fulcrum. 2 lbs. represents the weight of the valve and spindle. 4 lbs. represents the weight of the lever. 2" represents the diameter of the valve. 3.1416 represents the area of the valve in square inches. 50 lbs. represents the given pressure per square inch. 36.77 lbs. represents the weight of the weight. F FIRST OPERATION. 2" 2 lbs. weight of valve and spindle. Fig. 26 RULE.-Multiply the distance the center of the valve is from the fulcrum by the weight of the valve and spindle and call this product " Product No. 1." Example.-Let 2" equal distance of center of valve from fulcrum. Let 2 lbs. equal weight of valve and spindle. Then we have: 2X2=4" "Product No. 1." In the ordinary form, we have: 4 "Product No. 1. " 22 The Safety-valve Mathematically Considered. 63 F SECOND OPERATION. 4" 4 lbs. weight of lever. Fig. 27 RULE.-Multiply the distance the center of gravity of the lever is from the fulcrum by the weight of the lever and call this product " Product No. 2." Example.-Let 4" equal distance of center of gravity of the lever from the fulcrum. Let 4 lbs. equal weight of the lever. Then we have: 4X4-16, "Product No. 2." In the ordinary form, we have: 4 4 16 "Product No. 2. " THIRD OPERATION. As this operation embodies only the operations performed in the two preceding examples, a diagram is not necessary for illustration. RULE.-Add Products Nos. 1 and 2 together and call the answer "The Sum." Example.-Let 4 equal "Product No. 1." Let 16 equal "Product No. 2." Then we have: 4+16=20, "The Sum. " Putting it in the ordinary form, we have: 16 4 20 "The Sum. " FOURTH OPERATION. F 2' 3.1416 area of valve. 50 lbs. given pressure per sq. inch. Fig. 28 RULE.-Multiply the area of the valve by the distance of center of the valve from the fulcrum, then multiply the product by the given pressure per square inch, required to raise the valve, and call the last product "Product No. 3." 64 A Library of Steam Engineering. Example.-Let 3.1416 square inches equal area of the valve. Let 2" equal distance of center of valve from fulcrum. Let 50 lbs. equal given pressure per square inch required to raise the valve. Then we have: 3.1416×2×50-314.16, " Product No. 3." Putting it in the ordinary form, we have: 3.1416 2 6.2832 50 314.1600 "Product No. 3. " FIFTH OPERATION. NOTE. This operation can not be illustrated by diagram. RULE.-Subtract "The Sum" from "Product No. 3" and call the answer "The Remainder." Example.-Let 314.16 equal "Product No. 3." Let 20 equal "The Sum." Then we have: 314.16-20-294.16, "The Remainder." Putting it in the ordinary form, we have: 314.16 20 294.16 "Product No. 3." 66 'The Sum." "The Remainder. " NOTE.-The student will notice that "The Remainder" contains two decimals and that "The Sum" contains, none-the number 20 being a whole number. In subtraction the units in the sub- trahend must always be placed under the units in the minuend, the tens under tens, hundreds under hundreds, and so on. As there are no units in decimals, and as decimals begin with tenths which are followed toward the right by hundredths, thousandths and so on; the same care must be exercised in subtracting decimals as there is in the case of whole numbers, all the denomina- tions in the subtrahend must be placed under their respective denominations in the minuend. SIXTH OPERATION. The remainder 294.16 36.77 lbs. Fig. 29 RULE.-Divide "The Remainder" by the number of pounds contained in the weight, and the quotient will give the distance in inches the center of the weight will have to be placed on the lever to allow the valve to rise at the given pressure per square inch. The Safety-valve Mathematically Considered. 65 Example.-Let 294.16 equal "The Remainder." Let 36.77 equal the number of pounds contained in the weight. Then we have: 294.16-36.77-8 inches, the distance from the fulcrum the center of the weight will have to be placed to allow the valve to rise at a pressure of 50 pounds per square inch. Putting it in the ordinary form, we have: 36.77) 294.16 (8 inches 294.16 To put the entire operation in condensed form: Example.-Let 2 inches equal distance of center of valve from fulcrum. aborato Let 2 pounds equal weight of valve and spindle. Let 4 inches equal distance of center of gravity of lever from the fulcrum. Let 4 pounds equal weight of lever. Let 3.1416 square inches equal area of valve. Let 50 pounds equal pressure per square inch required to raise the valve. Let 36.77 pounds equal the weight of the weight. Then we have: First, the distance of the center of the valve from the fulcrum multiplied by the weight of the valve and spindle, 2X2 "Product No. 1." Second, the distance of the center of gravity of the lever from the fulcrum multiplied by the weight of the lever, 4X4" Product No. 2." 4 16 99 20 Third, Products Nos. 1 and 2 added together= "The Sum. Fourth, the distance of the center of the valve from the ful- crum multiplied by the area of the valve, and the pro- duct of this operation multiplied by the given pressure per square inch, 2x3.1416×50="Product No. 3." Fifth, "The Sum" subtracted from "Product No. 3, " 314.16— 20"The Remainder." Sixth, "The Remainder" divided by the number of pounds contained in the weight, 294.16-36.77-8 inches, the distance the weight must be placed from the fulcrum. Putting it in another form, we have: (2×3.1416×50) — (2×2)+(4×4) 36.77 8 inches. 314.16 294.16 5 Οι 66 A Library of Steam Engineering. Putting it in different form, we have: 3.1416×2×50=314.16 (2X2) + (4X4) = 20 Dividing by the weight, we have: 36.77) 294.16 (8 inches. 294.16 Performing the operation in the ordinary way, we have: 2X2= 4 "Product No. 1. " 4X4 16 "Product No. 2. " 20 "The Sum." 3.1416 Area of valve. 2" Distance of fulcrum from center of valve. 6.2832 50 Given pressure per square inch. 314.1600 20 66 Product No. 3. " 294.16 "The Sum." "The Remainder. " We now divide "The Remainder" by the number of pounds contained in the weight, as follows: 36.77) 294.16 (8 inches. The distance the weight must 294.16 be placed from the fulcrum to allow the valve to rise at a pressure of 50 pounds per square inch. TO DETERMINE THE AMOUNT OF WEIGHT REQUIRED TO BE PLACED A GIVEN DISTANCE FROM THE FULCRUM TO ALLOW THE VALVE TO RISE AT A GIVEN PRESSURE PER SQUARE INCH. RULE.-First, multiply the distance the center of the valve is from the fulcrum by the weight of the valve and spindle, and call the product "Product No. 1." Second, multiply the distance the center of gravity of the lever is from the fulcrum by the weight of the lever, and call the product" Product No. 2. " Third, add Products Nos. 1 and 2 together and call the answer " The Sum." Fourth, multiply the distance the center of the valve is from the fulcrum by the area of the valve and multiply the product by the pressure, per square inch, required to raise the valve, and call the last product" Product No. 3." Fifth, subtract" The Sum" from "Product No. 3" and call the answer "The Remainder." Sixth, divide" The Remainder" by the distance, in inches, the weight is required to be placed from the fulcrum, and the quotient will give the number of pounds required in the weight. The Safety-valve Mathematically Considered. 67 To determine the number of pounds required in the weight to be placed on the lever, 8 inches from the fulcrum, in the following diagram, to allow the valve to rise at a pressure of 50 pounds per square inch. F 2" 2 lbs. weight of valve and spindle. 4" X 4 lbs. weight of lever. 8" 2" diam. of valve. 2x 2x .7854 = 3.1416 area of valve. 50 lbs. pressure per sq. inch required. Fig. 30 F represents the fulcrum. 2" represents the distance of the center of the valve from the fulcrum. 4" represents the distance of the center of gravity of lever from the fulcrum. 8" represents the distance the weight is to be placed from the fulcrum. 2 lbs. represents the weight of the valve and spindle. 4 lbs. represents the weight of the lever. 2" represents the diameter of the valve. 3.1416 represents the area of the valve in square inches. 50 pounds represents the pressure per square inch required to be carried. FIRST OPERATION. F 2 2 lbs. weight of valve and spindle. Fig. 31 RULE.-Multiply the distance the center of the valve is from the fulcrum by the weight of the valve and spindle and call the product " Product No. 1." Example.-Let 2" equal the distance of the center of the valve from the fulcrum. Let 2 lbs. equal the weight of the valve and spindle. Then we have: 2X2-4, "Product No. 1." In the ordinary form, we have: 22 4 "Product No. 1" De 68 A Library of Steam Engineering. F SECOND OPERATION. 4" 4 lbs. weight of lever. Fig. 32 RULE.-Multiply the distance the center of gravity of the lever is from the fulcrum, by the weight of the lever, and call the product" Product No. 2." Example.-Let 4" equal distance of center of gravity of lever from the fulcrum. Let 4 lbs. equal weight of lever. Then we have: 4X4=16, "Product No. 2." 183068 ent THIRD OPERATION. RULE.-Add Products Nos. 1 and 2 together and call the answer "The Sum. دو bac 1." Example.-Let 4 equal " Product No. 1. Let 16 equal "Product No. 2." Then we have: 4+16-20, "The Sum." Putting it in the ordinary form, we have: 16 4 20 "The Sum.' 99 FOURTH OPERATION. F 2 3.1416 area of valve. 50 lbs. pressure required. Fig. 33 RULE.-Multiply the distance the center of the valve is from the fulcrum by the area of the valve, and multiply the product by the pressure per square inch required to raise the valve, and call the last product, " Product No. 3." Example. Let 2" equal distance of center of valve from the fulcrum. Let 3.1416 square inches equal area of the valve. Let 50 lbs. equal the given pressure. Then we have: 2x3.1416X50=314.16, "Product No. 3. " The Safety-valve Mathematically Considered. 69 Putting it in the ordinary form, we have: 3.1416 2 6.2832 50 314.1600 "Product No. 3. " FIFTH OPERATION. RULE.-Subtract "The Sum" from "Product No. 3," and call the answer "The Remainder. وو Example.-Let 314.16 equal "Product No. 3." Let 20 equal" The Sum." Then we have: 314.16-20-294.16, "The Remainder." Putting it in the ordinary form, we have: 314.16 20 294.16 "The Remainder. " un bean m SIXTH OPERATION. 294.16 the remainder. I 8" Fig. 34 RULE.-Divide "The Remainder" by the distance in inches the weight is required to be placed from the fulcrum and the quotient will give the number of pounds required in the weight to allow the valve to rise at a given pressure, which in this case is 50 pounds per square inch. Example.-Let 294.16 equal "The Remainder." Let 8" equal the distance the weight is to be placed from the fulcrum. Then we have: 294.16-8-36.77 pounds. Or, which is the same: 294.16 8 =36.77 pounds. 70 A Library of Steam Engineering. Putting it in the ordinary form, we have: 8) 294.16 (36.77 24 54 48 Pounds required in the weight placed 8 inches from the ful- crum to allow the valve to rise at a pressure of 50 lbs. per square inch. 4 202 61 56 56 56 NOTE.-The attention of the student is called to the last operation. It will be noticed that the dividend contains two decimals, while there is none in the divisor. In all cases of that kind whenever the first decimal is brought down and annexed to the remainder, a decimal point must be put after the last figure in the quotient, and all figures to the left of the decimal point are whole numbers, and all figures which follow to the right of the decimal point are decimals. Putting the foregoing operations in condensed form: Example.-Let 2 inches equal distance of center of valve from fulcrum. Let 2 pounds equal weight of valve and spindle. Let 4 inches equal distance of center of gravity of lever from the fulcrum. Let 8 inches equal distance the weight is to be placed on the lever from the fulcrum. Let 4 pounds equal weight of lever. Let 3.1416 square inches equal area of the valve. Let 50 pounds equal pressure per square inch required to raise the valve. Then we have: First, the distance of the center of the valve from the fulcrum multiplied by the weight of the valve and spindle, 2X 2="Product No. 1." Second, the distance of the center of gravity of the lever from the fulcrum multiplied by the weight of the lever, 4× 4" Product No. 2." Third, Products Nos. 1 and 2 added together equal "The Sum." Fourth, the distance of the center of the valve from the ful- crum, multiplied by the area of the valve, and the pro- duct of this operation multiplied by the given pressure per square inch, 2x3.1416X50" Product No. 3." Fifth, "The Sum" subtracted from "Product No. 3, " 314.16- "The Remainder." 20 Sixth, "The Remainder" divided by the distance in inches the weight is to be placed from the fulcrum, 29.416÷-8= 4 16 20 20 314.16 294.16 The Safety-valve Mathematically Considered. 71 36.77 pounds required in the weight 8 inches from the fulcrum to allow the valve to rise at a pressure of 50 pounds per square inch. Putting it in another form, we have: (2×3.1416×50)-(2x2)+(4X4) 36.77 lbs. 8 Putting it in different form, we have. 2x3.1416×50=314.16 (2x2)+(4X4)= 20 Dividing by the distance the weight 8) 294.16 is to be placed from the fulcrum. 36.77 lbs. Performing the operation by the ordinary method, we have: 3.1416 Area of valve. 2" Distance of center of valve from fulcrum. 6.2832.int 50 Given pressure per square inch. 314.1600 "Product No. 3." mon zavel and to divers 20 The sum to be subtracted. 294.16 "The Remainder." We now divide the remainder by the distance in inches the weight is to be placed on the lever from the fulcrum. Thus: 8) 294.16 (36.77 Pounds required in the weight. 24 54 48 61 56 56 56 Levis ent TO DETERMINE THE LENGTH OF THE SHORT ARM OF THE LEVER ALLOW THE VALVE TO RISE AT A GIVEN PRESSURE PER SQUARE INCH. TO RULE.-First, multiply the distance of the center of gravity of the lever from the fulcrum, by the weight of the lever, and call the product "Product No. 1." Second, multiply the distance the weight is from the fulcrum by the number of pounds contained in the weight, and call this product " Product No. 2." 72 A Library of Steam Engineering. Third, add Products Nos. 1 and 2 together and call the answer "The Sum." Fourth, multiply the area of the valve by the given pressure, per square inch, and deduct the weight of the valve and spindle from the product and call the answer "The Remainder." Fifth, divide" The Sum" by "The Remainder," and the quotient will give the required length of the short arm of the lever in inches. To determine the length required of the short arm of the lever in the diagram (Fig. 35), to allow the valve to rise at a pressure of 50 pounds per square inch. F 4" A 8" 2 lbs. weight of valve and spindle. 2.x 2 x .7854 = 3.1416 area of valve. 50 lbs. pressure per sq. inch required. 4 lbs. weight of lever. -2" diam. of valve. 36.77 lbs. Fig. 35 F represents the fulcrum. 4" represents the distance of center of gravity of the lever from the fulcrum. 8" represents the distance the weight is from the fulcrum. 2 lbs. represents the weight of the valve and spindle. 4 lbs. represents the weight of the lever. 2" represents the diameter of the valve. 3.1416 represents the area of the valve in square inches. 50 lbs. represents the pressure per square inch required to raise the valve. F FIRST OPERATION. 4" Λ 4 lbs. weight of lever. Fig. 36 RULE.-Multiply the distance of the center of gravity of the lever from the fulcrum by the weight of the lever and call the product "Product No. 1." Example.-Let 4" equal distance of center of gravity of the lever from the fulcrum. Let 4 lbs. equal weight of the lever. Then we have: 4X4-16, "Product No. 1." Putting it in the ordinary form: 4" 4 Distance of center of gravity of lever from fulcrum. Weight of lever. 16 "Product No. 1. " The Safety-valve Mathematically Considered. 73 F SECOND OPERATION. 8" 36.77 lbs. Fig. 37 RULE.-Multiply the number of pounds contained in the weight by the distance the weight is from the fulcrum and call this product" Product No. 2." Example. Let 8" equal the distance the weight is from the ful- crum. Let 36.77 lbs. equal the weight of the weight. Then we have: 36.77X8-294.16, "Product No. 2." Putting it in the ordinary form : bariupeЯ 36.77 Weight of the weight. 8" Distance of center of weight from ful- 294.16 crum. "Product No. 2. " THIRD OPERATION. RULE. Add Products Nos. 1 and 2 together and call the answer "The Sum." Example.-Let 16 equal "Product No. 1." Let 294.16 equal "Product No. 2." Then we have: 16+294.16-310.16, "The Sum." Putting it in the ordinary form : 294.16 "Product No. 2. " "Product No. 1. " 16 310.16 "The Sum. " FOURTH OPERATION. 2 lbs. weight of valve and spindle. 3.1416 area of valve. 50 lbs. given pressure per sq. inch. Fig. 38 RULE.-Multiply the area of the valve by the given pressure per square inch and then deduct the weight of the valve and spindle from the product and call the answer "The Remainder." 74 A Library of Steam Engineering. Example.-Let 50 lbs. equal the given pressure per square inch. Let 2 lbs. equal the weight of the valve and spindle. Let 3.1416 square inches equal the area of the valve. Then we have (3.1416 × 50)-2=155.08, "The Remainder." Putting it in the ordinary form : 3.1416 Area of valve. 50 Given pressure. 157.0800 2 Weight of valve and spindle. 155.08 66 The Remainder." FIFTH OPERATION. RULE.-Divide "The Sum" by "The Remainder" and the quotient will give the length of the short arm of the lever, or in other words, the distance the center of the valve must be from the fulcrum to allow the valve to rise at a pressure of 50 pounds per square inch. Example.-Let 155.08 equal "The Remainder." Let 310.16 equal "The Sum." Then we have: 310.16155.08-2 inches. Required length of short arm of the lever. Putting it in the ordinary form, we have: 155.08) 310.16 (2 inches. Required length of short 310.16 Putting the foregoing operation in condensed form: arm of lever. Example.-Let 4 inches equal distance of center of gravity of lever from the fulcrum. Let 4 pounds equal weight of lever. Let 8 inches equal distance of weight from fulcrum. Let 36.77 pounds equal weight of the weight Let 3.1416 square inches equal area of valve. Let 50 pounds equal the given pressure. Let 2 pounds equal the weight of the valve and spindle. Then we have: (4X4)+(36.77X8) -2 inches. Length of short. arm of lever. (3.1416×50)-2 Putting it in the ordinary form, we have: 4X4 16 36.77X8=294.16 310.16 "The Sum." This completes the operation above the line in the preceding exam- ple and the answer is called "The Sum." The Safety-valve Mathematically Considered. 75 The next thing is to perform the operation below the line. Thus 3.1416 Area of valve. 50 Given pressure. 157.0800 2 Weight of valve and spindle to be subtracted. 155.08 "The Remainder." This completes the operation below the line in the example. We now divide the answer obtained above the line by the answer obtained below the line. Thus : Or: "The Sum." "The Remainder." 310.16155.08-2 inches. 155.08) 310.16 (2 inches. Length of short 310.16 arm of lever. TO DETERMINE THE DIAMETER OF THE VALVE REQUIRED TO ALLOW THE VALVE TO RISE AT A GIVEN PRESSURE. RULE.-First, multiply the distance of the center of the valve from the fulcrum by the weight of the spindle and estimated weight of the valve and call the product "Product No. 1." Second, multiply the distance of center of gravity of the lever from the fulcrum by the weight of the lever and call the product "Product No. 2." Third, multiply the distance the weight is from the fulcrum by the number of pounds contained in the weight and call the product "Product No. 3." Fourth, add Products Nos. 1, 2 and 3 together and call the answer" The Sum." Fifth, multiply the given pressure per square inch by the distance of center of valve from fulcrum, then divide "The Sum" by the product and call the quotient" Quotient No. 1." Sixth, divide "Quotient No. 1" by .7854 and extract the square root of the last quotient and the answer will give the diameter, in inches, of the valve required. TO DETERMINE THE DIAMETER OF THE VALVE REQUIRED IN THE DIAGRAM (FIG. 39), TO ALLOW THE VALVE TO RISE A PRESSURE OF 50 POUNDS PER SQUARE INCH. AT F 2" 2 lbs. 4" 4 lbs. 50 lbs. given pressure per sq. inch. 8" 36.77 lbs. Fig. 39 76 A Library of Steam Engineering. F represents the fulcrum. 2" represents the distance of the center of the valve from the fulcrum. 4" represents the distance of the center of gravity of the lever from the fulcrum. F 8" represents the distance of the weight from the fulcrum. 2 lbs. represents the weight of the valve and spindle. 50 lbs. represents the pressure per square inch required to be carried. 2" FIRST OPERATION. 2 lbs. weight of valve and spindle. Fig. 40 RULE.-Multiply the distance of the center of the valve from the fulcrum by the weight of the valve and spindle and call the product "Product No. 1." Example.-Let 2 lbs. equal weight of valve and spindle. Let 2" equal distance of center of valve from fulcrum. Then we have: 2x2=4, "Product No. 1." F Or, which is the same thing: 2" 2 Distance of center of valve from fulcrum. Weight of valve and spindle. 4 "Product No. 1." SECOND OPERATION. 4" 4 lbs. weight of lever. Fig. 41 RULE.-Multiply the distance of the center of gravity of the lever from the fulcrum by the weight of the lever and call the product " Product No. 2." Example. Let 4 lbs. equal weight of lever. Let 4" equal distance of center of gravity of lever from the fulcrum. Then we have: 4X4-16, "Product No. 2." Or, putting it in the ordinary form: 4" Distance of center of gravity from fulcrum. 4 Weight of lever. 16 'Product No. 2." The Safety-valve Mathematically Considered. 77 F THIRD OPERATION. 8" 36.77 lbs. Fig. 42 RULE.-Multiply the number of pounds contained in the weight by the distance of the weight from the fulcrum and call the product "Product No. 3." Example.-Let 36.77 lbs. equal weight of the weight. Let 8" equal distance of the weight from the fulcrum. Then we have: 36.77X8-294.16, "Product No. 3." Putting it in the ordinary form, we have: 36.77 Weight of the weight. 8" Distance of weight from fulcrum. 294.16 'Product No. 3" FOURTH OPERATION. RULE.-Add Products Nos. 1, 2 and 3 together and call the answer "The Sum." Example.-Let 4 equal "Product No. 1." Let 16 equal "Product No. 2." Let 294.16 equal "Product No. 3." Then we have: 4+16+294.16-314.16, "The Sum." Putting it in the ordinary form, we have: 294.16 "Product No. 3." 16 "Product No. 2." 4 'Product No. 1." 314.16 "The Sum." FIFTH OPERATION. F 2 3.1416 the sum. 50 lbs. given pressure per sq. inch. Fig. 43 RULE.-Multiply the given pressure by the distance of the center of the valve from the fulcrum, and divide "The Sum" by the product and call the quotient "Quotient No. 1." 78 A Library of Steam Engineering. Example.-Let 2" equal the distance of center of valve from the fulcrum. Let 50 lbs. equal the given pressure per square inch. Let 314.16 equal "The Sum." Then we have: 314.16÷(50X2)=3.1416, "Quotient No. 1." In the ordinary form, we have: 50X2=100) 314.16 (3.1416 "Quotient No. 1." 300 141 100 416 400 160 100 600 600 SIXTH OPERATION. RULE.-Divide "Quotient No. 1" by .7854 and then extract the square root of the last quotient, the answer will give the diameter, in inches, of the required valve. Example. Let 3.1416 equal "Quotient No. 1." Let .7854 equal a constant. Then we have: quired valve. 3.1416.7854-2 inches. Diameter of the re- Putting it in the ordinary form, we have: .7854) 3.1416 (4 Square of the diameter of the valve. 3.1416 We now extract the square root of the quotient. Thus: 4 (2 inches. Diameter of required valve. 4 Putting the foregoing operations in condensed form: Example. Let 2 inches equal distance of center of valve from the fulcrum. Let 2 pounds equal estimated weight of valve and spindle. Let 4 pounds equal weight of the lever. Let 4 inches equal distance of center of gravity of lever from the fulcrum. The Safety-valve Mathematically Considered. 79 36.77 pounds equal weight of the weight. 8 inches equal distance of center of weight from fulcrum. 50 pounds equal given pressure per square inch. .7854 equal a constant. Then we have: (2×2)+(4×4)+(36.77×8)÷(50×2) -=2 inches. .7854 Performing the operation, we have: First, the weight of the valve and spindle multiplied by the distance of the center of the valve from the fulcrum. Thus: 2×2 equal "Product No. 1." Second, the weight of the lever multiplied by the distance of the center of gravity of the lever from the fulcrum. Thus: 4X4 equal "Product No. 2." Third, the weight of the weight multiplied by the distance of the center of the weight from the fulcrum. Thus: 36.77 x8 equal" Product No. 3." Fourth, Products Nos. 1, 2 and 3 added together: Thus: 4 "Product No. 1." 16 "Product No. 2." 294.16 "Product No. 3." 4 16 294.16 314.16 "The Sum." Fifth, the given pressure per square inch, multiplied by the distance of the center of the valve from the fulcrum, then "The Sum divided by the product. Thus: 50 2 "The Sum." 100) 314.16 (3.1416 "Quotient No. 1." 300 141 100 416 400 160 100 600 600 Or, in this manner: 314.16 "The Sum." -3.1416 "Quotient No. 1." 50×2 80 A Library of Steam Engineering. Sixth, "Quotient No. 1" divided by .7854, and the square root extracted of the last quotient. Thus: 3.1416.7854-2 inches. Diameter of valve. Performing the operation in the ordinary way, we have: .7854) 3.1416 (4) 3.1416 And the square root of the quotient is: 4 (2 inches. Diameter of the required valve. 4 TO DETERMINE THE PRESSURE, PER SQUARE INCH, REQUIRED TO RAISE A VALVE WITHOUT WEIGHING ANY OF ITS PARTS. -- RULE. First, multiply the number of cubic inches of wrought-iron in the lever by 2817, the product will give the number of pounds contained in the lever. Second, multiply the number of cubic inches of cast-iron in the weight by .2607, and the product will give the number of pounds contained in the weight Third, square the diameter of the spindle, and multiply the product by .7854 and multiply the second product by the length of the spindle, then multi- ply the last product by .2817, the answer will give the weight of the spindle in pounds. Fourth, immerse the valve in a small vessel of water, and from the rise of water in the vessel determine the number of cubic inches of water displaced by the valve, then if the valve is of brass, multiply the cubic inches of water displaced by .3194, and if it is of cast-iron multiply by .2607. DETERMINE THE PRESSURE PER SQUARE INCH REQUIRED TO RAISE THE VALVE SHOWN IN FIG. 44. FIRST OPERATION. Length of lever 36".-. End view -4" Spindle 6/ Length of Center of weight from fulcrum 30. 34" diam. of spindle. 18" Center of gravity. Diam. 4" of brass valve. A 1/2 <---12/' --- The valve immersed in water displaced 20 cubic inches. Fig. 44 24 *2/* 12" RULE.-Multiply the length of the lever, in inches, by the width, then multiply the product by the thickness, then multiply the last product by .2817, the answer will give the weight of the lever in pounds. The Safety-valve Mathematically Considered. 81 Example. Let 36" equal length of the lever. Let 2" equal width of the lever. Let 5 tenths of an inch equal thickness of the lever. Let .2817 of a pound equal weight of one cubic inch of wrought-iron. Then we have: 36X2X.5X.2817-10.1412 lbs. Weight of lever. Putting it in the ordinary form, we have: 36 2 72 .5 36.0 .2817 1 6902 8451 10.1412 lbs. Weight of lever. This we will transfer to the following diagram, and continue until we get a complete diagram. F 4" 18" λ 10.1412 lbs. weight of lever. 4" diam. of valve. Fig. 45 SECOND OPERATION. RULE.-Multiply the length of the cast-iron weight by its breadth, then multiply the product by the thickness of the weight, then deduct the area of the hole in the weight from the last product, then multiply the remainder by 2607, the answer will give the number of pounds contained in the weight. Example.-Let 12" equal length of weight. Let 12" equal depth of weight. Let 4" equal thickness of weight. Let 12" equal length of hole in the weight. Let 2" equal height of hole in the weight. Let 5 tenths of an inch equal width of hole in the weight. Let .2607 of a pound equal weight of a cubic inch of cast-iron. Then we have: Weight of weight. (12X12X4)-(12X2X.5)X.2607-147.0348 lbs. 6 82 A Library of Steam Engineering. Putting it in the ordinary form, we have: 12X12×4 576 12X2X.5= 12 Area of hole in the weight. 564 Number of cubic inches of iron in the weight. .2607 1 0428 15 642 130 35 147.0348 lbs. Weight of weight. We now have the weight of the lever and weight, which we trans- fer to the diagram (Fig.46). F 4" 18" A 147.0348 lbs. 10.1412 lbs. weight of lever. 4" Diam. of valve. Fig. 46 THIRD OPERATION. RULE.-Square the diameter of the spindle, then multiply the square of the diameter by .7854, then multiply the product by the length of the spindle, then multiply the last product by .2817, and the answer will give the weight of the spindle in pounds. Example.-Let 75 one hundredths of an inch equal diameter of spindle. Let .7854 equal a constant. Let 6 inches equal length of spindle. Let .2817 equal a constant. Then we have: .752X.7854X6X.2817-.7467092325 lbs. Which is nearly of one pound. Putting it in the ordinary way, we have: .75 .75 375 525 .5625 .7854 22500 28125 45000 Am't carried forward, .44178750 39375 The Safety-valve Mathematically Considered. 83 Am't brought forward, .44178750 6 2.65072500 .2817 18555075 2650725 21205800 5301450 .7467092325 lbs. Very nearly 3/4 of one pound,which for con- venience we will call it 34 of a pound, or, .75 lbs. We now have the weight of the lever, weight of weight, and weight of the spindle, which we transfer to the following diagram: F 4" 18" 10.1412 lbs. weight of lever. .75 lbs. weight of spindle. Fig. 47 147.0348 lbs. FOURTH OPERATION. RULE.-Multiply the diameter of the vessel of water by its diameter, and multiply the product by .7854, then multiply the product by the rise of water, in inches, caused by immersion of the valve, then multiply the last product by .3194 if the valve is of brass, and by .2607 if the valve is of cast-iron. Example.-Let 4.25 inches equal diameter of vessel of water. Let .7854 equal a constant. Let 1.41 inches equal the rise of water in the vessel. Let .3194 equal a constant for a brass valve. Then we have: Weight of valve. 4.25 X 4.25.7854X1.41.3194 6.38+ lbs. Putting it in the ordinary form, we have: 4.25 4.25 2125 850 17.00 Am't carried forward, 18.0625 84 A Library of Steam Engineering. Am't brought forward, 18.0625 .7854 722500 903125 1 445000 12 64375 14.18628750 1.41 1418628750 5 674515000 14 18628750 20.0026653750 .3194 800106615000 1800239883750 200026653750 6 00079961250 6.38885132077500 lbs. Weight of valve. As it is unnecessary to carry the operation farther than two decimals, all the decimals except two will be dropped, because even the third decimal represents but 8/1000 of a pound. We now have ascertained the weight of the lever, weight of the weight, weight of the spindle and weight of the valve, which we will transfer to the diagram (Fig. 48), and complete it for the purpose of determining the pressure per square inch required to raise it. ---- Distance of center of weight from fulcrum 30 F 6.38 wt. of valve. 4" dis. of valve from fulcrum. 18" dis. of center of 10.1412 lbs. wt. of lever. A gravity of lever .75 lbs. wt. of spindle. 4"x 4"x .7854-12.5664 area of valve. from fulcrum. .4" Diam. of valve. Fig. 48 FIRST OPERATION. त्रे 147.0348 lbs. LL F 4" .75 6.38 7.13 lbs. weight of valve and spindle. Fig. 49 RULE.-Multiply the weight of the valve and spindle by the distance the center of the valve is from the fulcrum, and call this product " Product No. 1." The Safety-valve Mathematically Considered. 85 Example. Let 7.13 lbs. equal weight of valve and spindle. Let 4" equal distance of center of valve from fulcrum. Then we have: 7.13X4-28.52, "Product No. 1." Putting it in the ordinary form, we have. 7.13 4 28.52 "Product No. 1." Istugira sa S SECOND OPERATION. F 18" 10.1412 lbs. weight of lever. Fig. 50 RULE.-Multiply the weight of the lever by the distance of the center of gravity of the lever from the fulcrum, and call this product "Product No. 2." Example.-Let 10.1412 lbs. equal weight of lever. Let 18" equal distance of center of gravity of lever from the fulcrum. Then we have: 10.1412X18-182.5416, "Product No. 2." Putting it in the ordinary way, we have: 10.1412 18 81 1296 101 412 182.5416 Product No. 2." THIRD OPERATION. Distance of center of weight from fulcrum 30" 147.0348 lbs. Fig. 51 RULE.-Multiply the weight of the weight by the distance its center is from the fulcrum, and call the product "Product No. 3." Example. Let 147.0348 lbs. equal weight of the weight. Let 30" equal distance of center of weight from the fulcrum. 86 A Library of Steam Engineering. Then we have: 147.0348X30=4411.044, "Product No. 3." Putting it in the ordinary form, we have: 147.0348 30 4411.0440"Product No. 3." FOURTH OPERATION. RULE.-Add Products Nos. 1, 2, and 3, together and call the answer "The Sum." Example. Let 28.52 equal "Product No. 1." Let 182.5416 equal "Product No. 2." Let 4411.0440 equal "Product No. 3." Then we have: Sum." 28.52 +182.5416+4411.04404622.1056, "The Performing the operation in the ordinary way, we have: 28.52 "Product No. 1." 182.5416 "Product No. 2." 4411.0440"Product No. 3." 4622.1056 "The Sum." FIFTH OPERATION. F 4" 12.5664 area of valve. Fig. 52 RULE.-Multiply the area of the valve by the distance of its center from the fulcrum, and call the product" Product No. 4." Example. Let 12.5664 square inches equal area of valve. Let 4 inches equal distance of center of valve from the fulcrum. Then we have: 12.5664X4 50.2656, "Product No. 4." Performing the operation in the ordinary way, we have: 12.5664 4 50.2656 "Product No. 4." SIXTH OPERATION. RULE.-Divide" The Sum" by "Product No. 4," and the quotient will give the pressure per square inch required to raise the valve as adjusted in Fig. 48. The Safety-valve Mathematically Considered. 87 S Example.-Let 4622.1056 equal, "The Sum." Let 50.2656 equal, "Product No. 4." Then we have: 4622.1056÷÷50.2656-91.95+ lbs. square inch required to raise the valve. Performing the operation, we have: 50.2656) 4622.1056 (91.95+lbs. 4523 904 98 2016 Pressure per Pressure per square inch required to raise the valve. 50 2656 47 93600 45 23904 2 696960 2513280 SIMPLE METHOD OF ADJUSTING SAFETY-VALVES. Let it be required to determine the pressure necessary to raise a valve adjusted according to the following diagram : F 2 2 lbs. weight of valve and spindle. 4" 4 lbs. weight of lever. 3.1416 area of valve. 8' 30.77 lbs. Fig. 53 RULE.-Take a bar of iron, of uniform section, at least twice the length of the lever of the safety-valve, balance the bar on the edge of a cold-chisel in a vise, and mark the point on the bar at which it balances, and call that "The Fulcrum." Then, to the right of that point, put another mark the same distance the valve is from the fulcrum; then balance the lever of the safety- valve, then measure the distance from the fulcrum to the point at which the lever balances and place a mark the same distance on the bar to the right of the point on which it balances, then measure the distance the weight is to be placed on the lever from the fulcrum, and lay off the same distance on the bar to the right of the point on which the bar balances; then lay off a distance to the left of the point on which the bar balances equal to the distance the valve is from the fulcrum. Then place the bar on the cold-chisel in the vise on the point at which it balances, and hang the valve and spindle, to the right on the bar from the point on which the bar balances, the same distance the valve is from the fulcrum when the valve is in its position; then hang the lever of the valve on the bar to the right of the point on which the bar balances, the same distance that the lever balances from its fulcrum; then hang the weight of the 88 A Library of Steam Engineering. safety-valve on the bar to the right of the point on which the bar balances the same distance the weight is to be placed on the lever of the valve; then hang weights on the opposite side of the bar, to the left of the point on which the bar balances, the same distance the valve is from the fulcrum, and continue to add weight at that point until the lever balances the valve, spindle, lever and weight on the opposite side of the bar; then remove and weigh the weights hung on the left half of the bar; then divide the number of pounds contained in the weights by the area of the valve and the quotient will give the number of pounds pressure per square inch required to raise the valve. We will now proceed to arrange the points mentioned in the rule, on a bar of iron (as shown in Fig. 54), preparatory to hanging the different parts of the safety-valve (in Fig. 53) on it. Center of gravity of lever. Fulcrum. 2" F 2" Distance from fulcrum. 4" 8" Counter weight distance from fulcrum. Valve and spindle distance from fulcrum. Safety valve weight distance from fulcrum. Fig. 54 We will now construct a diagram (Fig. 55), showing the counter- weight, and the different parts of the safety-valve (as shown in Fig. 53) hung in position. 2" Distance of valve and spindle from fulerum. F 2" 4" Distance of weight from fulcrum. 8" Distance of center of gravity from fulcrum. 3.1416 area of -valve. 1 157.08 lbs. 36.77 lbs. Lever of valve. Safety valve weight. Counter weight. 2 lbs. weight of valve and spindle. Fig. 55 It will be observed that each part of the safety-valve has the weight given in the diagram; that part of the diagram is entirely unnecessary to enable the student to determine the pressure per square inch required to raise the valve, and we only place them in the diagram for the purpose of making a calculation hereafter to prove the correct- ness of this method of determining the pressure required to raise the valve. We will now proceed to explain the modus operandi in detail. First, ascertain the distance the valve and spindle are from the fulcrum, then hang them at the same distance to the right of F on the bar. The Safety-valve Mathematically Considered. 89 Second, ascertain how far from the fulcrum the safety-valve lever will balance on a knife-edge, and hang the lever at the same distance to the right of F on the bar. Third, ascertain the distance the weight of the safety-valve is to be placed on the lever from the fulcrum, and hang it the same distance to the right of F on the bar. Fourth, hang weights to the left of F on the bar the same distance the valve and spindle hang to the right of F; and continue to add weight until the bar balances, and call these "the counter-weights." Fifth, remove the counter-weights and weigh them, then divide the weight of the counter-weights by the area of the valve, the quo- tient will give the pressure, per square inch, required to raise the valve. - Example. Let 157.08 pounds equal weight of counter-weights. Let 3.1416 square inches equal area of the valve. Then we have: 157.08-3.1416-50 lbs. Pressure required to raise the valve. Performing the operation in the ordinary way, we have: 3.1416) 157.0800 (50 lbs. Pressure required to raise the 157.080 0 valve, shown in Fig. 53. NOTE. In performing division with decimals, both the divisor and dividend should con- tain the same number of decimals; if either contains less than the other a sufficient number of ciphers should be annexed to the lesser to make the number of decimals in that equal to the number of decimals in the greater; for that reason, it will be observed, two ciphers have been added to the decimals in the dividend in the above example. The student will understand that there are two forces which ope- rate on a safety-valve, the one is downward and the other is upward. The downward pressure on the valve seat is produced by three forces: the weight of the valve and spindle, the pressure produced on the valve seat by the weight of the lever, and the pressure produced on the valve seat by the weight of the weight. The upward force consists of the pressure against the bottom of the valve, if this is greater than the combined downward forces the valve will rise, if it is less the valve will not rise. The student, therefore, must not get these forces con- founded with any pressure, strain or weight at the fulcrum, but his attention must be directed to the result produced at the end of the short arm of the lever at the center of the valve. In Fig. 55, the downward forces consist of the weights of the valve, spindle, lever and weight, and the upward force consists of the counter-weight, because it has a tendency to raise the valve, hence it is the same as though a total steam pressure of 157.08 pounds was under the valve. 90 A Library of Steam Engineering. TO DETERMINE THE ACTUAL PRESSURE PRODUCED ON THE VALVE SEAT BY THE VALVE, SPINDLE, LEVER AND WEIGHT. RULE.-Divide the distance of the lever from the fulcrum by the distance the valve is from the fulcrum, then multiply the weight of the lever by the quotient, the answer will give the total pressure in pounds produced down- ward at the valve by the weight of the lever. Second, divide the distance the weight is from the fulcrum by the distance the valve is from the fulcrum, then multiply the weight of the weight by the quotient, and the answer will give the total pressure produced downward at the valve by the weight of the weight. Third, as the valve and spindle are directly under the end of the short arm of the lever, the pressure produced downward at that point by the valve and spindle is equal to their combined weight, and requires no calculation. Fourth, add the pressure produced downward at the valve by the weight of the valve, spindle, lever and weight, and the sum will give the total press- ure produced downward at the end of the short arm of the lever, or the point at which the valve and spindle are hung. Example.-Let 2 inches equal distance of center of valve from the fulcrum. Let 2 pounds equal weight of valve and spindle. Let 4 inches equal distance of lever from the fulcrum. Let 4 pounds equal weight of lever. Let 8 inches equal distance of weight from the fulcrum. Let 36.77 pounds equal weight of the weight. Then we have: First, the weight of the valve and spindle on the valve seat equal: Second, the distance of the lever from the fulcrum divided by the distance the center of the valve is from the fulcrum, and the weight of the lever multiplied by the quotient: 4 -X4= 2 Third, the distance of the weight from the fulcrum divided by the distance the center of the valve is from the fulcrum, and the weight of the weight multiplied by the quotient: 8 -X 36.77= 2 Fourth, the pressure produced downward at the valve by the weight of the valve, spindle, lever and weight added together: 2 lbs. 8 lbs. 147.08 lbs. 157.08 lbs The Safety-valve Mathematically Considered. 91 It will be observed that in Fig. 55, the total pressure produced on the valve seat by the weight of the valve, spindle, lever and weight is 157.08 pounds, and that this pressure is downward two inches from the fulcrum to the right. It will also be observed that the weight. 157.08 pounds, two inches to the left of the fulcrum, while its strain is downward at that point, produces a strain upward of 157.08 pounds, two inches to the right of the fulcrum, and hence it exactly balances. the forces to the right of the fulcrum. We will now suppose the strain upward two inches from the fulcrum to the right, produced by the counter-weight to the left to be a total strain pressure of 157.08 pounds under the valve, by dividing that total strain by the area of the valve, we will have the pressure per square inch required to raise the valve. Example.-Let 157.08 pounds equal total pressure under the valve. Let 3.1416 square inches equal area of the valve. Then we have: 157.08 3.1416-50 lbs. per square inch. Now, as the 157.08 pounds total strain under the valve, is exactly counter-balanced by the weight of the valve, spindle, lever and weight, any increase in the total strain will raise the valve. HOW TO GRADUATE A SAFETY-VALVE LEVER. The method shown in Fig. 55 is a simple as well as an accurate method for graduating the lever of a safety-valve, so that the weight. can readily be adjusted to allow the valve to rise at any desired pres- sure, per square inch, and also to test the accuracy of the steam gauge. We will, therefore, proceed to show how to graduate a safety-valve lever by the use of the bar, as shown in Fig. 55. RULE.-First, multiply the desired pressure, per square inch, by the area of the valve, and the product will give the number of pounds of weight to be hung at the point where the counter-weight is in Fig. 55. In other words, hang the weight on the bar to the left of the fulcrum the same distance that the valve is from the fulcrum of the lever to be graduated. Second, move the weight of the safety-valve on the bar to a point where it will balance the counter-weight on the left of the fulcrum, and that will be the point at which the weight must be placed on the lever when the valve is in posi- tion on the boiler, to allow the valve to rise at the desired pressure. Third, mark the distance from the fulcrum on the lever of the valve at which the weight balanced the bar from the fulcrum of the bar, and mark the pressure, per square inch, at that point at which the valve will rise. these operations for all desired pressures. Continue Fourth, while performing the operation, the valve, spindle and lever must be kept constantly in their respective positions, as shown in Fig. 55. 92 A Library of Steam Engineering. TO DETERMINE THE PROPER AREA OF SAFETY-VALVES. The proper area of a safety-valve must be determined by the capacity of the boiler for generating steam, to which it is attached. In all cases, however, the area of the valve should be sufficient, under all circumstances, to relieve the boiler automatically of excess of pressure, and keep the pressure within the limit of perfect safety. In other words, all safety-valves should have an area large enough to prevent the accumulation of pressure in the boiler beyond that at which the valve is set to blow off. The size of the valve required depends in a great measure upon its construction; because the area of opening produced by the lift of the valve under steam pressure is the measure of the valve's efficiency. There are valves that will discharge double the amount of steam that other valves will of the same diameter. Great care should, therefore, be exercised in the selection of safety-valves, and under no circum- stances should any valve be selected that is in any way faulty in con- struction. And as to the size of the valve required, when selecting a lever-valve, scientific experiments and every-day practice have deter- mined that a lever-valve, to do its work effectively, should have an area of one square inch for every two square feet of the grate surface of the boiler to which it is to be attached; and it might as well be emphat- ically understood that no valve is a safety-valve unless it will, auto- matically, relieve the boiler and prevent the pressure of steam from crossing the danger line in the strength of the boiler. TO DETERMINE THE PROPER DIAMETER OF A SAFETY-VALVE. RULE.-First, divide the number of square feet of grate surface of the boiler by 2, and call the quotient "Quotient No. 1." Second, divide "Quotient No. 1" by .7854 and extract the square root of the last quotient, the answer will give the diameter of the valve. Example.-Let 20 square feet equal grate surface of the boiler. Let 2 equal a constant. Let .7854 equal a constant. Then we have: 20÷2 .7854 =3.56+ inches. Diameter of valve required. Performing the operation in the ordinary way, we have: 2) 20 .7854) 10.0000 (12.7323+ Continued on next page, 2 1460 7 854 The Safety-valve Mathematically Considered. 93 Brought from previous page, 21460 15708 57520 54978 25420 23562 18580 15708 28720 23562 We will now proceed to extract the square root of the last quo- tient. Thus: 12.7323 (3.56+inches. Diameter of valve required. 9 65) 373 325 706) 4823 3246 Another method, and one which will produce the same result is as follows: RULE.-Divide the number of square feet in the grate surface by the constant 1.5708, and extract the square root of the quotient, the answer will give the diameter of the valve. Example.-Taking the same number of square feet of grate surface taken in the preceding example. We have: 20÷1.5708-3.56+ inches. Diameter of valve. Performing the operation in the ordinary way, we have: 1.5708) 20.0000 (12.7323+ 15 708 4 2920 3 1416 1 15040 1 09956 50840 47124 37160 31416 57440 47124 94 A Library of Steam Engineering. Extracting the square root of the quotient, we have: 12.7323 (3.56+ inches. Diameter of valve. 9 65) 3 73 3 25 Comp Part off to door onstipa 706) 4823 4236 thes of Isonong, wou lliw W and brot Valves Generally. 95 CHAPTER III. VALVES GENERALLY. AREA OF OPENING-FLAT-SEATED VALVES. TO DETERMINE THE AREA OF OPENING PRODUCED BY THE LIFT OF A FLAT-SEATED VALVE. RULE.-Multiply the diameter of the valve by 3.1416, and then multi- ply the product by the lift of the valve, in hundredths of an inch, the answer will give the area of opening produced by the lift of the valve. Example.-Let 4 inches equal diameter of the valve. Let 3.1416 equal a constant. Let 25 one hundredths of an inch equal lift of the valve. Then we have: 4x3.1416X.25-3.1416 inches. Area of opening produced by the lift of the valve. The 3.1416 inches represent an opening one inch in width, and 3.1416 inches in length. TO DETERMINE THE SIZE OF A SQUARE HOLE FROM THE AREA OF OPEN- ING PRODUCED BY THE LIFT OF A FLAT-SEATED VALVE. RULE.-Extract the square root of the quotient in the above example, the answer will give the size of the hole in square form. Example.-Let 3.1416 equal number of square inches in area of valve opening. Then we have. 3.1416 (1.77+ inches square. opening produced by the 1 27) 214 189 347) 2516 2429 The size hole the area of lift of the valve. TO DETERMINE THE SIZE OF A ROUND HOLE FROM THE AREA OF OPENING PRODUCED BY THE LIFT OF A FLAT-SEATED VALVE. RULE.-Divide the area of opening produced by the lift of the valve by .7854, and then extract the square root of the quotient, the answer will give the diameter of the hole. 96 A Library of Steam Engineering. Example.-Let 3.1416 equal area of opening produced by the lift of the valve. Let .7854 equal a constant. Then we have: 3.1416.7854-2 inches. Performing the operation in the ordinary way, we have: .7854) 3.1416 (4 3.1416 Extracting the square root of the quotient, we have: 4 (2 inches. Diameter of round hole 3.1416 square inches will make. 4 AREA OF OPENING-BEVEL-SEATED SAFETY-VALVES. The rule for determining the area of the opening produced by the lift of bevel-seated valves differs materially from that for flat-seated valves. The reason is that as the angle of the seat deviates from that of 90 degrees to the center line of the valve's axis, the opening pro- duced by the lift becomes less in proportion to the lift, as the angle approaches a line parallel to the center line of the valve's axis. A flat-seated valve lifting a distance equal to one-quarter of its diameter, produces an area of opening equal to the area of the diam- eter of the valve; while a bevel-seated valve, until it rises above its seat, produces an area of opening less in proportion as the angle of the seat deviates from an angle of 90 degrees to the center line of the valve's axis. In other words, taking a flat-seated valve the opening produced by its lift is equal to the lift, while that of a bevel-seated valve is less than the lift in proportion as the bevel becomes greater. It will be observed, that as the angle of a bevel-seated valve ap- proaches a perpendicular line, the opening produced by the lift becomes less in proportion. The bevel might be made so great that the seat of the valve would become very nearly a cylinder, in which event the valve would have to lift above the seat before it would produce any percep- tible opening; and as the area of opening produced by the lift of the valve is governed by the opening produced by the lift, it will be seen that the greater the angle of the seat the less opening any given lift will produce. Take, for example, a flat-seated valve and a bevel- seated valve, each four inches in diameter, and the bevel-seated valve having its seat one inch in depth, and made at an angle of 45 degrees to the center line of the valve's axis, and each lifting one inch; the area of opening produced by the lift of the flat-seated valve would be 12.5664 square inches, while that of the bevel-seated valve would be but 10.4547+ square inches. We will now proceed to give the rule relating to bevel-seated valves. Valves Generally. 97 TO DETERMINE THE AREA OF OPENING PRODUCED BY THE LIFT OF ANY BEVEL-SEATED VALVE WHEN THE LIFT IS NOT ABOVE THE SEAT. RULE.—First, multiply the diameter of the valve in inches by the lift in hundredths of an inch, and then multiply the product by the sine of the angle of the seat, and call the last product "Product No. 1." Second, multiply the square of the lift in hundredths of an inch by the square of the sine of the angle of the seat, then multiply the product by the cosine of the angle of the seat, and call the last product "Product No. 2." Third, add Products Nos. 1 and 2 together and multiply the sum of the two by 3.1416 the product will give the area of opening produced by the lift of the valve. Example.-Let D equal diameter of the valve in inches. Let / equal lift of the valve in hundredths of an inch. Let s equal sine of the angle of the seat. Let c equal cosine of the angle of the seat. Let C equal 3.1416 a constant. A equal area of opening produced by the lift of the valve. Then we have: [ (DXIXs)+(12X82Xc) ]XC=A. Substituting figures for the values in the formula: Let 4 inches equal diameter of the valve. Let 12 one hundredths of an inch equal lift of valve. Let .707 equal sine of an angle of 45 degrees (bevel of seat). Let .707 equal cosine of an angle of 45 degrees (bevel of seat). Let 3.1416 equal a constant. Then we have: (4.12.707)+(.122.7072.707) X3.1416-1.08+ square inches.. Area of opening produced by the lift. Performing the operation in the ordinary way, we have: 4X.12X.707=.33936 "Product No. 1."" .12X.12X.707X.707X.707-.0050888626992 "Product No. 2 " The sum multiplied by 3.1416, .3444488626992×3.1416= 1.08212054705580672 square inches. Area of opening produced by lift of valve. 7 98 A Library of Steam Engineering. TO DETERMINE THE AREA OF OPENING PRODUCED BY THE LIFT OF A BEVEL-SEATED VALVE WHERE THE VALVE LIFTS ABOVE THE SEAT. RULE.-First, multiply the diameter of the valve in inches by the depth of the seat, in hundredths of an inch, then multiply the product by the sine of the angle of the seat, and call the last product "Product No. 1." Second, multiply the square of the depth of the seat in hundredths of an inch by the square of the sine of the angle of the seat, then multiply the product by the cosine of the angle of the seat, and call the last product "Product No. 2." Third, multiply that part of the lift of the valve above the seat, in hundredths of an inch, by the diameter of the valve, in inches, and call the product "Product No. 3." Fourth, add Products Nos. 1, 2 and 3 together and call the answer Sum." "The Fifth, multiply" The Sum" by 3.1416 and the product will give the area of opening, in square inches, produced by the lift of the valve. Example. Let D equal diameter of the valve in inches. Let d equal depth of the seat, in hundredths of an inch. Let s equal sine of the angle of the seat. Let c equal cosine of the angle of the seat. Let / equal lift of valve in hundredths of an inch. Let C equal 3.1416, a constant. Let A equal area of opening produced by lift of valve. Then we have: [(DXdXs)+(d²×8²xc)+(Dxl–d) ]×C=A. To simplify the operation we will put it in the following form: Taking a valve 2 inches in diameter and lifting .5 of an inch, with a depth of seat of .25 of an inch, and an angle of seat of 45 degrees. Then we have: DXdXs= 2X.25X.707 d2Xs2Xc=.252X.7072X.707.02208+ =.35350 "Product No. 1.". "Product No. 2." "Product No. 3." "The Sum." DX(l―d)=2X(.50-25) =.50 .87558 Then, multiplying the sum by 3.1416, we have: .87558 3.1416 525348 87558 350232 87558 2 62674 2.750722128 square inches. Area of opening produced by the lift of the valve. Valves Generally. 99 To simplify the operation still further we will put it in the fol- lowing form: Example.-Let 2 inches equal diameter of valve. Let 25 one hundredths of an inch equal depth of seat of valve. Let .707 equal sine of angle (45°) of seat. Let .707 equal cosine of angle (45°) of seat. Let 50 one hundredths of an inch equal lift of valve. Let 3.1416 equal a constant. Then we have: (2X.25×.707)+(.252.7072.707)+(2×.50—.25) X3.1416-2.75+. Performing the operation, we have: .25" Depth of seat. 2" Diameter of valve. .50 .707 Sine of the angle of the seat. .35350 "Product No. 1." Next we have: .25" Depth of seat. .25" Depth of seat. 125 50 Next we have: .0625 Square of depth of seat. .707 Sine of the angle of the seat. .707 Sine of the angle of the seat. 4949 4949 Square of the sine of the angle of the seat. .499849 .0625 Square of depth of the seat. 2499245 999698 2999094 .0312405625 .707 Cosine of the angle of the seat. Next we have: 2186839375 2186839375 .0220870776875 "Product No. 2." .50" Lift of valve. .25" Depth of seat subtracted from lift of valve. .25 Lift of valve above the seat. 2 Diameter of valve. .50 "Product No. 3." for M 100 A Library of Steam Engineering. Next, adding Products Nos. 1, 2 and 3 together, we have: 35350 "Product No. 1." .02208+ "Product No. 2." 50 66 'Product No. 3." .87558 "The Sum." Finally, multiplying "The Sum by the constant 3.1416, we have: .87558 3.1416 525348 87558 350232 87558 2 62674 2.750722128 square inches. Area of opening pro- duced by the lift of the valve. SINES AND COSINES. The following table contains the sines and cosines of the angles at which the seats of bevel-seated valves are commonly made; it is therefore given for convenient reference and the use of the student: Angle. Sine. Cosine. Angle. Sine. Cosine. 25° .42262 .90631 38° .61566 26° .78801 .43837 .89879 39° .62932 27° .77715 .45399 .89101 40° .64279 28° .76604 .46947 .88295 41° .65606 29° .75471 .48481 .87462 42° .66913 30° .74314 .5 .86603 43° .682 31° .73135 .51504 .85717 44° .69466 .71934 32° .52992 .84805 45° .70711 33 200 .70711 .54464 .83867 46° .71934 .69466 34° .55919 .82904 47° .73135 35° .682 .57358 .81915 48° .74314 36° .66913 .58779 .80902 49° .75171 .65606 37° .60182 .79864 50° .76604 .64279 As sines and cosines of angles enter largely into the science of steam and mechanical engineering, and particularly into that branch embraced in the steam boiler safety-valve, it is very important that the student should understand just what sines and cosines are. As the use of the diagram is the simplest method that can be employed, the attention of the student is directed to Fig. 56, and to a careful study of the same. Valves Generally. 101 It will be observed that the line C B represents the valve seat, which is made at an angle of 45°. Dotted line A B and dotted line from C to D represent the cosine, and the dotted line B D the sine of the angle. The angle of the seat of the valve being 45° the lines rep- resenting the sines and cosines are exactly the same length. It will also be observed that as the line representing the valve seat leaves the dotted line C E and approaches the perpendicular dotted line C F that the cosines become shorter and the sine becomes longer; and the opening produced by the lift of the valve becomes correspondingly less; hence it is, as previously stated, that as the angle of the valve seat deviates from the angle of 90° from the center line of the valve's axis, the opening produced by the lift becomes less. The perpendic- ular dotted line C F represents the center line of the valve's axis, and the horizontal dotted line C E has an angle of 90° from the perpendic- ular line C F, the angle at which flat-seated valves are made, and this line is called the radius, or, rather the distance from C to E on this line is called the radius. Fi A! Cosine Valve seat angle 45° C Cosine Sine B E D Fig. 56 By the use of the information thus obtained, and the employment of one of the simplest rules in trigonometry, we are enabled to formu- late a very simple rule for determining the area of opening produced by the lift of a bevel-seated valve, having its seat made at an angle of 45° to the center line of the valve's axis. Practice and scientific ex- periments-both have demonstrated that 45° is the proper angle at which the seats of all safety-valves should be made, and that has been adopted as the standard by the United States and British governments. Therefore, the following rule, based upon the standard of an angle of 45°, will be found practically correct without the use of sines or cosines, beside the confusion produced in the mind of the student by the use of sines and cosines is obviated. 102 A Library of Steam Engineering. TO DETERMINE THE AREA OF OPENING PRODUCED BY THE LIFT OF A BEVEL-SEATED VALVE HAVING ITS SEAT MADE AT AN ANGLE OF 45° TO THE CENTER LINE OF THE VALVE'S AXIS WITHOUT THE USE OF SINES OR COSINES. RULE.-First, add one-half of the lift of the valve, in hundredths of an inch, to the diameter of the valve, and multiply the sum by the constant 3.1416, and call the product "Product No. 1." Second, multiply the lift by .707, and call the product "Product No. 2." Third, multiply "Product No. 1" by "Product No. 2" and the product will give the area of opening produced by the lift of the valve. Example.-Let 4 inches equal diameter of the valve. Let 20 one hundredths of an inch equal lift of valve. Let 3.1416 equal a constant. Then we have (1 .20 4+ 2 X3.1416X.707X.20-1.82+ square inches. Area of opening produced by the lift of the valve. Performing the operation, we have: 2).20 inches. Lift of valve. .10 inches. One-half of lift of valve. inches. Diameter of valve. 4. 4.10 3.1416 2460 410 1640 410 12 30 12.880560 "Product No. 1." Next we have: .707 .20 Lift of valve. 66 .14140 Product No. 2." Finally we have: 12.880560 "Product No. 1." .14140 "Product No. 2." 515222400 12880560 51522240 1 2880560 1.82131118400 square inches. Area of opening produced by the lift of the valve. Valves Generally. 103 The next thing will be to illustrate by diagram how the area of opening produced by the lift of the valve is obtained, and why one-half of the lift is added to the diameter of the valve in per- forming the operation. This is shown in Fig. 57, in which the seat is made purposely large for greater convenience in illustrating the subject: 4 inches. SIXP DATRA Center line 1 Fig. 57 It will be observed that the opening produced by the lift is conic shaped, and that it forms the frustrum of a cone, of which A C and B D form the slant height, A B the diameter of the top, and CD the diameter of the base, and the perpendicular dotted lines L L represent the lift of the valve. If we add the diameter of the top to the diameter of the base and divide the sum by 2 we get the average diameter of the frustrum of the cone. Then, if we multiply the average diameter by 3.1416 we get the average circumference, then if we multiply the average circumfer- ence by the length of the slant side A C or B D we get the surface of the frustrum of the cone, and the surface represents the area of open- ing produced by the lift of the valve. It will also be observed that AC H and B D G form right angle triangles of equal sides, and that the horizontal distance from C to F and E to D is the same as the perpendicular distance from F to A and E to B, or H to F and G to E. Therefore, if we add one-half of the lift to the diameter of the valve A B, we have the average diameter of the frustrum of the cone, and this is done simply to obviate the unnec- essary labor of adding the diameter of the top of the cone to the diameter of the bottom and dividing the sum, because simply add- ing one-half of the lift of the valve to its diameter accomplishes the same result exactly, and has the advantage of being more simple and requiring much less labor. 104 A Library of Steam Engineering. To find the amount of opening the lift of the valve will produce, it must be observed that A H and B G each form the hypothenuse of a right angle triangle. Therefore, to get the length of the sides A Cor B D we multiply the hypothen use, or the lift of the valve which is the same thing in this case, by .707, and the product will give the length of the sides A C or B D, or in other words, it will give the opening produced by the lift of the valve. To illustrate suppose the valve (as shown in Fig. 57), was 4 inches in diameter and that its lift was of an inch, what is the area of opening produced by the lift. Then, according to rule, we add one-half of the lift, of an inch, which, expressed in decimals is .875 and .875÷2-.4375, and .4375 added to 4 inches (the diameter of the valve) will give an average diameter of 4.4375 inches, and this multiplied by 3.1416 will give the average circumference. Thus: 4.4375x3.1416-13.94+ inches, average circum- ference of the valve. Performing the operation, we have: 4.4375 3.1416 266250 44375 177500 44375 13 3125 13.94085000 inches. Average circumference of the valve. Then, if we multiply the average circumference of the valve by the length of the slant sides A C or B D of the conic shaped opening, we will get the area of opening, produced by the lift of the valve. The length of the slant side of the conic shaped opening is obtained by multiplying the lift.875 by .707, thus: .875X.707-.618625, decimals of an inch. Performing the operation, we have: .875 .707 6125 6125 .618625 Decimals of an inch, the length of the side of the cone and the actual opening pro- duced by the lift of the valve. Now, if we multiply this opening by the average circumference we will then have: .618625X13.94-8.62363250 square inches. Valves Generally. 105 Performing the operation, we have: .618625 13.94 2474500 5567625 1 855875 6 18625 8.62363250 square inches. Area of opening pro- duced by the lift of the valve. It will be observed that while the lift of the valve is .875 of an inch, the actual opening for the discharge of steam or liquid is from A to C or B to D, and that it amounts to .618625 of an inch-a fraction less than while the lift is ; and that the difference between the lift and the opening produced by it becomes greater as the line of the valve seat approaches the center line of the valve's axis perpendicularly. VALVE PROPORTIONS. It is very important that every engineer should understand not only how to proportion and adjust safety-valves, but he should under- stand how to proportion and adjust all valves connected with pipes employed to convey steam, water or other fluids, so as to offer as little resistance as possible. It frequently occurs that a valve is required to have too much lift for efficient performance of the work required of it. In such cases the valve should be replaced by one having an area sufficient to perform the functions required of it without the necessity of giving it a lift so great as to cause the valve to pound itself to pieces in a short time. It is therefore very essential that the engineer should understand how to determine the size of the valve required. LIFT OF FLAT-SEATED VALVES. TO DETERMINE THE LIFT OF A REQUIRED FLAT-SEATED VALVE TO PRO- DUCE AN AREA OF OPENING EQUAL TO THE AREA OF OPEN- ING PRODUCED BY THE LIFT OF A GIVEN VALVE. RULE.-Divide the lift of the given valve in hundredths of an inch, by the diameter of the required valve, and multiply the quotient by the diameter of the given valve. Example.-Let 25 one hundredths of an inch equal lift of the given valve. Let 4 inches equal diameter of the required valve. Let 3 inches equal diameter of the given valve. Then we have: .25 -X3=.1875 Decimals of an inch. Lift of the required valve. 106 A Library of Steam Engineering. Performing the operation, we have: 4.00).2500 (0.0625 2400 3 1000 .1875 Decimals of an inch. Lift of re- 800 2000 2000 quired valve. ANOTHER METHOD. RULE.-Multiply the lift of the given valve by its diameter, and divide the product by the diameter of the required valve and the quotient will give the lift of the required valve. Example.-Let 25 one hundredths of an inch equal lift of the given valve. Let 3 inches equal diameter of the given valve. Let 4 inches equal diameter of the required valve. Then we have: .25×3 4 -.1875 Decimals of an inch. Lift of the required valve. Performing the operation we have: .25 3 4.00) .750 (.1875 Decimals of an inch. Lift of the required valve. 400 3500 3200 3000 2800 2000 2000 LIFT OF FLAT SEATED VALVES REQUIRED TO PRODUCE A GIVEN AREA OF OPENING. TO DETERMINE THE LIFT OF A REQUIRED FLAT-SEATED VALVE THAT WILL PRODUCE AN AREA OF OPENING EQUAL TO THE AREA OF THE DIAMETER OF A GIVEN VALVE OR AREA OF CROSS SECTION OF A GIVEN PIPE. RULE.-Divide the area of the given valve, or area of cross section of the given pipe, by the diameter of the required valve and divide the quotient by 3.1416, and the last quotient will give the required lift of the required valve to produce an area of opening equal to the area of the diameter of a given valve, or the area of the diameter of a given pipe. Valves Generally. 107 Example.-What must be the lift of a required valve 5 inches in diameter to produce an area of opening equal to the area of a 3 inch valve or 3 inch pipe? Let 3 inches equal diameter of a given valve or pipe. Let .7854 equal a constant. Let 5 inches equal diameter of the required valve. Let 3.1416 equal a constant. Then we have: 3X3X.7854 5 854 ÷3.1416-45 Hundredths of an inch. Lift of re- quired valve. Or: 3X3X.7854 5×3.1416 45 Hundredths of an inch. Lift of required valve. Performing the operation, we have: 3 33 9 .7854 5) 7.0686 3.14160) 1.413720 (0.45 Hundredths of an inch. Lift of re- 1 256640 1570800 1570800 quired valve. It will be observed that the area of the diameter of a 3 inch valve or a diameter of a 3 inch pipe is 7.0686 square inches. Then, to determine whether or not a 5 inch valve, lifting 45 one hundredths of an inch, will produce an area of opening equal to the area of the diam- eter of a 3 inch valve or pipe, let the area of opening produced by a 5 inch valve, lifting 45 one hundredths of an inch, be determined. RULE.-Multiply the constant 3.1416 by the diameter of the valve and then multiply the product by the lift of the valve, and the last product will give the area of opening produced by the lift of the valve. Example.-Let 5 inches equal diameter of the valve. Let 3.1416 equal a constant. Let 45 one hundredths of an inch equal lift of valve. Then we have: 3.1416 5 Am't carried forward, 15.7080 108 A Library of Steam Engineering. Am't brought forward, 15.7080 .45 785400 6 28320 7.068600 Square inches, area of opening produced by the lift of the valve, and this is ex- actly the area of the diameter of a 3 inch valve or pipe. ANOTHER METHOD. RULE.-First, square the diameter of the pipe and multiply the prod- uct by .7854, the last product will give the area of cross section of the pipe, and we set that aside and call it "Product No. 1." Second, multiply the diameter of the given valve by 3.1416, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2" and the quotient will give the required lift of a given valve to produce an area of opening equal to the area of the diameter of a given pipe. Example.-Let 8 inches equal diameter of given pipe. Let .7854 equal a constant. Let 9 inches equal diameter of given valve. Let 3.1416 equal a constant. Then we have: 8X8X.7854 9x3.1416 1.77+ inches. A fraction over 14 inches; re- quired lift of the given valve. Performing the operation in the ordinary way, we have: 8 Diameter of the pipe. 8 Diameter of the pipe. 7854 64 Square of diameter of the pipe. A constant. 3 1416 47 124 50.2656 "Product No. 1." Next, multiplying the diameter of the given valve by 3.1416, we have: 9x3.1416-28.2744, "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 28.2744) 50.2656 (1.77+ inches. Required lift of given valve. 28 2744 21 99120 19 79208 2 199120 1 979208 Valves Generally. 109 DIAMETER OF FLAT-SEATED VALVES. TO DETERMINE THE REQUIRED DIAMETER OF A REQUIRED FLAT-SEATED. VALVE TO HAVE A GIVEN LIFT THAT WILL PRODUCE AN AREA OF OPENING EQUAL TO THAT PRODUCED BY THE GIVEN LIFT OF A GIVEN VALVE. RULE.-Multiply the diameter of the given valve by its lift, and divide- the product by the required lift of the required valve. Example. Let 2 inches equal diameter of the given valve. Let 15 one hundredths of an inch equal lift of given valve. Let 6 one hundredths of an inch equal required lift of required valve. Then we have: 2X.15 .06 -5 inches. Required diameter of required valve. Performing the operation, we have: .15 2 .06) .30 (5 inches. Required diameter of required 30 valve. LIFT OF BEVEL-SEATED VALVES.aibivib qiladi! The following rules are based upon the United States Standard valves; which valves have their seats made at an angle of 45 degrees to the center line of the valve's axis. TO DETERMINE THE LIFT OF A BEVEL-SEATED VALVE WHEN THE AREA OF OPENING REQUIRED AND THE DIAMETER OF THE VALVE ARE GIVEN. RULE.-First, multiply 3.1416 by the diameter of the valve and call the product "Product No. 1." Second, divide the given area of opening required by "Product No. 1," and divide the quotient by .707, the last quotient will give the required lift of the valve. Example.-Let 4.10 inches equal diameter of the valve. Let 3.1416 equal a constant. Let 1.821311184 square inches equal area of opening. required. Let .707 equal a constant. 110 A Library of Steam Engineering. Then we have: 1.821311184 .707-20 Hundredths of an inch. Lift required. 3.1416×4.10 Performing the operation in the ordinary way, we have: 3.1416 A constant. 4.10 Diameter of valve. 3141 60 12 5664 12.8805 60 "Product No. 1." Next, dividing the area of opening required by "Product No. 1," we have: 12.88056) 1.821311184 (.1414 ed 1 288056 5332551 5152224 1803277 1288056 5152224 5152224 Finally, dividing the quotient .1414 by .707, we have: .707) .1414 (.2 1414 Tenths, or 20 hundredths of an inch, lift required for a valve 4 inches in diame- ter to produce an area opening of 1.821311184 square inches. DIAMETER OF BEVEL-SEATED VALVES. TO DETERMINE THE DIAMETER OF A REQUIRED BEVEL-SEATED VALVE TO HAVE A GIVEN LIFT THAT WILL PRODUCE AN AREA OF OPENING EQUAL TO THAT PRODUCED BY THE GIVEN LIFT OF A GIVEN VALVE. RULE.-First, multiply the given lift of the required valve by .707 and call the product "Product No. 1." Second, divide the area of opening produced by the lift of the given valve by "Product No. 1" and call the quotient " Quotient No. 1." Third, divide "Quotient No. 1" by 3.1416, and deduct one-half of the lift of the required valve from the last quotient and the remainder will give the diameter of the required valve. Valves Generally. 111 Example.-Let .25 equal lift of required valve, in hundredths of an inch. Let 4 equal diameter of given valve, in inches. Let .20 equal lift of given valve, in hundredths of an inch. Let 3.1416 equal a constant. Let .707 equal a constant. Then we have: 4+(.20÷2)×3.1416X.20.707=1.821311184 square inches. Area of opening produced by the lift of the given valve. We will now proceed according to the rule. First, multiply the given lift of the required valve by .707, thus: .25X.707=.17675, "Product No. 1." Second, divide the area of opening, 1.821311184, produced by the lift of the given valve, by "Product No. 1." Thus: 1.821311184.17675 =10.304448, "Quotient No. 1." Third, divide "Quotient No. 1" by 3.1416 and deduct one-half of the lift of the required valve from the last quotient. Thus: (10.304448 3.1416)-(.25÷2)=3.155 inches, diameter of required valve. 112 A Library of Steam Engineering. CHAPTER IV. BOILER PLATE. All boiler plate employed in the construction of steam boilers should possess two qualities which are absolutely necessary to render the use of boilers safe. The first is tensile strength sufficient to withstand the strain put upon it by the pressure of steam; and the second is ductility to prevent cracking by unequal expansion and contraction, and by repeated heating and cooling. The latter quality is just as essential to safety in the use of steam boilers as the former; and therefore no plate of iron or steel should be put into any steam boiler that does not possess the requisite strength and ductility. To the absence of one of these essential qualities, and in many cases both, a large number of disastrous steam boiler explosions are directly attributable, resulting in terrible sacrifices of human life and destruc- tion of enormous amounts of property. An investigation into the cause of such disasters usually follows, and the very convenient conclusion is almost invariably reached that, "low water was the cause of the accident." Hence, the responsibility is fastened upon the engineer who had charge of the boilers, when in fact the blame should be put upon the boiler maker who built the boilers, or upon the inspector who inspected the material, for allowing such material to enter the construction of the boilers. It is very essential, therefore, that the boiler maker, as well as the engineer, should thoroughly understand the qualities necessary to render boiler plate fit for use in steam boilers, in order to guard against explosions from the employment of unfit and dangerous material. In order, then, to prevent the use of such material every plate should be tested before putting it into the boiler. The United States Govern- ment has fully recognized the importance of guarding against the use of bad material in the construction of marine boilers, and laws have been enacted to regulate the tensile strength and ductility of boiler material and to provide for testing every plate, subjected to tensile strain, employed in the construction of marine boilers. These laws also provide that boiler manufacturers shall, for marine purposes, use material containing a certain percentage of ductility for all boiler plate, based upon the tensile strength per square inch of the plate. And all Boiler Plate. 113 plates made for marine purposes are required to be stamped by the manufacturer, at their diagonal corners, about 4 inches from the edges, and at or near the center of the plate, with the tensile strength of the material, the name of the manufacturer, and the place where manufac- tured, and to have a coupon attached, to be cut off and used for testing the material in each plate, as shown in Fig. 58. SHEET WITH COUPON ATTACHED. John Smith Pittsburgh, Pa. T. S. 60,000. John Smith Pittsburgh, Pa. T. S. 60,000. Pittsburgh, Pa. John Smith T. S. 60,000. Fig. 58 The coupon is cut off for the use of inspectors in testing the material, and if the test made with the coupon, shows that the material in the plate possesses the required ductility, and the tensile strength stamped on the plate, the plate from which the coupon was cut is approved for use in a marine boiler; and the steam pressure per square inch allowed in such boiler is based upon the tensile strength of the material, the diameter of the boiler, and the manner in which the boiler is constructed. Preparatory to making the test with the coupon, the coupon is cut from the plate and cut out at the center to a width corresponding with the thickness of the material in the coupon. Test pieces inch and under, should be reduced, at point intended for breaking, to one inch in width. Test pieces over 5 inch thick, should be reduced in width at the center so as to approximate an area of cross section at point intended for breaking of of one square inch; but such reduced section should not be less than 35 inch in any case. 16 10 100 16 FORMS AND DIMENSIONS OF COUPONS. The following diagram (Fig. 59) shows the form and dimensions of a coupon employed in making a test of the material in the plate from which it was taken. 1" 10" Fig. 59 After the coupon has been cut to the proper width, the first thing to be done is to carefully measure the width and thickness at the small- est part, the part intended for breaking, and make a record of the dimensions. 8 114 A Library of Steam Engineering. The next step in the test is to place the coupon in the testing ma- chine, and pull the coupon apart, as shown in the following diagram (Fig. 60): Fig. 60 Dimensions of coupon at point of fracture before breaking. Width equal .50 of an inch; thickness equal .50 of an inch. Dimensions of coupon at point of fracture after breaking. Width equal .36 of an inch; thickness equal .34 of an inch. Strain at which the coupon parted at point of fracture, 15,000 pounds. TO DETERMINE THE TENSILE STRENGTH, PER SQUARE INCH, OF THE MATERIAL IN THE COUPON. RULE.-First, multiply the width of the sample, in hundredths of an inch, at point of fracture before breaking, by the thickness, in hundredths of an inch, at that point, and the product will give the area of the cross section of material at point intended for fracture, in hundredths of an inch. Second, divide the number of pounds of strain at which the coupon parted, by the area of the material before breaking, and the quotient will give the ten- sile strength, per square inch, of the material in pounds. Example.-Let 50 one hundredths of an inch equal width of mate- rial before breaking. Let 50 one hundredths of an inch equal thickness of material before breaking. Let 15,000 pounds equal strain at which the material parted. Then we have: 15000÷(.50X.50)=60000 lbs. of material per square inch. Tensile strength Performing the operation in the ordinary way, we have: .50 Width of material before breaking. .50 Thickness of material before breaking. .2500 Area of sample at point of fracture before breaking. And the strain at which the sample parted, divided by the area of sample at point of fracture, before breaking, we have .2500) 15000.0000 (60000 lbs. Tensile strength. 15000 0000 Boiler Plate. 115 It will be observed that four ciphers have been annexed to the dividend, to correspond with the number of decimals in the divisor, the reasons for which have been explained in a preceding chapter. TO DETERMINE DUCTILITY OF BOILER PLATE. RULE.-First, multiply the width of the sample before breaking, in hun- dredths of an inch, by the thickness, in hundredths of an inch, and the prod- uct will give the area of the material at point of fracture before breaking. Second, multiply the width of the sample at point of fracture after break- ing, in hundredths of an inch, by the thickness at point of fracture, in hun- dredths of an inch, and the product will give the area of material at point of fracture after breaking. Third, subtract the area of material at point of fracture after breaking, from the area of material at point of fracture before breaking, and call the product "The Remainder." Fourth, divide "The Remainder" by the area of material at point of fracture before breaking, and the quotient will give the percentage of ductility. Example.-Let .50 equal width of sample, in hundredths of an inch, before breaking. Let .50 equal thickness of sample, in hundredths of an inch, before breaking. Let .36 equal width of sample, in hundredths of an inch, after breaking. Let .34 equal thickness of sample, in hundredths of an inch, after breaking. Then we have: (.50x.50)-(.36.34) .50X.50 .51+ Per cent. ductility. Performing the operation in the ordinary way, we have: .50 Width of sample before breaking. .50 Thickness of sample before breaking. .2500 Area of sample before breaking. .36 Width of sample after breaking. .34 Thickness of sample after breaking. 144 108 .1224 Area of sample after breaking. And, deducting the area of the sample at point of fracture after breaking from the area of the sample before breaking, we have: 2500 .1276 "The Remainder." .1224 116 A Library of Steam Engineering. Then, dividing "The Remainder" by the area of sample at point of fraction before breaking, we have: .2500) .12760 (0.51+ Per cent. ductility. 12500 2600 2500 Giving the rule in more condensed form: RULE.-Subtract the area of sample at point of fracture, after breaking, from the area of sample intended for fracture, before breaking, and then divide "The Remainder" by the area of sample before breaking, and the quotient will give the percentage of ductility of the material. To make it still plainer: RULE.-Subtract the reduced area of sample from the original area, and divide "The Remainder" by the original area, the quotient will give the percentage of ductility. Example.-Let .96 equal original width of sample. Let .26 equal original thickness of sample. Let .81 equal reduced width of sample. Let .15 equal reduced thickness of sample. Then we have: .96X.26=.2496, Original area. .81X.15.1215, Reduced area. .2496) .12810 (0.513+ Per cent. ductility. 12480 3300 2496 8040 7488 THE PERCENTAGE OF DUCTILITY REQUIRED IN BOILER PLATE. No iron plates should be allowed to enter into the construction of any steam boiler, with a ductility of less than 25 per cent., and no steel plates should be allowed in the construction of any steam boiler, with a ductility of less than 50 per cent. for material having a tensile strength of 60,000 pounds, and one per cent. additional for every 1000 pounds additional tensile strength of the material. It will therefore be seen that all iron plates should have a ductility of not less than 25 per cent., and that steel plates should have a ductility of not less than 50 per cent., for 60,000 tensile strength and under, and one per cent. additional for each 1000 pounds in the tensile strength of the steel plates over 60,000 pounds tensile strength. And no steel plate should be used in any steam boiler with a tensile strength of less than 50,000 pounds per square inch. Steam Boilers. 117 CHAPTER V. STEAM BOILERS. SAFE-WORKING PRESSURE-SINGLE-RIVETED BOILERS. TO DETERMINE THE SAFE-WORKING PRESSURE FOR A CYLINDRICAL BOILER WITH SINGLE-RIVETED LONGITUDINAL SEAMS. RULE.-Multiply the thickness of material in the weakest plate in the shell of the boiler, in hundredths of an inch, by the tensile strength, in pounds per square inch, of the material in that plate, then divide the product by one-half of the diameter of the boiler in inches, and divide the quotient by 6, the factor of safety, and the last quotient will give the safe-working pressure, per square inch for a boiler with longitudinal seams single-riveted. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of plates. Let 40 inches equal diameter of the boiler. Let 6 equal a factor of safety. Then we have: (.25X60000)÷(40÷2) 6 -125 lbs. Safe-working pressure. Performing the operation in the ordinary way, we have; .25 60000 40÷2 20) 15000.00 (750 140 100 100 0 And, 750÷6=125 lbs. Safe-working pressure. BURSTING PRESSURE OF SINGLE-RIVETED BOILERS. TO DETERMINE THE BURSTING PRESSURE OF A BOILER WITH SINGLE- RIVETED LONGITUDINAL SEAMS. RULE.-Multiply the thickness of the weakest plate in the shell of the boiler, in hundredths of an inch, by the tensile strength in pounds per square 118 A Library of Steam Engineering. inch, and divide the product by one-half of the diameter of the boiler, in inches, then multiply the quotient by .56, the last product will give the bursting pressure per square inch. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of plates. Let 40 inches equal diameter of the boiler. Let .56 equal a constant. Then we have: .25 X 60000 402 X.56-420 lbs. Bursting pressure, per square inch, for single-riveted boilers. Performing the operation in the ordinary way, we have: .25 60000 40÷2= 20) 15000.00 (750 140 .56 100 45 00 100 375 0 420.00 lbs. Bursting pressure, per square inch, for single-riveted boilers. THICKNESS OF PLATE FOR SINGLE-RIVETED BOILERS. TO DETERMINE THE THICKNESS OF MATERIAL REQUIRED FOR SAFE- WORKING PRESSURE OF A BOILER, WITH SINGLE-RIVETED LONGI- TUDINAL SEAMS, WHEN THE DIAMETER, TENSILE STRENGTH OF MATERIAL AND WORKING PRESSURE ARE GIVEN. RULE.-Multiply the given pressure, per square inch, by 6, and mul- tiply the product by one-half of the diameter of the boiler, and divide the last product by the tensile strength of the material in pounds per square inch, the quotient will give the required thickness of the material in hundredths of an inch. Example.-Let 125 pounds per square inch equal the given pressure. Let 6 equal a constant. Let 40 inches equal diameter of the boiler. Let 60,000 pounds per square inch equal tensile strength of material. Then we have: 125×6×(40-2) =.25 Thickness of material required. 60000 Steam Boilers. 119 Performing the operation in the ordinary way, we have: 125 Given pressure per square inch. 6 A constant. 750 40-2- 20 One-half of the diameter of the boiler. Tensile strength of material. 60000) 150000 (0.25 Thickness of material required. 120000 300000 300000 DIAMETER OF SINGLE-RIVETED BOILERS. TO DETERMINE THE REQUIRED DIAMETER OF A BOILER, WITH SINGLE- RIVETED LONGITUDINAL SEAMS, WHEN THE WORKING-PRESSURE, THICKNESS AND TENSILE STRENGTH OF MATERIAL ARE GIVEN. RULE.-First, multiply the given thickness of the material, in hun- dredths of an inch, by the given tensile strength, per square inch, and call the product "Product No. 1." Second, multiply the given pressure, per square inch, by 6, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and multiply the quotient by 2, the last product will give the required diameter of the boiler in inches. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of material. Let 125 pounds equal given pressure per square inch. Let 6 equal a constant. Then we have: .25 X 60000 X2-40 inches. Required diameter of the boiler. 125×6 Performing the operation, we have: .25 Thickness of plate. 600 00 Tensile strength of plate. 15000.00 " Product No. 1." Next we have: 125 6 Pressure per square inch. Factor of safety. 750 "Product No. 2." 120 A Library of Steam Engineering. Finally, dividing "Product No. 1" by "Product No. 2," and multi- plying the quotient by 2, we have: 750) 15000 (20 1500 2 0 40 inches. Required diameter of the boiler. TENSILE STRENGTH OF MATERIAL FOR SINGLE-RIVETED BOILERS. TO DETERMINE THE REQUIRED TENSILE STRENGTH OF MATERIAL FOR A BOILER, WITH SINGLE-RIVETED LONGITUDINAL SEAMS, WHEN THE DIAMETER, THICKNESS OF MATERIAL, AND PRESSURE PER SQUARE INCH ARE GIVEN. RULE.-First, multiply the given pressure, per square inch, by 6, and call the product "Product No. 1." Second, multiply "Product No. 1" by one-half of the given diameter of the boiler, and call the product "Product No. 2." Third, divide "Product No. 2" by the given thickness of the material, in hundredths of an inch, and the quotient will give the required tensile strength of the material in pounds per square inch of section. Example. Let 125 pounds per square inch equal given pressure. Let 6 equal a constant. Let 40 inches equal given diameter of the boiler. Let 25 one hundredths of an inch equal given thickness of material. Then we have: 125×6× (40-2) .25 -60000 lbs. Required tensile strength per sec- tional square inch of material. Performing the operation in the ordinary way, we have: 125 6 750 20 .25) 15000.00 (60000 lbs. Required tensile strength per 150 00 00 sectional square inch of material. ΤΟ SAFE-WORKING PRESSURE OF DOUBLE-RIVETED BOILERS. CYLINDRICAL DETERMINE THE SAFE-WORKING PRESSURE FOR A BOILER, WITH DOUBLE-RIVETED LONGITUDINAL SEAMS, AND ALL HOLES IN THE PLATES DRILLED INSTEAD OF PUNCHED. oved ow zo RULE.-Multiply the thickness of the material in the weakest plate in the shell of the boiler, in hundredths of an inch, by the tensile strength of the Steam Boilers. 121 material, in pounds per square inch, then divide the product by one-half of the diameter of the boiler, in inches; then divide the quotient by 6, and add 20 per cent. to the last quotient, and the sum will give the safe-working press- ure for a cylindrical boiler, with longitudinal seams, double-riveted, and all holes in the plates drilled. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of the plates. Let 40 inches equal diameter of the boiler. Sove Let 6 equal a factor of safety. Let 20 per cent. equal amount to be added for double riveting. Then we have: (.25X60000) (402) X1.2-150 lbs. Pressure per square inch 6 allowable as safe- working pressure. Performing the operation in the ordinary way, we have: .25 60000 40÷220) 15000.00 6)750 125 1.2 250 125 Per cent. annexed to unity and multi- plying, is the same as adding the percentage. 150.0 lbs. Pressure allowable. SAFE-WORKING PRESSURE A SIMPLE RULE. A simple rule for determining the safe-working pressure for boilers with double-riveted longitudinal seams, and one which dis- penses with the trouble of adding the 20 per cent. to the pressure, and yet produces exactly the same result in every case, is as follows: RULE.-Multiply the thickness of material, in hundredths of an inch, in the weakest plate in the shell of the boiler, by the tensile strength of the material, per square inch; then divide the product by one-half of the diameter of the boiler, in inches, and then divide the quotient by 5, the last quotient will give the safe-working pressure, per square inch, in pounds 122 A Library of Steam Engineering. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of the material. Let 40 inches equal diameter of the boiler. Let 5 equal a factor of safety. Then we have: (.25X60000) (40-2) 5 150 lbs. Safe-working pressure. Performing the operation in the ordinary way, we have: .25 60000 40-÷2 20) 15000 5)750 150 lbs. Safe-working pressure. SAFE-WORKING PRESSURE-ANOTHER RULE. RULE.-Multiply the thickness of material, in hundredths of an inch, in the weakest plate in the shell of the boiler, by the tensile strength, per square inch, of the material, and divide the product by one-half of the diameter of the boiler, in inches, and multiply the quotient by .2 (two-tenths). Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of the material. Let 40 inches equal diameter of the boiler. Let 2 tenths equal a constant. Then we have: 25X60000 X.2=150 lbs. Safe-working pressure. 40:2 Performing the operation in the ordinary way, we have: .25 60000 Half diameter of boiler. 20) 15000.00 (750 140 .2 100 150.0 lbs. Safe-working pressure. 0 100 Steam Boilers. 123 SAFE-WORKING PRESSURE A SIMPLER RULE. Perhaps the simplest rule for determining the safe-working pressure for a boiler with double-riveted longitudinal seams is the following: RULE.-Multiply the thickness of material, in hundredths of an inch, in the weakest sheet in the boiler, by the tensile strength, and then divide the product by the diameter of the boiler in inches, and then multiply the quotient by 4, the last product will give the safe-working pressure per square inch. Example. Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of the material. Let 40 inches equal diameter of the boiler. Let .4 equal a constant. Then we have: ow then T 25 X 60000 40 X.4=150 lbs. Safe-working pressure, per square inch, for double riveting. Performing the operation in the ordinary way, we have: .25 Thickness of material. 60000 Tensile strength of material. Dividing by the diameter of the boiler. 40) 1500.00 (375 120 300 280 200 200 .4 A constant. 150.0 lbs. Safe-working pressure per square inch for double riveting. All of these rules give exactly the same result. The first rule for double-riveted boilers is the same as the rule for single-riveted boilers, except that 20 per cent. is added to the pressure allowed for single riveting when the longitudinal seams are double riveted. Therefore, whether we add 20 per cent. according to the first rule for double-riveted longitudinal seams, or divide by 5 and omit adding the 20 per cent. according to the second rule, or multiplying the quotient by .2 accord- ing to the third rule, or dividing with the diameter of the boiler instead of one-half of the diameter, and multiplying the quotient by .4 accord- ing to the fourth rule, it is the same in each case as the adding of 20 per cent. according to the first rule; hence, the student can select either of the foregoing rules with perfect safety, and with the assurance that each is absolutely correct. 124 A Library of Steam Engineering. BURSTING PRESSURE OF DOUBLE-RIVETED BOILERS. TO DETERMINE THE BURSTING PRESSURE OF A BOILER WITH DOUBLE-RIV- ETED LONGITUDINAL SEAMS, AND ALL HOLES IN PLATE DRILLED. RULE.-Multiply the thickness of the material in the weakest plate in the boiler, in hundredths of an inch, by the tensile strength in pounds per square inch, and divide the product by one-half of the diameter of boiler in inches, and multiply the quotient by .70, the last product will give the bursting pressure in pounds per square inch. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of material. Let 40 inches equal diameter of the boiler. Let .70 equal a constant. Then we have: .25 X 60000 40÷2 -X.70-525 lbs. - Pressure per square inch required to burst the boiler. Performing the operation in the ordinary way, we have: .25 60000 40÷2= 20) 15000.00 (750 140 .70 100 525.00 lbs. Bursting pressure. 100 SIMPLE RULE FOR DETERMINING BURSTING PRESSURE. A simpler rule for determining the bursting pressure, and one that gives exactly the same result, is as follows: RULE.-Multiply the thickness of material, in hundredths of an inch, of the weakest plate in the shell of the boiler by the tensile strength of the material, per square inch, and then divide the product by the diameter of the boiler, in inches, and then multiply the quotient by 1.4, the last product will give the bursting pressure, in pounds per square inch, for double-riveted longitudinal seams. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of material. Let 40 inches equal diameter of the boiler. Let 1.4 equal a constant. Steam Boilers. 125 Then we have: .25 X 60000 40 -X1.4 525 lbs. Bursting Pressure. Performing the operation in the ordinary way, we have: .25 60000 40) 15000.00 (375 120 1.4 300 150 0 280 375 200 525.0 lbs. 200 Bursting pressure per square inch. THICKNESS OF PLATE FOR DOUBLE-RIVETED BOILERS. TO DETERMINE THE THICKNESS OF MATERIAL REQUIRED FOR SAFE-WORK- ING PRESSURE OF A BOILER WITH DOUBLE-RIVETED LONGITU- DINAL SEAMS, WHEN THE DIAMETER, TENSILE STRENGTH OF MATERIAL AND WORKING PRESSURE ARE GIVEN. RULE.-Multiply the given pressure, per square inch, by 5, then multi- ply the product by one-half of the given diameter of the boiler, in inches, and then divide the last product by the given tensile strength of the material, and the quotient will give the thickness of the required material in hundredths of an inch. Example.-Let 150 pounds per square inch equal the given pressure. Let 5 equal a constant for double-riveted longitudinal seams. Let 40 inches equal given diameter of the boiler. Let 60,000 pounds per square inch equal tensile strength of the material. Then we have: 150X5X (40÷÷2) 60000 =.25 Thickness of required material in hundredths of an inch. Performing the operation in the ordinary way, we have: 150 5 750 40÷2= 20 Tensile strength of material. 60000) 150000 (0.25 Thickness of required material. 120000 300000 300000 126 A Library of Steam Engineering. DIAMETER OF DOUBLE-RIVETED BOILERS. TO DETERMINE THE REQUIRED DIAMETER OF A BOILER, WITH DOUBLE- RIVETED LONGITUDINAL SEAMS, WHEN THE WORKING PRESSURE, THICKNESS AND TENSILE STRENGTH OF MATERIAL ARE GIVEN. RULE.-First, multiply the given thickness of material, in hundredths of an inch, by the given tensile strength in pounds per square inch, and call the product "Product No. 1." Second, multiply the given pressure, per square inch, by 5, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and multiply the quotient by 2, the last product will give the required diameter of the boiler. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of material. Let 150 pounds per square inch equal given pressure. Let 5 equal a constant. Then we have: .25X60000 150X5 X2-40 inches. Required diameter of the boiler. Performing the operation, we have: .25 60000 15000.00 "Product No. 1." Next we have: 150 5 750 "Product No. 2." Finally we have: 750) 15000 (20 1500 2 0 40 inches. Required diameter of the boiler. TENSILE STRENGTH OF MATERIAL FOR DOUBLE-RIVETED BOILERS. TO DETERMINE THE REQUIRED TENSILE STRENGTH OF MATERIAL FOR A BOILER WITH DOUBLE-RIVETED LONGITUDINAL SEAMS, WHEN THE DIAMETER, THICKNESS OF MATERIAL, AND PRESS- URE PER SQUARE INCH ARE GIVEN. RULE.—First, multiply the given pressure, per square inch, by 5, and call the product "Product No. 1." Second, multiply" Product No. 1" by one-half of the given diameter of the boiler, and call the product "Product No. 2." Steam Boilers. 127 Third, divide "Product No. 2" by the given thickness of the material, in hundredths of an inch, and the quotient will give the required tensile strength, per square inch, of the material in pounds. Example. Let 150 pounds per square inch equal the given pressure. Let 5 equal a constant. Let 40 inches equal given diameter of the boiler. Let 25 one hundredths of an inch equal given thickness of material required. Then we have: 150X5X (40-÷2) .25 -60000 lbs. Tensile strength per square inch of material required. Performing the operation in the ordinary way, we have: 150 5 40-2= 750 20 .25) 15000.00 (60000 lbs. Required tensile strength of mate- 150 0000 rial per square inch. STRAIN ON LONGITUDINAL SEAMS. TO DETERMINE THE STRAIN PRODUCED ON LONGITUDINAL SEAMS, TENDING TO TEAR THEM ASUNDER, BY ANY GIVEN PRESSURE PER SQUARE INCH. RULE.-Multiply the given pressure, per square inch, in pounds, by one-half of the diameter of the boiler, in inches, and the product will give the strain, in pounds, on each longitudinal inch along the shell of the entire boiler as well as on the longitudinal seams. Example.-Let 150 pounds per square inch equal given pressure. Let 48 inches equal diameter of the boiler. Then we have: 150 Given pressure, 48÷2= 24 Half diameter of the boiler. 600 300 3600 lbs. Strain on each inch along the entire length of the boiler. This is the same as though the boiler was made of hoops an inch wide, the strain on each hoop tending to burst it would be 3600 pounds, if the hoops were 48 inches in diameter. 128 A Library of Steam Engineering. LONGITUDINAL STRAIN ON BOILERS. TO DETERMINE THE STRAIN IN A LONGITUDINAL DIRECTION ON EACH INCH IN THE CIRCUMFERENCE OF THE BOILER, IN A SHELL HAVING NO FLUES OR BRACES. RULE.-First, square the diameter of the boiler, in inches, and multiply the product by .7854, and then multiply the last product by the given press- ure, per square inch, and call the product "Product No. 1." Second, multiply 3.1416 by the diameter of the boiler, in inches, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2" and the quotient will give the strain in a longitudinal direction on each inch in the circle of the boiler. Example.-Let 48 inches equal diameter of the boiler. Let .7854 equal a constant. Let 150 pounds per square inch equal given pressure. Let 3.1416 equal a constant. Then we have: 48x48X.7854X150 3.1416×48 -=1800 lbs. Strain in a longitudinal direction on each inch in the circum- ference of the boiler. Performing the operation in the ordinary way, we have: 48 Diameter of boiler. 48 Diameter of boiler. 384 192 2304 .7854 9216 11 520 184 32 1612 8 1809.5616 150 90478 0800 Area of diameter of boiler. Pressure per square inch. 180956 16 271434.2400 "Product No. 1." Next, multiplying 3.1416 by the diameter of the boiler, we have: 3.1416 A constant. 48 Diameter of boiler. 25 1328 125 664 150.7968 "Product No. 2." Steam Boilers. 129 Finally, dividing "Product No. 1" by "Product No. 2," we have: 150.7968) 271434.2400 (1800 lbs. Strain in a longitudinal 150796 8 direction on each inch in the circum- ference of the boiler. 120637 44 120637 44 Supposing the boiler to be made of 151 staves, extending the entire length of the boiler, and 150 of them would be each one inch in width, and one would be .7968 of an inch in width-a fraction over of an inch-each of the staves or strips an inch wide would have a strain on them lengthwise of 1800 pounds, and the odd strip would have a strain in proportion to its width. The rule given is the general rule laid down for the guidance of engineers and boiler makers, and while it is absolutely correct, it is a long, tedious and very cumbersome rule. We will, therefore, give another rule which is also absolutely correct, and while it gives the answer exactly the same as the other rule in every case, it has a decided advantage on account of its brevity. SIMPLE RULE FOR DETERMINING LONGITUDINAL STRAIN ON BOILERS. RULE.-Divide the diameter of the boiler by 4, and multiply the quotient by the given pressure, per square inch, and the product will give the total strain in a longitudinal direction on each inch in the circumference of a boiler having no flues or braces. Example.-Let 4 equal a constant. Let 48 inches equal diameter of the boiler. Let 150 pounds per square inch equal given pressure. Then we have: 4)48 Diameter of boiler. 12 150 Given pressure per square inch. 600 12 1800 lbs. Strain in a longitudinal direction on each inch in the circumference of the boiler. WEIGHT OF STEAM DISCHARGED UNDER ANY GIVEN PRESSURE THROUGH A PIPE OR ORIFICE OF A GIVEN DIAMETER. A knowledge of the rule for determining the weight of steam a given pipe or orifice will discharge into the atmosphere, under any given pressure, in a given time, is very important to engineers, as it 9 130 A Library of Steam Engineering. frequently happens that owners of steam boilers furnish their neigh- bors or tenants, steam for power or heating purposes. It is, therefore, essential that the engineer should be able to determine the amount of coal required to generate the amount of steam required. We will, therefore, lay down a simple rule, which will be sufficiently accurate for all practical purposes. RULE.-Divide the steam pressure, per square inch, carried in the boiler by 70, and multiply the area of the orifice or cross section of the pipe, to be employed for conveying the steam, by the quotient, and the product will give the weight of steam discharged in pounds per second. Example.-Let 105 pounds equal steam pressure per square inch. Let 70 equal a constant. Let 1.5 inch equal diameter of pipe or orifice. Let .7854 equal a constant. Then we have: 1.52X.7854X(105÷70)=2.6507250 pounds. Weight of steam dis- charged per second. Performing the operation, we have: 70) 105 (1.5 The quotient. 70 350 350 Next we have: 1.5 1.5 75 15 novig 2.25 .7854 39270 15708 1 5708 1.767150 square inches. Area of orifice or cross section of pipe. Finally, multiplying the area of the orifice or cross section of the pipe by the quotient, we have: 1.767150 1.5 8835750 1 767150 2.6507250 lbs. Weight of steam discharged per second. Steam Boilers. 131 Then, to determine the weight of steam discharged per hour, we multiply the weight discharged per second by 3600, thus: 2.6507250 3600 1590 4350000 7952 1750 9542.6100000 lbs. Weight of steam discharged per hour into the atmosphere. This amount of steam does not represent the amount of steam a tenant actually uses, but it does represent the amount he can use if furnished steam through a 1 inch pipe. The cost of furnishing that amount of steam depends upon the efficiency of the boiler, and whether the feed water is heated before it enters the boiler. If the feed water is heated by exhaust steam, the amount of heat imparted to the water is clear gain. If the feed water is heated by live steam, the amount of heat imparted to it is imparted at the expense of the owner of the boiler. The ordinary two-flue boiler, with a well-proportioned furnace and stack, will evaporate, ordinarily, from 6 to 7 pounds of water per pound of good coal, with the temperature of feed water at 60° Fah- renheit, and 7 to 8 pounds with feed water at 200° Fahrenheit. A well-proportioned tubular boiler, with properly proportioned furnace and stack, will evaporate from 8 to 9 pounds of water per pound of good coal, with a feed water temperature of 60° Fahrenheit, and from 9 to 10 pounds with a feed water temperature of 200° Fahrenheit. instyled esastaid codo 132 A Library of Steam Engineering. CHAPTER VI. RIVETED JOINTS. DISTANCE OF RIVET HOLES FROM EDGE OF PLATE. The center of outer rows of rivet holes, for single or double riv- eting, should in no case be less than one and one-half times the diameter of the rivet holes from the edge of plate. Example.-Let 75 one hundredths of an inch equal diameter of rivet hole. Let 1.5 equal a constant. Then we have: .75 Diameter of rivet hole. 1.5 A constant. 375 75 1.125 inches. Distance of center of rivet hole from edge of the sheet. edge of the SHEARING STRENGTH OF RIVETS. The shearing strength of iron rivets of a given area of cross sec- tion is very nearly equal to the tensile strength of iron plate of equal area of cross section, and they are so taken in the examples. But the shearing strength of steel rivets employed with steel plates is from 15 to 20 per cent. less than the tensile strength of the plate; therefore, due allowance has been made in the rules and examples relating to steel plates and steel rivets. DISTANCE BETWEEN CENTERS OF ROWS OF RIVET HOLES FOR CHAIN RIVETING. TO DETERMINE THE DISTANCE BETWEEN CENTERS OF ROWS OF RIVET HOLES. RULE.-Multiply the diameter of rivet hole by the constant 2.5, and the product will give the distance between centers of rows of chain rivet holes. Example.-Let 75 one hundredths of an inch equal diameter of rivet hole. Let 2.5 equal a constant. Then we have: .75X2.5=1.875, or 13 inches. Distance between centers of rows of rivet holes for chain riveting. Riveted Joints. 133 Performing the operation, we have: .75 Diameter of rivet hole. 2.5 A constant. 375 150 1.875, or 13 inches. Distance between centers of rows of rivet holes for chain riveting. DISTANCE BETWEEN CENTERS OF ROWS OF RIVET HOLES FOR ZIG-ZAG RIVETING. TO DETERMINE THE DISTANCE BETWEEN CENTERS OF ROWS OF RIVET HOLES. RULE.-First, multiply the pitch from center to center of rivet holes by the constant 11, and add four times the diameter of rivet hole to the product, and call the sum "Sum No. 1." Second, add four times the diameter of rivet hole to the pitch from center to center of rivet holes, and call the sum "Sum No. 2." Third, multiply "Sum No. 1" by "Sum No. 2" and extract the square root of the product; then divide the square root by the constant 10, and the quotient will give the required distance between centers of rows of rivet holes. Example. Let 2.5885 inches equal pitch from center to center of rivet holes. Let 11 equal a constant. Let 625 one thousandths of an inch equal diameter of rivet hole. Let 4 equal a constant. Let 10 equal a constant. Then we have: (11X2.5885+4X.625)X(2.5885+4X.625) 10 mihivib yllsait =1.25 or 11 inches. Distance between centers of rows of rivet holes for zig-zag riveting. Performing the operation, we have: Then four times diameter .625 X4= of rivet hole. 2.5885 Pitch of rivet hole. 11 A constant. 25885 25 885 28 4735 2.500 30.9735 "Sum No. 1." 134 A Library of Steam Engineering. Next we have: Adding four times the .625 X4= diameter of rivet hole to the pitch. 2.5885 Pitch. 2.500 5.0885 "Sum No. 2." Next, multiplying "Sum No. 1" by "Sum No. 2" we have: 30.9735 5.0885 1548675 2477880 2 477880 154 8675 157.60865475 Product. Next, extracting the square root of the product, we have: 157.60865475(12.5542+ Square root. 1 22)57 44 245) 1360 1225 2505) 13586 12525 25104) 106154 100416 251082) 573875 502164 Finally, dividing the square root by the constant 10, we have: 10) 12.5542 (1.25542 Or a fraction over 14 inches, re- Cab 10 25 201 celost 20 quired distance between cen- ters of rows of rivet holes for zig-zag riveting. 55 50 54 50 42 40 45 221 Riveted Joints. 135 DIAMETER OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS- IRON PLATES AND IRON RIVETS. TO DETERMINE THE DIAMETER OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS-IRON PLATES AND IRON RIVETS. RULE.-Extract the square root of the thickness of the plate in decimals of an inch and multiply the square root by the constant 1.25, and the product will give the required diameter of rivet holes. Example. Let 25 one hundredths of an inch equal thickness of plate. Let 1.25 equal a constant. Then we have: (25) X1.25 .625, or § inch. Diameter of rivet holes. Performing the operation, we have: 25 (.5 Square root. .25 Next, multiplying the square root by the constant 1.25 we have: .5x1.25 .625, or ğ inch. Diameter of rivet holes. DIAMETER OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS- IRON PLATES AND IRON RIVETS. TO DETERMINE THE DIAMETER OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS-IRON PLATES AND IRON RIVETS. RULE.-Extract the square root of the thickness of the plate in decimals of an inch and multiply the square root by the constant 1.125, and the prod- uct will give the required diameter of rivet holes. Example.-Let 25 one hundredths of an inch equal thickness of plate. Let 1.125 equal a constant. 16 Then we have: (✓.25)×1.125=.5625, or inch. Diameter of rivet holes. Performing the operation, we have: .25 (.5 Square root. .25 Next, multiplying the square root by the constant 1.125, we have: .5X1.125 .5625, or inch. Diameter of rivet holes. 6 DIAMETER OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS- STEEL PLATES AND STEEL RIVETS. TO DETERMINE THE DIAMETER OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS-STEEL PLATES AND STEEL RIVETS. RULE.-Extract the square root of the thickness of the plate in decimals of an inch and multiply the square root by the constant 1.375, and the product will give the required diameter of rivet holes. 136 A Library of Steam Engineering. Example.-Let 25 one hundredths of an inch equal thickness of plate. Let 1.375 equal a constant. 16 Then we have: (.25)X1.375 .6875. or 1 inch. Diameter of rivet holes. Performing the operation, we have: .25 (.5 Square root. .25 Next, multiplying the square root by the constant 1.375, we have: .5X1.375 .6875, or 1 inch. Diameter of rivet holes. DIAMETER OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS- STEEL PLATES AND STEEL RIVETS. TO DETERMINE THE DIAMETER OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS STEEL PLATES AND STEEL RIVETS. RULE.-Extract the square root of the thickness of the plate and multi- ply the root by the constant 1.25, and the product will give the diameter of rivet holes. Example.-Let 25 one hundredths of an inch equal thickness of plate. Let 1.25 equal a constant. Then we have: (.25) X1.25 .625, or § inch. Diameter of rivet holes. Performing the operation, we have: .25 (.5 Square root. .25 Next, multiplying the square root by the constant 1.25, we have: .5X1.25 .625, or ğ inch. Diameter of rivet holes. PITCH OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS- IRON PLATES AND IRON RIVETS. о Fig. 61 Riveted Joints. 137 TO DETERMINE THE PITCH OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS IRON PLATES AND IRON RIVETS. RULE.-Divide the area of diameter of rivet hole by the thickness of the plate in decimals of an inch, then add the diameter of the rivet hole to the quotient, and the sum will give the required pitch from center to center of rivet holes. Example.-Let .625X.625X.7854 equal area of diameter of rivet hole. Let 25 one hundredths of an inch equal thickness of plate. Let 625 one thousandths of an inch equal diameter of rivet hole. Then we have: .625 X.625 X.7854 +.625-1.852+ inches. Pitch of rivet .25 holes from cen- ter to center. Performing the operation in the ordinary way, we have: .625 Diameter of rivet hole. .625 Diameter of rivet hole. 3125 1250 3750 .390625 .7854 1562500 1953125 3125000 2734375 .3067968750 Area of diameter of rivet hole. Then, dividing the area of diameter of rivet hole by the thickness of the plate in decimals of an inch, we have: .250000000).306796875 (1.227+ The quotient. 250000000 567968750 500000000 679687500 500000000 179687500 175000000 138 A Library of Steam Engineering. Finally, adding the diameter of rivet hole to the quotient, we have: 1.227 The quotient. .625 Diameter of rivet hole. 1.852 inches. Pitch from center to center of rivet hole. It will be noticed that seven ciphers have been annexed to the thickness of the plate in the divisor; that is done under the rule requiring an equal number of decimals in the divisor and dividend in performing the operation of division with decimal fractions, in order to readily determine the whole numbers in the quotient. PITCH OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS- IRON PLATES AND IRON RIVETS. TO DETERMINE THE PITCH OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS IRON PLATES AND IRON RIVETS. RULE.-Multiply the area of diameter of rivet hole by the number of rows of rivet holes (2), then divide the product by the thickness of the plate in decimals of an inch, and add the diameter of rivet hole to the quotient, the sum will give the required pitch of rivet holes in inches. Example. Let 625 one thousandths of an inch (inch) equal diameter of rivet hole. Let 2 equal number of rows of rivet holes. Let 3125 ten thousandths of an inch (inch) equal thickness of plate. Then we have: .625 X.625X.7854×2 +.625-2.5885 inches. Pitch of rivet holes .3125 Performing the operation, we have: .625 Diameter of rivet hole. .625 Diameter of rivet hole. 3125 1250 3750 .390625 .7854 1562500 1953125 3125000 2734375 from center to center Am't carried forward, .3067968750 Area of diameter of rivet hole. Riveted Joints. 139 .3067968750 Area of diameter of rivet hole. 2 Number of rows of rivet holes. .3125) .6135937500 (1.9635 Am't brought forward, Next, dividing by the thickness of plate. 3125 .625 Diameter of rivet hole. 30109 2.5885 inches. Required pitch from 28125 center to center of rivet holes. 19843 18750 10937 9375 15625 15625 NOTE. For double-riveted and triple-riveted laps, iron plate and iron rivets, the diameter of rivet holes should be one-sixteenth of an inch less than the single-riveted laps. PITCH OF RIVET HOLES FOR TRIPLE-RIVETED LAP JOINTS- IRON PLATES AND IRON RIVETS. TO DETERMINE THE PITCH OF RIVET HOLES FOR TRIPLE-RIVETED LAP JOINTS IRON PLATES AND IRON RIVETS. RULE.-First, multiply the area of diameter of rivet hole by the num- ber of rows of rivet holes and call the product "Product No. 1." Second, divide "Product No. 1" by the thickness of the plate and add the diameter of rivet hole to the quotient, and the sum will give the required pitch of rivets in inches. Example.-Let 875 one thousandths of an inch (inch) equal diameter of rivet hole. Let 3 equal number of rows of rivet holes. Let 75 one hundredths of an inch (inch) equal thick- ness of plate. Then we have: .875X.875X.7854X3 .75 +.875-3.2802 inches. Pitch from cen- ter to center of rivet holes. Performing the operation, we have: .875 Diameter of rivet hole. .875 Diameter of rivet hole. 4375 6125 7000 Am't carried forward, .765625 140 A Library of Steam Engineering. Am't brought forward, .765625 .7854 3062500 3828125 6125000 5359375 .6013218750 Area of diameter of rivet hole. 3 Number of rows of rivet holes. 1.8039656250 "Product No. 1." Dividing "Product No. 1" by the thickness of plate, we have: .75) 1.803965625 (2.4052 150 .875 Diameter of rivet hole. 303 3.2802 inches. Pitch from center to 300 center of rivet holes. 396 375 215 150 PITCH OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS- STEEL PLATES AND STEEL RIVETS. TO DETERMINE THE PITCH OF RIVET HOLES FOR SINGLE-RIVETED LAP JOINTS-STEEL PLATES AND STEEL RIVETS. RULE.-Multiply the area of the diameter of the rivet hole by .8, and divide the product by the thickness of the plate, and add the diameter of the rivet hole to the quotient, and the sum will give the required pitch from center to center of rivet holes. Example. Let 625 one thousandths of an inch (ğ inch) equal diameter of rivet hole. Let .8 equal a constant. Let 25 one hundredths of an inch equal thickness of plate. Then we have: .625 .625 .7854X.8 .25 +.625 1.60675 Pitch of riv- et holes. Performing the operation, we have: .625 Diameter of rivet hole. .625 Diameter of rivet hole. 3125 1250 Am't carried forward, .390625 3750 Riveted Joints. 141 Am't brought forward, .390625 .7854 1562500 1953125 3125000 2734375 .3067968750 Area of diameter of rivet hole. .8 A constant. .25).24543750000 (.98175 225 204 200 43 25 187 175 .625 Diameter of rivet hole. 1.60675 inches. Pitch of rivet holes. from center to cen- ter. 125 125 PITCH OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS- STEEL PLATES AND STEEL RIVETS. о Fig. 62 TO DETERMINE THE PITCH OF RIVET HOLES FOR DOUBLE-RIVETED LAP JOINTS STEEL PLATES AND STEEL RIVETS. RULE.-Multiply the area of diameter of rivet hole by the number of rows of rivet holes (2), then multiply the product by .80, then divide the last product by the thickness of the plate and add the diameter of rivet hole to the quotient, the sum will give the required pitch of rivet holes in inches. Example. Let 625 one thousandths of an inch ( inch) equal diameter of rivet hole. Let 2 equal number of rows of rivet holes. Let .80 equal percentage of shearing strength of rivets. Let 25 one hundredths of an inch equal thickness of plate. 142 A Library of Steam Engineering. Then we have: .625X.625X.7854X2X.80 +.625=2.588+ inches. Required pitch .25 of rivet holes. Performing the operation in the ordinary way, we have: .625 Diameter of rivet hole. .625 Diameter of rivet hole. 3125 1250 3750 .390625 .7854 1562500 1953125 3125000 2734375 .3067968750 Area of diameter of rivet hole. 2 Number of rows of rivet holes. .6135937500 .80 Percentage of shearing strength of rivets. Dividing by thickness of plate. .25).490875000000 (1.963+ 25 240 225 158 150 .625 Diameter of rivet hole. 2.588 inches. Required pitch of rivet holes. 87 75 PITCH OF RIVET HOLES FOR TRIPLE-RIVETED LAP JOINTS- STEEL PLATES AND STEEL RIVETS. TO DETERMINE THE PITCH OF RIVET HOLES FOR TRIPLE-RIVETED LAP JOINTS STEEL PLATES AND STEEL RIVETS. RULE.-First, multiply the area of diameter of rivet hole by the number of rows of rivet holes (3), then multiply the product by the shearing strength per sectional square inch of the rivet, which is .80 per cent. of the tensile strength of the plate, and therefore, .80 is used as a multiplier, then call the product "Product No. 1." Second, divide "Product No. 1" by the thickness of the plate, and add the diameter of rivet hole to the quotient, the sum will give the pitch, in inches, from center to center of rivet holes in each row. Riveted Joints. 143 Example.-Let 75 one hundredths of an inch equal diameter of rivet hole. Let 3 equal number of rows of rivet holes. Let .80 equal percentage of shearing strength of rivets as compared with tensile strength of plate. Let 5 tenths of an inch equal thickness of plate. Then we have: .75X.75X.7854×3×.80 5 +.75=2.87+ inches. Required pitch Performing the operation in the ordinary way, we have: .75 Diameter of rivet hole. .75 Diameter of rivet hole. of rivet holes. 375 525 .5625 .7854 22500 28125 45000 39375 .44178750 Area of diameter of rivet hole. 3 Number of rows of rivet holes. 1.32536250 80 1.0602900000 Percentage of shearing strength of rivets as compared with tensile strength of plate. Product No. 1." Next, dividing "Product No. 1" by the thickness of the plate and adding the diameter of rivet hole to the quotient, we have: .5) 1.06029 (2.12+ 10 5 640 .75 Diameter of rivet hole. 2.87 inches. Required pitch of rivet holes. 10 10 STRENGTH OF SINGLE-RIVETED LAP JOINTS-IRON PLATES AND IRON RIVETS. TO DETERMINE THE PERCENTAGE OF STRENGTH OF SINGLE-RIVETED LAP JOINTS-IRON PLATES AND IRON RIVETS. RULE.-First, subtract the diameter of rivet hole from the pitch, and divide the remainder by the pitch, and the quotient will give the percentage of strength of plate at joint. 144 A Library of Steam Engineering. ! Second, square the diameter of rivet hole and multiply by .7854, which will give the area of cross section of the rivet, and call the product "Product No. 1.' Third, multiply the pitch of rivets by the thickness of the plate in deci- mals of an inch, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2" and the quotient will give the percentage of strength of rivets at joint." PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 625 one thousandths of an inch equal diameter of rivet hole. Let 1.852 inches equal pitch from center to center of rivet holes. Let 25 one hundredths of an inch equal thickness of plate. Then we have: 1.852-.625 1.852 Performing the operation, we have: 1.852 .66+ Percentage of strength of plate at joint. Pitch of rivet holes. .625 Diameter of rivet hole. 1.852) 1.2270 (0.66+ Percentage of strength of plate 1 1112 at joint. 11580 11112 PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Next we have the percentage of strength of rivets at joint: .625X.625 X.7854 1.852X.25 Performing the operation, we have: .625 .6626+ Percentage of strength of rivets at joint. Diameter of rivet hole. .625 Diameter of rivet hole. 3125 1250 3750 .390625 .7854 1562500 1953125 3125000 2734375 "Product No. 1." .3067968750 Area of diameter of rivet hole. Riveted Joints. 145 Next, multiply the pitch of rivets by the thickness of the plate, we have: 1.852 Pitch of rivets. .25 Thickness of plate. 9260 3704 .46300 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: .46300).306796875 (0.6626+ Percentage of strength of 277800 rivets at joint. 289968 277800 121687 92600 290875 277800 STRENGTH OF DOUBLE-RIVETED LAP JOINTS-IRON PLATES AND IRON RIVETS. TO DETERMINE THE PERCENTAGE OF STRENGTH OF DOUBLE-RIVETED LAP JOINTS-IRON PLATES AND IRON RIVETS. RULE.-First, subtract the diameter of one rivet hole from the pitch, then divide the remainder by the pitch, and the quotient will give the percent- age of strength of the plate at joint. Second, multiply the area of diameter of one-rivet hole by the number of rows of rivets (2), and call the product "Product No. 1." Third, multiply the pitch from center to center of rivets by the thickness of the plate in decimals of an inch, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2," and the quotient will give the percentage of strength of the rivets at the joint as compared with the solid plate. PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 2.5881 inches equal pitch from center to center of rivet holes in each row. Let 625 one thousandths of an inch (inch) equal diameter of rivet hole. Let 2 equal number of rows of rivet holes in lap. Let 3125 ten thousandths of an inch equal thickness of plate. Then we have, first the percentage of strength of plate at joint: 10 2.5881-.625 2.5881 .75+ Percentage of strength of plate at joint. 146 A Library of Steam Engineering. Performing the operation, we have: 2.5881 Pitch of rivet holes. .625 Diameter of rivet hole. 2.5881) 1.96310 (0.75+ Percentage of strength of plate 1 81167 at joint. 151430 129405 PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Next we have the percentage of strength of the rivets at joint: .625 X.625 X.7854×2 2.5881.3125 Performing the operation, we have: .75+ Percentage of strength of rivets at joint. .625 Diameter of rivet hole. .625 Diameter of rivet hole. 3125 1250 3750 .390625 .7854 1562500 1953125 3125000 2734375 .3067968750 Area of diameter of rivet hole. 2 Number of rows of rivets. .6135937500 "Product No. 1." Next, multiplying the pitch of rivets by the thickness of the plate, we have: 2.5881 Pitch of rivets. .3125 Thickness of plate. 129405 51762 25881 77643 .80878125 "Product No. 2" Finally, dividing "Product No. 1" by "Product No. 2," we have: .80878125).6135937500 (.75+ Percentage of strength of 566146875 474468750 404390625 rivets at joint. Riveted Joints. 147 STRENGTH OF TRIPLE-RIVETED LAP JOINTS-IRON PLATES AND IRON RIVETS. TO DETERMINE THE PERCENTAGE OF STRENGTH OF TRIPLE-RIVETED LAP JOINTS IRON PLATES AND IRON RIVETS. RULE. First, subtract the diameter of rivet hole from the pitch, and divide the remainder by the pitch, and the quotient will give the percentage of strength of the plate at joint. Second, multiply the area of diameter of one rivet hole by the number of rows of rivets (3), and call the product "Product No. 1." Third, multiply the pitch of rivets from center to center by the thickness of material in the plate, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2," and the quotient will give the percentage of strength of rivets at joint. PERCENTAGE OF STRENGTH OF PLATE AT JOINT. an men Example. Let 875 one thousandths of an inch (inch) equal diameter of rivet hole. Let 3.2802 inches equal pitch of rivet holes from cen- ter to center. Let 3 equal number of rows of rivet holes. Let 75 one hundredths of an inch (inch) equal thick- ness of plate. Then we have, first the percentage of strength of plate at joint: 3.2802-.875 3.2802 -.7332+ Percentage of strength of plate at joint. Performing the operation, we have: 3.2802 Pitch of rivet holes. .875 Diameter of rivet hole. 3.2802) 2.40520 (0.7332+ Percentage of strength of plate at joint. 2 29614 109060 98406 106540 98406 81340 65604 PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Next we have the percentage of strength of rivets at joint: .875X.875X.7854x3 ==.7332+ 3.2802X.75 Percentage of strength of rivets at joint. 148 A Library of Steam Engineering. Performing the operation, we have: .875 Diameter of rivet hole. .875 Diameter of rivet hole. 4375 6125 Bo Jato ads most slod 7000 .765625 .7854 3062500 3828125 6125000 5359375 .6013218750 Area of diameter of rivet hole. 3 Number of rows of rivets. 1.8039656250 "Product No. 1." Next, multiplying the pitch of rivets by the thickness of the plate, we have: 3.2802 Pitch of rivets. 75 Thickness of plate. 164010 2 29614 2.460150 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 2.46015) 1.803965625 (.7332+ 1 722105 Percentage of strength of riv- ets as compared with the solid plate. 818606 738045 805612 738045 675675 492030 STRENGTH OF SINGLE-RIVETED LAP JOINTS-STEEL PLATES AND STEEL RIVETS. TO DETERMINE THE PERCENTAGE OF STRENGTH OF SINGLE-RIVETED LAP JOINTS-STEEL PLATES AND STEEL RIVETS. RULE.-First, subtract the diameter of rivet hole from the pitch and divide the remainder by the pitch, and the quotient will give the percentage of strength of plate at joint. Second, square the diameter of rivet hole and multiply by .7854, which will give the area of cross section of rivet, and multiply the area by .8, and call the product "Product No. 1." Riveted Joints. 149 Third, multiply the pitch of rivets by the thickness of the plate, in deci- mals of an inch, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2," and the quotient will give the percentage of strength of rivets at joint. PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 1.60675 inches equal pitch of rivet holes. Let 625 one thousandths of an inch equal diameter of rivet hole. Let .8 equal a constant. Let 25 one hundredths of an inch equal thickness of plate. Then we have first, percentage of strength of plate at joint: 1.60675-.625 1.60675 =.61+ Percentage of strength of plate at joint. Performing the operation, we have: 1.60675 Pitch of rivet holes. .625 Diameter of rivet hole. 1.60675) .981750 (0.61+ Percentage of strength of plate 964050 177000 160675 at joint. PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Next we have percentage of strength of rivets at joint: .625 X.625 X.7854X.8 .61+ 1.60675X.25 Percentage of strength of rivets at joint. Performing the operation, we have: ACEPPERTEL OF .625 Diameter of rivet hole. .625 Diameter of rivet hole. 3125 1250 3750 .390625 .7854 1562500 1953125 3125000 2734375 .3067968750 Area of diameter of rivet hole. .8 A constant. .24543750000 "Product No. 1.". 150 A Library of Steam Engineering. Next, multiplying the pitch of rivets by the thickness of plate, we have: 1.60675 Pitch of rivets. .25 Thickness of plate 803375 321350 .4016875 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 0.4016875) .24543750 (0.61+ Percentage of strength of riv- 24101250 ets at joint. 4425000 4016875 STRENGTH OF DOUBLE-RIVETED AND STEEL LAP JOINTS-STEEL PLATES RIVETS. OO O о о о Fig. 63 TO DETERMINE THE PERCENTAGE OF STRENGTH OF DOUBLE-RIVETED LAP JOINTS-STEEL PLATES AND STEEL RIVETS. RULE.-First, subtract the diameter of one rivet hole from the pitch, then divide the remainder by the pitch, and the quotient will give the percent- age of strength of the plate at joint. Second, multiply the area of diameter of one rivet hole by the number of rows of rivets (2), then multiply the product by .80, and call the last product "Product No. 1." Third, multiply the pitch from center to center of rivets by the thickness of material in the plate in decimals of an inch, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2" and the quotient will give the percentage of strength of the rivets at joint as compared with the solid plate. Riveted Joints. 151 PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 2.588 inches equal pitch from center to center of rivet holes in each row. Let 625 one thousandths of an inch (ğ inch) equal diameter of rivet hole. Let 2 equal number of rows of rivet holes in a double- riveted joint. Let .80 equal percentage of shearing strength of rivets as compared with tensile strength of plate. Let 25 one hundredths of an inch equal thickness of plate. Then we have first, the percentage of strength of plate at joint: 2.588-.625 2.588 =.75+ Percentage of strength of plate at joint Performing the operation in the ordinary way, we have: 2.588 Pitch of rivet holes. .625 Diameter of rivet hole. 2.588) 1.9630 (0.75+ Percentage of strength of plate 1 8116 15140 12940 at joint. PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Next, we have the percentage of strength of the rivets at joint: .625 X.625 X.7854X2X.80 2.588X.25 .75+ Percentage of strength of rivets at joint. Performing the operation in the ordinary way, we have .625 Diameter of rivet hole. Am't carried forward, .625 Diameter of rivet hole. 3125 1250 3750 .390625 .7854 1562500 1953125 3125000 2734375 .3067968750 Area of diameter of rivet hole. 152 A Library of Steam Engineering. Am't brought forward, .3067968750 2 Area of diameter of rivet hole. Number of rows of rivets. .6135937500 .80 .490875000000 Percentage of shearing strength of rivets. "Product No. 1." Then, multiplying the pitch of rivets by the thickness of the plate, we have: 2.588 Pitch of rivets. .25 Thickness of plate. 12940 5176 .64700 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: .64700).490875 (0.75+ 452900 Percentage of strength of rivets at joint. 379750 323500 It will be observed that the diameter of the rivet and rivet hole have been taken as being the same in the foregoing examples. The reason is that while the rivet before riveting is always smaller than the rivet hole, but after riveting, if the work is properly done, the rivet will fill the hole, and will therefore be equal in diameter; hence the strength of riveted joints is determined by the strength of the joints after completion. STRENGTH OF TRIPLE-RIVETED LAP JOINTS-STEEL PLATES AND STEEL RIVETS. Fig. 64 000 Riveted Joints. 153 TO DETERMINE THE PERCENTAGE OF STRENGTH OF TRIPLE-RIVETED LAP JOINTS-STEEL PLATES AND STEEL RIVETS. RULE.-First, subtract the diameter of rivet hole from the pitch, then divide the remainder by the pitch, and the quotient will give the percentage of strength of the plate at joint. Second, multiply the area of diameter of one rivet hole by the number of rows of rivets (3), then multiply the product by .80, and call the last product แ Product No. 1." Third, multiply the pitch of rivets from center to center by the thick- ness of material in the plate, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2" and the quotient will give the percentage of strength of the rivets at joint. PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 2.87 inches equal pitch of rivet holes. Let 75 one hundredths of an inch equal diameter of rivet hole. Then we have: 2.87-.75 2.87 .73+ Percentage of strength of plate at joint. Performing the operation in the ordinary way, we have: 2.87 Pitch of rivet holes. .75 Diameter of rivet hole. 2.87) 2.120 (0.73+ Percentage of strength of plate at 2009 1110 861 joint. PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Example.-Let 75 one hundredths of an inch equal diameter of rivet hole. Let 3 equal number of rows of rivets. Let .80 equal percentage of shearing strength of rivets as compared with the tensile strength of plate. Let 2.87 inches equal pitch of rivets from center to center. Let 5 tenths of an inch equal thickness of plate. Then we have: .75X.75 X.7854X3X.80 2.87X.5 73+ Percentage of strength of rivets at joint. 154 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: .75 Diameter of rivet hole. .75 Diameter of rivet hole. 375 525 .5625 .7854 22500 28125 45000 39375 .44178750 Area of diameter of rivet hole. 3 Number of rows of rivets. 1.32536250 .80 1.0602900000 Percentage of shearing strength of riv- ets as compared with tensile strength of plate. Product No. 1." Next, multiplying the pitch from center to center of rivets by the thickness of plate, we have: 2.87 Pitch of rivets. 5 Thickness of plate. 1.435 66 'Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 1.435) 1.06029 (.73+ Percentage of strength of rivets at joint. 1 0045 5579 4305 BUTT-STRAP JOINTS. Where single butt-straps are employed in butt joints the thickness of the strap must not be less than the thickness of the plate; where double butt-straps are employed the thickness of the straps must not be less than five-eighths of the thickness of the plate. Where single butt-straps are employed, the rule for determining the pitch for single, double, or triple riveting, and for determining the percentage of strength at the joints, is the same as for riveted lap joints. But for double butt-strapping the rule is different, on account of the rivets being in double shear instead of single shear, as in the case of lap joints or butt joints with single straps. In lap joints and single butt joints the rivets are subjected to being sheared through one cross section of the rivets, while in double butt-strapping the rivets are subjected to being sheared through two cross sections of the rivets. Hence, the difference between the rules for lap and single butt-strap joints and double butt-strap joints. Riveted Joints. 155 о о о 00 Fig. 65 SINGLE-RIVETED BUTT JOINT, WITH SINGLE BUTT-STRAP. PITCH OF RIVET HOLES FOR SINGLE-RIVETED DOUBLE BUTT-STRAP JOINTS- STEEL PLATES, STEEL BUTT-STRAPS AND STEEL RIVETS. о 19 gatbivib za 00 Fig. 66 SINGLE-RIVETED BUTT JOINT, WITH DOUBLE BUTT-STRAP. TO DETERMINE THE PITCH OF RIVET HOLES FOR SINGLE-RIVETED DOUBLE BUTT-STRAP JOINTS STEEL PLATES, STEEL BUTT-STRAPS AND STEEL RIVETS. RULE.-First, multiply the area of the diameter of rivet hole by 1.75, then multiply the product by .80, and call the product "Product No. 1." Second, divide "Product No. 1" by the thickness of plate, and add the diameter of rivet hole to the quotient, and the sum will give the required pitch in inches from center to center of rivets. Example.-Let 75 one hundredths of an inch equal diameter of rivet hole. Let 1.75 equal a constant for rivets in double shear. Let .80 equal percentage of shearing strength of rivets. Let 5 tenths of an inch equal thickness of plate. Then we have: .75X.75X.7854X1.75X.80 TASTE 5 +.75 1.987+ inches. Pitch of rivet holes from center to cen- ter. 156 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: .75 Diameter of rivet hole. .75 Diameter of rivet hole. 375 525 .5625 .7854 22500 28125 45000 39375 .44178750 Area of diameter of rivet hole. 1.75 A constant for rivets in double shear. 220893750 309251250 44178750 .7731281250 .80 Percentage of shearing strength of rivets. .618502500000 "Product No. 1." Next, dividing "Product No. 1" by the thickness of the material in the plate, and adding the diameter of rivet hole to the quotient, we have: .5) .6185 1.237 .75 Diameter of rivet hole. 1.987 inches. Pitch of rivet holes from center to center. PITCH OF RIVET HOLES FOR DOUBLE-RIVETED DOUBLE BUTT-STRAP JOINTS- STEEL PLATES, STEEL BUTT-STRAPS AND STEEL RIVETS. о о O O O о о 010 о Fig. 67 STAGGERED DOUBLE-RIVETED BUTT JOINT, WITH DOUBLE BUTT-STRAP. Riveted Joints. 157 о о о о о о о о о о о о 0000 Fig. 68 CHAIN DOUBLE-RIVETED BUTT JOINT, WITH DOUBLE BUTT-STRAP. TO DETERMINE THE PITCH OF RIVET HOLES FOR DOUBLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT- STRAPS AND STEEL RIVETS. RULE. First, multiply the area of diameter of rivet hole by the number of rows of rivet holes (2), then multiply the product by 1.75, on account of the rivets being in double shear; then multiply the last product by .80, and call the resulting product "Product No. 1.' Second, divide "Product No. 1" by the thickness of the plate and add the diameter of rivet hole to the quotient, and the sum will give the pitch of rivet holes from center to center in inches. Example.-Let 75 one hundredths of an inch equal diameter of rivet hole. Let 2 equal number of rows of rivets. Let 1.75 equal a constant for rivets in double shear! Let .80 equal percentage of shearing strength of rivets. Let 5 tenths of an inch equal thickness of plate. Then we have: .75X.75X.7854X2X1.75X.80 .5 +.75 3.22+ inches. Pitch of rivet holes from cen- ter to center. Performing the operation in the ordinary way, we have: .75 Diameter of rivet hole. .75 Diameter of rivet hole. 375 525 .5625 .7854 22500 28125 45000 Am't carried forward, .44178750 Area of diameter of rivet hole. 39375 158 A Library of Steam Engineering. Am't brought forward, .44178750 Area of diameter of rivet hole. 2 Number of rows of rivet holes. .88357500 1.75 Constant for rivets in double shear. 441787500 618502500 88357500 1.5462562500 .80 Percentage of shearing strength of rivets. 1.237005000000 "Product No. 1." Next, dividing "Product No. 1" by the thickness of material in the plate and adding the diameter of rivet hole to the quotient, we have: 5) 1.237005 2.47401 .75 Diameter of rivet hole. 3.22401 inches. Pitch of rivet holes from center to center. PITCH OF RIVET HOLES FOR TRIPLE-RIVETED DOUBLE BUTT-STRAP JOINTS- STEEL PLATE, STEEL BUTT-STRAPS AND STEEL RIVETS. 010 о о о о о О Fig. 69 TRIPLE-RIVETED BUTT-STRAP JOINT, WITH DOUBLE BUTT-STRAP. NOTE.-Frequently such joints have two rows of rivets, regularly spaced, and one row with double the space between the rivets from center to center; therefore, instead of using 3 as a mul- tiplier we use 2.5 to represent the rows of rivets where the third row is double spaced. TO DETERMINE THE PITCH OF RIVET HOLES FOR TRIPLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT- STRAPS AND STEEL RIVETS. RULE.-First, multiply the area of diameter of rivet hole by 2.5, then multiply the product by 1.75, on account of rivets being in double shear; then multiply the last product by .80, and call the resulting product "Product No. 1." Riveted Joints. 159 Second, divide "Product No. 1" by the thickness of the plate (not the straps), and add the diameter of rivet hole to the quotient, and the sum will give the pitch of rivet holes from center to center in inches. Example.-Let 1.3416 inches equal diameter of rivet hole. Let 2.5 equal number of rows of rivets. Let 1.75 equal a co-efficient for rivets in double shear. Let .80 equal percentage of shearing strength of rivets as compared with tensile strength of plate. Let 1.25 inches equal thickness of plate. Then we have: 1.3416×1.3416X.7854×2.5X1.75X.80 1.25 +1.3416-5.2997+inches. Required pitch of rivet holes from center to center. Performing the operation in the ordinary way, we have: 1.3416 Diameter of rivet hole. 1.3416 Diameter of rivet hole. 80496 13416 53664 40248 1 3416 1.79989056 .7854 719956224 899945280 1439912448 1 259923392 1.413634045824 Area of diameter of rivet hole. 2.5 Representing rows of rivets. 7068170229120 2 827268091648 3.5340851145600 1.75 A constant for rivets in double shear. 176704255728000 omg 2 47385958019200 3 5340851145600 6.184648950480000 .80 Percentage of shearing strength of rivets. 4.94771916038400000 "Product No. 1." 160 A Library of Steam Engineering. Next, dividing "Product No. 1" by the thickness of the plate, and adding the diameters of rivet hole to the quotient, we have: 1.3416 Diameter of rivet hole. 5.2997 inches. Required pitch of riv- 1.25) 4.947719 (3.9581 3 75 1 197 1 125 727 625 1021 1000 et holes from center to center. 219 125 The pitch, as shown in the example, is the required pitch for two rows of rivets, while the pitch in the third row is to be double that in the double rows. STRENGTH OF SINGLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT-STRAPS AND STEEL RIVETS. TO DETERMINE THE PERCENTAGE OF STRENGTH OF SINGLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT- STRAPS AND STEEL RIVETS. RULE. First, subtract the diameter of the rivet hole from the pitch, then divide the remainder by the pitch, and the quotient will give the percentage of strength of the plate at joint. Second, multiply the area of diameter of rivet hole by 1.75, on account of rivets being in double shear, then multiply the product by .80, and call the last product "Product No. 1." Third, multiply the pitch from center to center of rivets by the thickness of the plate, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2," and the quotient will give the percentage of strength of rivets at joint. PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 1.987 inches equal pitch of rivet holes from cen- ter to center. Let 75 one hundredths of an inch equal diameter of rivet hole. Then we have: 1.987-.75 .622+ Percentage of strength of plate at joint. 1.987 Riveted Joints. 161 Performing the operation in the ordinary way, we have: 1.987 Pitch of rivet holes. .75 Diameter of rivet hole. 1.987) 1.2370 (0.622+ Percentage of strength of plate 1 1922 at joint. 4480 3974 5060 3974 PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Example.-Let 75 one hundredths of an inch equal diameter of rivet. Let 1.75 equal a constant for rivets in double shear. Let .80 equal percentage of shearing strength of rivets. Let 5 tenths of an inch equal thickness of plate. Then we have: .75X.75X.7854X1.75X.80 1.987 X.5 .622+ Percentage of strength of rivets at joints.. Performing the operation in the ordinary way, we have: .75 Diameter of rivet hole. .75 Diameter of rivet hole. 375 525 .5625 .7854 22500 28125 45000 39375 .44178750 Area of diameter of rivet hole. 1.75 Constant for rivets in double shear. 220893750 309251250 44178750 .7731281250 .80 Percentage of shearing strength of .618502500000 rivets. 6. Product No. 1." 11 162 A Library of Steam Engineering. Next, multiplying the pitch of rivets from center to center by the thickness of the plate, we have: 1.987 Pitch of rivets. 5 Thickness of plate. .9935 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: .9935).6185025 (.622+ Percentage of strength of rivets 59610 22402 19870 25325 19870 at joint. STRENGTH OF DOUBLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT-STRAPS AND STEEL RIVETS. TO DETERMINE THE PERCENTAGE OF STRENGTH OF DOUBLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT- STRAPS AND STEEL RIVETS. RULE. First, subtract the diameter of rivet hole from the pitch, then divide the remainder by the pitch, and the quotient will give the percentage of strength of plate at joint. Second, multiply the area of diameter of rivet hole by the number of rows of rivets (2); then multiply the product by 1.75, on account of rivets being in double shear; then multiply the last product by .80, and call the resulting product "Product No. 1." Third, multiply the pitch of the rivets from center to center by the thickness of the plate, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2" and the quotient will give the percentage of strength of the rivets at joint. PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 3.22 inches equal pitch of rivet holes. Let 75 one hundredths of an inch equal diameter of rivet hole. Then we have: 3.22-.75 3.22 .76+ Percentage of strength of plate at joint. Performing the operation in the ordinary way, we have: 3.22 Pitch of rivet holes. .75 Diameter of rivet hole. 3.22) 2.470 (0.76+ Percentage of strength of plate 2 254 2160 at joint. 1932 Riveted Joints. 163 PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Example.-Let 75 one hundredths of an inch equal diameter of rivet hole. Let 2 equal number of rows of rivets. Let 1.75 equal a constant for rivets in double shear. Let .80 equal percentage of shearing strength of rivets. Let 3.22 inches equal pitch of rivets from center to center. Let 5 tenths of an inch equal thickness of plate. Then we have: .75X.75X.7854X2X1.75X.80 3.22X.5 .76+ Percentage of strength of rivets at joint. Performing the operation in the ordinary way, we have: .75 Diameter of rivet hole. .75 Diameter of rivet hole. 375 525 .5625 7854 22500 28125 45000 39375 .44178750 Area of diameter of rivet hole. 2 .88357500 1.75 Constant for rivets in double shear. 441787500 618502500 88357500 1.5462562500 .80 Percentage of shearing strength of rivets. 1.237005000000"Product No. 1." Next, multiplying the pitch of rivets from center to center by the thickness of plate, we have: 3.22 Pitch of rivets. 5 Thickness of plate. 1.610 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 1.610) 1.237005 (.76+ 1 1270 11005 Percentage of strength of rivets at joint. 9660 164 A Library of Steam Engineering. STRENGTH OF TRIPLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT-STRAPS AND STEEL RIVETS. TO DETERMINE THE PERCENTAGE OF STRENGTH OF TRIPLE-RIVETED DOUBLE BUTT-STRAP JOINTS-STEEL PLATES, STEEL BUTT-STRAPS AND STEEL RIVETS. RULE.-First, subtract the diameter of rivet hole from the pitch, and divide the remainder by the pitch, the quotient will give the percentage of strength of plate at joint. Second, multiply the area of diameter of rivet hole by 2.5, then multiply the product by 1.75, and multiply the last product by .80, and call the resulting product "Product No. 1." Third, multiply the pitch of rivets from center to center in inches, by the thickness of the plate in decimals of an inch, and call the product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2," and the quotient will give the percentage of strength of rivets at joint. PERCENTAGE OF STRENGTH OF PLATE AT JOINT. Example.-Let 5.2997 inches equal pitch of rivet holes from center to center. Let 1.3416 inches equal diameter of rivet hole. Then we have: 5.2997-1.3416 5.2997 .74+ Percentage of strength of plate at joint. Performing the operation in the ordinary way, we have: 5.2997 Pitch of rivet holes. 1.3416 Diameter of rivet hole. 5.2997) 3.95810 (0.74+ Percentage of strength of plate 3 70979 248310 211988 at joint. PERCENTAGE OF STRENGTH OF RIVETS AT JOINT. Example.-Let 1.3416 inches equal diameter of rivet hole. Let 2.5 equal number of rows of rivets. Let 1.75 equal a co-efficient for rivets in double shear. Let .80 equal percentage of shearing strength of rivets as compared with tensile strength of plate. Let 5.2997 inches equal pitch of rivets from center to center. Let 1.25 inches equal thickness of plate. Riveted Joints. 165 Then we have: 1.3416×1.3416X.7854X2.5X1.75X.80 5.2997×1.25 .74+ Percentage of strength of rivets at joint. Performing the operation in the ordinary way, we have: 1.3416 Diameter of rivet hole. 1.3416 Diameter of rivet hol. 80496 13416 53664 40248 1 3416 1.79989056 .7854 719956224 899945280 1439912448 1 259923392 1.413634045824 Area of diameter of rivet hole. 2.5 Number of rows of rivets. 7068170229120 2 827268091648 3.5340851145600 1.75 A co-efficient for rivets in 176704255728000 2 47385958019200 3 5340851145600 6.184648950480000 .80 double shear. Percentage of shearing strength of rivets to ten- sile strengh of plate. 4.94771916038400000 "Product No. 1." Next, multiplying the pitch by the thickness of plate, we have: 5.2997 inches. Pitch. 1.25 inches. Thickness of plate. 264985 105994 52997 6.624625 "Product No. 2." Next, dividing "Product No. 1" by Product No. 2," we have: 6.624625) 4.94771916 (.74+ Percentage of strength of 4 6372375 31048166 26498500 rivets at joint. 166 A Library of Steam Engineering. TABLE OF RIVETED JOINTS. IRON PLATES, IRON RIVETS, AND HOLES DRILLED. SINGLE RIVETED JOINTS. DOUBLE RIVETED JOINTS. Percent'ge of Strength of Rivets at the Joint. Percent'ge of Strength of Plate at the Joint. Breadth of Lap for Zigzag 77.9 77.9 75.8 75.8 74.2 74.2 72.9 72.9 71.8 71.8 70.9 70.9 70.2 70.2 69.5 69.5 68.9 68.9 22 ,, ,, د, ,, 1001000/000/000/000000/000/00 Riveting. L Breadth of*** Lap for Chain Riveting. Distance be- tween Centers of Rows of Rivet Holes for Zig- zag Riveting. Distance be- tween Centers of Rows of Rivet Holes for Chain Riveting. Pitch of Rivet Holes from Center to Center. Distance of Center of Outer Row of Rivet Holes from Edge of Sheet. Diameter of Rivet Holes. Thickness of Plate. Percent'ge of Strength of Rivets at the Joint. Percent'ge of Strength of Plate at the Joint. Breadth of Lap. Pitch of Rivet Holes from Center to Center. Distance of Center of Rivet Holes from Edge of Sheet. Diameter of Rivet Holes. 333 د, 22 11% 19% 116 1 19 د, "" 22 ,, "" 25" 2 22 د, I ,, د, 11 66 1 132 11% i 10 11920 ,, د, وو 1.8521 in. 1 in. 1.8754 " 1.9281 22 1.9976 " د, 2.0776 2.1646 " 2.2566" 2.3521 " 2.4503 "" 1222 00000 22 "2 ,, 213" co "" ,, ,, 22 ,, 22 66.2 66.2 63.3 63.3 61.1 61.1 59.3 59.3 57.8 57.8 56.6 56.6 55.6 55.6 54.8 66 54.8 54.0 54.0 "" 22 1 ,, ,, 1256 132 1ㅎ ​132 116 1 1/ د, " 2.5505 in. 1- 2.5885 " 2.6674" 2.7696" 2.8864 " 3.0130 " 3.1464 3.2848 " 3.4268 1 2 in. STEEL PLATES, STEEL RIVETS, AND HOLES DRILLED. 3300 SINGLE RIVETED JOINTS. DOUBLE RIVETED JOINTS. د, ,, 22 ,, د, د, 29 22 دو 66 116 1 / Percentage of Strength of Rivets at the Joint. Percentage of Strength of Plate at the Joint. Breadth of Lap for Zigzag Riveting. Breadth of Lap for Chain in. 75.8 73.4 71.5 71.5 70.0 70.0 22 68.7 68.7 67.6 67.6 66.7 66.7 66.0 66.0 65.3 65.3 Riveting.co Distance be- tween Centers of Rows of Rivet Holes for Zig- zag Riveting. Distance be- tween Centers of Rows of Rivet Holes for Chain Riveting. Pitch of Rivet Holes from Center to Center. Distance of Center of Outer Row of Rivet Holes from Edge of Sheet. Diameter of Rivet Holes. Thickness of Plate. Percentage of Strength of Rivets at the Joint. Percentage of Strength of Plate at the Joint. Breadth of Lap. Pitch of Rivet Holes from Center to Center. Distance of Center of Rivet Holes from Edge of Sheet. Diameter of Rivet Holes. in. 3 29 13 " I 1 " 416 42 د, وو د, 1,9% " ,, د, 9 22 22 22 19 11% 12 دو ,, 29 18014380000000 HALO22222 in. 2.5885 in. 17 99 دو ,, د, ,, 2.5881 " 2.6349 " 2.7084" 2.7992 " 2.9010" 3.0106 ,, 3.1259 " 3.2455 " 1 / 39 132 116 1/3/20 33 1 1/ 133 ,, ,, 60.1 57.6 in. 22 ,, 63.3 63.3 60.1 57.6 23 ,, 55.6 55.6 54.8 54.8 د, 52.7 52.7 51.6 51.6 50.6 50.6 49.8 49.8 22 ,, 22 22 1.8754 in. 1.8805 " 1.9180" 1.9745" 2.0419 " 2.1170 " 2.1974 دو 2.2816" 2.3688 "I د, د, 22 11% 1332 22 د, " I 66 T ,, 112 1 T ,, Thickness of Plate. Thickness of Plate. Boiler Flues. 167 CHAPTER VII. BOILER FLUES. There is no question connected with the science of steam engi- neering so little understood by engineers generally, as that of boiler flues. Until the experiments of Sir WILLIAM FAIRBAIRN demonstrated that a flue, 5 feet in length, of a given diameter and given thickness of material, would stand just double the pressure per square inch to collapse it that a flue 10 feet in length, of the same given diameter and same given thickness of material, would stand. Hence, in making cal- culations in regard to the safe-working or collapsing pressure of boiler flues, the length of the flue must be taken into consideration as well as the thickness of its material and its diameter. Therefore, as the length of the flue is increased the flue becomes proportionately weaker. In this respect it differs very materially from the shell of a boiler with its internal pressure. The power of the flue's resistance to collapse depends, in a great measure, upon the circularity of its form; for the reason that as the flue deviates in form from that of a perfect circle it becomes proportionately weak and less able to resist collapse; hence, the importance of making all boiler flues to conform as near as possible to that of a perfect circle. Carelessness in these matters has caused many serious accidents and the destruction of many human lives. The practice of making large boiler flues with the same thickness of material as that employed in making the shells of the boilers, has been another fruitful source of a large number of boiler accidents. And yet, the lack of information upon this subject among engineers and boiler makers is not to be wondered at, when the United States government itself had taken no notice of the length of boiler flues 16 inches and less in diameter, nor even the pressure of steam in that direction, until a few years ago. The only rule upon the subject, so far as flues 16 inches and less in diameter were concerned, was, "that all flues having a diameter of 16 inches shall have a thick- ness of of an inch, and in proportion for greater or less diameters." And this was without any regard to the length of the flue or to the pressure the flue was required to carry. To the credit of the govern- ment, however, it can be said that that error has been corrected, and most excellent rules governing such flues have been adopted. 16 168 A Library of Steam Engineering. Many rules relating to boiler flues have been promulgated by different authorities, and many of those rules are based entirely upon theory. It is, therefore, not always safe to follow such rules. The best teacher in all such matters is practice. Therefore, all of the rules that will be given here are such as are based upon actual experience and have been found safe in practice. It is an easy matter to make a flue that will be perfectly safe; but aside from the question of safety, the question of efficiency in the generation of steam must not be lost sight of. Therefore, while the flue may be made heavy enough to withstand any given pressure, yet its efficiency may be entirely destroyed. It is well known that thin material is a better conductor of heat than thick material. And here again is the danger of losing sight of the important question of safety. The student is, therefore, informed that the rules here laid down, are laid down with two important objects in view-safety and efficiency. And the conclusions herein reached are based entirely upon practical experience, so that the two things most essential-safety and efficiency-are kept constantly in view. SAFE-WORKING PRESSURE OF FLUES. TO DETERMINE THE SAFE-WORKING PRESSURE OF LAP-WELDED BOILER FLUES, NOT MADE IN SECTIONS. RULE. First, square the thickness of material in hundredths of an inch, and then multiply the product by the constant whole number 806300, and call the last product "Product No. 1. Second, multiply the diameter of the flue in inches by the length of the flue in feet, and then multiply the product by 3, and call the last product "Product No. 2." Third, divide "Product No. 1" by "Product No.2," and the quotient will give the safe-working pressure for a flue made in one length. Example. Let 35 one hundredths of an inch equal thickness of material. Let 806300 equal a constant. Let 16 inches equal diameter of the flue. Let 25 feet equal length of the flue. Let 3 equal a constant. Then we have: .35 X.35X806300 -82.31-lbs. 16×25×3 Safe-working pressure per square inch. Boiler Flues. 169 Performing the operation in the ordinary way, we have: .35X.35 .1225 The square of the thickness of material. 806300 A constant. 36 7500 735 0 9800 98771.7500 "Product No. 1." 16 Diameter of the flue in inches. 25 Length of the flue in feet. 80 32 400 3 A constant. 1200 "Product No. 2. Finally, dividing "Product No. 1" by "Product No. 2." we have: 1200) 98771.75 (82.31- lbs. 9600 2771 2400 3717 3600 1175 1200 THICKNESS OF MATERIAL OF FLUES. TO DETERMINE THE THICKNESS OF MATERIAL REQUIRED FOR ANY GIVEN PRESSURE, GIVEN DIAMETER AND GIVEN LENGTH OF FLUE, NOT MADE IN SECTIONS. RULE.-First, multiply the diameter of the flue in inches, by the length of the flue in feet, then multiply the product by the constant 3, and call the last product "Product No. 1." Second, multiply "Product No. 1" by the given pressure per square inch, and call the product "Product No. 2." Third, divide "Product No. 2" by the constant whole number 806300, and extract the square root of the quotient, the answer will give the thickness of material required in decimals of an inch. 170 A Library of Steam Engineering. Example.-Let 16 inches equal diameter of the flue. Let 25 feet equal length of the flue. Let 3 equal a constant. Let 82.31 pounds equal given pressure per square inch. Let 806300 equal a constant. Then we have: J16 X 16X25X3X82.31 806300 .35 Decimals of an inch. Thickness of mate- rial required. Performing the operation in the ordinary way, we have: 16 Diameter of flue. 25 Length of flue. 80 32 400 3 A constant. 66 1200 Product No. 1." 82.31 Pressure per square inch. 12 00 360 0 2400 9600 98772.00 "Product No. 2." Dividing "Product No. 2" by 806300, we have: 806300) 98772.00 (0.1225+ Square of thickness of material. 80630 0 1814200 1612600 2016000 1612600 4034000 4031500 Extracting the square root of the quotient, we have: 1225 (.35 Thickness of required material in 9 65) 325 hundredths of an inch. 325 Boiler Flues. 171 DIAMETER OF FLUES. TO DETERMINE THE DIAMETER REQUIRED FOR A FLUE NOT MADE IN SECTIONS, OF A GIVEN THICKNESS OF MATERIAL AND GIVEN LENGTH, TO CARRY A GIVEN SAFE-WORKING PRESS- URE PER SQUARE INCH. RULE.-First, multiply the square of the thickness of material in hun- dredths of an inch, by the constant whole number 806300, and call the product "Product No. 1." Second, multiply the length of the flue in feet, by the constant 3, and then multiply the product by the given safe-working pressure per square inch, and call the last product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter of the flue in inches. Example.-Let 35 one hundredths of an inch equal thickness of material. Let 806300 equal a constant. Let 25 feet equal length of flue. Let 3 equal a constant. Let 82.31 pounds per square inch equal given pressure.. Then we have: .35X.35X806300 25X3X82.31 16- inches. Diameter required. Performing the operation in the ordinary way, we have: .35 Thickness of material. .35 Thickness of material. 175 105 .1225 Next we have: Square of the thickness of material. 806300 A constant. 36 7500 735 0 9800 98771.7500 'Product No. 1." 25 Length of the flue. 3 A constant. 75 82.31 Given pressure per square inch. 411 55 5761 7 6173.25 "Product No. 2." 172 A Library of Steam Engineering. Then we have, "Product No. 1" divided by "Product No. 2:" 6173.25) 98771.75 (16-inches. The required diameter. 617325 37039 25 3703950 LENGTH OF FLUES. TO DETERMINE THE LENGTH REQUIRED FOR A FLUE NOT MADE IN SECTIONS, OF A GIVEN THICKNESS OF MATERIAL AND GIVEN DIAMETER, TO CARRY A REQUIRED WORKING PRESS- URE PER SQUARE INCH. RULE.-First, multiply the square of the given thickness of material in hundredths of an inch by the constant whole number 806300, and call the product "Product No. 1." Second, multiply the given diameter of the flue in inches by the constant 3, and then multiply the product by the given pressure per square inch, and call the last product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2" and the quotient will give the required length of the flue in feet. Example.-Let 35 one hundredths of an inch equal thickness of material. Let 806300 equal a constant. Let 16 inches equal diameter of the flue. Let 3 equal a constant. Let 82.31 pounds per square inch equal working pressure. Then we have: .35X.35X806300 -25- feet. Required length of the flue. 16×3×82.31 Next we have: .35 Thickness of material. .35 Thickness of material. 175 105 .1225 Square of thickness of material. 806300 A constant. 36 7500 735 0 98771.7500 "Product No. 1." 9800 Boiler Flues. 173 Performing the operation in the ordinary way, we have: 16 Diameter of flue. 3 A constant. 48 82.31 658 48 3292 4 3950.88 "Product No. 2." Then we have, "Product No. 1" divided by "Product No. 2:" 3950.88) 98771.75 (25- feet. Required length of the flue.. 79017 6 19754 15 19754 40 COLLAPSING PRESSURE OF FLUES. TO DETERMINE THE COLLAPSING PRESSURE OF FLUES NOT MADE IN SECTIONS. RULE.-First, multiply the square of the thickness of material in the flue, in hundredths of an inch, by the constant whole number 806300, and call the product "Product No. 1." Second, multiply the diameter of the flue, in inches by the length of the flue, in feet, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the collapsing pressure per square inch. Example.-Let 35 one hundredths of an inch equal thickness of material. Let 806300 equal a constant. Let 16 inches equal diameter of the flue. Let 25 feet equal length of the flue. Then we have: .35X.35 X 806300 16×25 -446.92+ lbs. Collapsing pressure per square inch. Performing the operation in the ordinary way, we have: .35 Thickness of material. .35 Thickness of material. 175 105 Am't carried forward, .1225 Square of thickness of material.. 174 A Library of Steam Engineering. Am't brought forward, Next we have: .1225 Square of thickness of material. 806300 A constant. 36 7500 735 0 9800 98771.7500 16 Diameter of flue. 25 Length of flue. 80 32 66 400 Product No. 2." Then we have, "Product No. 2" divided by "Product No. 1:" 400) 98771.75 (246.92+ lbs. Pressure per square inch required to collapse the flue. 800 1877 1600 2771 2400 3717 3600 1175 800 Much S Pumps. 175 CHAPTER VIII. PUMPS. HORSE POWER REQUIRED TO ELEVATE WATER. TO DETERMINE THE HORSE POWER REQUIRED TO ELEVATE A GIVEN QUANTITY OF WATER, TO A GIVEN HEIGHT, IN A GIVEN TIME. RULE.-First, multiply the area of cross section of the piston or plunger of the pump, in square inches, by the height, in feet, required to elevate the water, and divide the product by the constant 2.304, and the quotient will give the total weight of the column of water on the piston or plunger in pounds. Second, multiply the total weight of the column of water on the piston or plunger, in pounds, by the number of feet per minute the column of water is to move, and divide the product by 33000, and the quotient will give the horse power required to elevate the water, without including friction. Example.-Let 16 inches equal diameter of the piston or plunger. Let .7854 equal a constant. Let 100 feet equal perpendicular height of column of water. Let 2.304 equal a constant. Let 150 feet equal speed of column of water per minute. Let 33000 equal a constant. Then we have: [(16×16×.7854X100)÷2.304 ]X150 33000 Performing the operation, we have: =39.66+ Horse power required. 16 Diameter of piston or plunger. 16 Diameter of piston or plunger. 96 16 256 Square of diameter of piston or plunger. .7854 4 7124 39 270 157 08 201.0624 Area of piston or plunger. 176 A Library of Steam Engineering. Then, according to the rule, we multiply the area of the piston or plunger of the pump by the height of the column of water in feet, and divide the product by 2.304. Thus: 201.0624 Area of piston or plunger. 100 Height of column of water in feet. Total weight of water on piston or plunger. 2.3040) 20106.2400 (8726+ lbs. 18432 0 1674 24 1612 80 61 440 46 080 15 3600 13 8240 Next, proceeding according to the rule, we multiply the total weight of column of water on piston or plunger by the speed per min- ute the column is required to travel, and then divide the product by 33000. Thus : 8726 Total weight of column of water on piston or plunger. 150 436300 8726 Speed of column of water in feet per min- ute. 33000) 1308900 (39.66+ Horse power required. 99000 318900 297000 219000 198000 210000 198000 NOTE.-A pound of water one inch square will make a column 2.304 feet high. FEED PUMPS. Section 4418 of the Revised Statutes of the United States, requires that in the equipment of steam vessels, inspectors shall see "that ade- quate and certain provision is made for an ample supply of water to feed the boilers at all times, whether such vessel is in motion or not." The only way to provide an adequate supply of feed water with- out making the feed pumps unnecessarily large, is to regulate the size by the evaporative capacity of the boilers under the most favorable circumstances, which, in cylindrical two-flue marine boilers, has been Pumps. 177 found to be seven pounds of water per pound of coal, and in cylin- drical tubular land boilers ten pounds of water per pound of coal, as the average maximum evaporation in practice In the former the aver- age maximum amount of coal consumed per square foot of grate sur- face is found to be twenty pounds, while in land boilers the consump- tion of coal varies from 12 to 40 pounds per square foot of grate surface. All of which proves conclusively that the capacity of the feed pump should be based upon the evaporative capacity of boilers. As Section 12 of Rule II of the Rules and Regulations of the United States Board of Supervising Inspectors, provides that the feed water for high pressure boilers shall in no case be less than 180° Fahrenheit, nor less than 100° Fahrenheit for low pressure boilers, it will be safe to assume an evaporation of seven pounds of water per pound of coal for marine boilers, and a consumption of 20 pounds of coal per square foot of grate surface per hour, which determines the capacity of feed pump required. DIMENSIONS OF FEED PUMPS. TO COMPUTE THE DIMENSIONS OF FEED PUMPS. RULE. First, multiply the number of pounds of coal consumed per square foot of grate surface per hour, by the number of pounds of water evap- orated per pound of coal; then multiply the product by the number of boilers; then multiply that product by 27.64, the number of cubic inches in one pound of water, and the answer will give the number of cubic inches required for evaporation; then multiply the number of cubic inches required for evapora- tion by 1.12 (which is adding twelve per cent. for leakage, etc.), and call the last product "Product No. 1." Second, multiply the number of displacement strokes of the feed pump by 60, or if more than one pump, multiply the total number of displacement strokes of all the pumps by 60; then multiply the product by the length of stroke in inches; then multiply the last product by .7854, and call the resulting product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and extract the square root of the quotient, and the answer will give the required diameter of the plunger or piston of feed pumps. Example. Let 20 pounds equal amount of coal consumed per square foot of grate surface per hour. Let 7 pounds equal amount of water evaporated per pound of coal. Let 25 square feet equal grate surface for each boiler. Let 4 equal number of boilers. Let 27.64 cubic inches of water equal one pound. 12 178 A Library of Steam Engineering. Let 1.12 equal a constant (used as a multiplier is equiv- alent to adding 12 per cent). Let 60 equal a constant. Let 2 equal number of single-acting pumps. Let 20 equal number of displacement strokes for each pump per minute. Let 12 inches equal length of stroke of each Let .7854 equal a constant. pump. Then we have: 20X7X25X4X27.64X1.12 60×2×20×12X.7854 -4.37+ inches. Required diameter of plunger of each feed pump. Performing the operation, we have: 20 7 Pounds of coal consumed per square foot of grate surface per hour. Pounds of water evaporated per pound of coal. 140 25 Square feet of grate surface for each boiler. 700 280 3500 4 Number of boilers. 14000 Pounds of water evaporated per hour in 27.64 560 00 four boilers. Number of cubic inches in one pound of water. 8400 0 98000 28000 386960.00 7739 2000 38696 000 386960 00 1.12 Number of cubic inches of water evapo rated per hour. Multiplying by unity with per cent. an- nexed is equivalent to adding the per cent. 433395.2000 "Product No. 1." Performing the operation below the line in the above example, we have: 60 Number of minutes in one hour. 2 Number of pumps. 120 20 Number of displacement strokes per minute for each pump. Am't carried forward, 2400 Pumps. 179 Am't brought forward, 2400 12 Length of stroke of each pump. 4800 2400 28800 .7854 A constant. 11 5200 144 000 2304 00 20160 0 22619.5200 "Product No. 2." Next, dividing "Product No. 1" by "Product No. 2," we have: 22619.52) 433395.20 (19.16+ Square of diameter of plunger. 226195 2 207200 00 203575 68 3624 320 2261 952. 1362 3680 1357 1712 Finally, extracting the square root of the quotient, the square of the required diameter of the plunger, we have: 19.16 (4.37+ inches. Required diameter of plunger 16 83) 316 249 867) 6700 6069 of feed pumps. That the student may have a better understanding of the subject, the rule relating to feed pumps will be given in detail, and in a more simplified form. And for the purpose of proving the correctness of the rule already given, and for convenient illustration, the dimensions used in the previous examples will be again taken. RULE.-First, multiply the number of pounds of coal consumed per square foot of grate surface per hour by the number of pounds of water evap- orated per pound of coal, and call the product "Product No. 1." 180 A Library of Steam Engineering. Second, multiply the number of square feet of grate surface in one boiler, by the number of similar boilers to be supplied with the required feed pump, and call the product "Product No. 2." Third, multiply "Product No. 1," the number of pounds of water evap- orated per one square foot of grate surface per hour, by "Product No. 2," the number of square feet of grate surface in the four boilers, and the product will give the total amount of water required for evaporation, and we will call this product "Product No. 3." Fourth, as the pumps may not fill quite full during each stroke, and as there may be some leakage in pipes and boilers, and more or less water may pass off with the steam drawn from the boilers, we add 12 per cent. to the total amount of water required for evaporation; we then multiply "Product No. 3" by 1.12 and the product will give the total number of pounds of water required for evaporation with 12 per cent. added, and we will call this product "Product No. 4." Fifth, we now reduce the number of pounds of water required to cubic inches by multiplying "Product No. 4" by 27.64, the number of cubic inches contained in one pound of water, and the product will give the total amount of water in cubic inches required per hour, and we will call the product "Product No. 5." Sixth, divide "Product No. 5," the total amount of water in cubic inches required per hour by 60, and the quotient will give the total number of cubic inches of water required per minute, and we will call the quotient "Quotient No. 1." Seventh, divide "Quotient No. 1," the number of cubic inches of water required per minute, by the number of displacement strokes per minute of all of the pumps combined (2 pumps, 20 displacement strokes each, equal 40), and the quotient will give the number of cubic inches of water required for each stroke, and this quotient we will call "Quotient No. 2." Eighth, divide "Quotient No. 2," the number of cubic inches of water required for each stroke of the pump, by the length of the stroke in inches, and the quotient will give the number of cubic inches in the area of one inch in the length of the plunger, and call this quotient "Quotient No. 3." Ninth, divide "Quotient No. 3," the area of one inch in the length of the plunger, by .7854, and the quotient will give the square of the required diam- eter of the plunger of each pump, and call the quotient "Quotient No. 4." Tenth, extract the square root of "Quotient No. 4," the square of the diameter of pump plungers, and the answer will give the required diameter of each plunger. Pumps. 181 Example.-First, multiplying the number of pounds of coal (20) consumed per square foot of grate surface per hour, by the number of pounds (7) of water evaporated per pound of coal, we have: 20 7 140 "Product No. 1." Number of pounds of water evaporated per square foot of grate surface per hour. Second, multiplying the number of square feet of grate surface in one boiler (25) by the number of boilers (4), we have: 25 4 100 "Product No. 2." Square feet of grate sur- face in four boilers. Third, multiplying "Product No. 1" by "Product No. 2," we have: 140 100 14000 "Product No. 3." Pounds of water re- quired per hour for evaporation. Fourth, adding 12 per cent. to the amount of water required for evaporation, on account of leakage and pumps not filling quite full during each stroke, we multiply by unity with the per cent. annexed, which is equivalent to adding the per cent., and we have: 14000 1.12 280 00 1400 0 14000 15680.00 "Product No. 4." Total number of pounds of feed water required per hour. Fifth, reducing the number of pounds of feed water required to cubic inches by multiplying "Product No. 4" by 27.64, the number of cubic inches in a pound of water, we have: 15680 27.64 627 20 9408 O 109760 31360 433395.20 "Product No. 5." Total number of cu- bic inches of water required per hour. 182 A Library of Steam Engineering. Sixth, dividing "Product No. 5," the total number of cubic inches of water required per hour, by 60, we have: 60) 433395.20 (7223.25+ 420 133 120 139 120 195 180 152 120 "Quotient No. 1." Number of cubic inches of water re- quired per minute. 320 300 Seventh, dividing "Quotient No. 1," the number of cubic inches. of water required per minute, by 40 (20 for each pump), the number of displacement strokes of the pumps per minute, and we have: 40) 7223.25 (180.58+ 40 322 320 232 200 325 320 66 Quotient No. 2." Number of cu- bic inches of water required for each stroke of the pumps. Eighth, dividing "Quotient No. 2," the number of cubic inches of water required for each displacement stroke of the pumps, by 12 inches (the length of the stroke), we have: ber "Quotient No. 3." Number of cubic inches in the area of one inch in the length of plunger. 12) 180.58 (15.0483+ 12 60 60 488 58 48 100 96 40 36 Pumps. 183 Ninth, dividing "Quotient No. 3," the number of cubic inches in the area of one inch in the length of the plunger, by .7854, we have: .7854) 15.0483 (19.16+ 7 854 71943 ABO 7 0686 66 Quotient No. 4." The square of the required diameter of plunger. 12570 7854 47160 47124 Tenth, extracting the square root of "Quotient No. 4," the square of the required diameter of the plunger, we have: 19.16 (4.37+ inches. Required diameter of plunger. 16 83) 316 249 867) 6700 6069 MEDTEC OT OETUSE i 184 A Library of Steam Engineering. CHAPTER IX. SMOKE STACK PROPORTIONS. AREA OF STACK. TO DETERMINE THE AREA OF REQUIRED STACK. RULE.-Multiply the area of cross section of tubes or flues in the boiler by the constant 1.3, and the product will give the required area. Example.-Let 4 inches equal diameter of tubes. Let .7854 equal a constant. Let 40 equal number of tubes. Let 1.3 equal a constant. Then we have: 4X4X.7854x40x1.3-653.4528 square inches. Required area of stack. Performing the operation in the ordinary way, we have: 44 16 .7854 4 7124 7 854 12.5664 40 502.6560 1.3 150 79680 502 6560 653.45280 square inches. Required area of stack. DIAMETER OF SQUARE STACK. TO DETERMINE THE DIAMETER OF A SQUARE STACK. RULE.-Extract the square root of the required area in square inches of stack, and the answer will give the diameter of a square stack. Example.-Let 653.4528 square inches equal required area of stack. Then we have: a square stack. 653.4528-25.56+ inches. Required diameter of Smoke Stack Proportions. 185 Performing the operation, we have: 653.4528(25.56+ inches. Required diameter of 4 45) 253 225 505) 2845 2525 5106) 32028 30636 square stack. DIAMETER OF ROUND STACK. TO DETERMINE THE DIAMETER OF A ROUND STACK. RULE.-Divide the required area of the stack by .7854 and extract the square root of the quotient, the answer will give the diameter of a round stack. Example.-Let 653.4528 square inches equal required area of stack. Let .7854 equal a constant. Then we have: 653.4528 -28.84 inches. Required diameter of round stack .7854 Performing the operation, we have: .7854) 653.4528 (832 Square of required diameter. 628 32 25 132 23 562 15708 15708 Then, extracting the square root of the quotient (the square root of the required diameter), we have: 832 (28.84+ inches. Required diameter of 4 48) 432 round stack. 384 апоимнас 568) 4800 anotamos ens 4544 5764) 25600 23056 186 A Library of Steam Engineering. PERCENTAGE OF AIR SPACE FOR GRATE BARS. The required height of the stack depends, in a very large measure, upon conditions: First, no grate bars should be allowed in any boiler furnace burn- ing bituminous coal, with less than 60 per cent. of air space. All calculations based upon grate surface can not but be erroneous unless the air space is taken into account. Therefore, grate bars with con- tracted air space require a higher stack than grate bars with ample air space. Second, much depends upon the fuel used; wood requires less force of draft than lump coal, and lump coal less than nut, and nut less than slack. When we speak of 60 per cent. of air space, the student must understand that the air space is not governed by the width of the face of the bars, but the thickness of metal in the bars must be governed by the width of the air space required; therefore, if of an inch be required for air space, the thickness of metal in the bars must not be more than 42 one hundredths of an inch. PROPORTION AND ARRANGEMENT OF BREECHING. The breeching should contain an area of cross section of 15 per cent. larger than the area of cross section of the tubes or flues, but no larger, and in no case should the area be less than the area of cross section of the flues. In the construction of the breeching care should in all cases be taken to prevent the current of the gases of one boiler cutting across the current of the gases from any other boiler in a battery. The uptake for each boiler should be so constructed that the gases from the boilers will not come in contact with each other until they have been turned in the same direction toward the stack, and all sharp angles should be avoided. FORCE OF DRAUGHT REQUIRED. As the height of the stack should be governed by the force of draught required, and as practice has demonstrated that inch of water, as shown by a water-gauge, is required for bituminous lump coal to burn it to advantage, and of an inch for nut, not less than 1 of an inch should be provided for slack. HEIGHT OF STACK-BITUMINOUS LUMP COAL. TO DETERMINE THE HEIGHT OF STACK FOR BITUMINOUS LUMP COAL. RULE.-Divide the draught required in inches of water by the constant .0073, and the quotient will give the required height of the stack in feet. Smoke Stack Proportions. 187 Example.-Let 5 tenths of an inch of water equal force of required draught for lump coal. Let .0073 equal a constant. Then we have: .0073).5000 (68.5 feet. Required height of stack for bituminous lump coal. 438 620 584 360 365 HEIGHT OF STACK-BITUMINOUS NUT COAL. TO DETERMINE THE HEIGHT OF STACK FOR BITUMINOUS NUT COAL. Example.-Let 75 one hundredths of an inch of water equal force of required draught for nut coal. Let .0073 equal a constant. Then we have: .0073) .7500 (102.74- feet. Required height of stack for bituminous nut coal. 73 200 146 540 511 290 292 HEIGHT OF STACK-BITUMINOUS SLACK COAL. TO DETERMINE THE HEIGHT OF STACK FOR BITUMINOUS SLACK COAL. Example.-Let 1.125 inches of water equal force of required draught for slack coal. Let .0073 equal a constant. Then we have: .0073) 1.1250 (154+ feet. Required height of stack for bituminous slack coal. 73 395 365 300 292 188 A Library of Steam Engineering. CHAPTER X. SPECIFICATIONS AND PLANS FOR TUBULAR BOILERS. SPECIFICATIONS FOR BOILERS. There will be two horizontal return tubular boilers, each 16 feet in length from outside to outside of heads, and sixty inches in diam- eter, measured on the outside of the smallest ring, the first ring to extend 15 inches beyond the front head to form a smoke box. The boilers to be set in a battery with a dividing wall between them and so connected that each may be operated independent of the other, or that both may be operated together. 16 MATERIAL, SHELL AND HEADS. The shell plates are to be made of open hearth homogeneous flange steel, of an inch in thickness, and have a tensile strength of not less than 60,000 pounds, nor more than 65,000 pounds, per square inch of section. A test piece shall be cut from each plate, and each test piece shall show an elongation in testing of not less than 25 per cent. in 8 inches, or a reduction of area at point of fracture of not less than 50 per cent. The heads will be made of open hearth homo- geneous flange steel, of an inch in thickness. 10 TEST PIECES. The test section of one-half of the test pieces shall be eight inches long and of an inch wide for determining elongation and tensile strength, and one-half of the test pieces shall be made according to the rules of the United States Board of Supervising Inspectors of Steam Vessels, for determining reduction of area. RIVETS AND RIVET HOLES. All rivets will be made of the best quality of steel. All longitu- dinal seams will be double-riveted zig-zag, and placed above the fire line, the lower sheets will lap over the outer side of the upper sheets. The rivet holes for longitudinal seams are to be inch in diameter, and to have a pitch of 2 inches from center to center of rivet holes; and a distance of 14 inches from the center of the outer row of rivet holes to the center of the inner row of rivet holes. The center of outer row of rivet holes will not be less than 1 32 inches from the edge Specifications and Plans for Tubular Boilers. 189 of the sheet. All edges of sheets will be planed true and neatly and efficiently calked. All rivet holes for boilers will be fairly drilled, so that they will come fair without the use of a drift pin forcibly, and without the use of a reamer if such use will enlarge any rivet hole materially beyond the diameter herein specified. The circular seams will be single riveted, and the rivet holes for such seams will be inch in diameter, and the center of rivet holes will be placed not less than 1 inches from the edge of the sheet and have a pitch of 24 inches from center to center of rivet holes. In no case will In no case will any shoulder or off-set be allowed in any rivet hole; and in all cases must each rivet be driven so as to fill the rivet hole completely. BRACING. There will be six braces each 14 inches in diameter in the front head above the tubes, and attached to the head by heavy tee or angle-iron, and in such a manner as to relieve the tubes of any undue strain from the steam or hydrastatic pressure; and they will be attached to the shell by three inch rivets. The back head will have five such braces attached in a similar manner above the tubes; except that the braces on each side of the man hole will be attached to the head by means of crow feet, with two inch rivets; and all braces attached to the heads shall not be less than 3 feet in length. There will be two rods of 1 inch round iron, running from head to head below the tubes, and screwed tightly into each head and securely riveted. TUBES. The tubes are to be the standard American boiler tubes or their equivalent; and each boiler to contain 52 tubes 4 inches in diameter and 16 feet in length, and set in straight rows vertical and horizontal, with a clear space between the tubes of not less than 1 inch, and between the tubes and shell of the boiler a space of not less than 3 inches, and the tubes shall have a central vertical space of not less than 2 inches. MAN HOLES. There will be two man holes in each boiler, one in the back head above the tubes and one in the front head below the tubes. Each such man hole shall have an opening of not less than 11 x 15 inches; and each such man hole shall be flanged inwardly to a depth of not less than 1½ inches, and the faces of such flanges shall be planed or other- wise brought to a true surface, and each such man hole shall be pro- vided with the Eclipse man head (as shown in Figs. 72 and 73 of the plans). 190 A Library of Steam Engineering. WALL BRACKETS. There will be two cast-iron brackets securely riveted on each side of the boilers on a line with the upper row of tubes. Each bracket to be of sufficient size and weight to carry the boilers with perfect safety, and to have a width of not less than 9 inches, and project from the shell of the boiler a distance of not less than 12 inches. The forward brackets to be provided with a cast-iron bearing plate not less than 10 inches wide, 18 inches long and 2 inches thick. The back brackets to be provided each with similar plates and two wrought-iron rollers, each 10 inches in length and 1 inch in diameter. DOME. There will be a dome 30 inches in diameter and 30 inches in height, made of the same quality of material as that in the shell of the boiler. And the dome shall have a thickness of material of of an inch, and 16 shall be properly flanged and double riveted to the shell of the boiler. The dome head shall be of open hearth flange steel, and have a thick- ness of material of ğ of an inch, and to be bumped to a radius of not more than 20 inches. SMOKE BOX. The smoke box will be fitted with a collar riveted to the top to form a base for connecting the breeching. FIRE FRONT AND CASTINGS. The boilers will each be provided with a full, flush fire-front of neat design, fire doors, ash pit doors, a cleaning door 12 x 16 inches for back end, liners, grate bearers, anchor bolts, six buck staves and tee rods for the same, a full set of grate bars 4 feet in length, and having not less than 60 per cent. of air space. FITTINGS. The boilers will each be provided with one 3 inch spring-loaded safety-valve, made according to the rules of the United States Board of Supervising Inspectors, one 10 inch steam-gauge, one 2 inch asbestos blow-off valve, one 2 inch check-valve, one 14 inch water-gauge with x 12 inch glass, three inch gauge-cocks, all fitted to a Williams' safety water column (as shown in Figs. 74 and 75), or its equivalent, one set of fire irons, consisting of one hoe, one slice bar and one scraper. INSPECTION. Each boiler when completed, is to be subjected to a hydrastatic pressure of 185 pounds per square inch, in the presence of the inspector of any reliable boiler insurance company, and the certificate of inspec- tion of such company, and policy of insurance for five hundred dollars Specifications and Plans for Tubular Boilers. 191 to be furnished the purchaser of the boiler. All of the material and workmanship about the boilers to be first-class, and to be open to inspection, at all stages of construction, by any representative of the purchaser. 20 800 PLANS OF BOILERS. 00 bdooooooooop DA (PDAA Fig. 70. HORIZONTAL TUBULAR BOILER-FRONT VIEW. Fig. 71. HORIZONTAL TUBULAR BOILER-BACK VIEW. 80 FOLGER, CIT FOLGER 192 A Library of Steam Engineering. THE ECLIPSE MAN HOLE AND COVER. Fig. 72. REAR HEAD OF TUBULAR BOILER. Fig. 73. FRONT HEAD OF TUBULAR BOILER. THE PITTSBURGH SAFETY WATER COLUMN. Folger Cin. E CD I A I F Fig. 74. H Fig. 75. Figs. 74 and 75 show the general arrangement of the safety water column. Fig. 74 represents the column trimmed complete with water-gauge and gauge-cocks, ready for connection to the boiler. Specifications and Plans for Tubular Boilers. 193 Fig. 75 shows detailed sectional drawing of the column with work- ing parts. The steam connection is made at the top E, and the water connection at the lower end at F, or through the blow-off H, at the bottom. The column is placed on the boiler, so that the center of the lower gauge-cock is 2 or 2 inches above the tube or flue level. When the water falls in the boiler, it allows the seamless copper float A to drop to the lower gauge-cock, when the upper knocker B, on the float-rod D, strikes the lever C, opens the valve G, and allows the steam to escape through the whistle, and sound the low water alarm. In the same manner, when the water raises the float to the center of the upper gauge-cock, the lower knocker B raises the lever C, sounding the whistle for high water. II, Fig. 75, is where the water-gauge is attached, and that gauge indicates the exact height of the water in the boiler. The blow-off con- nection is made at H. The dotted lines representing the lever C in two different positions, represent the points at which the alarms are given for high and low water. 13 194 A Library of Steam Engineering. SINGLE SETTING, WITH FULL FRONT. PLAN A. Fig. 78 0000 M N Section on line A L K. Fig. 79 Pollo O Front Elevation. Fig. 81 1 Fig. 76 Fig. 77 Ground Plan. RS - B KW- 1 BIB H H Folger, Cin Longitudinal Sectional Elevation. SINGLE SETTING, WITH HALF ARCH FRONT. SECTION ON LINE W. Fig. 83 PLAN B. Fig. 80 T GROUND PLAN. 不 ​Fig. 82 BB 1 W V 1 V FolgerCIN FRONT ELEVATION. LONGITUDINAL SECTIONAL ELEVATION. Specifications and Plans for Tubular Boilers. 195 DIMENSIONS FOR SINGLE SETTING.-PLAN A. No. of A B C D E F G H I In. Ft. In. In. In. In. In. In. In. J K In. In. In. L M N O Р Q R S T In. In. In. In. In. In. In. In. No. of Fire Brick. Common Brick. Above Floor Level. 30 8 16 13 9 34 13 65 134 36 8 16 13 9 34 13 65 36 18 134 36 18 18 18 36 10 18 13 9 40 13 85 160 36 18 18 40 10 18 13 9 40 13 85 160 44 24 20 4446 64 13 38 36 18 20 70 13 44 36 20 22 70 13 44 42 20 22 74 13 48 42 22 24 4444 650 4000 800 4500 950 4900 1020 5600 42 10 18 13 9 40 13 85 160 44 24 20 6 76 13 50 42 22 24 4 1080 5800 44 10 18 13 13 12 18 48 12 20 48 14 20 54 12 54 14 60 12 66 12 24 66 14 24 66 16 24 2222 40 18 13 13 46 18 18 18 46 18 18 18 52 20 18 18 46 20 18 18 52 18 120 226 48 24 24 22 18 46 22 100 208 48 24 60 14 24 18 52 22 118 232 48 18 46 22 100 208 48 24 18 52 22 118 232 48 24 24 8 18 58 22 136 256 48 24 24 8 122 80 164 24 20 6 78 13 52 42 24 24 4 1150 5900 98 188 44 24 20 6 78 13 52 48 24 24 4 1300 6200 102 202 44 24 20 96 18 60 48 28 28 6 1500 8100 18 120 226 44 24 20 96 18 60 54 28 28 6 1650 9000 18 102 202 48 24 24 102 18 66 48 30 30 6 1650 9000 24 102 18 66 54 30 30 6 1850 9900 24 116 22 48 34 34 6 1900 12000 24 24 116 22 54 34 34 6 2075 12600 24 122 22 48 36 36 6 2075 12500 122 22 54 36 36 6 2275 13200 22 78 | 60 36 36 6 2475 13900 DIMENSIONS FOR SINGLE SETTING.-PLAN B. No. of No. of A B In. 30 36 C D E F G Ft. In. In. In. In. In. 8 16 13 9 34 13 8 16 13 9 34 13 H I J In. In. In. 60 129 36 60 129 36 K L M N In. In. In. 18 18 O P Q R S T In. In. In. In. In. In. In. Fire Brick. Common Brick. Above Floor Level. 4 64 13 38 36 18 20 18 18 4 70 13 44 36 20 36 10 18 13 9 40 13 80 155 36 18 18 4 70 13 44 42 20 2223 444 600 3250 760 3550 900 4100 40 10 18 13 9 40 13 80 155 44 24 20 6 74 13 48 42 22 24 4 980 4350 42 10 18 13 9 40 13 80 155 44 24 20 6 76 13 50 42 22 24 4 1040 4500 · 10 18 13 13 40 18 76 160 24 20 78 13 52 24 24 4 1100 4650 44 48 12 18 12 20 48 14 20 13 13 46 18 94 184 24 20 78 13 52 48 24 24 4 1250 5300 18 18 46 18 93 193 24 20 96 18 60 48 28 28 6 1450 7100 18 18 52 18 111 217 24 20 96 18 60 54 28 28 6 1600 7900 54 12 20 18 18 46 18 93 193 24 24 8 102 18 66 48 30 30 6 1600 8100 54 60 14 20 18 18 52 18 12 24 22 18 46 22 60 14 24 22 18 52 22 66 12 24 22 18 46 22 91 199 66 14 24 22 18 52 22 109 223 66 16 24 22 18 111 217 48 199 48 24 24 8 24 24 8 102 18 66 54 30 30 6 1800 9000 116 22 72 48 34 34 6 1850 11000 109 223 48 24 24 8. 48 24 48 24 24 58 22 127 247 48 24 24 8 ∞ ∞ ∞ ∞ 116 22 72 54 24 8 122 78 8 122 22 78 54 36 122 22 12888 34 34 6 2000 11900 48 36 36 6 2000 11700 36 6 2200 12300 78 60 36 36 6 2400 12900 196 A Library of Steam Engineering. Fig. 85 SINGLE SETTING, WITH FULL FRONT.-PLAN C. B I --- Fig. 84 四 ​B IDI A K FRSECTION ON LINE A B. ----J Fig. 87 7 LONGITUDINAL SECTIONAL ELEVATION. Fig. 86 N 口 ​D 18 0 ם ם Folger GIN FRONT ELEVATION. GROUND PLAN. SINGLE SETTING, WITH HALF ARCH FRONT.-PLAN D. Fig. 89 Fig. 91 E 1 K-- ANNARENKENNEKOERENNI TION ON А Б. ㅁ ​B B FRONT ELEVATION. O 1 k--- k -BI +P Fig. 88 A A--- B -D- M LONGITUDINAL SECTIONAL ELEVATION. Fig. 90 N GROUND PLAN, Folger-Cin Specifications and Plans for Tubular Boilers. 197 DIMENSIONS FOR SINGLE SETTING.-PLAN C. No. of No. of A B C D E F G H I J K L M N О P Q R S T U Ft. In. In. In. In. In. In. In. In. In. In. In. In. Ft. In In. In. In. In. In. In. In. Fire Brick Common Brick. Above Floor Level. 7 32 12 7 34 12 8 36 12 10 36 12 8 42 14 10 42 14 222244 10 44 14 2222222 20 16 45 44 20 16 48 47 20 16 48 47 20 16 48 47 20 16 48 47 7888∞ 32 64 85 26 19 11 6 20 40 12 16 36 34 4 600 6800 34 66 90 26 22 11 6 20 40 12 16 36 34 4 600 7500 36 68 92 26 22 12 6 24 40 12 16 36 34 650 7700 36 68 92 26 22 14 6 28 46 12 16 42 42 720 8500 42 74 98 27 21 12 8 24 40 12 16 36 34 730 8500 20 16 48 47 8 42 74 98 27 21 14 8 28 46 12 16 42 42 770 9600 24 16 48 47 10 44 76 100 27 21 15 0 28 46 12 16 42 42 880 10500 12 44 14 24 16 467/2 44 76 100 27 21 17 0 32 52 12 16 48 49 4 940 10800 14 44 14 24 16 48 461 44 76 100 27 21 19 0 36 58 12 16 54 84 4 1120 11500 12 48 16 14 48 16 24 12 54 16 24 15 54 16 12 60 18 14 60 18 24 16 60 18 26 15 66 16 66 16 72 9999 18 18 20 888 30 FFFFFF8228 24 24 24 2222222222 16 4512 48 88 103 26 21 17 2 32 52 12 20 48 49 4 1120 13600 16 451 48 88 103 26 21 19 2 36 58 12 20 54 84 4 1140 15700 20 50 481 54 94 112 26 24 17 6 32 52 12 48 49 1160 16200 20 50 481 10 54 94 112 26 24 20 8 36 56 16 20 54 90 4 1270 17500 20 50 4812 12 60 108 118 26 24 17 10 32 50 16 24 48 49 4 1400 20500 20 50 481 12 2 60 108 118 26 24 19 10 36 56 16 24 54 84 4 1500 23000 20 50 48 12 20 50 4812 12 20 50 48 20 50 48 88888 12 12 2222 60 108 118 26 24 22 0 40 56 16 24 54 96 4 1540 25300 66 114 124 26 24 21 2 36 56 16 24 54 90 1590 26000 66 114 124 26 24 22 2 40 56 16 24 54 96 1620 27000 72 120 130 26 24 22 6 40 56 16 24 54 96 4 1750 30000 DIMENSIONS FOR SINGLE SETTING.-PLAN D. A BCD Ft. In. In. In. 10 7780 32 34 14 36 14 36 4444 8 42 18 20 16 10 AE 222222 E F G HI J K L M N 0 P QRS TU V W XY Z Fire Brick. mon Brick. In. In. In. In. In. In. In. In. In. Ft. In In. In. In. In. In. In. In. In. In. In. In. In. No. of No. of Com- Above Floor Level. 14 20 16 46 20 16 20 16 14 20 16 42 18 20 16 46 10 44 18 24 16 46 6666666 45 46 45 46 45 46 45 46 45 78888 55555 32 64 73 34 66 75 36 68 77 36 68 77 8 42 74 83 45 8 42 74 83 45 10 44 76 85 12 44 18 24 16 46 14 44 18 12 48 19 442 10 24 16 46 44 10 24 16 50 481 44 76 85 76 48 88 93 14 48 19 24 16 50 481 48 88 93 16 16 566 15 66 24 12 54 19 24 20 50 15 54 19 24 20 50 12 60 21 24 20 468 14 60 21 24 20 463 16 60 21 26 20 20 47 481 BEET886 1010 008 26 20 26 20 26 20 26 20 27 19 27 19 222222 10 3 20 33 10 3 20 33 11 3 24 33 13 3 28 39 2222 16 36 34 4 36 9 24 12 600 6150 16 36 34 4 36 9 28 12 600 6200 16 36 34 36 28 12 15 650 6700 11 3 24 13 3 28 381 27 19 13 7 28 381 27 19 15 7 32 27 19 17 7 36 501 12121222 322 12 16 42 42 16 36 34 12 16 42 42 4 323 12 16 42 42 36 9 28 12 25 720 7050 4 323 124 32 16 11 730 7700 1214 32 16 25 770 8700 323 124 36 12 16 48 49 321 12 16 54 84 321 121 121 36 20 35 16 25 880 8800 940 9300 36 24 45 1120 9500 26 24 15 7 32 48 20 48 49 4 26 24 17 7 36 54 12 54 84 54 94 99 26 24 15 11 32 482 12 3614 121 36 361 121 436 20 35 1120 11100 24 45 1140 12900 48 49 4 34 141 42 20 35 1160 13200- 481 10 54 94 99 26 24 19 1 36 5212 16 54 90 4 34 1412 42 24 57 1270 14200 452 12 60 108 1013 26 2034 16 1 32 47 16 48 49 4 31 141 48 20 37 1400 16700 463 45 12 60 108 45 12 60 108 1013 26 208 18 1 36 53 16 24 54 84 4 30% 10134 26 203 20 3 40 53 16 24 54 96 4 301 1212 141 48 24 45 1500 19200 141 48 24 65 1540 21500 66 24 28 20 452 12 47 451% 12 66 114 66 114 108 26 21 19 5 36 522 16 24 54 90 4 31 108 26 21 20 5 40 72 24 30 20 48 462 12 72 120 115 284 1984 20 7 40 522 16 24 5212 16 54 96 4 31 141 54 24 141 54 24 65 57 1590 22100 1620 23100 24 54 96 4 272 19 54 24 65 1750 26000 198 A Library of Steam Engineering. SINGLE SETTING, WITH HALF ARCH FRONT AND BOILER RESTING ON FIRE FRONT AND BACK STAND. PLAN E. Fig. 93 I L. FRONT ELEVATION. Fig. 94 163100000 офо ΟΦΟ офо офооо 00 ΟΦΟ -24- G. L. FIS 22- KTP --------14- --- SECTION THROUGH C. D. LK L. ---- 15% k9 " 1912 +26- E R--- A नू " B C Fig. 92 FILLED WITH EARTH 16 0+ 21-1 9 K T G -21- T L-L- D SIDE ELEVATION SECTION THROUGH A. B. O- Fig. 95 2 4 GROUND PLAN. A 6O AIR SPACE k21" 17- AIR SPACE B 42 Folger CIN 下 ​ Specifications and Plans for Tubular Boilers. 199 Horse Power. Nom'l Rated SHELL. Diam Lgth In. Ft. Box, Inches. Depth Smoke SPECIFICATIONS FOR SINGLE SETTING.-PLAN E. THICKNESS BOILER PLATE Safe Working OF BOILER Homogeneous PLATE. Steel. Pressure for Boiler per DOME. Square Inch. Shell Heads Strength per Tensile Longitudinal Seams, Diam Hght In. In. Square Inch. Double Rivet'd In. In. Outlet, Size of Steam Inches. TUBES. Diam Lgth No. In. Ft. Square Feet. Total Heat- ing Surface, STACK. SHIPPING WEIGHT. Pounds Approximate. Diam Lgth Boiler Fixtures Boiler and In. Ft. Only. Only. Fixtures. 25 36 12 30 40 12 22 15 60,000 lbs. 166 lbs. 18 20 22 15 60,000 66 150 66 20 22 2/ 35 42 12 15 60,000 66 143 66 20 22 46 12 15 60,000 66 130 24 24 45 48 12 15 60,000 66 140 66 24 24 55 52 14 15 60,000 130 66 26 26 3 60 54 14 15 60,000 66 139 66 26 26 70 54 16 15 60,000 66 139 66 26 26 75 60 14 15 125 85 60 16 15 100 66 16 15 72 60,00005 60,000 66 126 66 28 28 4 66 126 66 28 28 43 60,000 66 136 66 16 15 60,000 66 125 66 121212 2223 ∞ ∞ ∞ ∞ 5 26 34 40 48 44 1222 05 12 318 18 35 2,970 1,830 4,800 3 12 404 20 40 3,441 2,159 5,600 12 464 22 40 3,760 2,240 6,000 42 3 12 491 22 40 4,345 2,355 6,700 3 12 551 24 40 5,045 2,355 7,400 35 31/ 46 31 2212 14 693 26 40 6,620 2,880 9,500 14 721 26 40 7,300 3,000 10,300 33 40 4 16 817 26 40 8,711 3,089 11,800 62 تجابر 14 940 30 40 9,180 3,420 12,600 52 4 82 1,045 30 30 422 64 4 16 1,265 30 30 4 16 1,578 40 16 30 40 10,643 3,457 14,100 36 40 40 13,294 15,900 4,206 17,500 4,700 20,600 FURNACE. DIMENSIONS FOR SINGLE SETTING.-PLAN E. SETTING. BOILERS. Back Pits. Soot and Width of Surface. in Grate Sq. Feet Number of Ash Pit. Length of Furnace. Width of Furnace. Length of of Apron. Length of Jamb. Thickness Front Wall U T S R Q P Bridge Wall. to Bottom of O Side Walls. Closing in of Boiler to Center of Center of FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. ZZ A B C D E F G H I J K L M N IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. SQ. FT. 25 36 12 21 7 2 6 61 8 16 4 7 6 10 46 75 85 7 11 2 92 14 3 2 10 3 4 3 6 3 10 9.44 30 40 12 23 7 2 6 6 2 2 16 4 7 10 10 10 6 10 5 10 5 9 12 7 3 3 2 3 3 93 4 3 4 4 0 4 2 11.11 35 42 12 23 6 62 8 16 4 8 0 11 06 115 115 10 12 7 72 33 53 40 46 12 25 7 26 6 2 8 16 4 8 16 4 8 4 11 4 11 7 7 16 16 0 12 7 1 2 11 4 45 48 12 25 7 26 6 3 6 16 4 8 6 11 9 7 26 26 1 12 7 1 2 11 4 55 52 14 27 8 7 7 6 3 9 18 4 9 0 12 37 66 46 3 14 6 11 4 60 54 14 27 8 7 7639 6 3 9 18 4 9 2 12 5 7 6 6 5 6 4 13 4 12 5 4 11 4 70 54 16 27 9 10 8 6 3 9 20 4 9 2 12 57 66 5 6 4 13 4 12 16 11 4 75 60 14 33 8 7 7 64 7 18 4 9 8 13 67 96 86 86 7 13 4 1 2 3 4 11 4 85 60 16 33 9 10 8 6 4 7 20 4 9 8 13 68 16 86 7 17 12 4 12 2 6 11 4 100 66 16 910 8 6 5 0 20 4 10 2 14 48 5 6 11 6 10 18 13 4 72 82228 83 6 4 4 4 4 12.83 2 3 6 4 10 4 8 14.58 8 4 10 4 10 15.28 04 10 5 4 16.66 4 4 10 5 6 18.05 125 72 16 33 9 10 8 6 5 6 20 4 10 8 14 7 8 78 67 67 27 1 16 11 4 10 2 4 6 24 11 6 0 56 54 85 04 10 5 6 4 10 6 0 4 10 6 0 22.22 45 46 6 24.88 7 7 0 29.50 20.83 21.52 200 A Library of Steam Engineering. Fig. 97 0 B 0 ! SINGLE SETTING, WITH HALF ARCH FRONT. PLAN F. FRONT ELEVATION. Fig. 98 <234 H-- SECTION THROUGH C. D. 72 1 -24-3 4-194 不 ​-D C မှ Fig. 96 k9- 932 -R ∞ FILLED WITH EARTH 16- O 21-17-9 Q 917-22- SIDE ELEVATION SECTION THROUGH A. B. Fig. 99 G ATR SPACE 4 AIR SPACE 2 17 SPACE GROUND PLAN. Folger, Cin. Specifications and Plans for Tubular Boilers. 201 in Grate Sq. Feet | Number of FURNACE. Surface. Soot and Width of Horse Power. Nom'l Rated SHELL. 25 Diam Lgth In. Ft. 36 12 15 30 40 12 15 Box, Inches. Depth Smoke Steel. Tensile Strength per In. In. Square Inch. Double Rivet'd SPECIFICATIONS FOR SINGLE SETTING.-PLAN F. THICKNESS BOILER PLATE Safe Working OF BOILER Homogeneous PLATE. Shell Heads Pressure for Boiler per Square Inch. Longitudinal Seams, DOME. Outlet, Size of Steam Inches. TUBES. Square Feet. Total Heat- ing Surface, STACK. SHIPPING WEIGHT. Pounds Approximate. Diam Hght In. In. Diam Lgth Diam Lgth Boiler Fixtures No. Boiler and In. Ft. In. Ft. Only. Only. Fixtures. 60,000 lbs. 166 lbs. 18 20 60,000 150 66 20 35 42 12 15 60,000 (6 143 66 20 2223 46 12 15 60,000 66 130 66 24 24 45 48 12 15 60,000 66 140 66 24 24 3 55 52 14 15 60,000 (( 130 66 28 26 3 60 54 14 15 60,000 66 139 66 28 26 70 54 16 15 60,000 (6 139 66 28 26 3 75 60 14 15 60,000 126 66 32 32 85 60 16 15 60,000 66 126 66 32 32 42 2223 ∞∞∞∞44 2호 ​26 3 12 318 18 35 2,970 1,830 4,800 2호 ​34 3 12 404 20 40 3,441 2,159 5,600 2호 ​40 3 12 464 22 40 3,760 2,240 6,000 42 3 12 491 22 40 4,345 2,355 6,700 48 3 12 551 24 40 5,045 2,355 7,400 44 3/ 14 693 26 40 6,620 2,880 9,500 31/1 46 31/ 14 721 26 40 7,300 3,000 10,300 40 4 16 817 26 40 8,711 3,089 11,800 62 3 14 940 30 40 9,180 3,420 12,600 52 4 16 1,045 30 40 10,643 3,457 14,100 100 66 16 15 60,000 (6 136 (6 125 72 16 15 60,000 66 125 32 32 43 64 32 32 5 82 4 16 1,265 36 40 13,294 4,206 17,500 4 16 1,578 40 40 15,900 4,700 20,600 SETTING. Back Pit. Ash Pits. Length of Furnace. Width of Furnace. Length of of Apron. Length of Jamb. Thickness Front Wall T S R P Side Walls. Closing in of Boiler to Center of Bridge Wall. to Bottom of O Center of DIMENSIONS FOR SINGLE SETTING.-PLAN F. A B Q D E F G H I T K L M N IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. 25 36 12 21 7 2 6 61 816 4 7 6 10 46 75 85 7 11 6 2 Q FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. SQ. FT. 92 14 3 2 103 43 63 10 U 9.44 30 40 12 23 7 2 6 6 2 2 16 4 7 10 10 10 6 10 5 10 5 9 12 7 3 3 2 3 3 93 43 44 04 2 11.11 35 42 12 23 7 26 6 62 8 16 4 8 0 11 0 6 11 5 11 5 10 12 7 3 7 2 3 3 5 3 8 3 6 4 64 4 4 4 12.83 40 46 12 25 7 2 6 62 8 16 4 8 4 11 7 7 16 160 12 7 12 52 11 4 2 3 6 4 10 4 8 14.58 45 48 12 25 7 26 6 63 6 16 4 8 6 11 97 26 26 1 12 7 4 12 2 11 4 8 4 10 4 10 15.28 55 52 14 27 87 7 63 9 18 4 9 0 12 37 66 46 3 14 9 4 12 64 11 4 04 10 5 4 16.66 60 54 14 27 8 7 7 63 9 18 4 9 2 12 57 66 56 4 13 8 12 54 11 4 4 4 10 5 6 18.05 70 54 16 27 9 10 8 6 3 9 20 4 9 2 12 5 7 6 6 5 6 4 13 8 4 12 16 11 4 2 04 10 5 6 20.83 75 60 14 33 8 7 7 6 4 7 18 4 9 8 13 67 96 86 7 13 8 4 12 34 11 4 2 24 10 6 0 21.52 85 60 16 910 8 64 7 20 4 9 8 13 68 16 86 7 17 12 4 12 2 6 11 4 2 100 66 16 33 910 8 6 5 0 20 4 10 2 14 48 5 6 11 6 10 18 13 4 72 125 72 16 33 9 10 8 6 5 6 20 4 10 8 14 78 67 27 1 16 11 4 10 2 4 6 2 4 11 6 0 56 54 85 4 4 10 6 0 22.22 45 46 6 24.88 7 7 0 29.50 BOILERS. 202 A Library of Steam Engineering. Fig. 101 K B SINGLE SETTING, WITH FULL ARCH FRONT. H G. L. FRONT ELEVATION. -S wwwww Fig. 102 6 SECTION THROUGH C, D, G. L. PLAN G. C R T -24-0 -234- 26- B 179 +-G Fig. 100 FILLED WITH EARTH D SIDE ELEVATION-SECTION THROUGH A. B. Fig. 103 SPACE NA G 4 ¥4 <9> 17 1 S AIR SPACE GROUND PLAN. 6 -10- 16- -17 99 Folger Cin a Specifications and Plans for Tubular Boilers. 203 Horse Power. Nom'l Rated SHELL. Diam Lgth In. Ft. Box, Inches. Depth Smoke OF BOILER Homogeneous PLATE. Steel. Tensile Shell Heads Strength per In. In. Square Inch. Double Rivet'd SPECIFICATIONS FOR SINGLE SETTING.-PLAN G. THICKNESS BOILER PLATE Safe Working Pressure for Boiler per Square Inch. Longitudinal Seams, DOME. Outlet, Size of Steam Inches. TUBES. Total Heat- Square Feet. ing Surface, STACK. SHIPPING WEIGHT. Pounds Approximate. Diam Hght Diam Lgth Diam Lgth Boiler Fixtures Boiler and No. In. In. In. Ft. In. Ft. Only. Only. Fixtures. 25 36 12 15 60,000 lbs. 166 lbs. 18 20 23 30 40 12 15 60,000 66 150 20 22 23 35 42 12 15 60,000 66 143 66 20 22 23 40 46 12 15 45 48 12 15 55 52 14 15 60,000 130 66 24 24 16 60,000 66 140 66 24 24 3 60,000 66 130 66 28 60 54 14 15 70 54 16 15 75 60 14 15 16 60,000 66 139 (C 28 26 60,000 66 139 28 26 32 60,000 66 126 66 32 32 85 60 16 15 16 100 66 16 15 125 72 16 15 |000000 212 60,000 60,000 60,000 66 126 66 32 32 412 66 136 66 32 66 125 66 B21171T 2223 627271 1212 0000444E 26 34 3 40 42 48 3 26 3½ 44 33 3/3/ 46 3 40 4 62 52 7171 0000 00 00 00 34 12 318 18 35 2,970 1,830 4,800 12 404 20 40 3,441 2,159 5,600 12 464 22 40 3,760 12 491 22 40 4,345 12 551 24 40 5,045 14 693 26 40 6,620 14 721 26 40 7,300 16 817 26 40 8,711 32 14 940 30 40 9,180 4 16 1,045 30 40 10,643 32 4/ 64 4 16 1,265 32 32 5 82 4 16 1,578 36 40 13,294 40 40 15,900 DIMENSIONS FOR SINGLE SETTING.-PLAN G. BOILERS. SETTING. A IN. 25 36 30 40 35 42 12 40 46 12 45 55 52 14 54 70 54 16 75 60 85 60 100 66 Boiler to Top 88 88 88 Boiler. Length of of Boiler. Diameter Boiler. Power of Rated Horse Nominal B D E F G H I J K L M N P Q R FT. 12 21 7 26 61 12 23 7 26 62 23 6 628 17 0 25 2 6 62 IN FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. 8 17 0 7 6 10 4 217 0 7 10 10 10 8 0 11 0 8 17 0 FT. IN. IN. IN. SQ. FT. 845 85 7 11 292 14 42 103 44 23 10 9.44 8 45 105 9 12 32 33 10 43 44 8 4 2 11.11 8 45 11 5 10 12 72 36 83 6 04 4 12.83 8 4 11 7 926 16 0 12 12 6 4 8 14.58 48 12 7 26 6 3 6 17 0 8 6 11 9 9 26 26 1 12 1 04 4 10 15.28 8 7 63 9 19 0 9 0 12 3 926 46 3 14 9 04 05 6 5 4 16.66 14 877 6 9 19 0 9 2 12 5 10 2 6 5 64 13 1 2 5 9 10 8 6 3 9 21 0 9 2 12 5 10 2 6 5 6 4 13 8 4 12 1 14 8 7 7 6 4 7 19 0 9 8 13 6 10 26 86 7 13 4 12 35 16 9 10 8 6 4 7 21 0 7 21 0 9 8 13 9 8 13 6 10 2 6 86 7 17 12 4 12 27 16 125 72 16 33 910 8 6 5 9108 0 21 0 10 2 14 4 11 2 6 11 6 10 18 13 4 7 6 6 21 0 10 8 14 7 11 2 7 27 1 16 11 4 10 2 46 676766 04 04 0 4 25 25 6 6 34 11 6 06 HOQ4+0 5 65 6 18.05 6 20.83 0 21.52 014 64 8 46 0 6 6 24.88 66 0 22.22 37 0 29.50 Furnace. Length of FURNACE. Back Pits. Soot and Width of Surface. in Grate Sq. Feet Number of Ash Pit. Length of Furnace. Width of A pron. Length of Thickness U T S FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. 2,240 6,000 2,355 6,700 2,355 7,400 2,880 9,500 3,000 10,300 3,089 11,800 3,420 12,600 3,457 14,100 4,206 17,500 4,700 20,600 204 A Library of Steam Engineering. 2334 00 DOUBLE SETTING, WITH HALF ARCH FRONT. PLAN H. Fig. 105 О 0 -+ B B FRONT ELEVATION. Fig. 107 Fig. 106 -V SECTION THROUGH G. D. -22- لي -4' 1342 D-- Fig. 104 C 114 R B T- 17- -21- 1-7-4 G+ S+- D FILLED WITH 二 ​1-6 EARTH 9 Q-17 22 1 + SIDE ELEVATION. SECTION THROUGH A. B... 1-7 U- 1-7 17 9 GROUND PLAN. Folger Cin Specifications and Plans for Tubular Boilers. 205. Horse Power. Nom'l Rated SHELL. Diam Lgth In. Ft. Box, Inches. Depth Smoke OF BOILER Homogeneous PLATE. Shell Heads Tensile Strength per In. In. Square Inch. Double Rivet'd SPECIFICATIONS FOR DOUBLE SETTING.-PLAN H. THICKNESS BOILER PLATE Safe Working Steel. Pressure for Boiler per Square Inch. Longitudinal Seams, DOME. Outlet, Size of Steam Inches. TUBES. Diam Lgth Total Heat- Square Feet. ing Surface, Diam Hght In. In. 110 120 54 140 54 150 60 170 60 200 66 250 72 NHOUN 52 14 15 82 60,000 lbs. 130 lbs. 28 26 14 15 5 16 60,000 66 139 66 28 26 16 15 60,000 66 139 66 28 14 15 16 60,000 66 126 66 32 32 16 15 60,000 66 126 66 32 32 16 15 16 15 60,000 60,000 66 136 66 66 125 66 32 32 2222222 5 5 26 5 32 10 10 10 60777 STACK. SHIPPING WEIGHT. Pounds Approximate. No. In. Ft. Diam Lgth In. Ft. Boilers Only. Fixtures Boiler and Only. Fixtures. 44 3 46 31/ 1212 14 1,386 34 40 13,240 6,160 19,400 14 1,442 36 40 14,600 6,500 21,100 40 4 16 1,634 36 40 17,422 6,678 24,100 62 3½ 14 1,880 40 40 18,360 7,340 25,700 52 4 16 2,090 40 40 21,286 7,414 28,700 64 4 16 2,530 44 40 26,588 8,412 35,000 82 4 16 3,156 48 40 31,800 10,000 41,800 DIMENSIONS FOR DOUBLE SETTING.-PLAN H. SETTING. FURNACE. Grate Surface. Square Feet in Number of Walls Between Thickness of of Furnace. Boilers and Sides Space Between Boilers. Middle Jamb. Thickness of Soot and Back P Width of Pits. of Ash Pits. Length of Furnaces. Width of Furnaces. Length of Aprons. Length of Walls. Boilers to Top Center of Side Walls. to Closing in of Center of Boilers Walls. Bottom of Bridge Front Walls to Outside Jambs. Thickness of Back End. Center of Boilers, I Floor Line to Front End. Center of Boilers, Floor Line to Walls. Height of Safety Valves. to Top of Floor Line Foundation. Width of Foundation. Length of Stack Saddles. Width of Front of Boilers. Riveted Domes to Center of Center of Inde- S Q R 立​ㄡ​ˊ Front of Boilers. pendent Domes to A Horse Power of Nominal Rated Boilers. Diameter of Boilers. Length of Dome Flanges. to Top of Top of Boilers Boilers. A B C D E F G H I J K L M N O P T U V W X IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. FT. IN FT. IN FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. SQ. FT. 110 52 14 27 120 54 14 27 140 54 16 27 150 60 14 33 170 60 16 33 87 7 639 18 4 15 9 12 3 7 6 6 4 6 3 14 9 4 12 6 4 11 4 2 4 0 4 10 5 4 29 6 17 33.32 8 7 7 639 18 4 16 1 12 5 7 6 6 5 6 4 13 9 10 8 6 3 9 20 4 16 1 12 5 7 6 6 5 6 4 13 8 4 12 5 8 4 4 11 4 2 4 4 4 10 5 6 2 7 6 17 36.10 12 1 6 11 4 2 5 0 4 10 5 6 111 6 17 41.66 87 7 64 7 18 4 17 1 13 6 7 9 6 86 7 84 12 3 4 11 4 2 5 2 4 10 6 02 3 6 17 43.04 910 8 6 4 7 20 4 17 1 13 6 8 1 6 8 6 7 17 12 4 12 2 6 11 4 2 5 4 4 10 6 0 2 1 17 44.44 200 66 16 33 910 8 6 5 0 20 4 18 1 14 4 8 5 6 11 6 10 18 13 4 7 2 5 6 5 4 8 5 45 46 62 7 6 17 49.76 250 72 16 33 9 10 8 6 5 6 20 4 19 1 14 7 8 6 7 2 7 1 16 11 410 2 4 6 2 411 6 0 77 0 6 17 59.00 BOILERS. 206 A Library of Steam Engineering. K-22 K- DOUBLE SETTING, WITH FULL FRONT. PLAN I. FRONT ELEVATION. 00 SECTION THROUGH C-D. Fig. 109 GROUND K15 -24"- -R- 21 9 K17 K Fig. 108 B FILLED WITH EARTH ·Q ---G++ SIDE ELEVATION SECTION THROUGH A-B. Fig. 111 +o- -Q Fig. 110 GROUND 22 Folger Ci GROUND PLAN 9-1 17--22 17 Specifications and Plans for Tubular Boilers. 207 SHIPPING WEIGHT. Pounds Approximate. STACK. OF BOILER Homogeneous PLATE. Steel. Shell Heads Strength per Tensile In. In. Square Inch. Double Rivet'd SPECIFICATIONS FOR DOUBLE SETTING.-PLAN I. THICKNESS BOILER PLATE Safe Working Pressure for Boiler per Square Inch. Longitudinal Seams, DOME. Outlet, Size of Steam Inches. TUBES. Square Feet. ing Surface, Total Heat- Diam Hght In. In. 121212222 00/0000/00 60,000 lbs. 60,000 130 lbs. 28 26 5 66 139 66 28 26 60,000 " 139 66 28 26 5 60,000 60,000 60,000 60,000 126 66 32 32 66 126 66 32 66 136 66 32 66 125 66 222 32 32 32 32 10 10 10 6777 Boiler Diam Lgth Diam Lgth Boilers Fixtures No. In. Ft. In. Ft. Only. Only. and Fixtures. 44 33 14 1,386 34 40 13,240 6,160 19,400 46 32 14 1,442 36 40 14,600 6,500 21,100 40 4 16 1,634 36 40 17,422 6,678 24,100 62 3 14 1,880 40 40 18,360 7,340 25,700 52 64 44 16 2,090 40 40 21,286 7,414 28,700 16 2,530 44 40 26,588 8,412 35,000 82 4 16 3,156 48 40 31,800 10,000 41,800 Box, Inches. Depth Smoke SHELL. Diam Lgth In. Ft. Horse Power. Nom'l Rated 110 52 14 15 120 54 14 15 140 150 60 54 16 14 15 15 170 60 16 15 200 66 16 15 250 72 16 15 FURNACE DIMENSIONS FOR DOUBLE SETTING.-PLAN I. SETTING. BOILERS. Grate Surface. Square Feet in Number of Boilers. Walls Between Thickness of of Furnace. Boilers and Sides Space Between Middle Jamb. Thickness of Soot and Back P Width of Pits. of Ash Pits. Length Width of Furnaces. of Furnaces. Length of Aprons. Length Outside Jambs. Thickness of P of Walls. Boilers to Top Center of Side Walls. to Closing in of Center of Boilers Walls. Bottom of Bridge O Front Walls to Back End. Center of Boilers, Floor Line to Safety Valves. to Top of Floor Line of Fire Fronts. to Top Floor Line Front End. Center of Boilers, Floor Line to Foundation. Width of Foundation. Length of Stack Saddles. Width of Front of Boilers. Riveted Domes to Center of Center of Inde- NÅ M A B C D E F G H I J K L Q R S T U V W X IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. FT. IN FT. IN FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. SQ. FT. 110 52 14 27 120 54 14 27 140 54 16 27 150 60 14 33 170 60 16 200 66 16 33 8 7 7 6 3 9 19 0 15 9 12 3 926 46 3 14 9 4 12 6 5 0 4 24 05 65 42 9 6 17 33.32 877 6 3 9 19 0 16 1 12 5 10 2 6 5 6 4 13 8 4 12 5 5 0 4 2 4 4 5 9 10 8 6 3 9 21 0 16 1 12 5 10 2 6 5 6 4 13 84 12 04 65 6 5 62 7 6111 6 17 36.10 6 17 41.66 877 64 7 19 0 17 1 13 6 10 2 6 8 6 7 13 0 66 02 3 6 17 43.04 910 8 6 4 7 21 0 17 1 13 6 10 2 6 8 6 7 17 12 4 12 2 7 0 4 2 5 6 6 02 1 6 17 44.44 910 8 6 5 0 21 0 18 1 14 4 11 2 6 11 610 18 13 4 7 2 5 6 6 4 8 5 4 6 06 62 7 250 72 16 33 910 8 6 5 6 21 0 19 1 14 7 11 27 2 7 1 16 11 410 2 4 6 3 411 6 0 6 37 02 75 6 17 49.76 6 17 59.00 Front of Boilers. pendent Domes to A Horse Power of Nominal Rated Boilers. Diameter of Boilers. Length of Dome Flanges. to Top of Top of Boilers Boilers. 208 A Library of Steam Engineering. TRIPLE SETTING, WITH HALF ARCH FRONT. PLAN J. B FRONT ELEVATION. Fig. 114 22 KISI P+ t- K-V- KAVI ニニニニニニ​サー​ナーーー ​SECTION THROUGH C-D. KS Fig. 113 ¢ Fig. 112 C 114 E --4-1-84- 00 GROUND -22 GROUND FILLED WITH EARTH 16- K -O- KQ-→ K1-79 --Tatt "*221 ++-+- SIDE ELEVATION SECTION THROUGH A-B.1 -G--- --Fig. 1-15- --G 0- 17 17 <17- GROUND PLAN. Folger, Cin AIR SPACE Specifications and Plans for Tubular Boilers. 209 Horse Power. Nom'l Rated SHELL. Diam Lgth In. Ft. Box, Inches. Depth Smoke OF BOILER Homogeneous PLATE. Steel. Shell Heads Strength per Tensile In. In. Square Inch. Double Rivet'd SPECIFICATIONS FOR TRIPLE SETTING.-PLAN J. THICKNESS BOILER PLATE Safe Working Pressure for Boiler per Square Inch. Longitudinal Seams, DOME. Outlet, Size of Steam Inches. TUBES. Total Heat- Square Feet. ing Surface, STACK. SHIPPING WEIGHT. Pounds Approximate. Diam Hght In. In. Diam Lgth Diam Lgth Boilers No. In. Ft. In. Ft. Only. Boilers Fixtures and Only. Fixtures. 165 180 54 225 60 255 60 300 66 375 72 16 NHHOOON 52 14 15 32 60,000 lbs. 130 lbs. 28 26 14 15 16 60,000 66 139 66 28 26 210 54 16 15 60,000 66 139 66 28 26 14 15 16 15 16 15 15 1212212 60,000 66 126 66 32 32 60,000 60,000 60,000 66 126 66 32 (6 136 66 32 32 66 125 66 32 32 2222222 7 32 6669888 46 40 62 82 FL248 44 32 33 1212 14 2,079 40 40 19,860 8,740 28,600 14 2,163 42 40 21,900 9,200 31,100 4 16| 2,451 42 40 26,133 9,467 35,600 3 14 2,820 46 40 27,540 10,760 38,300 52 4 16 3,135 46 40 31,929 10,871 42,800 64 4 16 3,795 50 40 39,882 12,518 52,400 4 16 4,734 54 40 47,700 14,000 61,700 FURNACE. DIMENSIONS FOR TRIPLE SETTING.-PLAN J. SETTING. BOILERS. Grate Surface. Square Feet in Number of Space Between Boilers. Walls Between Thickness of X IN. SQ. FT. Boilers and Sides. of Furnace. Middle Jambs. Thickness of Soot and Back P Width of of Ash Pits. Length Pits. Width of Furnaces. of Furnaces. Length of Aprons. Length Outside Jambs. Thickness of Back End. Center of Boilers, Floor Line to of Walls. Boilers to Top Center of Side Walls. to Closing in of Center of Boilers Walls. Bottom of Bridge Front Walls to Front End. Center of Boilers, Floor Line to Walls. Height of Safety Valves. to Top of Floor Line Foundation. Width of Foundation. Length of Stack Saddles. Width of Front of Boilers. Riveted Domes to Center of Front of Boilers. pendent Domes to Center of Inde- Horse Power of Nominal Rated Boilers. Diameter of Boilers. Length of Dome Flanges. to Top of Top of Boilers Boilers. U S P Q R O B C D E F G H I J K L M N T V W 63 9 18 4 8 7 7 9 10 8 8 7 7 63 9 18 4 63 9 20 4 64 7 18 4 9103 8 64 7 20 4 9 10 8 6 5 0 20 4 IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. 165 52 14 27 8 7 7 180 54 14 27 210 54 16 27 225 60 14 33 255 60 16 33 300 66 16 33 IN. IN. FT. IN FT. IN FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. 22 6 12 37 6 6 4 6 3 14 23 0 12 5 7 6 6 5 6 4 13 94 12 6 84 12 5 4 11 4 2 4 0 4 105 429 17 49.98 4 11 4 2 4 4 4 10 5 62 7 17 54.15 23 0 12 5 7 6 6 5 6 4 13 84 12 1 6 11 4 2 0 4 10 5 6 111 6 17 62.49 24 6 13 6 7 9 6 8 6 7 13 8 4 1 2 3 4 11 4 2 5 2 4 10 6 0 2 3 6 17 64.56 24 6 13 6 8 1 6 8 6 7 17 12 4 1 2 2 6 11 4 4 4 10 6 02 1 6 17 66.66 26 0 14 4 8 5 6 11 610 18 13 4 7 256 54 5 4 46 62 7 6 17 74.64 375 72 16 33 9 10 8 6 5 6 20 4 27 6 14 7 8 6 7 27 1 16 11 410 2 4 6 2 411 6 0 5 77 0 6 17 88.50 A 14 210 A Library of Steam Engineering. 22 -AIR SPACE 00 TRIPLE SETTING, WITH FULL FRONT. PLAN K. A B FRONT ELEVATION. Fig. 117 Fig. 118 1 OO B AIR SPACE H D C Fig. 116 T 6 6 K+-R- 24 2151 O E-II: SIDE ELEVATION 494 FILLED WITH EARTH 171 ←--Q--- SECTION THROUGH A-B.. Fig. 119 G SECTION THROUGH C-D. GROUND PLAN. 17 AIR SPACE Specifications and Plans for Tubular Boilers. 211 Horse Power. Nom'l Rated SHELL. Diam Lgth In. Ft. Box, Inches. Depth Smoke PLATE. Shell Heads Tensile Strength per In. In. Square Inch. Double Rivet'd SPECIFICATIONS FOR TRIPLE SETTING.-PLAN K. THICKNESS BOILER PLATE Safe Working OF BOILER Homogeneous Steel. Pressure for Boiler per DOME. Square Inch. Longitudinal Seams, Outlet, Size of Steam Inches. TUBES. Total Heat- Square Feet. ing Surface, STACK. SHIPPING WEIGHT. Pounds Approximate. Diam Hght Diam Lgth No. In. In. In. Ft. Diam Lgth In. Ft. Boilers Only. Boiler Fixtures and Only. Fixtures. 165 52 14 15 32 60,000 lbs. 130 lbs. 28 26 180 54 210 54 16 15 14 15 16 60,000 66 139 66 28 26 60,000 " 139 66 28 26 225 60 14 15 60,000 126 66 32 32 7 255 60 16 15 60,000 " 126 66 32 32 300 66 16 15 60,000 136 66 32 32 375 72 16 15 60,000 66 125 66 32 32 6699888 44 33 46 33 2012 14 2,079 40 14 2,163 42 40 40 4 16 2,451 42 40 62 31/1 14 2,820 46 40 52 4 16 3,135 46 40 31,929 64 4 16 3,795 50 82 4 16 4,734 54 40 2888888 40 19,860 8,740 28,600 21,900 9,200 31,100 26,133 9,467 35,600 27,540 31,929 10,760 38,300 40 10,871 39,882 12,518 42,800 47,700 14,000 61,700 12,518 52,400 BOILERS. DIMENSIONS FOR TRIPLE SETTING.-PLAN K. SETTING. FURNACE. A B C E F G H I J K L M N O P Q R S T U V W X IN. FT IN. ST. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. FT. IN FT. IN FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. FT. IN. IN. IN. SQ. FT. 165 52 14 27 8 7 7 6 3 9 19 0 22 6 12 3 9 26 46 3 14 9 4 12 6 5 0 4 24 05 6 5 4 29 6 17 49.98 180 54 14 27 8 7 7 6 3 9 19 0 23 0 12 5 10 2 6 5 6 4 13 8 4 12 5 5 0 4 2 4 4 5 6 5 62 7 6 17 54.15 210 54 16 27 9 10 8 6 3 9 21 0 23 0 12 5 10 2 6 5 6 4 13 8 4 12 1 7 0 4 25 05 65 6111 6 17 62.49 225 60 14 33 8 7 7 6 4 7 19 0 24 6 13 6 10 2 6 8 6 7 13 8 4 12 3 5 0 4 25 25 6 6 02 3 255 60 16 33 9 10 8 6 4 7 21 0 24 6 13 6 10 2 6 8 6 7 17 12 4 12 2 7 0 4 2 5 300 66 16 33 9 10 8 6 5 0 21 0 26 0 14 4 11 2 6 11 6 10 18 13 4 7 2 5 6 375 72 16 33 9 10 9 10 8 8 6 5 6 21 0 27 6 14 7 11 2 7 2 7 1 16 11 410 22 4 5 6 6 6 4 8 5 4 6 0 6 6 4 6 3 411 6 0 6 37 0 02 6966 8175 17 64.56 17 66.66 17 74.64 17 88.50 212 A Library of Steam Engineering. Fig. 120 ==== 11 11 H 11 1 11 of tot 101 1017 Fig. 121 կ 미 ​D 0 미 ​0 Folger Cin PLAN AND ELEVATION OF LEFT HAND ENGINE WITH TUBULAR BOILERS. Specifications and Plans for Tubular Boilers. 213 TH HOLIGH D D 0 0 Fig. 122 Fig. 123 O 881 PLAN AND ELEVATION OF RIGHT HAND ENGINE WITH TUBULAR BOILERS. Folger CIN 214 A Library of Steam Engineering. BOILERS. MATERIALS FOR TUBULAR BOILER SETTINGS. SINGLE SETTING. DOUBLE SETTING. TRIPLE SETTING. IN. FT. 36 x 8 7,600 660 36 x 10 8,600 750 70 8 780 60 7호 ​3 396 11,850 1,300 95 12 42/4 780 16,000 1,900 130 16 6 CO 1,200 31 450 13,250 36 x 12 9,800 820 78 10 4 492 1,500 15,000 1,650 106 13 5 900 18,000 2,200 140 18 72 1,300 120 15 6 990 20,000 2,500 160 20 40 x 12 10,000 850 80 10 4 510 15,500 1,690 124 15/ 61 1,014 20,500 2,500 170 21 42 x 12 10,300 850 83 101/ 41 510 16,000 1,700 128 16 6 1,020 22,000 2,500 175 2225 8 1,500 9 8 | 1,500 1,550 46 x 12 11,000 960 88 11 42 576 17,300 1,920 138 17 7 1,152 23,000 2,800 185 24 9 1,700 48 x 12 11,200 1,000 90 111 412 600 17,700 2,000 142 172 7 1.200 24,000 3,000 190 25 9 1,800 52 x 14 12,300 1,160 98 1231/ 5 10 696 19,250 2,320 154 191 72 1,392 26,000 3,500 210 26 10 2,000 54 x 14 12,400 1,260 100 12 5 10 756 19,500 2,520 156 19 8 1,512 27,000 3,800 212 27 11 2,300 54 x 16 13,200 1,350 106 131 54 810 20,500 2,700 164 201 81 1,620 28,000 4,000 220 28 11 2,450 60 x 14 12,700 1,260 102 123 541 756 20,000 2,500 160 20 8 1,500 27,500 3,850 218 27 11 2,300 60 x 16 14,100 1,380 113 14 52 828 22,000 2,760 176 22 9 1,656 30,000 4,100 230 30 12 2,500 66 x 16 72 x 16 15,250 1,750 122 14,850 1,680 119 15 151 6 1,008 6 1,050 23,400 3,360 187 24,250 3,500 194 2,100 33,500 5,250 266 34 14 The estimates of materials in the table, for double and triple setting, provide for placing boilers independently with dividing wall between them. 232 92 244 10 2,016 2,016 32,000 5,000 250 32 13 3,000 3,200 Specifications and Plans for Tubular Boilers. 215 HOW TO DRAUGHT PLANS AND SPECIFICATIONS FOR TUBULAR BOILERS. Care must be taken to build the side walls so that the longitudinal seams will be protected against the fire. Therefore, those seams should always be high enough to allow them to be covered without destroying any of the effective heating surface of the boiler. A careful study of the plans for various styles of settings of boilers, together with the tables and specifications, will enable the engineer of ordinary skill to draught his own plans and specifications. A variety of plans are given for the purpose of enabling engineers who desire to draught their own plans and specifications to select whichever plan may be found most suitable for their purpose. After having determined upon the plan desired, place a piece of tracing paper over the plan selected, and trace the different figures upon the tracing paper, omitting the letters. When completed, select from the proper tables, the figures that should be placed in the spaces occu- pied by the letters in the plan and transfer those figures to their proper places in the tracing, and when the places in the tracing, now occupied by letters in the plan, have been filled with the proper figures, the plan will be complete. By comparing the letters in each of the plans with those at the head of each of the columns in the table belonging to the plan selected, the figures in the table will be readily understood. Before transferring figures to the tracing paper containing the plan selected, the length and diameter of the boiler must first be determined. Then run down the columns A and B until the size of boiler selected is reached, and then transfer the length and diameter to the tracings, taking the plan from which the tracing was made as a guide as to where to place the figures. Having placed the length and diameter in the proper places on the tracing, proceed to transfer the figures in the succeeding columns to the right of the columns containing the length and diameter selected, and continue on the same line to the right until all of the figures represent- ing dimensions have been transferred to the tracing. The next thing to be done is to determine the tensile strength of the plate per square inch of section of which the boiler is to be constructed. As good material for boilers can be made of homoge- neous steel, ranging in tensile strength from 55,000 to 70,000 pounds per square inch, and of iron ranging from 50,000 to 60,000 pounds per square inch, no difficulty need be experienced in making the proper selection, if care is taken to prescribe the proper ductility for the metal selected. Homogeneous steel plates having a tensile strength of from 55,000 to 65,000 pounds per square inch, and iron plates having a ten- sile strength of from 50,000 to 60,000 pounds per square inch, are con- sidered better than boiler material having greater or less tensile strength. 216 A Library of Steam Engineering. But whatever tensile strength between the highest and lowest limit named is selected, steel should have a ductility of not less than 50 per cent., and iron plates should have not less than 25 per cent. This means that the test piece, when broken in a testing machine, should show a contraction of area at point of fracture of not less than 50 per cent. for steel, and not less than 25 per cent. for iron. Having determined upon the diameter and length of boiler; upon the thickness and tensile strength of material; and upon the number of boilers, the engineer is prepared to proceed with the draughting of his specifications. As a guide to construction of boilers, the kind of equipment to provide, attention is called to the specifications at the beginning of this chapter. These, however, can be varied according to circumstances, as they may arise. But the vital points enumerated should be strictly adhered to-that is tensile strength, ductility and thickness of material, and drilling all rivet holes instead of punching, and double riveting all longitudinal seams in every case where high pressure is desired. In other words, whenever a pressure is required to be carried in a boiler, which will produce a strain on the plates exceed- ing one-sixth of the tensile strength of the plates in the shell, it should be expressly stipulated in the specifications that all rivet holes for the boilers shall be fairly drilled, and that the longitudinal seams shall be double riveted. STRAIN ON LONGITUDINAL SEAMS. TO DETERMINE THE STRAIN ON EACH LONGITUDINAL INCH PRODUCED BY ANY GIVEN PRESSURE IN A BOILER. RULE.-Multiply the given pressure per square inch by one-half of the diameter of the boiler in inches, and the product will give the total strain on each inch in the length of the boiler. Example.-Let 150 pounds equal given steam pressure per square inch. Let 40 inches equal given diameter of the boiler. Then we have: 150 40÷2 20 3000 lbs. Tending to separate the material. If the boiler was made of hoops 40 inches in diameter, and 1 inch in width, there would be a strain of 3000 pounds on all parts of each hoop tending to tear it asunder. Now, suppose that the metal in the hoop was 25 one hundredths of an inch in thickness, and had a tensile strength of 60,000 pounds per square inch, what would be the actual strength of a strip of that material 1 inch in width and 25 one hundredths of an inch in thick- ness? Specifications and Plans for Tubular Boilers. 217 To determine the actual strength, the thickness of the material in decimals of an inch is multiplied by the tensile strength in pounds per square inch, and the product will give the actual strength of the material. Thus: .25 60000 15000.00 lbs. Actual strength of a strip 1 inch in width and 25 one hundredths of an inch in thickness. Therefore, if it is required to ascertain whether or not the longitu- dinal seams ought to be double riveted and all rivet holes drilled, we first divide the above answer by 6, and the quotient will give the total strain that ought to be allowed on each longitudinal inch in the boiler for single-riveted longitudinal seams. Thus: 6)15000 2500 lbs. Maximum strain for single riveting. Now, if we multiply the required steam pressure per square inch by one-half the diameter of the boiler, and find that the product is greater than the quotient in the above example, it will show that the longitudinal seams should be double riveted and all rivet holes in the boiler drilled. Performing the operation, we have: 150 Pressure per square inch. 20 One-half diameter of boiler. 3000 lbs. Actual strain per longitudinal inch in the boiler. This demonstrates that single riveting in longitudinal seams will not do. At the same time it may show that even double riveting will not do. If the actual strength of a strip of boiler plate be divided by 5, the quotient will give the total strain allowed on each longitudinal inch for double riveting. Thus: 5) 15000 3000 lbs. One-fifth of actual strength of plate. In this case, however, it is shown that double-riveted longitudinal seams will enable the boiler (40 inches in diameter, 60,000 pounds ten- sile strength of material, and 25 one hundredths of an inch thickness of material) to carry a safe-working pressure of 150 pounds per square inch. But if the quotient, as above shown, should be less than the product of the pressure multiplied by one-half of the diameter of the boiler, one of three things will have to be done if the safe-working pressure is to be a fixed amount, as in the above case; either the ten- sile strength of the material be required to be greater, or the thickness of the material be required to be greater, or else a smaller diameter of boiler will have to be selected. 218 A Library of Steam Engineering. DIAME R OF BOILER. TO DETERMINE THE REQUIRED DIAMETER OF BOILER. RULE.-Multiply the giver thickness of plate in decimals of an inch by the constant 192, and the product will give required diameter of boiler in inches. Example.-Let 375 one thousandths of an inch equal thickness of material. Let 192 equal a constant. Then we have: .375 192 Constant for double-riveted plate, 60,000 tensile strength. 750 33 75 375 72.000 inches. Required diameter of boiler. HORSE POWER OF BOILER. In practice, the number of horse powers required in a tubular boiler determines its diameter. For each horse power required the boiler should contain not less than 12 square feet of effective heat- ing surface. Two-thirds of the shell may be exposed to the heated gases of the furnace and be counted as effective heating surface, and two-thirds of the tube surface may also be computed as effective. Therefore, as capacity to do a given or a required amount of work is the first requisite in a boiler, it follows that pressure, diameter and length, with the necessary number of tubes, must be determined before tensile strength and thickness of boiler plate can be determined. THICKNESS OF PLATE IN SHELL FOR SINGLE RIVETING. TO DETERMINE THE THICKNESS OF PLATE FOR SHELL OF BOILER WITH SINGLE-RIVETED LONGITUDINAL SEAMS. RULE.-First, multiply the required steam pressure per square inch by the required diameter of the boiler in inches, then multiply the product by the constant 3, and call the last product "Product No. 1." Second, divide "Product No. 1" by the required tensile strength of plate per square inch, and the quotient will give the required thickness of plate for single-riveted longitudinal seams. Example.-Let 125 pounds equal required steam pressure per square inch. Let 72 inches equal required diameter of boiler. Let 3 equal a constant for single riveting. Let 60,000 pounds equal required tensile strength of plate per square inch. Then we have: 125×72X3 60000 --.45 Decimals of an inch. Thickness of plate required. Specifications and Plans for Tubular Boilers. 219 Performing the operation, we have: 125 Steam pressure in pounds. 72 Diameter of boiler in inches. 72 Diam 250 875 9000 3 A constant for single-riveted longitudinal seams. 27000 "Product No. 1." Next, dividing "Product No. 1" by the tensile strength of plate per square inch, we have: 60000) 27000.0(.45 Decimals of an inch. Thickness of plate 24000 0 3000 00 required. 3000 00 THICKNESS OF PLATE IN SHELL FOR DOUBLE RIVETING. TO DETERMINE THE THICKNESS OF PLATE FOR SHELL OF BOILER WITH DOUBLE-RIVETED LONGITUDINAL SEAMS. RULE.-First, multiply the required steam pressure per square inch by the required diameter of the boiler in inches, then multiply the product by the constant 2.5, and call the last product "Product No. 1." Second, divide "Product No. 1" by the required tensile strength of plate per square inch, and the quotient will give the required thickness of plate for double-riveted longitudinal seams. Example.-Let 125 pounds equal required steam pressure per square inch. Let 72 inches equal required diameter of boiler. Let 2.5 equal a constant for double riveting. Let 60,000 pounds equal required tensile strength of plate per square inch. Then we have: 125×72X2.5 60000 =.375 Decimals of an inch. Thickness of plate required. Performing the operation, we have: 125 Steam pressure in pounds. 72 Diameter of boiler in inches. 250 875 9000 2.5 A constant for double-riveted longitudinal 45000 18000 seams. 22500.0 "Product No. 1." 220 A Library of Steam Engineering. Next, dividing "Product No. 1" by the tensile strength of plate per square inch, we have: 60000) 22500.0 (.375 Decimals of an inch. Thickness 18000 0 of plate required. 4500 00 4200 00 300 000 300 000 TENSILE STRENGTH OF BOILER PLATE FOR SINGLE RIVETING. TO DETERMINE THE TENSILE STRENGTH OF PLATE FOR SHELL OF BOILER WITH SINGLE-RIVETED LONGITUDINAL SEAMS. RULE.-First, multiply the required steam pressure per square inch by the required diameter of the boiler in inches, then multiply the product by the constant 3, and call the last product "Product No. 1." Second, divide "Product No. 1" by the required thickness of plate in decimals of an inch and the quotient will give the required tensile strength, per sectional square inch of plate for single-riveted longitudinal seams. Example.-Let 125 pounds equal required steam pressure per square inch. Let 72 inches equal required diameter of boiler. Let 3 equal a constant for double riveting. Let 375 one thousandths of an inch equal required ten- sile strength per square inch of plate. Then we have: 125×72×3 .375 -72000 lbs. Required tensile strength per square inch of plate. Performing the operation, we have:. 125 Steam pressure in pounds. 72 Diameter of boiler in inches. 250 875 9000 3 A constant for single-riveted longitudinal 66 seams. 27000 Product No. 1." Next, dividing "Product No. 1" by the thickness of plate in dec- imals of an inch, we have: .375) 27000.000 (72000 lbs. Required tensile strength per square inch of plate. 2625 750 750 000 Specifications and Plans for Tubular Boilers. 221 TENSILE STRENGTH OF BOILER PLATE FOR DOUBLE RIVETING. TO DETERMINE THE TENSILE STRENGTH OF PLATE FOR SHELL OF BOILER WITH DOUBLE-RIVETED LONGITUDINAL SEAMS. - RULE. First, multiply the required steam pressure per square inch by the required diameter of the boiler in inches, then multiply the product by the constant 2.5, and call the last product "Product No. 1." Second, divide "Product No. 1" by the required thickness of plate in dec- imals of an inch, and the quotient will give the required tensile strength of plate per square inch for double-riveted longitudinal seams. Example. Let 125 pounds equal given pressure per square inch. Let 72 inches equal given diameter of boiler. Let 2.5 equal a constant. Let 375 one thousandths of an inch equal required thick- ness of plate. Then we have: 125X72X2.5 .375 -60000 lbs. Required tensile strength per square inch of plate. Performing the operation, we have: 125 72 250 875 9000 2.5 4500 0 18000 22500.0 "Product No. 1." Then, dividing "Product No. 1" by the thickness of plate, we have: .375) 22500.000 (60000 lbs. 2250 0 000 Then we have the following data for a boiler: Required tensile strength per square inch of plate. SINGLE-RIVETED LONGITUDINAL SEAMS. 125 pounds pressure per square inch. 72 inches diameter of boiler. 45 one hundredths of an inch thickness of plate. 60,000 pounds tensile strength per square inch of plate. 222 A Library of Steam Engineering. Or, we have: 125 pounds pressure per square inch. 72 inches diameter of boiler. 375 one thousandths of an inch thickness of plate. 72,000 pounds tensile strength of material. In the first set of items it will be observed that the material is too thick, and in the second, that the tensile strength of the material is too great, therefore the holes in the boiler should be drilled and the longitudinal seams double riveted. Hence we select the following data for specifications: DOUBLE-RIVETED LONGITUDINAL SEAMS. 125 pounds pressure per square inch. 72 inches diameter of boiler. 375 one thousandths of an inch thickness of plate. 60,000 pounds tensile strength per square inch of plate. The thickness of plate above selected is the least thickness that should be used. Hence, the rule requiring the diameter of the boiler in inches to be divided by the constant 192, gives the minimum thick- ness of plate that should be used for high-pressure boilers; and dividing the diameter of the boiler in inches by the constant 160, will give the maximum thickness that should be used in boilers to which the heat is applied to the outside of the shell. In no case, however, should the thickness of material in the shell of such boilers exceed inch. Even then, with very hot furnace fires, laps in seams exposed to the fire are liable to burn away and endanger the safety of the boiler. MAXIMUM THICKNESS OF BOILER PLATE. TO DETERMINE THE MAXIMUM THICKNESS OF BOILER PLATE FOR BOILERS TO WHICH THE HEAT IS APPLIED TO THE OUTSIDE OF THE SHELL. RULE.-Divide the diameter of the boiler in inches by the constant 160, and the quotient will give the maximum thickness of material in decimals of an inch. Example.-Let 40 inches equal diameter of the boiler. Let 160 equal a constant.. Then we have: 160) 40.0 (0.25 inch. Maximum thickness of material required. 32.0 800 8.00 Evaporative Tests of Steam Boilers. 223 CHAPTER XI. EVAPORATIVE TESTS OF STEAM BOILERS. HEAT UNITS. In entering upon the study of this subject, the student should familiarize himself with the meaning of the term "heat unit," as employed in connection with evaporative and calorimeter tests. Many text-book writers seem to regard temperature and heat units as synony- mous terms, and use them interchangeably. That is a grave mistake, as there is no similarity between them, because the heat units represent the amount of heat imparted to water or steam in raising either to a certain temperature as shown by a thermometer. For example, it requires 212.9 heat units to raise one pound of water from zero to 212° Fahrenheit, and 965.7 heat units additional to evaporate it to steam at a temperature of 212° Fahrenheit. So while the thermometer registers the sensible heat it does not register the work done or the number of heat units expended in doing the work. In evaporating one pound of water to steam from and at a temperature of 212° Fah- renheit, the thermometer remains stationary in registering 212°; yet in performing the work of evaporating the one pound of water at 212°, 965.7 units of heat are expended; add to that the 212.9 heat units required to raise the water to 212° Fahrenheit from zero, and we have a total of 1178.6 heat units expended in raising water from zero to 212° Fahrenheit, and evaporating it to steam at that temperature. Then, the heat units required to evaporate water to steam from and at any temperature of the water is called latent heat, or heat of vap- orization. The heat unit has been adopted as a standard for measuring work performed by the application of heat, and it represents the amount of heat required to raise one pound of water one degree of the Fahrenheit scale from or at a temperature of 32° Fahrenheit. It will be seen, by reference to the table of saturated steam, that when the temperature of 212° Fahrenheit has been reached, that the total heat units in one pound of steam increase at the rate of 305 one thousandths of a unit for every degree of temperature above 212° Fahrenheit. 224 A Library of Steam Engineering. HEAT UNITS IN STEAM. TO DETERMINE THE TOTAL HEAT UNITS IN ONE POUND OF STEAM EVAPORATED FROM WATER HAVING A TEMPERA- TURE OVER 212° FAHRENHEIT. RULE.-Subtract the temperature, 212° from the indicated temperature and multiply the remainder by 305 one thousandths, and then add the total number of heat units in one pound of steam at 212° (1178.6) to the product, and the sum will give the total heat of steam for any temperature above 212°. Example.-Let 331.805° equal indicated temperature in Fahrenheit degrees. Let 212° equal temperature of water at boiling point in Fahrenheit degrees. Let 305 one thousandths of a heat unit equal increase for each degree above 212° Fahrenheit. Let 1178.6 equal total heat units in one pound of steam at 212° Fahrenheit. Then we have: [(331.805-212) X.305]+1178.6=1215.14+ Total number of heat units Performing the operation, we have: in steam from water at 331.805° Fahrenheit. 331.805 Indicated temperature or sensible heat. 212 119.805 .305 Increase in heat units for each degree of temperature above 212° 599025 35 9415 36.540525 1178.6 1215.140525 Total heat units in steam at a tempera- ture of 331.805° Fahrenheit. FACTOR OF EVAPORATION. TO DETERMINE THE FACTOR OF EVAPORATION. RULE.-Subtract the number of heat units in one pound of the feed water due the temperature of that water, from the total number of heat units in one pound of steam due the temperature or pressure of steam under which the test is made, then divide the remainder by 965.7, and the quotient will give the factor of evaporation. Evaporative Tests of Steam Boilers. 225 Example. Let 1225.6417 equal total heat units according to the table in one pound of steam at 366.232° Fahrenheit, or at 151.304 pounds pressure by gauge. Let 62.011 equal heat units in feed water at 62° Fahrenheit. Let 965.7 equal heat required to evaporate one pound of water from and at 212° Fahrenheit. Then we have: 1225.6417-62.011 965.7 1.2+ Factor of evaporation. Performing the operation, we have: 1225.6417 62.011 965.7000) 1163.6307 (1.2+ Factor of evaporation. 965 7000 197 93070 193 14000 It must be borne in mind that the factor of evaporation varies with the temperature of the steam or temperature of the feed water; therefore, in order to reduce evaporative tests to the standard of from and at 212° Fahrenheit, the factor of evaporation must be obtained from the total heat units in the steam and the feed water, as shown during the test. Then when the factor of evaporation has been deter- mined, the number of pounds of water evaporated per pound of com- bustible matter in the coal, multiplied by the factor of evaporation, will give the number of pounds of water per pound of combustible matter that would have been evaporated had the feed water been injected into the boiler at a temperature of 212° Fahrenheit, and evaporated at that temperature. It will be observed that before the factor of evaporation can be determined, the total heat units in the steam per pound must first be ascertained, and that can readily be done by reference to the table of saturated steam. Take the average steam pressure per square inch, registered by the steam gauge during the test, and find in the table the total number of heat units in steam due that pressure, and subtract the heat units in one pound of water due the average temperature of the feed water from the total heat units in the steam per pound due the pressure under which the test was made, and divide the remainder by the con- stant 965.7, the quotient will give the factor of evaporation according to the rule previously given. 15 226 A Library of Steam Engineering. EVAPORATION FROM AND AT 212° FAHRENHEIT. TO DETERMINE THE EQUIVALENT EVAPORATION OF WATER TO STEAM, FROM AND AT A TEMPERATURE OF 212° FAHRENHEIT. RULE.-Subtract the heat units in one pound of water due the average temperature of the feed water injected into the boiler during the test, from the total heat units in one pound of steam due the average pressure per square inch shown by the steam gauge during the test, then divide the remainder by the constant 965.7, and the quotient will give the factor of evaporation. Next multiply the factor of evaporation by the number of pounds of water evapo- rated per pound of combustible matter in the coal, and the product will give the number of pounds of water per pound of combustible matter, that would have been evaporated had the feed water been injected into the boiler at a temperature of 212° Fahrenheit, and evaporated at that temperature. Example.-Let 1225.6417 equal total heat units (as shown by table) in one pound of steam at 366.232° Fahrenheit, or at 151.304 pounds gauge pressure. Let 62.011 equal heat units in feed water at 62° Fahrenheit. Let 965.7 equal heat units required to evaporate one pound of water from and at 212° Fahrenheit. Let 7 pounds equal weight of water evaporated per pound of combustible matter in the coal. Then we have: 1225.6417-62.011 965.7 X7-8.4+ Pounds of water per pound of combustible that would have been evaporated at a temperature of 212° from feed water at 212° Fahren- heit. Performing the operation, we have: 1225.6417 Heat units in steam. 62.011 Heat units in feed water. 965.7000) 1163.6307 (1.2+ Factor of evaporation. 965.7000 197 93070 193 14000 Then, multiplying the factor of evaporation by the number of pounds of water evaporated per pound of combustible matter in the coal, we have: 1.2 7 8.4 Pounds of water that would have been evapo- rated per pound of combustible if the feed water had had a temperature of 212° Fahrenheit and had been evaporated at that temperature. Evaporative Tests of Steam Boilers. 227 MOISTURE AND COMBUSTIBLE MATTER IN COAL-HOW DETERMINED. When evaporative tests are made it is usual to base the calcula- tions upon the combustible matter in the coal, in order to determine the efficiency of the boiler. Ordinarily this is done by placing in a bag exactly 100 pounds of coal of average moisture, taken from different parts of the pile from which the coal for making the evaporative test is to be taken. The weight of the coal must be exactly 100 pounds, exclusive of the weight of the bag in which it is placed. The bag is then tied up and placed in a suitable place during the test to thor- oughly dry the coal. After completing the test, the bag of coal is weighed again, and the number of pounds the coal weighs less than 100 represents the percentage of moisture in the coal. For example, if it is one pound, that is one per cent.; if it is two pounds, that is two per cent.; and so on. If it is one per cent., one per cent. is deducted from the total weight of the coal used during the test; if it is two per cent., two per cent. is deducted from the total weight of the coal used; and so on. After deducting the moisture, the weight of the ashes and clinkers is then deducted from the remainder, and that which then remains is termed combustible matter. The total number of pounds of water evaporated is then divided by the number of pounds of combustible matter in the coal consumed during the trial, and the quotient is taken as the number of pounds of water evaporated per pound of combustible matter. By dividing the total number of pounds of water evaporated by the total number of pounds of coal consumed during the trial, the quotient will give the number of pounds of water evaporated per pound of coal. Then, as the feed water is injected into boilers at various tempera- tures, and as the combustible matter in different coals varies materially, evaporative tests are usually reduced to the standard evaporation, as from and at 212° Fahrenheit. That is supposing the feed water to have been injected into the boiler at 212° Fahrenheit, and evaporated at that temperature. As a further illustration, let it be supposed that the evaporative test was made under an average pressure per square inch of 151.304 pounds, as shown by the steam gauge. By reference to the table, it will be found that the total heat units in one pound of steam at that pressure is 1225.6417, which means that it required 1225.6417 heat units to raise one pound of water from zero to 151.304 pounds pressure per square inch and evaporate it into steam at that pressure. Now, let it be supposed that the feed water had a temperature of 62° Fahrenheit, and to be absolutely correct, by reference to the table of properties of water it will be found that water at a temperature of 62° Fahrenheit contains 62.011 heat units; therefore, the number of heat 228 A Library of Steam Engineering. units required to evaporate water from a temperature of 62° Fahren- heit would be 62.011 less than it would be from zero. Hence, as the total heat units required to evaporate one pound of water from zero at a pressure by gauge of 151.304 pounds per square inch is 1225.6417, the heat units required to evaporate one pound of water under the same pressure from a temperature of 62° Fahrenheit would be 1225.6417 -62.011-1163.6307 heat units. Now, as it requires 1178.6 heat units to evaporate one pound of water from zero, at 212° Fahrenheit, and as it requires 212.9 heat units to raise water from zero to 212° Fahrenheit, it requires 212.9 heat units less to evaporate it to steam from and at 212° Fahrenheit than it does from zero. Hence, 1178.6-212.9-965.7 heat units required to evaporate one pound of water from and at a temperature of 212° Fahrenheit. Therefore, as it requires 1163.6307 heat units to evaporate one pound of feed water injected into a boiler at 62° Fahrenheit, under a pressure of 151.304 pounds by gauge, we multiply the number of heat units, 1163.6307, by the number of pounds of water actually evaporated per pound of combustible, and divide the product by 965.7, the num- ber of heat units required to evaporate one pound of water from a temperature of 212° Fahrenheit to steam at that temperature, and the quotient will give the number of pounds of water that would have been evaporated had the feed water been injected into the boiler at 212° Fahrenheit and evaporated at that temperature. Example. Let 1163.6307 equal heat units required to evaporate one pound of water from 62° Fahrenheit. Let 7 pounds equal weight of water actually evaporated per pound of combustible. Let 965.7 equal number of heat units required to evap- orate one pound of water from and at 212° Fahrenheit. Then we have: 1163.6307X7 965.7 Performing the operation, we have: 8.4+ Pounds of water per pound of combustible that would have been evaporated from and at a temperature of 212° Fahren- heit. 1163.6307 7 965.7000) 8145.4149 (8.4+ 7725 6000 419 81490 Pounds of water per pound of combustible that would have been evaporated from and at a temperature of 212° Fahrenheit. 386 28000 Now that the subject has been fully explained, the rule can be given in a brief and simplified form. Evaporative Tests of Steam Boilers. 229 REDUCTION OF TESTS TO STANDARD EVAPORATION OF FROM AND AT 212° FAHRENHEIT. TO REDUCE THE EVAPORATIVE TESTS TO THE STANDARD EVAPORATION OF FROM AND AT 212° FAHRENHEIT. RULE.-Subtract the heat units in the feed water due its temperature from the total heat units in the steam due the temperature of the steam, then divide the remainder by 965.7, and multiply the quotient by the actual num- ber of pounds of water evaporated per pound of combustible matter in the coal, and the product will give the evaporation as from and at 212° Fah- renheit. Taking feed water at an average temperature of 62° Fahrenheit, by reference to the table of properties of water, it will be found to con- tain 62.011 heat units per pound of water. Taking the average press- ure of the steam during the test at 151.304 pounds per square inch by the steam gauge, by reference to the table of properties of saturated steam, it will be found that the temperature due that pressure is 366.232° Fahrenheit, and that one pound of steam at that tempera- ture contains a total of 1225.6417 heat units, as shown by the table. Example.-Let 1225.6417 equal total heat units in steam at a press- ure of 151.304 pounds per square inch. Let 62.011 equal heat units in feed water at a tempera- ture of 62° Fahrenheit Let 965.7 equal heat units required to evaporate one pound of water from and at 212° Fahrenheit. Let 7 pounds equal weight of water actually evaporated per pound of combustible matter in the coal. Then we have: 1225.6417-62.011 965.7 Performing the operation, we have: |x7=8.4+ X7-8.4+ Pounds of water evapo- rated as from and at 212° Fahrenheit. 1225.6417 62.011 965.7000) 1163.6307 (1.2+ 965 7000 7 197 93070 8.4+ Pounds of water evaporated 193 14000 as from and at 212° Fah- renheit. NOTE.-Heat units are usually reckoned from 32° Fahrenheit, but for greater simplicity they have been taken at zero Fahrenheit, upon which the foregoing calculations are based. The tables of properties of saturated steam and properties of water are arranged upon the same basis. 230 A Library of Steam Engineering. Temp- erature in Number in Pounds as Fahrenheit of Units of Required for TABLE OF PROPERTIES OF SATURATED STEAM. PRESSURE PER SQUARE INCH. Total Pressure Pressure in Pounds NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Number of Units of Heat Weight of One Shown by Degrees. Heat in Evaporation, from a Steam Vacuum Gauge. Water. Called Latent Heat. Total Number of Units of Heat Contained in Steam. Cubic Foot of Steam in Number of Cubic Feet of Steam from One Decimals of a Pound. Cubic Foot of Water. 12345678 102. 102.086 1042.964 1145.050 .0030 20620.0 126.266 126.440 1026.010 1152.450 .0058 10720.0 141.622 141.877 1015.254 1157.131 .0085 7326.0 153.070 153.396 1007.229 1160.625 .0112 5600.0 162.330 162.722 170.123 170.577 1000.727 995.249 176.910 177.425 990.471 182.910 183.481 986.245 1169.726 1163.449 1165.826 .0163 3814.0 1167.896 .0189 .0214 .0137 4535.0 3300.0 2910.0 9 188.316 188.941 982.434 1171.375 .0239 2607.0 10 193.240 193.919 978.958 1172.877 .0264 2360.0 11 197.768 198.496 975.762 1174.258 .0289 2157.0 12 201.960 202.737 972.800 1175.537 .0313 1988.0 13 205.885 206.709 970.025 1176.734 .0337 1846.0 14 209.560 210.428 967.427 1177.855 .0362 1722.0 14.7 212.000 212.900 965.700 1178.600 .03797 1644.0 15 .304 213.025 213.939 964.973 1178.912 .0387 1612.0 16 1.304 216.296 217.252 962.657 1179.909 .0413 1514.0 17 2.304 219.410 220.409 960.450 1180.859 .0437 1427.0 18 3.304 222.378 223.419 958.345 1181.764 - .0462 1350.6 19 4.304 225.203 226.285 956.343 1182.628 .0487 1282.1 20 5.304 227.917 229.039 954.415 1183.454 .0511 1220.3 21 6.304 230.515 231.676 952.570 1184.246 .0536 1164.4 22 7.304 233.017 234.218 950.791 1185.009 .0561 1113.5 23 8.304 235.432 236.672 949.072 1185.744 .0585 1066.9 24 9.304 237.752 239.029 947.424 1186.453 .0610 1024.1 25 10.304 240.000 241.314 945.825 1187.139 .0634 984.8 26 11.304 242.175 243.526 944.277 1187.803 .0658 948.4 27 12.304 244.284 245.671 942.775 1188.446 .0683 914.6 28 13.304 246.326 247.748 941.321 1189.069 .0707 883.2 29 14.304 248.310 249.769 30 15.304 250.245 251.738 939.905 1189.674 .0731 938.925 1190.263 .0755 854.0 826.8 31 16.304 252.122 253.648 937.1878 1190.8358 .0779 801.2 32 17.304 253.952 255.512 935.8818 1191.3938 .0803 777.2 33 18.304 255.735 257.329 934.6088 1191.9378 .0827 754.7 34 19.304 257.476 259.103 933.3658 1192.4688 .0851 733.5 35 36 37 38 39 20.304 21.304 260.835 262.527 22.304 23.304 264.045 265.801 24.304 265.599 267.386 259.176 260.835 932.1523 1192.9873 .0875 713.4 930.9668 1193.4938 .0899 694.5 262.458 264.182 929.8068 1193.9888 .0922 676.6 928.6718 1194.4728 .0946 659.7 927.5608 1194.9468 .0970 643.6 40 41 42 43 28.304 44 29.304 25.304 267.120 268.938 26.304 268.611 270.460 27.304 270.073 271.954 271.507 273.417 272.915 274.855 926.4728 1195.4108 .0994 628.2 925.4058 1195.8658 .1017 613.4 924.3578 1196.3118 .1041 599.3 923.3323 1196.7493 .1064 586.1 922.3238 1197.1788 .1088 573.7 45 30.304 274.296 276.266 921.3343 1197.6003 .1111 561.8 46 31.304 275.652 277.651 920.3632 1198.0142 .1134 550.4 47 32.304 276.986 279.016 919.4052 1198.4212 1158 539.5 48 33.304 278.297 280.355 918.4662 1198.8212 .1181 529.0 49 34.304 279.585 281.672 917.5422 1199.2142 .1204 518.6 50 35.304 280.854 282.969 916.6316 1199.6006 .1227 508.5 51 36.304 282.099 284.243 915.7377 1199.9807 .1251 499.1 52 37.304 283.326 285.499 914.8557 1200.3547 .1274 490.1 Evaporative Tests of Steam Boilers. 231 Temp- erature in Number TABLE OF PROPERTIES OF SATURATED STEAM-Continued. PRESSURE PER SQUARE INCH. Total Pressure Pressure in Pounds NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Weight of One Number of Cubic Feet of Steam in Fahrenheit of Units of as Pounds Shown by Degrees. Heat in Number of Units of Heat Required for Evaporation, from a Steam Vacuum Gauge. Water. Called Latent Heat. Total Number of Units of Heat Contained in Steam. Cubic Foot of Steam in Decimals of a Pound. from One Cubic Foot of Water. 53 38.304 284.534 286.736 913.9871 1200.7231 .1297 481.4 54 39.304 285.724 287.952 913.1340 1201.0860 .1320 472.9 55 40.304 286.897 289.153 912.2906 1201.4436 .1343 464.7 56 41.304 288.052 290.335 911.4611 1201.7961 .1366 457.0 57 42.304 289.112 291.503 910.6407 1202.1437 .1388 449.6 58 43.304 59 44.304 290.316 292.654 291.425 293.790 909.8325 909.0346 1202.8246 1202.4865 .1411 442.4 .1434 435.3 60 45.304 292.520 294.911 908.2472 1203.1582 .1457 428.5 61 46.304 293.598 296.016 907.4713 1203.4873 .14793 422.0 62 47.304 294.663 297.108 906.7042 1203.8122 .15021 415.6 63 48.304 295.714 298.185 905.9477 1204.1329 .15248 409.4 64 49.304 296.752 299.249 905.2005 1204.4495 .15471 403.5 65 50.304 297.777 300.300 904.4621 1204.7621 .15697 397.7 66 51.304 298.789 301.338 903.7327 1205.0707 .15921 392.1 67 52.304 299.789 302.364 903.0116 1205.3756 .16147 386.6 68 53.304 300.776 303.377 902.2999 69 54.304 301.753 304.380 901.5947 70 55.304 302.718 305.370 900.8991 71 56.304 303.673 306.350 900.2101 72 57.304 304.617 307.320 899.5280 73 58.304 305.551 308.279 898.8537 74 59.304 306.474 309.228 898.1863 75 60.304 307.388 310.166 897.5269 76 61.304 77 62.304 309.184 78 63.304 79 64.304 308.290 311.092 312.011 310.069 312.920 310.945 313.821 896.8764 896.2301 895.5910 894.9571 80 65.304 311.812 314.712 894.3304 81 66.304 312.670 315.595 893.7092 82 67.304 313.520 316.468 893.0954 83 84 85 70.304 86 71.304 87 72.304 89 74.304 319.249 68.304 314.361 317.333 69.304 315.195 318.190 316.021 319.040 316.839 319.882 317.650 320.717 73.304 318.453 321.543 322.362 892.4871 891.8843 891.2862 1205.6769 .16372 381.3 1205.9747 .16598 376.1 1206.2691 .16817 371.2 1206.5601 .17038 366.4 1206.8480 .17259 361.7 1207.1327 .17481 357.1 1207.4143 .17704 352.6 1207.6929 .17923 348.3 1207.9684 .18142 344.1 1208.2411 .18360 340.0 1208.5110 .18579 336.0 1208.7781 .18797 332.1 1209.0424 .19015 328.3 1209.3042 .19232 324.6 1209.5634 .19454 1209.8201 .19674 317.3 1210.0743 .19887 313.9 1210.3262 .20105 310.5 320.9 90 75.304 320.039 323.176 91 76.304 320.821 323.981 92 77.304 321.597 324.781 887.2460 93 78.304 322.266 325.572 886.6896 94 79.304 95 80.304 323.884 96 81.304 323.128 326.358 327.136 324.634 327.909 886.1362 885.5887 97 82.304 325.378 328.675 885.0444 884.5052 98 83.304 326.114 329.433 883.9721 99 84.304 326.845 330.186 883.4421 100 85.304 101 327.571 330.935 86.304 328.291 331.678 102 87.304 329.005 332.414 882.9144 882.3909 881.8727 890.6938 1210.5758 .20321 890.1061 1210.8231 .20535 304.0 889.5251 1211.0681 .20753 300.8 888.9490 1211.3110 .20970 297.7 888.3758 1211.5518 .21183 294.7 887.8094 1211.7904 .21393 291.8 1212.0270 .21608 288.9 1212.2616 .21829 286.1 1212.4942 .22045 283.3 1212.7247 .22247 280.6 1212.9534 .22455 1213.1802 .22667 275.4 1213.4051 .22883 272.8 1213.6281 .23095 270.3 1213.8494 .23302 267.9 1214.0689 23510 265.5 1214.2867 .23717 263.2 307.2 278.0 103 88.304 329.714 333.145 881.3577 1214.5027 .23925 260.9 104 105 89.304 330.416 333.869 90.304 880.8481 1214.7171 .24132 258.7 331.113 334.587 880.3429 1214.9299 .24340 256.5 232 A Library of Steam Engineering. Temp- erature in Number Fahrenheit of Units of in as Pounds Shown by Degrees. from a Steam Water. Vacuum Gauge. 106 91.304 331.805 335.301 107 92.304 332.492 336.009 108 93.304 333.174 336.714 109 94.304 333.851 337.411 110 95.304 334.523 338.105 111 96.304 112 97.304 113 98.304 335.191 338.795 335.854 339.479 336.511 340.157 114 99.304 115 100.304 116 101.304 117 118 337.165 340.832 337.814 341.502 338.459 342.169 102.304 339.100 342.831 103.304 339.736 TABLE OF PROPERTIES OF SATURATED STEAM-Continued. PRESSURE PER SQUARE INCH. Total Pressure Pressure in Pounds NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Heat in Number of Units of Heat Required for Evaporation, Called Latent Heat. Total Number of Units of Heat Contained in Steam. of Cubic Feet of Weight of One Cubic Foot of Number Steam in Decimals Steam from One Cubic Foot of Water. of a Pound. 1215.1410 .24547 254.3 1215.3506 .24754 252.2 1215.5587 .24961 250.1 1215.7652 .25168 248.0 1215.9703 .25376 246.0 1216.1739 .25582. 244.0 1216.3760 .25788 242.0 1216.5768 .25994 240.1 1216.7762 .26199 238.2 879.8400 879.3416 878.8447 878.3542 877.8653 877.3789 876.8970 876.4198 875.9442 875.4721 875.0018 874.5352 343.488 874.0722 119 104.304 340.368 344.141 873.6120 120 105.304 340.995 344.789 873.1555 121 106.304 341.618 345.432 872.7027 122 107.304 342,238 346.073 872.2508 123 108.304 342.854 346.709 871.8027 124 109.304 343.466 347.343 871.3553 125 110.304 344.074 347.972 870.9118 126 111.304 344.678 348.596 870.4721 127 112.304 345.279 349.217 870.0342 1216.9741 .26405 236.3 1217.1708 .26611 234.5 1217.3662 .26816 232.7 1217.5602 1217.7530 .27224 229.3 1217.9445 .27428 227.6 1218.1347 .27628 226.0 1218.3238 .27828 224.4 1218.5117 .28027 222.8 1218.6983 .28227 221.2 1218.8838 .28426 219.7 1219.0681 .28625 218.2 1219.2512 .28824 216.7 .27020 231.0 128 113.304 345.876 349.835 869.5983 129 114.304 346.459 350.448 869.1663 130 131 132 117.304 116.304 346.227 115.304 347.059 351.059 868.7351 1219.4333 .29023 215.2 1219.6143 .29222 213.7 1219.7941 .29420 212.3 347.644 351.665 868.3079 352.267 867.8836 133 118.304 348.806 352.867 867.4601 134 119.304 349.382 353.463 867.0397 135 120.304 349.954 354.055 866.6223 1219.9729 .29618 210.9 1220.1506 .29816 209.5 1220.3271 .30013 208.1 1220.5027 .30209 206.7 -1220.6773 .30405 205.4 136 121.304 350.523 354.644 866.2068 137 122.304 351.089 355.230 866.7934 138 123.304 351.752 355.813 865.3820 139 124.304 352.211 356.392 864.9735 140 125.304 352.767 356.969 864.5661 141 126.304 353.319 357.541 864.1627 1220.8508 .30601 204.1 1221.0234 .30796 202.8 1221.1950 .30990 201.5 1221.3655 .31186 200.2 1221.5351 .31386 199.0 1221.7037 .31587 197.8 142 127.304 353.869 358.110 143 144 148 149 128.304 354.416 129.304 354.960 145 130.304 355.501 359.801 146 131.304 356.039 360.359 147 132.304 356.574 360.913 133.304 357.106 361.465 134.304 357.635 362.013 358.677 863.7613 863.3611 1221.8713 .31788 196.6 1222.0381 .31990 195.4 359.240 862.9640 862.5679 862.1740 861.7832 861.3934 861.0068 1222.2040 .32190 194.2 1222.3689 .32391 193.0 1222.5330 .32592 191.9 1222.6962 .32794 190.8 1222.8584 .32995 189.7 1223.0198 .33196 188.6 150 135.304 358.161 362.559 860.6213 151 136.304 358.683 363.100 860.2399 152 137.304 359.203 363.640 153 154 155 156 138.304 359.721 364.177 139.304 360.236 364.711 140.304 360.749 365.243 141.304 361.260 365.773 157 142.304 361.768 366.300 158 143.304 362.273 366.824 859.8588 859.4799 859.1031 858.7276 858.3533 1223.1803 .33400 187.5 1223.3399 .33580 186.4 1223.4988 33761 185.3 1223.6569 .33942 184.3 1223.8141 .34123 183.3 1223.9706 34304 182.3 1224.1263 .34485 181.3 857.9811 1224.2811 .34666 180.3 857.6112 1224.4352 .34847 179.3 Evaporative Tests of Steam Boilers. 233 Temp- erature in in as Pounds Shown by Degrees. Heat in from a Steam Vacuum Gauge. Water. 159 144.304 362.776 367.347 160 145.304 363.277 367.867 TABLE OF PROPERTIES OF SATURATED STEAM-Continued. PRESSURE PER SQUARE INCH. Total Pressure Pressure in Pounds NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Number Number of Units of Heat Fahrenheit of Units of Required for Evaporation, Called Latent Heat. Total Number of Units of Heat Contained in Steam. 1224.5885 Weight of One Cubic Foot of Steam Number of Cubic Feet of Steam in Decimals of a Pound. from One Cubic Foot of Water. 35028 178.3 1224.7410 .35209 177.3 857.2415 856.8740 161 146.304 363.774 368.383 856.5099 1224.8929 .35397 176.4 162 147.304 364.270 368.898 856.1461 1225.0441 .35585 175.5 163 148.304 364.764 369.410 855.7846 164 149.304 365.255 369.920 855.4243 165 150.304 365.744 370.428 855.0654 166 151.304 366.232 370.934 854.7077 167 152.304 366.717 371.438 854.3514 168 169 171 153.304 154.304 367.680 170 155.304 368.158 372.934 156.304 368.632 367.199 371.939 853.9974 372.437 853.6456 853.2942 373.427 852.9461 172 157.304 369.105 373.918 177 173 158.304 369.576 174 159.304 370.045 374.895 175 160.304 370.512 375.380 176 161.304 370.978 375.865 162.304 371.442 376.347 374.408 852.5995 852.2533 851.9094 851.5670 851.2239 1225.1946 .35773 174.6 1225.3443 .35961 173.7 1225.4934 .36149 172.8 1225.6417 .36337 171.9 1225.7894 .36525 171.0 1225.9364 .36714 170.1 1226.0826 .36903 169.2 1226.2282 .37092 168.4 1226.3731 .37272 167.6 1226.5175 .37452 166.8 1226.6613 .37632 166.0 1226.8044 .37812 165.2 1226.9470 .37992 164.4 1227.0889 .38172 163.6 850.8833 1227.2303 .38353 162.8 178 163.304 371.904 376.827 850.5441 1227.3711 .38534 162.0 179 164.304 372.364 377.305 850.2062 180 165.304 372.822 377.781 849.8698 181 166.304 373.275 378.255 849.5347 1227.5112 .38715 161.2 1227.6508 .38895 160.4 1227.7897 .39077 159.7 182 167.304 373.731 378.727 849.2011 183 168.304 374.183 379.197 848.8689 184 169.304 374.633 279.665 848.5380 185 170.304 375.081 380.131 848.2086 186 171.304 375.527 380.595 847.8805 187 172 304 375.971 381.056 847.5549 188 173.304 376.413 381.516 847.2297 189 174.304 376.853 190 175.304 377.291 381.974 846.9058 382.429 846.5844 191 176.304 377.727 382.883 846.2633 1227.9281 .39259 159.0 1228.0659 .39441 158.3 1228.2030 .39624 157.6 1228.3396 1228.4755 .39990 156.2 1228.6109 .40173 155.5 1228.7457 .40356 154.8 1228.8798 .40539 154.1 1229.0134 .40722 153.4 1229.1463 .40899 152.7 .39807 156.9 192 193 178.304 378.593 177.304 378.161 383.335 845.9437 383.785 845.6256 194 179.304 379.023 384.233 195 180.304 379.452 384.679 196 181.304 379.979 385.123 845.3089 844.9938 844.6801 197 198 199 200 205 206 209 210 182.304 380.305 385.567 183.304 380.729 386.008 184.304 381.152 185.304 381.573 386.887 201 186.304 381.992 387.324 202 187.304 382.410 387.760 203 188.304 382.827 388.194 204 189.304 383.242 388.627 190.304 191.304 384.066 389.485 207 192.304 384.475 389.912 208 193.304 384.883 390.337 194.304 385.288 390.759 195.304 385.671 391.179 844.3660 1229.2787 .41076 152.0 1229.4106 .41253 151.3 1229.5419 .41430 150.7 1229.6728 .41607 150.1 1229.8031 .41784 149.5 1229.9330 .41962 148.9 844.0543 1230.0623 .42140 148.3 386.449 843.7422 843.4326 1230.1912 .42318 147.7 1230.3196 .42496 147.1 843.1234 842.8148 1230.4474 .42667 146.5 1230.5748 .42838 145.9 842.5076 842.2010 1230.7016 .43009 145.3 1230.8280 .43180 144.7 383.655 389.057 841.8969 1230.9539 .43351 144.1 841.5942 841.2921 1231.0792 .43523 143.5 1231.2041 .43695 142.9 840.9914 840.6933 840.3967 1231.3284 .43866 142.3 1231.4523 1231.5757 .44039 141.8 .44211 141.3 234 A Library of Steam Engineering. CHAPTER XII. CALORIMETER TESTS. There is, perhaps, no operation connected with the science of steam engineering requiring more care and accuracy than that of mak- ing calorimeter tests. The rules and formulæ laid down in many excellent works on steam engineering are commendable for their accuracy, and in some cases, simplicity; yet few, if any, are calculated to reach the compre- hension or understanding of the student, unless he is proficient in the higher branches of mathematics. It will therefore be the purpose here to present this important subject in its utmost simplicity, in order that engineers of limited education may get the benefit of the science embraced therein. Evaporative tests, without the calorimeter tests, really prove but little, so far as the efficiency of a steam boiler is concerned. To deter- mine the real efficiency of a steam boiler the quality of the steam generated must be ascertained, as well as the quantity of water evapo- rated per pound of coal, or per pound of combustible. Without the calorimeter test the faulty constructed boiler may, if judged simply from the amount of water evaporated, show greater efficiency than the properly constructed boiler; because the imperfect or faulty boiler is liable to carry off more water with the steam than the properly con- structed boiler. Therefore, every pound of surplus water carried off with the steam, robs the boiler of the heat due the quantity of surplus water carried off with the steam without producing an equivalent in work performed. On the contrary, it may result largely detrimental to the performance of the engine supplied with steam from such a boiler. The main object then, is to get steam as dry as possible without superheating it, and the boiler that will furnish the driest steam and perform the greatest amount of work in proportion to the amount of fuel consumed is the best boiler, all other things being equal, such as proper setting of boilers, properly constructed furnaces and proper proportions of smoke stack, etc. One of the simplest devices for making calorimeter tests is that shown in Fig. 124. A represents the main steam pipe, and shows how the attachment is made. B represents a post for holding up the calorimeter pipe. The Calorimeter Tests. 235 portion of the pipe inside of the main steam pipe should be perforated with inch holes, and the pipe leading to the calorimeter or barrel should be made of inch gas pipe. The valve attached to the end of that pipe should have a petcock screwed into the top for the purpose A B Fig. 124. of opening it to allow any water in the pipe leading into the barrel to fall to the level of the water in the barrel, when the test has been completed. The barrel employed for this purpose should be in condition to absorb as little water as possible while making the test. The down pipe from the valve should have a small rubber hose attached to the end of the pipe, as shown in the diagram, and reach to within a short distance of the bottom of the barrel. The lower end of the hose should be closed, and the hose above that perforated all around with small holes. Before the water is put into the barrel, the barrel should be accu- rately weighed and the weight carefully noted, after which put an even number of pounds of water into the barrel for convenient calculation, and leave sufficient room for the desired amount of condensed steam. Fractions of pounds should be avoided as much as possible. Next, set the scales so that they will balance when five or six per cent. more water has been added in the form of condensed steam. Next, take the hose out of the barrel, open the valve, and let the steam blow through so as to heat the pipes thoroughly. While the steam is blowing through, take the temperature of the water in the barrel and carefully note the same. Shut the steam off and then insert. the hose into the barrel, then turn the steam on, and in a short time place the hand under the beam of the scales, to ascertain by lifting, if it is about to balance, and as soon as it is found that the scales are about to balance, shut the steam off and open the petcock and balance the scales. Then place the thermometer in the water, stir it around 236 A Library of Steam Engineering. and carefully observe the highest temperature reached. Note the and also note the weight of water in the barrel. same, We will now suppose the barrel to have contained 360 pounds of water before adding the condensed steam, and that the temperature of the water was 50° Fahrenheit. By reference to the table, it will be found that water at a temperature of 50° Fahrenheit contains 50.003 heat units per pound of water. Then we have: 360 pounds equal weight of water in the barrel before adding steam. 50.003 equal number of heat units per pound of water at a temperature of 50° Fahrenheit. Next, suppose the steam to have raised the temperature of the water in the barrel to 110° Fahrenheit. By referring to the table, it will be found that water, at a temperature of 110° Fahrenheit, contains. 110.110 heat units per pound of water. Then we have: 360 pounds equal weight of water in the barrel before adding steam. 50.003 equal number of heat units per pound of water at a temperature of 50° Fahrenheit, before adding steam. 110.110 equal number of heat units per pound of water at 110° Fahrenheit, the temperature after adding steam. Next, suppose that 20 pounds of steam and water have been added to the water in the barrel from the steam pipe, and we will have 380 pounds of water in the barrel. Then we have: 360 pounds equal weight of water in the barrel before adding steam. 50.003 equal number of heat units per pound of water at a temperature of 50° Fahrenheit, before adding steam. 110.110 equal number of heat units per pound of water at 110° Fahrenheit, the temperature after adding steam. 20 pounds equal weight of condensed steam and water added to water in the barrel. 380 pounds equal weight of water in the barrel after adding condensed steam and water. Calorimeter Tests. 237 Next, suppose the steam pressure per square inch indicated by the steam gauge during the trial to have been 100.304 pounds, by reference to the table, it will be found that the total heat per pound of dry steam at that pressure is 1216.9741 units. Then we have: 360 pounds equal weight of water in the barrel before adding steam. 50.003 equal number of heat units per pound of water at 50° Fahrenheit before adding steam. 110.110 equal number of heat units per pound of water at 110° Fahrenheit, the temperature after adding steam. 20 pounds equal weight of condensed steam and water added to water in the barrel. 380 pounds equal weight of water in the barrel after adding condensed steam and water. 1216.9741 equal number of heat units in one pound of dry steam at 100.304 pounds gauge pressure. PERCENTAGE OF WATER IN STEAM. TO DETERMINE THE PERCENTAGE OF WATER IN STEAM. RULE.-First, subtract the number of pounds of water contained in the barrel before the condensed steam was added from the total number of pounds in the barrel after the condensed steam was added, and multiply the remainder by the total heat units, as shown by the table, contained in one pound of steam due the pressure per square inch indicated by the steam gauge during the time the test was made, and the product will give the total heat units that would have been contained in the steam that has been discharged in the barrel if the steam had been dry, and call the product "Product No. 1." Second, multiply the number of heat units contained in one pound of the heated water in the barrel by the number of pounds of such water, and the product will give the total number of heat units contained in the heated water in the barrel, and call the product "Product No. 2." Third, multiply the number of heat units contained in one pound of the unheated water in the barrel by the number of pounds of that water, and the product will give the total number of heat units in the unheated water in the barrel, and call the product, "Product No. 3." Fourth, subtract the total heat units in the unheated water in the barrel from the total heat units contained in the water after being heated; in other words, subtract "Product No. 3" from "Product No. 2" and the remainder will give the number of heat units that have been added by the steam and water discharged into the barrel from the steam pipe, and call the remainder 'Remainder No. 1." 238 A Library of Steam Engineering. Fifth, subtract "Remainder No. 1" (the heat units that have been added to the original water in the barrel) from "Product No. 1" (the total heat units that would have been contained in the steam if the steam had been dry), and the remainder will show the difference in heat units between dry steam and the steam discharged in the barrel and call this "Remainder No. 2." Sixth, multiply the total heat units, as shown in the table, contained in one pound of steam due the pressure per square inch as shown by the gauge during the test, by the number of pounds of steam and water that have been added to the original water in the barrel, and the product will give the total number of heat units that would have been contained in the steam that has been discharged in the barrel if the steam had been dry, and call this product "Product No. 4." Seventh, divide "Remainder No. 2" (the difference in heat units between dry steam and the steam discharged in the barrel) by "Product No. 4," and multiply the quotient by 100, and the product will give the per cent. of water contained in the steam discharged into the barrel. Example.-Let 1216.9741 equal heat units in one pound of dry steam under 100.304 pounds gauge pressure. Let 380 pounds equal weight of water in the barrel after adding steam. Let 360 pounds equal weight of water in the barrel before adding steam. Let 110.110 equal heat units contained per pound of water after adding steam, temperature being 110° Fahrenheit. Let 50.003 equal heat units contained per pound of original water in the barrel at a temperature of 50° Fahrenheit. Let 20 pounds equal weight of steam and water added to original water in the barrel. Then we have: (1216.9741380—360)—(110.110×380-50.003×360) 1216.9741X20 Performing the operation, we have: 380 Pounds of water in barrel after adding steam. 360 Pounds of water in barrel before adding steam. 1216.9741 X100=2.04+ Per cent. of water in steam dis- charged in the barrel. 20 Pounds of water added by the steam to the water in the barrel. Heat units contained in one pound of dry steam at 100.304 pounds pressure 24339.4820 "Product No. 1." Total heat units that would have been contained in the steam discharged in the barrel if the steam had been dry. Calorimeter Tests. 239 Next we have: 110.110 380 8808 800 33033 0 41841.800 Next we have: 50.003 360 3000 180 15000 9 18001.080 Next we have: Heat units per pound of water in barrel after adding steam. Pounds of water in barrel after adding steam. 'Product No. 2." Total number of heat units in 380 pounds of water at a temperature of 110° Fahrenheit. Heat units per pound of water before adding steam. Pounds of water in barrel before adding steam. "Product No. 3." Total number of heat units in 360 pounds of water at a temperature of 50° Fahrenheit. 41841.800 "Product No. 2." 18001.080 "Product No. 3." 23840.720 "Remainder No. 1." The number of heat units which have been added to the water in the barrel by the steam and water dis- charged into it. Next we have: 24339.482 "Product No. 1." 23840.720 "Remainder No. 1." 498.762 "Remainder No. 2." Heat units, difference between dry steam and the steam discharged into the barrel. Next we have: 1216.9741 Heat units in one pound of dry steam at 100.304 pounds pressure. 20 Pounds of steam and water added to the original water in barrel. 24339.4820 "Product No. 4." Total heat units that would have been contained in the 20 pounds of steam discharged into the barrel if the steam had been dry. Finally, we have: 24339.482) 498.76200 (0.0204+ 486 78964 100 11 9723600 2.0400 Per cent. of water in the 9 7357928 steam discharged into the barrel. If the steam is superheated it will show a greater number of heat units per pound for a given pressure than is contained in the standard steam as shown in the table. 240 A Library of Steam Engineering. IIeat in Water. .089 .1811 32 32.000 TABLE OF PROPERTIES OF WATER FROM 32° TO 212° FAHRENHEIT. ELASTIC FORCE. Tem- perature In Pounds on the Square In Inches of Inch. in Fah- renheit Mercury. Degrees. NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Number of Units of Required for Total Number of Units of Heat Contained in Vapor. Weight of One Cubic Foot of Vapor in Decimals of a Pound. Number of Cubic Feet of Steam from One Cubic Foot of Water. .00030 208,080 Number of Units of Heat Evaporation, Called Latent Heat. 1091.700 1123.700 .092 .1884 33 33.000 1091.005 .096 .1960 34 34.000 1090.310 .100 .2039 35 35.000 1089.615 .104 2121 36 36.000 1088.920 1124.005 .00030 200,480 1124.310 .00031 193,180 1124.615 .00032 186,180 1124.920 .00033 179,380 .108 .2205 37 37.000 1088.225 1125.225 .00034 172,780 .112 .2292 38 38.000 1087.530 1125.530 .00036 166,380 .117 .2382 39 39.001 1086.834 1125.835 .00038 160,230 .122 .2476 40 40.001 1086.139 1126.140 .00040 154,330 .127 .2573 41 41.001 1085.444 .132 .2673 42 42.001 1084.749 .137 .2777 43 43.001 1084.054 .142 .2884 44 44.002 1083.358 .147 .2994 45 45.002 1082.663 .152 .3109 46 46.002 1081.968 1126.445 .00042 148,620 1126.750 .00043 143,220 1127.055 .00045 138,070 1127.360 .00047 133,120 1127.665 .00049 128,370 1127.970 .00050 123,840 .158 .3228 47 47.002 1081.273 1128.275 .00052 119,610 .164 .3351 48 48.003 1080.577 1128.580 .00054 115,490 .170 .3478 49 49.003 1079.882 .176 .3608 50 50.003 1079.187 .183 .3743 51 51.004 1078.491 .190 .3883 52 52.004 1077.796 .197 .4028 53 53.005 1077.100 .205 .4177 54 54.005 1076.405 .212 .4332 55 55.006 1075.709 .220 .4492 56 56.006 1075.014 .228 .4656 57 57.007 1074.318 .236 .4825 58 58.007 1073.623 .245 .5000 59 59.008 1072.927 .254 .5180 60 60.009 1072.231 .263 .5367 61 61.010 1071.535 .273 .5560 62 62.011 1070.839 .282 .5758 63 63.012 1070.143 .292 .5962 64 64.013 1069.447 1128.885 .00056 111,470 1129.190 .00058 107,630 1129.495 .00060 103,930 1129.800 .00062 100,330 1130.105 .00065 96,930 1130.410 .00067 93,680 1130.715 .00069 90,540 1131.020 .00071 87,500 1131.325 .00073 84,560 1131.630 .00076 81,740 1131.935 .00079 79,020 1132.240 .00082 76,370 1132.545 .00085 73,810 1132.850 .00088 71,330 1133.155 .00091 68,940 1133.460 .00094 66,630 302 .6173 65 65.014 1068.751 1133.765 .00097 64,420 .313 .6391 66 66.015 1068.055 .324 .6615 67 67.016 1067.359 .335 .6846 68 68.018 1066.662 .347 .7084 69 69.019 1065.966 .359 .7330 70 70.020 1065.270 .372 .7583 71 71.021 1064.574 .385 .7844 72 72.023 1063.877 .398 .8114 73 73.024 1063.181 .411 .8391 74.026 1062.484 .425 .8676 75 75.027 1061.788 .440 .8969 76 76.029 1061.091 .00127 49,570 1136.510 .00131 48,000 1136.815 .00135 46,510 1137.120 .00139 45,060 1134.070 .00100 62,290 1134.375 .00103 60,280 1134.680 .00107 58,340 1134.985 .00111 56,470 1135.290 .00115 54,660 1135.595 .00119 52,910 1135.900 1136.205 .00123 51,210 .455 .9271 77 77.030 1060.395 .470 .9583 78 78.032 1059.698 1137.425 .00143 43,650 1137.730 .00148 42,280 .486 .9905 79 79.034 1059.001 .502 1.023 80 80.036 1058.304 1138.035 .00153 40,960 1138.340 .00158 39,690 .518 1.056 81 81.037 1057.608 1138.645 .00163 38,480 535 1.091 82 82.039 1056.911 1138.950 .00168 37,320 Calorimeter Tests. 241 ELASTIC FORCE. In Pounds on the In Inches of Tem- perature in Fah- renheit Square Heat in Mercury. Degrees. Inch. Water. TABLE OF PROPERTIES OF WATER FROM 32° TO 212° FAHRENHEIT.-Continued. NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Number of Units of Number of Units of Heat Required for Evaporation, Called Total Number of Units of Heat Contained in Vapor. Weight Number of One of Cubic Cubic Foot of Feet of Steam Vapor in Decimals from One: Cubic of a Foot of Latent Heat. Pound. Water. .553 1.127 83 83.041 1056.214 1139.255 .00173 36,190 .571 1.163 84 84.043 1055.517 .590 1.201 85 85.045 1054.820 1139.560 .00178 35,100 1139.865 .00183 34,050 .609 1:240 86 86.047 1054.123 1140.170 .629 1.281 87 87.049 1053.426 1140.475 .00189 33,030 .00195 32,050 .650 1.323 88 88.051 1052.729 1140.780 .00201 31,100 .671 1.366 89 89.053 1052.032 1141.085 .00207 30,180 .692 1.410 90 90.055 1051.335 1141.390 .715 1.454 91 91.057 1050.638 .738 1.500 92 92.059 1049.941 .761 1.548 93 93.061 1049.244 .785 1.597 94 94.063 1048.547 .809 1.647 95 95.065 1047.850 .834 1.698 96 96.068 1047.152 .860 1.751 97 97.071 1046.454 .887 1.805 98 98.074 1045.756 .914 1.861 99 99.077 1045.058 .943 1.918 100 100.080 1044.360 .972 1.977 101 101.083 1043.662 .00213 29,290 1141.695 .00219 28,430 1142.000 .00226 27,600 1142.305 .00233 26,800 1142.610 .00240 26,020 1142.915 .00247 25,270 1143.220 1143.525 .00262 1143.830 .00270 23,140 1144.135 .00278 22,470 1144.440 .00286 21,830 1144.745 .00294 21,210 .00254 24,540 23,830 1.001 2.037 1.031 2.099 1.062 2.163 104 1.094 2.227 102 102.086 103 103.089 104.092 105 105.095 1042.964 1042.266 1041.568 1040.870 1145.050 .00302 1145.355 .00311 1145.660 .00320 19,500 1145.965 .00330 18,970 20,620 20,050 1.126 2.293 106 106.098 1040.172 1146.270 .00340 18,460 1.159 2.361 107 107.101 1039.474 1146.575 .00350 17,960 1.193 2.431 108 108.104 1038.776 1146.880 .00360 17,470 1.229 2.503 109 109.107 1038.078 1147.185 .00370 16,990 1.265 2.577 110 110.110 1037.380 1147.490 .00380 16,520 1.302 2.653 111 111.113 1036.682 1147.795 .00390 16,070 1.341 2.731 112 112.117 1035.983 1148.100 .00400 15,640 1.381 2.810 113 113.121 1035.284 1148.405 .00410 15,220 1.421 2.892 114 114.125 1034.585 1148.710 .00421 14,820 1.462 2.976 115 115.129 1033.886 1149.015 .00433 14,430 1.504 3.061 116 116.133 1033.187 1149.320 .00445 14,050 1.547 3.149 117 117.137 1032.488 1149.625 1.591 3.239 118 118.141 1031.789 1.636 3.331 119 119.145 1031.090 1.682 3.425 120 120.149 1030.391 1.730 3.522 121 121.153 1029.692 1.779 3.621 122 122.157 1028.993 1.828 3.723 123 123.161 1028.294 1.879 3.826 124 124.165 1027.595 1.931 3.933 125 125.169 1026.896 .00457 13,680 1149.930 .00470 13,320 1150.235 .00483 12,970 1150.540 .00496 12,630 1150.845 .00508 12,300 1151.150 .00521 11,980 1151.455 .00535 11,670 1151.760 .00549 1152.065 .00563 11,370 11,080 1.984 4.042 126 126.173 1026.197 2.039 4.153 127 127.177 1025.498 1152.370 .00578 10,800 1152.675 .00593 10,530 2.096 4.267 128 128.182 1024.798 1152.980 .00608 10,265 2.154 4.384 129 129.187 1024.098 1153.285 .00624 10,010 2.213 4.503 130 130.192 1023.398 1153.590 .00640 9,760 2.273 4.625 131 131.197 1022.698 1153.895 .00656 9,516 2.335 4.750 132 132.202 1021.998 2.398 4.878 133 133.207 1021.298 1154.200 .00673 1154.505 9,276 .00690 9,046 16 242 A Library of Steam Engineering. ELASTIC FORCE. In Pounds on the Square In Inches of Tem- perature in Fah- renheit Number of Units of Heat in Inch. Water. TABLE OF PROPERTIES OF WATER FROM 32° TO 212° FAHRENHEIT.-Continued. Mercury. Degrees. NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Number of Units of Heat Required for Evaporation, Total Number of Units of Heat Contained in Vapor. Weight of One Cubic Foot of Vapor in Decimals Number of Cubic Feet of Steam from One Called Latent Heat. of a Pound. Cubic Foot of Water. 2.461 5.009 134 134.212 1020.598 1154.810 .00707 8,826 2.526 5.143 135 135.217 1019.898 2.594 5.280 136 136.222 1019.198 2.663 5.420 137 137.227 1018.498 2.732 5.563 138 138.223 1017.797 2.803 5.709 139 139.239 1017.096 2.876 5.858 140 140.245 1016.395 2.952 6.011 141 141.251 1015.694 3.029 6.167 142 142.257 1014.993 3.108 6.327 143 143.263 1014.292 3.188 6.490 144 144.269 1013.591 3.270 6.657 145 145.275 1012.890 3.353 6.827 146 146.281 1012.189 3.438 7.001 147 147.287 1011.488 1155.115 1155.420 .00743 8,401 1155.725 .00761 8,191 1156.030 .00780 7,991 1156.335 .00799 7.798 1156.640 .00819 7,613 1156.945 .00839 7,433 1157.250 .00860 7,258 1157.555 .00881 7,088 1157.860 .00903 6,920 1158.165 .00925 6,755 1158.470 .00948 6,595 1158.775 .00971 6,440 .00725 8,611 3.526 7.179 148 148.293 1010.787 1159.080 .00993 6,290 3.615 7.361 149 149.299 1010.086 1159.385 .01016- 6,144 3.707 7.547 150 150.305 1009.385 1159.690 .01040 6,004 3.801 7.736 151 151.311 1008.684 1159.995 .01064 5,867 3.896 7.929 152 152.318 1007.982 1160.300 .01089 5,734 3.992 8.127 153 153.325 1007.280 1160.605 .01114 5,604 4.090 8.329 154 154.332 1006.578 1160.910 .01140 5,477 4.191 8.535 155 155.339 1005.876 1161.215 .01167 5,353 4.295 8.745 156 156.346 1005.174 4.400 8.959 157 157.353 1004.472 4.507 9.178 158 158.360 1003.770 4.617 9.401 159 159.367 1003.068 4.729 9.629 160 160.374 1002.366 1161.520 1161.825 1162.120 .01250 5.000 1162.435 .01279 4,888 1162.740 .01308 4,779 .01194 5,232 .01222 -5,114 4.843 9.861 161 161.381 1001.664 4.960 10.098 162 162.389 1000.961 5.079 10.340 163 163.397 1000.258 5.200 10.588 164 164.405 999.555 5.324 10.840 165 165.413 998.852 5.451 11.097 166 166.421 998.149 1163.045 .01338 4,673 1163.350 .01368 4,569 1163.655 .01399 4,467 1163.960 .01430 4,368 1164.265 1164.570 .01462 4,271 .01495 4,177 5.580 11.359 167 167.429 997.446 1164.875 5.711 11.627 168 168.437 996.743 5.845 11.900 169 169.445 996.040 5.981 12.178 170 170.453 995.337 -1165.180 1165.485 1165.790 .01631 3,826 .01528 .01562 4,085 3,996 .01596 3,910 6.120 12.461 171 171.461 994.634 1166.095 .01667 3,744 6.262 12.750 172 172.470 993.930 1166.400 .01704 3,664 6.408 13.045 173 173.479 993.226 6.555 13.345 174 174.488 992.522 6.704 13.651 175 175.497 991.818 6.857 13.963 176 176.506 991.114 7.013 14.281 177 177.515 990.410 7.172 14.605 178 178.524 989.706 1166.705 .01741 3,586 1167.010 .01779 3,510 1167.315 .01817 3,436 1167.620 .01855 3,365 1167.925 .01894 3,295 1168.230 .01934 3,226 7.335 14.935 179 179.533 989.002 1168.535 .01975 3,159 7.500 15.271 180 180.542 988.298 7.668 15.614 181 181.551 987.594 7.841 15.963 182 182.561 986.889 8.016 16.318 183 183.571 986.184 8.194 16.680 184 184.581 985.479 1168.840 .02017 3,093 1169.145 .02060 3,029 1169.450 .02104 2,966 1169.755 .02148 2.905 1170.060 .02193 2,846 Calorimeter Tests. 243 ELASTIC FORCE. Tem- perature In Pounds on the Number In Inches in Fah- of renheit of Units of Required for Square Heat in Mercury. Degrees. TABLE OF PROPERTIES OF WATER FROM 32° TO 212° FAHRENHEIT.-Continued. NUMBER OF BRITISH THERMAL UNITS IN ONE POUND FROM ZERO (FAHRENHEIT). Total Number of Units of Heat Contained Number of Cubic Feet of Steam Number of Units of Heat Weight of One Cubic Foot of Vapor in from One Inch. Water. Evaporation, Called Latent Heat. Decimals in Vapor. of a Pound. Cubic Foot of Water. . 8.375 17.049 185 185.591 984.774 1170.365 .02238 2.789 8.558 17.425 186 186.601 984.069 1170.670 .02284 2,733 8.745 17.807 187 187.611 983.364 1170.975 .02331 2,678 8.936 18.196 188 188.621 982.659 1171.280 .02379 2,624 9.132 18.593 189 189.632 981.953 1171.585 .02428 2,571 9.330 18.997 190 190.643 981.247 1171.890 9.532 19.408 191 191.654 980.541 9.738 19.827 192 192.665 979.835 9.947 20.253 193 193.676 979.129 10.160 20.687 194 194.686 978.424 10.377 21.129 195 195.697 977.718 .02470 2,519 1172.195 .02529 2,469 1172.500 .02580 2.420 1172.805 .02632 2,372 1173.110 .02685 2,325 1173.415 .02740 2,279 10.597 21.579 196 196.708 977.012 10.822 22.036 197 197.719 976.306 11.051 22.502 198 198.730 975.600 11.284 22.976 199 199.741 974.894 11.521 23.458 200 200.753 974.187 11.761 23.948 12.006 24.446 202 201 201.765 202.777 973.480 972.773 12.255 24.953 203 203.789 972.066 1173.720 .02796 2,234 1174.025 .02853 2,190 1174.330 .02910 2,147 1174.635 .02967 2,105 1174.940 .03025 2,064 1175.245 .03083 2,024 1175.550 .03142 1,985 1175.855 .03201 1,953 12.508 25.468 204 204.801 971.359 12.766 25.992 205 205.813 970.652 13.028 26.525 206 206.825 969.945 13.295 27.067 207 207.837 969.238 13.568 27.619 208 208.849 968.531 13.843 28.180 209 209.861 967.824 14.122 28.751 14.406 29.332 14.700 29.9218 210 210.874 211 211.887 212 212.900 967.116 966.408 965.700 1176.160 .03261 1,916 1176.465 .03323 1,880 1176.770 .03386 1,841 1177.075 .03450 1,809 1177.380 .03516 1,775 1177.685 .03584 1,741 1177.990 .03654 1,708 1178.295 .03725 1,676 1178.600 .03797 1,644 EVAPORATIVE AND CALORIMETER TESTS OF COIL BOILERS These tests were conducted by the Bureau of Steam Engineering of the United States Navy, under the direction of Commodore GEORGE W. MELVILLE, Engineer-in-Chief of the United States Navy and Chief of the Bureau, to whom the author is indebted for a copy of the official reports of these tests, which is here condensed and arranged expressly for this work, for the benefit and information of the student. of steam engineering, as well as to afford him an opportunity for making a practical application of the rules heretofore laid down in this chapter. 244 A Library of Steam Engineering. The first and second series of tests were made with the coil boiler shown in the following engravings: HO b b N N P Q INtay Hotta B B Feed Fig. 125. E O J Calorimeter Tests. 245 Fig. 126. Fig. 127. 1 1 CIN Folger C 246 A Library of Steam Engineering. The plant with which the tests were made consisted of the boiler shown in the engravings, including two cylindrical iron feed tanks of large dimensions. One was placed somewhat above the level of the ground and through its bottom the interior was connected by a suitable pipe with the feed pump. This was called the feed tank. Above and back of this tank was placed the measuring tank. When filled to the height of the lower sides of three rectangular slits cut equidistant in its circumference near the top, it was found to contain by the scales, 4413 pounds of water at a temperature of 56° Fahrenheit. Corrected for the mean temperature of the feed water during the test (50.4°) this weight becomes 4414.4 pounds, which is the weight used in the com- putations. All parts of these tanks, their pipes and cocks, and the feed apparatus throughout, were in full view; all leaks were therefore visible. There were but two, and the water from these were caught and returned to the tank. The supply of water came from a large elevated tank, kept full by a double-acting plunger steam pump. Forced draught was employed, and the air from a blower was dis- charged into an air-tight fire room which inclosed one-half of the cir- cumference of the boiler for a height of 7½ feet. The air duct entered on one side of the boiler, the lower edge of its section being at the level of the floor. The air lock was shut off from the fire room by a flap door opening upwards, and from the outside atmosphere by a sliding door. Within the air lock was the scale for weighing the coal. Two pyrometers were placed in the smoke pipe at right angles with each other and about two feet above the central drum of the boiler. The thermometer in the steam pipe stood in a bath of oil and was covered in that part projecting above the pipe. The calorimeter was of the ordinary barrel variety, fitted with an agitator for equalizing the temperature of the water within it. Steam came to it through a pipe about 10 feet in length, which was well pro- tected against loss of heat by radiation. It got its steam through a perforated tube inside of the steam pipe close to the boiler. The boiler has a plate-iron central drum, standing vertically in the center of the system, and extending from beneath the grate to about 11 feet 6 inches above the base line. It receives all the feed water from an inside pipe that passes through its bottom and extends to near the water line. The space above the water line forms practically all the steam room. Surrounding this drum, in nearly concentric circles, are eight separate coils or sections, varying equally in dimensions from 36 inches of the inner one to 110 inches of the outer one. Each sec- tion has 30 tubes or complete circles, making 240 in all. The tubes of each section are connected in half circles by screwed joints to two vertical headers, diametrically opposite to each other. The tubes are 2 inches in external diameter, and are set at an angle of about ten Calorimeter Tests. 247 degrees with the horizontal, to give direction to the current of circula- tion within them. The headers carrying the lowest ends of the half circles have a common connection at their bottom ends with a mani- fold, which at its inner end is connected with the central drum. Through this manifold the headers are supplied with water, the upper ends being closed. The headers, carrying the highest ends of the half circles, connect the manifold at their tops, through which all the steam generated passes into the upper end of the central drum. At their bottom ends they connect with a manifold, which serves as a mud drum, and as a blow-off for purging the boiler. The headers proper do not extend below the level of the circular tubes, the connections with the lower manifolds being through iron pipes 3 inches in diameter, screwed into the bottom ends of the headers and joined to the manifolds by shallow stuffing boxes. These manifolds are beneath the grate, and they and the headers are of cast-steel. The grate is an annulus, and has an area of 65.5 square feet. This was reduced to 53 square feet for the test by bricking, for convenience of manipulating the fires. The total heating surface, measured on the exterior of the tubes, headers and drum, is 2940 square feet; measured on the interior it is 2074.5 square feet. A portion of this heating surface was, however, covered by the bricks laid on the grate, so that the area of these super- fices was reduced to 2473.5 and 2060 square feet, respectively. These are the figures used to represent the heating surface in making the various computations. About 30 per cent. of this heating surface is above the level of the mean water line. The greatest diameter of the outer casing is 123 inches; the diam- eter of the outer casing around the drum above the tubes is 54 inches; and the diameter of smoke pipe is 40 inches. The height from base to top of drum casing is 11 feet 8 inches, and from base to top of casing around tubes is 8 feet 2 inches. The light iron casing that envelopes the furnace and tubes is lined at the furnace level with fire bricks, and above that with curved tile hollowed in the back. Then we have: 2240 Weight of boiler exclusive of smoke pipe, 111998 tons. Weight of boiler with water to first gauge, 13134 tons. Weight of boiler with water to second gauge, 13128 tons. 240 1906 The following is an analysis of the coal used in making the test: Fixed carbon, 70.67 per cent. Volatile matter, 25.35 per cent. Sulphur, 0.53 per cent. Moisture, 1.35 per cent. Ashes, 2.10 per cent. 248 A Library of Steam Engineering. EVAPORATIVE TESTS. Two tests each of twelve hours' duration were made. Beginning with cold water in the boiler, a wood fire was started and continued until steam at 160 pounds had blown off for about ten minutes. The embers were then hauled out, and with 200 pounds of wood, a new fire was started; the height of the water in the gauge glass marked, and the trial held to have commenced. The blower was run slowly until the coal was well ignited, and then at full speed. The record of the water used was made when that emptied from the measuring tank had been pumped into the boiler. At the end of the test the water remaining in the feed tank, after bringing the height in the gauge glass to the original place, was deducted from the quantity of record. The feed water temperatures were obtained from the feed tank. The coal was filled into a barrel on the platform scales in the air lock to a fixed weight, tallied and dumped, without wheeling, into the fire room. The air pressure was maintained constant by slight variations in the speed of the blower engine. When nearing the end of a run, the fire was allowed to burn down, the blower being gradually slowed until nothing was left but clinkers giving off but little heat. When no more steam formed, the trial came to an end. The steam generated was blown into the atmosphere through a screw top valve, the hand wheel of which was in charge of an attend- ant, who regulated the flow to maintain a constant pressure of 160 pounds by the gauge. The data of the runs of record will be found in the following tables. Calorimeter Tests. 249 TABLE OF EVAPORATIVE TESTS. FIRST EVAPORATIVE TEST. SECOND EVAPORATIVE TEST. Number. Reference TIME. Boilers. Gauge at Pressure by Steam Barometer Pounds in Inches Pounds of Pounds of TEMPERATURE. FAH. of of Coal Dry Mercury. Consumed Refuse. Water Fed to Boiler. Feed Water Steam by Ther- mometer Uptake. In Atmos- phere. Fire Room. Mercury. ti at Boiler. in Inches of Air Pressure REVOLUTIONS OF ENGINE AND BLOWER. Engine. Blower. 1234 12:00 160 29.40 *80 13,243.2 52.00 360 788 66 1:00 160 29.40 3,000 17,657.6 51.00 360 826 67 2:00 160 29.40 17,657.6 51.00 358 841 69 3:00 160 29.40 17,657.6 51.00 358 823 69 5 4:00 160 29.35 6,250 22,072.0 51.00 357 924 69 5:00 160 29.40 22,072.0 51.00 360 912 68 7 6:00 160 29.40 7,500 22,072.0 50.5 359 835 68 77 7:00 160 29.40 17,657.6 50.25 360 821 68 9 8:00 160 29.40 8,000 22,072.0 50.5 366 927 65 10 9:00 160 29.40 17,657.6 50.00 370 897 60 11 10:00 160 29.40 22,072.0 50.00 369 922 61 12 10:35 160 29.40 8,000 1,590 8,828.8 50.00 367 820 62 222222222222 70 70 71 74 75 74 7 731707174 250 500 300 600 302 604 294 588 285 570 285 570 275 550 280 560 277 554 275 550 285 570 200 400 Means and Totals. 29.3958 32,830 1,590 220,720.0 50.7 362 861 66 72.5 275.666 551.333 12345678 10:00 160 29.45 *80 4,414.4 50.00 362 860 57 74 11:00 160 29.50 22,072.0 50.00 364 910 63 75 12:00 160 29.45 9,250 17,657.6 50.00 366 920 69 77 1:00 160 29.45 22,072.0 50.25 370 867 77 2:00 160 29.40 22,072.0 51.00 368 972 72 3:00 160 29.40 9,750 22,072.0 51.00 365 925 73 80 4:00 160 29.45 22,072.0 50.50 367 930 71 81 5:00 160 29.45 22,072.0 50.50 367 920 67 80 6:00 160 29.45 10,000 17,657.6 50.00 367 908 67 79 10 7:00 160 29.45 22,072.0 50.00 367 905 68 78 11 8:00 160 29.45 17,657.6 49.75 367 850 65 75 12 9:00 160 29.45 22,072.0 49.25 368 885 67 75 13 9:33 160 29.45 8,112 1,798 Means and Totals.. 29.446 37,192 1,798 7,201.6 49.00 241,164.8 61 75 222222222221 270 540 293 586 287 574 275 555 262 524 280 560 260 520 265 530 272 544 260 520 264 528 260 520 ... 50.15 270.666 541.333 *WOOD. 400 pounds of wood were used at the beginning of each test, and an allowance made of of a pound of coal for each pound of wood consumed. 366.5 903.666 66.846 77.307 250 A Library of Steam Engineering. CALORIMETER TESTS. Calorimeter tests were made with this boiler at regular hourly intervals during the evaporative tests-twelve during the first trial and eleven during the second, making twenty-three in all; and the quality of steam was calculated from the following formula: 1/ W W (h'—h)-(T—h') Q equals quality of steam, dry saturated steam being unity. W equals weight of water originally in calorimeter. w equals weight of water added to W by condensed steam. T equals total heat of water due to observed pressure. 7 equals latent heat of steam at observed pressure. h equals total heat of water of initial temperature t in calorimeter. h' equals total heat of water of final temperature. NOTE. The greater the weight of condensed steam (w) the smaller will be the percentage of error in the value of Q, other things being equal. The importance or weight of an error in any one of the observed readings, the others being correct, increases in the following order: P, W, t, t', w. Fig. 128. Π Π Calorimeter Tests. 251 TABLE OF CALORIMETER TESTS. CALORIMETER TESTS DURING CALORIMETER TESTS DURING FIRST EVAPORATIVE TEST. SECOND EVAPORATIVE TEST. Test. Number of TIME. Average Pressure of Steam in WEIGHTS. TEMPERATURE. FAH. Pounds Calo- At At Absolute. rimeter. W+b. W. W+b+w. w. t Q. t. Tem- perature of Steam by Ther- mometer. Beginning. End. P. b. Fah. 3 5 9 10 11 1284ECTSPOT 11:16 11:31 174.4 82 396.8 314.8 414.6 17.8 114.0 54.0 .93710 360 12:14 12:33 174.4 407.7 325.7 425.8 18.1 114.3 53.3 .98132 360 1:18 1:35 174.4 406.9 324.9 425.7 18.8 113.0 53.0 .90730 358 2:14 2:35 174.4 407.9 325.9 426.3 18.4 111.2 52.2 .91469 358 3:20 3:37 174.4 404.2 322.2 422.1 17.9 110.0 52.2 .90777 357 4:25 4:43 174.4 404.3 322.3 423.3 19.0 111.7 50.8 .88590 360 7 5:16 5:31 174.4 405.0 323.0 423.6 18.6 110.6 51.2 .90765 359 6:15 6:30 174.4 404.4 322.4 424.4 20.0 114.8 52.5 .87128 360 7:20 7:39 174.4 404.4 322.4 423.2 18.8 110.2 51.6 .86654 365 8:20 8:36 174.4 405.0 323.0 424.7 19.7 111.8 51.5 .84973 370 9:12 9:26 174.4 403.4 326.4 428.2 19.8 110.2 51.8 .81730 369 12 9:55 10:10 174.4 82 415.6 333.6 436.1 20.5 113.2 52.0 .85969 367 Percentage of moisture in the steam, 100 (1-Q)-11.615 Mean values of Q and temperature. .88385 362 9 10 11 12345678QON 9:55 10:14 174.4 82 413.0 331.0 433.4 20.4 112.2 50.3 .89604 362 10:15 11:13 174.4 410.0 328.0 429.3 19.3 113.8 53.7 .89029 364 12:00 12:17 174.4 405.5 323.5 424.6 19.1 112.0 54.0 .84256 366 1:00 1:20 174.4 407.0 325.0 426.8 19.8 113.8 53.8 .84818 370 2:00 2:20 174.4 405.0 323.0 424.0 19.0 112.2 55.0 .83051 368 3:00 3:18 174.4 405.0 323.0 424.6 19.6 112.8 53.2 .84258 365 4:00 4:15 174.4 405.4 323.4 423.0 17.6 110.0 52.7 .92232 367 4:57 5:14 174.4 408.5 326.5 427.5 19.0 111.4 53.0 .86624 367 6:25 6:40 174.4 82 430.0 348.0 449.0 19.0 109.5 54.3 .873C5 367 7:18 7:32 174.4 82 405.0 323.0 424.6 19.6 112.0 52.2 .84627 367 8:25 8:42 174.4 82 405.0 323.0 423.6 18.6 110.8 52.6 .87363 367 Mean values of Q and temperature. .86651 366.66 *Percentage of moisture in the steam, 100 (1- Q)=13.349 There is no reason why the test should show such a large percentage of moisture in the steam for this boiler, unless it be the sudden opening of the throttle valve, causing the lifting of a quantity of water from the boiler and discharging it into the calorimeter. 252 A Library of Steam Engineering. From the foregoing tables the following data are obtained, from which the computations for the potential evaporation are made: Average steam pressure (absolute) P. Average temperature of feed water.... (a) Number of pounds of water vaporized, W₁XQ. (b) Number of pounds of water carried over with the steam, W₁(1-Q) Total heat of steam at pressure P.. Total heat of water at temperature t₁.. (c) Units of heat required to vaporize one pound of water from a temperature t₁ and under a pressure P..... (c) Units of heat required to raise the temperature of one pound of water from t₁ to the temperature due to the pressure P...... (d) Units of heat required to vaporize one pound of water from and at a temperature of 212° Fahrenheit, and under atmospheric pressure. 174.4 50.4° 404,232.0 57,652.4 1,194.88° 18.504° 1,176.376 324.196 965.80 ..475,528,823.2320 Total heat required to vaporize the water aXc.... Total heat required to raise the temperature of the water bXc'. (e) Total heat obtained from the fuel as measured by the steam discharged... (f) Units of heat obtained per pound of fuel.... (g) Units of heat obtained per pound of combustible...... 18,690,677.4704 ..494,219,500.7024 7,058.0603 7,416.9268 C f Potential evaporation per pound of fuel, from a temper- ature t, and under a pressure P. 5.99983 C g Potential evaporation per pound of combustible, from a temperature t₁ and under a pressure P...... 6.30489 d f Equivalent potential evaporation per pound of fuel, from and at a temperature of 212° Fahrenheit, and under atmospheric pressure.. 7.30799 d g Equivalent potential evaporation per pound of combus- tible, from and at a temperature of 212° Fahrenheit, and under atmospheric pressure. NOTE. All weights are given in pounds and all temperatures in degrees Fahrenheit. 7.67957 TOTAL QUANTITIES. Duration of test, in hours, two of twelve hours each. 24 Fuel consumed.. 70,022 Refuse from fuel, in dry ashes, dust and clinkers. 3,388 Combustible consumed... 66,634 Water fed to boiler, by tank measurement, Wi Per cent. of the fuel in dry refuse, etc..... 1. 461,884.6 4.8384 Calorimeter Tests. 253 AVERAGE QUANTITIES. 50.4° 364.33° 882.833° Temperature of feed water, t₁ Temperature of steam, by thermometer. Temperature of uptake... Temperature of atmosphere. Temperature of fire room.... Barometer, in inches of mercury..... 72.48° 75° Pressure of steam at boiler, in pounds, per square inch above a perfect vacuum, 14.7+ pressure by gauge, in pounds, P, Pressure of steam at engine, in pounds, per square inch above a perfect vacuum... Air pressure, in inches of water, in fire room. Air pressure, in inches of water.. Revolutions of blowing engines, per minute. Revolutions of blower, per minute... 29.4205 174.4 2 273.166 546.333 Pounds of Pounds of RATES OF COMBUSTION. Fuel. Combustible. Amount consumed per hour... 2,917.5833 2,776.375 Amount consumed per hour per square foot of grate surface (53).... 55.0487 52.3844 Amount consumed per hour per square foot of heating surface (exterior).. VAPORIZATION IN POUNDS OF WATER. 1.1795 1.1224 Per Pound of Fuel. Per Pound of Combustible. Apparent evaporation, by tank measurement, from a temperature t₁ and under a pressure P... Equivalent apparent evaporation, from and at 212° Fahrenheit, and under atmospheric pressure...... Actual evaporation, into steam of quality Q, from a temperature t, and under a pressure P.. 6.5963 6.9318 8.0345 8.4429 5.7729 6.0665 7.0314 7.3890 Equivalent actual evaporation from and at 212° Fahren- heit, and under atmospheric pressure... .. Potential evaporation, or evaporation had all the heat obtained from fuel been utilized in converting the water in boiler into dry saturated steam from a tem- perature t, and under a pressure P...... Equivalent potential evaporation from and at 212° Fah- renheit, and under atmospheric pressure.. 5.9986 6.3037 7.3063 7.6779 THIRD AND FOURTH SERIES OF TESTS. The coil boiler employed in making these evaporative and calori- metric tests is shown in Figs. 128 and 129. This boiler is 11 feet 5 inches long, 7 feet 9 inches in breadth, and 12 feet 1 inches in extreme height. It has 2026.75 square feet of heat- ing surface, of which 9.43 per cent. is superheating surface. The grate area is 47 square feet, and the ratio of grate to heating surface is 1 to 43.12. The boiler was supplied with rocking grates, but they were - 254 A Library of Steam Engineering. little used during the tests. The estimated weight of the boiler empty is 9.75 tons, and the actual weight of water to the second gauge is 1.8 tons. Forced draught was employed, the air ducts entering the ash- pit on one side. During the tests these air ducts were shut off when the fire doors were opened. The plant used for the tests contained two large tanks, the meas- uring tank holding, when filled to the lower edge of the wier, 8990.15 pounds of water, corrected for the temperature of the feed. Their contents at different levels were ascertained by weighing into them successive 400 pound quantities, the level after each weighing being marked on a batten. The calorimeter was of the ordinary barrel variety. Its steam came through a 1 inch pipe about 8 feet long, well covered. Steam entered this pipe through a perforated tube which entered the main pipe near the boiler. A wood fire was started and continued until steam at 160 pounds had blown off for some minutes. The embers were then hauled out and a new fire started, the height of the water in the glass gauge marked, and the trial held to commence. The wood was entered in the table at four- tenths its weight. The speed of the blower was gradually increased until 2 inches of water column was attained. The record of the water was made when the measuring tank was emptied. At the end of the test the water remaining in the feed tank, after bringing that in the boiler to the original height in the glass, was deducted from the quantity last recorded. The feed water temperatures were taken from the feed tank. The coal was filled into a barrow on a platform scale to a fixed weight (300 pounds net) and tallied. When nearing the end of the run the firing was discontinued, the blower gradually slowed until the fire had so far burned out that no steam was formed, and the trial then came to an end. The steam generated was blown into the atmosphere through a screw stop-valve, which was controlled by an attendant, who regulated the flow to maintain a constant pressure of 160 pounds by gauge. The calorimeter tests were made with regularity; the water in the barrel was well agitated and the top of the latter was covered. The hose was heated before being placed in the barrel, and accuracy in weight obtained by the use of counter-weights of one tenth of a pound each. The data of the runs of record will be found in the following tables: Calorimeter Tests. 255 FIRST EVAPORATIVE TEST. SECOND EVAPORATIVE TEST. Number. Reference Steam Pressure Barometer Pounds in Inches of Pounds of Pounds of Water Fed TABLE OF EVAPORATIVE TESTS. TEMPERATURE.-FAH. Air Revo- TIME. Pressure lutions of by Gauge Feed of Fuel Steam by Dry to In Water Thermome- at Boilers. Mercury. Consumed Refuse. Boilers. t1. ter at Boiler. Uptake. Atmos- phere. in Inches of Water. Blowing Engine. #160 1 8:19 160 30.15 4,200 8,990.15 58 9 10 2345678-O 9:19 160 30.15 2,400 8,990.15 58 514 995 60 10:19 160 30.15 1,200 8,990.15 58 515 1,070 62 11:19 160 30.15 2,100 17,980.30 58 500 915 12:19 160 30.15 1,500 8,990.15 58 506 1,055 64 1:19 160 30.15 1,500 8,990.15 58 510 1,040 65 2:19 160 30.15 1,800 17,980.30 58 515 1,035 67 3:19 160 30.15 2,100 8,990.15 58 520 1,065 67 4:19 160 30.15 1,800 17,980.30 58 500 1,025 69 5:19 160 30.15 2,100 8,990.15 58 475 1,035 71 11 6:19 160 30.15 1,200 17,980.30 58 480 1,005 69 12 7:50 160 30.15 3,457 1,840.75 58 825 64 222222222222 380 Means and Totals.. 20,060 136,693.00 503.5 1,005.5 65.54 * 160 1 8:07 160 30.05 8,990.15 58 3,600 J 880 60 3 7 9 10 11 12 28466TBOOT 2 9:07 160 30.05 2,100 3,990.15 58 430 875 66 10:07 160 30.05 1,500 8,990.15 58 865 67 11:07 160 30.05 2,100 17,980.30 58 390 925 68 12:07 160 30.05 1,800 8,990.15 58 380 925 68 1:07 160 30.05 1,800 17,980.30 58 385 925 2:07 160 30.05 2,400 8,990.15 58 420 925 70 3:07 160 30.05 2,100 8,990.15 58 434 1,025 70 4:07 160 30.05 2,100 17,980.30 58 445 925 5:07 160 30.05 1,800 8,990.15 58 390 940 6:07 160 30.05 2,100 17,980.30 58 394 925 69 6:45 160 30.05 2,870 Means and Totals. 23,560 8,990.151 360.00 114,202.40 58 ... 2 2222222222 885888RRES : 380 407.55 926 68 *Wood. 256 A Library of Steam Engineering. During the trial seventeen calorimetric tests were made, and the quality of steam was calculated by the following formula: 1/W (h'-h)-(T-h' 00 Q equals quality of steam, dry saturated steam being unity. W equals weight of water originally in calorimeter. w equals weight of water added to W by condensed steam. T equals total heat of water due to observed pressure of steam. 7 equals latent heat of steam at observed pressure. h equals total heat of water of initial temperature t in calorimeter. h' equals total heat of water of final temperature t' in calorimeter. Π C0000004300000000 Cowo00 CO20000 Cooooooooooo00 000000000000 wwwwcOOT O 101TOO O O O O O O Q Qon FALLOOOOWW Ο G Fig. 129. Folger. Cin. The observed data and computed results are given in the following table: Calorimeter Tests. 257 TABLE OF CALORIMETER TESTS. Tem- perature of Steam by Ther- mometer. °Fah. Average TIME. Pressure of WEIGHTS. Steam in TEMPERATURE. OFAH. At Beginning. Pounds At Absolute. Calo- rimeter. Q. W+b. W. End. W+b+w. w. t t. P. b. Number of Test. FIRST TEST OF SECOND TRIAL SECOND TEST OF SECOND TRIAL 9:00 9:25 174.81 72 407.2 335.2 427.7 20.5 128.0 56.0 1.0917 514 9:30 9:50 174.81 72 404.2 332.2 420.7 16.5 115.6 56.9 1.0797 515 10:30 10:45 174.81 72 405.0 333.0 420.2 15.2 112.7 56.2 1.1420 500 11:27 11:45 174.81 72 404.5 332.5 420.2 15.7 111.1 56.5 1.0446 506 1:15 1:30 174.81 72 406.5 334.5 422.1 15.6 111.0 56.5 1.0839 515 3:00 3:15 174.81 72 406.2 334.0 422.2 16.0 112.0 56.5 1.0483 520 4:25 4:40 174 81 72 406.0 334.0 421.7 15.7 110.8 56.1 1.0529 475 5:30 5:48 174.81 72 404.4 332.0 420.2 15.8 111.9 56.4 1.0572 480 Percentage of moisture in the steam, 100 (1-Q)=-7.5 Mean values of Q and temperature...... 1.07504 503.125 12345678 9 12348 9:23 9:36 174.81 71.2 400.2 329.0 414.7 14.5 112.7 60.0 1.09308 430 10:20 10:32 174.81 71.2 400.3 329.1 415.2 14.9 113.1 60.9 1.04277 390 11:20 11:35 174.81 71.2 401.0 329.8 416.4 15.4 112.5 60.4 .99850 380 12:15 12:25 174.81 71.2 401.8 330.6 416.9 15.1 112.5 60.4 1.02770 385 1:30 1:40 174.81 71.2 401.2 330.0 417.1 15.9 115.1 60.6 1.05980 420 2:40 2:50 174.81 71.2 400.4 329.2 415.1 14.7 113.0 61.0 1.04445 434 3:40 3:50 174.81 71.2 400.0 328.8 414.9 14.9 113.1 61.0 1.03880 445 4:40 4:50 174.81 71.2 400.1 328.9 415.5 15.4 113.6 60.8 1.01377 390 5:40 5:49 174.81 71.2 400.0 328.8 416.1 16.1 116.2 60.7 1.02370 394 Percentage of moisture in the steam, 100 (1-Q)=-3.8 Mean values of Q and temperature.. 1.038063 407.55 17 258 A Library of Steam Engineering. From the foregoing tables the following data are obtained, from which the computations for the potential evaporation are made: Average steam pressure (absolute) P.. Average temperature of the feed water, t₁.. (a) Number of pounds of water vaporized, W₁XQ.. (b) Number of pounds of water carried over with the steam W₁ (1-Q)... 1 Total heat of steam at pressure P.. .... Total heat of water at temperature t₁...... (c) Units of heat required to vaporize one pound of water from a temperature t₁ and under a pressure P………….. (c) Units of heat required to raise the temperature of one pound of water from t₁ to the temperature due to the pressure P………….. ..... (d) Units of heat required to vaporize one pound of water from and at a temperature of 212° Fahrenheit, and under atmospheric pressure. 174.81 58° 280,822.4 0° 1,194.962 26.12 1,168.962 316.785 965.8 Total heat required to vaporize the water aXc.... Total heat required to raise the steam from the temperature of saturation to actual temperature... 328,275,000 18,209,252.1 (e) Total heat obtained from the fuel, as measured by the steam discharged.. (f) Units of heat obtained per pound of fuel.... 346,484,252.1 7,595 (g) Units of heat obtained per pound of combustible...... 8,817.96 f Potential evaporation per pound of fuel from a tempera- ture t, and under a pressure P. C .... 6.4972 C g Potential evaporation per pound of combustible from a temperature t₁ and under a pressure P... 7.5434 d f Equivalent potential evaporation per pound of fuel from and at a temperature of 212° Fahrenheit, and under atmospheric pressure..... d g Equivalent potential evaporation per pound of combus- tible from and at a temperature of 212° Fahrenheit, and under atmospheric pressure... NOTE. All weights are given in pounds and all temperatures in degrees Fahrenheit. 7.8639 9.1302 TOTAL QUANTITIES. Duration of test in hours.. 24.15 Fuel consumed... 45,620 Refuse from fuel in dry ashes, dust and clinkers.. Combustible consumed... 6,327 39,293 Water fed to boiler, by tank measurement, W1... Per cent. of the fuel in dry refuse, etc.... 280,822.4 13.87 Calorimeter Tests. 259 AVERAGE QUANTITIES. Temperature of feed water t₁.... Temperature of steam, by thermometer.. Temperature of uptake... Temperature of atmosphere.. Temperature of fire room. Barometer, in inches of mercury.. 58° 458.053° 965.909° 66.682° 30.10 Pressure of steam at boiler in pounds, per square inch above a perfect vacuum, 14.7+ pressure by gauge in pounds, P...... Pressure of steam at engine, in pounds, per square inch above a perfect vacuum.... Air pressure, in inches of water. Revolutions of blowing engines, per minute. 174.7 2 RATES OF COMBUSTION. Amount consumed per hour.. Amount consumed per hour per square foot of grate surface.... Amount consumed per hour per square foot of heating surface (exterior). VAPORIZATION IN POUNDS OF WATER. Apparent evaporation, by tank measurement, from a temperature t, and under a pressure P. Equivalent apparent evaporation, from and at 212° Fah- renheit, and under atmospheric pressure... Actual evaporation into steam of quality Q, from a tem- perature t₁ and under a pressure P..... Equivalent actual evaporation, from and at 212° Fah- renheit, and under atmospheric pressure. Potential evaporation, or evaporation had all the heat obtained from fuel been utilized in converting the water in the boiler into dry saturated steam from a temperature t₁ and under a pressure P...... Equivalent potential evaporation from and at 212° Fah- renheit, and under atmospheric pressure.. - 382 Pounds of Fuel. Pounds of Combustible. 1,889.026 1,627.039 40.192 34.617 .93204 .80278 Per Pound of Fuel. Per Pound of Combustible. 6.156 7.147 7.450 8.650 6.156 7.147 7.450 8.650 6.49747 7.54344 7.86325 9.12915 NOTE. The heat units in the foregoing computations are reckoned from 32° Fahrenheit. Peabody's tables were employed in making these tests. EVAPORATIVE AND CALORIMETER TESTS OF LAND BOILERS. A practical test of two land boilers, in which the author partici- pated, is here illustrated for the information of the student of steam engineering. This test was not made to determine the efficiency of the boilers, but to determine the economy and efficiency of the Hawley Down Draft Furnace, which was attached to the boilers. Three things were to be determined: The efficiency of the furnace in increasing the horse power of the boilers; the efficiency of the furnace in increas- ing the economy of the boilers; and the efficiency of the furnace in preventing the formation of smoke; in all of which the furnace showed 260 A Library of Steam Engineering. a most remarkable performance, as shown by the following data of the test made. The trial covered a period of ten hours and five minutes, beginning at nine o'clock, A. M., and ending at seven o'clock and five minutes, P. M., during which the following observations were carefully made: Pressure of atmosphere by barometer. Boiler pressure by gauge. Force of chimney draft by gauge connected with breeching at the base of the chimney. Temperature of waste gases by pyrometer inserted in breeching over the boilers. Temperature of atmosphere, in open air and in boiler room, by thermometers suitably located. Temperature of feed water by thermometers, with their bulbs suspended in brass wells containing oil; each thermometer being in- serted in a feed pipe close to the boiler. Quality of steam was determined by a Barrus Universal Calorim- eter, closely connected with and piped from a nozzle located over the center of the steam drum of the boilers. All instruments, except the barometer, were observed and read- ings taken every fifteen minutes during the trial. The barometer was read hourly. The calorimetric condition of the steam was taken every five minutes. The scales employed for weighing coal, and scales employed for weighing feed water, were carefully tested before commencement of the trial and after its completion. A steam gauge was employed that accurately registered the steam pressure. The accumulation of water from the drip box of the calorimeter was accurately weighed from time to time. The steam discharged per hour, by the orifice in the heat gauge of the calorimeter, was computed by the following formula: AXPXS 70 ×C=X. Pounds of steam discharged per hour. A equals area of orifice in heat gauge of calorimeter, and in this case equaled .012271875 square inch. P equals absolute pressure due the temperature of the steam shown by the upper thermometer of the calorimeter. S equals number of seconds in one hour. 70 equals a constant. C equals .9682, a constant. X equals number of pounds of steam discharged per hour. Putting the formula in words we have the following rule: Calorimeter Tests. 261 RULE.-Multiply the area of the orifice in the heat gauge of the calo- rimeter by the absolute steam pressure per square inch (gauge and atmospheric pressure added together) due the temperature of the steam shown by the upper thermometer of the calorimeter, then multiply the product by 3600, the number of seconds in one hour, then divide the last product by the constant 70, and finally, multiply the quotient by the constant .9682, and the product of this operation will give the number of pounds of steam discharged per hour by the orifice of the heat gauge of the calorimeter. Example.-Let 125 one thousandths of an inch (inch) equal diameter of orifice of the heat gauge. Let .7854 equal a constant. Let 83.76 pounds equal absolute pressure per square inch due the average temperature of steam shown by the thermometer of the heat gauge. yante sa Let 3600 equal number of seconds in one hour. Let 70 equal a constant. Let .9682 equal a constant. Then we have: .1252.7854×83.76×3600 X.9682-51.1819566 lbs. 70 Performing the operation, we have: Steam discharged per hour by orifice of heat gauge of calorimeter. .125 Diameter of orifice. .125 Diameter of orifice. 625 250 125 .015625 Square of diameter of orifice. .7854 A constant. 62500 78125 125000 109375 .0122718750 Area of orifice. 83.76 Steam pressure per square inch. 736312500 859031250 368156250 981750000 1.027892250000 3600 Number of seconds in one hour. 616 735350000000 3083 676750000 3700.412100000000 "The Product." 262 A Library of Steam Engineering. Next, dropping the ciphers in the decimals and dividing the product by 70, we have: 70) 3700.4121 (52.863+ "The Quotient." 350 200 140 604 560 441 420 212 210 21 Finally, multiplying the quotient by the constant .9682, we have: 52.863 .9682 105726 422904 3 17178 47 5767 51.1819566 Pounds of steam discharged per hour by orifice of heat gauge of calorimeter. Whence the percentage of moisture in steam intercepted by the drip box of the calorimeter becomes: W X100 M, in which X W equals weight of water drawn from the drip box per hour. X equals weight of steam discharged by orifice of heat gauge per hour. M equals percentage of moisture of steam. Putting the formula in words, we have the following rule: RULE.-Divide the weight of water in pounds, intercepted by the drip box per hour, by the total weight of steam in pounds discharged by the orifice of the heat gauge per hour, and multiply the quotient by 100, and the product will give the percentage of the moisture in the steam. Example.-Let 8 tenths of a pound equal weight of water inter- cepted by drip box per hour. Let 51.1819 pounds equal weight of steam discharged by orifice of heat gauge per hour. Let 100 equal a constant. Calorimeter Tests. 263 Then we have: .8 51.1819 X100-1.563 Percentage of moisture in steam received by the drip box Performing the operation, we have: 51.1819).800000 (0.01563+ "The Quotient." 511819 2881810 2559095 3227150 3070914 1562360 1535457 Multiplying "The Quotient" by 100, we have:uciam e T .01563 100 1.56300 Percentage of moisture in steam re- ceived by the drip box. The formula for computing the moisture remaining in the steam after leaving the drip box, and indicated by the thermometer of the heat gauge, is given by Barrus, the inventor of the calorimeter employed in this trial, to be: T-°7° C O percentage of moisture, when T equal normal reading of lower thermometer. 7° equal reading of lower thermometer while operating boilers. C equal 21.2 a co-efficient for temperature between 310° and 320° Fahrenheit, given by upper thermometer. Putting the formula in words, we have the following rule: RULE.-Subtract the reading of the lower thermometer while operating the boilers from the normal reading of the lower thermometer, and divide the remainder by the constant, which in this case is 21.2, and the quotient will give the percentage of moisture in steam after leaving drip box. Example. Let 266° equal normal temperature by lower thermom- eter. Let 265.6 equal average temperature by lower thermom- eter. Let 21.2 equal a co-efficient for temperatures between 310° and 320°. 264 A Library of Steam Engineering. Then we have: 266-265.6 21.2 -.018+ of one per cent. Performing the operation, we have: 266 Normal temperature by lower thermometer. 265.6 Average temperature by lower thermom- eter. .4 "The Remainder." Next, dividing The Remainder by the co-efficient 21.2, we have: 21.2) .400 (0.018+ of one per cent. 212 1880 1696 MOISTURE IN COAL-HOW DETERMINED. The moisture in the coal was determined by carefully weighing 100 pounds of coal in a bag at the commencement of the trial, and placing it on the shell of one of the boilers to remain during the trial to be thoroughly dried. At the end of the trial it was again weighed and reduction in weight determined and carefully noted. The smoke stack was frequently noticed during the trial, especially during the firing of the furnaces and the breaking up of the fires, and during the several tests scarcely any smoke was emitted from the stack, and at no time was there any black or dense smoke whatever, notwith- standing the crowding of the boilers to their utmost capacity. KIND OF BOILERS-HORIZONTAL TUBULAR. Number of boilers, 2. Diameter of boilers, 72 inches. Length of boilers, 18 feet. Number of tubes in each boiler, 80. Diameter of tubes, 4 inches. Length of tubes, 18 feet. Length of grate surface, 4 feet. Width of grate surface (both boilers), 13.83 feet. Area of grate surface, 55.32 square feet. Air space in grate surface, 52.80 per cent. Diameter of stack (sheet iron), 48 inches. Height of stack above grates, 88 feet. Effective heating surface-two-thirds of shell and two-thirds of tube surface being computed as effective-2460 square feet. Ratio of effective heating surface to one foot of grate surface 44.76 square feet. Nominal horse power of both boilers, 204 horse power. Calorimeter Tests. 265 CALORIMETER EXPERIMENTS. Duration of calorimeter experiments, 8.25 hours. Average temperature by upper thermometer, 315° Fahrenheit. Average temperature by lower thermometer, 265° Fahrenheit. Normal temperature by lower thermometer, 266° Fahrenheit. Total moisture intercepted by drip box, 6.62 pounds. 6.62 -=.8+ lbs. Average moisture intercepted by drip 8.25 box per hour. .02789225 X 83.76×3600 X.9682-51.18+ lbs. Average weight of steam 70 escaped through orifice of heat gauge per hour. .8 X100=1.56+ per cent. Moisture received by drip cup 51.18 per hour. 266-265.6 21.2 -.018+ of 1 per cent. Moisture in steam after leaving drip box and entering heat gauge. 1.56+.018=1.578 per cent. Total entrainment of water in steam. OBSERVED DATA AND COMPUTED RESULTS. Duration of trial, 10.083 hours. Average pressure by gauge, 87.70 pounds. Average atmospheric pressure by barometer, 14.49 pounds. Average absolute steam pressure, 103.19 pounds. Average temperature of feed water, 90.27° Fahrenheit. Total weight of coal fired on grate, 16925 pounds. Moisture in coal, 1 per cent. Total weight of dry coal fired, 16,755.75 pounds. Total weight of clinkers and ashes, 1351 pounds. Total weight of combustible consumed, 15,404.75 pounds. Average condition of steam (dry steam equals 1) .9842. Total weight of water evaporated, 140,033 pounds. CAPACITY RESULTS. 1214.5-90.27 965.7. 140033.9842-137820.5 lbs. Water actually evaporated corrected for quality of the steam X137820.5=160436.8+ lbs. Equivalent evaporation as from and at 212° Fah. 160436.8 10.083 =15911.6+ lbs. Equivalent evaporation per hour from and at 212° Fahrenheit. 266 A Library of Steam Engineering. 160436.8 16755.75 ECONOMIC EVAPORATION. 137820.5 16755.75 =8.225+ lbs. Water actually evaporated per pound of dry coal from actual steam press- ure and temperature of feed water. 9.575+ lbs. Equivalent evaporation of water per pound of dry coal from and at 212° Fahrenheit. 160436.8 15404.7 10.41+ lbs. Equivalent evaporation of water per pound of combustible from and at 212° Fahrenheit. on 15911.6 2460 -=6.46+ lbs. RATE OF EVAPORATION Water evaporated from and at 212° Fahrenheit per square foot of effective heating surface per hour. COMMERCIAL HORSE POWER. The Centennial standard of horse power for a boiler is the evapo- ration of 30 pounds of water per hour per horse power, at a steam pressure of 70 pounds per square inch, and feed water at a temperature of 100° Fahrenheit. To reduce this to equivalent evaporation as from and at 212° Fahrenheit, the standard adopted by the American Society of Mechanical Engineers, proceed according to the following rule: - RULE. From the total heat units in steam at 70 pounds pressure, sub- tract the heat units in the feed water at 100° Fahrenheit, then divide the remainder by 965.7, and multiply the quotient by 30 (the Centennial standard), and the product will give the commercial standard evaporation from and at 212° Fahrenheit for one horse power. Example.-Let 1210.2423 equal heat units in steam at 70 pounds pressure. Let 100.08 equal heat units in feed water at 100° Fah- renheit. Let 965.7 equal a constant. Then we have: 1210.2423-100.08 965.7 X30=34.48770+lbs. Water evaporated per com- Performing the operation, we have: mercial horse power per hour, from and at 212° Fahrenheit. 1210.2423 Am't carried forward, 1110.1623 100.08 Calorimeter Tests. 267 Am't brought forw'd, 965.7) 1110.1623 (1.14959+ 965 7 144 46 9657 47 892 38 628 9 2643 86913 57300 48285 30 34.48770 lbs. Water evaporated per com- mercial horse power per hour, from and at 212° Fahrenheit. 90150 86913 The Centennial standard, when reduced to equivalent evaporation as from and at 212° Fahrenheit, is very nearly 34.5 pounds of water evaporated per horse power per hour, which is the standard commer- cial horse power. Applying that standard to the case in point, we have: 15911.6 34.5 =461.2+ Horse power developed by the boil- ers in the case under considera- tion. Nominal horse power of the boilers, on the basis of 12 square feet of effective heating surface per horse power: 2460 --205 Nominal horse power of the two boilers. 12 Commercial horse power developed over nominal horse power of the boilers: 461.2-205 205 X100-125- per cent. Comparisons were made with an evaporative and calorimeter test, made with a battery of two tubular boilers, the exact duplicate of these, except that those with which comparisons were made were equipped with the ordinary furnace, and in point of economy, those equipped with the Hawley furnace showed a reduction of 35.6 per cent. in the cost of fuel, and an increase in commercial horse power developed, as compared with the other boilers, of 66.5 per cent., and showed an entire abatement of the smoke nuisance; all of which was accomplished with the use of a very inferior quality of bituminous slack and nut coal. In order that the student may get a clear idea of the construction of this furnace, the following engravings are provided for his careful study: 268 A Library of Steam Engineering. Fig. 130. HAWLEY DOWN DRAFT WATER TUBE GRATE. Fig. 131. TUBULAR BOILER. FOLGEREN Calorimeter Tests. 269 O AA A A 10 Fig. 132. HAWLEY DOWN DRAFT FURNACE FRONT. Fig. 133. TUBULAR BOILER. OOO Folger, Eng. HAWLEY FURNACE. FOLGER C 270 A Library of Steam Engineering. HAWLEY FURNACE. HAWLEY FURNACE. Fig. 134. NATIONAL WATER TUBE BOILER. Fig. 135. HEINE BOILER. Folger Cin. Folger Eng H Calorimeter Tests. 271 Fig. 136. STIRLING BOILER. HAWLEY FURNACE. CO 1:0 FOLGER 272 A Library of Steam Engineering. 0000 HAWLEY FURNACE HAWLEY FURNACE Fig. 137. MUNROE WATER TUBE BOILER. Fig. 138. BABCOCK & WILCOX BOILER FOLGER IN FOLGER LIN Calorimeter Tests. 273 Manhole Smoke Pipe 4 HAWLEY FURNACE Water Grate Grate HAWLEY FURNACE He Steam Safety Valve Feed Water Level Line of Diaphragm forming downward water passage Folger Cin Fig. 139. PECK BOILER. Fig. 140. CALDWELL STANDARD BOILER. Folger Cin. Manhole 18. 274 A Library of Steam Engineering. HAWLEY, FURNACE, H Fig. 141. CAMPBELL & ZELL WATER TUBE BOILER. Fig. 142. GALLOWAY BOILER. Folger, Cin HAWLEY FURNACE Folger, Cin Calorimeter Tests. 275 TEMPERATURE OF A FURNACE. The approximate temperature of a furnace can be obtained in a very simple manner by placing a short, thick piece of iron weighing about 20 pounds in the furnace fire in such a position as to enable it to absorb the greatest heat of the furnace, and allow it to remain a sufficient length of time. Then procure a quantity of water of about three times the weight of the iron. Weigh the iron and water, each, accurately; and just before removing the iron from the furnace take the temperature of the water, then immerse the iron in the water and take the temperature of the water after immersing the iron, and note the highest temperature reached-bearing in mind that the specific heat of iron is about one-ninth that of water-and we are ready. TO COMPUTE THE HEAT OF A BOILER FURNACE. RULE.-Subtract the temperature of the water before immersing the iron in it from the highest temperature after immersing the iron; then multiply the remainder by the number of pounds of water; then multiply the product by the constant whole number 9, and call the last product "Product No. 1." Next, divide "Product No. 1" by the number of pounds contained in the piece of iron, and add the highest temperature of the water to the quotient, the sum will give the temperature of the furnace approximately, or near enough for practi- cal purposes. Example.-Let 140° equal temperature of water after immersing heated iron. Let 45° equal temperature of water before immersing heated iron. Let 40 pounds equal weight of water. Let 9° equal specific heat of water, 9 times that of iron. Let 15 pounds equal weight of iron. Then we have: (140-45)X40X9 15 +140-2420° Approximate temperature of the furnace. Performing the operation in the ordinary way, we have: 140 45 95 40 3800 9 34200 "Product No. 1." 276 A Library of Steam Engineering. Next, dividing "Product No. 1" by 15, the number of pounds contained in the piece of iron, and adding to the quotient 140°, the highest temperature of the water after immersing the iron, we have: 15) 34200 (2280 30 42 30 120 120 0 140 2420° Approximate temperature of the furnace. Speed of Pulleys and Gearing. 277 CHAPTER XIII. SPEED OF PULLEYS AND GEARING. DIAMETER OF PULLEYS. TO DETERMINE THE DIAMETER OF A REQUIRED PULLEY. RULE.-Multiply the number of revolutions per minute of the given pul- ley by its diameter, in inches, and divide the product by the number of revo- lutions the required pulley is required to make per minute, and the quotient will give the required diameter of the required pulley in inches. Example. Let 175 equal number of revolutions of given pulley per minute. Let 36 inches equal diameter of given pulley. Let 420 equal number of revolutions of required pulley per minute. Then we have: 175×36 420 -15 inches. Required diameter of required pulley. Performing the operation, we have: 175 Revolutions of given pulley. 36" Diameter of given pulley. 1050 525 Revolutions of required pulley. 420) 6300 (15 inches. Required diameter of required 420 2100 2100 pulley. It will be observed that it makes no difference as to which pulley is the driven pulley or which is the driver, always multiply the num- ber of revolutions per minute of the known pulley by its diameter, in inches, and divide the product by the number of revolutions the other pulley is required to make, and the quotient will give the required diameter of the other pulley in inches. 278 A Library of Steam Engineering. SPEED OF PULLEYS. TO DETERMINE THE SPEED OF A REQUIRED PULLEY. RULE.-Multiply the number of revolutions of the pulley whose speed is known by its diameter in inches, and divide the product by the diameter, in inches, of the pulley whose speed is unknown, and the quotient will give the number of revolutions per minute of the pulley whose speed is unknown. Example.-Let 175 equal number of revolutions per minute of pulley whose speed is known. Let 36 inches equal diameter of pulley whose speed is known. Let 15 inches equal diameter of pulley whose speed is unknown. Then we have: 175×36 15 -420 Revolutions of pulley whose speed was unknown. Performing the operation, we have: 175 Revolutions of pulley whose speed is 36" known. Diameter of pulley whose speed is known. 1050 525 Diameter of pulley whose speed is unknown. 15) 6300 (420 Revolutions per minute of pulley 60 whose speed was unknown. 30 30 0 SPEED OF LAST DRIVEN PULLEY IN A SERIES. TO DETERMINE THE SPEED OF H IN FIG. 143. Driver B D F O 36" Dia. A C O E G 〃 8 125 rev. per min. 30" Dia. 9 34" Dia. 107 28" Dia., Fig. 143 HOO O11 12 Speed of Pulleys and Gearing. 279 RULE.-First, multiply the diameters, in inches, of all of the driving pulleys together, A, C, E and G (Fig. 143), and multiply the product by the number of revolutions of the driver A, and call the last product "Product No. 1." Second, multiply the diameters, in inches, of all of the driven pulleys together, B, D, F and H, and call this product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the number of revolutions per minute of H, the last driven pulley. Example.-Let 36" equal diameter of the driver A. Let 30" equal diameter of C. Let 34" equal diameter of E. Let 28" equal diameter of G. Let 125 equal number of revolutions per minute of the driver A. Голов Let 8" equal diameter of driven pulley B. Let 9" equal diameter of D. Let 10" equal diameter of F. Let 12" equal diameter of H. Then we have: 36X30X34X28X125 8X9×10×12 Performing the operation, we have: =14875 Revolutions per minute of driven pulley H. 36" A. 30" c. 1080 34" E. 4320 3240 36720 28" G. 293760 73440 to adte insid 1028160 125 Revolutions of A. 5140800 2056320 1028160 128520000 "Product No. 1." 280 A Library of Steam Engineering. Multiplying the diameter of the driven pulleys together, we have: 8" B. 9" D. 72 10" F. 720 12" H. 1440 720 8640 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 8640) 128520000 (14875 Revolutions per minute of driven pulley H. 8640 42120 34560 75600 69120 64800 60480 43200 43200 Another method is to determine the number of revolutions of each succeeding pulley separately, thus: 125 Revolutions of A per minute. 36" Diameter of A. 750 375 Diameter of B= 8") 4500 562.5 30" Revolutions of B and C per minute. Diameter of C. Diameter of D= 9") 16875.0 1875 34" Revolutions of D and E per minute. Diameter of E. 7500 5625 Diameter of F= 10″) 63750 Am't carried forward, 6375 Revolutions of F and G per minute. Speed of Pulleys and Gearing. 281 Revolutions of F and G per minute. Diameter of G. Am't brought forward, 6375 28" 51000 12750 Driver H Diameter of H= 12") 178500 14875 Revolutions of H per minute. SPEED OF LAST DRIVEN PULLEY IN A SERIES. TO DETERMINE THE SPEED OF A (FIG. 144) WITH H AS A DRIVING PULLEY. COR F G 12' 10' 14.875 rev. 28" E 34" C O A 30" 36" Fig. 144 In this case H, F, D and B are the driving pulleys, and G, E, C and A are the driven pulleys. RULE.-First, multiply the diameters, in inches, of all of the driving pulleys together, and then multiply the product by the number of revolutions per minute of the driver, which, in this case is H, and call the last product "Product No. 1." Second, multiply the diameters, in inches, of all the driven pulleys together and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the number of revolutions of the last driven pulley in the series. Example.-Let 12" equal diameter of driver H. Let 10" equal diameter F. Let 9" equal diameter D. Let 8" equal diameter B. Let 14875 revolutions equal speed of driving pulley H. Let 28" equal diameter of driven pulley G. Let 34" equal diameter E. Let 30" equal diameter C. Let 36" equal diameter A. Then we have: 12X10X9X8X14875 28X34X30X36 -125 revolutions. Speed of driven pulley A. 282 A Library of Steam Engineering. Performing the operation, we have: 12" Driver. 10" Driver. 120 9" Driver. 1080 8" Driver. 8640 14875 Revolutions of H. 43200 60480 69120 34560 8640 128520000 "Product No. 1." Next, multiplying the diameter of the driven pulleys together, we have: 28" Driven. 34" Driven. 112 84 952 30" Driven. 28560 36" Driven 171360 85680 1028160 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 1028160) 128520000 (125 revolutions. Speed of driven pulley A. 1028160 2570400 2056320 5140800 5140800 DIAMETER OF LAST DRIVEN PULLEY IN A SERIES. TO DETERMINE THE DIAMETER OF THE LAST DRIVEN PULLEY. - (FIG. 143). RULE. First, multiply the diameters, in inches, of all the driving pulleys together, and multiply the last product by the number of revolutions Speed of Pulleys and Gearing. 283 per minute of the driver A, and call the product of this operation "Product No. 1." Second, multiply all the given diameters, in inches, of the driven pulleys together, and multiply the last product by the number of revolutions of the pulley whose diameter it is required to determine, and call the product of this operation "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter in inches of H, the last driven pulley. Example.-Let 36 inches equal diameter of driver A. Let 30 inches equal diameter of C. Let 34 inches equal diameter of E. Let 28 inches equal diameter of G. Let 125 equal number of revolutions per minute of driver A. Let 8 inches equal diameter of driven pulley B. Let 9 inches equal diameter of D. Let 10 inches equal diameter of F. Let 14875 equal number of revolutions of the last driven pulley H. Then we have: 36X30X34X28X125 8X9X10X14875 -12 inches. Required diameter of the last driven pulley H. Performing the operation, we have: 36" Diameter of A. 30" Diameter of C. 1080 34" Diameter of E. 4320 3240 36720 28" Diameter of G. 293760 73440 1028160 125 Revolutions of A. 5140800 2056320 1028160 128520000 "Product No. 1." 284 A Library of Steam Engineering. Next, multiplying the known diameters of the driven pulleys together, and multiplying the last product by the revolutions of the pulley whose diameter it is required to determine, and we have: 8" Diameter of B. 9" Diameter of D. 72 10" Diameter of F. 720 14875 Revolutions of pulley whose speed is unknown. 3600 5040 5760 2880 720 10710000 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 10710000) 128520000 (12 inches. Required diameter of the last 10710000 21420000 21420000 driven pulley H. DIAMETER OF INTERMEDIATE DRIVING PULLEYS. TO DETERMINE THE DIAMETER OF THE INTERMEDIATE DRIVING PULLEY E. Driver A 36" 125 rev. per min. BOX C E H 9 G 30" 10 O, 12 28 !! 14.875 rev. per min. Fig. 145 RULE.-First, multiply the diameters, in inches, of all the driven pul- leys together, and multiply the last product by the number of revolutions per minute of the last driven pulley, and call the product of this operation "Product No. 1." Second, multiply the known diameters, in inches, of all the driving pul- leys together, and multiply the last product by the number of revolutions per minute of the first driving pulley, and call the product of this operation "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter, in inches, of the intermediate driving pulley E. Speed of Pulleys and Gearing. 285 Example.-Let 8" equal diameter of first driven pulley B. Let 9" equal diameter of D. Let 10" equal diameter of F. Let 12" equal diameter of H. Let 14875 equal number of revolutions per minute of last driven pulley H. Let 36" equal diameter of first driving pulley A. Let 30" equal diameter of C. Let 28" equal diameter of G. Let 125 equal number of revolutions per minute of first. driving pulley A. Then we have: 8X9X10X12X14875 36X30X28X125 -34 inches. Required diameter of driv- ing pulley F. Performing the operation, we have: 8" Diameter of B. 9" Diameter of D. 72 10" Diameter of F. 720 12" Diameter of H. 1440 720 8640 14875 Revolutions of last driven pulley H. 43200 60480 69120 34560 8640 128520000 "Product No. 1." Next, multiplying the known diameters, in inches, of the driving pulleys together, and multiplying the last product by the revolutions per minute of the first driving pulley, A, we have: 36" Diameter of A. 30" Diameter of C. 1080 28" Diameter of G. 8640 2160 Am't carried forward, 30240 286 A Library of Steam Engineering. Am't brought forward, 30240 125 Revolutions of first driving pulley A. 151200 60480 30240 3780000 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 3780000) 128520000 (34 inches. Required diameter of driving 11340000 15120000 15120000 pulley E. DIAMETER OF INTERMEDIATE DRIVEN PULLEYS. TO DETERMINE THE DIAMETER OF INTERMEDIATE DRIVEN PULLEYS. Driver O 36" A O C 125 rev. per min. 30/ F H E G O 10 34 28 77 12 14.875 rev. per min. Fig. 146. RULE. First, multiply the diameters, in inches, of all the driving pulleys together, and multiply the last product by the number of revolutions per min- ute of the driver A, and call the product of this operation "Product No. 1." Second, multiply the known diameters, in inches, of all the driven pulleys together, and multiply the last product by the number of revolutions per min- ute of the last driven pulley H, and call the last product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter, in inches, of the intermediate driven pulley D. Example.-Let 36" equal diameter of first driving pulley A. Let 30" equal diameter of C. Let 34" equal diameter of E. Let 28" equal diameter of G. Let 125 equal number of revolutions per minute of first driving pulley A. Let 8" equal diameter of first driven pulley B. Let 10" equal diameter of F. Let 12" equal diameter of H. Let 14875 equal number of revolutions per minute of last driven pulley H. Speed of Pulleys and Gearing. 287 Then we have: 36x30x34x28x125 8x10x12x14875 -9 inches. Required diameter of driven pulley D. Performing the operation, we have: 36" Diameter of A. 30" Diameter of C. 1080 34" Diameter of E. 4320 3240 36720 28" Diameter of G. 293760 73440 1028160 125 Revolutions of first driving pulley A. 5140800 2056320 1028160 128520000 "Product No. 1." Next, multiplying the known diameter, in inches, of the driven pulleys together, and multiplying the last product by the revolutions per minute of the last driven pulley H, we have: 8" Diameter of B. 10" Diameter of F. 80 12" Diameter of H. 160 80 960 14875 Revolutions of last driving pulley H. 4800 6720 7680 3840 960 14280000 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 14280000) 128520000 (9 inches. Required diameter of 128520000 driven pulley D 288 A Library of Steam Engineering. GENERAL RULE. In all cases where there is a series of pulleys, the product of the diameters of the complete set of pulleys multiplied by the revolutions per minute of the first driving pulley or last driven pulley, must be divided by the product of the diameters of the incomplete set of pul- leys multiplied by the revolutions per minute of the first driving pulley or last driven pulley. In other words, the complete set of pulleys must always be divided by the incomplete set of pulleys; always performing the operation with the complete set first. If the driving pulleys are complete and the driven pulleys are incomplete, multiply the diameters, in inches, of the driving pulleys and the revolutions per minute of the first driving pulley together, and divide the product by the diameters, in inches, of the driven pulleys and the revolutions per minute of the last driven pulley multiplied together. In all cases containing a series of pulleys the product of the diameters in inches of the driving pulleys multi- plied together, must be multiplied by the number of revolutions of the first driving pulley; and the product of the diameters, in inches, of the driven pulleys must be multiplied by the last driven pulley. DIAMETERS OF INTERMEDIATE DRIVING AND DRIVEN PULLEYS. TO DETERMINE THE RESPECTIVE DIAMETERS OF B AND C (FIG. 147). Driver O A 36 // 125 rev. per min. D E 9 O, G H 10 O 34" 12 28" Fig. 147 14,875 rev. per min. RULE.--First, multiply the known diameters, in inches, of the given driven pulleys together, and multiply the last product by the number of rev- olutions per minute of the last driven pulley H, and call the product "Product No. 1." Second, multiply the known diameters, in inches, of the given driving pul- leys together, and multiply the last product by the number of revolutions per minute of the first driving pulley A, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," then assume any convenient diameter for required driven pulley, and multiply the quotient by the assumed diameter, in inches, of the required driven pulley, and the product will give the required diameter, in inches, of the required driving pulley. Example.-Let 9" equal diameter of driven pulley D. Let 10" equal diameter of F. Let 12" equal diameter of H. Speed of Pulleys and Gearing. 289 Let 14875 equal number of revolutions per minute of H. Let 36" equal diameter of driving pulley A. Let 34" equal diameter of E Let 28" equal diameter of G. Let 125 equal number of revolutions per minute of A. Let 8" equal assumed diameter of the required 'driven pulley. Then we have: 9x10x12x14875 X8-30 inches. =30 36x34x28X125 Performing the operation, we have: 9" Diameter of D. 10" Diameter of F. 90 12" Diameter of H. 180 90 1080 14875 Revolutions of H. 5400 7560 8640 4320 1080 Required diameter of required driving pulley. 16065000 "Product No. 1.". Next, multiplying the known diameters, in inches, of the driving pulleys together, and multiplying the last product by the number of revolutions per minute of the first driving pulley A, we have: 19 36" Diameter of A. 34" Diameter of E. 144 108 1224 -28" Diameter of G. 9792 2448 34272 125 Revolutions of A. 171360 68544 4284000 "Product No 2." 34272 290 A Library of Steam Engineering. Next, dividing "Product No. 1" by "Product No. 2," we have: 4284000) 16065000 (3.75 The Quotient. 12852000 32130000 29988000 21420000 21420000 Finally, multiplying the quotient by the assumed diameter of the required driven pulley, we have: 3.75 8" Assumed diameter of required driven pul- ley B. 30.00 inches. Required diameter of required driving pulley C. DIAMETERS OF DRIVING AND DRIVEN PULLEYS ON INTERMEDIATE SHAFT TO MAKE A GIVEN NUMBER OF REVOLUTIONS. TO DETERMINE THE DIAMETERS OF DRIVEN AND DRIVING PULLEYS FOR INTERMEDIATE SHAFT REQUIRED TO MAKE A GIVEN NUMBER OF REVOLUTIONS PER MINUTE (FIG. 148). Shaft No. 4 Shaft No. 5 Driver Shaft No. 1 O A 36" Dia. 125 rev. per min. Shaft No. 2.0 Shaft No. 3 D F H O,, E O, G 9 10 12 34" Dia. 28" 14.875 rev. per min. Rev. of shaft 562.5 per min. Fig. 148 RULE.-First, multiply the number of revolutions of the driving shaft by the diameter, in inches, of the driving pulley on that shaft, then divide the last product by the number of revolutions per minute of the shaft whose diameters of pulleys it is required to determine, and the quotient will give the diameter, in inches, of the driven pulley for that shaft. Second, multiply the product of the diameters of all the driven pulleys succeeding the shaft on which it is required to place the pulleys whose diameters it is required to determine, and multiply the product by the number of revolutions per minute of the last driven pulley, and call the last product "Product No. 1." Third, multiply the product of the diameters of all the driving pulleys succeeding the shaft on which it is required to place the pulleys whose diameters it is required to determine, and multiply the product by the number of revolu- tions of the shaft on which the required pulleys are to be placed, and call the last product "Product No. 2." Fourth, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter, in inches, of the driving pulley for the shaft on which the required pulleys are to be placed. Speed of Pulleys and Gearing. 291 Applying the rule to a series of driving and driven pulleys, as shown in Fig. 148, proceed as follows to determine the required diam- eter of the driven pulley B on shaft No. 2: RULE.-Multiply the number of revolutions per minute of shaft No. 1 (the driving shaft) by the diameter, in inches, of driving pulley A on that shaft, and divide the product by the number of revolutions per minute shaft No. 2 is required to make, and the quotient will give the required diameter of the driven pulley B on that shaft. Example.-Let 125 equal number of revolutions per minute of shaft No. 1. Let 36 inches equal diameter of driving pulley A. Let 562.5 equal number of revolutions per minute shaft Mode No. 2 is required to make. Then we have: 125 X36 562.5 -8 inches. Required diameter of driven pul- ley B on shaft No. 2. Performing the operation, we have: 125 Revolutions of shaft No. 1 per minute. 36" Diameter of driving pulley A. 750 375 Revolutions of shaft No. 2 per 562.5) 4500.0 (8 inches. Required diameter of driven minute. 4500.0 pulley B on shaft No. 2. Next, in determining the diameter of the driving pulley Con shaft No. 2 (Fig. 148), proceed as follows: RULE.-First, multiply the diameter of all of the driven pulleys together, D, F, and H, located to the right in the engraving, and multiply the last product by the number of revolutions per minute the last driven pulley H is required to make, and call the last product "Product No. 1." Second, multiply the diameters of the driving pulleys together, E and G, and multiply the product by the required number of revolutions per minute shaft No. 2 is required to make, and call the last product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter of driving pulley C on shaft No. 2. Example.-Let 9" equal diameter of driven pulley D. Let 10" equal diameter of F. Let 12" equal diameter of H. Let 14875 equal revolutions per minute of H. Let 34" inches equal diameter of driving pulley E. Let 28" inches equal diameter of G. Let 562.5 equal revolutions per minute of shaft No. 2. 292 A Library of Steam Engineering. Then we have: 9X10X12X14875 34x28×562.5 -=30 inches. Required diameter of driving pulley C on shaft No. 2. Performing the operation, we have: 9" Diameter of D. 10" Diameter of F. 90 12" Diameter of H. 180 90 1080 14875 Revolutions of shaft No. 4. 5400 7560 8640 4320 1080 Next we have: 16065000 "Product No. 1." 34" Diameter of E. 28" Diameter of G. 272 68 952 562.5 Revolutions of shaft No. 2. 4760 1904 5712 4760 535500.0 "Product No. 2. " Finally, dividing "Product No. 1" by "Product No. 2," we have: 535500) 16065000 (30 inches. Required diameter of driving 1606500 0 SPEED OF GEARING. pulley C on shaft No. 2. The rules relating to the speed of pulleys relate also to the speed of gearing with one notable exception, which is explained further on. The diameter of the gear wheels must be taken at the pitch line when diameters are employed in making calculations. The number of teeth, however, may be taken instead of diameters. The exception in the Speed of Pulleys and Gearing. 293 rule mentioned above occurs when the driving and driven shafts are to be placed a given distance apart. With this single exception, the rules laid down for determining the speed of pulleys are applicable in deter- mining the speed of gearing. Hence, it is unnecessary to reiterate them here. DIAMETER OF GEAR WHEELS. TO DETERMINE THE REQUIRED DIAMETER OF THE DRIVING WHEEL. RULE.-First, multiply the distance, in inches, the shafts are apart from center to center, on which the wheels are to run, by 2, and call the product "Product No. 1." Second, add the number of revolutions per minute of the driving and driven wheels together, and divide the sum by the number of revolutions per minute of the driven wheel, and call the quotient "Quotient No. 1." Third, divide "Product No. 1" by "Quotient No. 1," and the quotient of this operation will give the diameter of the pitch line of the driving wheel in inches. Example.-Let 18 inches equal distance required between centers of wheels. Let 2 equal a constant. Let 150 equal revolutions of driving wheel per minute. Let 30 equal revolutions of driven wheel per minute. Then we have: 18X2 (150+30)÷30 -6 inches. Required diameter of driving wheel at the pitch line. Performing the operation, we have: 18 Distance between centers. 2 A constant. 36" Product No. 1." This product represents the combined diameter of both wheels. Next, adding the number of revolutions of both wheels together and dividing the sum by the revolutions of the driving wheel, we have: 150 Revolutions of driving wheel. 30 Revolutions of driven wheel. 30) 180 (6 "Quotient No. 1." 180 Finally, dividing "Product No. 1" by "Quotient No. 1," we have: 6)36 6 inches. Required diameter of the driving wheel at the pitch line. 294 A Library of Steam Engineering. TO DETERMINE THE REQUIRED DIAMETER OF THE DRIVEN WHEEL. RULE.-First, multiply the distance, in inches, the wheels are to run apart, from center to center, by 2, and call the product "Product No. 1." Second, add the number of revolutions per minute of the driving and driven wheels together, and divide the sum by the number of revolutions per minute of the driving wheel, and call the quotient "Quotient No. 1." Third, divide "Product No. 1" by "Quotient No. 1," and the quotient of this operation will give the diameter of the driven wheel at the pitch line in inches. Example.-Let 18 inches equal distance required between centers of wheels. Let 2 equal a constant. Let 150 equal revolutions of driving wheel per minute. Let 30 equal revolutions of driven wheel per minute. Then we have: 18×2 (150+30)÷150 -30 inches. Required diameter of driven wheel at the pitch line. Performing the operation, we have: 18 Distance between centers. 2 A constant. 36 "Product No. 1." This product represents the combined diameter of both wheels. Next, adding the number of revolutions of both wheels together, and dividing the sum by the revolutions of the driven wheel, we have: 150 Revolutions of driving wheel. 30 Revolutions of driven wheel. 150) 180 (1.2 "Quotient No. 1." 150 300 300 Finally, dividing "Product No. 1" by "Quotient No. 1," we have: 1.2) 36.0 (30 inches. 36 Required diameter of the driven. wheel at the pitch line. 0 It will be noticed that there is very little difference in the form of the above examples, and they are therefore here given for comparison. To determine diameter of driving wheel: 18x2 (150+30)÷30 -6 inches. Speed of Pulleys and Gearing. 295 To determine diameter of driven wheel: 18X2 (150+30)÷150 -30 inches. It will be observed that the only difference is in the divisors, by which the sum below the line is divided. In the first case, in order to determine the required diameter of the driving wheel, we divide by the revolutions of the driven wheel. In the second case, in order to determine the required diameter of the driven wheel, we divide by the revolutions of the driving wheel. Therefore, in all cases where the diameter of one wheel is required, we must divide the sum below the line by the revolutions of the other wheel. However, but one of the above operations is necessary for determining the required diameter of both wheels. After the diameter of one of the wheels is determined- and it may be either-the diameter of the other may be determined as follows: RULE.-Multiply the number of revolutions of the wheel whose diameter has been determined, according to any of the foregoing rules, by the diameter of such wheel, and divide the product by the number of revolutions of the other wheel, and the quotient will give the required diameter of the other wheel. Example.-Let 6 inches equal the ascertained diameter of one of the wheels. Let 150 equal revolutions of wheel whose diameter has been ascertained. Let 30 equal revolutions of wheel whose diameter is to be determined. Then we have: 150×6 30 --30 inches. Required diameter of the other Performing the operation, we have: On the other hand: 150 6 wheel. 30) 900 (30 inches. Required diameter of the 90 0 other wheel. Let 30 inches equal diameter of driven wheel. Let 30 equal revolutions per minute of driven wheel. Let 150 equal revolutions of driving wheel. Then we have: 30x30 -6 inches. Required diameter of driving wheel. 150 296 A Library of Steam Engineering. Performing the operation, we have: 30 30 150) 900 (6 inches. Required diameter of driving 900 wheel. DIAMETER O. GEAR WHEELS-SIMPLE METHOD. TO DETERMINE THE REQUIRED DIAMETER OF THE DRIVING WHEEL. RULE. First, multiply the distance, in inches, between the centers of the shafts by 2, and multiply the product by the number of revolutions of the driven wheel per minute, and call the last product "Product No. 1." Second, add the number of revolutions per minute of both wheels together and call the total "The Sum." Third, divide "Product No. 1" by "The Sum," and the quotient will give the required diameter of the driving wheel in inches. Example. Let 18 inches equal distance between centers of the shafts. Let 2 equal a constant. Let 30 equal number of revolutions per minute of driven wheel. Let 150 equal number of revolutions per minute of driving wheel. Then we have: 18×2×30 150+30 -6 inches. Required diameter of driving wheel. Performing the operation, we have: Next we have: 18" 2 - Distance between centers of shafts. A constant. 36" Combined diameter of both wheels. 30 Revolutions per minute of driven wheel. 1080 "Product No. 1." 150 Revolutions per minute of driving wheel. 30 Revolutions per minute of driven wheel. 180 "The Sum." Finally, dividing "Product No. 1" by "The Sum," we have: 180) 1080 (6 inches. Required diameter of driving wheel. 1080 Speed of Pulleys and Gearing. 297 TO DETERMINE THE REQUIRED DIAMETER OF THE DRIVEN WHEEL. RULE.-First, multiply the distance, in inches, between the centers of the shafts by 2, and multiply the product by the number of revolutions of the driving wheel per minute, and call the product "Product No. 1." Second, add the number of revolutions per minute of both wheels together, and call the answer "The Sum." Third, divide "Product No. 1" by "The Sum," and the quotient will give the required diameter of the driven wheel in inches. Example.-Let 18 inches equal distance between centers of the shafts. Let 2 equal a constant. Let 150 equal number of revolutions per minute of the driving wheel. Let 30 equal number of revolutions per minute of the driven wheel. Then we have: 18×2×150 150+30 -30 inches. Required diameter of driven wheel. Performing the operation, we have: Next we have: 18" Distance between centers of shafts. 2 A constant. 36" Combined diameter of both wheels. 150 Revolutions per minute of driving wheel. 1800 36 5400 "Product No. 1." 150 Revolutions per minute of driving wheel. 30 Revolutions per minute of driven wheel. 180 "The Sum." Finally, dividing "Product No. 1" by "The Sum," we have: 180) 5400 (30 inches. Required diameter of driven wheel. 540 0 298 A Library of Steam Engineering. CHAPTER XIV. ELECTRICITY. The great field now occupied by this comparatively newly devel- oped science makes it imperative for steam engineers to study and understand it, or be relegated to the lower ranks of steam engineering. In order, then, to afford an opportunity to the engineer and student of steam engineering to become acquainted with some of the rudiments and elementary principles of this important science, so as to better prepare them to engage in the study of the higher and more intricate branches of electrical engineering, this chapter will be devoted to a short and simple elucidation of a few of the more plainer principles upon which this science has been founded. The first and most important step for the student to take is to thoroughly acquaint himself with the meaning of the terms employed in connection with this branch of steam engineering, as well as with the laws governing the measurement of resistance, pressure, quantity and power developed by electricity. DEFINITION OF TERMS-ELECTRICAL UNITS. The following resolutions were adopted by the Electrical Congress at Paris, in 1881: First. In electrical measurements the three fundamental units shall be adopted: centimeter, gramme and second (c. g. s.). Second. The practical unit, the ohm and the volt, will preserve their actual value: 109 for the ohm, and 108 for the volt. Third. The unit of resistance, the ohm, will be represented by a column of mercury of one square millimetre section at the tempera- ture zero centigrade the length of this column is approximately 1.05 metres. Fourth. The current produced by a volt through an ohm shall be called an ampere. Fifth. The unit of quantity, the coulomb, shall be the quantity of electricity defined by the condition that an ampere gives a coulomb per second. Sixth. A farad, the unit of capacity, shall be the capacity defined by the condition that a coulomb in a farad gives a volt. Electricity. 299 ELECTRICAL RESISTANCE. Electrical resistance is that property of conductors by virtue of which they tend to reduce the intensity of a current passing through them. The practical unit of resistance is the ohm, which, according to Ohm's Law, may be stated as follows: The strength of the current varies directly as the electro-motive force, and inversely as the total resistance. Or, adopting the units employed by practical electricians, it may be stated in this way: The number of amperes of current flowing through a circuit is equal to the number of volts of electro-motive force divided by the number of ohms of resistance in the entire circuit. First. The resistance of a conducting wire is directly propor- tional to its length. Second. The resistance of a conducting wire is inversely propor- tional to the area of its cross section, and therefore in the usual round wires is inversely proportional to the square of its diameter: Third. The resistance of a wire of a given length and given thickness depends upon the material of which it is made. ELECTRO-MOTIVE FORCE. Electro-motive force is the name given to that which moves or tends to move electricity from one place to another. For brevity, it is sometimes written E. M. F. The practical E. M. F. unit is called the volt. A mechanical horse power is 33,000 pounds lifted the height of one foot in one minute's time. An electrical horse power is 746 watts. A watt is one ampere mul- tiplied by one volt. A DYNAMO. NAMES OF THE DIFFERENT PARTS. The names of the different parts of the dynamo, as shown in Fig. 149, are as follows: A A, field magnets; B, armature; C, commutator; D D, bearings; E, pulley; F, switch; G G, brushes and brush holder; H H, binding posts; I I, oil cups; J, base frame and yoke; K K, pole pieces. These parts, assembled together in one machine, comprise a dynamo or motor. HOW TO BUILD A DYNAMO. TO PRODUCE A TWELVE VOLT MACHINE. First, procure a complete set of castings, then drill, tap and fit all the parts neatly and accurately; then assemble the field magnets by 300 A Library of Steam Engineering. securing the two castings in an upright position to the base frame, with large, square-headed bolts going through the bottom of the base into the field castings. Next get the armature shell ready. Saw twenty-four notches, about one-eighth of an inch deep, in the edge of flanges in line with one another. In these notches insert thin pieces of fibre the width of the flanges, and about one-eighth of an inch projection above the edge of the flanges. K Lul E H F H J Fig. 149. Then wind soft iron wire, of about No. 12 gauge, on the arma- ture shell, first soldering the inner end of the wire to the iron. Wind the shell full until the wire comes up to the outer edges of the flange, keeping the surface of the wire as level as possible. Solder the outer end of the wire to the outer surface of the wire to keep it in place. Now insulate the entire surfaces of the iron armature core with about two layers of thin cloth, coating the cloth with shellac or glueing the edges down to make it stick in place. The ends of the armature core and a portion of the shaft, about three-fourths of an inch from the ends of the armature core, should also be insulated. This insu- lation is placed on to keep the wire from coming in metallic contact with any portion of the frame. Electricity. 301 TO WIND THE ARMATURE. After having the armature core thoroughly insulated, the next thing to be done is to wind the armature core with about one pound of No. 18, B. & S. gauge, double cotton-covered wire. Start the wire on top of armature between space divided by fibre projections, allowing the end of the wire to project at least 6 inches. The end can be drawn around one of the fibre projections in order to keep it in place. Start the wire on the left side of the space, and carry it by one hand along the surface of the armature core and through the space between the two fibre wedges at the opposite end of the core corre- sponding with the space in which the coil is started. With the other hand grasp the armature core, and as the wire is carried across the end of the armature core, the armature is turned through a half revolution, bringing the opposite side of the core uppermost. The wire is then laid between the two pieces of wedges diametrically opposite those embracing the wire. On the opposite side of the armature the wire is carried across the commutator end of the armature core, and the arma- ture is returned to the position of starting by returning to its first position, and the wire is laid alongside of the portion first laid on The wire is carried lengthwise around the armature in this manner until the space between the fibre projections is filled with one layer. This will bring eight turns around the entire armature to each layer. In winding, the wire should be kept in equal proportions on the opposite side of the shaft, in order to keep the armature in balance. After one layer of wire has been put on in this manner, the operation must be continued until four layers have been put on, which will make four wires deep and eight wires wide, which consti- tutes one coil. Bring the two ends out, allowing at least 6 inches on each wire, and cut the wire off the spool, and to that end attach a card marking it "outside end of first coil." The two ends may be twisted together, so as to hold them in place. Now proceed to wind on the next coil. Revolve the armature half way and commence the coil alongside the first on the left side at the opposite side of the armature where the ends of the first coil come out. Wind this in the same manner as the first, or No. 1 coil, was wound: bring the ends out on the left-hand side and attach a card on the out- side end, marking it "outside end of second coil." Next, proceed to wind coil No. 3. Turn the armature half-way and again commence on the left-hand side of coil No. 1. Wind in the sain manner in which Nos. 1 and 2 were wound, and bring the ends out on the same side of the armature No. 1 commenced on. This end must be marked "outside end of third coil." 302 A Library of Steam Engineering. Next, begin to wind coil No. 4. Turn the armature over, and begin in the same manner in which Nos. 1, 2, and 3 coils were begun. Begin on the left-hand side in space, wind the same as the others, bring the last end out, cut it off and twist ends together, marking the outside end, "outside end of fourth coil." It must be borne in mind that each coil should have eight com- plete turns and four complete layers, making the wind on the armature four layers deep and eight layers wide. In winding the coils, care must be taken to keep the wires divided. on equal sides of the shaft at the end, in order to keep the armature in electrical and mechanical balance. TO HANDLE THE ARMATURE. When winding the armature, take a small box, saw notches in the ends and rest the shafts in the notches, so the armature core can be turned to bring it in the necessary position while winding. After the armature is all wound, tie two cords on the center of the armature, about 11 inches apart, under these cords slip a thin piece of mica or cloth-mica is the best; then put five or six narrow strips of thin copper under these cords and over the mica. The cord is employed to hold the mica bands in position. The next thing to be done is to band the armature. This is done by wrapping fine brass wire over the top of the mica, and continuing until the wrapped wire has reached a width of 1 inch, or 14 inches. The little cleats of thin copper under the brass wire must then be bent up and over the top of the brass wire, and hammered down slightly. Then solder the whole thing; that is, solder the brass wire together, so as to have a firm band of wire about 1 inches wide. It should be well insulated underneath with mica or cloth, to keep the band from coming in contact with the armature wire. TO MAKE AND CONNECT THE COMMUTATOR. First, take the commutator shell, the piece that goes on the end that clamps the commutator segments in place, true up and bevel the inside, so as to fit the commutator segments before the segments are sawed out. The end of the commutator shell opposite the one that has the shoulder extending to it, should be tapped, and the hexagon nut should be bored and tapped to screw into the place which is used. to bind the segments together; then drill a hole in the center of the shell for the shaft. After these parts are all fitted, separate the segments by running a saw in the slots. There will be, when separated, twelve pieces, each of which should be numbered, so as to assemble side by side in their required positions. Electricity. 303 Before placing segments on the shell it will be necessary to insert a narrow piece of mica between each segment. The mica should be about one-thirty-second of an inch thick. Take a small piece of wood, or other material, and assemble the seg- ments around it in their proper order; then place a soft rubber band around the segments to hold them in a circle in their proper position; then insert a slip of mica between each segment, so that they are all thoroughly insulated and do not come in metallic contact with one another. Then take the diameter of the segments as they are, and take two pieces of wood and halve them out so as to encircle the entire com- mutator; clamp the two pieces of wood around the segments by put- ting the whole into a vise, and drawing them tightly together. Then bind the two ends of the wood tightly together with two pieces of wire. The wood is used to handle the segments after they are in their proper position, and should not exceed three-eighths of an inch in thickness. Next, insulate the core of the commutator shell with thick cloth; then insulate each end of the shell on the inside with thick cloth, which is done to keep the segments from coming in metallic contact with the outer portion of the shell; in short, each segment must be insulated from its neighbor and also from the commutator shell. Next, make a small key-seat in the commutator where it goes on the shaft, then drill a hole in the shaft and insert a small metal point to serve as a key. This is done to hold the commutator in proper position after it is mounted to the armature shaft. Before placing the commutator on the shaft, it should be placed in a lathe and the entire surface trued up. After the commutator is mounted the wires should be soldered in their respective segments; there should be a groove filed in each segment at the shoulder, the wires embedded in the grooves and then soldered. Care must be taken to get good metallic contact in each case. TO CONNECT THE ARMATURE. Connect the outside end of the first coil to the inside end of the alternate coil on the right-hand side; then the outside of the last named coil to the inside of the next alternate coil on the right-hand side, and so on until the twelve coils are connected to their respective segments in the commutator, giving the commutator segments what is termed a "lead" of one segment. TO WIND THE FIELD MAGNETS. Cut two fibre heads for each magnet, 34 inches in diameter, with a hole in each, sufficiently large to allow these heads to slip over the cast- iron field core and fit tight. Slip one head to each end of the field core, and insulate the space between them with thin cloth; then drill a small hole in one of the heads next to the pole piece end for the 304 A Library of Steam Engineering. inner wire to project through. Put the inner wire through this hole, and insulate the portion of wire projecting through the head with a small rubber tube. Next wind the field coil full-eight layers deep in all-with No. 16 B. & S. double cotton-covered wire. Next drill a hole in the opposite side of the same fibre head that the first wire went through. This hole should be near the outer edge of fibre head, and the last end of the wire should be brought through this hole. The outer field magnet should be wound exactly like the first. The field should be unscrewed from the base when ready for wind- ing, and replaced when both are wound, after which connect the two inside ends together. Care must be taken to get good metallic contact. One of the outside terminals must be connected to the brush holder, and the other brush holder must be connected to one binding post, and the other outside field connection must be connected to the opposite binding post. After winding the fields and armature, they should be well coated, when finished, with shellac varnish. This will prevent them from absorbing moisture, and also will insulate and tend to preserve them. TO INSULATE THE BRUSH HOLDERS. The brush holders must be insulated from the frame, with paper, rubber or fibre washer, placed one on each side of the frame. There should also be a small fibre bushing fitted into the frame where the brush holder is attached. In short, all parts of the machine in circuit with the wires should be thoroughly insulated. TO MAKE THE BRUSHES. Take about fourteen strips of No. 30 spring copper, three-eighths of an inch wide and three inches long. One end of these should be soldered together so as to make a brush that will give sufficient elas- ticity. To clamp the brushes, take a thin piece of brass the width of the brush holder, drill one hole in each end and attach it to the brush holder with two machine screws, which will serve as a clamp for hold- ing the brushes in position. TO OPERATE THE MACHINE. First run the machine as a motor, or send a current through the fields to charge them. Before an electric current can be generated by rotating the armature, the field magnets must contain some magnetism, no matter how small an amount. This is known as residual magnet- ism, and it must be apparent in a machine before it is possible to gen- erate electricity by cutting the lines of force at right angles in rotating the armature. When the armature is rotated, a feeble current is set in it, and passes through the field magnets, fully exciting them, and up Electricity. 305 as the armature continues to rotate the current is built up until the machine gives out its normal capacity. Set the upper brush and shift it backward and forward until the spark is cut down to the lowest possible minimum. The brushes should not bear too hard on the commutator, but still have sufficient pressure to insure a good electrical contact. It will be well to use a little light grade of oil on the commutator occasionally, such as sperm oil; but it will be found better to use a heavier grade of oil on the bearings. This machine, when run as a dynamo, will operate ten 6 candle- power lamps, at 3500 revolutions per minute, when loaded at 12 volts pressure, giving a quantity of 15.55 amperes. When run as a motor it will require 12 volts pressure to develop its full power, which will be a trifle less than one-fourth horse power. Five cells of battery will run it very well; still it will require eight to bring it up to its full power. 50 VOLTS, E. M. F. This machine can be wound for 50 volts potential. The wind- ing is done in precisely the same manner as described for a 12 volt machine, with the exception that there must be twenty-six layers of No. 24 wire on the fields, and six layers of No. 22 wire on the armature. Eight turns to each layer. 110 VOLTS, E. M. F. The coupling of the armature is precisely the same as in the 12 volt machine, with the exception that the 12 volt machine is coupled in what is known as "series" winding, while the 50 and 110 volt machines are coupled in what is known as shunt" winding. In the 50 and 110 volt machines, the two inner ends of the field coils are connected together. The two outside ends are connected to the two binding posts, H H, as shown in Fig. 149. Each of the two outside ends is connected to its respective binding post separately. The two- wires leading to the brushes, G G, should also be connected to the bind- ing posts, H H, one wire to each post, bringing the outside of one field coil, A, and the wire from one brush holder, G, together at one post, H, on the left-hand side, and bringing the outside wire of the other field coil A, to the wire leading from the brush holder on the right-hand side to the binding post H, on the right-hand side. The two wires that lead from the machine to lamps are con- nected to the binding posts, H H. The switch, F, must connect in the circuit anywhere, so as to open the circuit or close it when the current. is to be turned off. This machine is known as "shunt wound," and at 3500 revolu- tions per minute, will give an electro-motive force of 50 volts at 5. 20 306 A Library of Steam Engineering. amperes, or 110 volts at 3 amperes, with a capacity to operate five 16 candle-power (50 or 110 volts) lamps, requiring about one-half horse power. Or it can be operated as a one-fourth horse power motor when connected in a 50 or a 110 volt circuit. MULTIPLE SYSTEM OF COUPLING. To connect the lamps to this machine, either 12, 50 or 110 volts, stretch two wires from the respective binding posts at any desired point, keeping the wires separate. The wire leading from one side of the machine is known as the positive or outgoing wire, and the wire returning on the other side of the machine is known as the negative or returning wire. To connect lamps one wire of the lamp should be connected to the positive or outgoing wire, and the other wire of the lamp should be connected to the negative or returning wire. The lamps can be connected in this manner until all the lamps in the cir- cuit that the machine will carry are connected. This manner of coup- ling lamps in the circuit is known as "multiple coupling." To burn lamps of less voltage they can be connected in what is known as "series." The current goes from lamp No. 1 to No. 2, from No. 2 to No. 3, and so on; the last wire to the last lamp being connected to the negative or wire returning to the machine. In the series method of coupling, five lamps requiring 10 volts each, coupled in series, would be equal to one lamp of 50 volts current in multiple. In the series coupling, the entire current of the machine passes through each lamp; the lamps being all in direct series with the machine. In the multiple coupling, each lamp takes only its por- tion of the current; that is to say, if two lamps are coupled in mul- tiple, the current is split twice; if five lamps, the current is split five times; and so on. The Steam Engine Indicator. 307 CHAPTER XV. THE STEAM ENGINE INDICATOR. The steam engine indicator is employed as an instrument of com- munication with the interior of the steam engine while in operation. By its instrumentality we are enabled to diagnose the condition of the various organs of the engine's interior, with even more accuracy than a skillful physician can the condition of the various organs of the interior of the human body. It not unfrequently happens that even the most skillful physicians are unable to determine the cause, nature, or location of a disease, ex- cept by a post mortem examination. While, with the indicator in the hands of a skillful engineer, the cause, nature and location of any disease of the steam engine's interior can be determined with even more accuracy than can be done by any ocular inspection. Any defect in the construction, movement, or setting of the valves, can be easily and readily detected. Any defect or deficiency can be located and determined with the utmost accuracy and precision; and hence the skillful engineer will be able to prescribe, with certainty, a remedy that will cure or correct any defect or deficiency in the steam engine pointed out on the diagram of the steam engine indicator. The indicator not only points out and locates any defects in the interior working parts of the steam engine, but it gives the quality, quantity, and temperature of steam in the cylinder; the mechanical force exerted by it during every part of the piston's motion, and the quantity of water consumed in performing any given amount of work. The transmission of all this information to the engineer by the indicator is not to satisfy mere idle curiosity, but for the important purpose of determining the economy of the steam engine; for, after all, the engine that will do the greatest amount of work at the least expense is undoubtedly the best engine. It is, therefore, of the utmost importance that the working parts of each engine should be so pro- portioned and so adjusted as to produce the best possible results. In this question of economy the whole construction of the steam engine is involved; and to determine the question with accuracy it is necessary to ascertain just what is going on in the interior of the engine when it is in operation, as it can be determined in no other way; and from the information recorded by the indicator the engi- 308 A Library of Steam Engineering. neer is enabled to determine the nature and location of defects and to apply the remedy, in such a way that the best possible results, according to the principle upon which the engine is constructed, may be produced. The qualifying remark is made for the reason that there is a vast difference in the construction of the various types of steam engines, and no adjustment of their working parts will place them exactly on the same level in point of economy; but what can be done with the indicator is to enable the engineer to adjust and proportion the valves of each particular type of engine, so as to produce the best possible results for that particular engine, so far as the principle of its con- struction will admit. It can not, therefore, be too strongly urged upon the engineer, as well as upon the student of steam engineering, to become thoroughly familiar with the use of the steam engine indicator. DESCRIPTION OF THE INDICATOR. In imparting the information relating to the construction of the indicator, the Tabor indicator has been selected for illustration. Not because of its real or supposed superiority, nor to give it any advantage, over other first-class instruments; but because: First, for the lack of space to give a description of each particular make; and, second, because it embodies all of the essential features necessary to perform the work required of such instruments. It is, therefore, no disparagement what- ever to any other first-class indicator that it, instead of any other make, has been selected. The student will, therefore, understand that in studying this description he is simply studying the construction of all first-class indicators and not any particular make. One of the features of the indicator here described is the means employed to communicate a straight line movement to the pencil, as shown in Figs. 150, 151 and 152. Fig. 150 E Fig. 151 FRONT VIEW OF STANDARD INSTRUMENT AND BACK VIEW OF PENCIL MECHANISM. The Steam Engine Indicator. 309 The stationary plate A, Fig. 150, and B, Fig. 152, is attached, in an upright position, to the cover of the steam cylinder. This plate contains a curved slot, which serves as a guide and controls the motion of the pencil bar. The side of the pencil bar is provided with a roller E B D Fig. 152 SECTIONAL VIEW OF STANDARD INSTRUMENT. which turns on a pin, and fitted so as to roll freely from end to end of the slot with little lost motion. The radius of the curve in the slot is such, and the pin carrying the roller is attached to such a point in the pencil bar, that the end of the pencil bar which carries the pencil, moves up and down in a straight line, when the roller is moved from one end of the slot to the other. The steam cylinder of this indicator and the base of the paper drum are made in one casting. Inside of the steam cylinder C is a movable lining within which the piston of the indicator works. This inside cylinder is attached by means of a screw thread at the bottom, and openings on the opposite sides at the tops are provided for the insertion of a tool for screwing it in or out. Openings through the sides of the outer cylinder are provided to allow the steam which leaks by the piston to escape. The pencil mechanism is carried by the cover of the outside cylinder. The cover proper is stationary, but a swivel plate, which extends over nearly the whole of the cover, has the pencil mechanism attached to it. By means of this swivel plate, the pencil bar may be turned into contact with the paper drum, as is done when taking a diagram. The pencil mechanism is attached to the swivel by means of the vertical plate containing the slot, and a small standard placed on the opposite side of the swivel for connecting the back link D (Figs. 150, 151 and 152). The slotted plate is backed by a plate of similar size, which serves to receive the pressure brought to bear on the pencil when taking diagrams, and to keep the pencil bar in place. The pencil mechanism consists of three pieces-the pencil bar, the piston-rod link and the back link. The two links are parallel with each other during every part of the stroke of the pencil bar. The 310 A Library of Steam Engineering. connection between the piston and the pencil mechanism is made by means of a steel piston rod. At the upper end where it passes through the cover it is hollow, and has an outside diameter of three-sixteenths of an inch; at the lower end it is solid and its diameter is reduced. It connects with the piston through a ball and socket joint. The socket forms an independent piece, which fits into a square hole in the center of the piston, and is fastened by means of a central stem provided with a screw, which passes through a hole and receives a nut applied from the under side. The nut has a flat-sided head, so that it may be readily operated with the fingers. A number of shallow grooves are cut around the outside of the piston to serve as water packing, so called. The springs used in this indicator are of the duplex type, being made of two spiral coils of wire, with fittings, as shown in Fig. 153. 200000000 The springs are adjusted under steam pressure, and are therefore correct only when used for steam engines. If required for water or other purposes, either special springs should be obtained, adjusted for the purpose for which they are required, or they should be tested at the time they are to be used and the actual scale of spring determined. It should be borne in mind that springs become impaired by long usage, and their scale changes in consequence. Therefore, for important work, the accuracy of the spring should always be tested by comparison on the spot with a reliable steam gauge, employing, as nearly as possible, the conditions under which the instrument was used. For steam work they may tested by attaching to the main steam pipe, a half-inch pipe fitted with a globe valve, a tee for the attachment of the indicator, another tee for the attachment of the steam gauge, and finally a small drip valve. By keeping the drip valve slightly open, and regulating the globe valve, any desired pressure in the apparatus can be secured. Fig. 153 TABLE OF SPRINGS AND MAXIMUM PRESSURES FOR THE SAME. be Scale of Spring. Maximum safe press- ure to which a Spring can be subjected. Scale of Spring. Maximum safe press- ure to which a Spring can be subjected. 8 10 40 100 10 15 48 120 12 20 50 125 16 24 60 150 20 40 64 160 24 48 80 200 30 70 100 250 32 75 The Steam Engine Indicator. 311 The paper drum E (as shown in Fig. 151) turns on a vertical steel shaft, secured at the lower end to the frame of the indicator. The drum is supported at the bottom by a carriage, which has a long ver- tical bearing on the shaft. It is guided at the top by the same shaft, which is prolonged for that purpose, the drum being closed at the top and provided with a central bearing. The drum is held in place by a close fit, and is easily removed by hand when desired. Stops are pro- vided on the inside of the drum at the bottom, with openings in the carriage to correspond, so as to prevent the drum from slipping. These are so placed that the position of the drum may be changed so as to take diagrams in the reverse position of the pencil mechanism when so desired. The drum is made of thin brass tubing, so as to be ex- tremely light. Steel clips are attached to the drum for holding the paper. The drum carriage projects below the lower end of the drum, where it is provided with a groove for the reception of the driving cord. The drum spring, by which the backward movement of the drum is accomplished, consists of a flat spiral spring, placed in a cavity under the drum carriage encircling the bearing. It is attached at one end to the frame below and at the other end to the drum carriage. In its normal position, the drum carriage is kept against a stop, by means of the pull of the spring. By loosening a thumb screw which encircles the shaft and holds the drum carriage down to place, the carriage may be lifted so as to clear the stop, and the tension on the spring may then be adjusted. This is done by simply winding or unwinding as may be desired. A carrier pulley serves to operate the driving cord from any direction. A single pulley is mounted within a circular perpendicular plate, the center of which coincides with the center of the driving cord, and with the circumference of the pulley. The plate can be turned about its center so as to swing the pulley into any desired angular position, and thereby lead the cord off in any desired direction. The plate is held by a circular frame, which also serves as a clamp, and the pulley is fixed in position by the use of the same nut which secures the frame to the pulley arm. A ratchet is cut on the edge of the drum carriage, and a pawl is provided, attached to a frame, so as to engage in it whenever it is desired to stop the motion of the drum without unhooking the driving cord. The indicator is attached by means of a coupling having but one thread. The indicator cock is provided with a stop, so as to turn only the 90 degrees needed for opening and closing. The pressure of the pencil on the paper drum is regulated by means of a screw, which passes through a projection on the slot plate, and strikes against a small slot provided for that purpose, which stands on the frame. This 312 A Library of Steam Engineering. screw is operated by a handle, which is of sufficient size to be readily worked by the fingers, and which also serves as a handle for turning the pencil mechanism back and forth, as is done in the act of taking diagrams. The end of the pencil bar is shaped in the form of a thin tube for the reception of the pencil, lead or metallic marking point. MANAGEMENT AND CARE OF THE INDICATOR. The indicator should at all times be kept in good working order, and special attention should be directed to the condition of the cylinder and piston. These should particularly be kept free from any accumu- lation of dirt, for a little accumulation of dirt in the cylinder produces distortions in the diagrams and makes them almost valueless. The same result is produced by insufficient lubrication of the piston. The piston is easily removed by simply unscrewing the cover and lifting the whole from its place. After that is done introduce a cloth or cotton waste into the cylinder and clean it thoroughly. Then clean the out- side of the piston and lubricate the cylinder and piston before replacing the piston. This should be done at frequent intervals, especially when the indicator is used continuously for a long test; in fact, the piston should be lubricated at least once for every half dozen diagrams taken. It will not do to use ordinary oil, but the best cylinder oil should be used. After cleaning the inside of the indicator and lubricating the piston and cylinder, the parts may afterward be lubricated without disconnecting them. Care must be taken to have perfect freedom in the pencil movement else the diagrams will be defective. The pencil bar, when not encumbered by a spring should, upon being raised to its highest position, drop like a dead weight to its lowest position with perfect freedom. The surface of the roller and the sides of the slot should be kept clean, but should not be lubricated. The roller should be kept free and the pin oiled. The pivots at the joints in the pencil mechanism should be oiled from time to time. Examination should be made to see that a suitable amount of side play exists in these joints to insure their working with perfect freedom. All the working parts requiring lubri- cation except the inside of the cylinder should be lubricated with watch oil. Due attention must be given to the paper drum as well as to the pencil mechanism. The paper drum is in constant motion when the indicator is set to work, and is therefore subjected to much wear. The drum carriage should be taken down from time to time and the central bearing cleaned. The top bearing of the drum and the carriage bearing The Steam Engine Indicator. 313 must also be cleaned; and all of these parts should be frequently lubricated. The number of spring to be used in any particular case depends mainly on the boiler pressure; as a general rule, however, the lightest spring that can safely be employed is preferable, because the lighter the spring the higher will be the diagram, and hence the more accurate the measurements can be made. For the purpose of comparison it will be advantageous to employ the same spring, where admissible, in taking diagrams from various engines and under various conditions, as such comparisons can be made without making allowance for change in the tension of the spring. A No. 40 spring is a good standard in such cases for pressures below 80 or 90 pounds. For low-pressure cylinders in compound engines it is customary to employ springs below No. 20. For engines of high rotative speed it is well to employ springs of high tension to insure a smooth line in the diagram. Springs are readily changed by removing the cover of the cylinder, loosening the screw beneath the piston, unscrewing the piston from the spring and the spring from the cover, and placing the spring desired. When the ball and socket end of the piston rod is introduced into the square hole in the center of the piston, care must be taken to have it set fairly in the hole before the screw is applied. If such care is not taken, the corners may catch and cause derangement. To obtain well-defined diagrams, the pencil lead should be fre- quently sharpened. Very satisfactory results are obtained by the employment of a metallic marking point instead of a lead pencil, using chemically prepared paper on the drum. A piece of ordinary brass wire, suitably sharpened, will answer for the marking point. It requires less frequent sharpening than the lead pencil and the general character of the work done by it is much superior. It must be borne in mind, however, that lines made by a metallic point on chemically prepared paper will disappear in a short time. Therefore, in order to preserve the diagrams, the lines should be traced over with a pencil or a pen within a few days after they are made. The tension on the drum spring should be adjusted so that the minimum force exerted by the spring may be sufficient to keep the cord taut. This, however, must be increased according to increased rotative speed, but under no circumstances should there be enough ten- sion to make it difficult to pull the cord with the thumb and forefinger. The lightest tension consistent with good work is the most desirable. The tension on the drum spring may be adjusted by removing the paper drum, then loosening the thumb-screw which encircles the central shaft, lifting the drum carriage so as to clear the stop, and then winding the carriage in the desired direction. 314 A Library of Steam Engineering. HOW TO INDICATE A STEAM ENGINE. The first thing to be done is to provide means for attaching the indicator to the cylinder of the engine; and, second, to provide means for giving motion to the paper drum. To attach the indicator a hole should be drilled at each end of the cylinder, and tapped to fit a half-inch steam pipe, to which to connect the indicator cock. In horizontal engines the cylinder itself, instead of the head, should be drilled and tapped for the attachment of the indicator steam pipe. Care must be taken to drill the holes so that they can not be interfered with by the piston of the engine; and wherever the pipes are attached they should communicate freely with the steam in the cylinder throughout the entire length of the stroke of the piston; they should be short and free from any unnecessary bends. The pipes, bends, cocks or valves connecting the indicator with the cylinder should be so constructed and attached as to form a free and unobstructed passage of the steam in the engine cylinder to the piston of the indicator. The best results are not obtained by connecting the two ends of the cylinder with one indicator applied at the center, for the reason that it makes the pipes too long for accurate work, and necessitates the employment of extra or additional bends, all of which should be avoided, especially in indicating high-speed engines. In all such cases the indicator should be placed as near the cylinder as possi- ble so as to reduce the length of the steam pipe to its minimum. If but one indicator is available it should be used first at one end, then at the other. Should there be a necessity for placing the indicator at the Fig. 154. THREE-WAY COCK. center of the cylinder, large pipes and easy bends will, to a great extent, avoid the errors due to long pipes. In such cases a three-way cock, such as shown in Fig. 154, should be employed. HOW TO ATTACH THE INDICATOR. In drilling and tapping holes in a cylinder care should be exercised that the chips do not enter it unless they can be removed. A simple The Steam Engine Indicator. 315 method for preventing chips going into the cylinder is to admit steam into it while the work of drilling and tapping the holes is being done, so that the chips will be blown out as fast as they are made. Before attaching the indicator the cock should be opened and the pipes blown out and thoroughly cleared of all loose material that may have collected or lodged in them. The motion to be given to the paper drum is one that coincides with the motion of the piston, except that it is on a reduced scale. This motion may be obtained in various ways, one of which is shown in Fig. 155. REDUCING LEVER. E Fig. 155. The reducing lever E consists of a strip of pine board three or four inches in width, about one and one-half times the stroke of the engine in length. It is hung by a screw or small bolt to a wooden frame attached overhead. At the lower end a connecting rod C D, is attached, about one-third as long as the stroke, one end to the lever and the other to a stud screwed into the cross head. The lever E should stand in a vertical position when the piston is at the middle of its stroke. The connecting rod at C, when the piston is in the middle of its stroke, should be as far below a horizontal line as it would be above it at either end of the stroke. The cords which operate the paper drum may be attached to a screw inserted in the lever E near the point of its suspension; but a better plan is to provide a segment A B, the center of which coincides with the point of suspension, and allows the cord to pass around the circular edge. The distance from edge to center should bear the same proportion to the length of the reducing lever that the desired length of the diagram bears to the length of the stroke of the engine. For an engine having a stroke of 316 A Library of Steam Engineering. 48 inches, the lever should be 72 inches in length, and the connecting rod C D 16 inches in length; in which case, to obtain a diagram 4 inches long, the radius of the segment would be required to be 6 inches. It is, however, immaterial as to the actual length of the diagram, but 4 inches is usually satisfactory, but for high speeds it may be advan- tageously reduced to 3 inches. The cords should leave the segment in a line parallel with the axis of the cylinder. The pulleys over which they pass should incline from a vertical plane and point to the indi- cators wherever located. If the indicators and reducing levers can be placed so as to be in line with each other, the pulleys may be dis- pensed with, and the cords carried directly from the segment to the instruments, a longer arc being provided for that purpose. The carrier pulley on each indicator should be adjusted so as to point in the direc- tion in which the cord is received. In arrangements of this kind the reduced motion is not mathe- matically correct, because the leverage is not constant at all points of the stroke. Pantagraph motions have been devised for overcoming these defects. Two forms have been successfully used, which, if well made, well cared for, and properly handled, will reproduce the motion on the reduced scale with perfect accuracy. H LAZY TONGS. Fig. 156 Fig. 156 represents the manner of attaching the lazy tongs, when the indicators are applied to the side of the cylinder. It works in a horizontal plane. The pivot end is supported by a post B, erected in front of the guides, and the working end receives motion from an iron attached to the cross head. By adjusting the post to the proper height, and at a proper distance in front of the cross head, the cords may be carried from the cord pin E (Fig. 157) to the indicator without the use of carrier pulleys. As it is not always convenient to adjust the posi- tion and height of the post in this manner, a second post is required for the attachment of a pulley over which the cord passes. The posts can be erected on a board base, which can be fastened to the floor, and the posts rigidly held in place by wooden braces. The Steam Engine Indicator. 317 Fig. 157 shows a perspective view of the lazy tongs on a larger scale, detached from its working position on the engine. It consists of strips of wood put together at the pivot ends with hollow pins and large riveted heads. The end B is attached to the stationary post by B Fig. 157. an iron stud, to which it is firmly held by means of a nut having flat sides, so as to be worked with the fingers. The end A drops into a hole in the iron attached to the cross head of the engine. The cord pin E is attached to a cross bar C D, which may be moved into different positions with relation to the center B, by changing the screws C and D which hold it in place. Holes are provided in the cross bars for various positions of the cord pin, corresponding to the various posi- tions of the cross bar, this pin being always placed in line between the points A and B. The holes are one-half inch apart, and a change of one hole produces a change of one-forty-eighth in the length of the motion of the cord pin. In the hole nearest the center B the reduc- tion of motion is three-forty-eighths, or one-sixteenth; in the second hole it is four-forty-eighths, or one-twelfth; in the third hole it is five- forty-eighths, or a fraction over one-tenth; in the fourth hole it is six- forty-eighths, or one-eighth; and so on. The joints of the lazy tongs can be readily tightened when they become loose by upsetting the rivet heads. In using the instrument the joints should be kept well lubricated. A good plan is to keep the whole apparatus when not in use in a bath of oil. THE PANTAGRAPH. E D C B Fig. 158. 318 A Library of Steam Engineering. Fig. 158 shows a simple form of pantagraph, for use when the indicators are attached to the side of the cylinder. The working end A receives motion from the cross head, and the foot piece B is attached to the floor. The cord pin D is fixed in line between the pivot B, and the working end A, and the pulleys E, attached to the block G, guide the cords to the indicators. The indicator rigging that gives the best results at high speeds is a plain, reducing lever, provided at the lower end with a slot that rides on a stud screwed into the cross head. The length of the lever should be two and one-half times the length of the stroke. Whatever plan is adopted for an indicator rig, long stretches of cord should be avoided. If the motion must be carried a long distance it is desirable to use a fine annealed wire in place of the cord, except at the point where it passes over the pulley. Braided linen cord, a little in excess of one-sixteenth of an inch in diameter, is a suitable material for indicator work. MANNER OF TAKING DIAGRAMS. Stretch a blank card smoothly around the paper drum and fasten the ends with the spring clips. Attach the driving cord so that the motion of the drum will be central. Open the cock to admit steam to the indicator till the parts have become heated. Shut the steam off, bring the pencil or marking point in contact with the paper, adjust the top screw and trace a fine line upon the card. This is the atmos- pheric line. Open the cock, and after two or three revolutions, apply the pencil to the card and take the diagram. If it is desired to ascer- tain the condition of the valve adjustment the pencil need only be applied while the engine is making one revolution. But to determine the power it should be applied a longer time so as to obtain a number of diagrams superimposed on the same card. The fluctuations in the admission of steam, produced by governors which do not regulate closely, are so common that this course should always be pursued to obtain average results. After the diagram has been traced and the cock has been closed, the pencil should be applied lightly to the paper to see whether or not the position of the atmospheric line has been changed. If a new line has been traced it is evidence of error or derangement, and the operation should be repeated on a new card. In adjusting the valves the boiler pressure should be observed, and the changes that are made before taking a diagram should be noted on the card for reference. If a series of diagrams are being taken for power, they should be numbered in order and the number of revolutions per minute noted on every card. If tests are to be made for power used by machines or tenants, a number of diagrams should be taken under each The Steam Engine Indicator. 319 condition and the results averaged. It is well, in such cases, to mark each card of a set by some letter, and on the first of the set specify the machines in operation at the time. When taking diagrams care should be taken to see that the pressure of the pencil on the paper is just sufficient to make a legible mark, and no more; greater pressure only increases friction and produces inaccuracy in the diagram. Accumulation of water in the indicator is a hindrance to the taking of an accurate diagram; for that reason the indicator should be thoroughly heated and freed from any accumula- tion of water before taking a diagram. If much water is entrained in the steam the cylinder cocks should be kept slightly open while taking diagrams, otherwise they are likely to be distorted. When taking diagrams from steam engines the height of the barometer, or pressure of the atmosphere, should always be carefully noted. This is necessary when the economy of the engines is to be determined, and it is desirable in all cases to know how much the exhaust pressure is above zero. Even at the sea level the pressure is constantly changing, and there are many engines working in places far above the sea level where the atmospheric pressure is always less, in some cases very much below 14.7 pounds per square inch, or 29.9 inches of mercury. Care should therefore be exercised in this respect, as many engineers are inclined to ignore this fact. All steam gauges in ordinary use indicate pressures above the atmospheric pressure Vacuum gauges indicate the pressure below the atmospheric pressure. Neither kind, however, show the pressure above zero, or total pressure, and to arrive at this the pressure of the atmos- phere must be added to the steam gauge pressure in the first case, or the amount of vacuum subtracted from the atmospheric pressure in the second. THE ESSENTIAL FEATURES OF THE INDICATOR DIAGRAM. The shape of the figure traced upon the indicator card depends upon the manner in which the steam pressure acts in the cylinder. If the steam be admitted at the beginning, and exhausted at the end of the stroke, and admission continued from one end of the stroke to the other, the shape of the diagram will be nearly rectangular. If the admission continues through only a part of the stroke, the diagram will assume a shape similar to that in Fig. 159. Fig. 159 has been selected to illustrate the essential features in the indicator diagram, because it exhibits clearly all the operations effected by the pressure that commonly take place in a steam engine cylinder. This diagram shows that the admission of steam commences at A and ends at D; the cut off commences at C and becomes complete at D; 320 A Library of Steam Engineering. expansion occurs from D to E; the release or exhaust begins at E and continues to the point H; the compression of the exhaust steam com- mences at G and ends at the admission point A. The line A B is called the admission line; BC the steam line; D E the expansion line; F G the exhaust or back pressure line-or in the case of condensing engines, the vacuum line; H A the compres- sion line; and J I the atmospheric line. The curve which joins two adjacent lines, represents the action of the steam when one operation changes to another and can not properly be classed with either line. The point of cut off, D, lies at the end of admission; the point of release, E, at the beginning of the exhaust; the point of compres- sion, H, at the end of the exhaust. The proportion of the whole B GH Fig. 159. +A length of the diagram borne by the distance of the point D from the admission end, represents the proportion of the stroke completed at the point of cut off; so also in the case of the point of release, and in that of compression for the uncompleted portion of the stroke. The pressure at the points of cut off, release, and compression, are the heights of these various points above the atmospheric line measured on the scale of the spring. There are three principal objects for which the indicator diagram is used: First, to serve as a guide in setting the valves of an engine. Second, to determine the indicated power developed by an engine. Third, to determine, in connection with a feed-water test showing the amount of steam consumed, the economy with which an engine works. A WELL-FORMED DIAGRAM. Fig. 159 shows the general features of a well-formed diagram, the attainment of which should be the aim in setting the valves of an engine. The admission of steam is prompt, line perpendicular to the atmospheric line. making the admission The initial pressure is The Steam Engine Indicator. 321 fully maintained up to the point where the steam begins to be cut off. The somewhat early release secures a free exhaust and uniformly low back pressure, and the exhaust valve closes before the return stroke is completed, thus providing for compression. These are the first require- ments to be met in producing an economical engine. DERANGEMENT OF VALVE GEAR-HOW REVEALED. Derangement in the valve gearing is revealed in the diagram by tardy admission or release; by low initial pressure or high back press- ure; or by absence of compression; either one of which causes an increased consumption of steam for performing the same amount of work. The angular position of the eccentric controls all the movements of the valves, but improper lengths of the eccentric rods, or improper proportions of lap and lead, or wrong position of the eccentric, will produce some of the faults mentioned. REGULATING THE EXHAUST. In regulating the exhaust of an engine, the utility of employing compression should not be overlooked. In the first place, it serves to overcome the momentum of the reciprocating parts of the engine, and to reduce the strain upon the connections caused by the sudden appli- cation of the pressure at admission. In the second place, compression is desirable on the ground of economy in the consumption of steam. It fills the wasteful clearance spaces of the cylinder with exhaust steam instead of filling them with live steam from the boiler. Compression produces a loss by the increased back pressure which it creates, but the loss is more than counterbalanced by the gain resulting from the reduc- tion of clearance waste. But it is not advisable to compress above the boiler pressure. PIPES AND PASSAGES FOR ADMISSION AND EXHAUST OF STEAM. The valves being in proper adjustment the indicator diagram shows whether or not the pipe and passages for the admission and exhaust of steam are of sufficient size. In automatic cut-off engines the admission line should be parallel with the atmospheric line, and the initial pressure should not be more than three pounds less than the boiler pressure. The back pressure should not, in any engine, exceed one pound, where the exhaust proceeds directly to the atmosphere. Much can often be learned by applying the indicator to the steam and exhaust pipes, using the same mechanism for driving the paper drum as that used when the indicator is operated at the cylinder. 21 322 A Library of Steam Engineering. EXAMINATION OF VALVES OF ENGINES LONG IN USE. Before making adjustments upon engines that have been long in use, it should be ascertained whether a valve, which should travel in a different place, has worn to a shoulder upon its seat. If changed under such circumstances, loss from leakage may follow, and be sufficient in amount to more than counterbalance the saving that might otherwise result. USEFULNESS OF THE INDICATOR. The indicator is useful in determining the amount of power devel- oped by the engine. The diagram reveals the force of the steam at every point of the stroke; the power is computed from the average amount of this force, which is independent either of the adjustment of the valves, the form of the diagram or of any condition upon which economy depends. The diagram gives what is termed the indicated power of an engine, which is the power exerted by the steam. The indicated power consists of the net power delivered, and in addition, that consumed in propelling the engine itself. The indicator is useful in determining in connection with a feed- water test, the number of pounds of steam consumed by an engine per indicated horse power per hour. This quantity forms a measure of the performance of an engine, and when compared with the performance of the best of its class, shows the economy with which the engine works. The amount of steam consumed is usually found by weighing the feed water before it is supplied to the boiler. This requires the erection of a weighing apparatus, which generally consists of two tanks and a platform scales. One tank is placed on platform scales elevated above the other tank, the latter tank being larger than the former. The water is drawn into the first tank and weighed; it is then discharged into the second tank and from this it is pumped into the boiler. Feed-water tests are of no value when made by measuring the water fed to the boiler unless leakage of water from the boiler, if any exist, is allowed for. Strict attention should always be given to this important matter, and the rate of leakage should be determined by observing the fall of water in the gauge when no steam is being drawn from the boiler, a constant pressure being maintained. A portion of the feed water consumption of an engine may be found by computation from the diagram without the aid of a feed- water test. The manner of making this computation will be found. in the chapter on "The Buckeye Engine." If it were not for the losses produced by leakage and cylinder condensation, the whole amount of feed water consumed might be determined in this manner. Leakage The Steam Engine Indicator. 323 of steam often occurs and cylinder condensation is inevitable, yet the extent to which these losses act is not revealed by any marked effect produced upon the lines of the diagram. The measurement of the consumption of steam by diagram can not, therefore, be taken to show actual performance without allowance for these losses. But diagrams taken in connection with feed-water tests will reveal the extent of the losses produced by leakage and cylinder condensation. These losses are represented by that part of the feed water consumption which remains after deducting the steam computed from the diagram, or accounted for by the indicator as it is termed. The loss by condensa- tion is nearly constant for different engines working under similar conditions, and an allowance may therefore be made for its amount. The other leakage, depending upon the wearing surface of valves, piston and cylinder, is variable in different cases. The fact of the presence of such leakages may be detected by a trial under boiler pressure with the engine at rest, the leakage being revealed by escape at the indicator cock or exhaust pipe. The amount of this leakage may be determined by computing that part of the loss not covered by condensation. In other words, in the case of leaking engines, when the indicator and feed-water tests show that there is more loss than is produced in good practice by condensation, the excess represents the probable amount of loss by leakage. A valuable use for the indicator is thus found in con- nection with feed-water tests. To make it available in practice, tables A, B and C are appended showing the percentages of loss that occur from cylinder condensation. TABLE A. PERCENTAGE OF LOSS BY CYLINDER CONDENSATION TAKEN AT CUT OFF IN SIMPLE ENGINES. Percentage of Stroke Completed at Cut Off. Percentage of Feed Water Consumption Accounted for by the Indicator Diagram. Percentage of Feed Water Consumption Due to Cylinder Condensation. 5 10 58 66 42 34 15 71 29 20 74 26 30 78 22 40 82 18 50 86 14 The quantities in Table A apply to the type of unjacketed sim- ple engines, in common use, having cylinders exceeding 20 inches in diameter. 324 A Library of Steam Engineering. TABLE B. CONDENSATION TAKEN AT PERCENTAGE OF LOSS BY CUT OFF IN THE CYLINDER HIGH PRESSURE CYLINDER- Percentage of Stroke Completed at Cut Off. IN COMPOUND ENGINES. Percentage of Feed Water Consumption Accounted for by the Indicator Diagram. Percentage of Feed Water Consumption Due to Cylinder Condensation. 10 74 15 76 20 78 30 82 40 85 50 88 26 24 22 18 15 12 The quantities in Table B apply to compound engines of the best class, having steam jacketed cylinders. TABLE C. PERCENTAGE OF LOSS BY CYLINDER CONDENSATION TAKEN AT CUT OFF IN THE HIGH PRESSURE CYLINDER-IN Percentage of TRIPLE EXPANSION ENGINES. Stroke Completed at Cut Off. Percentage of Feed Water | Percentage of Feed Water Consumption Accounted for by the Indicator Diagram. Consumption Due to Cylinder Condensation. 22 20 16 13 10 15 20 30 40 50 78 80 84 87 90 The quantities in Table C apply to triple expansion engines of the best class, having steam jacketed cylinders, supplied with dry but not superheated steam. As an example of the use of the tables, Table A will be taken as an illustration of a test made on a 16 inch non-condensing engine of the automatic cut-off four-valve type. A feed-water test gave a con- sumption of 32.61 pounds per indicated horse power per hour. The cut off was 23.7 per cent., and the steam accounted for by the indicator was 69.2 per cent. Referring to Table A it appears that good practice calls for a considerable larger per cent. to be accounted for. At 23.7 per cent. cut off the amount is about 75 per cent. A leakage test of the valves and piston showed that the steam valves were tight, and that the piston was practically tight. The exhaust pipes, however, leaked badly, a considerable blow of steam issuing from the exhaust pipes when pressure was applied to them. Evidently leakage was the The Steam Engine Indicator. 325 cause of the low percentage of feed water accounted for by the diagram, its extent being represented by the difference between 69 per cent. and 75 per cent. The leaking valves were repaired, after which a feed-water test gave a consumption of 29.37 pounds per indicated horse power per hour. The cut off was 22.3 per cent. of the stroke, and the per- centage of steam accounted for by the diagram was brought up to 74.5 per cent. or practically the figure given in the table. COMBINING THE DIAGRAMS FROM COMPOUND ENGINES. There are certain losses in simple or non-compound engines which compounding corrects in a great part, but in turn introduces other losses, which it is desirable to reduce to the least amount possible. These are the losses between the two cylinders, and consists of conden- sation in the passages, pipes and receiver-if one is used-friction in the parts and pipes, and expansion of steam taking place between the two cylinders without doing useful work. The extent of these losses can be shown by combining the diagrams from the two cylinders and drawing in the hyperbolic curve. This curve should just touch the expansion line of the high-pressure diagram at the point where the exhaust from that cylinder begins, and the space between the curve and both diagrams below this point and also the space between the two diagrams represent the loss between the two cylinders. To correctly combine the two diagrams the clearance in each cylinder should be known and accounted for, as well as the piston displacement, and the relative length of the two diagrams when com- bined is the ratio of the volume of the two cylinders, that is, the piston displacement plus the clearance at one end. It is best to decide on the total length of the low-pressure diagram first, and a length that can easily be divided into one hundred parts will be found most convenient, as percentages of this length can be easily measured; 10 inches for a scale of tenths, or 12 inches for a scale of eighths, are fair lengths. Decide on the scale of pressure to which the two diagrams are to be plotted-usually it will be most convenient to take the scale of the original low-pressure diagram-draw in the atmospheric and vacuum lines and erect perpendiculars at the two extremes of the combination diagram, one of which is the clearance line. All measurements of distances should be made from the clearance line, and all measurements of pressure from the atmospheric line. Divide each original diagram into any desired equal parts, say ten, as that is a good number. Find the volume of piston displacement of the low-pressure cylinder, and add to it the volume of clearance; the total length of the diagram rep- resents this total volume. Divide the clearance by the total volume, and the quotient will give the percentage this clearance bears to the 326 A Library of Steam Engineering. whole length. Set off this distance from the clearance line, and divide the remainder-representing the piston displacement-into the same number of equal parts that the original diagram is divided into. If the scale selected is the same as that of the original diagram, transfer the pressures directly with a pair of dividers from the lines on the diagram to the corresponding lines on the combination; draw in the connecting portions of the diagram, and the result will be an elongated diagram from the low-pressure cylinder, or as if it had been taken with the same spring as before, but with greatly enlarged paper drum. Find the total volume of the high-pressure cylinder and divide by the total volume of the low-pressure cylinder, and the quotient will give the percentage of length of the diagram, which should be meas- ured from the clearance line. Divide the clearance volume of the high- pressure cylinder by the total volume of the low-pressure cylinder, and measure off the percentage of length as before from the clearance line. Divide the remaining length of the high-pressure diagram-rep- resenting the piston displacement-into the same number of equal parts as the original diagram, and transfer the pressure from the lines. on the original diagram to the corresponding lines on the combination and the new scale of pressures. If the original high-pressure diagram was taken with a No. 40 spring, and the combination diagram made to a scale of 10 pounds per inch, the new diagram will be four times as high as before, although it may be shorter. Next draw the hyperbolic curve as shown in Fig. 162. T H.P. Cyl. Schie 40 1 + L10 91 81 71 61 5 41 31 2 Fig. 160. I L.P. Cyl. Scale 10 6] 7: 8: 91 Fig. 161. 10 21 31 4: 5) 140 45 The Steam Engine Indicator. 327 High-pressure cylinder, diameter 30 inches. Low-pressure cylinder, diameter 50 inches. Boiler pressure, 82 pounds. Revolutions of engine per minute, 56. Vacuum, 12 pounds. Clearance Line Initial Pressure No. 586 Atmosphere 21 31 5. 61 7 81 166 110 Vacuum 10 20 30 40 50 60 70 80 90 100 Fig. 162. Fig. 162 is a combination of 160 and 161, taken from a tandem compound engine. In order to make the explanation clearly understood, the diagrams in Figs. 160, 161 and 162 have the construction lines shown, and the necessary calculations are given below: The combination diagram shown in Fig. 162, was drawn 12 inches long over all, and by photographing reduced to its present length. Diameter of high-pressure cylinder, 30 inches. Diameter of low-pressure cylinder, 50 inches. Stroke of piston, 72 inches. Diameter of piston rod, both cylinders, 6 inches. VOLUME OF CYLINDERS. TO DETERMINE THE VOLUME OF HIGH AND LOW PRESSURE CYLINDERS. RULE.-First, determine the area of the diameter of the cylinder; then determine the area of the diameter of the piston rod; then subtract the area of the piston rod from the area of the cylinder, and multiply the remainder by the length of the stroke of the piston, in inches, and the product will give the volume of piston displacement for that stroke. The cards being taken from the piston rod end. Second, add the volume of clearance to the volume of piston displacement, and the sum will give the total volume of the cylinder. 328 A Library of Steam Engineering. VOLUME OF HIGH-PRESSURE CYLINDER. (30X30X.7854)-(6.25 x 6.25X.7854) X72-48684.9825 cubic inches. Volume of piston displacement of high- 48684.9825 pressure cylinder. Clearance volume of high-pressure cylinder. 2545 Total volume of high-pressure cylinder. 51229.9825 Volume of piston displace ment of high-pressure cylinder. VOLUME OF LOW-PRESSURE CYLINDER. (50X50X.7854)-(6.25 x 6.25X.7854) X72-139163.0625 cubic inches. Volume of piston displacement of low- 139163.0625 pressure cylinder. Clearance volume of low-pressure cylinder. 7673 Total volume of low-pressure cylinder. 146836.0625 Volume of piston displace- ment of low-pressure cylinder. COMPUTING THE HORSE POWER OF AN ENGINE FROM THE INDICATOR DIAGRAM. The work done by the steam in the cylinder of an engine is meas- ured by the product of the force exerted on the piston, into the dis- tance through which the piston moves, and is usually expressed by the term foot pounds. If, for example, a force of 33 pounds per square inch on a piston having an area of 100 square inches is employed to drive the piston 100 times over a stroke of 4 feet, the work done by the steam amounts to 33X100X100X4=1,320,000 foot pounds. The amount of horse power which the steam develops is the foot pounds of work done in a minute divided by 33,000. In the example given, the horse power developed when 100 strokes are made per minute is 1,320,000 divided by 33,000. Thus: 1,320,000÷33000-40 horse power. The force exerted by the steam upon the piston is given by the indicator diagram, but as it varies in amount in different points of the stroke, it is necessary to determine the equivalent force which, acting constantly, would produce the same result. This is done by comput- ing from the diagram what is termed the MEAN EFFECTIVE PRESSURE. The product of the mean effective pressure, expressed in pounds per square inch; the area of the diameter of the cylinder, expressed in square inches; the length of the stroke, expressed in feet; and the number of strokes per minute, which is twice the number of revolu- tions of the crank per minute, gives the number of foot pounds of work performed per minute. This result, divided by 33,000 gives the amount of horse power developed. To compute from the diagram the mean effective pressure, two lines are drawn perpendicular to the atmospheric line, one at each end The Steam Engine Indicator. 329 of the diagram, and the intermediate space divided into 10 equal parts, with a perpendicular line at each point of division, as shown in Fig. 163. A ready method of performing the division is to lay upon the diagram a scale of 10 equal parts, the total length of which may be a small amount in excess of the length of the diagram. It is so placed in a diagonal position that the extreme points on the scale lie upon the two outside perpendicular lines. The desired points may then be dotted with a sharp pencil opposite the intermediate divisions on the scale, as shown in Fig. 163. The points where the lines of division cross the diagram should be dotted; and in locating these points they C D Fig. 163. E F A H B should be so placed that the area of the figure enclosed by a straight line joining them is exactly equal to the area enclosed by the curved line of the diagram. The proper location can be readily determined by the eye. Fig. 163 shows the extreme perpendicular lines A B and C D, the intermediate lines of division, the points of intersection, and those points which require special location, as for example, the one at E, which is so placed that the area enclosed by the dotted line E F and E G is equal to that enclosed by the diagram from F to G. The determination of the mean effective pressure now consists in finding the average length of the various perpendicular lines included between the points of intersection measured on the scale of the spring. This may be done by measuring each line with the scale and averaging the results. A better and quicker method is to employ a strip of paper and mark one after another of the various distances on its edge, mak- ing a mechanical addition and requiring only a final measurement with the scale of the spring and then dividing it by the number of spaces in the diagram. The proper course to pursue is to lay the edge of the paper on the first line and mark the distance from A to H, 330 A Library of Steam Engineering. shown in the diagram (Fig. 163), starting from the end of the measuring paper. Transfer the paper to the last line, and add to the first meas- urement the distance from I to D. Mark off from the end of the paper one-half the sum of these two distances, and from the middle point continue the addition of the intermediate nine divisions. When all have been marked, measure with the scale of the spring, from the end of the paper to the last addition, and divide the result by 10. This gives the mean effective pressure. It is essential that one-half the sum of the first and last distances on the diagram be taken, and this, together with the sum of distances of intermediate lines, be divided by 10. It is an error to divide by 11, as is sometimes done. EXPANSION LINE IN NON-CONDENSING ENGINES CARRYING A LIGHT LOAD. Fig. 164. In non-condensing engines carrying a light load the expansion line often extends below the atmospheric line, as shown in Fig. 164. In computing the mean effective pressure in such cases, the distances below the atmospheric line must be deducted from those above it. The sum of the first and last distances here becomes the difference between the first and last. PLANIMETERS. When the mean effective pressure on a large number of diagrams is desired, much time and labor may be saved by the employment of a planimeter, an instrument designed to measure the areas of irregular figures. It is operated by moving a tracer, with which it is fitted, over the line of the diagram, and it records the area upon a graduated wheel. One of these devices is the Coffin Averaging Instrument, shown in Fig. 165. In using this planimeter the grooved metal plate I is first con- nected to the board upon which the apparatus is mounted, in the position shown in the engraving, being held in place by a thumb-screw applied from the back side. The indicator card is then placed under The Steam Engine Indicator. 331 the clamps C and K, which may be sprung away from the board a suf- ficient amount to allow the card to be introduced, and the card is moved toward the left into such a position that the atmospheric line is near to and parallel with the lower edge of the stationary clamp C, while the extreme left-hand end of the diagram is even with the perpendicu- lar edge of the clamp. The movable clamp, K, which is fastened at the bottom to a sliding plate, is then moved toward the left until the vertical beveled edge just touches the extreme right-hand end of the diagram. The diagram shown in the engraving represents the proper O D K Fig. 165. location which should exist when these preliminary adjustments have been completed. The slide at the bottom of clamp K fits closely, so that the application of a slight pressure is required to displace it. The beam of the instrument is next placed on the board with the pin at the lower end resting in the groove I, and the weight Q applied to the top of the pin so as to keep it securely in place. The tracer O is moved to the right-hand end of the diagram and set at the point D on the line of the diagram, where the clamp K and the diagram touch each other. Here a slight indentation is made in the paper by press- ing the finger on the top of the tracer, and this serves as a starting point. The graduated wheel is next turned so as to bring its zero mark to the zero mark on the vernier. The instrument is now ready for operation. The tracer O is carefully moved over the line of the dia- gram, in the direction of motion of the hands of a watch, and con- tinued until a complete circuit is made and the tracer finally reaches the starting point D. Keeping the eye on the wheel, the tracer is now 332 A Library of Steam Engineering. moved upward by sliding it along the edge of the clamp K until the reading on the wheel returns to zero. Another light indentation is made in the paper to mark the new position which the tracer occupies. This point is represented at A in the engraving. The instrument is now moved away, the clamp pushed back, and the distance between the two points D and A is measured, by employing a scale correspond- ing to the number of the spring used in the indicator. The distance thus found is the mean effective pressure, expressed in pounds per square inch of piston. This planimeter determines the desired result without computa- tion, but it may be used also for determining the area enclosed by the diagram. This area is given by the reading on the graduated wheel, when the circuit of the diagram has been made and the tracer reaches the starting point D. The wheel has fifteen main divisions, each of which represents one square inch of area. Each division has five sub- divisions, each subdivision representing one-fifth, or two-tenths of a square inch of area. The vernier scale enables the subdivisions to be used to fifteenths, each of these fifteenths, therefore, representing two- one-hundredths of a square inch. Having obtained the area in this manner, the mean effective pressure may be computed by dividing the number of the spring representing the pressure per inch in height by the length of the diagram, in inches, and multiplying the quotient by area, in square inches. the In first placing the indicator card under the clamps, care must be observed that the ends of the diagram set a little way from the edge of the clamp, so as to allow for one-half of the diameter of the tracer, and to bring the center of the tracer over the center of the line of the diagram. Another devise for measuring diagrams is the Amsler Polar Plani- meter, as shown in Fig. 166. This instrument does not give the mean effective pressure directly, but it determines the area of the diagram, and from this area the mean effective pressure is computed in the manner above pointed out. The polar planimeter consists of two arms A and B, and the meas- uring wheel C. To operate it a piece of smooth, hard paper is laid on the table, and the instrument placed upon it, with the needle point A pressed into the board. This point serves as a center about which the apparatus is turned. The indicator card is laid under the tracer B and held either by tacks, which fasten it to the table, or what is quite sufficient, by the pressure of the fingers. The tracing point is set on the line of the diagram-say, near the middle of the steam line-and a slight indentation made in the paper to serve as a starting point. The graduated wheel is set at the zero mark. The tracer is then moved The Steam Engine Indicator. 333 over the line of the diagram in the direction of motion of the hands of a watch, finally making a complete circuit returning to the starting mark. The number of divisions and fractions of a division, shown on the wheel at the point opposite the stationary zero mark, indicates the area of the diagram traced. The wheel has ten main graduations, each of which represents one square inch of area. Each main division is subdivided ten times, and each subdivision represents one-tenth of a square inch of area. A stationary vernier scale is placed beside the graduated edge of the wheel, and serves to indicate the smaller frac- tions, viz., hundredths. To read the vernier, the To read the vernier, the eye is run along the B Fig. 166. stationary scale until a line of division is found which is just opposite- a division on the wheel. The number of the division on the vernier, reckoned from the zero mark, is the number of hundredths sought. If, for example, the reading of the area is two main divisions on the wheel, and four of the subdivisions and the line of coincidence on the vernier is No. 7 reckoned from zero, the area sought is 2.47 square inches. To reduce this to mean effective pressure, two perpendicular lines are drawn, one through each terminal point of the diagram, and the length. of the diagram is measured by measuring the distance between these two perpendiculars. Suppose this distance is 3.78 inches, and suppose the number of the spring employed is 40. Then the mean effective pressure is found by dividing 40 by 3.78 and multiplying the quotient by 2.47, the product will be 26.13 pounds per square inch, the mean effective pressure. In working the polar planimeter care must be taken to place the diagram so that the two arms are not brought too near each other at one end of the course, nor yet carried too far apart at the other end. 334 A Library of Steam Engineering. At a point midway between the two extremes the two arms should lie about parallel to each other. As an example of the manner of computing the horse power of an engine, suppose an engine has a cylinder 15 inches in diameter, a pis- ton rod 2 inches in diameter, a stroke of 2 feet, and engine running at a speed of 135 revolutions per minute. Suppose the indicator dia- grams show a mean effective pressure of 36 pounds per square inch, this being the average of the indications at the two ends. The area of the cylinder to be used is the net area obtained by deducting one- half the area of the piston rod. The area of a cylinder 15 inches in diameter is 15X15X.7854-176.7150 square inches. The area of the piston rod, 2 inches in diameter, is 2.5X2.5.7854-4.9+ square inches. One-half the area of the piston is 2) 4.9 2.45 square inches. The net area of the cylinder is 176.7150 Area of cylinder. 2.45 Half the area of the rod. 174.2650 Net area of cylinder. The speed of the piston in feet per minute is 135X2.5X2=675 feet. The horse power developed is 174.265X36X675 33000 128.3+ Horse power. RULE.-Multiply the net area of the diameter of the cylinder by the mean effective pressure per square inch, and multiply the product by the number of feet the piston traveled per minute, and divide the last product by 33,000, and the quotient will give the horse power of the engine, as shown in the preceding and in the following examples. Net area of cylinder in square inches mean effective pressure X piston speed in feet per minute 33000 -Horse power. When an engine has more than one cylinder, the power developed in each cylinder is computed according to the rule given, and the results added together. U. S. Marine Boiler Inspection and Engineers' License Laws. 335 CHAPTER XVI. UNITED STATES MARINE BOILER INSPECTION AND ENGI- NEERS' LICENSE LAWS. In entering upon this subject the student is warned against the practice of committing to memory so-called examination questions, and also in his wasting time in learning how to answer them; because the time spent in that direction neither tends to aid him in passing a successful examination before the inspection officers, nor to make him a safe or a practical engineer. The examination of engineers under the steam boiler inspection. laws of the United States, or of any of the States which have enacted laws upon that subject, are not based upon any fixed set of questions, nor upon any particular part of the profession of steam engineering; but they reach into and cover the whole field, and especially that relat- ing to the management of steam boilers with safety to human life. This, then, is the sole object; it is the entire object in the passage of all laws relating to the construction, inspection and management of steam boilers, and to the examination and licensing of engineers. The United States Government has set the example on this conti- nent in the enactment of laws requiring the inspection of steam boilers and appurtenances therewith connected, and the examination and licensing of engineers. Since then a number of States have followed the example of the Federal Government in providing, by law, for the examination and licensing of engineers, and some of the States have also included the inspection of steam boilers in their statutory enactments. It is, there- fore, safe to predict that ere long every State in the Union will have similar laws placed upon its statute books. As such laws are enacted for the sole purpose of protecting life and property, they necessarily exclude incompetent persons from hav ing the care or management of steam boilers and steam engines. It is therefore, of the utmost importance that all persons who desire to follow the profession of steam engineering should fit themselves for successful examination by the inspection officers. Aside from the requirements of the law, it is the duty of every engineer to acquire the knowledge and experience so indispensable in the prevention of steam boiler 336 A Library of Steam Engineering. accidents. Ignorance on the part of the boiler maker who constructs the boiler, or on the part of the engineer under whose care and man- agement it is placed, is no excuse for the destruction of human life, even in the absence of any law excluding ignorant or incompetent persons from having the care or management of steam plants. There- fore, to enable engineers, and those who desire to become engineers, to take charge of and manage steam boilers and steam engines without hazard to human life, and to pass a successful examination before inspection officers, this chapter is devoted to the practical applica- tion of the rules and formulæ contained in the chapters which pre- cede it. As those laws will be taken as the standard by the various States in the enactment of inspection and license laws, and by all inspectors appointed and operating under them, the student will have the assur- ance, if he acquires the information contained in this work, that he will have no difficulty in passing a successful examination if he is a practical engineer; and in addition thereto he will have the satisfac- tion of knowing how to manage steam plants without endangering life or property. In order to give full and complete information in regard to the requirements of the United States laws upon this subject, every sec- tion of the statutes, and every rule of the United States Board of Supervising Inspectors of Steam Vessels, requiring any mathematical calculations to be made by the United States inspectors of boilers, by candidates for the office of inspector, by marine boiler makers, and by engineers in their examination for engineers' license, will be separately and fully treated under this head. QUALIFICATIONS OF INSPECTORS OF BOILERS. Section 4415 of the Revised Statutes of the United States provides that an inspector of boilers shall be "a person of good character and suitable qualifications and attainments to perform the services required of inspectors of boilers, who, from his knowledge and experience of the duties of an engineer employed in navigating vessels by steam, and also of the construction and use of boilers, and machinery, and appur- tenances therewith connected, is able to form a reliable opinion of the strength, form, workmanship, and suitableness of boilers and machinery to be employed without hazard to life, from imperfections in the ma- terial, workmanship, or arrangements of any part of such apparatus for steaming." It is not only important that inspectors should thoroughly under- stand all the rules and regulations governing the construction and inspection of steam boilers to enable them to perform the duties pre- U. S. Marine Boiler Inspection and Engineers' License Laws. 337 scribed by law, but under department regulations, no person can be appointed inspector without first passing a most rigid examination, and no person can pass the required examination without a thorough knowledge of the requirements of the law as laid down and explained in this chapter. QUALIFICATIONS OF ENGINEERS. Section 4441 of the Revised Statutes of the United States provides that, "whenever any person applies for authority to perform the duties. of engineer of any steam vessel, the inspectors shall examine the applicant as to his knowledge of steam machinery, and his experience as an engineer, and also the proofs which he produces in support of his claim; and if, upon full consideration, they are satisfied that his character, habits of life, knowledge, and experience in the duties of an engineer are all such as to authorize the belief that he is a suitable and safe person to be intrusted with the powers and duties of such a station, they shall grant him a license, authorizing him to be employed. in such duties for the term of one year, in which they shall assign him to the appropriate class of engineers; but such license shall be sus- pended or revoked upon satisfactory proof of negligence, unskillful- ness, intemperance, or the willful violation of any provision of this title." It will be observed that the knowledge required on the part of inspectors of boilers and on the part of engineers covers completely the entire field of steam engineering so far as it relates to the safety of human life. In order, then, that an inspector shall have the qualifica- tions prescribed by law he must be "able to form a reliable opinion of the strength, form, workmanship and suitableness of boilers and machinery to be employed without hazard to life from imperfections in the material, workman- ship, or arrangements of any part of such apparatus for steaming." It will also be noticed that the knowledge and experience required on the part of engineers are so interwoven with those required by inspectors, that the qualifications of persons desiring to become chief engineers of steam vessels require a knowledge and experience little short, if anything, of those required by inspectors. It is, therefore, impossible to separate the two, so far as the requirements of the United States laws are concerned, and the decisions and rules of the department having charge of the steam vessel inspection service require that no person shall be appointed inspector of boilers who has not held a license as chief engineer for a period of not less than three years, in addition to the qualifications prescribed by law. Therefore, the student of steam engineering, the inspector of boilers and the engineer are equally interested in the instructions given in this chapter, as neither ean perform the duties. 22 338 A Library of Steam Engineering. prescribed by law without the knowledge here given, nor perform the duties of an engineer without hazard to life or property. In order, then, that each may become familiar with the law, as well as to enable them to obtain the information each is required to possess, each section of the statutes, and each rule of the Board of Supervising Inspectors upon which any calculation is made, will be cited. STRENGTH OF STEAM BOILERS. [Section 4415 of the Revised Statutes of the United States.] BURSTING PRESSURE OF SINGLE-RIVETED BOILERS. RULE. For boilers with single-riveted longitudinal seams in the cylin- drical shell, multiply the thickness of material in the weakest plate, in deci- mals of an inch, by the tensile strength, in pounds per square inch; then divide the product by one-half of the diameter of the boiler, in inches, and multiply the quotient by the constant .56, the product will give the bursting pressure in pounds per square inch. Example. Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of weakest plate. Let 40 inches equal diameter of the boiler. Let .56 equal a constant for single-riveted longitudinal seams. Then we have: .25 X 60000 40÷2 X.56-420 lbs. Bursting pressure per square inch. .25 Performing the operation in the ordinary way, we have: Thickness of material in weakest plate. 60000 Tensile strength of material per square inch. Dividing by one-half of 40÷2= 20) 15000.00 (750 the boiler's diameter. 140 .56 A constant. 100 4500 100 3750 0 420.00 lbs. Bursting pressure per square inch. BURSTING PRESSURE OF SINGLE-RIVETED BOILERS SIMPLE RULE. Another rule is as follows, and obviates the necessity of dividing the diameter of the boiler: U. S. Marine Boiler Inspection and Engineers' License Laws. 339 RULE.-Multiply the thickness of material in the weakest plate, in deci- mals of an inch, by the tensile strength, in pounds per square inch; then divide the product by the diameter (instead of half diameter as in the previ ous case) of the boiler, in inches, and multiply the quotient by 1.12, the product will give the bursting pressure in pounds per square inch. Taking the same boiler employed in the previous example, we have: .25 60000 Dividing by diameter of boiler. 40) 15000.00 (3 75 120 1.12 300 750 280 375 375 200 200 420.00 lbs. Bursting pressure per square inch. BURSTING PRESSURE OF DOUBLE-RIVETED BOILERS. What amount of pressure per square inch is required to burst a boiler with longitudinal seams double riveted? RULE-Multiply the thickness of material in the weakest plate, in deci- mals of an inch, by the tensile strength, in pounds per square inch, and divide the product by one-half of the diameter of the boiler, in inches, and multiply the quotient by the constant .70; the product will give the bursting pressure in pounds per square inch. Example.-Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of weakest plate. Let 48 inches equal diameter of the boiler. Let .70 equal a constant for double-riveted longitudinal seams. Then we have: .25 60000 Dividing by one-half of 48-2= 24) 15000.00 (625 the diameter of the boiler. 144 .70 60 437.50 lbs. 48 Bursting pressure per square inch. 120 120 Putting the operation in condensed form, we have: .25 X 60000 48÷2 X.70-437.50 lbs. Bursting pressure per square inch. 340 A Library of Steam Engineering. BURSTING PRESSURE OF DOUBLE-RIVETED BOILERS SIMPLE RULE. To obviate the necessity of dividing the diameter of the boiler, we perform the operation according to the following rule: RULE.-Multiply the thickness of material in the weakest plate, in deci- mals of an inch, by the tensile strength, in pounds per square inch, and divide the product by the diameter of the boiler, in inches; then multiply the quotient by the constant 1.4; the product will give the bursting pressure, per square inch, for boilers with double-riveted longitudinal seams. Taking the same boiler employed in the previous example, we have: .25 X 60000 48 X 1.4-437.50 lbs. Bursting pressure per square inch. Performing the operation in the ordinary way, we have: .25 Thickness of material. 60000 Tensile strength of material. Dividing by diameter of boiler. 48) 15000.00 (312.5 144 1.4 A constant. 60 125 00 48 3125 120 437.50 lbs. Bursting pressure per square inch. 96 240 240 SAFE-WORKING PRESSURE OF MARINE BOILERS. [Section 4433 of the Revised Statutes of the United States.] This section of the Revised Statutes relates to boilers made of plates. It therefore does not include pipe or sectional boilers. Its provisions are as follows: 'SEC. 4433. The working steam pressure allowable on boilers constructed of plates inspected as required by this title, when single riveted, shall not produce a strain to exceed one-sixth of the tensile strength of the iron or steel plates of which such boilers are constructed; but where the longitudinal laps of the cylindrical parts of such boilers are double riveted, and the rivet holes for such boilers have been fairly drilled instead of punched, an addition of twenty per centum to the U. S. Marine Boiler Inspection and Engineers' License Laws. 341 working pressure provided for single riveting may be allowed: Provided, that all other parts of such boilers shall correspond in strength to the additional allowances so made; and no split caulking shall in any case be permitted." The portion of this section relating to the drilling of rivet holes. means that all rivet holes for boilers requiring the addition of twenty per cent. provided by law must be drilled. To emphasize this inter- pretation of the law, the Board of Supervising Inspectors of Steam Vessels has provided, by rule, that the additional twenty per cent. in the working steam pressure shall be allowed only where the longitu- dinal laps of the cylindrical part of the boilers are double riveted, and "all the rivet holes have been fairly drilled instead of punched." It will therefore be observed that a boiler, in order to be allowed the additional twenty per cent., must have all of its rivet holes drilled as well as to have its longitudinal laps double riveted. Therefore, if a boiler has its longitudinal seams double riveted, and the holes in such seams have been drilled, it will not be allowed the additional twenty per cent. if the other holes have been punched instead of drilled. The student will therefore bear this in mind, as he proceeds with his studies of the requirements of the United States laws. SAFE-WORKING PRESSURE OF SINGLE-RIVETED BOILERS. What is the safe-working pressure for cylindrical boilers, or boilers having cylindrical shells, where the longitudinal seams are single riveted? RULE.-Multiply the thickness of material in the weakest plate, in deci- mals of an inch, by the tensile strength, in pounds per square inch, and divide the product by the radius, or one-half of the diameter of the boiler, in inches, and then divide the quotient by 6 (which is called the factor of safety), and the last quotient will give the pressure, per square inch, allowable. Example.-Let 25 one hundredths of an inch equal thickness of material. Let 60,000 pounds per square inch equal tensile strength of material. Let 40 inches equal diameter of the boiler. Let 6 equal a factor of safety. Then we have: (.25X60000) (40÷2) 6 -125 lbs. Safe-working pressure per square inch. 342 A Library of Steam Engineering. Putting the operation in the ordinary form, we have: .25 Thickness of material. 60000 Tensile strength of material. Dividing by one-half of the 40÷2= 20) 15000.00 (750 The quotient. diameter of the boiler. 140 100 100 0 Then dividing the quotient by 6, the factor of safety, we have: 6)750 125 lbs. Safe-working pressure per square inch The law provides that no greater strain than one-sixth of the ten- sile strength of the material will be allowed. The examples given hardly make it clear that the pressure given in the answer produces no greater strain than that limited by law. This point will therefore require explanation. The meaning of the law is that but one-sixth of the actual strength of the material will be allowed on the material. Material, then, having a tensile strength of 60,000 pounds per square inch of section, would, if it were one-fourth of an inch in thickness, and one inch in width, have an actual strength of 15,000 pounds, or one-fourth of 60,000 pounds; and such material would be allowed a strain of 2500 pounds, one-sixth of its actual strength. In other words, the law means one- sixth of the actual strength of the material in the plates-the strength of the seams is not taken into consideration except so far as limiting the pressure is concerned; but the law means the solid part of the sheets, and not the seams, when it refers to one-sixth of the tensile strength of the plates. In order to make this plain to the student, a mathematical demon- stration will be made. ACTUAL STRENGTH OF BOILER PLATE. RULE.-Multiply the tensile strength of the material, per square inch, by the thickness of the plates, in decimals of an inch, and the product will give the actual strength of the material one inch in width. Example.-Let 65,842 pounds equal tensile strength per square inch of the material. Let 25 one hundredths of an inch equal thickness of the plates. U. S. Marine Boiler Inspection and Engineers' License Laws. 343 Then we have: 65842 .25 329210 131684 16460.50 lbs. Strength of material one inch in width and 25 one hundredths of an inch in thickness. Taking the material employed in the boiler in the example pre- ceding the last, in order to prove the correctness of the rule given for such cases: Example. Let 25 one hundredths of an inch equal thickness of material. Let 60,000 pounds per square inch equal tensile strength of material. Then we have: .25 60000 15000.00 lbs. Tensile strength of material one inch in width and 25 one hun- dredths of an inch in thickness. This material, then, would be allowed a strain of one-sixth of 15,000 pounds; and one-sixth of 15,000 pounds is 6) 15000 2500 lbs. Now, if we divide one-sixth of the strength of the material one inch in width and 25 one hundredths of an inch in thickness by one- half of the diameter of the boiler in inches, we have the pressure per square inch that will produce a strain on the plates equal to one-sixth of the tensile strength of such plates. Thus: Taking a boiler 40 inches in diameter, we have: 40÷2-20 inches. One-half of the diameter of the boiler. 20) 2500 (125 lbs. Pressure per square inch on a boiler 20 50 40 100 40 inches in diameter that will produce a strain on the material having a tensile strength of 15,000 pounds equal to one-sixth of that. strength. 100 344 A Library of Steam Engineering. STRAIN PRODUCED ON THE PLATES BY THE PRESSURE PER SQUARE INCH IN THE BOILERS. RULE.-Multiply the given pressure, in pounds per square inch, by one- half of the diameter of the boiler, in inches, and the product will give the exact strain on each inch in the entire length of the boiler. Example. Let 125 pounds per square inch equal given pressure. Let 40 inches equal diameter of the boiler. Then we have: 125 lbs. Given pressure per square inch. 402 20 inches. Half diameter of boiler. 2500 lbs. Strain produced on each inch in the length of the boiler. To allow such a strain, the material would have to be six times stronger per square inch than 2500 pounds. Hence: 2500×6=15000 lbs. Required tensile strength of mate- rial one inch in width and 25 one hundredths of an inch in thickness. THICKNESS OF MATERIAL REQUIRED. How thick would the material have to be with a tensile strength of 60,000 pounds per square inch, to have a tensile strength of 15,000 pounds in a width of one inch? RULE.-Divide the tensile strength of the material one inch in width by the tensile strength of the material per square inch, and the quotient will give the thickness required in decimals of an inch. Example. Let 15,000 pounds equal tensile strength of material one inch in width. Let 60,000 pounds equal tensile strength of the same material one square inch. Then we have: 15000 60000 =.25 Decimals of an inch. Thickness of material required. Performing the operation in the ordinary way, we have: 60000) 15000.0 (0.25 Decimals of an inch. Thickness of 12000 O material required. 3000 00 3000 00 To sum the matter up briefly, the student is informed that the safe-working pressure allowable, under the law, for single-riveted boilers, is simply a pressure per square inch that will produce a strain on the material in the boiler not exceeding a strain equal to one-sixth of the strength of that material. This must not, however, be confounded U. S. Marine Boiler Inspection and Engineers' License Laws. 345 with the strength of the material per square inch-that is, if it was one inch thick-because the law simply means one-sixth of the strength of the material just as it is in the boiler, no matter what its thickness may be; and when adding twenty per cent. for double riveting, the law simply allows an addition of twenty per cent. on that allowed for single-riveted boilers. SAFE-WORKING PRESSURE OF DOUBLE-RIVETED BOILERS- HOLES DRILLED. [Section 4433 of the Revised Statutes of the United States.] What is the safe-working pressure for a boiler with double-riveted longitudinal seams and all the rivet holes drilled? RULE.-Multiply the thickness of material in the weakest plate in the boiler, in decimals of an inch, by the tensile strength, in pounds per square inch, and divide the product by one-half the diameter of the boiler, in inches; then divide the quotient by 6, and then add twenty per cent. to the last quotient; the sum will give the pressure, per square inch, allowable under the United States rule. Example.-Let 25 one hundredths of an inch equal thickness of material. Let 60,000 pounds per square inch equal tensile strength of material. Let 40 inches equal diameter of the boiler. Let 6 equal a factor of safety. Let 20 per cent. equal amount of pressure to be added for double riveting. Then we have: {(.25 x 60000) + 6 } x 6X1.2-150 lbs. Safe-working pressure for double riveting. Performing the operation in the ordinary way, we have: .25 Thickness of material. 60000 Tensile strength of material. Dividing by one-half of the 40-2= 20) 15000.00 diameter of the boiler. Factor of safety. 6) 750 125 lbs. Pressure for single riveting. 1.2 Multiplying by unity and the required per- centage annexed is equivalent to add- ing 20 per cent. 250 125 150.0 lbs. Safe-working pressure for double- riveted longitudinal seams and all rivet holes in the boiler drilled. 346 A Library of Steam Engineering. SAFE-WORKING PRESSURE OF DOUBLE-RIVETED BOILERS SIMPLE RULE. The simplest rule for calculating the safe-working pressure pre- scribed by the United States laws for double-riveted boilers is as follows: RULE.-Multiply the thickness of material in the weakest plate in the boiler, in decimals of an inch, by the tensile strength of the material, in pounds per square inch, and divide the product by the diameter of the boiler, in inches, and then multiply the quotient by .4; the product will give the safe-working pressure, per square inch, for a boiler with double-riveted longitudinal seams, if all the rivet holes have been drilled, as required by law. Example.-Let 25 one hundredths of an inch equal thickness of material. Let 60,000 pounds per square inch equal tensile strength of material. Let 40 inches equal diameter of the boiler. Let .4 equal a constant. Then we have: .25 X 60000 X.4-150 lbs. Safe-working pressure per 40 square inch. Performing the operation in a simple way, we have: 25 Thickness of material. 60000 Tensile strength of material. Dividing by the diameter of the boiler. 40) 15000.00 (375) 120 .4 A constant. 300 150.0 lbs. Safe-working pressure per 280 square inch. 200 200 TENSILE STRENGTH OF BOILER PLATE. [Section 4430 of the Revised Statutes of the United States; and Section 3, of Rule 1, of the United States Board of Supervising Inspectors of Steam Vessels.] RULE. Divide the breaking strain of sample piece, in pounds, by its area of cross section at point of fracture before it is broken, and the quotient will give the tensile strength of the material in pounds per square inch. Example.-Let 15,525 pounds equal the strain required to pull the sample in two. Let 97 one hundredths of an inch equal width of sam- ple at point intended for fracture. Let 26 one hundredths of an inch equal thickness of sample at point intended for fracture. U. S. Marine Boiler Inspection and Engineers' License Laws. 347 Then we have: 15525 -=61558+lbs. Tensile strength per square inch. .97X.26 Performing the operation in a simple way, we have: .26 26 .97 582 194 .2522 Area of sample piece before breaking. Next, dividing the strain at which the sample broke by the area, we have: .2522) 15525.0000 (61558+ lbs. Tensile strength per square- inch. 15132 3930 2522 14080 12610 14700 12610 20900 20176 DUCTILITY OF BOILER PLATE. [Sections 4430 and 4431 of the Revised Statutes of the United States; and Section 6, of Rule 1, of the United States Board of Supervising Inspectors of Steam Vessels.] Section 6, of Rule 1, of the Board of Supervising Inspectors, pro- vides that the ductility of boiler plate shall be as follows: "Iron of 45,000 pounds tensile strength shall show a contraction of area (at point of fracture) of fifteen (15) per cent., each additional 1,000 pounds tensile strength shall show one (1) per cent. additional contraction of area, up to and including 55,000 T. S. Iron of 55,000 T. S. and upward, showing twenty-five (25) per cent. reduction of area, shall be deemed to have the lawful ductility. All steel plate of one- half inch thickness and under, shall show a contraction of area of not less than fifty (50) per cent. Steel plate over one-half inch in thick- ness up to three-quarters inch in thickness, shall show a reduction of not less than forty-five (45) per cent. All steel plate over three- fourths inch thickness, shall show a reduction of not less than forty (40) per cent." 348 A Library of Steam Engineering. "" The terms "contraction of area" and "reduction of area are synonymous terms, and they are employed interchangeably in the law; and what is meant is the contraction of the material at point of frac- ture. All iron or steel plate having the quality of being ductile will contract in dimensions at the point at which they are pulled apart; and the extent, amount or per cent. (all of which are the same) of such reduction or contraction of the material at the point at which it is broken, represents its ductility. Iron or steel having no ductility will show no contraction in size of sample at point of fracture. On the other hand, the more ductility the material has the greater will be the contraction and elongation of the sample employed in testing. If the material is very ductile it will stretch considerably before breaking, and consequently reduce considerably at the point where it is pulled apart, while material that has no ductility will not stretch, neither will it be reduced at point of fracture by pulling it apart. The importance, then, of having boilers made of material of the greatest possible amount of ductility becomes apparent, when it is known that boiler material not having the proper amount of ductility is liable to crack and produce dangerous results. And for that reason no material ought to be allowed in any steam boiler with less ductility than that prescribed by the United States laws; and engineers placed in charge of the building of new boilers should see to it that they are constructed in accordance with these laws in every particular, and more especially for land purposes, where there is no law governing the con- struction of steam boilers, because, so far as marine boilers are concerned, unless they are made strictly up to the very letter of the law they will be allowed reduced pressure or be condemned. DUCTILITY OF BOILER PLATE-HOW DETERMINED. RULE. First, multiply the width of the sample by its thickness, in deci- mals of an inch, at the smallest point (the point intended for fracture before breaking) and call the product "the area of sample before breaking." Second, multiply the width of the sample by its thickness, in decimals of an inch, after breaking at the point of fracture, and call the product "the area of sample after breaking." Third, subtract "the area of sample after breaking" from "the area of sample before breaking," and then divide the remainder by "the area of sam- ple before breaking," and the quotient will give the percentage of ductility of the material. Example. Let 97 one hundredths of an inch equal original width of sample. Let 26 one hundredths of an inch equal original thick- ness of sample. U. S. Marine Boiler Inspection and Engineers' License Laws. 349 Let 88 one hundredths of an inch equal reduced width of sample. Let 18 one hundredths of an inch equal reduced thick- ness of sample. Then we have: (.97X.26) (.88X.18) .97X.26 .37+ Percentage of ductility. Performing the operation in the ordinary way, we have: .97X.26 .88X.18 Dividing the remainder by the area of sample before breaking. .2522 Area of sample before breaking. .1584 Area of sample after breaking. .2522)-.0938.0 (0.37+ Percentage of ductility. 756 6 181 40 176 54 The material, it will be observed, shows a reduction of area or a duc- tility-which is the same thing in law-of thirty-seven per cent. The material being steel, and having a tensile strength of 60,000 pounds per square inch, the test shows that it is too hard for boiler purposes, and therefore liable to crack when subjected to unequal expansion and con- traction, as well as to repeated heating and cooling; and hence it would be condemned by the inspectors for marine purposes. It is, therefore, well to remark here, that any material that is not good enough for marine boilers ought not be allowed in the construction of any other class of boilers. And yet, material condemned by the United States. inspectors for use in marine boilers is, with few exceptions, used in the construction of land boilers. SAFE-WORKING PRESSURE BASED UPON DATA FOR ANY GIVEN MARINE BOILER. [Section 4433 of the Revised Statutes of the United States; Sections 3 and 6, of Rule 1, and Section 3, of Rule 2, of the United States Board of Supervising Inspectors of Steam Vessels.] What would be the working pressure allowable, per square inch, in a boiler 48 inches in diameter, made of steel plates 26 one hun- dredths of an inch in thickness, samples of which, one inch by 26 one hundredths of an inch, broke in testing, at a strain of 15,452 pounds, and reduced in dimensions at point of fracture to 85 one hundredths of an inch in width, and to 15 one hundredths of an inch in thickness; the boiler being made with double-riveted longitudinal and all of the rivet holes in the boiler drilled? seams, Section 4433 of the Revised Statutes prescribes the working steam pressure allowable, and limits the pressure to produce a strain on the 350 A Library of Steam Engineering. material in the shell of the boiler not to exceed one-sixth of the tensile strength of the material for single-riveted longitudinal seams, but allows an addition of twenty per cent. for double-riveted longitudinal seams, in case all the rivet holes in the boiler have been drilled. Section 3, of Rule 1, of the United States Board of Supervising Inspectors relates to the testing of boiler plate, and reads as follows: "SEC. 3. To ascertain the tensile strength and other qualities of iron plate, there shall be taken from each sheet to be used in shell or other parts of boiler, which are subject to tensile strain, a test piece prepared in form according to the following diagram: 10 inches in length, 2 inches in width-cut out in the center in the manner indicated in Fig. 167. IRON PLATE. FORM OF TEST PIECE FOR DETERMINING DUCTILITY AND TENSILE STRENGTH. k- 116 k 10" Fig. 167 V. "All sample pieces of iron plate five-sixteenths (5) of an inch thick and under shall be one inch wide at reduced section; plate over five-sixteenths (15%) of an inch thick shall be reduced in width at cen- ter to an aggregate area approximating four-tenths (1) of one square inch, but such reduced area shall, in no case, exceed 45 nor be less than 35 one hundredths of an inch, and the force at which the piece can be parted in the direction of the fibre or grain (when of iron) represented in pounds avoirdupois, in proportion to the ratio of its area, shall be deemed the tensile strength per square inch of the plate from which the sample was taken; and should the tensile strength ascertained by the test equal that marked on the plates from which the test pieces were taken, the plates must be allowed to be used in the construction of marine boilers. Provided, always, that the plate possesses homogeneousness, toughness, and the ability to with- stand the effect of repeated heating and cooling; but should these tests prove any plate to be overstamped (all plates are required to be stamped with the tensile strength of the material), such plate must be rejected as failing to have the strength stamped thereon. But nothing herein. U. S. Marine Boiler Inspection and Engineers' License Laws. 351 shall be so construed as to prevent the manufacturer from restamp- ing such plate, and all other plates in the lot, at the lowest tensile strength indicated by the deficient sample, provided such restamping is done previous to the use of the plates in the manufacture of marine boilers. When more than one sample shall be tested from one sheet, the sample showing the lowest tensile strength shall be allowed as the tensile strength of the plate." PERCENTAGE OF DUCTILITY OF BOILER PLATE, TENSILE STRENGTH AND SAFE-WORKING PRESSURE. The portion of Section 3, of Rule 2, relating to steam pressure allowable for boilers with double-riveted longitudinal seams reads as follows in regard to the drilling of all the rivet holes: "The pressures allowable on boilers when all the rivet holes have been fairly drilled instead of punched, and the longitudinal laps of their cylindrical parts double riveted." We will now proceed to answer the question contained at the com- mencement of this subject, in accordance with the law: RULE.-First, ascertain the percentage of ductility of the material by subtracting the reduced area of sample test pieces showing the least reduction of area at point of fracture from the original area at same point, and then divide the remainder by the original area, the quotient will give the of ductility of the material. If a reduction of 50 per cent. is shown, the mate- rial will be allowed, and we proceed with our calculations. per cent. Second, ascertain the tensile strength of the material by dividing the lowest strain at which any of the samples broke by the area of the sample before breaking, and the quotient will give the tensile strength. Third, multiply the thickness of the weakest plate in the boiler, in hun- dredths of an inch, by the tensile strength of the material, in pounds per square inch; then divide the product by one-half of the diameter of the boiler, in inches; then divide the quotient by 6; then add 20 per cent. to the last quotient, and the sum will give the working pressure allowable. Example. Let .85"X.15" equal dimensions of sample at point of fracture after breaking. Let 1.00"X.26" equal dimensions of sample at point of fracture before breaking. Let 15,452 pounds equal strain at which sample broke. Let .26" equal thickness of weakest plate in the boiler. Let 48" equal diameter of the boiler. Let 6 equal a constant. Let 20 equal per cent. of pressure to be added for double riveting. 352 A Library of Steam Engineering. Then we have: 1.00X.26.2600 Area of sample before breaking. .85X.15.1275 Area of sample after breaking. Dividing remainder by area of sam- .2600).1325.0 (0.51- Per cent. ductility. ple before breaking. 1300 0 25 00 26 00 It is now demonstrated that the material possesses the lawful duc- tility—not less than fifty per cent.-and therefore it will be allowed in the construction of marine boilers. We will therefore next proceed to determine the tensile strength of the material per square inch, in order to determine the working pressure allowable. Then, we divide the strain at which sample broke by the area of sample at point of fracture before breaking, thus: 1.00X.26.2600) 15452.0000 (59430+ lbs. Tensile strength per 13000 24520 23400 11200 10400 8000 7800 square inch. 2000 Having ascertained the tensile strength of the material per square inch, we now proceed to determine the working pressure allowable under the law, thus: 59430.26 ÷6-107.31 lbs. Pressure for single-riveted 48÷2 boiler. Next we have: 107.31 .20 21.4620 lbs. 20 per cent. of 107.31. Then, adding the twenty per cent. to the pressure that would be allowed for single riveting, we have: 107.31 lbs. Pressure for single riveting. 21.462 20 per cent. of 107.31. 128.772 lbs. Working pressure allowable in the boiler. U. S. Marine Boiler Inspection and Engineers' License Laws. 353 STEEL PLATE. FORM OF TEST PIECE FOR DETERMINING ELONGATION AND TENSILE STRENGTH. 6' 6" 不 ​Fig. 168 Section 3, of Rule 1, provides that sample pieces of steel plate, for determining the percentage of elongation, shall be made similar in form to that shown in Fig. 168; but all test pieces one inch, and over one inch, in thickness shall have the straight part of the reduced section 8 inches in length; and that the length of the reduced section of samples having a thickness of less than one inch shall be governed by the following rule: LENGTH OF REDUCED SECTION OF TEST PIECE. RULE.-Multiply the width by the thickness, and the product by 8, the last product will give the required length of reduced width of sample. The width of reduced part is fixed at one inch, and the elongation under the test shall be not less than 25 per cent. Example.-Let 98 one hundredths of an inch equal width of sample. Let 30 one hundredths of an inch equal thickness of sample. Let 8 equal a constant. Then we have: .98X.30X8=2.3520 inches. Required length of reduced width of sample. ELONGATION OF TEST PIECE. RULE.-Divide the amount of elongation by the original length of the reduced width of test piece, and multiply the quotient by 100, and the product will give the percentage of elongation. 23 354 A Library of Steam Engineering. Example.-Let 75 one hundredths of an inch equal amount of elongation. Let 3 inches equal original length of reduced width of test piece. Let 100 equal a constant. Then we have: .75 X100-25 Percentage of elongation. 3 SPECIAL TEST OF STEEL PLATE. The last paragraph of Section 3, of Rule 1, provides: "That where contracts for boilers for ocean-going steamers require a test of material in compliance with the British Board of Trade, British Lloyds or Bureau Veritas rules for testing, the inspectors shall make the tests in compliance with the following rules: "Steel plates shall, in all cases, have an ultimate elongation not less than twenty per cent. in a length of 8 inches. It is to be capable of being bent to a curve of which the inner radius is not greater than one and one-half times the thickness of the plates after having been heated to a low cherry red, and quenched in water of 82° Fahrenheit." The tests are to be made by United States inspectors, and the test pieces are to be made according to the following form: 6' 6' 2'' Approxi- mately 116- Approxi- mately S' 91/2" Fig. 169 K PERCENTAGE OF ELONGATION. RULE.-Subtract the length of sample piece before breaking from the length of sample piece after breaking; then divide the remainder by the length of sample piece before breaking, and multiply the quotient by 100; the product will give the per cent. of elongation. Example. Let 9.6 inches equal length of sample piece after breaking. Let 8 inches equal length of sample piece before breaking. U. S. Marine Boiler Inspection and Engineers' License Laws. 355 Then we have: +10.2 9.6-8 X100-20 Per cent. elongation. 8 Performing the operation, we have. 9.6 8 8) 1.6 .2 100 20.0 Per cent. elongation. matielonens BUMPED HEADS FOR BOILERS. WORKING STEAM PRESSURE. Section 17, of Rule 2, of the United States Board of Supervising Inspectors, provides that the working pressure allowable on bumped heads shall be according to the following rule: RULE.-Multiply one-sixth of the tensile strength, per square inch, of the plate by the thickness of the plate, and divide the product by six-tenths of the radius to which the head is bumped, which will give the pressure, per square inch, of steam allowed. Example.-Let 60,000 pounds equal tensile strength of plate per square inch. Let 6 equal a constant. Let 25 one hundredths of an inch equal thickness of plate. Let 24 inches equal radius to which the head is bumped. Let .6 equal a constant. Then we have: (60000÷6)X.25 or morrego edy 247.6 -=173.61+1bs. Pressure per square inch allowable. Performing the operation, we have: 6) 60000 10000 .25 ered ou ey Am't carried forward, 500 00 2500.00 2000 O 356 A Library of Steam Engineering. Dividing by six-tenths 24X.6= 14.4)2500.00(173.61+lbs. Pressure per square inch of the radius to which the head is bumped. allowable. 144 1060 1008 520 432 880 864 160 144 WORKING STEAM PRESSURE SIMPLE RULE. RULE.-First, multiply the thickness of plate, in decimals of an inch, by the tensile strength in pounds per square inch, and call the product "Product No. 1." Second, multiply the radius, in inches, to which the head is bumped by 6, and multiply the product by .6, and call the last product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the pressure, per square inch, allowable. Example. Let 25 one hundredths of an inch equal thickness of plate. Let 60,000 pounds equal tensile strength of plate per square inch. Let 24 inches equal radius to which the head is bumped. Let 6 equal a constant. Let 6 equal a constant. Then we have: .25X60000 =173.61+lbs. Pressure per square inch allow 24X6X.6 able. Performing the operation, we have: Next we have: .25 Thickness of plate. 60000 Tensile strength of plate. 15000.00 "Product No. 1." 24 Radius to which the head is bumped. 6 A constant. 144 .6 A constant. 86.4 "Product No. 2." U. S. Marine Boiler Inspection and Engineers' License Laws. 357 Finally, dividing "Product No. 1" by "Product No. 2," we have: 86.4) 15000.00 (173.61+lbs. Pressure per square inch allowable. 864 6360 6048 3120 2592 5280 5184 960 864 THICKNESS OF MATERIAL REQUIRED. RULE.-Multiply the radius, in inches, to which the head is bumped by .6, and multiply the product by the required pressure, in pounds per square inch, and call the last product "Product No. 1." Second, divide the tensile strength of the plate, in pounds per square inch, by 6, and call the quotient "Quotient No. 1." Third, divide "Product No. 1" by "Quotient No. 1," and the quotient of this operation will give the required thickness of material in decimals of an inch. Example.-Let 24 inches equal the radius to which the head is bumped. Let .6 equal a constant. Let 173.61 pounds equal the required pressure per square inch. Let 60,000 pounds equal the tensile strength of plate per square inch. Let 6 equal a constant. Then we have: t 24X.6X173.61 60000:6 .25 Decimals of an inch. Required thickness of Performing the operation, we have: plate. 24 Radius to which the head is bumped. .6 A constant. 14.4 173.61 Required pressure per square inch. 69 444 694 44 1736 1 2499.984" Product No. 1." 358 A Library of Steam Engineering. Next we have: A constant. 6) 60000 Tensile strength of plate in pounds. 10000 "Quotient No. 1." Finally, dividing "Product No. 1" by "Quotient No. 1," we have: 10000) 2499.984 (.2499+ Decimals of an inch. Required thickness 2000 O of plate. 499 98 400 00 99 984 90 000 9 9840 9 0000 THICKNESS OF MATERIAL REQUIRED SIMPLE RULE. RULE.-Multiply the required pressure, in pounds per square inch, by the radius, in inches, to which the head is bumped; then multiply the product by the constant 6; then multiply the product of that operation by .6; then divide the last product by the tensile strength, per square inch, of the plate, and the quotient will give the required thickness of material for the head. Example.-Let 175 pounds equal required pressure per square inch. Let 24 inches equal radius to which the head is bumped. Let 6 equal a constant. Let 6 equal a constant. Let 60,000 pounds equal tensile strength of plate per square inch. Then we have: 175X24X6X.6 60000 =.252 Decimals of an inch. Required thickness of plate. Performing the operation, we have: 175 24 700 350 4200 6 25200 .6 Am't carried forward, 15120.0 U. S. Marine Boiler Inspection and Engineers' License Laws. 359 Dividing by tensile strength of 60000) 15120.0 (0.252 Decimals of an inch. Required plate per square inch. thickness of plate. 12000 0 3120 00 3000 00 120 000 120 000 RADIUS TO WHICH THE HEAD IS REQUIRED TO BE BUMPED. RULE.-First, multiply the thickness of material for the head, in deci- mals of an inch, by the tensile strength, in pounds per square inch, and call the product "Product No. 1." Second, multiply the required pressure, in pounds per square inch, by 6; then multiply the product by .6, and call the last product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the radius, in inches, to which the head is required to be bumped. Example.-Let 252 one thousandths of an inch equal thickness of plate. Let 60,000 pounds equal tensile strength of plate per square inch. Let 175 pounds equal the required pressure per square inch. Let 6 equal a constant. Let .6 equal a constant. Then we have: .252 X 60000 175X6X.6 --24 inches. Radius to which the head is required to be bumped. Performing the operation, we have: .252 60000 Thickness of plate. Tensile strength of plate. 15120.000 "Product No. 1." Next we have: 175 Pressure per square inch. 6 A constant. er men 1050 .6 A constant. 630.0 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 630) 15120 (24 inches. Radius to which the head is required to be bumped. 1260 2520 2520 360 A Library of Steam Engineering. HYDROSTATIC PRESSURE FOR MARINE BOILERS. [Section 4418 of the Revised Statutes of the United States.] The hydrostatic test is made with water, instead of steam pressure, and it is applied to boilers to discover any weakness or defect that could not be discovered by an ocular inspection of the boiler. The boiler is pumped full of water until a certain pressure in the boiler is reached beyond that allowed as a safe-working pressure, and if no defect or weakness is shown, the boiler is considered safe and the working steam pressure prescribed by law is allowed. Section 4418 of the Revised Statutes provides that: "All boilers used on steam vessels and constructed of iron or steel plates, inspected under the provisions of Section 4430, shall be subjected to a hydrostatic test, in the ratio of one hundred and fifty pounds to the square inch to one hundred pounds to the square inch of the work- ing steam pressure allowed." HYDROSTATIC PRESSURE FOR SINGLE-RIVETED BOILERS. RULE.-Multiply the thickness of material, in decimals of an inch, by the tensile strength of the material, in pounds per square inch; then divide the product by one-half the diameter of the boiler, in inches; then divide the quo- tient by 6; then divide the last quotient by 2, and then add the quotient of the last operation to the preceding quotient, and the sum will give the hydrostatic pressure in pounds per square inch. Example.-Let 60,000 pounds per square inch equal tensile strength of material. Let 25 one hundredths of an inch equal thickness of material. Let 40 inches equal diameter of the boiler. Let 6 equal a constant. Let 2 equal a constant. Then we have: .25 60000 40÷2= 20) 15000.00 6) 750 2) 125 62.5 187.5 lbs. Hydrostatic pressure. U. S. Marine Boiler Inspection and Engineers' License Laws. 361 HYDROSTATIC PRESSURE FOR SINGLE-RIVETED BOILERS-SIMPLE RULE. RULE.-Multiply the thickness of material, in hundredths of an inch, by the tensile strength of the material, in pounds per square inch; then divide the product by the diameter of the boiler, in inches; and then multiply the quotient by .5; the last product will give the hydrostatic pressure in pounds per square inch. Taking the boiler in the previous example, we have: .25 60000 40) 15000.00 (375 120 .5 300 187.5 lbs. Hydrostatic pressure. 280 200 200 HYDROSTATIC PRESSURE FOR DOUBLE-RIVETED BOILERS. RULE.-Multiply the thickness of material in the weakest plate, in deci- mals of an inch, by the tensile strength of the material, in pounds per square inch; then divide the product by one-half of the diameter of the boiler, in inches; then divide the quotient by 6; then add 20 per cent. of the last quotient to the quotient; then divide the sum by 2; then add the last quotient to the sum, and the last sum will give the hydrostatic pressure per square inch required. Example. Let 25 one hundredths of an inch equal thickness of weakest plate. Let 60,000 pounds per square inch equal tensile strength of plate. Let 40 inches equal diameter of the boiler. Let 6 equal a constant. Let 20 per cent. equal additional safe-working pressure for double riveting. Let 2 equal a constant Then we have: .25 60000 Thickness of weakest plate. Tensile strength of material. 40÷2 20) 15000.00 6) 750 Am't carried forward, 125 lbs. Working pressure for single riveting. 362 A Library of Steam Engineering. Am't brought forward, Adding 20% 125X.20= 125 lbs. 25 lbs. Working pressure for single riveting. Twenty per cent. of working pressure. 2) 150 lbs. 75 lbs. Working pressure for double riveting. One-half of the working pressure. 225 lbs. Hydrostatic pressure required. HYDROSTATIC PRESSURE FOR DOUBLE-RIVETED BOILERS-SIMPLE RULE. RULE.-Multiply the thickness of material in the weakest plate, in deci- mals of an inch, by the tensile strength of the material, in pounds per square inch; then divide the product by the diameter of the boiler, in inches; then multiply the quotient by .6; the last product will give the hydrostatic pressure in pounds per square inch. Taking the boiler employed in the previous example, we have: .25 X 60000 40 .6=225 lbs. Hydrostatic pressure per square inch. Performing the operation in the ordinary way, we have: Thickness of material. 60000 Tensile strength of material. .6 A constant for double riveting. .25 Dividing by diameter of boiler. 40) 15000.00 (375 120 300 225.0 lbs. 280 200 200 Hydrostatic pressure per square inch. HYDROSTATIC PRESSURE FOR SINGLE AND DOUBLE RIVETED BOILERS. The rule for determining the hydrostatic pressure for both sin- gle and double riveted longitudinal seams in condensed form is as follows: RULE.-Add one-half of the working pressure allowable to the working pressure, and the sum will give the hydrostatic pressure required. Example.-Let 150 pounds per square inch equal the working press- ure for either single or double riveted longitudinal seams. Then we have: 2) 150 225 lbs. Hydrostatic pressure required. 75 U. S. Marine Boiler Inspection and Engineers' License Laws. 363 REDUCTION OF WORKING STEAM PRESSURE. [Section 4418 of the Revised Statutes of the United States.] Under authority of Section 4418 of the Revised Statutes, inspec- tors may at any time cut down the working steam pressure to the limit of safety, for any deterioration in the material, or other causes of weakness produced by the use of the boiler for a long period, or on account of negligence on the part of the engineer in failing to take proper care of the boiler, and thus causing weakness or deterioration in the material. Hence, at inspections a lower hydrostatic pressure is applied than would be in the case of a new boiler, and the working steam pressure allowed is always two-thirds of the hydrostatic pressure. Therefore, while in theory the working steam pressure is based upon the strength of the material in the boiler, and the manner in which the boiler is constructed, yet in practice the working steam pressure is based upon the hydrostatic pressure that has been applied, and this hydrostatic pressure is always based upon the strength of the boiler. THICKNESS OF MATERIAL FOR MARINE BOILERS LIMITED. [Section 4434 of the Revised Statutes of the United States.] 30 This section of the Revised Statutes of the United States limits the thickness of material in the shell of boilers, to which the heat is applied to the outside of the shell, to 30 (1) one hundredths of an inch for all steamers on the Mississippi River and its tributaries, and to boilers on all other steamers to one-half inch. THICKNESS OF MATERIAL REQUIRED FOR SINGLE-RIVETED BOILERS. RULE.-Multiply the given pressure, in pounds per square inch, by 6; then multiply the product by one-half of the diameter of the boiler, in inches; then divide the last product by the tensile strength of the material, in pounds per square inch, and the quotient will give the thickness of material required in decimals of an inch. Example.-Let 125 pounds per square inch equal the given pressure. Let 6 equal a constant. Let 40 inches equal diameter of the boiler. Let 60,000 pounds per square inch equal tensile strength of material. Then we have: 125X6X(40-2) 60000 =.25 inches. Thickness of material required. 364 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 125 Working steam pressure. 6 A constant. 750 Multiplying by one-half of 40÷2- 20 diameter of boiler. Dividing by tensile strength of 60000) 15000.0 (0.25 Decimals of an inch. Thickness of material. material required. 12000 0 3000 00 3000 00 THE WORKING STEAM PRESSURE BASED ON THE HYDROSTATIC PRESSURE. [Section 4418 of the Revised Statutes of the United States.] In no case will a working steam pressure per square inch be allowed greater than two-thirds of the hydrostatic pressure that has been ap- plied. It sometimes happens that the required amount of hydrostatic pressure can not be obtained on account of leakages in valves and joints; and even in such cases only two-thirds of the hydrostatic press- ure obtained will be allowed as a working steam pressure. If the owner of the boiler desires the full amount of steam pressure to which his boiler would otherwise be entitled, he must get his valves and joints tight enough to withstand the increased pressure that the hydrostatic pressure will produce, and then the hydrostatic pressure must again be applied, and if the required pressure in making the hydrostatic test is obtained the boiler will be allowed the amount of steam pressure to which it would otherwise be entitled, but not until then. U. S. Marine Boiler Inspection and Engineers' License Laws. 365 45,000 Diameter Thickness Tensile Strength. of of 1-6 Boiler. Plates. Pressure. 7,500 20 per ct. Addition'l 1-6 1-6 9,166.6 Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l .1875 78.12 93.74 .21 87.5 105.00 .23 36 .25 104.16 86.8 97.21 95.83 114.99 106.47 124.99 .26 108.33 129.9 INCHES .29 120.83 .3125 130.2 .33 137.5 .35 145.83 104.16 95.48 116.65 106.94 127.76 117.12 115.74 138.88 120.37 144.44 144.99 134.25 161.11 156.24 144.67 173.6 165.00 152.77 183.32 174.99 114.57 128.3 UNITED STATES TABLE OF PRESSURES ALLOWABLE ON MARINE BOILERS. 50,000 Tensile Strength. 8,333.3 55,000 Tensile Strength. 60,000 Tensile Strength. 1-6 10,000 65,000 Tensile Strength. 1-6 10,833.3 70,000 Tensile Strength. 1-6 11,666.6 20 per ct. Addition'l 145.82 Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 104.16 124.99 116.66 139.99 126.38 112.84 135.4 121.52 151.65 136.11 163.33 140.54 127.77 153.32 138.41 166.09 149.07 178.88 127.31 152.77 138.88 166.65 150.46 180.55 162.03 194.43 132.4 158.88 144.44 173.32 156.48 187.77 168.51 202.21 147.68 177.21 161.11 193.33 174.53 209.43 187.90 225.48 159.14 190.96 173.6 208.32 188.07 225.68 202.5 243.04 168.05 201.66 183.33 219.99 198.61 238.33 213.88 256.65 .375 156.25 187.5 162.03 194.43 173.61 208.33 190.97 178.23 213.87 194.44 233.32 210.64 252.76 226.84 272.20 229.16 208.33 249.99 225.69 271.82 243.05 291.66 .1875 74.01 88.89 82.23 98.67 90.46 108.54 98.68 118.41 106.9 128.28 115.13 138.16 .21 82.89 99.46 92.1 110.52 101.31 121.57 110.52 132.62 119.73 143.67 128.93 154.71 .23 90.78 108.93 100.87 121.04 110.96 133.15 121.05 145.26 131.13 157.35 141.22 169.46 .25 38 98.68 118.41 109.64 131.56 120.61 144.73 131.57 157.88 142.54 171.04 153.5 184.20 .26 102.63 123.15 114.03 136.83 125.43 150.51 136.84 164.2 148.24 177.88 159.64 191.56 INCHES .29 114.47 .3125 137.36 127.19 123.35 148.02 137.00 152.62 139.91 167.89 152.63 183.15 165.35 198.42 178.06 213.67 164.46 150.76 180.91 164.47 197.36 178.17 213.08 191.88 230.25 .33 130.26 156.31 144.73 173.67 159.2 191.04 173.68 208.41 188.15 225.78 202.62 243.14 .35 138.15 165.78 153.5 184.21 168.85 202.62 184.21 221.05 199.56 239.47 214.91 257.89 .375 148.00 177.60 164.73 197.67 180.91 217.09 197.36 236.83 213.81 256.57 230.26 276.31 .1875 70.31 84.37 78.12 93.74 85.93 103.11 93.75 112.5 101.56 121.87 109.37 131.24 .21 78.75 94.50 87.49 104.98 96.24 115.48 105.00 126.00 113.74 136.48 122.49 146.98 .23 86.25 103.5 95.83 114.99 105.41 126.49 115.00 138.00 124.58 149.49 134.16 160.99 .25 93.75 112.5 104.16 124.99 114.58 40 137.49 125.00 150.00 135.41 162.49 145.83 174.99 .26 97.5 117.00 108.33 129.99 119.16 142.99 130.00 156.00 140.83 168.99 151.66 181.99 INCHES .29 108.75 130.5 .3125 117.18 140.61 120.83 130.2 144.99 132.91 159.49 145.00 174.00 157.08 188.49 169.16 202.99 156.24 143.22 171.86 156.25 187.45 169.27 203.12 182.29 218.74 123.75 148.5 137.49 164.98 151.24 181.48 165.00 198.00 178.74 214.48 192.49 230.98 .35 .375 131.25 157.5 140.62 168.74 145.83 174.99 160.41 192.49 175.00 210.00 189.58 227.49 204.16 244.99 156.24 187.48 171.87 206.24 187.5 225.00 203.12 243.74 218.74 262.48 366 A Library of Steam Engineering. UNITED STATES TABLE OF PRESSURES ALLOWABLE ON MARINE BOILERS.-Continued. 45,000 Diameter Thickness of Tensile Strength. 50,000 Tensile Strength. 55,000 Tensile Strength. of 1-6 7,500 1-6 8,333.3 1-6 9,166.6 60,000 Tensile Strengt 1-6 10,000 65,000 Tensile Strength. 1-6 10,833.3 70,000 Tensile Strength. 1-6 11,666.6 Boiler. Plates. Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'i Pressure. 20 per ct. Addition'l .1875 66.96 80.35 74.40 89.28 81.84 98.20 89.28 107.13 96.72 116.06 104.16 124.99 .21 75.00 90.00 83.32 99.99 91.66 109.99 100.00 120.00 108.33 129.99 .23 82.14 98.56 91.23 105.51 100.39 120.46 109.52 131.42 118.65 142.38 116.66 139.99 127.77 153.32 42 .25 89.28 107.13 99.2 119.04 .26 92.85 111.42 103.17 123.8 109.12 130.94 113.49 119.04 142.84 128.96 154.75 138.88 166.65 136.18 123.8 148.56 134.12 160.94 144.44 173.32 INCHES .29 103.57 124.28 115.07 138.08 126.57 151.85 138.09 165.7 149.6 179.52 161.11 193.33 .3125 111.6 133.92 124.00 148.8 136.4 163.68 148.74 178.56 161.2 193.44 173.61 208.23 .33 117.85 141.42 130.94 157.12 144.04 .35 125.00 150.00 138.88 166.65 .375 133.92 160.7 148.8 172.84 152.77 183.32 166.66 199.99 178.56 163.68 196.40 157.14 188.56 170.23 204.27 183.33 219.99 180.55 216.66 194.44 233.32 178.57 214.28 193.45 232.14 208.33 249.99 .1875 63.92 76.7 71.02 85.22 78.12 93.74 85.22 102.26 92.32 110.78 99.42 119.3 .21 71.59 85.9 79.54 95.44 87.49 104.98 95.45 114.54 103.4 124.08 111.36 133.63 .23 78.4 94.08 87.12 104.54 44 .25 85.22 102.26 .26 88.63 106.35 INCHES .29 98.86 118.63 94.69 113.62 98.48 118.17 109.84 131.80 .3125 106.53 127.83 .33 112.5 135.00 95.83 114.99 104.54 125.44 104.16 124.99 113.63 136.35 108.33 129.99 118.18 141.81 120.83 144.99 118.36 142.03 130.2 156.24 142.04 124.99 149.98 137.49 164.98 150.00 113.25 135.9 121.96 146.35 123.1 147.72 123.56 159.07 128.02 153.62 137.87 165.44 131.81 158.17 142.79 171.33 153.78 184.53 170.44 153.88 184.65 165.71 198.85 180.00 162.49 194.98 174.99 209.98 .35 119.31 .375 143.17 132.57 159.08 145.83 127.81 153.37 142.04 170.44 156.24 174.99 159.09 190.9 172.34 206.8 185.6 222.72 187.48 170.45 204.54 184.65 221.58 198.86 238.63 .1875 61.14 73.36 67.93 81.51 74.72 89.66 81.51 97.81 88.31 105.97 95.1 114.12 .21 68.47 82.16 76.08 91.29 83.69 100.42 91.3 109.56 98.91 118.69 106.52 127.82 .23 75.00 90.00 83.33 100.00 91.66 109.99 100.00 120.00 108.33 129.99 116.66 139.99 .25 81.51 46 97.81 90.57 108.68 99.63 119.55 108.69 130.42 117.75 141.3 126.8 152.16 .26 84.78 101.73 94.2 113.04 103.62 124.34 113.44 135.64 122.46 146.95 131.88 158.25 INCHES .29 94.56 113.47 105.07 126.00 115.57 138.68 126.09 151.3 136.59 163.92 147.1 176.52 .3125 101.9 122.28 113.21 135.85 124.54 149.44 135.86 163.03 147.19 176.62 158.51 190.21 .33 107.6 129.12 119.56 .35 114.13 136.95 126.8 .375 122.28 146.73 143.47 131.52 157.82 143.97 152.16 139.49 167.38 152.17 135.86 163.03 149.45 179.34 163.04 155.43 172.16 182.6 164.85 195.64 176.62 211.94 186.51 167.39 200.86 197.82 177.53 213.03 190.21 228.25 U. S. Marine Boiler Inspection and Engineers' License Laws. 367 Diameter Thickness of of 1-6 45,000 Tensile Strength. 7,500 50,000 Tensile Strength. UNITED STATES TABLE OF PRESSURES ALLOWABLE ON MARINE BOILERS.-Continued. 55,000 Tensile Strength. 65,000 Tensile Strength. 1-6 8,333.3 1-6 9,166.6 60,000 Tensile Strength. 1-6 10,000 1-6 10,833.3 70,000 Tensile Strength. 1-6 11,666.6 Boiler. Plates. Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per et. Addition'l .1875 58.59 70.30 65.1 78.12 71.61 85.93 78.12 93.74 84.63 101.55 91.13 109.35 .21 65.62 78.74 72.91 87.49 80.2 96.24 87.49 104.98 94.79 113.74 102.08 122.49 .23 71.87 86.24 79.85 95.82 87.84 105.4 95.83 114.99 103.81 124.57 111.8 133.16 48 .25 78.12 93.74 86.8 104.16 95.48 114.57 104.16 124.99 112.84 135.4 121.52 145.82 .26 81.25 97.50 90.27 108.32 99.3 119.16 108.33 129.99 117.36 140.83 126.38 151.65 INCHES .29 90.62 108.74 100.69 120.82 110.76 132.91 120.83 144.99 130.9 157.08 140.97 169.16 .3125 97.65 117.18 108.56 130.2 119.35 143.22 130.21 156.25 141.05 169.26 151.9 182.28 .33 103.12 123.74 114.58 137.49 126.04 151.24 137.5 165.00 148.95 178.74 160.41 192.49 .35 109.37 131.24 121.52 145.82 133.67 160.4 145.83 174.99 157.98 189.58 170.13 204.15 .375 117.18 140.61 130.2 156.24 143.22 171.86 156.25 187.50 169.27 203.12 182.29 218.74 .1875 52.08 62.49 57.87 69.44 63.65 76.38 69.44 82.44 75.23 90.27 81.01 97.21 .21 58.33 69.99 64.81 77.77 71.29 85.54 77.77 93.32 84.25 101.1 90.74 108.88 .23 63.88 76.65 70.98 85.17 78.08 93.69 85.18 102.21 92.28 110.73 99.38 119.25 .25 54 69.44 83.32 77.16 92.59 84.87 101.84 92.59 111.10 100.3 120.36 108.02 129.62 26 72.22 86.66 80.24 96.28 88.27 105.92 96.29 115.54 104.31 125.17 112.44 134.8 INCHES .29 80.55 96.66 89.5 107.40 98.45 118.14 107.41 128.88 116.35 139.62 125.3 150.36 .3125 86.8 104.16 96.44 115.72 106.09 127.30 115.55 138.66 125.38 150.45 135.03 162.03 20 .00 91.66 109.99 101.84 122.22 112.03 134.43 122.22 146.66 .35 97.22 116.66 .375 104.16 108.02 129.62 118.82 124.99 115.74 138.88 127.31 142.58 152.77 138.88 129.62 155.54 166.65 132.4 158.88 140.43 168.51 151.23 181.47 150.46 180.55 162.03 194.43 142.59 171.10 .1875 46.87 56.24 52.08 62.49 57.29 68.74 62.5 75.00 67.7 81.24 72.91 87.49 .21 52.5 63.00 58.33 69.99 64.16 76.99 69.99 84.00 75.83 90.99 81.66 97.99 .23 57.5 69.00 63.88 76.65 70.27 84.32 76.66 91.99 83.05 99.66 89.44 107.32 60 .25 62.5 75.00 69.44 83.32 76.38 91.65 83.33 99.99 90.27 108.32 97.22 116.66 .26 65.00 78.00 72.22 86.66 79.44 95.32 86.66 103.99 93.88 112.65 101.11 121.33 INCHES .29 72.5 87.00 80.55 96.66 88.61 106.33 96.66 .3125 78.12 93.74 86.8 104.16 95.48 114.57 .33 82.5 99.00 91.66 109.99 100.83 120.99 .35 87.5 105.00 97.22 116.66 .375 93.75 112.5 104.16 124.99 112.77 104.18 124.99 112.95 135.54 121.52 109.99 132.00 119.16 142.99 128.33 106.94 128.32 116.66 133.99 126.38 151.65 136.11 114.58 137.49 125.00 150.00 135.41 162.49 115.99 104.72 125.66 135.32 145.82 153.99 163.33 145.83 174.99 368 A Library of Steam Engineering. of of 1-6 Boiler. Plates. Pressure. 7,500 20 per ct. Addition'l 1-6 50,000 Tensile Strength. 8,333 3 UNITED STATES TABLE OF PRESSURES ALLOWABLE ON MARINE BOILERS.-Continued. 45,000 Diameter Thickness Tensile Strength. 65,000 Tensile Strength. 1-6 55,000 Tensile Strength. 9,166.6 60,000 Tensile Strength. 1-6 10,000 1-6 10,833.3 70,000 Tensile Strength. 1-6 11,666.6 Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l .1875 42.61 51.13 47.34 56.8 52.07 62.49 56.81 68.17 61.55 73.86 66.28 79.53 .21 47.72 57.26 53.00 63.63 58.33 69.99 63.63 76.35 68.93 82.71 74.24 89.08 .23 52.27 62.72 58.00 69.69 63.88 76.65 69.69 83.62 75.5 90.6 81.31 97.57 .25 56.81 68.17 63.13 75.75 69.44 83.32 75.75 90.90 82.07 98.48 88.37 106.04 66 .26 59.09 70.9 65.65 78.78 72.22 86.66 78.78 94.53 85.35 102.42 91.91 110.29 INCHES .29 65.90 79.08 73.23 87.87 80.55 96.66 87.87 105.44 95.2 114.24 102.52 123.02 .3125 71.00 85.2 78.91 94.69 86.89 104.16 94.69 113.62 102.58 .00 75.00 90.00 83.33 99.99 91.66 109.99 99.99 120.00 108.33 123.09 110.47 132.56 129.99 116.66 139.99 .35 79.56 95.47 88.38 106.05 97.22 116.66 106.00 127.27 114.89 137.86 123.73 148.47 .375 85.22 102.26 94.69 113.62 104.16 124.99 113.62 136.34 123.1 147.72 132.57 159.08 .1875 39.06 46.87 43.4 52.08 47.74 57.28 52.08 62.49 56.42 67.70 60.76 72.91 .21 43.75 52.5 48.6 58.33 53.47 64.16 58.33 69.99 63.19 75.82 68.05 81.66 .23 47.91 57.49 53.24 63.88 58.56 70.27 63.88 76.65 62.21 83.05 74.53 89.43 .25 52.08 62.49 72 57.87 69.44 63.65 76.38 69.44 83.32 75.22 90.26 81.01 97.21 .26 54.16 64.99 60.18 72.21 66.2 79.44 72.22 86.66 78.24 93.88 84.25 101.10 INCHES .29 60.41 72.49 67.12 80.54 73.84 88.60 80.55 96.66 87.26 104.71 93.98 112.77 .3125 65.10 78.12 72.33 86.8 79.57 95.48 86.8 104.16 94.03 112.83 101.27 121.52 .33 68.75 82.5 76.38 91.65 84.02 100.82 91.66 109.99 99.3 119.16 106.94 128.32 .35 72.91 87.49 81.01 97.21 89.11 106.93 97.22 116.66 105.32 126.38 113.42 136.1 .375 78.12 93.74 86.8 104.16 95.48 114.57 104.16 124.99 112.84 135.43 121.52 145.82 .1875 36.05 43.21 40.06 48.07 44.07 52.87 48.07 57.68 52.08 62.49 56.08 67.29 .21 40.38 48.45 44.87 53.84 49.35 59.22 53.84 64.60 58.33 69.99 62.82 75.38 .23 44.23 53.07 49.14 58.96 54.05 64.86 58.95 70.76 63.88 76.65 68.80 82.56 .25 48.07 78 57.68 53.41 64.09 58.76 70.5 64.4 76.92 69.44 83.32 74.78 89.73 .26 50.00 60.00 55.55 66.66 66.11 73.33 66.66 79.99 72.22 86.66 77.77 93.32 INCHES .29 55.76 66.91 61.96 74.35 68.16 81.79 74.35 89.22 80.55 96.66 86.75 104.1 .3125 60.09 72.1 66.77 80.12 73.45 88.14 80.12 96.14 86.8 104.16 93.48 112.17 .33 63.46 76.15 70.51 84.61 77.56 93.07 84.61 101.53 91.66 109.99 98.71 118.45 .35 67.3 80.76 74.78 89.73 82.26 98.71 89.74 107.68 97.22 .375 72.11 86.53 80.12 96.14 88.14 105.76 96.15 115.38 104.16 116.66 104.70 124.99 125.64 112.17 134.6 U. S. Marine Boiler Inspection and Engineers' License Laws. 369 Diameter Thickness 45,000 Tensile Strength. 55,000 Tensile Strength. UNITED STATES TABLE OF PRESSURES ALLOWABLE ON MARINE BOILERS.-Continued. 50,000 Tensile Strength. 65,000 Tensile Strength. of of 1-6 Boiler. Plates. Pressure. 7,500 20 per ct. Addition'l 1-6 8,333.3 1-6 9,166.6 60,000 Tensile Strength. 1-6 10,000 1-6 10,833.3 70,000 Tensile Strength. 1-6 11,666.6 Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l Pressure. 20 per ct. Addition'l .1875 33.48 40.17 37.2 44.68 40.92 49.1 44.64 53.56 48.36 58.03 52.08 62.49 .21. 37.5 45.00 41.66 49.99 45.83 54.99 50.00 60.00 54.16 64.99 58.33 69.99 .23 41.02 49.22 45.63 54.75 50.19 60.22 54.75 65.71 59.32 71.18 63.65 76.38 84 .25 44.64 53.56 49.6 59.52 54.56 65.47 59.52 71.42 64.48 77.37 69.44 83.32 .26 46.42 55.7 51.58 61.89 56.74 68.08 61.9 74.28 67.05 80.46 72.22 86.66 INCHES .29 51.78 62.13 57.53 69.03 63.29 75.94 69.04 82.84 74.8 89.76 80.55 96.66 .3125 55.8 66.96 62.00 74.4 68.2 81.84 74.4 89.28 80.6 96.72 86.8 104.16 .33 58.92 70.7 65.47 78.56 72.02 86.42 78.57 94.28 85.11 102.13 91.66 109.99 .35 62.5 75.00 60.44 83.32 76.38 91.65 83.33 99.99 90.27 108.32 97.22 116.66 .375 66.96 80.35 74.4 89.28 81.84 98.2 89.28 107.13 96.72 116.06 104.16 124.99 .1875 31.25 37.5 34.72 41.66 38.19 45.82 41.66 49.99 45.13 54.15 48.68 58.33 .21 35.00 42.00 38.88 46.65 42.77 51.32 46.66 55.99 50.55 60.66 54.44 65.32 .23 38.33 45.99 42.59 51.10 46.85 56.22 51.11 61.33 55.37 66.44 59.62 71.54 .25 90 41.66 49.99 46.29 55.54 50.92 61.1 55.55 66.66 60.18 72.21 64.81 77.77 .26 43.33 51.99 48.14 57.76 52.96 63.55 57.77 69.32 62.59 75.1 67.4 80.88 INCHES .29 48.33 57.99 53.7 64.44 59.07 70.8 64.44 77.32 69.81 83.77 75.18 90.21 .3125 52.08 62.49 57.86 69.43 63.65 76.38 69.44 83.32 75.23 90.27 81.01 97.21 .33 55.00 66.00 61.11 73.33 67.22 80.66 73.33 87.99 79.44 95.32 85.55 102.66 .35 58.33 69.99 64.81 77.77 71.29 85.54 77.77 93.32 84.25 101.1 90.72 108.88 .375 62.5 75.00 69.44 83.32 76.38 91.65 83.33 99.99 90.27 108.32 97.22 116.66 .1875 29.29 35.14 32.55 39.06 35.8 42.96 39.06 46.87 42.31 50.77 45.57 54.68 .21 32.81 39.37 36.45 43.74 40.1 48.12 43.75 52.5 47.39 56.86 51.04 61.24 .23 35.93 43.11 39.93 47.91 43.92 52.7 47.91 57.49 51.9 62.28 55.9 67.08 .25 96 39.06 46.87 43.4 52.08 47.74 57.28 52.08 62.49 56.42 67.67 60.76 72.91 .26 40.62 48.74 45.14 54.16 49.65 59.58 54.16 64.99 58.78 70.53 63.19 75.82 INCHES .29 45.31 54.37 50.34 60.4 55.38 66.45 60.41 72.49 65.45 78.54 70.48 84.57 .3125 48.82 58.58 54.25 65.1 59.67 71.6 65.1 78.12 70.52 84.62 75.95 91.14 .33 51.56 61.87 57.29 68.74 63.02 75.62 68.75 82.5 74.47 89.36 80.2 96.24 .35 54.68 65.61 60.76 72.91 66.83 80.19 72.91 87.49 78.99 94.78 85.06 102.07 .375 58.58 70.29 65.1 78.12 71.61 85.93 78.12 93.74 84.63 101.55 91.14 109.6 24 370 A Library of Steam Engineering. FLAT SURFACES IN STEAM BOILERS. Sections 6 and 7 of Rule 2, of the United States Board of Super- vising Inspectors of Steam Vessels, regulates the thickness of boiler plate for flat surfaces, diameters of stay bolts, pitch of stay bolts, and distances between braces, the kind of stay bolts and braces, and man- ner of attaching the same, the manner of testing steel bars intended for use as stay bolts, and the steam pressure allowable for flat surfaces. ENGINEERS REQUIRED TO CALCULATE STRAINS ON STAYS AND BRACES. Not only are inspectors required to thoroughly understand the rules for making the various calculations to enable them to determine the steam pressure allowable in all cases, but, under Section 5, of Rule 5, of the Board of Supervising Inspectors, no person can obtain a license as an engineer, or have the grade of his license raised, unless he can mathematically determine the strain to which stay bolts and braces are subjected by both steam and hydrostatic pressure. The lack of information on this important subject, on the part of many engineers and boiler makers, has been a most fruitful source of serious steam boiler accidents. It is, therefore, of vital importance that those who build, as well as those under whose care and manage- ment they are placed, should fully understand the rules governing the construction of flat surfaces and steam pressure allowable. AREA OF SURFACE STAYED BY BOLT. Before entering upon a study of the rules governing the construc- tion of flat surfaces, it will be well for the student to familiarize him- self with the rules which will enable him to determine the extent or area of surface stayed by any given bolt or stay. His attention is therefore invited to a careful study of the following rules and illus- trations: RULE.-Multiply the distance, in inches, from center to center of bolts by itself, or, in other words, square the distance of bolts from center to center, and the product will give the number of square inches contained in the area of surface stayed by the bolt or stay. Example.-Let 5 inches equal distance from center to center of stay bolts. Then we have: 5X5-25 square inches. Area of surface to be stayed by each bolt. To illustrate this more clearly, we will take a section from the flat surface of a boiler, as shown in Fig. 170. U. S. Marine Boiler Inspection and Engineers' License Laws. 371 It will be observed that whatever pressure per square inch is put upon the section, as shown in Fig. 170, the load is distributed equally 10" 10' 5" 5" 50 5" Ө O 5" 5" 5" 179 5 Fig. 170 5" between the center and the outer bolts, and the natural conclusion would be that the center bolt bears one-half of the entire strain put upon the whole surface 10"x10", as shown in the diagram. But that is not true, as the center bolt bears but one-fourth of the entire strain, as shown in Fig. 171. 10" 5" K-- 10" Fig. 171 57 о 5" 10" 5' 107 372 A Library of Steam Engineering. It will be observed that the width and breadth of the inner dia- gram, as shown in Fig. 171, are just one-half of those of the outer diagram, while it has an area of but one-fourth of that of the entire diagram, hence it bears but one-fourth of the entire strain; it there- fore follows that the center bolt bears but one-fourth of the entire strain, and the other bolts combined bear three-fourths of the strain. on that particular section. The question now arises, how much of the entire strain does each of the outer bolts bear? This is illustrated in Fig. 172: 10" --2--- 5" ----2-½'' - T Ө C 21/2" B 21½" 22" 5' C 212" K- K--2-- 10" 5' Ө B 21/2" C K--22' 5" 17/18 A О 5" B 1 5" 5" 11/17 不一​不 ​不一​不 ​*-7/7-* Bè 2" C K--22'-*- 10" Fig. 172 With a pressure of 100 pounds on the square inch, there would be a total pressure on the entire surface of the diagram, as shown in Fig. 172, of 10,000 pounds, including the space occupied by the bolts or parts of bolts within the diagram, which space should be deducted to obtain perfect accuracy, for the reason that there is no strain or pressure on the space occupied by the bolts. The area of the entire surface is 10"x10"-100 square inches, and that multiplied by 100 pounds, pressure per square inch, gives the total strain on the entire surface: 100X100-10,000 pounds; and this divided by 4, gives the strain borne by the center bolt: 10,000÷4= 2500 pounds, total strain on A in the diagram. Diagrams B B B B, as shown in Fig. 172, are each but one-half the size of A, and C C C C are each but one-half the size of B, or one- fourth the size of A. Therefore, it will be observed, that B B B B, U. S. Marine Boiler Inspection and Engineers' License Laws. 373 each having but one-eighth of the entire surface, each bears but one- eighth of the entire pressure, and that the four combined bear one-half of the entire pressure. C C C C, each containing but one-sixteenth of the entire surface, each bears but one-sixteenth of the entire pressure, and consequently each corner bolt bears but one-sixteenth of the entire strain, while each outer center bolt bears one-eighth of the entire strain. To secure perfect accuracy in determining the amount of strain produced by a given pressure per square inch on the center bolt, the area of cross section of the bolt must be deducted from the area of diagram A, while one-half of the area of the cross section of the bolts in diagrams B B B B, and one-fourth of the area of the cross section of the bolts in diagrams C C C C, must be deducted, as these propor- tions represent the amount of area of the bolts embraced in their respective diagrams, as will be seen by reference to Fig. 172. CONSTRUCTION OF FLAT SURFACES AND STEAM PRESSURE ALLOWABLE. PLATES NOT OVER 7-16 OF AN INCH THICK. STRAIN ON BOLTS LIMITED TO 6000 POUNDS PER SQUARE INCH OF SECTION. Furnaces, fire boxes, back connections and other flat surfaces; screw stay bolts and nuts, or plain bolts with single nut and socket, or riveted head and socket, or screw stay bolts, ends riveted, pitch limited to 10 inches; screw stay bolts without socket, not allowed where salt water is used in boilers. DIAMETER OF STAY BOLTS. RULE.-First, multiply the given steam pressure per square inch by the square of the distance between centers of stay bolts, and divide the last product by 6000, and call the answer "The Quotient." Second, divide "The Quotient" by .7854, and extract the square root of the last quotient; the answer will give the required diameter of stay bolt at the bottom of thread in decimals of an inch. Example.-Let 165.67 pounds equal given steam pressure per square inch. Let 4 inches equal pitch of stay bolts from center to center. Let 6000 equal a constant. Let .7854 equal a constant. Then we have: 165.67X4X4) 6000 .7854 .75+ Decimals of an inch. Diam- eter of bolt at bottom of thread. 374 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 165.67 Given pressure per square inch. 4X4= 16 Square of distance between centers of bolts. 994 02 1656 7 Dividing by the constant. 6000) 2650.72 (0.44178666+ "The Quotient.” 2400 0 250 72 240 00 10 720 6 000 4 7200 4 2000 52000 48000 40000 36000 40000 36000 40000 36000 Next, dividing "The Quotient" by .7854, we have: .7854).44178666 (.5625- The last quotient. 39270 49086 47124 19626 15708 39186 39270 Finally, extracting the square root of the last quotient, we have: .5625 (.75 Decimals of an inch. Diameter of bolt at bottom of thread. 49 145) 725 725 U. S. Marine Boiler Inspection and Engineers' License Laws. 375 PITCH OF BOLTS BASED ON DIAMETER OF BOLTS AND STEAM PRESSURE. RULE.-First, multiply the area of cross section of the stay bolt at the bottom of thread by 6000, and call the last product "Product No. 1." Second, divide "Product No. 1" by the given steam pressure per square inch, and extract the square root of the quotient; the answer will give the required distance, in inches, from center to center of stay bolts. Example.-Let 75 one hundredths of an inch equal the diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 165.67 pounds per square inch equal given steam pressure. Then we have: 75 .75X.75X.7854×6000 165.67 -4 inches. Distance of bolts from center to center. Performing the operation in the ordinary way, we have: .75 Diameter of bolt. .75 Diameter of bolt. 375 525 .5625 Square of diameter of bolt. .7854 A constant. 22500 28125 45000 39375 .44178750 Area of cross section of bolt at bot- tom of thread. 6000 A constant. 2650.72500000"Product No. 1." Next, dividing "Product No. 1" by the given steam pressure (165.67 pounds per square inch), we have: 165.670) 2650.725 (16+ The quotient. 1656 70 994 025 994 020 Finally, extracting the square root of the quotient, we have: 16 (4 inches. Distance of bolts from center to center. 16 376 A Library of Steam Engineering. PITCH OF BOLTS BASED ON THICKNESS OF PLATE AND STEAM PRESSURE. RULE.-Multiply the constant whole number 28,672 by the square of the thickness of plate, in decimals of an inch; then divide the product by the given steam pressure per square inch, and extract the square root of the quotient; the answer will give the distance of stay bolts from center to center in inches. Example.-Let 28,672 equal a constant. Let 25 one hundredths of an inch equal thickness of plate. Let 112 pounds per square inch equal given steam pressure. Then we have: 28672X.25.25 112 -4 inches. Distance of bolts from center to center. Performing the operation in the ordinary way, we have: 28672 A constant. .25X.25= .0625 Square of thickness of plate. 14 3360 57 344 1720 32 Dividing by the given pressure. 112) 1792.0000 (16 The quotient. 112 672 672 Finally, extracting the square root of the quotient, we have: 16 (4 inches. Distance of bolts from center to center. 16 THICKNESS OF PLATE BASED ON PITCH OF BOLTS AND STEAM PRESSURE. RULE.-Multiply the given steam pressure per square inch by the square of the distance of stay bolts from center to center, in inches; then divide the product by the constant 28,672, and extract the square root of the quotient; the answer will give the required thickness of plate in decimals of an inch. U. S. Marine Boiler Inspection and Engineers' License Laws. 377 Example.-Let 165.67 pounds per square inch equal given steam pressure. Let 4 inches equal distance of stay bolts from center to center. Let 28,672 equal a constant. Then we have: 165 165.67X(4X4) 28672 .30+ Decimals of an inch. Thickness of plate required. Performing the operation in the ordinary way, we have: 165.67 Given steam pressure. 4X4= 16 Square of distance between centers of bolts. 994 02 1656 7 Dividing by the constant. 28672.00) 2650.72.00 (0.0924+ The quotient. 2580 48 00 70 24 000 57 34 400 12 89 6000 11 46 8800 Finally, extracting the square root of the quotient, we have: .0924 (.30+ Decimals of an inch. Thickness of 9 60) 24 ANOTHER RULE. plate required. RULE.-First, multiply the given steam pressure per square inch by the square of the distance of stay bolts from center to center, and divide the product by the constant 112, and call the answer "The Quotient." Second, extract the square root of "The Quotient," and multiply the root by the constant .0625, and the product will give the required thickness of plate in decimals of an inch. Example.-Let 175 pounds equal steam pressure per square inch. Let 4 inches equal distance of stay bolts from center to center. Let 112 equal a constant. Let .0625 equal a constant. 378 A Library of Steam Engineering." Then we have: (175X42) 112 -X.0625 .3125 Decimals of an inch. Thick- ness of plate. Performing the operation in the ordinary way, we have: 4X4= 175 Steam pressure in pounds. 16 Square of distance between centers of bolts. 1050 175 A constant. 112) 2800 (25 "The Quotient." 224 560 560 Next, extracting the square root of "The Quotient," we have: 25 (5 Square root. 25 Finally, multiplying the square root by the constant .0625, we have: .0625 5 .3125 Decimals of an inch. Thickness of plate. STEAM PRESSURE BASED ON DIAMETER AND PITCH OF BOLTS. RULE.-First, multiply the square of the diameter of the stay bolt, in decimals of an inch, by .7854, and then multiply the product by the constant whole number 6000, and call the last product "Product No. 1." Second, obtain the number of square inches in the area stayed by the bolt by squaring the distance from center to center of the bolts, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2" (the surface to be stayed), and the quotient will give the safe-working pressure per square inch. Example.-Let 75 one hundredths of an inch equal diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 4 inches equal distance of stay bolts from center to center. U. S. Marine Boiler Inspection and Engineers' License Laws. 379 Then we have: 75X.75X.7854×6000 -=165.67+ lbs. Safe-working pressure 4X4 per square inch. Performing the operation in the ordinary way, we have: .75 Diameter of bolt. .75 Diameter of bolt. 375 525 .5625 Square of diameter of bolt. .7854 A constant. 22500 28125 45000 39375 .44178750 2650.72500000 Area of cross section of bolt at bottom of thread. 6000 A constant. 'Product No. 1." Next we have; 4 Distance between centers of bolts. 4 Distance between centers of bolts. 16 "Product No. 2." Number of square inches in the area of surface stayed by bolt. Then, dividing "Product No. 1" by "Product No. 2," the area of surface stayed by bolt, we have: 16) 2650.725 (165.67+ lbs. Safe-working pressure per square inch. 16 105 96 90 08 80 107 96 112 112 5 STEAM PRESSURE BASED ON THICKNESS OF PLATE AND PITCH OF BOLTS.. RULE.-Multiply the constant whole number 28,672 by the square of the thickness of plate, in decimals of an inch, and divide the product by the square of the distance, in inches, of the stay bolts from center to center; the quotient will give the safe-working pressure in pounds. 380 A Library of Steam Engineering. Example.-Let 28,672 equal a constant. Let 25 one hundredths of an inch equal thickness of plate. Let 4 inches equal pitch of stay bolts from center to center. Then we have: 28672X.25X.25 =112 lbs. Safe-working pressure. 4X4 Performing the operation in the ordinary way, we have: 28672 A constant. .25X.25= .0625 Square of thickness of plate. 14 3360 57 344 1720 32 Dividing by square 4X4 16) 1792.0000 (112 lbs. Safe-working pressure. of distance between centers of bolts. 16 111 19 16 32 22 32 STEAM PRESSURE-UNITED STATES GOVERNMENT RULE. The United States government rule is as follows for material not over seven-sixteenths inch thick: RULE.-Multiply the constant whole number 112 by the square of the thickness of plate, in sixteenths of an inch, and divide the product by the square of the distance from center to center of stay bolts, in inches; the quotient will give the safe-working pressure. Example.-Let 112 equal a constant. Let of an inch equal thickness of plate. Let 4 inches equal distance of stay bolts from center to center. Then we have: 1=16 Next we have: 112×4×4 =112 lbs. Safe-working pressure. 4X4 U. S. Marine Boiler Inspection and Engineers' License Laws. 381 Performing the operation in the ordinary way, we have: 112 A constant. 4X4= 16 Square of thickness of plate. 672 112 Dividing by square of dis- 4X4 16) 1792 (112 lbs. Safe-working pressure. tance between centers of bolts. 16 19 16 32 32 This rule looks very simple until we consider the fact that boiler plate is seldom made in sixteenths of an inch, but nearly always in decimals of an inch; it then becomes apparent that the rules which precede the above are much simpler and much easier comprehended. CONSTRUCTION OF FLAT SURFACES AND STEAM PRESSURE ALLOWABLE. PLATES OVER 7-16 OF AN INCH THICK. STRAIN ON BOLTS LIMITED TO 6000 POUNDS PER SQUARE INCH OF SECTION. Flat surfaces of furnaces, fire boxes and back connections; screw stay bolts and nuts, or plain bolts with single nut and socket, or riveted head and socket, or screw stay bolts, ends riveted, pitch limited to 10 inches; screw stay bolts, without socket, not allowed where salt water is used in boilers. DIAMETER OF STAY BOLTS. RULE.-First, multiply the given steam pressure per square inch by the square of the distance between centers of stay bolts, and divide the last product by 6000, and call the answer "The Quotient." Second, divide "The Quotient" by .7854, and extract the square root of the last quotient; the answer will give the required diameter of stay bolt at the bottom of thread. Example.-Let 175 pounds equal given steam pressure per square. inch. Let 8 inches equal pitch of stay bolts from center to- center. Let 6000 equal a constant. Let .7854 equal a constant. Then we have: 175X82 .7854-1.54+ inches. Diameter of bolt at bot- 6000 tom of thread. 382 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 175 Steam pressure per square inch. 8X8= 64 Square of distance between centers of bolts. 700 1050 Dividing by the constant. 6000) 11200 (1.8666+ "The Quotient." 6000 52000 48000 40000 36000 40000 36000 40000 36000 Next, dividing "The Quotient" by .7854, we have: 7854) 1.8666 (2.3766+ The last quotient. 1 5708 29580 23562 60180 54978 52020 47124 48960 47124 Finally, extracting the square root of the last quotient, we have: 2.3766 (1.54+ inches. Diameter of bolt at bottom of thread. 1 25) 1 37 1 25 304) 1266 1216 PITCH OF BOLTS BASED ON DIAMETER OF BOLTS AND STEAM PRESSURE. RULE. First, multiply the area of cross section of the stay bolt at the bottom of thread by 6000, and call the last product "Product No. 1." Second, divide "Product No. 1" by the given steam pressure per square inch, and extract the square root of the quotient; the answer will give the required distance, in inches, from center to center of stay bolts. U. S. Marine Boiler Inspection and Engineers' License Laws. 383 Example. Let 1.54 inches equal diameter of the bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant, c Let 175 pounds per square inch equal given steam pressure. Then we have: 1.542X.7854×6000 175 -8 inches. Distance of bolts from center to center. Performing the operation, we have: 1.54 Diameter of bolt. 1.54 Diameter of bolt. 616 770 154 2.3716 Square of diameter of bolt. .7854 A constant. 94864 118580 189728 1 66012 1.86265464 Area of cross section of bolt at 0 bottom of thread. 6000 A constant. 11175.92784000 "Product No. 1." Next, dividing "Product No. 1" by the given steam pressure (175 pounds per square inch), we have: 175) 11175.9278 (63.8624+ The quotient. 1050 675 525 1509 1400 1092 1050 427 350 778 700 384 A Library of Steam Engineering. Finally, extracting the square root of the quotient, we have: 63.8624 (7.99+ inches. Distance of bolts from center 49 to center. 149) 1486 1341 1589) 14524 14301 PITCH OF BOLTS BASED ON THICKNESS OF PLATE AND STEAM PRESSURE. RULE.-Multiply the constant whole number 30,720 by the square of the thickness of plate, in decimals of an inch; then divide the product by the given steam pressure per square inch, and extract the square root of the quotient; the answer will give the distance of stay bolts from center to center in inches. Example.-Let 30,720 equal a constant. Let 75 one hundredths of an inch equal thickness of plate. Let 156.73 pounds per square inch equal given pressure. Then we have: 30720X.75X.75 156.73 -10.5 inches. Distance of bolts from cen- ter to center. Performing the operation in the ordinary way, we have: .75X.75= 30720 A constant. .5625 Square of thickness of plate. 15 3600 61 440 1843 20 15360 0 Dividing by the given 156.73) 17280.0000 (110.25+ The quotient. pressure. 15673 16070 15673 39700 31346 83540 78365 U. S. Marine Boiler Inspection and Engineers' License Laws. 385 Next, extracting the square root of the quotient, we have: 110.25 (10.5 inches. Distance of bolts from cen- 1 205) 1025 1025 ter to center. THICKNESS OF PLATE BASED ON PITCH OF BOLTS AND STEAM PRESSURE. RULE.-Multiply the given steam pressure per square inch by the square of the distance of stay bolts from center to center, in inches; then divide the product by the constant 30,720, and extract the square root of the quotient; the answer will give the required thickness of plate in decimals of an inch. Example.-Let 156.72 pounds per square inch equal given steam pressure. Let 10.5 inches equal distance of stay bolts from center to center. Let 30,720 equal a constant. Then we have: 156.72×10.52 30720 --.75- Decimals of an inch. Thickness of plate required. Performing the operation in the ordinary way, we have: 156.72 Given steam pressure. 10.5X10.5= 110.25 Square of distance between centers of bolts. 7 8360 31 344 1567 2 15672 15360 0000 0 1918 3800 00 Dividing by the 30720.0000) 17278.3800.0 (0.5624+ The quotient. constant. 1843 2000 00 75 1800 000 61 4400 000 13 7400 0000 12 2880 0000 Next, extracting the square root of the quotient, we have .5624 (.75- Decimals of an inch. Thickness of 49 145) 724 725 plate required. 25 386 A Library of Steam Engineering. STEAM PRESSURE BASED ON DIAMETER AND PITCH OF BOLTS. RULE.-Multiply the area of cross section of the bolt at the bottom of thread by 6000; then divide the product by the square of the distance from center to center of stay bolts, and the quotient will give the steam pressure per square inch allowable. Example.-Let 1.75 inches equal diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 10.375 inches equal pitch of bolts from center to center. Then we have: 1.752X.7854× 6000 10.375 X 10.375 =134+lbs. Steam pressure per square inch. Performing the operation in the ordinary way, we have: 1.75 Diameter of bolt. 1.75 Diameter of bolt. 875 1 225 175 3.0625 Square of diameter of bolt. .7854 A constant. 122500 153125 245000 2 14375 2.40528750 Area of cross section of bolt at bot- tom of thread. 6000 A constant. 10.375×10.375-107.640625) 14431.72500000 (134+ lbs. Steam pressure per 10764 0625 square inch. 3667 66250 3229 21875 438 443750 430 562500 STEAM PRESSURE BASED ON THICKNESS OF PLATE AND PITCH OF BOLTS. RULE.-Multiply the constant whole number 30,720 by the square of the thickness of plate, in decimals of an inch, and divide the product by the square of the distance, in inches, of the stay bolts from center to center; the quotient will give the safe-working pressure in pounds. U. S. Marine Boiler Inspection and Engineers' License Laws. 387 Example.-Let 30,720 equal a constant. Let 75 one hundredths of an inch equal thickness of plate. Let 10.5 inches equal pitch of stay bolts from center to center. Then we have: 30720X.75 X.75 10.5 X 10.5 =156.73+ lbs. Safe-working pressure. Performing the operation in the ordinary way, we have: 30720 A constant. .75X.75= .5625 Square of thickness of plate. 15 3600 61 440 1843 20 15360 0 Dividing by square 10.5X10.5=110.25) 17280.0000 (156.73+lbs. Safe-working of distance between pressure. centers of bolts. 11025 62550 55125 74250 66150 81000 77175 38250 33075 CONSTRUCTION OF FLAT SURFACES AND STEAM PRESSURE ALLOWABLE. PLATES OVER 7-16 OF AN INCH THICK. STRAIN ON BOLTS LIMITED TO 6000 POUNDS PER SQUARE INCH OF SECTION. Flat surfaces other than furnaces, fire boxes and back connections; stay bolts, with ends threaded, having nuts on both inside and outside of plate. DIAMETER OF STAY BOLTS. RULE.-First, multiply the given steam pressure per square inch by the square of the distance between centers of stay bolts, and divide the last product by 6000, and call the answer "The Quotient." Second, divide "The Quotient" by .7854, and extract the square root of the last quotient; the answer will give the required diameter of stay bolt at the bottom of thread. 388 A Library of Steam Engineering. Example.-Let 175 pounds equal given steam pressure per square inch. Let 14 inches equal pitch of stay bolts from center to center. Let 6000 equal a constant. Let .7854 equal a constant. Then we have: 175 X 142 6000 .7854-2.69+ inches. Diameter of bolt at bottom of thread. Performing the operation in the ordinary way, we have: 175 Steam pressure per square inch. 14X14 196 Square of distance between centers of 1050 1575 175 bolts. Dividing by the constant. 6000) 34300 (5.7166+ 30000 43000 42000 10000 6000 40000 36000 40000 36000 "The Quotient." Next, dividing "The Quotient" by .7854, we have: .7854) 5.7166 (7.2785+ The last quotient. 5 4978 21880 15708 61720 54978 67420 62832 45880 39270 U. S. Marine Boiler Inspection and Engineers' License Laws. 389 Finally, extracting the square root of the last quotient, we have: 7.2785 (2.69+ inches. Diameter of bolt at bot- tom of thread. 4 46) 327 276 529) 5185 4761 PITCH OF BOLTS BASED ON DIAMETER OF BOLTS AND STEAM PRESSURE. RULE.-First, multiply the area of cross section of the stay bolt at the bottom of thread by 6000, and call the last product "Product No. 1." Second, divide "Product No. 1" by the given steam pressure per square inch, and extract the square root of the quotient; the answer will give the required distance, in inches, from center to center of stay bolts. Example.-Let 2.07 inches equal diameter of the bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 140 pounds per square inch equal given steam pressure. Then we have: 2.072X.7854×6000 140 12 inches. Required pitch of bolts. Performing the operation in the ordinary way, we have: 2.07 Diameter of bolt. ow eyew ventbro 2.07 Diameter of bolt. gatoriore etsta 1449 414 or 4.2849 Square of diameter of bolt. .7854 A constant. 171396 214245 thelioga en 342792 2 99943 3.36536046 Area of cross section of bolt at bot- tom of thread. 6000 A constant. 20192.16276000 "Product No. 1." 390 A Library of Steam Engineering. Next, dividing "Product No. 1" by the given steam pressure (140 pounds per square inch), we have: 140) 20192.16276 (144+ The quotient. 140 619 560 592 560 Finally, extracting the square root of the quotient, we have: 144 (12 inches. Distance of bolts from center to center. 1 22) 44 44 PITCH OF BOLTS BASED ON THICKNESS OF PLATE AND STEAM PRESSURE. RULE.-Multiply the constant whole number 35,840 by the square of the thickness of plate, in decimals of an inch; then divide the product by the given steam pressure per square inch, and then extract the square root of the quotient; the answer will give the distance of stay bolts from center to center in inches. Example.-Let 35,840 equal a constant. Let 75 one hundredths of an inch equal thickness of plate. Let 140 pounds per square inch equal given steam pressure. Then we have: 35840X.75X.75 140 -12 inches. Distance of bolts from center to center. Performing the operation in the ordinary way, we have: 35840 A constant. .75X.75 .5625 Square of thickness of plate. 17 9200 71 680 2150 40 Dividing by the given steam 140) 20160.0000 (144 The quotient. pressure. jod fathed to no 17920 0 140 616 560 560 560 U. S. Marine Boiler Inspection and Engineers' License Laws. 391 Next, extracting the square root of the quotient, we have: 144 (12 inches. Distance of bolts from center to 1 center 22) 44 44 THICKNESS OF PLATE BASED ON PITCH OF BOLTS AND STEAM PRESSURE. RULE.-Multiply the given steam pressure per square inch by the square of the distance of stay bolts from center to center, in inches; then divide the product by the constant 35,840, and extract the square root of the quotient; the answer will give the required thickness of plate in decimals of an inch. Example.-Let 140 pounds per square inch equal given steam pressure. Let 12 inches equal distance of stay bolts from center to center. Let 35,840 equal a constant. Then we have: 140×12X12 35840 .75 Decimals of an inch. Thickness of plate required. Performing the operation in the ordinary way, we have: Rod Dividing by the constant. 140 Given steam pressure. 12X12 144 Square of distance between centers of 560 560 140 bolts. 35840) 20160.0 (0.5625 The quotient. 17920 O 2240 00 2150 40 aer 89 600 71 680 17 9200 17 9200 Next, extracting the square root of the quotient, we have .5625 (.75 Decimals of an inch. Thickness of 49 145) 725 plate required. 725 392 A Library of Steam Engineering. STEAM PRESSURE BASED ON DIAMETER AND PITCH OF BOLTS. RULE.-Multiply the area of cross section of the bolt at the bottom of thread by 6000; then divide the product by the square of the distance from center to center of stay bolts, and the quotient will give the steam pressure per square inch allowable. Example.-Let 2.07 inches equal diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 14 inches equal pitch of bolts from center to center. Then we have: 2.072X.7854× 6000 =103+ lbs. Steam pressure per square 14X14 inch. Performing the operation in the ordinary way, we have: 2.07 Diameter of bolt. 2.07 Diameter of bolt. 1449 4 14 4.2849 Square of diameter of bolt. .7854 A constant. 171396 214245 342792 2.99943 3.36536046 6000 Area of cross section of bolt at bottom of thread. A constant. Dividing by square 14X14= 196) 20192.16276000 (103+ lbs. Steam pressure per of distance be- tween centers of bolts. 196 square inch, 592 588 STEAM PRESSURE BASED ON THICKNESS OF PLATE AND PITCH OF BOLTS. RULE.-Multiply the constant whole number 35,840 by the square of the thickness of plate, in decimals of an inch, and divide the product by the square of the distance, in inches, of the stay bolts from center to center; the quotient will give the safe-working pressure in pounds. U. S. Marine Boiler Inspection and Engineers' License Laws. 393 Example.-Let 35,840 equal a constant. Let 75 one hundredths of an inch equal thickness of plate. Let 14 inches equal pitch of bolts from center to center. Then we have: 35840X.75X.75 14X14 =102.85+ lbs. Safe-working pressure. Performing the operation in the ordinary way, we have: 35840 A constant. .75X.75= .5625 Square of thickness of plate. 179200 71680 215040 179200 Dividing by square of 14X14= 196) 20160.0000 (102.85+ lbs. Safe-working distance between cen- ters of bolts. pressure. 196 560 392 1680 1568 1120 980 Taking the same plate and placing the same bolts at a distance of 12 inches from center to center, and determining the safe-working pressure, we have: 35840X.75X.75 12×12 =140 lbs. Safe-working pressure. Performing the operation in the ordinary way, we have: 35840 .5625 179200 71680 215040 179200 Dividing by square of 12X12= 144) 20160.0000 (140 lbs. Safe-working pressure. distance between centers of bolts. 144 576 576 0 394 A Library of Steam Engineering. CONSTRUCTION OF FLAT SURFACES AND STEAM PRESSURE ALLOWABLE. PLATES OVER 7-16 OF AN INCH THICK. STRAIN ON BOLTS LIMITED TO 6000 POUNDS PER SQUARE INCH OF SECTION. Flat surfaces other than furnaces, fire boxes and back connec- tions; stay bolts with double nuts, and a washer at least one-half the thickness of plate, and in size equal to two-fifths of pitch of stay bolts, or plates fitted with double angle iron, riveted to plate with leaf at least two-thirds of the thickness of plate, and depth at least one-fourth of the pitch. DIAMETER OF STAY BOLTS. RULE.-First, multiply the given steam pressure per square inch by the square of the distance between centers of stay bolts, and divide the last product by 6000, and call the answer "The Quotient." Second, divide "The Quotient" by .7854, and extract the square root of the last quotient; the answer will give the required diameter of stay bolt at the bottom of thread. 8.400 Example.-Let 175 pounds equal given steam pressure per square inch. Let 15 inches equal pitch of stay bolts from center to center. Let 6000 equal a constant. Let .7854 equal a constant. Then we have: 175X 152 6000 .7854-2.88+ inches. Diameter of bolt at bottom of thread. Performing the operation in the ordinary way, we have: 175 Steam pressure per square inch. 15X15 225 Square of distance between centers of bolts. 875 A New Ti 350 350 Dividing by the constant. 6000) 39375 (6.5625 "The Quotient." 36000 33750 30000 37500 From stea 041, Sica Jadi 01/09 36000 15000 12000 30000 30000 U. S. Marine Boiler Inspection and Engineers' License Laws. 395 Next, dividing "The Quotient" by .7854, we have: .7854) 6.5625 (8.34+ The last quotient. 62832 27930 23562 43680 31416 Finally, extracting the square root of the last quotient, we have: 8.34 (2.88+inches. Diameter of bolt at bottom of 4 48) 434 384 568) 5000 4544 thread. PITCH OF BOLTS BASED ON DIAMETER OF BOLTS AND STEAM PRESSURE. RULE.-First, multiply the area of cross section of the stay bolt at the bottom of thread by 6000, and call the last product "Product No. 1." Second, divide "Product No. 1" by the given steam pressure per square inch; then extract the square root of the quotient, and the answer will give the required distance, in inches, from center to center of stay bolts. Example. Let 2.88 inches equal diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 146 pounds per square inch equal given steam pressure Then we have: 2.882X.7854×6000 146 16.36+ inches. Distance of bolts from center to center. Performing the operation in the ordinary way, we have: 2.88 Diameter of bolt. 2.88 Diameter of bolt. Am't carried forward, 2304 2304 576 8.2944 Square of diameter of bolt. 396 A Library of Steam Engineering. Am't brought forward, 8.2944 Square of diameter of bolt. .7854 A constant. 331776 414720 663552 5 80608 6.51442176 Area of cross section of bolt at bottom of thread. 6000 A constant. 39086.53056000 "Product No. 1." Next, dividing "Product No. 1" by the given steam pressure (146 pounds per square inch), we have: 146) 39086.5305 (267.7159+ The quotient. 292 988 876 1126 1022 1045 1022 233 146 870 730 1405 1314 Finally, extracting the square root of the quotient, we have: 267.7159 (16.36+ inches. Distance of bolts from center 1 26) 167 156 323) 1171 969 3266) 20259 19596 to center. PITCH OF BOLTS BASED ON THICKNESS OF PLATE AND STEAM PRESSURE. RULE.-Multiply the constant whole number 51,200 by the square of the thickness of plate, in decimals of an inch; then divide the product by the given steam pressure per square inch, and extract the square root of the quotient; the answer will give the distance of stay bolts, in inches, from center to center. U. S. Marine Boiler Inspection and Engineers' License Laws. 397 Example.-Let 51,200 equal a constant. Let 75 one hundredths of an inch equal thickness of plate. Let 146 pounds per square inch equal given steam pressure. Then we have: 151200X.752 146 14+ inches. Distance of bolts from center to center. Performing the operation in the ordinary way, we have: .75X.75= 51200 A constant. .5625 Square of thickness of plate. 25 6000 102 400 3072 00 25600 0 Dividing by the given pressure. 146) 28800.0000 (197+ The quotient. 146 1420 1314 1060 1022 Finally, extracting the square root of the quotient, we have: 197(14+ inches. Distance of bolts from cen- ter to center. 1 24) 97 96 THICKNESS OF PLATE BASED ON PITCH OF BOLTS AND STEAM PRESSURE. RULE.-Multiply the given steam pressure per square inch by the square of the distance of stay bolts from center to center, in inches; then divide the product by the constant 51,200, and extract the square root of the quotient; the answer will give the required thickness of plate in decimals of an inch. Example.-Let 128 pounds per square inch equal given steam pressure. Let 15 inches equal distance of stay bolts from center to center. Let 51,200 equal a constant. 398 A Library of Steam Engineering. Then we have: 128X15X15 51200 .75 Decimals of an inch. Thickness of plate required. Performing the operation in the ordinary way, we have: Dividing by the constant. 128 Given steam pressure. 15X15 225 Square of distance between centers of bolts. 640 256 256 51200) 28800.0 (0.5625 The quotient. 25600 0 3200 00 3072 00 128 000 102 400 25 6000 25 6000 Next, extracting the square root of the quotient, we have: .5625 (.75 Decimals of an inch. plate required. 49 145) 725 725 Thickness of STEAM PRESSURE BASED ON DIAMETER AND PITCH OF BOLTS. RULE.-Multiply the area of cross section of the stay bolt at the bottom of thread by 6000; then divide the last product by the square of the distance, in inches, from center to center of the stay bolts; the quotient will give the steam pressure in pounds per square inch. Example.-Let 2.88 inches equal diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 15 inches equal pitch of bolts from center to center. Then we have: 2.882X.7854× 6000 15X15 173.71+ lbs. Steam pressure per square inch. U. S. Marine Boiler Inspection and Engineers' License Laws. 399 Performing the operation in the ordinary way, we have: 2.88 Diameter of bolt. 2.88 Diameter of bolt. 2304 2.304 5 76 8.2944 Square of diameter of bolt. .7854 A constant. 331776 414720 663552 5 80608 6.51442176 Area of cross section of bolt at bot- tom of thread. 6000 A constant. Dividing by 15X15 225) 39086.53056000 (173.71+ lbs. Steam pressure per square of dis- tance of bolts square inch. between cen- ters. 225 1658 1575 836 675 1615 1575 403 225 STEAM PRESSURE BASED ON THICKNESS OF PLATE AND PITCH OF BOLTS. RULE.-Multiply the constant whole number 51,200 by the square of the thickness of plate, in decimals of an inch, and divide the product by the square of the distance, in inches, of the stay bolts from center to center; the quotient will give the steam pressure per square inch allowable. Example. Let 51,200 equal a constant. Let 75 one hundredths of an inch equal thickness of plate. Let 15 inches equal pitch of stay bolts from center t、, center. Then we have: 51200X 752 =128 lbs. Steam pressure per square inch. 15X15 400 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 51200 A constant .75X.75 .5625 Square of thickness of plate. Hod 20 taleta 25 6000 102 400 3072 00 25600 0 Dividing by square 15X15 225) 28800.0000 (128 lbs. Steam pressure per square of distance between centers of bolts. inch. 225 630 450 1800 1800 CONSTRUCTION OF FLAT SURFACES AND STEAM PRESSURE ALLOWABLE. STEEL STAY BOLTS EXCEEDING 1 1-4 INCHES AND NOT EXCEEDING 2 1-2 INCHES IN DIAMETER. Steel stay bolts exceeding 14 inches and not exceeding 21 inches in diameter at bottom of thread, are allowed a strain of 8000 pounds per square inch of cross section. DIAMETER OF STAY BOLTS. RULE.-First, multiply the given steam pressure per square inch by the square of the distance between centers of stay bolts, and divide the last product by 8000, and call the answer "The Quotient." Second, divide "The Quotient" by .7854, and extract the square root of the last quotient; the answer will give the required diameter of stay bolts at the bottom of thread. Example. Let 175 pounds equal given steam pressure per square inch. Let 15 inches equal pitch of bolts from center to cen- ter. Let 8000 equal a constant. Let .7854 equal a constant. Then we have: 175 X 152 .7854-2.5 inches. Diameter of bolt at bot- 8000 tom of thread. U. S. Marine Boiler Inspection and Engineers' License Laws. 401 Performing the operation in the ordinary way, we have: Dividing by the constant. 175 Steam pressure per square inch. 15X15 225 Square of distance between centers of bolts. 875 350 350 8000) 39375 (4.9218+ "The Quotient." 32000 73750 72000 17500 16000 15000 8000 70000 64000 Next, dividing "The Quotient" by .7854, we have: .7854) 4.9218 (6.2666+ The last quotient. 4 7124 20940 15708 52320 47124 51960 47124 48360 47124 Finally, extracting the square root of the last quotient, we have: 6.2666 (2.5+ inches. Diameter of bolt at bottom of thread. 4 45) 226 225 STEAM PRESSURE BASED ON DIAMETER AND PITCH OF BOLTS. RULE.-Multiply the area of cross section of the stay bolt at the bottom- of thread by 8000; then divide the product by the square of the distance, in inches, from center to center of stay bolts, and the quotient will give the steam pressure per square inch allowable. 26 402 A Library of Steam Engineering. Example. Let 2.5 inches equal diameter of bolt at the bottom of thread. Let .7854 equal a constant. Let 8000 equal a constant. Let 15 inches equal pitch of bolts from center to center. Then we have: 2.52X.7854X8000 =174.53+ lbs. Pressure per square inch. 15X15 Performing the operation in the ordinary way, we have: 2.5 Diameter of bolt. 2.5 Diameter of bolt. 125 50 6.25 Square of diameter of bolt. .7854 A constant. 2500 3125 5000 4 375 4.908750 Area of cross section of bolt at bottom of thread. 8000 A constant. Dividing by square 15X15 225) 39270.000000 (174.53+ lbs. Steam pressure per of distance be- tween centers of bolts. square inch. 225 1677 1575 1020 900 1200 1125 750 675 STEEL STAY BOLTS EXCEEDING 2 1-2 INCHES IN DIAMETER. Steel stay bolts exceeding a diameter of 2 inches at the bottom of thread, are allowed a strain of 9000 pounds per square inch of cross section. DIAMETER OF STAY BOLTS. RULE.-First, multiply the given steam pressure per square inch by the square of the distance between centers of stay bolts, and divide the last product by 9000, and call the answer "The Quotient." U. S. Marine Boiler Inspection and Engineers' License Laws. -403 30 Second, divide "The Quotient" by .7854, and extract the square root of the last quotient; the answer will give the required diameter of stay bolt at the bottom of thread. Example.-Let 208.81 pounds equal given steam pressure per square inch. Let 16 inches equal pitch of bolts from center to center. Let 9000 equal a constant. Let .7854 equal a constant. Then we have: 208.81X162 9000 .7854-2.74+ inches. Diameter of bolt at bot- tom of thread. Performing the operation in the ordinary way, we have: 208.81 Steam pressure per square inch. 16X16= 256 Square of distance between centers of bolts. 1252 86 10440 5 41762 Dividing by the constant. 9000.00) 53455.36 (5.9394+ "The Quotient." 45000 00 8455 360 8100 000 355 3600 270 0000 85 36000 81 00000 4 360000 3 600000 Next, dividing "The Quotient" by .7854, we have: .7854) 5.9394 (7.5622+ The last quotient. 5 4978 44160 39270 48900 47124 17760 15708 20520 15708 404 A Library of Steam Engineering. Finally, extracting the square root of the last quotient, we have: 7.5622 (2.75-inches. Diameter of bolt at bottom 4 of thread. 47) 356 329 545) 2722 2725 STEAM PRESSURE BASED ON DIAMETER AND PITCH OF BOLTS. RULE. Multiply the area of cross section of the stay bolt at the bottom of thread by 9000; then divide the product by the square of the distance, in inches, from center to center of stay bolts, and the quotient will give the steam pressure per square inch allowable. Example.-Let 2.75 inches equal diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 9000 equal a constant. "Let 16 inches equal pitch of bolts from center to center. Then we have: 2.752X.7854× 9000 -208.81+ lbs. Steam pressure per square 16×16 inch. Performing the operation in the ordinary way, we have: 2.75 Diameter of bolt. 2.75 Diameter of bolt. 1375 1 925 550 7.5625 Square of diameter of bolt. .7854 A constant. 302500 378125 605000 5 29375 5.93958750 Area of cross section of bolt at bottom of thread. 9000 A constant. Am't carried forward, 53456.28750000 U. S. Marine Boiler Inspection and Engineers' License Laws. 405 Dividing by 16x16=256) 53456.28750000 (208.81+ lbs. Steam pressure per square of distance between centers of bolts. 512 2256 2048 2082 2048 348 square inch. 256 FIRE PUMPS. Section 4471 of the Revised Statutes of the United States provides that, "Every steamer exceeding 200 tons burden, and carrying passen- gers, shall be provided with two good double-acting fire pumps, to be worked by hand. Each chamber of such pumps, except pumps upon steamers in service on the 28th day of February, 1871, shall be of suf- ficient capacity to contain not less than 100 cubic inches of water." SIZE OF PUMP REQUIRED UNDER THE PROVISIONS OF SECTION 4471 OF THE REVISED STATUTES. RULE.-Assume any convenient and practical length of stroke of pump; then divide the number of cubic inches (100) of water required by the length of the stroke of the pump in inches; then divide the quotient by .7854, and extract the square root of the last quotient; the answer will give the required diameter of the plunger. Example. Let 8 inches equal assumed length of stroke of pump. Let 100 cubic inches equal capacity of chamber of pump. Let .7854 equal a constant. Then we have: 100-8 .7854 -3.98+ inches. Required diameter of plunger of pump. Performing the operation in the ordinary way, we have: 8) 100 .7854) 12.5000 (15.91+ square inches. Square of required diam- 7854 4 6460 eter of plunger. 3 9270 71900 70686 12140 7854 406 A Library of Steam Engineering. Next, extracting the square root of the quotient, we have: 15.91 (3.98+ inches. Required diameter of plunger of pump. 9 69) 691 621 8102 788) 7000 6304 HORSE POWER OF MARINE ENGINES. Under the regulations of the Treasury Department, inspectors are required to enter in a book, containing particulars of inspection, the horse power of the engines of each steamer inspected by them. As the use of the indicator is not required, the actual horse power of each engine must be determined by computation, which may be done by the following rule: RULE.-Multiply the area of the diameter of the cylinder by the average (or, in other words, the mean effective) pressure per square inch, throughout the stroke; then multiply the product by the number of feet the piston travels per minute, and divide the last product by 33,000. Example.-Let 24 inches equal diameter of cylinder. Let .7854 equal a constant. Let 150 pounds equal initial pressure per square inch. Let one-half equal point of stroke of piston at which steam is cut off. Let 5 feet equal length of piston stroke. Let 33,000 equal a constant. MEAN EFFECTIVE PRESSURE. The first thing is to determine the average (or mean effective) pressure per square inch throughout the stroke, and that is done by the aid of the following simple diagram: 150 lbs. 150 lbs. 150 lbs. 150 lbs. 120 lbs. 100 lbs. 85.68 lbs. 75 lbs. 1 2 3 4 5 6 7 Fig. 173 U. S. Marine Boiler Inspection and Engineers' License Laws. 407 (5) (6) (7) 5) 150 6) 150 7) 150 (8) 8) 150 30 25 21.42+ 18.75 4 4 4 4 120 100 85.68 75.00 pawfo The diagram represents a steam engine cylinder. The figures at the bottom of the diagram represent different parts of the stroke of the piston, and the figures at the top represent the steam pressure at the different parts of the stroke. It will be noticed that the pressure (150 pounds) remains the same during one-half of the stroke, when it begins to fall. The steam is cut off at four-eighths, or one-half stroke, and the steam does its work expansively during the remainder of the stroke; but the pressure per square inch is decreased, as shown by the figures to the right of the point of cut off, at the top of the diagram, and these figures are obtained in the following manner: The steam being cut off at 4, and when the piston gets to 5, we desire to determine the pressure at that point, we say of 150; at 6 we say of 150; at 7 we say 4 of 150; at 8 we say of 150. of 150 is: 5) 150 30 4 120 lbs. Pressure at 5% of the stroke. 4 of 150 is: 6) 150 25 4 100 lbs. Pressure at of the stroke. of 150 is: 7) 150 21.42+ 4 85.68 lbs. Pressure at % of the stroke. of 150 is: 8) 150 18.75 4 75.00 lbs. Pressure at of the stroke. It makes no difference at what part of the stroke the steam is cut off, the operation is the same. The figure at the bottom of point of cut off is always taken as the numerator, and all succeeding figures as denominators, while the pressure at the top of the point of cut off is 408 A Library of Steam Engineering. always taken as the dividend. So that the pressure at point of cut off is always divided by the figures at the bottom succeeding the figures at which the steam is cut off, and the quotient of each operation is mul- tiplied by the figure at the bottom of the diagram at the point of cut off; in this way the pressure for each point in the stroke is obtained, after which all of the pressures on top of the diagram are added together, and the sum is divided by the number of divisions in the diagram, which in the present case is 8. Thus we have: 150 1st part of stroke. 150 2nd part of stroke. 150 3rd part of stroke. 150 4th part of stroke. 120 5th part of stroke. 100 6th part of stroke. 85.68 7th part of stroke. 75 8th part of stroke. 8) 980.68 122.585 lbs. Average pressure throughout the stroke. We are now prepared to determine the horse power of the engine. Proceeding according to the rule, we have: 24X24X.7854X122.585X2X5×18 33000 =302.4887+ Horse power of engine. It will be noticed that the figure 2 is used in the example, and will, therefore, require explanation. As the piston makes two strokes during each revolution of the crank, the piston travels twice the length of one stroke, which in this case is 10 feet for each revolution of the crank, or twice 5 feet for each revolution of crank. Hence, the use of the figure 2. Performing the operation in the ordinary way, we have: 24 Diameter of cylinder. 24 Diameter of cylinder. Am't carried forward, 96 48 576 Square of diameter of cylinder. .7854 A constant. 4 7124 54 978 452.3904 Area of diameter of cylinder. 392 70 U. S. Marine Boiler Inspection and Engineers' License Laws. 409 Am't brought forward, 452.3904 Area of diameter of cylinder. 122.585 Average pressure throughout the stroke. 2 2619520 36 191232 226 19520 904 7808 9047 808 45239 04 55456.2771840 2 Number of strokes of piston to one rev- olution of crank. 5 Length of stroke. 110912.5543680 554562.7718400 18 Number of revolutions of crank. 4436502 1747200 5545627 718400 33000) 9982129.8931200 (302.48+ Horse power of engine. 99000 82129 66000 161298 132000 292989 264000 SAFETY-VALVES. [Sections 4418, 4419, 4436 and 4437 of the Revised Statutes of the United States. Section 24, of Rule 2; Section 5, of Rule 5 and Section 14, of Rule 9, of the United States Board of Supervising Inspectors.] Section 4418 of the Revised Statutes requires inspectors to "satisfy themselves that the safety-valves are of suitable dimensions, sufficient in number and well arranged; and that the weights of the safety- valves are properly adjusted, so as to allow no greater pressure in the boilers than the amount prescribed in the certificate of inspection." Section 4419 gives the inspectors authority to take one of the safety- valves "wholly from the control of all persons engaged in navigating a steam vessel" if they find it necessary to do so. Section 4436 pro- vides that "every boiler shall be provided with a good, well-constructed safety-valve or valves, of such number, dimensions and arrangements as shall be prescribed by the Board of Supervising Inspectors." Sec- tion 4437 reads as follows: 410 A Library of Steam Engineering. CC SEC. 4437. Every person who intentionally loads or obstructs, or causes to be loaded or obstructed, in any manner, the safety-valve of a boiler, or who employs any other means or device whereby the boiler may be subjected to a greater pressure than the amount allowed by the certificate of the inspectors, or who intentionally deranges or hinders the operation of any machinery or device employed to denote the state of the water or steam in any boiler, or to give warning of approaching danger, or who intentionally permits the water to fall below the prescribed low water line of the boiler, and every person concerned therein, directly or indirectly, shall be guilty of a misde- meanor, and shall be fined two hundred dollars, and may also be im- prisoned not exceeding five years." Section 24, of Rule 2, of the Board of Supervising Inspectors, relates to the construction of lever safety-valves, and to the required capacity of lever and spring-loaded valves. Section 5, of Rule 5, provides that "no person shall receive license (as an engineer) who is not able to determine the weight necessary to be placed on the lever of a safety-valve (the diameter of the valve, length of lever and distance of valve from fulcrum being given) to withstand any given pressure of steam in a boiler, or who is not able to figure and determine the strain brought on the braces of a boiler with a given pressure of steam, the position and distance apart of braces being known; such knowledge to be determined by an examina- tion in writing, and the report of examination filed with the applica- tion in the office of the local inspectors, and no engineer, or assistant engineer, now holding a license, shall have the grade of the same raised without possessing the above qualifications." Section 14, of Rule 9, "in case it comes to the knowledge of in- spectors that steam has been carried in any boiler in excess of that allowed by law, such inspectors are advised to report the facts to the United States District Attorney for prosecution under Section 4437 of the Revised Statutes, and in addition thereto require the owners to place on the boilers a lock-up safety-valve that will prevent the carry- ing of any excess of steam, and such valve shall be exclusively under the control of the inspectors." LEVER SAFETY-VALVES. Section 24, of Rule 2, of the Board of Supervising Inspectors, provides, among other things, the following: "Lever safety-valves to be attached to marine boilers shall have an area of not less than one square inch to two square feet of the grate surface in the boiler, and the seats of all such safety-valves shall have an angle of inclination of 45 degrees to the center line of their axis." UJ. S. Marine Boiler Inspection and Engineers' License Laws. 411 CO AREA OF A LEVER SAFETY-VALVE. RULE.-Divide the number of square feet of the grate surface of the boiler by 2, and the quotient will give the required area of a lever safety-valve in square inches. Example.-Let 4 feet equal width of grate surface. Let 5 feet equal length of grate surface. Let 2 equal a divisor. Then we have: 4X5 80 10 square inches. Required area of lever safety 2 valve. Performing the operation in the ordinary way, we have: 4 Width of grate surface. Dividing by the divisor. 2) 20 5 Length of grate surface. 10 square inches. Required area of lever safety-valve. DIAMETER OF LEVER SAFETY-VALVE. RULE.-First, multiply the length of the grate surface, in feet, by the width of the grate surface, in feet, and divide the product by 2; the quotient will give the required area of the valve in square inches. Second, divide the required area of the valve by .7854, and extract the square root of the quotient; the answer will give the required diameter of the valve in inches. Example.-Let 5 feet equal length of grate surface. Let 4 feet equal width of grate surface. Let 2 equal the divisor. Let .7854 equal a constant. Then we have: (5X4)-2 .7854 3.56+ inches. Required diameter of valve. Performing the operation in the ordinary way, we have: 5 Length of grate surface. 4 Width of grate surface. 2) 20 Am't carried forward, 10 elf foom Stored 412 A Library of Steam Engineering. Am't brought forward, .7854) 10.0000 (12.7323+ The quotient. 7 854 2 1460 15708 57520 54978 25420 23562 18580 15708 28720 23562 Next, extracting the square root of the quotient, we have: 12.7323 (3.56+ inches. Required diameter of 9 65) 373 325 706) 4823 4236 valve. SPRING-LOADED SAFETY-VALVES. The third paragraph of Section 24, of Rule 2, reads as follows: "Any spring-loaded safety-valve constructed so as to give an in- creased lift by the operation of steam, after being raised from its seat, or any spring-loaded safety-valve constructed in any other manner so as to give an effective area of opening equal to that of the aforemen- tioned spring-loaded safety-valve may be used in lieu of the common lever weighted valve on all boilers on steam vessels, and all such spring- loaded safety-valves shall be required to have an area of not less than one square inch to three square feet of the grate surface of the boiler; and each spring-loaded safety-valve shall be supplied with a lever that will raise the valve from its seat a distance of not less than that equal to one-eighth of the diameter of the valve opening; and the seats of all such safety-valves shall have an angle of inclination to the center line of their axis of 45 degrees. But in no case shall any spring-loaded safety-valve be used in lieu of the lever safety-valve without the ap- proval of the Board of Supervising Inspectors." AREA OF SPRING-LOADED SAFETY-VALVES. RULE.-Divide the number of square feet of the grate surface of the boiler by 3, and the quotient will give the area of the valve in square inches. U. S. Marine Boiler Inspection and Engineers' License Laws. 413 Example.-Let 5 feet equal length of grate surface. Let 4 feet equal width of grate surface. Let 3 equal a divisor. Then we have: 5X4 -6.6666+ square inches. Required area of valve. 3 Performing the operation in the ordinary way, we have: 5 4 3) 20.0000 6.6666+ square inches. Required area of valve.. DIAMETER OF SPRING-LOADED SAFETY-VALVES. RULE. First, multiply the length of the grate surface, in feet, by the width of the grate surface, in feet, and divide the product by 3; the quotient will give the required area of the valve in square inches. Second, divide the required area of the valve by .7854, and extract the square root of the quotient; the answer will give the required diameter of the valve in inches. Example.-Let 5 feet equal length of grate surface. Let 4 feet equal width of grate surface. Let 3 equal the divisor. Let .7854 equal a constant. Then we have: (5X4)÷3 2.91 inches. Required diameter of valve. .7854 Performing the operation in the ordinary way, we have: 5 4 3) 20.0000 .7854) 6.6666 (8.4881+ The quotient. 6 2832 38340 31416 69240 62832 64080 62832 12480 7854 414 A Library of Steam Engineering. Next, extracting the square root of the quotient, we have: 8.4881 (2.91+ inches. Required diameter of valve. 4 49) 448 441 581) 781 581 PRACTICAL ADJUSTMENT OF SAFETY-VALVES. As the adjustment of safety-valves, mathematically, has been exhaustively treated in a preceding chapter, it will not be necessary to go extensively into the subject here; hence, the instructions will be confined to the illustration of the more complicated questions connected with safety-valves, met in practice by inspectors and engineers. In illustrating this subject, the valves heretofore employed were small and made simple for the purpose of enabling the student to more easily understand the rules and examples given. The rules laid down here will be exactly the same as those given in the fore part of this work, but they will now be applied to the safety-valves in general use; there- fore, the information that will now be imparted will be strictly prac- tical, while that previously given was merely theoretical. The student will therefore have an opportunity to make a practical application of the information given in the theory of the adjustment of safety-valves mathematically. PRESSURE PER SQUARE INCH REQUIRED TO RAISE A VALVE. [Section 4418 of the Revised Statutes of the United States.] RULE.-First, multiply the weight of the valve and spindle by the dis- tance of the center of the valve from the fulcrum, and call the product "Product No. 1." Second, multiply the weight of the lever by the distance of its center of gravity from the fulcrum, and call the product "Product No. 2." Third, multiply the weight of the weight by the distance of its centers from the fulcrum, and call the product "Product No. 3." Fourth, add Products Nos. 1, 2 and 3 together, and call the answer "The Sum." Fifth, square the diameter of the valve and multiply the square of its diameter by .7854, and multiply the product by the distance of the center of the valve from the fulcrum, and call the last product "Product No. 4." Sixth, divide "The Sum" by "Product No. 4," and the quotient will give the pressure per square inch required to raise the valve. U. S. Marine Boiler Inspection and Engineers' License Laws. 415 Example.-Let 4 inches equal distance of center of valve from fulcrum. Let 9 pounds equal weight of valve and spindle. Let 20 inches equal distance of center of gravity of lever from fulcrum. Let 35 pounds equal weight of lever. Let 32 inches equal a distance of weight on lever from fulcrum. Let 180 pounds equal weight of the weight. Let .7854 equal a constant. Let 4 inches equal diameter of the valve. The valve and attachments being arranged as shown in Fig. 174. F 4" 9 lbs. 4" Diameter of valve. Then we have: 20" A 35 lbs. 32" 40" Λ 180 lbs. Fig. 174 (9X4)+(35X20)+(180×32) 4X4X.7854X4 -129.23+ lbs. Pressure per square inch required to raise the valve. Performing the operation in the ordinary way, we have: Next we have: 9X4= 36 "Product No. 1." 35X20 180X32 700 "Product No. 2." 5760 "Product No. 3." 6496 "The Sum." 4 Diameter of valve. 4 Diameter of valve. 16 Square of the diameter of valve. .7854 A constant. 47124 7854 12.5664 Area of valve. 4 Distance of center of valve from the fulcrum. 50.2656 "Product No. 4." 416 A Library of Steam Engineering. Finally, dividing "The Sum" (6496) by "Product No. 4" (50.2656), we have: 50.2656) 6496.0000 (129.23+ lbs. Pressure required to raise 5026 56 the valve. 1469 440 1005 312 464 1280 452 3904 11 73760 10 05312 1 684480 1 507968 DISTANCE FROM FULCRUM A GIVEN WEIGHT IS REQUIRED TO BE PLACED ON THE LEVER TO ALLOW THE VALVE TO RISE AT A GIVEN PRESSURE. [Section 4418 of the Revised Statutes of the United States, and Section 5, of Rule 5, of the United States Board of Supervising Inspectors.] RULE. First, multiply the area of the valve by the distance of the center of the valve from the fulcrum; then multiply the product by the given pressure, in pounds per square inch, and call the product "Product No. 1." Second, multiply the weight of the valve and spindle by the distance of the center of the valve from the fulcrum, and call the product "Product No. 2.' Third, multiply the weight of the lever by the distance of its center of gravity from the fulcrum, and call the product "Product No. 3." Fourth, add Products Nos. 2 and 3 together, and call the answer “The Sum;" then subtract "The Sum" from "Product No. 1," and call the answer "The Remainder." Fifth, divide "The Remainder" by the number of pounds contained in the weight, and the quotient will give the distance from the fulcrum, in inches, the weight is required to be placed to allow the valve to rise at a given pressure per square inch. Example.-Let 3 inches equal diameter of the valve. Let .7854 equal a constant. Let 3 inches equal distance of center of valve from the fulcrum. Let 125 pounds equal given pressure per square inch. Let 73 pounds equal weight of valve and spindle. Let 28 pounds equal weight of lever. Let 18 inches equal distance of center of gravity of lever from the fulcrum. Let 140ğ pounds equal weight of the weight. U.S. Marine Boiler Inspection and Engineers' License Laws. 417 Reducing the vulgar fractions to decimals we have them placed with their respective whole numbers in their proper positions, as shown in Fig. 175. F 3.57 18.125" 7.375 lbs. weight of 28.75 lbs. weight of lever. valve and spindle. 3.5" Diameter of valve 125 lbs. pressure per square inch at which the valve is required to rise. Fig. 175 140.625 lbs. Then, following the rule, we have: (3.5X3.5X.7854X3.5X125)-(7.375X3.5)+(28.75 X 18.125) 140.625 =26.04+ in. Distance the center of the weight is to be placed from the fulcrum. Performing the operation in the ordinary way, we have: 3.5 Diameter of valve. 3.5 Diameter of valve. 175 10 12.25 Square of diameter of valve. .7854 A constant. 4900 6125 9800 8575 9.621150 Area of valve. 3.5 Distance of center of valve from fulcrum 4 8105750 28 863450 33.6740250 125 Given pressure per square inch. 168 3701250 673 480500 3367 40250 4209.2531250 "Product No. 1." no alt to 27 418 A Library of Steam Engineering. Next, multiplying the weight of the valve and spindle (7.375 pounds) by the distance of the center of the valve from the fulcrum (3.5 inches), we have: 7.375 3.5 3 6875 22 125 25.8125 "Product No. 2." Next, multiplying the weight of the lever (28.75 pounds) by the distance of the center of gravity of the lever from the fulcrum (18.125 inches), we have: 28.75 18.125 14375 5750 2 875 230 00 287 5 521.09375 "Product No. 3." Adding "Product No. 2" to "Product 25.8125 No. 3." 'Product No. 2." 546.90625 "The Sum." Next, subtracting "The Sum" from "Product No. 1," we have: 4209.253125 "Product No. 1." 546.90625 "The Sum." 3662.346875 "The Remainder." Finally, dividing "The Remainder" by the number of pounds contained in the given weight, we have: 140.625000) 3662.346875 (26.04+ inches. Distance the center of 2812 50000 849 846875 843 750000 6 09687500 5 62500000 weight is required to be placed from the fulcrum. BE PLACED A NUMBER OF POUNDS TO BE CONTAINED IN A WEIGHT TO GIVEN DISTANCE FROM THE FULCRUM TO ALLOW THE VALVE TO RISE AT A GIVEN PRESSURE IN POUNDS PER SQUARE INCH. [Section 4418 of the Revised Statutes of the United States, and Section 5, of Rule 5, of the United States Board of Supervising Inspectors.] RULE.-First, multiply the area of the valve by the distance of the center of the valve from the fulcrum; then multiply the product by the given pressure in pounds per square inch, and call the last product "Product No. 1." U. S. Marine Boiler Inspection and Engineers' License Laws. 419 Second, multiply the weight of the valve and spindle by the distance of the center of the valve from the fulcrum, and call the product "Product No. 2." Third, multiply the weight of the lever by the distance of its center of gravity from the fulcrum, and call the product "Product No. 3." Fourth, add Products Nos. 2 and 3 together, and call the answer "The Sum;" then subtract "The Sum" from "Product No. 1," and call the answer "The Remainder." Fifth, divide "The Remainder" by the distance, in inches, the weight is to be placed on the lever from the fulcrum; the quotient will give the number of pounds to be contained in the weight. Example.-Let 4 inches equal diameter of the valve. Let .7854 equal a constant. Let 4 inches equal distance of center of valve from the fulcrum. Let 150 pounds per square inch equal given pressure. Let 10 pounds equal weight of the valve and spindle. Let 423 pounds equal weight of the lever. 16 Let 20 inches equal distance of center of gravity of lever from the fulcrum. even Let 40 inches equal distance of center of weight from the fulcrum. Reducing the vulgar fractions to decimals we place them with their respective whole numbers, and the whole numbers not having any decimals, in their respective positions, as shown in the following diagram: F 4.5" 10.5 lbs. weight of 20.1875" A 40" valve and spindle. 42.125 lbs. weight of lever. 4.25" Diameter of valve. 150 lbs. pressure per square inch at which the valve is required to rise. Then we have: Fig. 176 (4.25X4.25X.7854X4.5X150)-(10.5X4.5)+(42.12520.1875) 40 =216.95+ lbs. Weight required Performing the operation in the ordinary way, we have: 4.25 Diameter of valve. 4.25 Diameter of valve. 2125 850 17 00 Am't carried forward, 18.0625 Square of diameter of valve. €420 A Library of Steam Engineering. Am't brought forward, 18.0625 Square of diameter of valve. .7854 A constant. 722500 903125 1 445000 12 64375 14.18628750 Area of valve. 4.5 Distance of center of valve from fulcrum. 7 093143750 56 74515000 63.838293750 150 Given pressure per square inch. 3191 914687500 6383 8293750 9575.744062500 "Product No. 1." Next, multiplying the weight of valve and spindle (10.5 pounds) by the distance of the center of the valve from the fulcrum (4.5 inches), we have: 10.5 4.5 5 25 42 0 47.25 "Product No. 2." Next, multiplying the weight of the lever (42.125 pounds) by the distance of the center of gravity of the lever from the fulcrum (20.1875 inches), we have: 42.125 20.1875 210625 294875 3 37000 4 2125 842 50 850.3984375 "Product No. 3." Adding "Product No. 2" to "Product No. 3." 47.25 "Product No. 2." 897.6484375 "The Sum." Next, subtracting "The Sum" from "Product No. 1," we have: 9575.7440625 "Product No. 1." 897.6484375 "The Sum." 8678.0956250 "The Remainder." U. S. Marine Boiler Inspection and Engineers' License Laws. 421 Finally, dividing "The Remainder" by the distance, in inches, the center of the weight is to be placed on the lever from the fulcrum, we have: 40.000000) 8678.095625 (216.95+ lbs. Weight required. 8000 0000 678 09562 400 00000 278 095625 240 000000 000138 0956250 36 0000000 2 09562500 furre mich et 2 00000000 SIZE OF A CUBIC CAST-IRON WEIGHT REQUIRED TO BE PLACED A GIVEN DISTANCE FROM THE FULCRUM TO ALLOW THE VALVE TO RISE AT A GIVEN PRESSURE PER SQUARE INCH. [Section 4418 of the Revised Statutes of the United States; and Section 5, of Rule 5, of the United States Board of Supervising Inspectors.] RULE.-First, multiply the area of the valve by the distance of the center of the valve from the fulcrum; then multiply the product by the given pressure in pounds per square inch, and call the last product "Product No. 1." Second, multiply the weight of the valve and spindle by the distance of the center of the valve from the fulcrum, and call the product "Product No. 2." Third, multiply the weight of the lever by the distance of its center of gravity from the fulcrum, and call the product "Product No. 3." Fourth, add Products Nos. 2 and 3 together, and call the answer "The Sum;" then subtract "The Sum" from "Product No. 1," and call the answer "The Remainder." Fifth, divide "The Remainder" by the distance, in inches, the weight is to be placed on the lever from the fulcrum; the quotient will give the number of pounds to be contained in the weight. Sixth, divide the number of pounds to be contained in the weight by the constant .2607, and extract the cube root of the quotient; the answer will give the size, in inches, of a cubic weight required to be placed a given distance from the fulcrum to allow the valve to rise at a given pressure. Example.-Let 4 inches equal diameter of the valve. Let .7854 equal a constant. Let 4 inches equal distance of center of valve from the fulcrum. Let 150 pounds per square inch equal given pressure. 422 A Library of Steam Engineering. Let 10 pounds equal weight of the valve and spindle. Let 40 pounds equal weight of the lever. Let 20 inches equal distance of center of gravity of lever from the fulcrum. Let 40 inches equal distance of center of weight from the fulcrum. Let 2607 ten thousandths of a pound equal weight of a cubic inch of cast-iron. Then we have: (4X4X.7854x4x150)-(10X4)+(40×20) -167.496 lbs. Weight required. 40 Then 167.496.2607-8.6+ inches. Dimensions of cubic weight required. Performing the operation in the ordinary way, we have: 4 Diameter of valve. 4 Diameter of valve.- 16 Square of diameter of valve. .7854 A constant. 4 7124 7 854 12.5664 Area of valve. 4 50.2656 150 Given pressure per square inch. 2513 2800 5026 56 7539.8400 "Product No. 1." Next, multiplying the weight of the valve and spindle (10 pounds) by their distance from the fulcrum (4 inches), we have: 10 4 66 40 Product No. 2." Next, multiplying the weight of the lever (40 pounds) by the distance of the center of gravity of the lever from the fulcrum (20 inches), we have: 40 800 "Product No. 3." 20 U. S. Marine Boiler Inspection and Engineers' License Laws. 423 Next, adding Products Nos. 2 and 3 together, we have: 800 "Product No. 3." 40 "Product No. 2." 840 "The Sum." Next, subtracting the "The Sum" from "Product No. 1," we have: 7539.84 "Product No. 1." 840 "The Sum." 6699.84 "The Remainder." Next, dividing "The Remainder" by the distance the center of the weight is to be placed from the fulcrum, we have: 40) 6699.84 (167.496 Number of pounds to be con- tained in the weight. 40 269 240 299 280 198 160 384 360 240 240 Next, dividing the number of pounds contained in the weight by 2607, we have: .2607) 167.4960 (642.485+ Number of cubic inches to be 156 42 contained in the weight. 11 076 10 428 6480 5214 12660 10428 22320 20856 14640 13035 424 A Library of Steam Engineering Finally, extracting the cube root of the quotient, we have: 642.485 (8.6+ inches. Cubic weight required, 512 8X8X8= 8X8X300-19200 130 485 8X6X30 6×6= 1440 36 20676 1240 56 Then a safety-valve adjusted according to the following diagram, will rise at a pressure of 150 pounds per square inch. F 4/ 20" A40 lbs. 10 lbs. 4" Diameter of valve 40" -8.6- 8.6 8.6 -8.61 Fig. 177 k--8.6"- PRESSURE REQUIRED TO RAISE A VALVE HAVING A SPHERICAL WEIGHT OF GIVEN DIAMETER ATTACHED TO LEVER. As inspectors of marine boilers are often required to determine the pressure per square inch required to raise the valve attached to a steam boiler under pressure, the following diagrams will show how those valves are found in practice, and the rules will show how to determine the pressure per square inch required to raise them. adu gaibinib F 3.5" 18" 36" 8 lbs. weight of A30 lbs. weight of lever. valve and spindle. 3.5" Diameter of valve. Spherical weight 8 inches in diameter cast-iron. Fig. 178 RULE.-Determine the number of pounds contained in the spherical cast-iron weight by multiplying the cube of its diameter by the decimal .5236; then multiply the product by .2607, and the last product will give the num- ber of pounds contained in the weight. Thus: 8X8X8X.5236X.2607 69.889+ lbs. Weight of spherical weight. U. S. Marine Boiler Inspection and Engineers' License Laws. 425 Performing the operation in the ordinary way, we have: 8 Diameter of spherical weight. 8 Diameter of spherical weight. 64 Square of diameter of spherical weight. 8 Diameter of spherical weight. 512 Cube of diameter of spherical weight. A constant. .5236 1 0472 5 236 261 80 268.0832 Number of cubic inches contained in the .2607 weight. Weight of a cubic inch of cast-iron. 18765824 16 084992 53 61664 69.88929024 pounds. Weight of spherical weight. We are now prepared to make the following diagram and proceed to determine the pressure per square inch required to raise the valve: F 3.5" 18" 36" 8 lbs. Approximate weight of valve and spindle. A30 lbs. weight of lever. 3.5" Diameter of valve 69.889+ lbs. 3.5x3.5x.7854-9.62115 square inches, area of valve. eat molt ingles Fig. 179 - RULE. First, multiply the weight of the valve and spindle by the distance of the center of the valve from the fulcrum, and call the product "Product No. 1." Second, multiply the weight of the lever by the distance of the center of gravity of the lever from the fulcrum, and call the product "Product No. 2." Third, multiply the weight of the weight by its distance from the fulcrum, and call the product "Product No. 3." Fourth, add Products Nos. 1, 2 and 3 together, and call the answer "The Sum." Fifth, multiply the area of the valve by the distance of its center from the fulcrum, and call the product "Product No. 4." Sixth, divide "The Sum" by "Product No. 4," and the quotient will give the pressure per square inch required to raise the valve. 426 A Library of Steam Engineering. Example.-Let 8 lbs. equal weight of valve and spindle. Let 3.5" equal distance of center of valve from the fulcrum. Let 30 lbs. equal weight of lever. Let 18" equal distance of center of gravity of lever from the fulcrum. Let 69.889 lbs. equal weight of weight. Let 36" equal distance of centers of weight from the fulcrum. Let 9.62115 square inches equal area of valve. Then, we have: (8X3.5)+(30x18)+(69.889x36) =91.58+ lbs. Pressure per square 9.62115×3.5 inch required to raise the valve. Performing the operation in the ordinary way, we have: 3.5 Distance of center of valve from the fulcrum. 8 Weight of valve and spindle in pounds. 28.0 "Product No. 1." 30 Weight of lever. 18 Distance of center of gravity of lever from the fulcrum. 240 30 540 "Product No. 2." 69.889 Weight of weight. 36 Distance of center of weight from the fulcrum. 419 334 2096 67 2516.004 "Product No. 3." Adding Products Nos. 1, 2 and 3 together, we have: 28 540 "Product No. 1." 66 'Product No. 2." 2516.004 "Product No. 3." 3084.004 "The Sum." Next, multiplying the area of the valve by the distance of the center of the valve from the fulcrum, we have: 9.62115 3.5 Area of the valve. Distance of center of valve from the fulcrum. 4 810575 28 86345 33.674025 "Product No. 4." U. S. Marine Boiler Inspection and Engineers' License Laws. 427 Finally, dividing "The Sum" by "Product No. 4," we have: 33.674025) 3084.004000 (91.58+ lbs. 3030 66225 53 341750 Pressure per square inch required to raise the valve. 33 674025 19 6677250 16 8370125 2 83071250 2 69392200 It will be observed that the weight of the valve and spindle of a safety-valve attached to a steam boiler under pressure, can be obtained only by approximation, but the difference between the approximate weight of the valve and spindle and the actual weight of the same, can affect the result, at most, but a fractional part of a pound per square inch, and hence the rule in this case is sufficiently accurate to enable inspectors to determine whether the safety-valves have been tampered with, which is the real object of these rules. F Determine first the number of pounds contained in the weight. 4" 12 lbs. weight of valve and spindle. -4" Diameter of valve. 4x4x.7854-area of valve. 20" 40 lbs. weight of lever. 118 38" 6" Cast-iron Fig. 180 12" PRESSURE REQUIRED TO RAISE VALVE WITH CAST-IRON CONIC WEIGHT OF GIVEN DIMENSIONS. RULE.-First, subtract the diameter of the top from the diameter of the bottom, and divide the remainder by 2; then square the quotient, and call the product "Product No. 1." Second, square the length of the side of the weight, and call the product "Product No. 2." Third, subtract" Product No. 1" from "Product No. 2;" then extract the square root of the remainder, and the answer will give the length of the weight. Fourth, multiply the diameter of the bottom by the diameter of the top, and add the product and the square of the top and the square of the bottom together, and call the answer "The Sum." Fifth, multiply "The Sum" by .7854; then multiply the product by the length of the weight, and divide the last product by 3; the quotient will give the number of cubic inches contained in the weight. Sixth, multiply the number of cubic inches contained in the weight by 2607, and the product will give the number of pounds contained in the weight. 428 A Library of Steam Engineering. Example. Let 12" equal diameter of bottom of weight. Let 6" equal diameter of top of weight. Let 2 equal a constant. Let 8" equal length of the side of the weight. Let .7854 equal a constant Let 3 equal a constant. Let 2607 ten thousandths of a pound equal weight of a cubic inch of cast-iron. First, subtracting the diameter of the top from the diameter of the bottom, and dividing the remainder by 2, and squaring the quotient, we have: 12 Diameter of bottom of weight. 6 Diameter of top of weight. Dividing remainder by 2) 6 The remainder. 3 Squaring the quotient. 3 Square of the quotient. 9 "Product No. 1." Second, squaring the length of the side of the weight, we have: 8 Length of side of weight. 8 Length of side of weight. Square of length of side of weight. 64 "Product No. 2." Third, subtracting "Product No. 1" from "Product No. 2," and extracting the square root of the remainder, we have: 64 "Product No. 2." 9 "Product No. 1." 55 (7.4162-inches. Length through center of weight. 49 144) 600 576 1481) 2400 1481 14826) 91900 88956 148322) 294400 296644 JJ. S. Marine Boiler Inspection and Engineers' License Laws. 429 Fourth, multiplying the diameter of the bottom by the diameter of the top, and adding the product and the square of the diameter of the bottom together, we have: Next we have: Next we have: 12 Diameter of bottom of weight. 6 Diameter of top of weight. 72 Product of diameter of bottom multiplied by diameter of top. 6 Diameter of top of weight. 6 Diameter of top of weight. 36 Square of diameter of top of weight. 12 Diameter of bottom of weight. 222 22 12 Diameter of bottom of weight. 24 12 144 Square of diameter of bottom of weight. Then we have the three products added together, thus: 72 Product of diameter of bottom multiplied by top of weight. 36 Square of diameter of top of weight. 144 Square of diameter of bottom of weight. 252 The Sum." Fifth, multiplying "The Sum" by .7854, and multiplying the product by the length of the weight, and dividing the last product by 3, we have: 252 .7854 1008 1 260 20 16 176 4 197.9208 7.4162 3958416 1 1875248 1 979208 79 16832 1385 4456 3) 1467.82023696 489.2734+ Number of cubic inches contained in the weight. 430 A Library of Steam Engineering. Sixth, multiplying the number of cubic inches contained in the weight by .2607, we have: 489.2734 .2607 34249138 29 356404 97 85468 127.55357538 pounds. Weight of conic-shaped weight. Having ascertained the number of pounds contained in the weight, we are enabled to complete the diagram preparatory to determining the pressure per square inch required to raise the valve. F 4" 12 lbs. weight of valve and spindle. 4" Diameter of valve. 4x4x.7854-12.5664, area of valve. 20" 38" 40 lbs. weight of lever. 127.55 lbs. Fig. 181 We now proceed according to the rule laid down in the preceding case of the valve with spherical weight. Example.-Let 12 lbs. equal weight of valve and spindle. Let 4" equal distance of center of valve from the ful- crum. Let 40 lbs. equal weight of lever. Let 20" equal distance of center of gravity of lever from the fulcrum. Let 127.55 lbs. equal weight of weight. Let 12.5664 square inches equal area of valve. Then we have: (12×4)+(40×20)+(127.55x38) 12.5664×4 113+ lbs. Pressure per square inch required to raise the valve. Performing the operation in the ordinary way, we have: 12 X 4 48 40×20 800 127.55X38-4846.9 "Product No. 1." "Product No. 2." "Product No. 3." 5694.9 "The Sum." U. S. Marine Boiler Inspection and Engineers' License Laws. 431 Then, multiplying the area of the valve by the distance of its cen- ter from the fulcrum, and dividing "The Sum" by the product, we have: 12.5664X4 50.2656) 5694.9000 (113+ lbs. Pressure per square inch required to raise the valve. 5026 56 668 340 502 656 165 6840 150 7968 PRESSURE REQUIRED TO RAISE A VALVE WITH TWO OR MORE CAST-IRON WEIGHTS OF GIVEN NUMBER OF POUNDS IN EACH. Safety-valves with two or more weights, as shown by the following diagram, are frequently found in practice. F 4" 20" 24" 30" 37 12 lbs. weight of valve and spindle. 4" Diameter of valve. 4x4x.7854-12.5664, area of valve. A45 lbs! weight of lever. 35 lbs. 50 lbs. 120 lbs. Fig. 182 RULE.-First, multiply the weight of valve and spindle by the distance of the center of the valve from the fulcrum, and call the product "Product No. 1." Second, multiply the weight of the lever by the distance of its center of gravity from the fulcrum, and call the product "Product No. 2." Third, multiply the weight of the first weight by its distance from the fulcrum, and call the product "Product No. 3." Fourth, multiply the weight of the second weight by its distance from the fulcrum, and call the product "Product No. 4." Fifth, multiply the weight of the last weight by its distance from the ful- crum, and call the product "Product No. 5." Sixth, add "Products Nos. 1, 2, 3, 4 and 5 together, and call the answer The Sum." Seventh, multiply the area of the valve by the distance of its center from the fulcrum, and call the product "Product No. 6." Eighth, divide "The Sum" by "Product No. 6," and the quotient will give the pressure per square inch required to raise the valve. 0 432 A Library of Steam Engineering. Example. Let 12 lbs. equal weight of valve and spindle. Let 4" equal distance of center of valve from the ful- crum. Let 45 lbs. equal weight of lever. Let 20" equal distance of center of gravity of lever from the fulcrum. Let 50 lbs. equal weight of first weight. Let 24" equal distance of first weight from the fulcrum. Let 35 lbs. equal weight of second weight. Let 30" equal distance of second weight from the fulcrum. Let 120 lbs. equal weight of third or last weight. Let 37" equal distance of third or last weight from the fulcrum Let 12.5664 square inches equal area of the valve. Then we have: (12X4)+(45X20)+(50X24)+(35X30)+(120X37) =151.95+ lbs. Pressure per square 12.5664X4 inch required to raise the valve. Performing the operation in the ordinary way, we have: First, 12X 4 48 "Product No. 1." Second, 45 X 20 900 Third, 50x24 1200 Fourth, 35X30 1050 Fifth, 120×37-4440 66 Product No. 2." "Product No. 3." "Product No. 4." 66 'Product No. 5." Sixth, 7638 "The Sum." Seventh, 12.5664X4 50.2656 "Product No. 6." Finally, dividing "The Sum" by "Product No. 6," we have: 50.2656) 7638.0000 (151.95+ lbs. 5026 56 2611 440 2513 280 Pressure per square inch required to raise the valve. 98 1600 50 2656 47 89440 45 23904 Jonore and see 2 655360 2513280 U. S. Marine Boiler Inspection and Engineers' License Laws. 433 Fig. 183 Fig. 184 THE UNITED STATES LEVER SAFETY-VALVE. Fig. 185 Fig. 186 SPECIFICATIONS FOR UNITED STATES LEVER SAFETY-VALVE. Section 24, of Rule 2, of the United States Board of Supervising Inspectors provides, that all lever safety-valves to be attached to marine boilers must be made according to the following specifications: All lever safety-valves, to be attached to marine boilers, shall have an area of not less than one square inch to two square feet of grate surface of the boiler or boilers to which they are attached; and the seats of all such valves shall have an angle of inclination of 45 degrees. to the center line of their axis. All the points of bearing on the lever must be in the same plane. The distance of the fulcrum from the bearing of the lever on the spindle directly over the center of the valve must in no case be less than the diameter of the valve opening. The length of the lever should not exceed the distance of the ful- crum from the center of the valve multiplied by 10. The width of the bearings of the fulcrum must not be less than three-fourths of one inch. The length of the fulcrum links should not be less than 4 inches. The lever or fulcrum points must be made of wrought iron or steel; and the knife-edge fulcrum points and bearings for the points must be made of steel and hardened. 28 434 A Library of Steam Engineering. The valve seat and bushings for the valve spindle must be made of composition (gun metal) when the valve is intended to be attached to a boiler using salt water; but when the valve is intended to be attached to a boiler using fresh water, and generating steam of a high pressure, the parts named, with the exception of the bushings for the spindle, may be made of cast-iron. The valve must be guided by its spindle, both above and below the ground seat, and above the lever, through supports made of com- position (gun metal), or bushed with it. The spindle should fit loosely in the bearings or supports. When the valve is intended to be attached to the boilers of steamers navigating rough waters, the fulcrum links may be connected directly with the spindle of the valve; provided always that the knife-edged fulcrum points are made of steel and hardened, and that the vertical movement of the valve is unobstructed by any lateral movement. this case the short end of the lever should be attached directly to the valve casing. In SPRING-LOADED SAFETY-VALVES. S Fig. 187. Drip Pipe Folger Cin. U. S. Marine Boiler Inspection and Engineers' License Laws. 435 The engravings represent a spring-loaded valve, made in accord- ance with the requirements of the United States laws. Quite a num- ber of this kind of valves are made by different manufacturers, all differing in some degree from each other, and yet made in compliance Fig. 188. E D L Folger. Cin D 51 D N B R Fig. 189. Fig. 190. with the requirements of the law. It is provided with an overhang- ing lip, the effect of which is to increase the area presented to the operation of the steam after the valve is raised from its seat, and cause an increased lift in the movement of the valve from its seat. This movement is sudden, and if the valve was held in position by a lever and weight instead of a spring, the tendency would be to break the lever the first time the valve raised from its seat. Fig. 187 shows the valve and parts in vertical section. Fig. 188 shows the arrangement of the link D to which the lever L (Figs. 187 and 189) is attached; which lever is employed to raise the valve from 436 A Library of Steam Engineering. its seat, as required by law. Fig. 190 is a plan showing the brass bush B, with nickel seat N, and adjustable ring R, all of which are also shown in cross section in Fig. 187. The plan of the adjustable ring shows the vertical notches around its circumference, and by which it is turned up or down on the bush B, as shown in Fig. 187, by the medium of any sharp-pointed instrument, inserted from the outside of the valve case through the hole now occupied by the pin A, which pin is merely used to hold the ring in place after it has been properly adjusted. The object of the adjustable ring is to increase or diminish the size of the steam-escape orifice around the circumference of the inside of the overhanging lip of the valve head, and which is accomplished by screwing the ring up or down on the bush B; the reason for this regulation is to adjust the closing point so as to pre- vent too great a reduction of pressure in the boiler before the valve seats. By screwing the adjustable ring upward the seating of the valve is prolonged, and the steam pressure can be made to fall considerably below the point at which the valve was forced from its seat. By screw- ing the adjustable ring downward the valve can be made to seat within a fraction of the pressure required to raise it. It is good practice, how- ever, to adjust the ring so as to allow the valve to seat within two or three pounds of the pressure required to raise it. The bearing of the lower end of the spindle S, it will be observed, is considerably below the seat of the valve; this is to prevent tipping of the valve and cause it to come fairly to its seat after being lifted or raised from it. The spindle near the lower end is provided with a collar, by which the valve may be forced from its seat should it become fast or stick to the seat; this is also a requirement of the law, because oftentimes safety-valves become fast and stick to their seats, and no lifting of the spindle will cause them to rise. It is, therefore, provided by law that all safety-valves shall be so constructed that they can be forced from their seats by means of levers, or other suitable arrange- ments. The United States marine lever safety-valve is provided with a pin which passes through the lower end of the spindle and through a hub on the valve containing the lower end of the spindle. DIMENSIONS OF STEAM DRUM LEGS. Section 4435, of the Revised Statutes of the United States provides, that where two or more boilers have water connections, "there shall also be provided a similar steam connection, having an area of opening into each boiler of at least one square inch for every two square feet of effective heating surface contained in any one of the boilers so con- nected, half the flue and all other fire surface being computed as effective." U. S. Marine Boiler Inspection and Engineers' License Laws. 437 DIAMETER OF STEAM DRUM LEGS OR OPENING INTO EACH BOILER. RULE.-Divide the number of square feet of effective heating surface by 2; then divide the quotient by .7854, and extract the square root of the last quotient. Example.-Let 44 inches equal diameter of the boilers. Let 3.1416 equal a constant. Let two-thirds of circumference of boilers equal portion exposed to heat. Let 30 feet equal length of boilers. Let 2 equal number of flues in each boiler. Let 16 inches equal diameter of each flue. Let 144 square inches equal number of square inches in a square foot. First, determine the number of square feet of effective heating surface. RULE. First, multiply the constant 3.1416 by the diameter of the boiler, in inches; then divide the product by 3, and multiply the quotient by 2; the last product will give two-thirds of the circumference of the boiler. Second, multiply the length of the boiler, in feet, by 12, and the product will give the length of the boiler in inches. Third, multiply two-thirds of the circumference of the boiler, in inches, by the length, in inches, and the product will give the effective heating surface of the shell in square inches; the last product we will call "Product No. 1." Fourth, as but half the inner surface of each flue is to be computed as effective, we multiply the constant 3.1416 by the diameter of the flue in inches; the product will give the circumference of the flue. Fifth, multiply the length of the flue, in feet, by 12, and the product will give the length of the flue in inches. Sixth, multiply the circumference of the flue, in inches, by the length, in inches, and the product will give the entire surface of one flue, or one-half of each of the two flues, in square inches. Seventh, add the number of square inches in the shell of the boiler exposed to the heat to the number of square inches in the upper half of each flue in square inches, and the sum will give the entire number of square inches of effective heating surface. Eighth, divide the entire effective heating surface in square inches by 144, and the quotient will give the number of square feet of effective heating surface. Ninth, divide the number of square feet of heating surface by 2; then divide the quotient by .7854, and extract the square root of the last quotient; the answer will give the diameter of the legs of the steam drum or opening into each boiler. 438 A Library of Steam Engineering. Then we have: 3.1416X44-138.2304 inches. Circumference of boiler. First, And, 3) 138.2304 Second, Third, Fourth, Fifth, Sixth, Seventh, 46.0768 2 92.1536 inches. Two-thirds of the circumfer- ence of boiler. 30X12=360 inches. Length of boiler. 92.1536 360 Two-thirds of the circumference of the boiler in inches. Length of boiler in inches. 5529 2160 27646 08 33175.2960 square inches. Effective heating surface of shell of boiler. 3.1416×16-50.2656 inches. Circumference of one flue, or two half flues. 30×12-360 inches. Length of flue. 50.2656 Circumference of flue in inches. 360 Length of flue in inches. 3015 9360 15079 68 18095.6160 square inches. Effective heating sur- face in both flues. 33175.2960 square inches. Effective heating surface in shell 18095.6160 square square inches. of boiler. Effective heating surface in flues. 51270.9120 square inches. Total effective heating surface in boiler. Eighth, 144) 51270.9120 (356+ square feet. Total heating surface of boiler. 432 807 720 870 864 U. S. Marine Boiler Inspection and Engineers' License Laws. 439 Ninth, dividing the number of square feet of effective heating surface by 2, and then dividing the quotient by .7854, we have: 2) 356 .7854) 178.0000 (226+ Last quotient. 157 08 20 920 15 708 5 2120 4 7124 Finally, extracting the square root of the last quotient, we have: 226 (15+ inches. Required diameter of legs of 1 25) 126 125 steam drum. BOILER FLUES. [Rules based upon the requiremen.s of the Rules and Regulations of the United States Board of Supervising Inspectors.] To obviate extreme thickness of material in boiler flues, and avoid impairing the efficiency of the flues in the generation of steam, flues are made in sections in nearly all cases where high pressure of steam and long flues are required. In this manner the thickness of the material can be reduced to its minimum. To accomplish the same object various kinds of flues have been introduced; notably the cor- rugated and ribbed flues, which flues have become familiar to engi- neers and boiler makers. Corrugated flues should have their ends plain not to exceed 6 inches, according to the United States Rules, and, for large flues, the corrugations should not be less than 1 inches deep, and the flues must conform practically to a true circle; and may be allowed a work- ing steam pressure according to their thickness of material and diameter of flue without regard to length, as prescribed in Section 14, of Rule 2, of the Board of Supervising Inspectors. SAFE-WORKING STEAM PRESSURE ALLOWABLE FOR CORRUGATED AND RIBBED FLUES. RULE.-Divide the constant whole number 14,000 by the mean diameter of the flue, in inches, and multiply the quotient by the thickness of material in decimals of an inch; the product will give the working steam pressure allowable. 440 A Library of Steam Engineering. Example.-Let 14,000 equal a constant. Let 40 inches equal mean diameter of the flue. Let 5 tenths of an inch equal thickness of material in the flue. Then we have: 14000 40 X.5-175 lbs. Pressure per square inch allowable. Performing the operation in the ordinary way, we have: 40) 14000 (350 120 5 Thickness of material. 200 175.0 lbs. Pressure per square inch 200 allowable. 0 THICKNESS OF MATERIAL REQUIRED IN CORRUGATED FLUES FOR ANY REQUIRED PRESSURE PER SQUARE INCH. RULE.-Multiply the pressure required per square inch by the mean diameter of the flue, in inches, and divide the last product by the constant whole number 14,000; the quotient will give the thickness of material required in decimals of an inch. Example.-Let 175 pounds per square inch equal the required pressure. Let 40 inches equal mean diameter of the flue. Let 14,000 equal a constant. Then we have: 175×40 14000 .5 Decimals of an inch. Thickness of material required. Performing the operation in the ordinary way, we have: 175 Pressure per square inch 40 Diameter of the flue. 14000) 7000.0 (0.5 Decimals of an inch. Thickness of 7000 0 material required. Ribbed flues, in regard to pressure per square inch allowable, and thickness of material required, are governed by the same rules that govern corrugated flues. U. S. Marine Boiler Inspection and Engineers' License Laws. 441 PLAIN CYLINDRICAL FURNACE FLUES. [Section 15, of Rule 2, of the United States Board of Supervising Inspectors.] PRESSURE ALLOWABLE FOR PLAIN CYLINDRICAL FURNACE FLUES. RULE.-First, multiply the constant whole number 89,600 by the square of the thickness of the material in the flue, in decimals of an inch, and call the last product "Product No. 1." Second, multiply the diameter, in inches, by the length of the flue, in feet, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the steam pressure allowable per square inch. Example.-Let 89,600 equal a constant. Let 5 tenths of an inch equal thickness of material. Let 40 inches equal diameter of the flue. Let 7 feet equal the length of the flue. Then we have: 89600X.5X.5 40×7 -=80 lbs. Steam pressure per square inch allowable. Performing the operation in the ordinary way, we have: 89600 A constant. 5X5= .25 Square of thickness of material. 448000 179200 22400.00 " Product No. 1." Next we have: 40 Diameter of flue in inches. 7 Length of flue in feet. 280 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 280) 22400 (80 lbs. Pressure per square inch allowable. 2240 0 FURNACE FLUES WITH ADAMSON AND ANGLE-IRON RINGS. Where plain cylindrical furnace flues are made in sections, and flanged to a depth of not less than 2 inches, and substantially riveted together with wrought-iron rings between such flanges, and such rings having a thickness of not less than one-half inch and a width of not less 442 A Library of Steam Engineering. than 2 inches; or in lieu thereof, angle-iron rings are employed, and such rings having a thickness of not less than double the thickness of material in the flue and a depth of not less than 2 inches, and sub- stantially riveted in position with wrought-iron thimbles between the Fig. 191 ADAMSON RING. FLANGED JOINT REQUIRED BY RULE 2, SECTION 15. Fig. 192 ANGLE-IRON STRENGTHENING RING REQUIRED BY RULE 2, SECTION 15. inner surface of such rings and the outer surface of the flue, at a dis- tance from the flue not to exceed 2 inches, with rivets, having a diam- eter of not less than one and one-half times the thickness of material in the flue, and placed apart at a distance not to exceed 6 inches from center to center at the outer surface of the flue, the distance between the flanges or the distance between such angle-iron rings shall be taken as the length of the flue in determining the pressure allowable. WORKING STEAM PRESSURE ALLOWABLE. RULE. First, multiply the constant whole number 89,600 by the square of the thickness of material in the flue, in decimals of an inch, and call the product "Product No. 1." Second, multiply the diameter of the flue, in inches, by the distance between the flanges of the longest section, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the pressure per square inch allowable. Example. Let 89,600 equal a constant. Let 5 tenths of an inch equal thickness of material in the flue. Let 40 inches equal diameter of the flue. Let 5 feet equal distance between flanges or angle-iron rings. U. S. Marine Boiler Inspection and Engineers' License Laws. 443 Then we have: 89600X.5X.5 --112 lbs. Pressure per square inch allowable. 40×5 Performing the operation, we have: .5 .5 .25 Square of thickness of material in the flue. Then, multiplying the constant whole number (89,600) by the square of the thickness of material in the flue (.25), we have: Next we have: 89600 .25 4480 00 17920 O 22400.00 Product No. 1." 40 Diameter of flue in inches. 5 Distance in feet between rings of longest section. 200 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 200) 22400 (112 lbs. Pressure per square inch allowable. 200 240 200 400 400 THICKNESS OF MATERIAL REQUIRED FOR A PLAIN FURNACE FLUE OF A GIVEN DIAMETER, GIVEN LENGTH AND GIVEN STEAM - PRESSURE PER SQUARE INCH. RULE. First, multiply the given diameter of the flue, in inches, by the length of the flue, in feet; then multiply the product by the given pressure per square inch, and call the last product "Product No. 1." Second, divide "Product No. 1" by the constant whole number 89,600, and extract the square root of the quotient; the answer will give the thickness of material in decimals of an inch. Example.-Let 40 inches equal diameter of the flue. Let 4 feet equal length of the flue. Let 140 pounds per square inch equal given pressure. Let 89,600 equal a constant. 444 A Library of Steam Engineering. Then we have: 40×4X140 89600 5 Decimals of an inch. Thickness of material required. Performing the operation in the ordinary way, we have: 40 Diameter of flue in inches. 4 Length of flue in feet. 160 140 Pressure per square inch. 6400 160 22400 Product No. 1." Next, dividing "Product No. 1" by the constant 89,600, we have: 89600) 22400.0 (0.25 The quotient. 17920 O 4480 00 4480 00 Finally, extracting the square root of the quotient, we have: 25 (.5 Decimals of an inch. Thickness of material .25 required. DIAMETER REQUIRED OF A FURNACE FLUE OF A GIVEN LENGTH, GIVEN THICKNESS OF MATERIAL AND GIVEN STEAM PRESSURE PER SQUARE INCH. RULE. First, multiply the constant whole number 89,600 by the square of the thickness of material, in decimals of an inch, and call the last product "Product No. 1." Second, multiply the given pressure per square inch by the length of the flue in feet, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter of the flue in inches. Example. Let 89,600 equal a constant. Let 5 tenths of an inch equal thickness of material. Let 140 pounds per square inch equal given pressure. Let 4 feet equal length of the flue. U. S. Marine Boiler Inspection and Engineers' License Laws. 445. Then we have: 89600X.5X.5 140×4 -40 inches. Required diameter of the flue. Performing the operation in the ordinary way, we have: Next we have: 5X.5= 89600 A constant. .25 Square of thickness of material. 4480 00 17920 0 22400.00 "Product No. 1." 140 Given pressure per square inch. 4 Length of flue in feet. 560 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 560) 22400 (40 inches. Required diameter of the flue. 2240 0 LENGTH OF A FURNACE FLUE, OR LENGTH BETWEEN THE FLANGES OF SUCH A FLUE MADE IN SECTIONS, WHEN THE THICKNESS OF MATERIAL, STEAM PRESSURE PER SQUARE INCH, AND DIAMETER OF THE FLUE ARE GIVEN. square RULE.-First, multiply the constant whole number 89,600 by the of the thickness of material, in decimals of an inch, and call the product "Product No. 1." Second, multiply the given pressure per square inch by the diameter of the flue, in inches, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required length of the flue, or longest section between the flanges, in feet. Example.-Let 89,600 equal a constant. Let 5 tenths of an inch equal thickness of material. Let 140 pounds per square inch equal given pressure. Let 40 inches equal diameter of the flue. Then we have: 89600X.5X.5 4 feet. Length of flue or longest section.. 140×40 446 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 89600 A constant. Next we have: 5x5= .25 Square of thickness of material. 4480 00 17920 0 22400.00 Product No. 1." 140 Given pressure per square inch. 40 Diameter of the flue. 5600 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 5600) 22400 (4 feet. Length of flue or longest section. 22400 HALF-ROUND IRON STRENGTHENING RINGS FOR FLUES. [Section 15, of Rule 2, of the United States Board of Supervising Inspectors.] Fig. 193 HALF-ROUND IRON STRENGTHENING RING REQUIRED BY RULE 2, SECTION 15. The strengthening rings may be made of half-round iron, as shown in Fig. 193, containing an area of cross section of not less than 9.6 times the thickness of material in the flue, and held in position around the flue with thimbles, at a distance from the surface of the flue not to exceed 2 inches, and substantially riveted with rivets, spaced not more than 8 inches from center to center at the surface of the flue, for rivets having a diameter of not less than seven-eighths of an inch; and not more than 6 inches from center to center for rivets having a diam- eter of not less than three-quarters of an inch; and not more than 4 inches from center to center for rivets having a diameter of not less than five-eighths of an inch; and no such rivets shall be allowed hav- ing a diameter of less than five-eighths of an inch. REQUIRED DIAMETER OF THE HALF-ROUND IRON STRENGTHENING RING. RULE.-Multiply the constant 9.6 by the thickness of the material in the flue, in decimals of an inch, and multiply the product by 2; then divide the last product by .7854, and extract the square root of the quotient; the answer will give the required diameter of the half-round iron strengthening ring. U. S. Marine Boiler Inspection and Engineers' License Laws. 447 Example. Let 9.6 equal a constant. Let 5 tenths of an inch equal thickness of material in the flue. Let 2 equal a constant. Let .7824 equal a constant. Then we have: 9.6X.5X2 .7854 3.5-inches. Required diameter of half- round iron strengthening ring. Performing the operation in the ordinary way, we have: 9.6 A constant. 5 Thickness of material. 4.80 2 A constant. .7854) 9.6000 (12.22+ The quotient. 7 854 1 7460 1 5708 17520 15708 18120 15708 Next, extracting the square root of the quotient, we have: 12.22 (3.5-inches. Required diameter of half- round iron strengthen- ing ring. 9 65) 322 325 REQUIRED DIAMETER OF HALF-ROUND IRON STRENGTHENING RING SIMPLE RULE. RULE.-Multiply the constant 24.5 by the thickness of material in the flue, in decimals of an inch, and extract the square root of the product. Example. Let 24.5 equal a constant. Let 5 tenths of an inch equal thickness of material in the flue. Then we have: 24.5.5-3.5 inches. Required diameter of half- round iron ring. Performing the operation in the ordinary way, we have: 24.5 A constant. 5 Thickness of material. 12.25 Product. 448 A Library of Steam Engineering. Next, extracting the square root of the product, we have: 12.25 (3.5 inches. Required diameter of half- 9 round iron ring. 65) 325 325 WORKING STEAM PRESSURE ALLOWABLE FOR FLUES HAVING HALF- ROUND IRON STRENGTHENING RINGS. RULE.-First, multiply the constant whole number 89,600 by the square of the thickness of material in the flue, in decimals of an inch; then multiply the product by .75, and call the last product "Product No. 1." Second, multiply the diameter of the flue, in inches, by the distance from center to center of strengthening rings, in feet, and call the product "Prod- uct No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the working steam pressure per square inch allowable. Example.-Let 89,600 equal a constant. Let 5 tenths of an inch equal thickness of material in the flue. Let .75 equal a co-efficient. Let 40 inches equal diameter of the flue. Let 3 feet equal distance from center to center of strength- ening rings. Then we have: 89600X.5X.5X.75 40×3 -140 lbs. Working steam pressure per square inch allow- able. Performing the operation in the ordinary way, we have: 89600 A constant. 5X5= .25 Square of thickness of material. 4480 00 17920 O 22400.00 .75 Next we have: 1120 0000 15680 000 16800.0000 "Product No. 1." 40 Diameter of flue in inches. 3 Length of flue in feet. 120 "Product No. 2." U. S. Marine Boiler Inspection and Engineers' License Laws. 449 Finally, dividing "Product No. 1" by "Product No. 2," we have: 120) 16800 (140 lbs. Working steam pressure per square inch allowable. 120 480 480 0 THICKNESS OF MATERIAL REQUIRED FOR FLUES REINFORCED WITH HALF-ROUND IRON STRENGTHENING RINGS. RULE.-First, multiply the given pressure, in pounds per square inch, by the diameter of the flue, in inches, and multiply the product by the distance between the strengthening rings of the flue, in feet, and call the last product "Product No. 1." Second, multiply the constant 89,600 by the co-efficient .75, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and extract the square root of the quotient; the answer will give the required thickness material in decimals of an inch. Example.-Let 140 pounds equal given pressure per square inch. Let 40 inches equal diameter of the flue. Let 3 feet equal distance from center to center of strength- ening rings. Let 89,600 equal a constant. Let .75 equal a co-efficient. Then we have: 140X40X3 =.5 89600X.75 Decimals of an inch. Required thick- ness of material for flue. Performing the operation in the ordinary way, we have: 140 lbs. Given pressure per square inch. Next we have: 40 inches. Diameter of flue. 5600 3 feet. Distance between strengthening rings. 16800 "Product No. 1." 89600 A constant. .75 A co-efficient 4480 00 62720 0 67200.00 Product No. 2. 29 450 A Library of Steam Engineering. Next, dividing "Product No. 1" by "Product No. 2," we have: 67200) 16800.0 (0.25 The quotient. 13440 0 3360 00 3360 00 Finally, extracting the square root of the quotient, we have: .25 (.5 Decimals of an inch. Thickness of mate- 25 rial required. DISTANCE FROM CENTER TO CENTER OF HALF-ROUND IRON STRENGTHENING RINGS. RULE.-First, multiply the constant 89,600 by the co-efficient .75, and then multiply the product by the square of the thickness of material in the flue, in decimals of an inch, and call the last product "Product No. 1." Second, multiply the given pressure, in pounds per square inch, by the diameter of the flue, in inches, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the distance from center to center of half-round iron strengthening rings. Example.-Let 89,600 equal a constant. Let .75 equal a co-efficient. Let 5 tenths of an inch equal thickness of material in the flue. Let 140 pounds equal given pressure per square inch. Let 40 inches equal diameter of the flue. Then we have: 89600X.75X.5X.5 -3 feet. Distance from center to center of strengthening rings. 140×40 Performing the operation in the ordinary way, we have 5X.5= 89600 A constant. .75 A co-efficient. 4480 00 62720 0 67200.00 .25 Square of the thickness of materia 3360 0000 13440 000 16800.0000 "Product No. 1." U. S. Marine Boiler Inspection and Engineers' License Laws. 451 Next, multiplying the given pressure per square inch by the diam- eter of the flue, we have: 140 lbs. Given pressure. 40 inches. Diameter. 5600 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 5600) 16800 (3 feet. Distance from center to center of strengthening rings. 16800 REQUIRED DIAMETER FOR A FLUE SUPPORTED WITH HALF- ROUND IRON STRENGTHENING RINGS. RULE.-First, multiply the constant 89,600 by .75, and then multiply the product by the square of the thickness of material in the flue, and call the last product "Product No. 1." Second, multiply the given pressure per square inch by the distance from center to center of reinforcement rings, in feet, and call the product "Product No. 2." Third, divide "Product No. 1" by "Product No. 2," and the quotient will give the required diameter of the flue in inches. Example.-Let 89,600 equal a constant. Let .75 equal a co-efficient. Let 5 tenths equal thickness of material in the flue. Let 140 pounds equal given pressure per square inch. Let 3 feet equal distance from center to center of rein- forcement rings. Then we have: 89600X.75 X.5X.5 140×3 -40 inches. Required diameter of flue. Performing the operation in the ordinary way, we have: 89600 A constant. .75 A co-efficient. 448000 627200 6720000 .5X.5=___.25 Square of thickness of material. 3360 0000 13440 000 16800.0000 "Product No. 1." 452 A Library of Steam Engineering. Next, multiplying the given pressure per square inch (140) by the distance from center to center of reinforcement rings in feet (3), we have: 140 3 420 "Product No. 2." Finally, dividing "Product No. 1" by "Product No. 2," we have: 420) 16800 (40 inches. Required diameter of flue. 1680 0 In all cases where the half-round iron ring is employed to strengthen flues, the area of cross section and diameter of such rings must be in accordance with the thickness of material in the flues, and such area of cross section, or diameter, must be determined by the first rule given under the head of "half-round iron strengthening rings for flues." The reason for basing the area or diameter of such rings upon the thickness of material in the flues is that the steam pressure is based upon the thickness of material in the flue. RIVETED AND LAP-WELDED FLUES MADE IN SECTIONS. Section 8, of Rule 2, of the United States Board of Supervising Inspectors, provides that the following tables shall include all riveted and lap-welded flues exceeding 6 inches in diameter, and not exceed- ing 40 inches in diameter, not otherwise provided for by law. And all such flues shall be made in sections, according to their respective diam- eters, not to exceed the lengths prescribed in the table; and such sec- tions shall be properly fitted one into the other, and substantially riveted, and the thickness of material required for any such flue of any given diameter shall, in no case, be less than the least thickness pre- scribed in the table for any such given diameter; and all such flues may be allowed the prescribed working steam pressure, if, in the opinion of the inspectors, it is deemed safe to make such allowance; and inspectors are therefore required, from actual measurement of each flue, to make such reduction from the prescribed working steam pressure for any material deviation in the uniformity of the thickness of material, or for any material deviation in the form of the flue from that of a true circle, as in their judgment, the safety of navigation may require. A small amount of material has been added to the thickness of material for flues over 6, and not over 10, inches in diameter, in order to prevent flattening too easily when using the hammer to remove the scale. All other flues in the table are made according to rule: U. S. Marine Boiler Inspection and Engineers' License Laws. 453 TABLE OF DIMENSIONS REQUIRED AND PRESSURES ALLOWABLE. SAFE-WORKING PRESSURE FOR CYLINDRICAL LAP-WELDED AND RIVETED BOILER FLUES MADE IN SECTIONS. Greatest Length of f Sections Allowable 5 FEET. 3 FEET. Least Thickness of f Material Allowable .18" .20" .21" .21" .22" .22" .23" .24" .25" .26" .27" .28" .29" .30" .31" .32" .33" Over Over Diameters of Flues Over 6" 7" 8" 9" and not and not and not and not over over over Over Over Over Over Over Over Over Over Over Over Over 10" Over Over 11" 12" 13" 14" 15" 16" 17" 18" 19" and not and not and not and not and not and not and not and not and not and not 20" and not 21" and not 7" 8" 9" over 10" over over over over over 11" 12" 13" 14" Thickness of Material Required. .18 inch. ... Over 20" over 21" over 22" Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure 189 over over over Over Over 22" and not over 15" 16" 17" 18" 19" 23" .19 (C 194 ... ... .20 66 199 184 ... .21 66 204 189 179 174 .22 66 194 ... 184 179 172 158 .23 66 199 189 184 179 165 152 .24 66 204 194 189 187 172 158 147 .25 66 199 194 195 179 165 153 143 .26 66 204 199 203 186 172 159 148 139 .27 66 ... 204 211 193 178 165 154 145 136 .28 66 200 185 172 160 150 141 134 .29 66 207 191 178 166 155 146 138 131 .30 66 198 ... 184 172 161 152 143 135 129 .31 66 205 190 178 166 156 148 140 133 126 .32 66 ... ... 196 183 171 161 153 145 137 131 125 .33 66 202 189 177 167 157 149 141 135 129 122 .34 66 194 182 172 162 154 146 139 132 126 .35 66 200 187 177 167 159 150 143 137 130 .36 66 .. 193 ... 182 172 163 154 147 141 134 .37 (6 187 177 168 159 151 145 138 .38 66 182 ... 172 163 155 149 142 .39 176 167 159 152 146 .40 66 172 163 156 150 .41 66 167 160 153 .42 66 164 157 .43 66 ... 160 454 A Library of Steam Engineering. 451 TABLE OF DIMENSIONS REQUIRED AND PRESSURES ALLOWABLE.-Continued. SAFE-WORKING PRESSURE FOR CYLINDRICAL LAP-WELDED AND RIVETED BOILER FLUES MADE IN SECTIONS. Greatest Length of f Sections Allowable Least Thickness of Material Allowable 30 INCHES. .34" .35" .36" .37" .38" .39" .40" .41" .42" .43" .44" .45" .46" .47" .48" .49" .50" Over Over Over Over Over Over Over 23" 24" 25" 26" 27" 28" 29" Diameters of Flues and not and not and not and not and not and not and not over over over over Over over 24" 25" 26" 27" 28" 30" Thickness of over 29" 38" Over 38" and not and not over 39" Pounds Pounds Pounds Pounds Pounds Pounds Tounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Pounds Material Required. Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure Pressure .34 inch. 121 ... Over 30" Over 31" Over 32" Over 33" Over 34" Over Over Over 35" 36" 37" Over 39" and not and not and not and not and not and not and not and not over Over over Over over over Over over over 31" 32" 33" 35" 36" 37" 40" .35 66 125 120 .36 (C 129 123 119 .37 CC ... 132 127 122 117 .38 (6 136 130 125 121 116 .39 66 139 133 129 124 119 115 .40 66 143 ... 137 132 127 123 118 115 .41 66 ... 147 141 135 130 126 121 118 114 .. .42 151 144 139 133 129 124 120 117 112 .43 66 154 147 142 137 132 127 123 119 115 112 .44 66 157 151 145 140 135 130 126 122 118 115 110 .45 66 155 148 143 138 133 129 124 121 117 113 110 .46 66 151 146 141 136 132 127 124 120 116 112 109 .47 66 149 144 139 135 130 126 122 118 115 112 .48 66 109 147 142 138 133 129 125 120 117 114 .49 (6 111 108 ... 145 140 135 131 127 123 120 116 .50 66 113 111 108 143 138 134 130 126 122 119 116 .51 66 113 110 107 141 137 133 129 125 121 118 .52 CC 115 112 109 140 135 131 127 123 120 .53 66 117 114 111 138 134 129 126 .54 CC 123 119 116 113 137 132 128 125 122 .55 66 118 116 135 131 127 124 121 118 .56 133 130 .57 66 126 123 120 132 128 125 122 .58 66 131 127 59 66 124 130 .60 66 126 129 U. S. Marine Boiler Inspection and Engineers' License Laws. 455 SAFE-WORKING PRESSURE FOR ANY DIAMETER OF FLUE OVER 10 INCHES. RULE.-Multiply the thickness of material, in hundredths of an inch, by the constant whole number 43, and divide the product by one-half of the diameter of the flue, in inches; the quotient will give the pressure per square inch allowable. Thickness of material to be used as whole numbers. Example.-Let 30 equal the number of hundredths of an inch in thickness of material. Let 43 equal a constant. Let 16 inches equal diameter of flue. Then we have: 30X43 16÷2 -161+ lbs. Pressure per square inch allowable. Performing the operation in the ordinary way, we have: 30 Thickness of material. 43 A constant. 90 120 16÷2= 8) 1290 161+ lbs. Pressure per square inch allowable. INCREASE IN LENGTH OF SECTIONS ALLOWABLE BY REDUCTION IN PRESSURE Riveted and lap-welded flues of any thickness of material, diam- eter and length of sections prescribed in the table, may be made in sections of any desired length, exceeding the maximum length allowed by the table, by reducing the prescribed pressure in proportion to the increased length of section according to the following rule: RULE.-Multiply the pressure in the table allowed for any prescribed thickness of material and diameter of flue by the greatest length, in feet, of sections allowable for such flue, and divide the product by the desired length of sections, in feet, from center line to center line of rivets in the circular seams of such sections; the quotient will give the working steam pressure allowable. Example.-Taking a flue in the table, 24 inches in diameter, required to be made in sections not exceeding 2.5 feet in length, and having a thickness of material of 44 one hundredths of an inch, and allowed a pressure of 157 pounds per square inch, and it is desired to make this flue in sections 5 feet in length. Then we have: 157×2.5 --78.5 lbs. Pressure allowable. 5 456 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 157 Given pressure. 2.5 Given length of sections. 785 314 Dividing by desired length of sections. 5) 392.5 78.5 lbs. Pressure allowable. THICKNESS OF MATERIAL REQUIRED FOR FLUES AND TUBES NOT OTHERWISE PROVIDED FOR. [Section 9, of Rule 2, of the United States Board of Supervising Inspectors.] Tubes and flues not exceeding 6 inches in diameter, and made of any prescribed length, and lap-welded flues required to carry a working steam pressure not to exceed 60 pounds per square inch, and having a diameter not exceeding 16 inches, and a length not exceeding 18 feet; and lap-welded flues required to carry a steam pressure exceeding 60 pounds per square inch, and not exceeding 120 pounds per square inch, and having a diameter not exceeding 16 inches, and a length not exceeding 18 feet, and made in sections not exceeding 5 feet in length, and fitted properly one into the other and substantially riveted; and all such tubes and flues shall have a thickness of material according to their respective diameters, as prescribed in the following table: Outside Diameter, Inches. Thickness of Material, Inch. Outside Diameter, Inches. Thickness of Material, Inch. Outside Diameter, Inches. Thickness of Material, Inch. 2 141234 141234 HHHH22223 1 11 13 .072 31 .072 34 .083 331 141234 .120 .120 .120 1/3/20 .095 4 .134 12 .095 4 .134 .095 .148 .109 .165 15 .109 7 .165 16 .109 .165 PREPRAE9: .180 10 .203 11 .220 .229 13 .238 14 .248 .259 .270 LAP-WELDED FLUES. LAP-WELDED FLUES NOT EXCEEDING 6 INCHES IN DIAMETER. [Section 10, of Rule 2, of the United States Board of Supervising Inspectors.] Lap-welded flues not exceeding 6 inches in diameter may be made of any required length without being made in sections. And all such lap-welded flues and riveted flues not exceeding 6 inches in diameter, may be allowed a working steam pressure not to exceed 225 pounds per square inch, if deemed safe by the inspectors. U. S. Marine Boiler Inspection and Engineers' License Laws. 457 LAP-WELDED FLUES EXCEEDING 6 INCHES IN DIAMETER AND NOT EXCEEDING 16 INCHES IN DIAMETER AND NOT EXCEEDING 18 FEET IN LENGTH. [Section 11, of Rule 2, of the United States Board of Supervising Inspectors.] Lap-welded flues exceeding 6 inches in diameter and not exceed- ing 16 inches in diameter and not exceeding 18 feet in length, and required to carry a steam pressure not exceeding 60 pounds per square inch, shall not be required to be made in sections. [Section 12, of Rule 2, of the United States Board of Supervising Inspectors.] Lap-welded and riveted flues exceeding 6 inches in diameter and not exceeding 16 inches in diameter and not exceeding 18 feet in length, and required to carry a steam pressure exceeding 60 pounds per square inch and not exceeding 120 pounds per square inch, may be allowed, if made in sections 5 feet in length, and properly fitted, one into the other and substantially riveted. LAP-WELDED AND RIVETED FLUES EXCEEDING 6 INCHES IN DIAMETER AND NOT EXCEEDING 40 INCHES IN DIAMETER. [Section 13, of Rule 2, of the United States Board of Supervising Inspectors.] Riveted and lap-welded flues exceeding 6 inches in diameter and not exceeding 40 inches in diameter, required to carry a working steam pressure per square inch exceeding the maximum steam pressure pre- scribed for any such flue in the table of Section 8, of this rule, shall be constructed under the provisions of Section 15, of this rule (Rule 2), and limited to the working steam pressure therein provided for fur- nace flues; but in no case shall the material in any such riveted or lap- welded flue be of less thickness for any given diameter than the least thickness prescribed in the aforementioned table for flues of such diameters. COPPER STEAM PIPES. THICKNESS OF MATERIAL REQUIRED. [Section 38, of Rule 2, of the United States Board of Supervising Inspectors.] All copper steam pipes shall be flanged to a depth of not less than four times the thickness of the material in the pipes, and all such flanging shall be made to a radius not to exceed the thickness of mater- ial in such pipes. And all such pipes shall have a thickness of material according to the working steam pressure allowed on the boilers, and such thickness of material shall be determined by the following rule: 458 A Library of Steam Engineering. RULE.-Multiply the working steam pressure allowed the boiler, in pounds per square inch, by the diameter of the pipe, in inches; then divide the product by the constant whole number 8000, and add .0625 to the quo- tient; the sum will give the thickness of material required in decimals of an inch. Example.-Let 175 pounds equal working steam pressure per square inch allowed the boiler. Let 5 inches equal diameter of the pipe. Let 8000 equal a constant. Let .0625 equal a constant. Then we have: 175X5 8000 +.0625.1718+ Decimals of an inch. Thickness of material required. Performing the operation in the ordinary way, we have: 175 Steam pressure. 5 Diameter of pipe. 8000) 875.0 (0.1093+ The quotient. 800 0 75 000 72 000 3 0000 2 4000 Next, adding the constant .0625 to the quotient, we have: .1093 .0625 .1718 Decimals of an inch. Thickness of material required. UNITED STATES MARINE EXAMINATION QUESTIONS AND ANSWERS. FOR ENGINEERS AND INSPECTORS. Write application for appointment to the position of local inspector of boilers of steam vessels at... In which give your name in full, age, native or naturalized citizen, experience, if any, in the construction of steam machinery, and in the care and management of .such; and such other experience and knowl- edge required by Section 4415 of the Revised Statutes of the United States, and the qualifications prescribed by it and Section 4416 Revised Statutes. U. S. Marine Boiler Inspection and Engineers' License Laws. 459 Question. Reduce the following vulgar fractions to decimal frac- 16 15 32 tions: 29 32 19 64 Answer. .375 .625 .875 .0625 .9375 .21875 .90625 .296875 .015625 Q. Extract the square root of 3429904. A. 1852. Q. Extract the square root of 3461460.25. A. 1860.5. Q. A. Extract the cube root of 6486889625. 1865. Q. Extract the cube root of 6691435355.584. A. 1884.4. Q. Define the term "tensile strength as employed in Section 3, of Rule 1, of the Board of Supervising Inspectors. In other words, what is meant by the tensile strength of boiler plate? A. Tensile strength is the adhesive quality of the plate tending to resist a strain to separate it; and it means the total strain, in pounds, required to separate a square inch of cross section of the material. Q. How is the tensile strength of boiler plate determined? A. Tensile strength in boiler plate is determined by breaking a sample in a machine made for that purpose, and dividing the strain at which the sample parted by the area of sample at point of fracture before breaking. Example.-Let .26 equal thickness of sample in hundredths of an inch. Let .96 equal width of sample in hundredths of an inch. Let 14,976 equal breaking strain in pounds. Then we have: 26 .96 156 234 .2496) 14976.0000 (60000 lbs. 14976 0000 Q. How is the ductility of boiler plate determined? A. The ductility is obtained by subtracting the area of sample at point of fracture, after breaking, from the area of sample, before breaking, and dividing the remainder by the area of sample before breaking. Example.-Let .26.96 equal area of sample before breaking. Let .18X.80 equal area of sample after breaking. 460 A Library of Steam Engineering. Then we have: .26X.96=.2496 .18X.80 .1440 .2496).1056.0 (.42+ Percentage of ductility. 998 4 57 60 49 92 Q. Determine the bursting pressure per square inch of a boiler having a diameter of 42 inches and a thickness of material of .25 of an inch, made of wrought iron plates having a tensile strength of 55,000 pounds per square inch, with longitudinal seams double riveted and holes drilled? [Section 4415 Revised Statutes.] A. Multiply the thickness of the weakest plate in the shell of the boiler by its tensile strength per square inch, and divide the product by one-half of the diameter of the boiler, in inches, and multiply the quotient by .70 for double-riveted longitudinal seams; by .65 for stag- gered-riveted longitudinal seams; and .56 for single-riveted longitu- dinal seams. Example. Let .25 equal thickness of weakest plate in hundredths of an inch. Let 55,000 equal tensile strength per square inch of weakest plate. Let 42 equal diameter of the boiler in inches. Then we have: .25 55000 125000 125 21) 1375000 (654.76+ lbs. Would be the bursting pressure if there were no seams in the boiler. 126 115 105 100 84 160 147 130 126 U. S. Marine Boiler Inspection and Engineers' License Laws. 461 As the seams in this boiler are double-riveted we multiply the above quotient by .70, and we have the following: 654.76 .70 458.3320 lbs. Bursting pressure. Q. Determine the bursting pressure per square inch of a boiler having a diameter of 48 inches, and a thickness of material of .26 of an inch, made of homogeneous steel plates having a tensile strength of 60,000 pounds per square inch, with longitudinal seams single riv- eted and holes drilled? [Section 4415 Revised Statutes.] A. The rule for this is the same as that for the preceding ques- tion, only .56 is used as a constant instead of .70 Example.-Let .26 equal thickness of weakest plate in hundredths. of an inch. Let 60,000 equal tensile strength of weakest plate per square inch. Let 48 inches equal diameter of boiler in inches. Then we have: .26 60000 24) 15600.00 (650 lbs. Would be the bursting 144 pressure if there were no seams in the shell of the boiler. 120 120 But as the boiler is single-riveted we multiply the above quotient by .56, and we have the following: 69599 3900 3250 364.00 lbs. Bursting pressure. Q. Determine the pressure per square inch allowable on a new boiler to which the heat is applied on the outside of the shell, to be employed on the Mississippi River and its tributaries, which boiler is 40 inches in diameter, containing two flues 16 inches in diameter each, and shell made of homogeneous steel plates .26 of an inch thick, 60,000- pounds tensile strength, longitudinal seams double riveted and holes- drilled? [Section 4434 Revised Statutes.] 462 A Library of Steam Engineering. A. That boiler would not be allowed for the reason that you can not get two 16 inch flues in a 40 inch boiler and have 3 inches space between and around the flues as required by Section 4434. This, of course, is a catch question, but any man who is well posted will not be caught by it. The thickness of material in flues is not given, neither is the length, and a candidate ought to detect that. Q. Determine the pressure per square inch allowable for steel plates .375 of an inch thick, having a tensile strength of 70,000 pounds, intended for a new boiler to be employed on the Mississippi River and its tributaries, which boiler is required to be 46 inches in diameter, with longitudinal seams single-riveted and holes punched, and containing two flues each 16 inches in diameter? [Section 4434 Revised Statutes.] A. The pressure allowable would be 190.21+ pounds if it were not for the law prohibiting the use of material over .30 of an inch thickness; and again, it will be seen that the thickness of the flues is not given, therefore it is impossible, without that information, to deter- mine the pressure allowable, and no inspector would be allowed to determine the pressure except by taking the strength of the flues into consideration as well as the shell. Q. Determine the pressure per square inch allowable on a boiler 48 inches in diameter, with plates .26 inch in thickness, and a tensile strength of 58,000 pounds per square inch, with holes drilled and long- itudinal seams single riveted? [Section 4433, Revised Statutes, Rule 2, Section 3.] A. This question is put direct except that part of it relating to the holes being drilled, but longitudinal seams are single riveted, that is designed to catch the candidate, as he may allow 20 per cent. on account of the holes being drilled. Therefore the question should be answered as follows: .26 58000 208000 130 24) 15080.00 (628.33+ and 628.33÷6=104.72+ lbs. The pressure allowable. 144 68 48 200 192 80 72 80 72 U. S. Marine Boiler Inspection and Engineers' License Laws. 463 Q. Determine the pressure per square inch allowable on a boiler 40 inches in diameter, with plates .24 of an inch thick, and a tensile strength of 60,000 pounds per square inch, with holes punched and longitudinal seams double-riveted? [Section 4433 Revised Statutes, Rule 2, Section 3.] A. 120 pounds. No allowance is made on account of longitudinal seams being double riveted, because the holes are punched. Q. Determine the pressure per square inch allowable on a required cylindrical boiler, 43 inches in diameter, longitudinal seams double riveted, holes drilled, boiler to be made of homogeneous steel plates, .2549" in thickness, samples of which .2549" x .98" separated in testing, at the smallest point, under a strain of 14,996 pounds, and reduced in dimensions, at point of fracture to .89" x .18"? [Section 4433 Revised Statutes, Rule 1, Sections 3 and 6; and Rule 2, Section 3.] A. This is intended for a catch question, in which the candidate is required to demonstrate his ability to determine the tensile strength of boiler plate, ductility of boiler plate and the safe-working pressure, or rather the pressure allowable under the law. If he understands his business he will discover that the ductility of the material is but a fraction over 31 per cent., when Rule 1, Section 6, requires not less than 50 per cent., otherwise the boiler would be allowed a pressure according to the following operation: .2549X.98.249802, area of sample. And breaking strain of sam- ple (14,996 pounds) divided by area of cross section of sample, equals: (14996.249802) X.2549 43×2 X1.2=142.34+ lbs. Q. Determine the hydrostatic pressure per square inch to be applied to a boiler 42 inches in diameter, with plates .26 of an inch in thickness, and a tensile strength of 65,000 pounds per square inch, longitudinal seams double riveted and holes drilled. Boilers to be employed on a passenger steamer? [Rule 2, Section 4.] To determine the hydrostatic pressure the working pressure must first be obtained, and the whole to be performed as follows: 65000X.26 42×2 X1.2=160.94+ lbs. Working pressure. And one-half of the working pressure (80.47) added to the working pressure (160.94) equals 241.41 pounds, the hydrostatic pressure. Q. Determine the steam pressure per square inch allowable (based on the diameter of stay bolts) on a fire-box boiler of the locomotive type with stay bolts in flat surface 15 of an inch in diameter, and placed 5 inches apart from center to center. The bolts at the bottom of the thread being of an inch in diameter? [Rule 2, Section 6.] 464 A Library of Steam Engineering. A. RULE.-Multiply the area of the diameter of cross section of the stay bolt at the bottom of the thread by 6000, and divide the product by the number of square inches in the surface stayed by the bolt; the quotient will give the pressure allowable per square inch. Example.-Let 75 one hundredths of an inch equal diameter of bolt at bottom of thread. Let .7854 equal a constant. Let 6000 equal a constant. Let 5 inches equal distance from center to center of bolts. Then we have: .75 X.75X.7854 X 6000 5X5 --106.029 lbs. Pressure allowable. Q. Determine the required diameter of stay bolts placed 5 inches apart from center to center, to be allowed a pressure per square inch in the boiler of 150 pounds? [Rule 2, Section 6.] A. Multiply the pressure per square inch by the number of square inches of surface to be stayed; then divide the product by 6000, and divide the quotient by .7854, and extract the square root of the last quotient; the answer will give the diameter of the bolt at bottom of thread required for a given pressure. Taking the previous case, we have: 106.029 X 25 6000X.7854 .75 Diameter of bolt at bottom of thread. Q. Determine the distance from center to center, stay bolts of an inch in diameter at the bottom of the thread, have to be placed in the flat surface of a boiler, to permit a pressure of 175 pounds per square inch to be carried in such boiler? [Rule 2, Section 6.] A. To determine the distance from center to center of stay bolts, multiply the area of bolt at bottom of thread by 6000; then divide the product by the required pressure per square inch, and extract the square root of the quotient; the answer will give the distance in inches from center to center of stay bolt. Taking the previous case, we have: .75X.75X.7854 X 6000 106.029 5 inches. Distance from center to center of stay bolt. Q. Determine the width (for testing) of a sample of boiler plate having a thickness of .24 at the point intended for fracture? [Rule 1, Section 3.] U. S. Marine Boiler Inspection and Engineers' License Laws. 465 A. RULE.-Divide the area of one-fourth of a square inch by the given thickness, and the quotient will give the width required. Example.-Let 25×100 equal one-fourth of a square inch. Let .24 equal given thickness of material. Then we have: 100X.25 =1.04+ .24 Q. Determine the pressure per square inch allowable on a corru- gated flue 38" in diameter, with a thickness of material of one-half inch? [Rule 2, Section 14.] A. 184.20+ pounds. Q. Determine the pressure allowable on a cylindrical, lap-welded flue, 8 feet long, 38 inches in diameter, made of material one-half inch in thickness? [Rule 2, Section 15.] A. 73+ pounds. Q. Determine the thickness of material required for a cylindrical, lap-welded flue 6 feet long, 40 inches in diameter, to carry a steam pressure of 90 pounds per square inch? [Rule 2, Section 10.] A. Multiply the diameter of the flue, in inches, by its length, in feet; then multiply the product by the pressure per square inch; then divide the last product by 89,600, and extract the square root of the quotient. 40×6×90 89600 49+ Thickness of material. Q. Determine the diameter of a lever safety-valve, for a boiler having a grate surface 4 feet in length and 4 feet in width. [Rule 2, Section 24.] A. 3.59+ Diameter of valve. Q. Determine the pressure per square inch required to raise a safety-valve of the following description: Example.-Let 4 inches equal diameter of valve. Let 9 pounds equal weight of valve and spindle. Let 4 inches equal length of short arm of lever. Let 20 inches equal distance of center of gravity of lever from fulcrum. Let 40 inches equal length of long arm of lever. Let 32 inches equal distance of center of weight from fulcrum. Let 18 pounds equal weight of lever. Let 150 pounds equal weight of weight. [Section 4418 Revised Statutes.] 30 466 A Library of Steam Engineering. A. 103.37+ pounds. Q. Determine diameter of piston and length of stroke of fire pump required for passenger steamers. [Section 4471 Revised Statutes.] A. Divide the number of cubic inches (100) required by the statute by the length of the stroke of the pump; then divide the quo- tient by .7854, and extract the square root of the last quotient; the answer will give the diameter of piston or plunger of pump required. Then we have: 100 8 .7854-3.98+ inches. Diameter of plunger or piston. Length of stroke, 8 inches. Q. Determine the diameter of steam drum legs for a battery of two cylindrical boilers, each boiler being 42 inches in diameter and 30 feet in length, containing two flues, each 15 inches in diameter, with the boilers so set that two-thirds of the outer surface of the shell is exposed to the heat of the furnace. [Section 4435 Revised Statutes.] A. Divide the number of square feet of effective heating surface in one of the boilers by 2, and divide the quotient by .7854, and extract the square root of the last quotient; the answer will give the diameter of each leg of the steam drum. The answer is obtained as follows: 3.1416 Constant employed to obtain circumference. Diameter of flues. Am't carried forward, 15 15 7080 31 416 47.1240 Circumference. 2 Flues. 94.2480 3.1416 42 Circumference of two flues. Constant employed to obtain circumference. Diameter of boiler. 6 2832 125 664 131.9472 94.2480 3) 226.1952 Circumference of boiler in inches. Circumference of two flues in inches. Total circumference of boiler and two flues in inches. 75.3984 % of circumference of boiler and flues in 2 12) 150.7968 12.5664 inches. 2% of circumference of boiler and flues in inches. 2% of circumference of boiler and flues in feet. U. S. Marine Boiler Inspection and Engineers' License Laws. 467 Am't brought forward, 12.5664 2% of circumference of boiler and flues in feet. 30 Length of boiler and flues. 2) 376.9920 Number of square feet of heating surface. .7854) 188.4960 Area of steam drum leg. .240 Square of diameter of leg. The square root of 240 is obtained as follows: 240 (15.49+ inches. Diameter of steam drum leg. 1 Answer to above question. 25) 140 125 304) 1500 1216 3089) 28400 27801 Q. Determine the actual horse power of a pair of high pressure engines, on a steamer, with cylinders each 24 inches in diameter, 5 feet stroke, crank making 25 revolutions per minute, valve cutting off at three-quarters of the stroke, initial pressure 150 pounds, clearance not taken into consideration. [Section 4415 Revised Statutes.] A. Multiply the area of the diameter of the cylinder by the aver- age pressure per square inch throughout the stroke (after deducting loss due to friction); then multiply the product by the number of feet the piston travels per minute; then divide the last product by 33,000; the answer will give the horse power of one engine. In this case there are two engines; therefore the horse power of one must be multiplied by two; and in this case the answer is 591.62+ horse power. Example.- 150 150 150 150 150 150 128.58 112.5 150 150 150 1 2 3 4 5 6 7 8 150 (150-7)X6=128.58 150 (150-8)X6=112.5 150 128.58 112.5 8) 1141.08 142.635 lbs. Average pressure throughout the stroke. Let friction equal 2.5+ (142.635+14.7-2.5) X.075=14.112+ 468 A Library of Steam Engineering. Then we have: 24X24X.7854X (142.635-14.112) X5×2×25 And 33000 -=440.47+ horse power. 440.47X2-880.94 horse power of two engines. Q. Determine the nominal horse power of a pair of high pressure engines of the same dimensions as those above? [Section 4415.] A. Multiply the square of the diameter of the cylinder by the velocity (250 feet per minute), and divide the product by 1000, the quotient will give the nominal horse power of the engine. EXPLANATION. The rule for determining the nominal horse power of an engine is based on an average pressure throughout the stroke of 60 pounds. per square inch, and a velocity of piston of 250 feet per minute. Therefore, taking the pair of engines referred to above, we have: 242X250 1000 X2 288 Horse power for both engines, or 144 horse power each. Q. Determine the nominal horse power of a cylindrical horizontal flue boiler, 24 feet in length, 48 inches in diameter, containing two- flues, each flue having a diameter of 16 inches; two-thirds of the shell being exposed to the heat of the furnace, and one-half of each flue being effective heating surface? [Section 4415 Revised Statutes.] A. 25.13+ nominal horse power. This answer is based on 12 square feet of effective heating surface to the horse power in the boiler. Q. Determine the bursting pressure of a steam engine cylinder having an inside diameter of 24 inches, a thickness of metal of 11 inches, and a tensile strength of 25,000 pounds per square inch? [Sec- tion 4415 Revised Statutes.] A. 2604.16+ pounds. Example.- 25000X1.25 24÷2 --2604.16+ lbs. The bursting pressure per square inch. 16 Q. How much lap must a slide valve have with a travel of 4 inches and a lead of of an inch of an engine having a length of stroke of 5 feet, required to cut off when three-fourths of the stroke- is reached? [Section 4415 Revised Statutes.] A. Deduct from the entire length of the stroke of the engine the length of the stroke that is to be made before the steam is cut off; then divide the remainder by the full stroke, and extract the square root of the quotient, and multiply this root by half the entire travel of the valve, and from this travel subtract half the lead; the remainder will U. S. Marine Boiler Inspection and Engineers' License Laws. 469 give the lap required to cut off steam at any given point in the stroke of the engine. Then we have: 5-3.75 .0625 .5 and .5×(4÷÷2)=1, and 1 =.96875 Lap required to 5 2 cut off at three- quarter stroke. Q. A steam pipe having an inside diameter of 4 inches, being supplied with a flat-seated valve 5 inches in diameter, what must be the amount of lift of the valve to produce an area of opening equal to the area of cross section of the opening of the steam pipe? [Section 4415 Revised Statutes.] A. Divide the area of the pipe by the circumference of the valve, the quotient will give the required lift of the valve. 42 X.7854 5×3.1416 .8 of an inch. Lift required. Q. Determine the thickness of material required for a cylindrical boiler having a diameter 44 inches, longitudinal seams single riveted, tensile strength of material 60,000 pounds per square inch, and to carry a pressure of 113.63 pounds per square inch? [Section 4415 Revised Statutes.] A. Multiply the given pressure by the radius of the diameter of the boiler, then multiply the product by 6, and divide the last product by the given tensile strength of material; the quotient will give the thickness required in hundredths of an inch for single-riveted longi- tudinal seams. Then we have: 113.63X 44 2 X 660000-.2499+ of an inch. Thickness of mate- rial required. Q. Determine the thickness of material required for a cylindrical boiler, 40 inches in diameter, made of material having a tensile strength of 65,000 pounds per square inch, holes drilled and longitudinal seams double riveted, and required to carry a steam pressure of 150 pounds per square inch? [Section 4415 Revised Statutes.] A. Multiply the given pressure by the radius of the given diam- eter of boiler, and multiply the product by 6; then divide the last product by the given tensile strength, and multiply the quotient by .833; the answer will give the thickness in hundredths of an inch. 150X (40-2) X6 X.833 .2307+ of an inch. Thickness of material required. 65000 470 A Library of Steam Engineering. Q. Determine the tensile strength required for a boiler 42 inches in diameter, made of plates .26 of an inch in thickness, longitudinal seams single-riveted, and to carry 125 pounds to the square inch? [Sec- tion 4415 Revised Statutes.] A. Multiply the given pressure by 6, and multiply the product by the radius of the diameter of the boiler, then divide the product by the thickness of the material; the quotient will give the tensile strength required. 125X6X(42-2) .26 =60576+ lbs. Q. Determine the diameter of a boiler required to carry a steam pressure of 150 pounds per square inch, made of material .25 of an inch thick, double riveted, holes drilled and tensile strength 60,000 pounds per square inch? [Section 4415 Revised Statutes.] A. Multiply the thickness of the material by the tensile strength and call this "Product No. 1;" then multiply the given pressure by 6 and call this “Product No. 2;" then divide "Product No. 1" by "Prod- uct No. 2," and multiply the quotient by 2; add 20 per cent. to the last product, and the answer will give the diameter of the boiler. Then we have: .25 X 60000 150×6 2) +20 X2 +20 per cent. 40 inches. Diameter of boiler. A United States Butte Ship. 471 CHAPTER XVII. A UNITED STATES BATTLE SHIP. Figs. 194 and 195 represent the general arrangement of machinery of a triple-screw, protected cruiser of 7350 tons. The boilers and machinery of this ship are designed to propel the vessel at sea at a maintained speed of 21 knots per hour. To attain that speed her motive machinery has been constructed so as to be capable of developing from 20,000 to 21,000 indicated horse power. The power of this vessel has been divided into three parts, each being developed by a separate, triple-expansion engine, driving its own Screw. In case of accident to any of her engines, this cruiser would still have a reserve power sufficient to drive her at a good rate of speed. In fact, it would be almost impossible to imagine or conceive a combination of circumstances or accidents that would render her entirely helpless. A leading feature of the design is, that by means of a clutch coupling, either propeller can be disconnected from its engine and left free to revolve, thus retarding the way of the ship but very slightly, when she is being propelled by one or two engines. A still further advantage is, that in moderate cruising, say with one-third power, a few boilers can be used with the high steam pressure for which they are intended, and one engine driven at the full power for which it is designed; by this means the power will be obtained economically, instead of wastefully, as it is when a large engine is running with low power. It is estimated that with one-third power and one screw, the ship can be driven about 15 knots per hour; with two screws and two-thirds power, from 18 to 19 knots per hour-the screws not in use being allowed to revolve freely in either case. The foregoing are the reasons for adopting the plan of three screws instead of two, and not because three screws possess any other advan- tage over two. On the contrary, it was supposed by the designers, that so far as speed is concerned, that two screws would have a slight advantage over three, by the application of the same power. Experiments have demonstrated that when three propellers are placed abreast of each other, the efficiency of the center one is greatly impaired by the interference of the side ones by the water flowing to 472 A Library of Steam Engineering. Fig. 194. Fig. 195. General Arrangement of Machinery. A United States Battle Ship. 13 473 LOOKING AFT. Fig. 198 O SIDE OF STE FORWA CENTER OF STRUT POST Fig. 196 94 93 92 91 full 90 89 88 87 86 85 84 8,8 82 81 80 Folger Cin SECTION THROUGH CENTER SHAFT. Fig. 197 100 99 98 97 96 95 LOOKING FORWARD. Fig. 199 474 A Library of Steam Engineering. 474 W A Fig. 200., Fig. 201. Fig. 202. Through Line No. 54. Through Line No. 37 Through Line No. 29. Fig. 203 Fig. 204, Through Lines 72 & 82. Through Lines 63 & 72. A United States Battle Ship. 475 STARBOARD ENGINE 아이 ​LOOKING AFT Fig. 208. Fig. 205. Fig. 206 # о O STARBOARD ENGINE LOOKING FORWARD Fig. 207. 476 A Library of Steam Engineering. it. For this reason, the side propellers have been placed in this cruiser about fifteen feet forward of the center one, thus avoiding, as much as possible, working the latter in the race of the former. It will also be noticed, that in looking at the ship from aft (Fig. 199), the propellers are not in the same horizontal line, but that the side ones are placed as high as their diameters will permit; also, that to bring these propellers clear of the ship's sides, the side shafts incline outwards from the center line of the ship 4 degrees, as shown in Fig. 196, and they also incline slightly upwards. The shaft of the center propeller is in the central longitudinal plane of the ship, and inclined slightly downwards. The central screw has four blades, and about ten per cent. more pitch than the side ones, as it works in more or less disturbed water. The side screws are three bladed, and the blades of all the propellers are adjustable, and are set to the pitch, which, upon trial, is found to be most efficient. MAIN ENGINES. The main engines have cylinders of 42, 59 and 92 inches diameter, with a common stroke of 42 inches. They are arranged in three water- tight compartments, each being complete in itself with all of its auxil- iaries, and entirely independent of the others, so that one or two might be completely disabled without in the least interfering with the work- ing of the other. Figs. 205, 206, 207 and 208 show the plan, sectional elevation and end views of one of the set of triple-expansion engines. AIR PUMPS AND CONDENSERS. The air pumps are two vertical, single-acting, lifting pumps for each engine, the pumps being 22 inches in diameter by 20 inch stroke. They are driven by two simple engines, with cylinders 7 inches in diameter and 12 inch stroke, arranged to exhaust into the condenser or into either receiver. Each engine is geared to make two and one- half revolutions for each double stroke of the pump to which it is attached. There are three condensers of the type shown in the engravings, with a cooling surface in each of 9474 square feet. MAIN AND AUXILIARY BOILERS. Steam is supplied by eight main, and two auxiliary boilers, all built of milled steel. Six of the main boilers are each 15 feet 6 inches in diameter outside, and 21 feet 3 inches in length; and two are each 11 feet 8 inches in diameter, and 18 feet 8 inches in length. The shells of the large boilers are 124 inches in thickness; those of the A United States Battle Ship. 477 Fig. 209. Air Pumps and Engines. 13 Fig, 210. section on of 2 Fig. 213 рестом Fig. 211 Fig. 212 478 A Library of Steam Engineering. Fig. 216. Collarples. Fig. 218 Suction en IE Aux Mack 107 Mara Lagun Fig. 214 251 Fig. 215. Fig. 221. Fig. 220. Fig. 219. Fig. 217 1:19 Innexe lunges of lubes Juction on CD Condenser. Drain Fig. 222. Section en ty Fig. 224. Fig. 223. A United 479 States Battle Ship. Diameter of Boiler Outside,- Diameter of Furnace Inside,- -15 ft. 6 in. 3 ft. 3 in. Grate Surface,- -175.5 sq ft Main Boiler-Six of This Kind. Ratio of Heating to Grate Surface,338 Calorimeter,- ---22.3 Ratio of Grate Surface to Calorimeter,- 7.9 Boiler Pressure,- No. of Tubes,- Fig. 225. 160 lbs 1128/312 stay 6 BW G (816 ord'n'y 12 BW G. Length of Grate, 6 ft. 9 in. HEATING SURFACE. Tubes,-- -5177.00 sq. ft. Furnaces,- 375.49 Comb. Chambers,- Total,- 380.00 -5932.4 66 "1. 46 " 68 66 Fig. 226. 00 0000 0000 000000000 00000 4 ++ 4 $ $. + .4 ي P 4. * P du 6 f. 念 ​5 0 6 + 。 + + $ -+ 6- 6 0- ? G @ & -· 19 0 G + + → 480 A Library of Steam Engineering. Diameter of Boiler Outside, Diameter of Furnace Inside,- Length of Grate,- ft. 8 in. 3 ft. 6 in 6 ft. 0 in. Grate Surface, Main Boiler No. of Tubes,- HEATING SURFACE. Tubes,- -2497.0 sq. ft. Two of This Kind. Calorimeter, Furnaces,- Comb. Chambers,- Total,- 171.2 202.0 -2870.2 66 66. 64 " 66 Fig. 228 Boiler Pressure; Ratio of Heating to Grate Surface,- 632 -84 sq. ft. -34.2 6 B. W. G 156 stay 1476 ord'n'y 12 B. W. G. 6.47 for one end. .160 lbs. Ratio of Grate Surface to Calorimeter,-6.5 Fig. 227. • + -- 中 ​& A United States Battle Ship. 481 Auxiliary Boiler. Two of This Kind. Ahole, Than Length of Grate,- Diameter of Boiler Outside,- Diameter of Furnace Inside,- 10 ft. 1 in. Grate Surface,- -32 sq. ft., one boiler. 2 ft. 9 in. 5 ft. 10 in. Ratio of Heating to Grate Surface,- Calorimeter,- -30.3 -59 HEATING SURFACE. Ratio of Grate Surface to Calorimeter,- Zilerived Tubes,-- 824.66 sq. ft. Boiler Pressure,- 5.4 150 lbs Bull straps thick, Furnaces,- 60.0 " 66 72-72 -5- No. of Tubes,- Fig. 231 Comb. Chamber,- 84.0 "6. 31 Total, -968.66 $6 66 216 58 stay 6 B. W. G 158 ord'n'y 12 B. W. G. Fig. 229. Fig. 230. + + 000 0000 000400. D ( 0 0 0 0 0 0 0- Area of slots in top or dry pipe to be qual to the area through stop rate Brass p 5 25 BWG. thick 3 Θ 19--9- 10. O 31 482 A Library of Steam Engineering. smaller are 1 inches in thickness. The auxiliaries are each 10 feet 125 inches outside diameter, and 8 feet 6 inches in length, and the shells are of an inch thick. All of the boilers are constructed to carry a safe-working pressure of 160 pounds per square inch. All of the main boilers are of the Scotch type, double ended, and with 24 inch steel tubes. Each large boiler has eight corrugated steel furnaces of 3 feet 3 inches inside diameter; and each small boiler has four such furnaces 3 feet 6 inches inside diameter. The heating surface of each of the large boilers is 5932 square feet, and of each small boiler 2870 square feet, and of each auxiliary boiler 968 square feet, making a total of 43,269 square feet. The total grate surface is 1285 square feet-each large boiler having 175 square feet, and each small boiler 84 square feet, and each auxiliary boiler 32 square feet. The auxiliary boilers are placed above the pro- tected deck, as shown in Fig. 195. The main boilers are arranged in groups of two, in four separate water-tight compartments, with five athwartship fire rooms and three smoke pipes. The forced draught is on the closed fire room system; each fire room being supplied with centrifugal fan blowers. The feed water is supplied by a main and auxiliary feed pump in each working fire room. There are evaporators to make up the loss of fresh water to the boilers, and to supply the distillers, reversing and turning engines, fire and bilge pumps, ice machines, steam ash hoists, capstan engine, and all other auxiliaries fitted in the most modern ships. The Corliss Engine. 529 E E Fig. 290. 34 530 A Library of Steam Engineering. Second. The employment of a rotating valve, capable of being cheaply fitted and easily maintained, as well as being easily moved and conveniently operated by mechanism entirely outside of the steam space. Third. The employment of a single eccentric to oscillate a wrist plate with connections direct with each of the four valves in such a manner as to cause a rapid opening and closing of the valves, and yet hold each valve open and nearly still at either end of its range, by swinging the line of connection nearly into line between the centers, and causing nearly a full port opening to be maintained during a proper interval, and thus giving a free and complete steam supply as well as a free exhaust. Fourth. A simple and effective method for detaching the steam valves from the driving mechanism, and causing a rapid and certain closure at the proper time, so as to produce any required expansion of steam in the cylinder. Fifth. A direct connection of the governor with the steam-valve mechanism, so as to produce the required ratio of expansion of steam in the cylinder, thus adjusting the power of the engine to correspond with the amount of work to be done, and reduce to a minimum the variations of engine speed caused by changing loads. Sixth. Making adjustments so as to throw the least possible work on the regulating mechanism, and thus give the governor the greatest possible sensitiveness and accuracy of action. The main points then, are the rapid and wide opening of the steam and exhaust valves; the shortness and directness of the ports and the resulting small clearances; the quickness of closure of the steam valves; the adaptation of the main valves to the functions of a cut-off valve; the connections of the governor to the cut-off gear in such a manner as to determine the point of cut-off without being itself hampered by the connections; the location of the exhaust ports at the underside of the cylinder, so as to drain the cylinder thoroughly at the end of each stroke. Using Figs. 291, 292 and 293 for illustration, the action of the Cor- liss engine may be described as follows: MECHANISM FOR OPERATING THE VALVES. The valves are driven by the cam rod A, through the wrist plate B, vibrating on a pin C, projecting from the cylinder. Rods D DDD, having right and left screw adjustments on their ends, take motion from pins E E E E, on the wrist plate B, and transmit it to the steam valve bell crank F F, and to the exhaust levers G G, moving them with a peculiar varying motion, in such a manner as to open and close the The Corliss Engine. 531 Folger Gi O O B E M TIN E 1 1 Fig. 261. Fig. 292. Fig. 293. Forqan oo 532 A Library of Steam Engineering. ports H H H H rapidly (Fig. 295), and to hold them open when the valves I I I I are off the ports, in such a manner as to give the least possible loss of pressure during admission and exhaust of steam. The rods leading to the steam valves, as shown in Figs. 291 and 293, are fitted with catches, or hooks J J, which may be disengaged as the valves open, at any desired point within about half-stroke, and the time of the disengagement is determined by the rotation of the cams K K, on the valve and stems behind the bell cranks, which cams are rotated by the governor L, through the rods M M. The slowing of the engine, in consequence of reduced steam press- ure or of increased load, causes the catches J J, to hold their contact longer, and the steam to be admitted through a greater part of the stroke of the piston; and in case of increased pressure or decreased load causes the catches J J to disengage earlier during the stroke of the piston. When the catches are disengaged, the steam valves are closed by the vacuum dash pots N N, attached to the vertical rods O O, con- nected to the bell cranks F F. THE DASH POT. The dash pots are attached to the pedestals P P, under the cylinder, above the floor, in a position easily accessible. In the lower chamber of the dash pot there is a vacuum which causes the valves to close promptly. Small openings are provided for the escape of air which may leak into the chambers, and these openings are provided with spring valves which close automatically. In the upper chamber of each dash pot the compression of air is regulated by valves, which per- mit the quiet seating of the dash-pot plungers. LIMITATION OF ENGINE SPEED. Aside from the commendable feature of the Corliss engine embrac- ing the four valves-located one at each corner, reducing clearance, permitting the steam to flow naturally through the cylinder, that is, in at the top, and with all water out at the bottom-perhaps no feature has contributed more to the success of this engine than the long strokes. and moderate speed formerly invariably employed. Unfortunately, however, of late years the tendency has been to drift into too high rotative speed, and therefore too frequently at the expense of economy. Too high rotative speed means a larger percentage of clearance, an increase of friction, increased steam consumption, increase in the con- sumption of lubricants, and increased wear generally over long strokes and moderate rotative speeds. The student of engineering must not conclude that high speeds are an objection, on the contrary they are desirable; it is only extreme The Corliss Engine. 533 rotative speeds that are objected to. It must be borne in mind that whatever loss of steam results from clearance is lost each stroke, and that this loss increases in proportion as the number of revolutions of the engine are increased. Clearance therefore means waste space in an engine; and if the percentage of waste space in an engine of 2 foot stroke would be four per cent., the same amount of waste space in an engine of 4 foot stroke would be but two per cent. FRICTION. Again, in the matter of friction, assuming a main bearing to be 8 inches in diameter, the frictional surfaces in one revolution would move 8×3.1416-25.1348 inches, or an amount equal to the circumfer- ence of the bearing. Hence, whether the stroke of the engine be 2 feet or 4 feet, the main bearing friction remains the same; but the 4 foot stroke engine does twice the work in one stroke that the 2 foot stroke engine does. Therefore, a 4 foot stroke engine making 100 revo- lutions per minute would be doing the same amount of work that a 2 foot stroke engine would do making 200 revolutions per minute, but the latter would subject the main bearings to twice the amount of friction that the former would in doing the same amount of work. Furthermore, the wear and tear of the engine depends largely upon the number and suddenness of reversals of the direction of the motion of its reciprocating parts; and hence, the greater the number within a given time the greater the wear and tear within that time. From what has been said on this subject the student must not be led to understand that the high rotative speed engines are objected to as such. On the contrary, many of them have been brought to the very highest stage of perfection, and they have their advantages as well as disadvantages. But what it is desired to impress upon his mind is the three cardinal points which do, or at least ought to, govern in the building of steam engines: Safety, economy and cost. First, every engine should be so constructed and so run that its speed will not menace life or property. Second, every engine should be constructed and run at a rate of speed to produce the greatest attainable amount of economy. Third, every engine should be constructed at a cost which will not be at the expense of safety or economy in running it. But in the latter lies the great fault with the construction of many steam engines. Their builders, in their eagerness to outstrip their competitors, either lose sight of or refuse to recognize the two most important points connected with the construction and running of steam engines-safety and economy. Every engine should be run within the limit of absolute safety, just as a steam boiler is run, or at least ought to be run. A steam boiler that would burst at a pressure 534 A Library of Steam Engineering. of 437.5 pounds, is allowed, under the laws of the United States, a working steam pressure of 125 pounds per square inch; which, if based upon the bursting pressure of the boiler, is a factor of safety of 31; and no engineer who understands his business would be fool-hardy enough to exceed the limit. Yet, in running engines, the factor of safety is seldom taken into consideration, and in many cases they are run at as high a rate of speed as they can be made to run without con- sidering the question of safety or economy. In order that the student may get a correct idea of a well propor- tioned and constructed Corliss engine a description will be given here of one of that type of engines. THE LANE & BODLEY COLUMBIAN CORLISS ENGINE. This engine is produced in honor of the Columbian Exposition, and is taken as a sample of first class Corliss engines, because of the familiarity of the author with its design and construction. This company, believing that modern progress had outgrown the forms and details of construction, which were doubtless admirable in 1849, and being desirous to exhibit an engine which would show marked improvement over the Corliss engine of forty years ago, designed and constructed the Columbian Corliss Engine. Modern conditions of steam pressure, speed, and continuity of service demanded by electric light, the manufacture of artificial ice and street railway service, required recognition. In the older forms of Corliss engines the details depended for their form and construction largely upon convenience and cheapness of production. For example, the main parts consisted of a cylinder, under each end of which was bolted a leg or pedestal. To the cylinder was bolted a girder containing the slides, the girder form- ing the connection between the cylinder and main shaft pedestal, or pillow block, to which the other end of the girder was bolted. This method of construction permitted comparatively simple and light separate parts, each of which could be used for either right or left- hand engines; and for the comparatively slow speeds and low steam pressures formerly employed, they answered their purpose quite well. Severer modern service, howeve requires the elimination of all unnec- essary joints. The Columbian Corliss, it will be noticed by reference to Fig. 294, consists of two main parts-the cylinder and frame. The cylinder in the larger sizes being bolted directly to the foundation without the interposition of pedestals or legs, and in the smaller sizes the pedestals being cast on the cylinder. These pedestals are of box form in cross section, giving two vertical walls of metal for the direct support of the weight of the cylinder, and in addition thereto presenting to view The Corliss Engine. 535 ப Fig. 294. Fig. 297. Fig.295. DLL Fig. 296. 536 A Library of Steam Engineering. plain surfaces, easily kept clean, and affording no recesses for the lodg- ment of dirt. The frame has cast on its outer end the main bearing with its pedestal. This construction prevents the possibility of loose- ness and springing in three useless joints, besides presenting a much neater appearance. It will be noticed that the term "frame" has been substituted for that of "girder." In the older form of construction it was customary, and quite proper, owing to the comparatively light service required, to make an unsupported span from the cylinder to the main pillow block; but the severe modern service requiring a sup- port at the end of the slide, the term "girder" in such case becomes inappropriate, and hence the substitution of the term "frame" in its stead. THE CYLINDER. The cylinder, as shown in Figs. 291 and 294, is fitted with circular bonnets and circular corners of large radius on top at each end; this, for many reasons, is considered an improvement over the square bon- nets and square-cornered cylinder. The steam is properly guided by the interior curved surface into the ports. The absence of sharp cor- ners on bonnets and cylinder ends renders cleaning easier and avoids the probability of bruised corners. The hole is round, and the circular bonnets and cylinder ends are in good form, besides the rounded cylin- der is stronger than the square. The iron top cast on the cylinder is one of the peculiar features of this engine, and gives it a handsome finish. LAGGING FOR CYLINDERS. The use of wood lagging for the top of Corliss engines is not good practice. The shrinking, swelling and warping of the wood soon ren- ders the cylinder unsightly in appearance, all of which is avoided by having the top in one piece cast with the cylinder. STEAM AND EXHAUST VALVES. The steam and exhaust valves, shown in Fig. 295, are placed in their proper positions in their relation to each other. Fig. 296 is a cross section through the center of the cylinder, and Fig. 297 is a cross section through steam and exhaust valves. STEAM AND EXHAUST CHESTS. The steam chest A, shown in Fig. 298, is much larger than such chests are usually made, as will be seen by comparison with B, as shown in Fig. 299. The object of the enlarged steam chest is to pro- vide a reservoir for steam near the cylinder, and beyond all contracted The Corliss Engine. 537 openings, which results in securing a better admission and steam line, shown by indicator diagrams. The exhaust chest C, in Fig. 298, is separated from the bottom of the cylinder, and the exhaust steam with its reduced temperature, is thereby prevented from robbing the walls of the cylinder of heat. A C B D Fig. 298. Fig. 299. The cylinders are counter bored at each end to such an extent as to allow the piston ring to overrun the shoulder about one-half the width of the ring. The cylinders are not steam jacketed unless spec- ially ordered by purchasers, owing to contradictory results obtained from experiments seeking to establish the value of steam jackets. Experiments recently made by men competent to obtain correct results, show that in the same engine, when worked under varying conditions of steam pressure, load, and cut off, the jacket is sometimes a source of gain and sometimes a source of loss. Assuming the correctness of these experiments, it is deemed wise to save annoyance and expense while awaiting further enlightenment on this subject. CYLINDER HEADS. The cylinder heads of this engine are scraped metal to metal, and a perfectly steam tight joint is produced without packing of any sort. Over the back cylinder head is fitted a plain, polished cast-iron cover, secured in position by one hexagonal head cap screw in the center. The sides of the cylinder are lagged with polished wood or iron, behind which mineral wool or other suitable non-conducting material is packed. FRAMES. The frame of an engine of this type is subjected to complex strains, as it combines slides for the cross head, main bearing for the crank shaft, and seats for the governor and rocker arm. 538 A Library of Steam Engineering. The strains between the cylinder and main bearing are tension, compression, bending horizontally and vertically, and twisting. Frames for engines of this type are usually in cross section, somewhat as shown in Figs. 300, 301 and 302. Fig. 300. CC Fig. 301. Fig. 302. Fig. 300 is of proper form to resist tension, compression, and hor- izontal bending. Fig. 301, or channel section, resists tension, compression, and ver- tical bending satisfactorily. Fig. 302, or box section, is a form capable of properly resisting all the strains to which it is constantly subjected in service; and this has been adopted in the Columbian Corliss Engine. It must be borne in mind that if the inherent stiffness of this box section is to be retained, no side, top or bottom openings can be permitted. The form of this style of frame, in addition to its other qualities, affords no convenience for the collection of dirt or rubbish. THE SLIDES. The slides for the cross head are cast in one piece with the frame. > In the history of the Corliss engine various forms for the slides have been adopted, among which are the forms represented by Figs. 303, 304 and 306; 305 being an exaggeration merely of 304. A Folger Cin. Fig. 303. Fig. 304. Fig. 305. Fig. 306. The flat slide (Fig. 303), for distribution of lubricant and to obviate excessive friction, is a most excellent form; its shortcoming, however, is in the absence of any means for taking up the side wear on the inside guide plunger of the cross head. The Corliss Engine. 539 The V shaped slides, shown in Fig. 304, is not a bad form unless the angles are permitted to run to extremes in the direction shown in Fig. 305. No side motion is permitted, but for even distribution of lubri- cants it is not so well adapted, as the oil tends to leave the upper edges. The friction is inclined to be excessive, and by referring to Fig. 305, which shows the V slides in exaggerated form, it readily may be imagined that a downward pressure of the cross head in the angle would produce such friction as to render it beyond the power of the steam pressure to move it. Fig. 306, or bored guide, possesses all of the advantages of the flat or the V guides, and some peculiar to itself. It approximates the flat guide near enough to admit of perfect oil distribution, and the cross head has no tendency to wedge itself in as in the V form (Fig. 305); it affords a lateral support for the cross head when both are worn; it mallows the cross head to turn slightly and adjust itself to the align- ment of the crank pin should the main shaft be slightly out of level; and finally, and most important, it can be made in perfect alignment with the center of the cylinder. The pressure of the cross head on the slide when the engine runs over on the outward stroke is downward upon the bottom slide only, for both outward and in ward stroke of the cross head. This down- toward pressure tends to deflect the frame downward vertically, which in o this engine is resisted by a support, the slide pedestal A (Fig. 306), cast in one piece with the frame, and extending entirely across the frame and downward to the foundation to which it is securely bolted by widely separated bolts. It is this pedestal which makes the term "girder" for the new design inappropriate. The slide pedestal, as well as the main bearing pedestal, is of box section, conforming in style with that of the cylinder, each presenting two vertical walls of metal for the direct support of the loads they are designed to sustain. When the engine runs under on the outward stroke the pressure of the cross head on the slide is upward against the top slide for both the outward and the inward stroke of the piston. This tends to deflect the frame vertically upward, and to spring the top slide in that direction. In some engines there is no special pro- vision against this. But in the Columbian Corliss Engine, at 3 (Fig. 306), at the slide end a massive brace is cast, which ties the overhang- ing top slide down to the bottom slide, thence by means of the slide pedestal A (Fig. 306) the strain is transmitted to the foundation, and entirely preventing the springing of frame and slide. At each end of the slide is cast a pocket for the reception of such oil as may be forced off of the slide by the motion of the cross head. The slide length is such that the cross head overruns on each end, which prevents the wearing of shoulders on the slides. 540 A Library of Steam Engineering. The main pedestal bearing on the foundation is narrow and very long, extent of bearing upon the foundation being secured by increase of length rather than width. The holding down bolts are two in num- ber and in the line of the length of the engine. By this it is sought to avoid the possibility of springing the frame, as may be done with four widely separated foundation bolts in a wide base, resulting in throwing the main shaft bearing out of level. The main bearing is full babbitt lined to the extreme ends, there being neither in this nor any other babbitted bearing on the engine the iron lips at each end, sometimes employed, and which by presenting a different metal inter- fere with the uniformity of wear. The babbitt is thoroughly planed after being cast, and then truly bored and scraped to a bearing. On each side of the main shaft is a babbitt-lined quarter box, each adjusted by wedges extending the full length of the main bearing, and having studs extending upwards through the cap, on which are nuts, by means of which they may be raised. Since the quarter boxes receive nearly the full pressure of the steam acting on the piston, it is desirable to back them up their full length with wedges rather than to receive this pressure on the points of set screws, sometimes used for this purpose. The bottom bearing for the shaft is a part of the frame itself, rigidity in so important a bearing being imperative, and those constructed of four quarter boxes, adjustable in all directions, are the cause of much care and trouble, owing to the absence of this rigidity. The cap is pro- vided with large oil pockets, and is so constructed as to admit of easy inspection and of feeding the shaft by hand while the engine is in operation. PISTON PACKING. Fig. 308. Fig. 307. Fig. 309. The packing employed in the piston is of the Babbitt & Harris patent, which has been used for seventeen years in connection with these engines. This packing is composed of a narrow ring and is self- adjusting. Its construction and arrangement are shown in Figs. 307, 308 and 309. The Corliss Engine. 541 TO ADJUST THE CROSS HEAD. The requirements of a good cross head are, besides that of being of sufficient strength, that it should possess ample bearing surface upon the slides; that the distribution of the thrust from the connecting rod should be equally distributed over the surfaces; that the adjust- ment for wear should be such that it can be taken up evenly over the entire surface; that the slippers can be easily removed without unship- ping the cross head or connecting rod; that the connecting rod can be readily detached, and that the lubrication of pin and surfaces shall be even and continuous. A A Fig. 310. Fig. 310 shows a cross head designed to fulfill the foregoing enum- erated requirements. The first is satisfied by the large surface of the slippers A A, lined with the best Babbitt metal, hammered in place before being planed. The equal distribution of pressure over this surface is obtained by having the wrist pin centrally located, longi- tudinally and vertically, in the cross head, and by having the cross head bear for its full length and breadth in any condition of wear upon the shoes, which are wedges and may be forced evenly forward by means of a single screw in each, located at the end of and in the middle of each shoe, the shoe being cramped in position when properly adjusted by means of cap screws. These long, tapered shoes, adjusted longitudinally, evenly taking up the wear, are superior to two or more wedges or set screws, each separately adjusted, by which one part of the cross-head shoe may sustain the whole load. By moving the clamp screws either top or bottom shoe may be readily removed endwise without disturbing any of the other parts. A tapped hole in the side of the cross head is provided for the insertion of a stud for operating the parallel motion of an indicator. 54215 A Library of Steam Engineering CONNECTING RODS.' The connecting rod is of the solid end type; the adjustment is effected by a block with one taper face; its position is changed by slack- ening one and tightening the other screw, as shown in Figs. 311 and 312. This method securely locks the wedge. Fig. 311. Fig. 312. The boxes have a nearly full support by the wedge. The crank pin boxes are of bronze, lined with the best of Babbitt metal. The cross head end is shown in Fig. 312, and the crank end in Fig. 311. The cross head end has solid phosphor-bronze boxes. VALVES AND VALVE RODS. Fig. 313. www Fig. 314. Fig. 315. Fig. 316. The valves are of the original Corliss type, with cylindrical ends of unusual length. The stems are of phosphor-bronze or steel, extend- ing clear through the length of the valve, with a bearing in the back The Corliss Engine. 543 bonnet. Spiral springs are set into the blades of the valve stems to allow some compensation. The results establish the durability of such construction, as well as its superiority over the ordinary tee head valve stem, notwithstanding the increased cost of the former over the latter. The construction of these valves are shown in Figs. 313, 314, 315 and 316. THE GOVERNOR. Fig. 317. A centrifugal governor is essentially a balance, the centrifugal force of balls being opposed by a resistance, usually consisting of weights or springs; it differs in one important particular, however, in that it should not be astatic, as all delicate balances are-that is, tend to remain in any position when the balancing is effected-but should have definite positions corresponding to definite loads. As a change in speed of the engine has to occur before its governor will respond, that governor is the best which controls the fluctuations within the narrowest limits. To accomplish this result a number of conditions have been satisfied, of which the following are the most important: First, it should be 544 A Library of Steam Engineering. able to act upon slight variations in the speed of the engine. Second, it should be able to act quickly. Third, it should not be sensitive to rhythmical resistance induced by any throttling or by the cut-off mechanism. To satisfy the first and third conditions it is necessary that the changes due to alterations in speed should be much greater than those due to variations in resistance. For the second, all causes which retard a ready adjustment of balance between the centrifugal force and resistance should be reduced to a minimum; these are, prin- cipally, friction inertia, or momentum, of the moving parts, either of the governor or of the mechanism it may operate. The governor with which this engine is fitted is shown in Fig. 317. It is designed to satisfy the conditions above enumerated, and it is extremely simple; the centrifugal ends of two balls situated upon the upper ends of bell crank levers are resisted by a spring at the inner ends of the same; by this mechanism the centrifugal and resisting forces can be most accurately adjusted and regulated. The speed of the governor, about 200 revolutions per minute, produces a great change of centrifugal force for a slight variation in speed; in fact, this dif- ference varies as the square of the number of revolutions for a given fluctuation of speed, so that as compared with that of a slow running governor, say at 60 revolutions per minute, under otherwise similar conditions, its ability to be affected by changes of speed is as 100 to 9, with a corresponding insensibility to varying internal resistances. Fric- tion also increases the difference between the maximum and the mini- mum speed of the engine. The use of small balls, which a high speed of governor renders possible, allows quick action, as a given force will move them through a given space quicker than it will heavier balls. THE VALVE MOTION. Fig. 318. The Corliss Engine. 545 The valve motion of this engine is noticeable for its large bearings and pins. This is an important feature, for the reason that in most engines of this class these joints have been the first parts to wear loose. The strap ends are lined with phosphor-bronze bushings, which are Fig. 319. split on top, and can be closed up by set screws, the wear being taken up entirely around the circle. Another important feature of the valve motion of this engine is the double adjustment for the steam rods. In some engines these rods are simply screwed into the strap ends, and rigidly attached at 35 546 A Library of Steam Engineering. the crab-claw end, so that the only adjustment possible is some mul- tiple of the pitch of the screw. With the wide face crab claw, as shown in Fig. 318, there is sufficient room for a tapped hole at both ends, permitting the smallest desirable adjustment. The illustration, shown in Fig. 319, represents the construction of the bonnet on the valve motion side, and shows a very important improvement; it extends into the recess made in the crank A, which operates the valve, thus forming a rigid bearing for the crank to swing upon. This obviates the overhung bearing, whereby excessive lateral strain is thrown on the valve stem and bearings, and rapidly wearing them out. By this construction the entire strain on the valve stem is a torsional one, which is easily transmitted. By removing the back bonnets and the keys which secure the cranks to their stems, the valves can be removed without disarranging the valve motion. THE DASH POT. F. G B C A E D Fig. 320. Fig. 320 represents the vacuum dash pot. The dash pot has two cylinders, a close-fitting piston F, working in the upper, and a plunger G, working in the lower cylinder. The two chambers have independent duties, exactly opposite in time and character. On the up stroke the lower cylinder affords resistance gradually by the attenuation of the air, and on the down stroke the upper cylinder produces resistance by the compression of contained air. The perfection of the operation of the dash pot depending wholly upon the perfection of alternately created vacuum and compression, The Corliss Engine. 547 means are provided on both cylinders to prevent the opposite condi- tions obtaining in their cylinders accidentally. The result then obtained is a vacuum in the lower cylinder, and the unbalanced atmos- pheric pressure on the area of the plunger, causing the latter to drop quickly on the release of the crab claw, and at a point in the descent the piston closes an air opening A, compressing the air remaining in the upper cylinder, bringing the parts gradually to rest. A small check valve at C, shown on the right side of the illustration (Fig. 320), is placed near the bottom of the vacuum cylinder, and allows any air to escape which may have been taken in from leakage during the up stroke. This valve will require some attention. The spring B should be just stiff enough to hold its leather face against the seat, with the least possible pressure, so as to not allow any air to accumulate and destroy the vacuum. To prevent noisy contact of the piston against the bottom of the dash pot, if not previously air-cushioned, there is a leather washer secured to its under surface. In this leather there is a small portion which is shaved thin, and so cut as to form a flat valve over a small hole drilled to the piston, through which oil can work at each stroke to lubricate the vacuum plunger, which requires a certain amount of oil to make it air tight. The drain pipe D, in the bottom of the vacuum cylinder, is used to drain away surplus oil while the engine is standing, but must always be kept closed while the engine is running. The regulating valve E, which connects with the dash cylinder, is used to adjust the amount of air-cushioning. The piston F should be cushioned sufficiently to prevent it from striking the bottom, and yet letting it settle without a jar. Too much cushion prevents the piston reaching its lower position until the valve rod pushes it down. The double jump sometimes occurs when the check valve C becomes misplaced, or its spring bent, so that air can leak in and destroy the vacuum. In starting the engine the dash pot is apt to work irregular at first, on account of the air being cold and the oil stiff. This may require opening up of the regulating valve E. But after the engine has run awhile the valve will have to be throttled to obtain the required amount of cushion. PILLOW BLOCKS. It will be observed that the main pillow block (Fig. 321) is prac- tically solid, with the exception of the side boxes. These boxes are backed by two heavy wedges, having bolts extending through the cap, 548 A Library of Steam Engineering. Fig. 321. Fig. 322. by which means the necessary adjustments are easily made. The main bearing pedestal has a large base similar to those under the slides and cylinder. The whole is substantially and conveniently arranged, and the construction is among the best found in modern practice. COMPOUND ENGINE. Compound engine, is a term applied to an engine which has two cylinders, arranged to use steam successively. A compound engine may consist of two entire engines, driving different loads; or two cylinders with one piston rod, acting on one crank; or two cylinders, pistons and rods coupled to cranks on the same shaft. The first appli- cation of the system was made many years ago, but without any The Corliss Engine. 549 marked success. It was, however, generally abandoned, until 1854, when it was applied to marine purposes. CONDENSERS. Compound engines are generally supplied with condensers, though not necessary to the attainment of some valuable advantages. The best results, however, are attained with high pressure of steam and a condenser; the functions of which is to create a vacuum in the exhaust end of the cylinder, and thus prevent the pressure of the atmosphere from acting on that side of the piston and producing a pressure counter to the steam pressure on the opposite side. To the successful opera- tion of a condenser a large supply of water is necessary for condensing the exhaust steam quickly. An air pump in connection with a con- denser is also a necessity. A simple form of condenser with air pump is shown in Fig. 323. This is what is termed a belted air pump and condenser. The lower part of the column is the condensing chamber; the valve shown being on the injection pipe. The air pump is vertical, single acting with double acting discharge. The comparative advantages of the condensing engine over the non-condensing is shown in the table which follows: FIRST-CLASS ENGINE PERFORMANCE. A valuable table of first-class engine performance is introduced here for the benefit of the engineer and student of steam engineering, as it contains the essence of all the principles of steam as applied to work in the cylinder. The two outside columns denote initial pressure, or the pressure at which the steam is admitted to the cylinder, and following the piston to any one of the several points of "cut off," as noted, and on a line with the number of pounds of initial pressure will be found the mean effective pressure in the columns marked M. E. P., the column marked N. C. for non-condensing, and C for condensing engines. is In the column marked "terminals" will be found the pressure of steam at the end of stroke. This pressure, however, is noted for an absolute vacuum as a base, and if the pressure on a steam gauge wanted, subtract 14.7 from the number given in the table. Should the number in the table be less than 14.7, then there will be a vacuum in the cylinder equal to the difference between 14.7 and the tabular number. In the M. E. P.'s the condensing column (C) shows 12 pounds greater M. E. P. than the non-condensing (N. C.). This is due to the 550 A Library of Steam Engineering. Fig. 323. AIR PUMP AND CONDENSER. The Corliss Engine. 551 action of the condenser, which should maintain a vacuum of 12 pounds; showing this difference in favor of the condensing engine. The four columns of water consumption denote pounds of water for each horse power for each hour of running. The two columns "theoretical," however, deal with steam as a perfect gas; a condition, by the way, that has never yet been attained, as there is always more or less water carried with the steam; hence the figures given in the "practical" columns are the very best that we can hope to attain in ordinary practice. An example of the uses of the table may be seen by taking 80 pounds initial pressure, and cut off at one-fifth of stroke, which gives 35.02 pounds mean effective pressure (N. C.), and a terminal of 20.39 pounds pressure absolute; or, in other words, 5.69 pounds by the gauge, and a water consumption of 24.3 pounds per horse power per hour, showing very good work. The same steam in a condensing (C) engine would give 47.02 M. E. P., and a water consumption of 19.8 pounds per horse power per hour. This table has been calculated with great care, and with a reason- able allowance for clearance, compression, etc., and represents what is possible to be done by good design, workmanship, erection and subsequent handling. The figures relating to condensing engines are the more likely to be deviated from, as there are other things to be considered, and over which the engine has no control, but they will not be found far out of truth. On the whole, the table sets up a standard of good work, and in practice should be approached as nearly as possible by all who own or handle steam. Tests may be made to prove the efficiency of any engine by weighing the feed water delivered to the boiler for a number of hours and dividing the whole amount by the number of hours of the test. This will give the rate per hour, which, divided by the aver- age indicated horse power for the same time, will give the rate of water consumption. Great care must be taken that no leaks or wastes of any kind exist between the weighing apparatus and engine cylinder-no steam to be drawn from the boiler for any other purpose than to drive the engine. Even the feed pump should be a power pump and not a steam pump if driven by the same boiler, as its supply will be unduly charged to the engine. The boiler should also be known to be in good operation, clean, and not likely to be foaming or priming. Neglect in this particular would destroy all value of the test, unless an expert calorimeter test were made to determine the percentage of entrained water in the steam. 552 A Library of Steam Engineering. CUT-OFF. INITIAL PRESSURE. 1 4 CUT-OFF. WATER CONSUMPTION. Theo- Practical retical. N.C. C. N.C. C. TERMINALS....... WATER CONSUMPTION. M. E. P. Practical Theo- retical. N. C. C. TABLE SHOWING FIRST CLASS ENGINE PERFORMANCE. 15 100 CUT-OFF. TERMINALS...... M. E. P. N. C. C. TERMINALS... WATER CONSUMPTION. M. E. P. Practical Theo- retical. N. C. C. TERMINALS.... WATER CONSUMPTION. M. E. P. Practical Theo- retical. N. C. C. N.C. C. N.C. C. N.C. C. N.C. C. INITIAL PRESSURE. 1 10 CUT-OFF. 40 3.65 15.65 6.41 69.4 20.0 51.4 14.3 45 5.42 17.42 7.00 52.0 19.9 38.5 14.2 50 7.19 19.19 7.59 43.0 19.6 31.9 14.0 55 8.96 20.96 8.17 37.9 19.2 28.1 13.7 60 10.73 22.73 8.76 33.1 19.0 25.3 13 6 65 12 50 24 50 9.35 31.4 18.8 23.3 13.4 70 14 27 26 27 9 93 29.4 18 6 21.8 13.3 75 16.04 28 04 10.52 27.8 18.5 20.6 13.2 80 17.81 29 81 11.11 26.6 18.2 19.7 13.0 85 19.58 31.58 11.70 25.6 18.1 19.0 12.9 90 21.36 33.36 12.28 24.8 17.9 18.4 12.8 95 23.13 35.13 12.87 24.2 17.8 17.9 12.7 100 24.90 36.90 13.46 23.6 17.6 17.5 12.6 N.C. C. N.C. C. 55 9.05 21.05 9.07 40.7 20.5 31.3 15.2 13.46 25.46 11.79 34.4 21.5 27.9 16.3 17.34 29.34 14.49 33.7 22.0 27.2 17.2 40 11.32 23.32 9.87 36.0 20.2 27.7 15.0 16.15 28.15 12.87 32.6 21.1 25.7 16.0 20.39 32.39 15.81 31.6 21.8 25.3 17.0 45 13.59 25.59 10.72 32.9 20.0 25.3 14.8 18.85 30.85 13.94 30.5 20.9 24.0 15.8 23.45 35.45 17.13 29.8 20.6 24.0 16.9 50 15.86 27.86 11.55 30.4 19.8 23.4 14.7 21.54 33.54 15.00 28.8 20.6 22.7 15.6 26.50 38.50 18.45 28.4 21.3 22.9 16.7 18.12 30.12 12.38 28.7 19.7 22.1 14.6 24.24 36.24 16.08 27.6 20.5 21.7 15.5 29.56 41.56 19.77 27.3 21.2 22.0 16.6 60 20.39 32.39 13.20 27.4 19.6 21.1 14.5 26.93 38.93 17.15 26.5 20.3 20.9 15.4 32.61 44.61 21.09 26.4 21.0 21.3 16.4 65 22.66 34.66 14.03 26.4 19 3 20.3 14.3 29.63 41.63 18.23 25.7 20.1 20.2 15.2 35.67 47.67 22.41 25.8 20.8 20.8 16.3 70 24.92 36 92 14.86 25.3 19.3 19.5 14.2 32.32 44.32 19.31 24.9 19.9 19.6 15.1 38.72 50.72 23.73 25.2 20.6 20.4 16.1 75 27.19 39.19 15.69 24.4 19.0 18 8 14.1 35.02 47.02 20.39 24.3 19.8 19.1 15.0 41.78 53.78 25.05 24.8 20.4 20.0 16.0 80 29.46 41.46 16.51 23.9 18.8 18.4 13.9 37.71 49.71 21.46 23.7 19.7 18.7 14.9 44.83 56.83 26.37 24.3 20.3 19.6 15.9 85 31.72 43.72 17.34 23.4 18.6 18 0 13.8 40.41 52.41 22.54 23.4 19.5 18.4 14.8 47.89 59.89 27.69 24.0 20.3 19.3 15.8 90 33.93 45 93 18.17 22.9 18.6 17.6 13.8 43.10 55.10 23.62 23.0 19.4 18.1 14.8 50.94 62.94 29.01 23.6 20 2 19.0 15.8 95 36 26 48.26 19.00 22.0 18.4 17.3 13.6 45.80 57.80 24.70 22.6 19.4 17.8 14.7 54.00 66.00 30.33 23.1 20.1 18.7 15.7 100 3 10 CUT-OFF. 40 20.75 32.75 17.11 32.9 22.9 27.0 18.2 45 24.13 36.13 18.67 31.1 22.8 25.5 18.1 50 27.50 39.50 20.24 29.6 22.5 24.3 17.8 55 30.87 42.87 21.80 28.4 22.2 23.3 17.6 60 34.24 46 24 23.37 27.4 22.0 22.5 17.5 65 37.61 49 61 24.94 26.7 21.9 21.9 17.4 70 40.98 52.98 26.51 26.1 21.7 21.4 17.2 75 44.35 56.35 28.07 25.6 21.5 21.0 17.0 80 47.72 59.72 29.64 25.1 21.3 20.6 16.9 85 51.09 63.09 31.20 24.6 21.2 20.2 16.8 90 54.46 66.46 32.77 24.2 21.0 19.9 16.7 95 57.83 69.83 34.33 23.9 20.9 19.6 16 6 100 61.20 73.20 35.90 23.7 20.8 19.4 16.5 35 100 CUT-OFF. 41 CUT-OFF. 10 1 CUT-OFF. 23.70 35.70 19.80 33.0 23.9 27.5 19.3 26.22 38.22 22.44 33.0 24.6 28.3 20.3 30.50 42.50 27.78 32.5 26.0 28.5 22.3 40 27.32 39.32 21.60 31.6 23.5 26.3 18.9 30.08 42.08 24.49 32.7 24.3 26.9 20.1 34.75 46.75 30.33 31.5 25.8 27.6 22.1 45 30.94 42.94 23.42 30.4 23.3 25.3 18.8 33.95 45.95 26.55 30.4 24.0 25.8 19.9 39.00 51.00 32.88 30.7 25.6 26.9 21.9 50 34.56 46.56 25.23 29 3 23 1 24.4 18.6 37.81 49.81 28.60 29.2 23.8 25.0 19 7 43.25 55.25 35.43 30.0 25.4 26.3 21.7 55 38.18 50.18 27.04 28.3 22 8 23.6 18.4 41.68 53.68 30.66 28.8 23.6 24.4 19.5 47.50 59.50 37.98 29.4 25.2 25.8 21.5 60 41.80 53.80 28.85 27.5 22.6 22 9 18.4 45 54 57.54 32.71 28.2 23.4 23.9 19.3 51.75 63.75 40.52 28.8 25.0 25.3 21.3 65 45.42 57.42 30.66 26.8 22.5 22.3 18.2 49.41 61.41 34.77 27.6 23.2 23.4 19.2 56.00 68.00 43.07 28.4 24.8 24.9 21.2 70 49.05 61.05 32.47 26.2 22.4 21.8 18.1 53.27 65.27 36.82 27.1 23.0 23.0 19.0 60.25 72.25 45.61 27.9 24.7 24.5 21.0 75 52.68 64.68 34.28 25.7 22.3 21.4 17.9 57.14 69.14 38.88 26.7 22.9 22.6 18.8 64.50 76.50 48.16 27.6 24.6 24.2 21.0 80 56.31 68.31 36.09 25 3 22.2 21.1 17.9 61 00 73.00 40.93 26.1 22.8 22.2 18 8 68.75 80.75 50.70 27.2 24.5 23.9 20.9 85 59.94 71.94 37.90 25.0 22.2 20.8 17.8 64.87 76.87 42.99 25.8 22.6 21.9 18 7 73.00 85.00 53.25 27.0 24.4 23.7 20.8 90 63 57 75.57 39.71 24.7 22.0 20.6 17.7 68.73 80.73 45.04 25.5 22.5 21.6 18.6 77.25 89.25 55.79 26.8 24.2 23.5 20.7 95 67.20 79.20 41.52 24 5 21.9 20.4 17.6 72.60 84.60 47.10 25.2 22.4 21.4 18.5 81.50 93.50 58.34 26.6 24.1 23.3 20.6 100 The Corliss Engine. 553 ECONOMY IN THE GENERATION AND USE OF STEAM. It is a question whether or not the term "horse power" should be applied to boilers, as it is likely to fail in giving any idea of their ability or capacity. The standard adopted by the judges at the Centennial Exposition, of 30 pounds of water per hour evaporated at 70 pounds pressure, with feed water at 100° Fahrenheit, and rating this as a "horse power," is a fair one for both engines and boilers, and has been favorably received by engineers and steam users; but as the same boiler may be made to do more or less work with less or greater degree of economy, it should also be required that the rating of a boiler be based on the amount of water it will evaporate at its most economical rate. If all the data in connection with a boiler are known it is easy to compute the number of pounds of water it will evaporate in an hour; and if we know the water consumption per horse power per hour of the engine that is used in connection with this boiler, it is then possible to say what the horse power of this boiler is under the circumstances. If a certain boiler with one kind of setting, one grade of fuel, and excellent draft would develop, say 100 horse power, the same boiler in another setting, with another grade of fuel and poor draft, might do well if it did one-half as much work. Take the position, that the boiler is 100 horse power, and is devel- oping that amount of work through some ordinary slide-valve engine that uses 50 pounds of water per horse power per hour; with a good Corliss engine that is using but 25 pounds per hour, the above boiler would develop 200 horse power. There is no trouble whatever to find engines running, some using more than 50 and others less than 25 pounds of water per horse power per hour. A comparative record of tests, on excellent authority, showing the relative efficiency and amount of water used per horse power per hour by a number of ordinary slide-valve engines is here given: Lbs. of water per H. P. per hour 32.34 33.65 35.52 38.83 46.35 58.67 51.00 56.09 66.81 Relative efficiency.. .715 .681 .651 .595 .499 .494 .453, .412 .346 The relative efficiency is compared with the attainable results of Corliss engine taken at 1000. Thus the amount of water used per horse power per hour gives convincing evidence of the difference in engines, and is the only true basis of comparison. in The coal required per horse power per hour is evidently dependent every case upon the economy of the engine and boiler jointly. 554 A Library of Steam Engineering. With an evaporation of 9.25 pounds of water per pound of coal, and thirty pounds of water per horse power in the engine, there would be required per horse power per hour 3.24 pounds of coal. Such boiler performance, however, is not common in average practice; so, generally, a low cost of power in fuel is due to using an excellent engine in con- nection with only a fair boiler duty. An example of this may be seen in a printed report of a test of two steam plants-we will call them "A's" and "B's." Said circular is issued by "A," who claims the highest efficiency, and calls especial attention to his engine as using the least coal: Water used per horse power per hour.. Coal used per horse power per hour... Relative efficiency.. A. B. 43.84 lbs. 36.64 lbs. · 3.65 lbs. 4.07 lbs. 1.000 lbs. 0.897 lbs. Which of these engines is the more economical-the one that uses but 3.65 pounds of coal, or the one that uses but 36.64 pounds of water? We can easily see from the figures that it was the boiler that made A's engine use the less coal, and the engine appears in a bad light, as it was not near so efficient in duty as B's, when rated by the only true standard-that of water consumption. Had the engines been changed, each to the other's boilers, then both A and B would have been sur- prised. Presuming an engine to be mechanically perfect-that is, tight valves, piston packing and joints-its economical performance depends on four things: A quick and full opening of the admission valve, allowing full boiler pressure to be admitted to the cylinder, an instan- taneous closure of the same at the proper time, and this time to be varied, automatically, according to the varying loads and steam press- ure; a late but perfect exhaust opening, and an exhaust closure early enough to secure necessary cushion for piston. The reverse of this is a sluggish opening of steam valve, with a decreased speed of move- ment, and at no time allowing full boiler pressure to the cylinder, an equally imperfect valve closure, and always late, with no variation whatever; an early, cramped and sluggish exhaust opening, combined with a like exhaust closure. Such will be found in the ordinary slide- valve engine. SLIDE VALVE AND CORLISS ENGINE DIAGRAMS. The indicator diagrams, shown in Fig. 324, one over the other, are those of a Corliss and an ordinary slide-valve engine; the dotted line the Corliss, and the solid line the slide valve. Drawn thus is graphic- The Corliss Engine. 555 ally shown the relative performance of the two engines, cylinders same size, doing the same work, and using the same boiler pressure- say 82 pounds. In the Corliss diagram we see the steam is promptly admitted at A to nearly full boiler pressure, and that pressure con- tinued until nearly a sufficient quantity has been taken to do the work of that stroke, then the valve is instantly closed at B, and no more steam is taken from the boiler. The steam now confined in the cylinder continues to drive the piston onward by its expansive force, increasing in volume to fill the cylinder, and the pressure reduced, as shown by the curved line B C. At the point C the piston has about completed its stroke, and the pressure has fallen to about 8½ pounds, B { 1 2 3 Fig. 324. and the difference in pressure between B and C has been utilized in producing power; its vitality is expended, leaving nothing to be lost, and only a mere breath to be exhausted in the air. For the return stroke steam is admitted to the other end of the cylinder, and the same order of things takes place. From this plan of stopping the flow, or cutting off the passage of steam from the boiler to cylinder, we get the name of "cut-off engine." The fact must be noted, that if the steam valve should remain open too long, the speed of the engine would be increased; and if closed too soon, or before steam enough had entered the cylinder to do the work, the engine would run slower, and the speed change with every change of load or work. To control such variation a governor is used, and by its action the steam valves are closed at exactly, the right time, changing as often as necessary, thus making the engine an "automatic cut-off engine." In the ordinary slide-valve engine the action of the valve is the same under all changes of boiler pressure or load on engine, and the speed is controlled by "throttling," that is, contracting or enlarging the opening through which the steam passes on its way from the boiler to the engine. Referring again to the diagram and tracing the solid 556 A Library of Steam Engineering. line, we find that at the outset we are far short of getting boiler press- ure at 1. This is a loss, as can be seen by comparing the height of the line at A B with that of 1 2. Thus steam is blown into the cylinder, wearing itself out, as it were, until the piston reaches 2, when the valve closes and expansion begins and continues to point 3, when of necessity the exhaust must be opened to clear the cylinder for the return stroke, and right here we have a pressure of some 25 pounds to throw away-a cylinder full of 25-pound steam thrown away! This wearing the steam out by tortuous and insufficient openings through which it must pass in order to reach a moving piston is a great loss of pressure, and is plainly shown by the diagram, as is also the loss of the use of the pressure contained in the cylinder full of steam, which is thrown away for want of opportunity to take advan- tage of its expansive nature. The pressure thus wasted, stroke after stroke, day after day, must be generated by the burning of fuel, and for this fuel the users of such engines are obliged to pay, and not infre- quently are blind to the fact that it is thrown away. There are other points which might be named in this connection, but it may be stated, as the result of a large number of experiments, carefully made with different engines, working under various condi- tions, for the purpose of comparison from a working stand point, that the Corliss engine uses an average of 25 pounds of water per horse power per hour, and the ordinary slide valve or throttling engine uses an average of 40 pounds of water per horse power per hour. Or, in other words, the Corliss engine will do as much work on 5 pounds of coal as an ordinary slide valve can do on 8 pounds. This being the case, it naturally follows that a smaller boiler can be used to generate a given power, and make a saving in cost of outfit in this particular. Presuming that 15 square feet of heating sur- face in a boiler were required per horse power for a slide-valve engine, then as a Corliss engine will effect a saving of one-third, 10 square feet per horse power will be sufficient for its needs. Again, should a certain boiler be fully worked to supply a certain slide-valve engine, a Corliss could displace this engine, and the boiler would then be able to do one-half more work, and as easily as before. FEED WATER HEATING. In the preceding pages boiler performance has been discussed, and it was stated that a horse power consists of 30 pounds of water evaporated per hour, with the feed water delivered at a temperature of 100° Fahrenheit, and converted into steam at 75 pounds pressure. This point should always be taken into consideration in deter- mining the evaporation of steam boilers. It is plainly unfair to com- The Corliss Engine. 557 pare the performance of a boiler which is fed with water at 60° Fah- renheit with one that is supplied at a temperature of 212° Fahrenheit; therefore, it naturally follows that the temperature of the water must be noted when testing a boiler, and then calculate what the amount would have been had the feed been at 212° Fahrenheit. This computation would be a tedious one, and beyond the abil- ity of the average engineer; but the work is very much simplified by a table, which has been prepared from the formula used in such cases. In using the table the temperature of the feed water, the rate of coal consumption and the steam pressure in boiler, should be known. By reference to the table a factor will be obtained which, multiplied by the pounds of water evaporated per pound of coal, will give the number of pounds of water that would be evaporated had the feed water been injected into the boiler at a temperature of 212° Fahrenheit. Thus, suppose a boiler carrying 80 pounds of steam were fed direct from a spring or well at 50° Fahrenheit, and on weighing this water as pumped to the boiler, and also the coal burned during the time of test, the boiler is found evaporating only 7 pounds of water to one of coal. This would look wasteful, but by using the table it is found that there could have been evaporated 9 pounds had the feed been heated to 212° Fahrenheit. (73X1.204-9.03.) The use of heaters for raising the temperature of feed water is becoming general, yet there seems to be a great variety of opinions as to their use and action, when the numberless forms and claims made for each are considered. Those that make use of the exhaust steam from engine, pumps, etc., for heating are the most common, as they make use of what would otherwise go to waste, and they are highly to be commended. Two forms of these may be given notice as thoroughly practical and effective: A jet, or open heater, consisting of a vessel into which the exhaust steam is thrown; also a jet or spray of water, which in falling takes up more or less of the heat of the steam while passing through it, and is drawn off at the bottom by the pump to sup- ply the boiler. Another form is the surface or closed heater, with a series of copper, brass, or other pipes, around which the exhaust steam circulates, while the water is forced through the pipes on its way to the boiler, and is under boiler pressure while in the heater. The open heaters are very effective in a general way, but they must not be, in any case, expected to heat water over 208° or 210° Fahren- heit. The reason for this is that water vaporizes or boils at 212° Fah- renheit, when heated to that point in a vessel with free outlet; there- fore, it would break into steam and pass out with the exhaust if heated to that point in an open heater." CC 558 A Library of Steam Engineering. TABLE OF FACTORS OF EVAPORATION. BOILER PRESSURE IN POUNDS PER SQUARE INCH. TEMPERATURE OF FEED. 30 40 50 60 70 80 90 100 110 120 130 140 150 40 32985 32 Degrees 1.207 66 66 1.188 1.211 1.214 1.199 1.203 1.206 1.192 1.196 1.217 1.209 1.212 1.214 1.220 1.223 1.225 1.227 1.228 1.231 1.233 1.234 1.236 1.217 1.219 1.221 1.223 1.224 1.226 1.228 1.199 1.201 1.204 1.206 1.208 1.210 1.212 1.214 1.216 1.217 66 1.178 1.182 1.185 1.188 1.191 1.194 1.196 1.198 1.200 1.202 1.204 1.205 1.207 70 66 1.167 1.171 1.175 80 66 1.157 1.161 1.165 90 66 1.147 1.151 100 66 1.136 1.140 110 66 1.126 1.130 120 66 1.116 1.126 130 66 1.105 1.109 1.113 140 66 1.095 1.099 150 66 160 66 170 66 180 39 190 66 1.043 200 66 210 66 39 1.032 1.022 1.178 1.181 1.183 1.168 1.170 1.173 1.177 1.179 1.181 1.154 1.157 1.160 1.162 1.165 1.167 1.169 1.171 1.144 1.147 1.150 1.152 1.151 1.156 1.158 1.160 1.133 1.136 1.139 1.142 1.144 1.146 1.148 1.150 1.120 1.123 1.129 1.131 1.134 1.136 1.138 1.116 1.118 1.121 1.123 1.125 1.127 1.102 1.105 1.108 1.110 1.113 1.115 1.117 1.085 1.088 1.092 1.095 1.098 1.100 1.102 1.104 1.106 1.074 1.078 1.081 1.084 1.087 1.090 1.092 1.094 1.096 1.064 1.067 1.071 1.074 1.077 1.079 1.081 1.084 1.086 1.053 1.057 1.060 1.064 1.066 1.069 1.071 1.073 1.075 1.047 1.050 1.053 1.056 1.058 1.061 1.063 1.065 1.036 1.040 1.043 1.045 1.048 1.050 1.052 1.026 1.029 1.032 1.035 1.037 1.040 1.042 1.185 1.188 1.190 1.191 1.193 1.195 1.196 1.175 1.183 1.185 1.186 1.172 1.174 1.176 1.162 1.164 1.165 1.129 1.131 1.152 1.153 1.140 1.141 1.143 1.133 1.119 1.120 1.122 1.108 1.110 1.098 1.100 1.101 1.087 1.089 1.155 1.145 1.134 1.124 1.112 1.113 1.103 1.091 1.092 1.077 1.067 1.079 1.080 1.082 1.068 1.070 1.072 1.054 1.056 1.044 1.058 1.059 1.061 1.046 1.047 1.049 1.051 The Corliss Engine. 559 PERCENTAGE OF SAVING IN FUEL BY USING HEATERS. (BOILER PRESSURE 90 POUNDS.) PRIMARY TEMPERATURE OF FEED WATER. WHEN DELIVERED TO BOILER. 32° 40° 50° 60° 70° 80° 90° 100° TEMPERATURE 60 Degrees 2.3 1.7 0.8 70 66 3.2 2.5 1.7 0.8 80 66 4.0 3.4 2.5 1.7 0.8 90 66 4.9 4.2 3.4 2.5 1.7 0.8 100 6. 5.7 5.0 4.2 3.4 2.5 1.7 0.8 110 66 6.6 5.9 5.0 4.2 3.4 2.5 1.7 0.8 120 66 7.4 6.7 5.9 5.0 4.2 3.4 2.5 1.7 130 66 8.3 7.6 6.7 5.9 5.0 4.2 3.4 2.5 140 66 9.1 8.4 7.6 6.7 5.9 5.0 4.2 3.4 150 66 9.9 9.3 8.4 7.6 6.7 5.9 5.0 4.2 160 66 10.8 10.1 9.3 8.4 7.6 6.7 5.9 5.0 170 66 11.6 11.0 10.1 9.3 8.4 7.6 6.7 5.9 180 66 12.5 11.8 11.0 10.1 9.3 8.4 7.6 6.7 190 13.3 12.6 11.8 11.0 10.1 9.3 8.4 7.6 200 14.2 13.5 12.6 11.8 11.0 10.1 9.3 8.4 210 66 15.0 14.3 13.5 12.6 11.8 11.0 10.1 9.3 220 33 15.9 15.2 14.3 13.5 12.6 11.8 11.0 10.1 560 A Library of Steam Engineering. The closed, or coil heater, can be used to heat water to 225° or 230° Fahrenheit, but would make a back pressure of 4 to 6 pounds on the engines, exhausting into it, and this would be an objection in most cases, yet situations exist where such back pressure is maintained for other reasons, and advantage could be taken of it. A temperature of 212° to 214° Fahrenheit can be easily attained in a closed heater with- out noticeable resistance to the engines, provided the heater is large enough and properly adapted to its work. It is never profitable to have a small heater of any kind. Still another form of heater may be used which takes advantage of the lost heat in another direction, and that is by way of the stack. Great care must be used in applying such, as it may result in any- thing but satisfaction. Where the boiler is small and the fire has to be forced, and a very hot breeching and stack is the result, then this form may be used to some profit. The proper remedy in such cases is, by all means, a larger boiler; yet sometimes this is not expedient, and a coil of iron pipe may be placed in the stack or breeching, and the feed water forced through it to the boiler with a check valve beyond it. This should never be used unless the escaping gases are at 600° Fahrenheit, or over, and should not result in cooling down to less than 450° Fahrenheit, and should always have water passing through while the fires are strong. The value of the increased temperature of the water fed to boilers may readily be shown, and the percentage of gain computed by using the steam and water tables to be found in the chapters on evaporative and calorimeter tests. A pound of water taken at 32° Fahrenheit, and raised to 210° Fahrenheit will have required 178 units of heat. To evaporate this pound to steam at 90 pounds will require a total of 1182 heat units. Now, 178 heat units is 15 per cent. of the 1182 required, and may be supplied by a heater using the waste from the engine, thereby saving 15 per cent. of the fuel burned. The ultimate value of a heater using exhaust steam can not be more than 15 per cent. in economy, and that when water is taken at 32° and delivered at 212° Fahrenheit. A fair result will be when water is received at 60° and delivered at 210° Fahrenheit, making a saving of 12.6 per cent., with boiler pressure at 90 pounds. TO PUT AN ENGINE ON THE DEAD CENTER. There are a number of ways of putting an engine on a dead cen- ter, and several of which will be here described; and the descriptions here given are applicable to other engines as well as the Corliss. First. Where an engine is constructed with an ordinary bed plate, and the bed plate has been truly planed, the simplest and most accurate The Corliss Engine. 561 manner of putting the engine on the dead center, is by taking a sur- face gauge, placing it on the bed plate immediately in front of the cen- ter of the main shaft, and setting the point of the needle of the sur- face gauge to the center of the main shaft. Then move the gauge toward whichever center it is desired to place the engine, and revolve the crank on the main shaft until the center of the crank pin comes fairly to the point of the needle of the surface gauge, then the engine is exactly on the dead center. Second. Another method is to place some stationary object close to the rim of the fly wheel or main driving wheel, in which station- ary object the point of one of the legs of the compasses can be put, as shown in Fig. 325 3 2 A 00 Fig. 325. B is an improvised rest for holding the compasses while making the marks 1 and 2 on the rim wheel A. This rest is made in two parts-ordinary boards. One is nailed to the floor and B is nailed in an upright position to it. The next step to be taken is to turn the engine toward the center on which it is desired to place it, until the cross head has traveled to within one or two inches of the end of its stroke, then stop and place a mark on the cross head and one of the guides, so that both will form a continuous line. Then place one of the points of the compasses C at D, in the board or rest B, and with the other point of the compasses scribe A an arc on the rim of the wheel, as shown in Fig. 325. The mark at 1, is the first mark; but it was made in the position mark 2 is in, as shown in the diagram, after it was made the wheel was revolved into the position shown. 36 562 A Library of Steam Engineering. After making mark 1, the compasses must neither be opened nor closed, but kept set just as they were when mark 1 was made. After making the first mark on the wheel, the wheel is again revolved until the cross head has traveled to the end of its stroke, and back until the mark on the cross head and guide come squarely line and line, and then stop. Now take the compasses and put another mark on the wheel similar to the first, but the last mark will be at 2, as shown in Fig. 325. After the second mark has been made take another pair of com- passes-as the one employed in making marks 1 and 2 on the rim of the wheel must not be disturbed, but kept in their original position- and find the center between marks 1 and 2, and put a third mark on the rim of the wheel, the third mark will be at 3, as shown in the diagram. Again place the original compasses in the original center D, in the board B, and turn the wheel A back until the upper point of the compasses is reached by mark 3; and the engine will be on the dead center. Before beginning operations to put the engine on the center, the engineer must be careful to take up all lost motion between the crank pin and piston head. The next method is illustrated in Figs. 326 and 327. Procure a planed board a little longer than the stroke of the engine, the lower edge of the board should be true, so that it can be leveled in position by a spirit level. Draw a line square across the center of the board at A (Fig. 326), then draw two lines B and B' square across the board at a distance from A equal to exactly one-fourth of the length of the stroke of the engine. If the stroke of the engine is, say 4 feet, the distance from line A to line B must be 1 foot, and the distance from line A to line B' must be 1 foot, and the two lines B and B' will be 2 feet apart. This is done so as to get the lines B and B' midway between the center of the main shaft and the center of the crank pin when the pin is on the center. The distance from line A to line G on either side represents the distance from the center of the main shaft to the center of the crank pin when the pin is on the center, and the distance from lines G to G represents the length of the stroke of the engine. Next, fasten the board to the ceiling or joist overhead, in any manner most convenient, but so as to bring the face of the board E, as shown in Fig. 327, in perpendicular line with the center of the crank pin D, measuring from the face of the crank to the outer end of the crank pin. While attaching the board in position care must also be taken, by the use of a plumb line, to get the center line A on the board E (Fig. 326), in perpendicular line with the center of the main shaft. As the plumb line will hang some distance from the face of the crank, a square should be placed on the face of the crank to the center of the The Corliss Engine. 563 main shaft, and the board E moved about until the plumb line touches the square. When the board E is in that position it should also be level. If the board is in its proper position, the line A, in Fig. 326, will be in perpendicular line with the center of the main shaft, the board will be exactly level and the face of the board will be in perpen- dicular line with the center of the crank pin. In short, the board will be in the position shown in Figs. 326 and 327. We are now prepared to put the engine on either dead center. G B A B E E G H D LL F D 1 H Fig. 326. Fig. 327. The next thing to be done is to make an improvised trammel F, as shown in Fig. 327, by driving two pointed nails through the board, one at each end. Then place the point of the lower nail into the cen- ter of the main shaft, and press the point of the upper nail into the board E, anywhere on line B or B', whichever side the crank is to be placed on the center, and remove the point of the lower nail from the center of the main shaft, keeping the point of the upper nail in the trammel in the center into which it was pressed while the lower nail was in the center of the main shaft; turn the engine over until the crank pin center comes to the point of the lower nail in the tram- mel, as shown in Fig. 327, and the engine will be exactly on the dead center. To put the engine on the other dead center, put the point of the lower nail of the trammel in the center of the main shaft, and press the point of the upper nail of the trammel into the line B or B', 561 A Library of Steam Engineering. whichever is opposite to the line first employed, and remove the point of the lower nail of the trammel from the center of the main shaft, keeping the upper nail in the center into which it was pressed; turn the engine over until the center of the crank pin comes to the point of the lower nail in the trammel, and the engine will be exactly on the dead center on that side of the main shaft. HOW TO SET THE VALVES OF A CORLISS ENGINE. Take off the back bonnets of the cylinder, and notice that the lines marked on the ends of the valves and on the cylinder are the lines of opening, lap and lead. The lines of lap and lead are to be followed when the engine is not to be indicated for valve adjustment. On the back of the hub of the wrist plate will be found a center. line, and a corresponding line on the hub of the trunnion which sup- ports the wrist plate. When these two lines coincide the wrist plate will be in its central position. On each side of the center line of the wrist plate trunnion will be found another line which marks the vibra- tion of the wrist plate, and when the center of the wrist plate coincides with either of these lines the wrist plate is in its extreme position. First. Place the wrist plate in its central position, and by means of the adjusting nuts make the lengths of the valve connections so that each steam and exhaust valve may have the proper lap. Then connect the eccentric hook to its pin on the wrist plate, and turn the eccentric loose around the shaft, and, if necessary, adjust the length of the eccentric rod so that the center mark on the wrist plate will vibrate to the extreme lines of travel marked on the trunnion. Second. Place the crank on either dead center, and turn the eccen- tric on the shaft in the direction in which the engine is to run, enough more than one-quarter of a revolution ahead of the crank to show the proper opening or lead of the steam valve nearest the piston. Then tighten the set screw on the eccentric, and turn the shaft with eccen- tric one-half revolution in the same direction until the engine is on the other center, and notice if the lead is the same on the valve nearest to the piston as it was on the other valve; if not, adjust it by making its connection either longer or shorter as may be required. Third. To adjust the cams for disengaging the steam-valve crab claws, let governor balls remain in their lowest position, and move the wrist plate to either extreme of its travel, and hold it in this position; then lengthen or shorten the governor rod for the steam valve at the opposite end, which will then be open so that the steel cam on the cam collar will touch the curved arm of the crab claw; move the wrist plate to the other extreme of its travel and adjust the other governor rod in the same manner. See that the adjustable brass cam on the The Corliss Engine. 565 cam collar is placed in the proper position for preventing the hooking up of the crab claw, this being a safety stop to act if the governor ceases to rotate. Fourth. To prove the correctness of the cut off adjustment, raise the balls of the governor to about where they would be at work, or to medium height, and block them there. Then with the eccentric con- nected to wrist plate, turn the engine shaft slowly in the direction in which it is intended to run, and when the crab claw is detached by the cam, measure the distance the cross head has moved from its extreme position Continue to turn the shaft in the same direction, and when the other crab claw is detached by the cam, measure the distance the crab claw has moved from this extreme position, and if the cut off is equalized these two distances will be the same. If they are not, adjust the lengths of the governor rods until the points of cut off measure alike. In the belted governor the lever on the side of the governor column carries a cam, upon which the weight of the balls rest. This will drop out of the way when the engine is up to speed, and will allow the stop motion cams to become operative and stop the engine in case the governor belt breaks. Fifth. To start the engine, the lever on the side of the governor column must be raised so as to allow the crab claws to hook up. Sixth. To adjust the dash pot rods, first shorten them so that the crab claw can not hook up when the wrist plate is moved from one extreme position to the other. Then with the wrist plate moved to one of its extreme travel marks, lengthen the dash-pot rod at the same end of the cylinder until the crab claw will just hook up. Then be sure to tighten the lock nut securely, as the working loose of this rod will cause the breaking of the valve bonnet. Throw the wrist plate to the other extreme mark, and adjust the other dash pot rod in the same manner. TO ADJUST THE GOVERNOR. If the governor "hunts" or dances, slacken the jam nuts under the balls, and screw the balls down on the arms one-half or a whole turn, then tighten the jam nuts. The engine will now run faster. Then to recover correct speed slacken up slightly on the spring. Repeat this operation until "hunting" ceases. To decrease the speed of the engine slack off on the spring, and at the same time screw the balls further out on the arms-not too much or the governor will "hunt." To increase the speed of the engine tighten the spring-not too much or the governor will "hunt"-at the same time screw the balls down further on the arms. 566 A Library of Steam Engineering. Regulations for speed can be made only to a limited extent. Engi- neers must therefore be careful not to go to extremes. If the governor does not act quick enough, screw the balls further in on the arms and loosen up on the spring. If carried too far the governor will "hunt." In making changes do not give more than one turn to the balls each time, then run the engine and observe the result. Be careful to have the balls the same distance from the arm pins so as to evenly balance each other. The original position for balls at stated speed for the engine is 5ğ inches from the center of balls to the center of arm pins. "Hunting" occurs when the spring is too tight, or when the balls are too light, or when the balls are too far out on the arms for the ten- sion of the spring. Lack of sensitiveness occurs when the spring is too slack, or when the balls are too far down on the arms. The best regulation takes place when the spring and length of arms are such as to just prevent "hunting"-that is, when a slight increase of either will produce "hunting." The governor may sometimes "hunt" when starting up, but if once steadied by hand until speed is obtained, it will continue in good shape after adjustment to load, the slight change of which it will readily detect. Beware of any tendency to tinker. Make no adjustment except after careful consideration. CARE OF THE GOVERNOR. A governor will be efficient only when kept in perfect condition. Its lubrication, cleanliness and proper adjustment is essential to the performance of its functions. Friction of the moving parts is the chief cause of defective regulation, and any tendency to sluggishness of action should receive immediate attention. The governor should be washed out weekly with benzine or coal oil. REYNOLDS CORLISS ENGINE. TO ADJUST THE VALVES OF THE ENGINE. The working edges of the valves and ports are shown by radial lines on the ends of valves and valve chests, at the side of the cylinder opposite the wrist plate. Both steam and exhaust valves indicate lap when the lines on the valves are nearer the center of cylinder than the lines on the chests. The Corliss Engine. 567 Fig. 329 shows wrist plate central for adjusting valve connections. Three marks on back of hub of wrist plate D, and one mark on wrist plate stand, which is bolted to the cylinder, show how eccentric con- nection is to be adjusted so that the wrist plate will travel correctly when in motion. To set valves, as in Fig. 329, place the center mark on wrist plate hub even with mark on wrist plate stand, and then adjust length of valve connections so that steam valves A, and exhaust valves B, will have lap according to columns in tables; the lap being given in parts of an inch opposite the size of cylinder. BINNER-END-CO-MIL Fig. 328. REYNOLDS CORLISS ENGINE. Fig. 330 shows position of wrist plate D when the engine crank is on the center and eccentric set to give steam valves proper lead. Exhaust valves will be correct if they have been set according to Fig. 329, and will need no further attention. Put crank on center and then move eccentric so that steam valves will have lead according to table; the lead being given in parts of an inch opposite the size of cylinder. TABLE FOR SETTING VALVES. Diameter of Cylinder. Lap of Steam Valves. Lap of Exhaust Valves. Lead of Diameter Steam Valves. of Cylinder. Lap of Steam Valves. Lap of Exhaust Valves. Lead of Steam Valves. 8 24 10 3 16 26 5 16 5 3 16 12 16 32 32 28 5 16 14 4 16 32 30 5 16 16 16 32 32 18 16 32 34 20 16 32 36 110010000 16 1 16 16 22 3 16 32 64 568 A Library of Steam Engineering. O O O о B O ⚫ D Fig. 329. D Fig. 330. B O O O The Corliss Engine. 569 TO ADJUST THE LENGTH OF DASH-POT RODS H. When rod is down as far as it will go, the shoulder E on brass hook should just clear the steel block F on valve arm, as shown in cut, leaving the clearance below block, as shown at G. This adjust- O о O f G ي حا لها H Fig. 331. ment of rod H must be made when the wrist plate D is at its extreme throw, as shown by the mark on the back hub. This adjustment of rod H is very important. If the rod is too short the steam valves will not open; if too long the rod will be bent or the bonnets broken, or both. 570 A Library of Steam Engineering. CHAPTER XX. THE ATLAS AUTOMATIC ENGINE. As this engine presents many novel features, its construction and the operation of the valve gear will be here explained and illustrated for the information of engineers and students of steam engineering. It is automatic in its operation of the main valve, and accom- plishes the distribution of steam with a single slide valve. In steam engines of moderate size the means almost universally employed to effect the distribution of steam is the plain slide valve. This device, on account of its simplicity, comparatively small size and little cost, and disposition to remain tight under wear, has caused it to be generally preferred to other kinds of valves. But when of dimen- sions suitable for large engines, or when working under high pressure, it has the disadvantage of being forced by the steam so strongly against its seat as to cause great friction, rapid wear and the absorption of a considerable amount of power to operate it. To obviate this defect many devices have been resorted to for relieving the valve of the pressure of steam, or for balancing the pressure by causing it to act equally upon opposite sides. Most of these balanced slide valves, however, have proven unsatisfactory in practice, being cumbersome, expensive, and difficult to keep in order, or requiring undesirable and costly modifications in the form of cylinder; so that the plain valve for a long time seemed likely to hold its own against them. The advent of the single valve automatic engine, however, ren- dered some form of balance valve a necessity. In this style of engine the valve has not only to distribute the steam, but is made to regu- late the motion of the engine, by being given a variable travel con- trolled by a governor. Though it is possible to make a governor pow- erful enough to operate an unbalanced slide valve, it can not be made to regulate very closely. The ordinary defects of the unbalanced slide valve are therefore intensified in the automatic engine, the earlier cut off and expansion of steam causing a greater difference on and under the valve. The friction causes the valve to stick at the end of its travel, and the springs or balls of the governor yielding, the ports are held open longer than the relations of load and speed require. Variations in boiler pressure also interfere with governing. Under higher pressure the engine will increase its speed, and on account of the varying resistance, The Atlas Automatic Engine. 571 Fig. 332. THE ATLAS AUTOMATIC ENGINE. A E G TF 572 土 ​C D- PLAN. Fig. 333. GROUND LINE END ELEVATION. The Atlas Automatic Engine. 573 MATERIAL FOR FOUNDATIONS. ENGINES. BRICKS. SAND, Bushels. CEMENT, Barrels. LIME, Barrels. SULPHUR, Pounds. 7 x 10 8 x 10 1,400 14 11 14 1,400 14 14 9 x 12 2,000 20 20 10 x 12 11 x 14 12 x 14 2,000 20 20 ... 3,300 33 13 30 3,300 33 13/ 30 Fig. 334. ENGINE FOUNDATION TEMPLATE. 574 A Library of Steam Engineering. SPECIFICATIONS. 6 CYLINDER. DIMENSIONS OF HORSE POWER AT GIVEN REVOLUTIONS PER MINUTE, WITH 40 POUNDS MEAN EFFECTIVE PRESSURE. TWO BAND WHEELS. DIAMETERS. FLOOR SPACE REQUIRED. SHIPPING Approximate. WEIGHT. Standard Diameter. Stroke. Indicated Revolu- Indicated Horse Inches. Inches. Power. tions per Minute. Horse Power. Inches. Revolu- Diameter. Width of tions per Minute. Main Steam Exhaust Length. Width. Pounds Belt. Shaft. Inches. Inches. Pipe. Pipe. Inches. Inches. Feet. Inch. Feet. Inch. Approxim'te 7 O02244 240 24 77771 12 1212 7171 223344 19922 ** 0 0 9 334455 222 8877 ∞ ∞ 1,900 1,900 3,400 3,400 4,800 4,800 330088 DIMENSIONS OF FOUNDATIONS. of Notch. Shaft to Edge Center of of Notch. Engine to Edge Center of Top of Notch. Bottom of Foundation to Center of Shaft. Foundation to Bottom of Foundation. Top of Shaft to End of Center of Bottom to Top of Foun- dation. Top of Foun- 2 Ꮓ X T N L of Cylinder. dation to Center M dation. to Top of Foun- Floor Line Floor Line. Foundation to Bottom of tion at Top.. Side of Founda- Center to tion at Top. Side of Founda- Center to Width of Foundation at Top. Length of Foundation at Top. Center to T I IN. INCH. IN. INCH. INCH. FT. IN. 0 13 4 10 12 111 0 13 4 10 12 11/ 6 9 5500 12121212 14 12/7/7 14 12/7/ 19/ 192 40228 00 11222121 8484 FT. INCH. INCH. 222288 20 20 188 2 11 18/ 2 11 18 11/ 6/1 1212 63 17 12 17/ 12 12121212 F INCH. 13 INCH. 124 131 134 FT. IN. E tion at Bottom. Side of Founda-A tion at Bottom. Side of Founda- Center to Width at Bottom. of Foundation Length at Bottom. of Foundation ENGINE. 5509 15 15 16 16 IN. 7171 2222222 0 0 044 667798 H414 9 1 94 0 0055 12222 O055 112222 B C 4000000 IN. 19 9 1 A SIZE. 3 FT. 7 x 10 8 x 10 9 x 12 10 x 12 11 x 14 12 x 14 178800 + N 6 6 6 6 The Atlas Automatic Engine. 575 Fig. 335. Fig. 336. Fig. 337. --B 576 A Library of Steam Engineering. one. the engine will often run faster under a heavy load than under a light All of these defects, however, are entirely overcome in the Atlas engine, by a peculiar form of balanced valve and automatic shaft governor, all of which are exceedingly simple in their construction as well as in their operation. The valve consists of but two principal parts, only one of which is in operation when the engine is running. This is clearly shown in Figs. 335, 336 and 337. Fig. 335 is a perspective view of the valve and hood complete. Fig. 336 is a section through the valve and hood on the line A B, in Fig. 335. Fig. 337 is a perspective view of the valve and seat with hood detached. The cylinder ports and valve seat are exactly the same as for an ordinary slide valve. Upon the seat is placed a slide valve a little narrower than the length of the ports, as shown in Fig. 337, but it has a length, width of exhaust cavity, and laps, the same as an ordinary slide valve adapted to the seat. The exhaust cavity is usually of such a width as to leave no inside lap. It will also be observed that the valve is so constructed that the force of the exhaust steam is discharged against the hood instead of against the valve, as it does in the case of an ordinary slide valve. The valve is made trapezoidal in section, and finished and scraped true on all of its working surfaces. The inclined sides have an angle of about 20 degrees with the base, and the height is such as to make the end area somewhat greater than that of the steam port in the seat. The ends are cored out to secure lightness, and to obtain a uniform metal thickness throughout, so as to prevent distortion of shape by changes of temperature. The stem is attached by passing it through a lug, which is then clamped between a collar or nut on the end of the stem or valve rod. The hood is placed over the valve to relieve it from steam pressure. This hood is carefully fitted to the valve and seat by scraping, and it is so shaped and strengthened that it will bear great pressure without distortion. Across the inside of the hood recesses are formed to cor- respond, in width and position, with the ports in the valve seat. The valve being slightly narrower than the length of the ports, the ends of the ports and the recesses join, thus forming passages entirely around the valve, and balancing it against whatever pressure there may be in the ports. To prevent lateral movement of the hood, a cap screw is put through each corner into the valve seat, bottoming before it is quite screwed down, so as to leave a small space between the head of the screw and the hood, permitting the latter to rise a short distance from the seat. On the back of the hood, kept in place by studs, are a num- ber of helical springs, which, bearing against the inside of the steam The Atlas Automatic Engine. 577 chest cover, hold the hood to the seat. The tension of the springs is only sufficient to hold the hood under working conditions. Should over pressure occur in the cylinder, from water or otherwise, the springs will yield, and the hood immediately rising from the seat, opens communication between the cylinder and the exhaust port, and relieves the pressure. 24 From a superficial view it might be supposed that the making of the valve narrower than the length of the ports, would sacrifice some of the port area, but upon closer examination it will be readily seen that such is not the case. The area of the cross section of the valve being greater than the area of the port, and the recess in the hood corresponding in size and position with the port, it is obvious that when the valve has traveled so far as to uncover the width of the port the full area will be exposed, and the passage under the hood exceeding that area will cause no obstruction. Another very import- ant effect of the mutual action of the valve and hood is at first glance not so obvious, but is readily shown. This valve opens and closes much more rapidly than is possible for a plain slide valve having the same motion upon the seat. the seat. When a plain slide valve opens, the area of the opening is only such part of the full area of the port as the width of the opening is of the full width of the port. Thus if the port be an inch wide, and 4 of an inch be uncovered, the port will be open of its area. With the hooded valve, the recesses coinciding with the ports, the opening is not merely along the face, but around the whole perimeter of the end of the valve; and as the area of the end of the valve is made greater than the area of the port, its peri- meter will be much greater than the end of the length of the port- usually about three times as great-although the width of the valve is less than the port length. Compared with the plain valve, open 24, the hooded valve in like position would be open. This property renders this style valve peculiarly adapted for automatic governing, where, under light load, the travel of the valve is much reduced and the width of port opening very small. It combines the large port area, with rapid opening and closing, essential to economy in high speed engines. Besides it is perfectly balanced, and there is nothing to cause wear and leakage except its weight, which is very light in proportion to its size. When wear does occur, the peculiar shape of the valve renders it a very simple matter to refit it as tight as when new. Scrape off the face of the hood where it bears on the seat enough to let it down on the valve again; then scrape the valve, hood and seat to a good bearing, and it is done. The tapering sides facilitate the fitting. • All single valve automatic engines operate their valves by means of a movable eccentric, whose throw and position relative to the crank are made to vary by means of a centrifugal governor, whose action 37 578 A Library of Steam Engineering. determines the amount of port opening and time of closure necessary to maintain the required speed under the load carried. No detail of the steam engine has been the subject of more study and inventive skill than this "shaft governor." None has yet been devised better adapted to its purpose or more efficient than the one used in connec- tion with the valve here described. In a manner peculiar to itself it combines the centrifugal principle, common to all shaft governors, with the inertia of a revolving wheel as a regulating element. This is effected without any intricacy or complication of parts. The result secured is a combination of great sensibility and power. Fig. 338. 2 A a Fig. 339. O Ο $20 Fig. 340. The illustrations clearly show the construction of the governor. It is composed of but seven essential parts, three of which are dupli- cated, making ten in all. The hub or sleeve A (Fig. 338) is bored to fit the engine shaft, and turned on the outside. It has one long and two short arms, with wrists a¹ a² a³. Opposite the long arm is a projecting lip at, which is finished to a radius whose center is in the wrist a³. The eccentric B (Fig. 341), is a flanged ring of sufficient size to surround the shaft and admit of a lateral motion sufficient to permit the amount of eccentricity required. This will be half the full travel of the valve it is required to operate, or half the sum of the widths of the steam ports and steam lap of the valve. From one side of the eccentric projects an arm like the long arm of the hub A (Fig. 338), having an eye on the end to fit over the wrist a³ (Fig. 340). The dis- tance from the center of the eccentric to the center of the eye, is half the steam lap of the valve shorter than the distance from the center of The Atlas Automatic Engine. 579 the sleeve to the center of the wrist a³. Therefore, when mounted on the wrist and engine shaft in its most central position, the eccentric will still have a throw sufficient to move the valve back and forth from line to line of the steam ports, but not to uncover them. Oppo- site the arm described the eccentric carries a lug piece B (Fig. 344), so formed as to embrace the lip of the hub for support and guidance, having two eyes, i i, for connection with the next piece. The dead wheel C (Figs. 348, 363 and 365), is a four-armed pulley, having its hub bored to fit loosely the turned part of the hub A (Fig. 338); its rim is provided with two lugs, h h,' to connect it with the Fig. 343. Fig. 341. shaft B Eccen tric Fig. 342. lug pieces of the eccentric and on two opposite arms, lugs and seats to receive the following pieces: The weights D D1 (Figs. 350, 351, 363 and 365), each having a lug d (Figs. 350 and 351), for attaching by means of a pin to the lugs at the ends of the arms of the dead wheel; a wrist d¹ d¹, as shown in Figs. 350 and 351, for connection with the hub A (Figs. 363 and 365), and a recess and pin d², as shown in Figs. 350 and 351, for attachment to the following pieces: The springs E E¹ (Figs. 363 and 365) are helices of tempered steel; they are each provided with a connecting rod e (Figs. 357 and 358), and washer and nut for attaching and adjusting the tension. When in place the springs rest on the seats on the dead wheel arms, and are attached by their rods to the weights D D¹, which they tend to draw inward. The weight links F F1, shown in Figs. 352, 353, 363 and 365, are short bars with an eye in each end adapted to connect the hub A and the weights D D¹, by means of the wrists a¹ d¹ and a² d² (Figs. 338, 580 A Library of Steam Engineering. Fig. 346. i PART OF B. Fig. 345. Fig. 344. C Fig. 348. Fig. 350. D D Fig. 351. d 2 G K G Fig 347. 1 Fig. 349. Fig 352. Fig. 353 LL F LL F The Atlas Automatic Engine. 581 339, 350 and 351.) The weights, springs and links are duplicated only for the sake of symmetry and balance. The pairs act as one. The eccentric link G (Figs. 346, 347, 363 and 365) is like the weight link in appearance; it connects the eccentric B and the wheel C, by means of the lugs h¹ 1 (Fig. 363), or h (Fig. 365.) The parts already described are shown assembled in Figs. 363 and 365. They are so connected together that the joints may work freely. Wear may be taken up where necessary, and means are provided for lubricating. a1-2 Fig. 354. a 3 O d Fig. 355. i Fig. 356. O O Fig. 359 O e e O a3 Fig. 362. d 2 Fig. 360. h Fig. 357. Fig. 361. Fig. 358. When mounted upon the engine shaft, the eccentric is placed in line with the valve-slide wrist to which it is to be connected by means of the usual straps and rod, and the hub A is keyed to the shaft in such position that the wrist a³ will lie in the same plane as the crank pin on the opposite side of the shaft. It will be observed that the weights D D¹ and their links F F1 form knee or "toggle" joints, con- necting the hub A and wheel C, so that when the weights swing out- ward or inward, the wheel is compelled to turn slightly around the hub. Or conversely, turning the wheel through a small angle, will swing the weights out or in, according to the direction of the rotation. The wheel being connected to the eccentric by means of the link G, 582 A Library of Steam Engineering. its rotation about the shaft causes the center of the eccentric to be car- ried across the shaft, so as to diminish its eccentricity as the weights move outward, or increase it as they move inward. These parts are capable of two relative positions. The first is shown in Fig. 363, which is right for a left-hand engine running over. In this position, if the wheel be turned to the right, the weights will be drawn towards the center of the wheel; if turned to the left, the weights will be car- ried outward until the set screws outside of the weights come in con- tact with the wheel rim, stopping further motion. But if the set screws be removed the wheel may be turned further to the left until the weight link passes a radial position, when, the motion continuing, the weight will be again drawn inward, finally reaching the position Crank Pin D 5h1 C C D1 a3 Fig. 363. Fig. 364. shown in Fig. 363. This, however, can not be done without removing the eccentric link connecting lugs hi (Fig. 363). That having been done, the parts may be brought to the position shown, and the link replaced so as to connect the other pair of lugs, h¹ i¹. This reverses the governor, and the engine will now run under. The springs are of such strength, and are given such initial ten- sions as will just balance the centrifugal force developed in their inner position, and revolving at the number of revolutions per minute the engine is intended to run; and so proportioned that an additional revolution per minute is required to increase the force of the weights sufficiently to carry them to their outer position. Thus a variation of one revolution per minute would be sufficient to readjust the posi- tion of the eccentric from that required for the full valve opening, or full load, to that for the least opening, or no load. In this respect the weights and springs act in this governor precisely as they do in most The Atlas Amuc Engine. 583 other shaft governors. But in this the weights can not move except by dragging themselves around the dead wheel. If they move at mod- erate speed, the wheel being in perfect balance and the friction incon- siderable, little resistance is offered, but the inertia of the wheel will prevent any sudden or violent movement. Other governors, not hav- ing this feature, are prone to "kick" at times with dangerous violence, a fault their makers attempt to correct by the application of friction plates or cataracts, either of which has the effect of making the gov- ernor sluggish in its action. Another very important advantage is secured from the inertia of this dead wheel: By reason of inertia a body in motion offers a resistance proportioned to its weight to any increase or diminution of its motion. Thus a wheel running loose on Orank Pin E c Fl D D1 EL B (a3) Fig. 365. Fig. 366. a revolving shaft tends to revolve at the same speed as the shaft, so long as that speed is uniform, but if it change, the wheel resisting a change of speed, will not immediately take up an increase of speed in the shaft but will fall behind. If the shaft suffer a fall in speed, the wheel tending to uniform motion will overrun it. In the governor these influences affect the dead wheel, which is here so connected that action set up by a change of speed is made to adjust the admission of steam to meet and correct that change. The dead wheel, therefore, acts to reinforce the action of either of the weights, or the springs, as occasion requires. The latter may therefore be made much lighter and smaller than usual in other governors. In fact the dead wheel may be regarded as the real governing agent, the centrifugal part of the governor merely indicating the speed to be maintained To show the power with which the dead wheel can act to move the eccentric and valve, let it be assumed, for example, that the weight of the rim is 500 pounds, and 584 A Library of Steam Engineering. the rim is 36 inches in diameter, and that the standard speed is to be 200 revolutions per minute. Under such conditions the energy stored in the revolving rim would be 7670.6 foot pounds. If a resistance were thrown on the engine sufficient to reduce the speed to 199 rev- olutions per minute, at the latter speed the stored energy would be 7594.1 foot pounds, a difference of 76.5 foot pounds. The eccentric corresponding to that size of wheel requires a movement of its center of but two inches to carry the action of the valve through its entire range. Therefore, the force available for this, developed by a change of rate of one revolution per minute, or one-half of the per cent., amounts to a pressure of 459 pounds. The great energy with which ATMOS. 120 revolutions. Steam, 70 lbs. Fig. 367. LINE Spring, 40. the wheel tends to maintain its speed, is of course available likewise to overcome any resistance to motion of the valve or rod. Hence the sensitiveness of this governor to variations of load and its ample power to meet them. TO SET THE VALVES OF THE ENGINE. First. Turn the engine in the direction in which it is to run, until the center of the crank pin comes to the line joining the centers of the shaft and cylinder; that is, place it on the inner center. Second. Observe that the center of the pin in the long arm of the hub on which the eccentric swings, is on the same center line extended beyond the shaft. It is obvious that if the center of the eccentric is brought to the same line, the eccentric will be in position of least throw. Third. Mark the position of the governor spring seats, so they can be returned to the same place; then remove the springs from the governor, being careful that the shaft does not turn, and roll the dead wheel backward until the center of the eccentric comes to the line as described. The weights will then be out near the dead wheel rim. Block them in this position so there can be no movement. The Atlas Automatic Engine. 585 Fourth. The steam chest cover having been removed, adjust the valve stem by means of nuts at the yalve cross head, so that the valve will lap at the outer steam port of an inch. Fifth. Without moving the blocking from the governor, turn the engine in the direction in which it is to run-to the outer center-and observe that the lap of the valve over the inner steam port is the same as it was on the outer steam port; if it is not, adjust the length of the valve stem and the position of the governor until the lap of ports is equalized. Then turn the crank, being on either center, unblock the weights and allow them to come in until the end of the valve comes "line to line" with the port, and turn out the stop screws in the back ATMOS. 120 revolutions. Steam, 62 lbs. Fig. 368. LINE Spring, 40. of the weight until they prevent the weights from going out any fur- ther by coming in contact with the dead wheel rim. This completes the valve setting, and the steam chest cover may now be put on. Sixth. Roll the dead wheel forward until the weights come in as far as they will, and replace the springs, adjusting them to the marks, so that they will have the same tension as before. The speed of the engine may be varied to a limited extent by increasing the tension of the springs for greater, and diminishing it for less speed. But the springs must never be compressed so much that the spires will come together and prevent the weights from going out, and the tension must always be sufficient to overcome all fric- tion and give the weights a lively return when they have been thrown out. The keyway for securing the governor hub is on the same side of the shaft as the long arm. The key should be fitted side wise, but should bear lightly at the top and bottom, otherwise the sleeve may be sprung out and prevent the wheel from turning. Figs. 367 and 368 are fair average samples of indicator diagrams taken from this engine. 586 A Library of Steam Engineering. Fig. 369. CYCLOIDAL ENGINE. The Atlas Automatic Engine. 587 THE CYCLOIDAL ENGINE. This engine is provided with four valves-two steam and two exhaust valves-of the multi-ported type. They have a very short travel, open and close quickly, and the steam valves may be adjusted for any port opening by a very little motion of a simple cycloid. The steam valves are controlled by a shaft governor. The exhaust valves are operated by a fixed eccentric. The steam valves are opened by a cycloid, but they are closed by steam pressure. The mechanism by which these valves are operated is plainly shown in Figs. 370 and 371. Fig. 372 is a cross section through the valves and cylinder, and shows the valves and mechanism by which they are operated. Fig. 373 is perspective view of the compound cycloidal engine. Figs. 374 and 375 show the construction of the valves, and Fig. 376 a longitudinal section of the compound engine. Fig. 370. 588 A Library of Steam Engineering. Fig. 371. Fig. 372. The Atlas Automatic Engine. 589 Fig. 373. PERSPECTIVE VIEW OF COMPOUND ENGINE. Fig. 374. 10 Fig. 375. 10 590 A Library of Steam Engineering. 590 Fig. 376. SECTIONAL VIEW OF COMPOUND ENGINE. The Brown Engine. 591 CHAPTER XXI. THE BROWN ENGINE. The Brown engine is one of the leading American steam engines, with features peculiar to itself, and may, therefore, properly be classed among those which distinctly differ from all other steam engines. As many of these engines are in use, in this and other countries, it is of the utmost importance to engineers and students of steam engineering that they should thoroughly understand the anatomy of this import- ant combination of American mechanism. In order then that the infor- mation so essential to engineers in the prosecution of their profession may be imparted in its simplest form, a plain and simple description will be here given of the important details of this engine, using for the purpose of convenient and intelligent illustration, the Greenwald Brown Engine. The engineer who will study the anatomy of this engine will obtain accurate information, not only as to the details and principles of construction of this particular engine, but as to all engines of this class, no matter by whom built. The valves of the Brown engine are plane plates, sliding upon plane surfaces, presenting opposing surfaces so distributed that no shoulders are worn thereon by the variable travel of the cut-off valves. The closure of these valves is made instantaneous and positive by the action of the steam pressure upon the area of the valve stems, the shock of the closing movement being noiselessly received, after ports are closed, by air cushions provided for the purpose. This arrange- ment secures positive action of the cut-off valves at any desired rate of piston speed. As the steam pressure within the steam chests and the cylinder is identical at the time of disengagement, there is no fric- tion of the valve against the seat to retard its proper closure. Each valve has several ports or openings, a suitable number being provided to insure prompt and ample area to maintain the closest approximation to the boiler pressure within the cylinder, up to the point of cut off, in case of the steam valves; and in case of the exhaust valves, the openings in them afford the greatest possible freedom from back or counter pressure on the exhaust side of the piston. Each valve is entirely independent of the others, both in regard to actuation and adjustment; hence the most precise adjustment pos- sible in modern automatic cut-off engines can be made. 592 045 ② Fig. 377. THE GREENWALD BROWN ENGINE, The Brown Engine. 593 The valves are actuated by a cam shaft common to this class of engines, which shaft has a coincident rotary motion with the main crank shaft. The steam valves are moved by separate eccentrics hav- ing sufficient angular advance to insure a quick opening movement. The exhaust valves derive their movement from separate wall cams, the peculiar shape of which give to the exhaust valve ports a very quick opening on the exhaust side of the piston, the valves remaining at rest during the greater part of the piston's stroke, and closing quickly just as the stroke is completed, remaining so with the valve stationary during the following steam stroke of the piston. In order to convey a clear idea of the valve mechanism, reference will be had to Figs. 378, 379, 380, 381, 382 and 383: Fig. 378 is a side view, with the covers of the steam chest removed, and showing the side shaft, valve rigging and cut-off shaft. Fig. 379 is a sectional plan, showing the steam chest, exhaust chamber under the cylinder, side shaft, and coupling clutch, gears for governor, and main shaft attachment. Fig. 380 is a sectional end elevation, showing the steam chest and exhaust chamber, with the valves and their connections. Fig. 381, shows a general arrangement from valves to main shaft Upon reference to the figures it will be noticed that there are four valves-one steam and one exhaust valve at each end of the cylinder. The steam valve works vertically, and in Fig. 380 is shown open admitting steam. The exhaust valve in the chamber under the cyl- inder moves horizontally, and is shown closed. The valve rods are connected to the valves by means of brass flanged nuts; and in case of removal of any valve or its attachments at any time, and to verify the lengths of the valve rods, a gauge e g and s g is provided for a guide, as shown in Figs. 382 and 383, the short side X being for that pur- pose. To test the length of the steam valve rod, the rod is screwed into the nut and the valve pushed up in the chest as far as it will go, and the length X (Fig. 383), from the chest to the bottom of the drip cup Q, of the gauge s g, is the correct length. For the exhaust, the valve is also pushed in as far as it will go, and the rod screwed in; and the length X (Fig. 382) of the gauge e g, from the cap of the exhaust chamber to the center of the strap Z, is the proper length of the rod. The full length of the gauges e g and s g gives the position of the edge. of the valves and edge of ports, when opening, measuring from the points named. In referring to Figs. 379 and 381 it will be observed that the side shaft is driven from the main shaft by a train of gears. The gear shaft R (Fig. 379) carries a miter wheel, which engages with its mate 38 594 A Library of Steam Engineering. B O O Fig. 378 O A B J O H O Felger Gir Fig. 379 Fig. 380 Y Fig. 382 FG T Fig. 384 MAIN SHAFT() L K Fig. 381 N E.G. Z U Fig. 383 HH W R M S.G The Brown Engine. 595 on the side shaft running the entire length of the engine, and on which is secured the eccentrics and cams to move the valves. In the arrangement shown, the engine is adjusted to run under on the out- ward stroke. To run the reverse way, or over on the outward stroke, it is necessary to change the position of the miter wheel on the side shaft A to the outer end of the shaft, just opposite to the position shown in the engraving; for it must be borne in mind that the side shaft must always run in the direction indicated by the arrows. This shaft also has a bevel wheel K, which gears with a pinion on the gov- ernor spindle. Near the governor spindle, on the shaft A, is the clutch coupling V, which is so constructed that the cylinder end of the shaft A may be disengaged from the crank end while the engine is at rest, and revolves by turning the hand wheel H in the direction shown by the arrows. By this arrangement the valve rigging can be moved for the purpose of making adjustments, or for warming the cylinder. In Fig. 380, A is the side shaft; B, steam eccentric, having arm C secured to lever D, which vibrates on E as the shaft A revolves. As the lever D is raised by the movement of the eccentric B, it engages with a steel catch on the dog F. This dog is hung in the center of the bridle G by a steel pin, and is free to move. When the lever D engages the dog F it lifts the bridle G through the dog, and therefore in open- ing moves upward in the guide W, carrying the valve rod and valve attached to the top of the bridle G. The upward movement of the valve will continue until the tail of the dog F comes in contact with the tripping disc and the arm I, when the dog is tripped, and the valve having no support, falls and closes. The tripping disc is attached to the upper end of the cut-off arm I, and is readily adjusted by means of a screw and jam nut, as shown in Fig. 380. The arm I is set-screwed to the cut-off shaft J, which shaft is oscillated by the governor. Motion is imparted to the governor through the bevel wheels K and L, as shown in Fig. 379 and 381. The governor spindle is hollow and carries the weights or balls on short arms, connected to the gov- ernor rod running down through the spindle and attached at the lower end of the bridle M, as shown in Fig. 384, and the bridle M is secured to arm N, which arm is set-screwed to cut-off shaft J. By the move- ment of the governor balls the rod is made to move up and down, thus imparting an oscillating motion to the arm N, cut-off shaft J, and arm I, with its tripping disc. When the tendency of the governor is to speed up, the arm I moves from the cylinder towards the dog F, thus cutting off earlier; and when the tendency of the governor is to slow down, the arm I moves toward the cylinder and from the dog F, thus cutting off later. 596 A Library of Steam Engineering. The exhaust valves are operated by the cams O O on the side shaft A. The pin P, with its steel roller moving in the groove S (Fig. 380), causes the arms T and U to vibrate back and forth, giving motion to the valves. Portions of this groove are concentric with shaft A. In this part of its path the valves are motionless; but during the other part of the revolution of the side shaft the travel of the valves is quick and full in opening or closing. As the valve rod moves on a straight line, and the arm T on an arc of a circle, it is provided with a sliding steel block which compensates and corrects this movement. The exhaust chambers are provided with false seats, and the steam valve seats being part of the steam chest they are readily accessible. The gears, cams, eccentrics and cut-off arms are all secured with set screws, and are therefore easily adjusted. In addition to the gauge employed for verifying the length of rods and locating the steam valves, there are scribe marks 1 and 2 on bridle G (Figs. 380 and 383.) Mark 1 should appear just at the top edge of guide W for correct length of valve rod, and at Q when the port is just opening, and also indicates the amount of lap the valve has if it properly covers the seat. TO SET THE VALVES. 16 In setting the valves see that the miter wheels H H are in their proper positions and secured, so as to turn the side shaft A in the direc- tion indicated by the arrows. Then put the engine on the dead center toward the cylinder or head end, and then see that the clutch V is in engagement, making the two sections of side shaft A practically one shaft and secure the clutch. Now turn the steam eccentric at the head end, in the direction the side shaft should run, until the lower mark on the bridle G (Fig. 383) has come of an inch above the upper edge of the guide W, then secure this eccentric firmly to the shaft by tight- ening its set screw. With the engine in the same position, adjust the exhaust valve at the opposite or crank end of the cylinder. The crank should be turned in the same direction as that in which the steam eccentric was turned, until the full length of the exhaust gauge e g (Fig. 382), and the distance from the exhaust chamber to the center of the wrist Z coincide, thus causing the exhaust valve to be at the com- mencement of its opening movement. Next, put the engine on the crank end dead center, and set the other valves in the same manner. TO SET THE TRIPS. Turn the side shaft by the hand wheel H (Figs. 378 and 379) until the steam eccentric is at its full throw, or full port opening of the valve, and secure, by the set screws, the arm I, with its tripping disc, The Brown Engine. 597 to the cut-off shaft J, so that it will trip the valve at that point. If the steam valves do not close at the same piston travel on each end, the difference can be corrected by adjusting the discs on the arm I (Fig. 379.) When the engine is running, notice that when the steam valve is closed, that the upper mark on the bridle G is just visible at the top of the guide W. If it is not visible, it can be made to show above or disappear below the guide, by turning the air cocks in the dash pot. If this does not correct the trouble, see if there is anything under the piston of the dash pot to prevent seating. This concludes the valve setting. Of course, for absolute correct and nice adjustment, the use of the indicator will be required. The speed of the engine can be varied by adding to or taking weight from the governor balls, or moving the sliding weight on the arm Y in or out (Fig. 379); adding weight to governor balls, or mov- ing weight on arm Y out, increases the speed of the engine, while taking weight from the governor balls, or moving the weight on arm Y in, decreases the speed of the engine. THE CROSS HEAD. HEPO Fig. 385. Fig. 385 is a side view and Fig. 386 is an end view partly in sec- tion. It will be observed that the wrist pin of the cross head, as shown in Fig. 385, is exactly in the center of the bearings, and hence the angular thrust of the connecting rod produces a uniform pressure on the entire surface of the bearings. The shoes of the cross head are adjustable, as shown in Fig. 386, by means of wedges. 598 A Library of Steam Engineering. 0 0 Fig. 386. Fig. 387. The Brown Engine. 599 THE MAIN BEARING. Fig. 387 represents the main bearing. It is provided with quarter boxes, which are adjustable by wedges and draw screws for taking up lost motion. This bearing also has a large base, similar to those under the slides and cylinder. Fig. 388 is a plan and elevation of this engine. E A ---G- D Fig. 388. PERFORMANCE OF THE ENGINE. To show the performance of this engine in every day practice, the following indicator diagrams are given: The frictional diagram (Fig. 389) was taken with a Thompson indicator from a Greenwald Brown engine, at the power station of the Mount Auburn Electric Railway, Cincinnati, Ohio. Cylinder, 42 inch stroke and 18 inches in diameter. Boiler pressure, 60 pounds. Revolutions per minute, 70. Spring, 30. 600 A Library of Steam Engineering. Fig. 389. The diagram (Fig. 390) was taken with a Thompson indicator, from a Greenwald Brown engine, at The Windisch-Muhlhauser Brew- ing Co.'s Brewery, Cincinnati, Ohio. Cylinder, 42 inch stroke and 20 inches in diameter. Boiler pressure, 80 pounds. Revolutions per minute, 60. Spring, 40. Fig. 390. STEAM AND EXHAUST PORTS. AREA OF STEAM PORTS. RULE.-Multiply the area of cross section of bore of cylinder by the number of feet the piston travels per minute, and divide the product by the constant 6000, and the quotient will give the required area of steam port. Example.-Let 12 inches equal diameter of bore of cylinder. Let .7854 equal a constant. Let 90 equal number of revolutions of crank per minute. Let 3 feet equal length of stroke of piston. Let 2 equal number of strokes of piston for each revo- lution of crank. Let 6000 equal a constant. The Brown Engine. 601 Then we have: 122X.7854X90X3X2 6000 -10.178784 square inches. Required area of steam port. Performing the operation, we have: 12 inches. Diameter of bore of cylinder. 12 inches. Diameter of bore of cylinder. "Product No. 1." 24 12 144 Square of diameter of bore of cylinder. .7854 31416 31 416 7854 "Product No. 2." 113.0976 square inches. Area of cross section of bore 90 "Product No. 3." 10178.7840 of cylinder. Number of revolutions of crank per minute. 3 feet. Length of stroke. "Product No. 4." 30536.3520 2 Number of strokes of piston for each revolution of crank. 6000) 61072.7040 (10.178784 square inches. Required area 6000 10727 6000 47270 42000 52704 48000 47040 42000 50400 48000 24000 24000 of steam port. AREA OF STEAM PORTS-RULE IN DETAIL. RULE.-First, square the diameter of the cylinder in inches, and call the product "Product No. 1." Second, multiply "Product No. 1" by the constant .7854, and call the product "Product No. 2." 602 A Library of Steam Engineering. Third, multiply "Product No. 2" by the number of revolutions of the crank per minute, and call the product "Product No. 3." Fourth, multiply "Product No. 3" by the length of the stroke of the piston in feet, and call the product "Product No. 4." Fifth, multiply "Product No. 4" by the number of strokes (2) of the pis- ton for one revolution of the crank, and call the product "Product No. 5." Sixth, divide the last product ("Product No. 5") by the constant 6000, and the quotient will give the area or required number of square inches for steam port. AREA OF EXHAUST PORTS FOR ENGINES HAVING EXHAUST VALVES. RULE.-Perform the operation precisely in the manner prescribed for steam ports, until the last product is reached, then, instead of using 6000 as a divisor, use the constant 4000 as a divisor, and the quotient will give the required area in square inches for exhaust ports. Taking the last product in the preceding example, and we have: 4000) 61072.7040 (15.268176 square inches. Required area of 4000 21072 20000 10727 8000 27270 24000 32704 32000 7040 4000 30400 28000 24000 24000 each exhaust port. AREA OF EXHAUST PORTS-SIMPLE RULE. RULE.-Multiply the area of steam ports, as determined by the preceding rule, by 6, and divide the product by 4, and the quotient will give the required area of exhaust ports. Example.-Let 10.178784 square inches equal required area of exhaust port, as determined by the preceding rule. Let 6 equal a constant. Let 4 equal a constant. The Brown Engine. 603 Then we have: 10.178784 6 4) 61.072704 15.268176 square inches. Required area of ex- DIMENSIONS OF STEAM AND EXHAUST PORTS. haust ports. RULE.-Assume any convenient length, in inches, for steam port, accord- ing to the diameter of the cylinder, and divide the area as determined by the preceding rules by the assumed length of port, and the quotient will give the required width of steam or exhaust port. Example. Let 10.178784 inches equal required area of steam ports. Let 10 inches equal assumed length of steam ports. Then we have: 10.178784 10 -1.0178784 inches. Required width of steam port.. Performing the operation in the ordinary way, we have: 10) 10.178784 (1.0178784 inches. Required width of steam port. 10 17 10 78 70 38 87 80 84 80 40 40 2008 78 70 We now have a steam port: Width, 1.0178784 inches; length, 10 inches. The same rule applies to exhaust ports. Example.-Let 15.268176 inches equal area of exhaust ports. Let 10 inches equal assumed length of exhaust ports. 604 A Library of Steam Engineering. Then we have: 10) 15.268176 (1.5268176 inches. Required width of ex- 10 52 haust ports. 50 320 26 20 888 68 60 81 80 17 10 76 70 60 60 ENGINES HAVING SEVERAL STEAM AND EXHAUST PORTS. The Brown engine, it will be observed, has three steam and three exhaust ports for each end of the cylinder, in such a case each port should have an area of but one-third of that required for engines hav- ing but one steam and one exhaust port for each end of the cylinder, as, for instance, the Corliss engine. In the case of engines like the Brown engine, for example, where three steam and three exhaust ports are employed for each end of the cylinder, or, for that matter, any other number of ports more than one, the area required for one port is divided by the number of ports contained for the admission or the exhaust of steam. Taking the area of steam ports, as shown in the preceding examples, 10.178784, and dividing the same by the number contained at either end of the cylinder (3), we have: 3) 10.178784 3.392928 square inches. Area required for WIDTH OF STEAM PORTS. each steam port. Now, as the ports of the Brown engine are necessarily shorter than in cases where they extend across the cylinder, we will assume the required length to be 7 inches, then we divide the area by the assumed required length, and we have: 7) 3.392928 .484704 Decimals of an inch. Width of each steam port. The Brown Engine. 605 WIDTH OF EXHAUST PORTS. After having determined the width of steam ports, and the exhaust ports are of the same length, divide the width of the steam ports by 4 and multiply the quotient by 6. Example. Let .484704 decimals of an inch equal width of steam port. Let 4 equal the divisor. Let 6 equal the multiplier Then we have: 484704 4 X6-.727056. Decimals of an inch. Required width of exhaust ports. Performing the operation in the ordinary way, we have: 4).484704 .121176 6 .727056 Decimals of an inch. Required width. of exhaust ports. 606 A Library of Steam Engineering. CHAPTER XXII. THE BUCKEYE ENGINE. This engine presents many novel and interesting features worthy of consideration by the student of steam engineering. Many of these features are peculiarly its own, and tend to place it in a class separate and distinct from other leading American steam engines. As this engine has come into general use it is a matter of importance to engi- neers to understand its construction and the adjustment of the valve gear, as well as the running gear generally. It is a high speed engine and automatic in its valve movements. In order that the student may have a clear conception of the anatomy of this engine, a detail descrip- tion of its various parts, with illustrations, are here given. thly Portugal All PUTTING THE ENGINE TOGETHER. SHOP MARKS. rts capable of being wrongly put together or misadjusted, are adjus d in the shop as correctly as possible and marked, so that in case a adjustment has been disturbed, by shipment or otherwise, it can readily be restored. THE MAIN ECCENTRIC. The main eccentric is marked to indicate its angular position on the shaft by a chisel mark on its hub next to the governor eccentric, and a corresponding mark on the shaft. But these marks being invis- ible when the governor eccentric is in position, the engineer should make marks on the eccentric and its strap to coincide when the crank is placed on its outer dead center. The outer dead center is preferred to the other because in that position the mark on the eccentric may be located near its longest radius and be in convenient position for observation. When set to the marks the leads, as shown by marks, will be from inch to 5 inch, and compressions from 13 inches to 5 inches, according to size of engine, generally about one-eighth stroke. THE GOVERNOR CONTAINING WHEEL. The governor containing wheel is similarly marked on its back hub, and when this mark coincides with a corresponding mark on the O (LEGO GORSER (O (IN,O, क Fig. 391. BUCKEYE ENGINE-BACK VIEW. 607 608 A Library of Steam Engineering. shaft, and other adjustments are correct, the latest cut off will gen- erally be at about five-eighths of the stroke, and the earliest somewhere inside of the first inch or two of stroke, which, when non-condensing and under ordinary conditions as to pressure, friction, etc., will hold the engine from running objectionably above speed under its lightest possible load and highest pressure. This is the safety test, the failure of which is corrected by slightly advancing the governor containing wheel on the shaft. On the other hand, the latest cut off will give maximum guaranteed power, with margin for regulation. To obtain greater maximum power, the governor containing wheel may sometimes be set a little back of the marks, as for instance, when disconnection from a considerable portion of load is impossible, or when the pressure carried is lower than usual, but in all cases of such adjustment the safety test should be applied. THE ECCENTRIC RODS. The eccentric rods are marked to indicate the distance they enter the bases of the eccentric straps, and are just visible when correct. These and other marks should always be followed except when performance tests, hereinafter given, show reasons for departure from them. ADJUSTMENTS INDEPENDENT OF MARKS. It is very essential that every engineer should thoroughly under- stand the adjustment of any engine under his charge, by means of inspection of the performance of the parts concerned, independent of shop marks, as such marks may become obscured, obliterated, or tam- pered with, or may not indicate the best possible adjustment for exist- ing conditions. THE MAIN VALVE. The main valve differs from the common slide valve in that it delivers steam to the cylinder from its interior through ports in its face, and exhausts steam from its cylinder at its ends into and through the containing chest, hence its movement is the reverse of that of the common slide valve; it admits steam to the cylinder by moving toward the end to which steam is to be admitted, and from the end from which steam is to be exhausted, and consequently, the openings for admission are not visible on removal of the chest cover, but are shown by marks on the cylinder and valve in line with the borders of their respective ports. The marks on the valve, however, represent its ports only at the ends where narrowest, and ignore a certain amount of extra graduated lead introduced to moderate the shock of induction at high pressure; hence when lead is spoken of, that shown by the marks is meant. The Buckeye Engine. 609 C Fig. 392. PLAN. O O 1. 39 610 A Library of Steam Engineering. Fig. 393. FRONT ELEVATION. The Buckeye Engine. 611 The exhaust opening, however, is visible like the admission of the common slide valve, and for which it must not be mistaken. TO SET THE MAIN VALVE. If the main eccentric is not correctly set the valve can be set only to give equal openings for admission at the two ends, regardless of the ED O O O 0 O O O 11 11 11 11 11 " Fig. 394. END ELEVATION. position of the piston; these once equalized will remain so in all posi- tions of the eccentric, though equalization should be tested after the eccentric is set by noting whether compressions are equal. TO SET THE ECCENTRIC. If the engine runs over-that is, top of driving wheel from the cylinder-place the crank on its outer dead center, otherwise, on its inner dead center. In either case the projecting side of the eccentric should be upward but inclined toward the crank sufficiently to cause the valve to show the desired lead; but if the engine is so conditioned, 612 A Library of Steam Engineering. as to size and attachments, that it can be readily revolved, the com- pression should next be noted, both as to equality and amount; if unequal they should be equalized even at the expense of unequal leads, since compression with high speed and high duty engines is a much more important factor in its effects in the running qualities of the engine than lead. THE AMOUNT OF COMPRESSION. The amount of compression may be, as before stated, about one- eighth of the stroke, though with the high speed short stroke styles, more may be frequently given to advantage. TO ADJUST FOR AMOUNT OF COMPRESSION. To adjust for the amount of compression, shift the eccentric for- ward for more or backward for less compression; but any considerable change of this kind will be accompanied by a slight loss of equality, owing to the fact that the piston travels faster when approaching the end of its travel farthest from the shaft than at the other end. THE ADJUSTMENT FOR EQUALIZING COMPRESSIONS. The adjustment for equalizing compressions is to either lengthen or shorten the eccentric connection by adjusting the rod out of or into the stroke, or to shift the member clamped to the valve stem, so as to throw the valve toward the end at which the compression is least. In chosing between the above adjustments select that which will cause the rocker arm to be nearest vertical at its end movement, bearing in mind, however, that if the cut off has been equalized the first named adjustment will not disturb that equality, while the last named shift- ing member on stem will. TO SET THE GOVERNOR WHEEL. Place the crank on one of its dead centers, preferably the inner one, and set the governor so that its eccentric will be on its inner dead center also, as nearly as can be determined by inspection. The shop marks will then coincide, or nearly so, unless the eccentric has been connected to the levers a half turn, relatively to the wheel, from its position when marked, in which case the marks will be diametrically opposite each other, hence, the need for observing the above rule rather than placing entire reliance on the marks. CUT OFF EQUALIZATION. It remains now only to equalize the cut off as between the two strokes, and it should be understood at the outset, that this can not be done perfectly for all points within the range of variation, on account The Buckeye Engine. 613 of unequal piston velocity at corresponding points in outer or inner strokes, due to the virtual shortening of the connecting rod when deflected from parallelism with the center line. But if equalized at or near quarter stroke, it will be practically equal throughout its proper working range, the inequality becoming apparent only at very early cuts, when that at the front or shaft end will be latest. ADJUSTMENT FOR EQUALIZATION. Block the governor wheels about half way out, turn the engine over and note when, in each stroke, the cut off takes place, measured from its beginning. If found unequal change the length of the con- nections between the governor eccentric and valve stem, in such a way as to throw the valve toward the end at which the latest cut off takes place. The adjustment may be made at the junction of the eccentric, straps, and rod, or in the connection between the upper cut-off rocker arm and the valve stem. If the latter place is chosen, two different constructions may be encountered. Originally the connections screwed out of or into the neck of a bronze box, fitted to a pin in the rocker arm above mentioned, and was secured by means of a lock nut. Lat- erally the box and rod are made in one piece, and the adjustment is made by screwing the ball and socket joint to and fro on the rod, securing with lock nut as before. CHOICE OF METHODS OF THE ADJUSTMENT. Choice of methods of the adjustment should be made by noting which will cause the pin in the upper cut-off rocker arm to vibrate most nearly equally each way past the pin in the main arm near which it works. Any noticeable inequality in that respect should be corrected by adjustment at the eccentric strap junction first, and cut off equalized at the other place afterward. TO DETERMINE POINT OF CUT OFF OF VALVE CLOSURE. If there is no steam, remove the cover plates of the balance pis- tons, when the closure may be seen, or if not visible up to closure, it may be felt for by means of a piece of wire at its end like a chisel, and it should be considered closed when slight traces of an opening can still be felt. If there is steam available remove the indicator plugs, admit enough steam by the throttle to blow a little at the opened orifices, but not enough to move the engine, turn the engine over slowly until the point of cut off is indicated by the stoppage of the blow, or preferably by its partial stoppage, since if entirely stopped, the point may have been passed by a small but unknown amount. This is perhaps the most exact method, especially if the engine is stopped 614 A Library of Steam Engineering. in each stroke, when a slight forward strain on the eccentric by its balance weight will stop the blow, and a push in the contrary direc- tion will start it again. Having determined the points of cut off and found them unequal, proceed as before directed; but repeated trials may be avoided by plac- ing the cross head at the mean of the two unequal points, preferably near the latest, and adjusting as required. When an engine is too large, or otherwise so conditioned as to preclude turning over, loosen the governor wheel and revolve it on the shaft. With the back cover plates removed note how far the valve, when at each extreme of its travel, must move to close the open port. That distance should be the greatest at the stem end by an amount equal to one-twelfth of the travel of the valve. This displacement from symmetrical movement is necessary to compensate for the dis- tortion of piston travel before explained. ADJUSTMENTS BY INDICATOR CARDS. When cards for careful adjustment are taken, it is important that the movement rig be substantially correct, and that such slight dis- tortions as may take place shall be the same at each end, so as not to introduce apparent inequality when none exist. The rigs now, and for a number of years, furnished with these engines are correct enough for the purpose, though the following test, if carefully made, would show slight distortion. With engine on one dead center, and indicator properly coupled up, apply the pencil to the paper and make a short vertical mark, such as the spring will permit without undue force. Then turn the engine till the cross head moves one inch and make another mark, and so on throughout the stroke, ending with the other dead center. On measuring the scale so produced the end divisions will be found slightly larger than the others, but if the distortion is the same at each end no harm will result, but in order that they shall be equal it is necessary that the angle formed by the chord, with the center line of its actuating lever, shall be 90°—a right angle—at mid stroke. Any deviation from this condition will diminish the divisions at one end and shorten the diagram. DISTORTION FROM ELASTICITY OF THE CORD. This is to be apprehended only at high speeds-200 revolutions and upward per minute-and is present when a diagram taken at full speed is longer or differentially located on the paper than when taken at slow speed. To avoid this use a well-stretched cord, having as little elasticity as possible, or, better still, a light wire; and, in either case, give the drum springs as much tension as may be thought safe. The Buckeye Engine. 615 The indicator should be attached by means of inch pipe fittings and a three-way cock, so that diagrams from each end can be taken in quick succession; and when taking cards to test cut off equilization the cock should be reversed several times, applying the pencil each time, so as to avoid being deceived by changes of load. If this method results in a multiplicity of lines, so much the more evident is the need of it, and if the diagram is not thought sufficiently reliable, continue taking until one is obtained that covers a period of sufficient uniform load to be reliable. TEST DIAGRAMS. Fig. 395. Fig. 395 represents perfect performance as to equality of the sev- eral functions, and in nearly perfect performance in other respects as is practically attainable, especially at high rotative speeds. The dotted compression and exhaust lines represent about the amount of allowable variations in these functions to meet different conditions, mainly rotative speed, as the higher the speed the greater the compres- sion that may be needed to give best running qualities, by which test a greater amount even than that shown by the dotted lines is some- times found to be advantageous. Fig. 396. 616 A Library of Steam Engineering. Fig. 396 shows a "slow" main eccentric, about such as might be produced if the eccentric had no angular advance; that is, if it was at its mid travel when the crank was on the dead centers. The remedy is to advance the eccentric until satisfactory performance is attained- good diagram and running qualities. Fig. 397. Fig. 397 shows a fast eccentric-that is too much angular advance. The remedy is to set it backwards until satisfactory diagrams are obtained, though, as a matter of fact, if the displacement from correct and equal adjustment were that of the eccentric only, the distorted result would show slight inequality. A B Fig. 398. Fig. 398 shows displacement of the main valve, giving great ine- quality in its functions. The remedy is to make the proper adjust- ment for equalization, according to the method already explained, throwing the valve more toward the end from which diagram A was produced. The Buckeye Engine. 617 A B Fig. 399. Fig. 399 shows unequal cut off. The remedy is to make adjust- ments as previously explained, throwing the cut-off valve toward the end from which the largest diagram B was taken. THE GOVERNOR. The following names of details of the governor, as shown in Fig. 400, are given for convenient reference: The levers or weight arms a a are called levers for convenience. The weights A A, clamped on the levers a a, determine speed, the heavier they are the slower the speed. The lever pivots bb are studs screwed to arms of the containing wheel on which the levers move freely. The links B B couple the free ends of the levers a a to ears on the sleeve of the governor eccentric C. The governor eccentric C is free to turn on the shaft, and is advanced 90° by the outward movement, or expansion of the levers to the outer extreme of their range of movement. The main springs F F are of steel wire, tempered. They are attached adjustably to the arm of the containing wheel by means of the tension screws c c. The tension screws c c are the means by which the tension of the springs is adjusted. The spring clips d d are clamped on levers a a, and are provided with slots or eyes into which the springs F F are hooked, and they are adjustable along the levers to a certain extent. The outer lever stops e e are cylinders of wood fitted to sockets in the outer caps of the links B B, and they limit the expansion of the levers by coming in contact with the inner surface of the wheel rim; but when the governor works properly they seldom if ever strike. 618 A Library of Steam Engineering. The inner lever stops ff are blocks of wood on which the levers rest when not expanded. The auxiliary springs P P are employed to counteract a tendency to "race" when the tension of the main springs is great enough to give close regulation at light and medium loads. They should be of just sufficient force to start the levers out at very near the running speed, G P a "g A b d A C 9 B -di PP G Fig. 400. for without them the levers would not start out until running speed was exceeded. With sufficiently strong main springs the auxiliary springs are not needed, but the strains and pivot pressure resulting from such spring strength are objectionable. When used the tension of the main springs should always be so high as to require their use to prevent "racing," otherwise speed will be too slow at heavy loads. They have contact with the fingers p p when the levers are about half way out. The guide rollers G G are introduced in most high speed engines to restrain the springs from bowing outward from centrifugal force, The Buckeye Engine. 619 and so changing the direction of pull on the clips as to disturb regula- tion; but they are seldom needed at speed below 240 revolutions per minute. CLOSE REGULATION. Ignoring all frictional or other disturbing elements, perfect regu- lation would be secured by such an amount of initial tension on the main springs F F that if the arms could move inward until their center of force coincided with the center of the shaft, or a line joining the center of the shaft and the pivots of the levers, in other words, the zero of centrifugal force, the springs would at the same time have their tension just relaxed; that is, arrive at their zero of force. Then it is evident that as the levers moved outward at a constant speed, cen- trifugal force and spring force, would increase in the same ratio and be in equilibrium with each other at the same speed at all points within the range of the movement of the levers. In other words, the regulation would be perfectly isochronous. But suppose the spring tension to be less than this, so that as the levers moved inward, the zero of spring force would be reached before that of centrifugal force; then as they moved outward, the spring force would increase more rapidly than the centrifugal force, at a constant speed, and a constantly increasing speed would be required to keep the two forces in equilibrium, and the amount the speed would have to increase in order to carry the levers from their inner to their outer limit of movement would be the meas- ure of the governor's variation. Thus, if 100 revolutions per minute. be required in a given time to start the lever outward, and 105 in the same time to expand them to their outer limit, the extreme variation would be 5 per cent., which would be tolerably close regulation, since in practice the changes of load seldom cover more than half that range. TO SECURE CLOSE REGULATION. It follows that when too much change of speed accompanies changes of load or of pressure, the remedy is to increase the spring tension until the regulation is satisfactory. This operation is accom- panied by increase or outward shifting of nuts or inward shifting of spring clips to prevent increase of speed. If the equilibrium becomes unstable and "racing" makes its appearance, as shown by the levers periodically moving outward and inward without load fluctuations sufficient to produce such movements, the tension must be relaxed until equilibrium is restored; but if "racing" makes its appearance only when heavily loaded, it may be cured by giving the auxiliary springs more power by means of greater reach of fingers, or by opening them slightly, but not so as to continue in contact with the fingers much beyond mid movement. 620 A Library of Steam Engineering. FRICTIONAL DISTURBANCE-CENTRIPETAL FRICTION. The friction of the straps of the governor eccentric, and of the valve and gear operated by it, is centripetal in its tendency, that is, tending toward the center. It tends to hold the eccentric backward and to pull the levers inward centripetally, like the action of the main springs, thus accelerating the speed slightly. The stronger the main springs the less this acceleration, but it is always present to some extent, and it varies throughout the range of movement of the levers, being greatest when they are at or nearest their inner stops, when the forces in equilibrium are weakest and the angle of the links and eccen- tric ears most acute, and next greatest when levers are most expanded, when link and ear angle is most obtuse but forces in equilibrium greatest; and least at or near mid movement, when the link and ear angle is a right angle or a little more. But for this inequality of frictional effect it would call for only slightly less spring tension, but as it diminishes from inner extreme to mid movement, or slightly beyond, and increases from that point outward, it would be counteracted by diminishing tension for the inner half of movement and increased tension for outer half, so that if tension is sufficient to give close regulation throughout outer half of lever movement it will be slightly in excess for inner half, and the result will be that on starting the engine the levers will not start out until proper speed is exceeded, and stable regulation is possible only with loads too light to require at any time later quarter-stroke cut off, hence the functions and needs of the auxiliary springs. TO CHANGE SPEED. от O Scales Fig. 401. For any considerable change of speed different weights should be attached, the weight of which may be found as follows: Remove one lever with its weight attached, suspend or support the pivot end at the pivot hole, and allow the weighted end to rest on a fulcrum on scales, as shown in Fig. 401, at a measured distance from one end. The Buckeye Engine. 621 The point selected is immaterial, but when once selected the same point must rest on the scales during subsequent weighings. A point which comes about central over the inner stop block may be pre- ferred as coinciding more nearly with the center of gravity of the attached weights. Weigh the weighted lever thus independently sup- ported at its pivot, multiply the weight by the square of the present number of revolutions per minute, and divide the product by the square of the desired number of revolutions per minute; the quotient will give the required weight of weighted lever, weighed as shown in the diagram MINOR CHANGES OF SPEED INCREASE. Increase of spring tension may be tried, if room for it exists on the tension screws, and if, when the desired increase has been ob- tained, there is shown no tendency to "race" at any load, and at starting the levers start out before proper speed is exceeded, the cor- rect adjustment has been made, and the regulation will be closer than before. But if the necessary tension can not be applied, proceed as follows: Add a little spring tension and also shift the spring clips from the lever pivots about twice as much as the added tension. The stability and closure of regulation will be about as before, for while increase of tension tends toward instability of equilibrium, shifting the clips from the lever pivots, as above described, tends toward increased stability and more speed variation. Another method is to shift the weights toward the pivots of the levers, if the space in that direction is not thereby too nearly exhausted and the weights still have good bearings on their inner stops. TO DECREASE SPEED. Spring tension may be reduced, but to retain as close regulation as before, the spring clips should at the same time be shifted toward the lever pivots twice as much as the tension is reduced. Decreased speed may also be accomplished by shifting weights from pivots if room exist for such movement ers, REVERSAL OF DIRECTION OF MOTION. Place lever pivots in the idle holes, or holes occupied by guide roll- with levers turned end for end and other side out, so that after motion is reversed the pivot ends of levers will still go ahead of the free ends, and the weights will rest on the stops as before. Reset con- taining wheel and eccentric according to directions already given. 622 A Library of Steam Engineering. VALVE ARRANGEMENT. Fig. 402 shows a central section through the cylinder and the right- hand end of the valve and chest, the other being a section through one of the balance pistons. The line of section for valve and chest is indicated in Fig. 403 by dotted lines b bƒƒ b' b'. The steam enters at D (Fig. 402) into passage a a,, which lead to balance pistons d, through which it passes to the interior I I of the valve, as shown by the arrows, in which chambers the boiler pressure is constantly maintained when the engine is at work. The balance pistons are packed in their orifices, with steam metal spring rings and followers F, and also fitted to work steam tight on faces on the cover plates of the valve, thus forming a steam tight communication between passages a a and the moving valve. Coiled steel springs E serve to hold the pistons to their seats on the valve when the steam is shut off. From the interior of the valve the steam is admitted to the cylin- der through ports in its faces, as they are alternately brought by its movement to coincide with the cylinder ports. The valve is shown as just beginning to admit steam at the left-hand end, as indicated by the arrows; while at the other end the cylinder port is shown partly open for the exhaust, which passes into the chest and thence into the pipe K at its bottom, its course also being indicated by arrows. The cut-off valve is formed by two plates, shown at cc (Fig. 402) and v v (Fig. 404), and rigidly connected by rods h h h'h' (Fig. 404). These plates work on seats surrounding the valve ports, which valve ports they alternately cover at times relative to the piston travel deter- mined by the governor. The area of the balance piston is such as to hold the valve seat against the force tending to throw it off, due to the pressure in the valve and cylinder ports. This tendency is variable, being greatest at the moment of induction and least after cut off, while the counter- acting force due to the area of the balance pistons is constant, and con- sequently in excess, except during induction. To counteract such excess, shallow recesses corresponding to the cylinder ports in shape and area are formed in the valve seats near their inner margins. These recesses, or relief chambers," as they are called, one filled with steam pressure from the interior of the valve through small holes, as shown at f (Fig, 402), at the right-hand end, while the port at the same end is open for exhaust, and relieved of such pressure by the movement of the valve, as shown at the left a little before induction. By this device the average pressure of contact is reduced to about what is needed to insure wear enough to keep the surface in good condition. 66 Channels ee (Fig. 402) are cut across the valve faces near their ports, to prevent the steam in the ports from acting on any larger area -through any slight imperfections of fit-than is embraced in the bal- The Buckeye Engine. 623 H K K J These pressure in excess of that in the chest can be found in them. conditions are attended to in the construction of these engines, but since wear from the valve forces reduces their depth, cases may occur of their becoming inadequate from long wear, so as to require deepen- ing or otherwise enlarging. H ance pistons, and thus throw the valve from its seat. To perform this function they require to be of such capacity, and the free escape of steam that may leak into them to be so well provided for, that no G Fig. 402. 624 A Library of Steam Engineering. 2 V G V Fig. 403. SECTION ON LINE B B (FIG. 402). The Buckeye Engine. 625 L t t v A V V' R v h h h V V h V υ Fig. 404. SECTION ON LINE C C (FIG. 402). 40 626 A Library of Steam Engineering. Fig. 405 represents a cross section of the girder bed and compound rocker arm. The rocker shaft a vibrates in adjustable bearings bb; the main rocker arm A is clamped to this shaft by bolts c c, and trans- mits motion from the eccentric rod to the valve stem. Through the center of the arm A there is a long bearing, carrying the cut-off rocker shaft B, with its arms C and D. This shaft receives motion from the b A H Fig. 405. cut-off eccentric rod on the pin E; the cut-off valve thus derives a motion compounded from the movements of both eccentrics, and therefore has a constant length of travel on, and relative to, the main valve. RUDIMENTS AND ANALYSES OF INDICATOR DIAGRAMS. For convenient reference the following names are given to the different parts of a diagram, as shown in Figs. 406, 407, 408 and 409: THE STEAM LINE. The steam line, or initial pressure line, A B (Fig. 406), is formed while steam is being admitted to the piston. In diagrams from automatic cut-off engines, shown in Fig. 406, the steam line should theoretically be horizontal, and should show a pressure nearly or quite equal to that in the boiler, but in practice, the port and steam pipe areas and valve travel necessary to attain these conditions at the high piston and rotative speeds required for electric generation, and some other purposes, would involve greater loss from friction, condensation, leakages, etc., than would result from several pounds' loss of average pressure during admission. The Buckeye Engine. 627 F A B" B B G V E C D V Fig. 406. DIAGRAMS FROM THROTTLING ENGINES. In diagrams from throttling engines, on the other hand, as shown in Fig. 407, the greater the fall of pressure during admission, the bet- ter the economy indicated, since other things being equal, this fall of pressure, which may be called initial expansion, leads to a correspond- ingly diminished terminal pressure. A G V LL F E B C 200 D V Fig. 407. THE POINT OF CUT OFF. The point of cut off B, being anticipated by more or less fall of pressure, it is customary to locate it at the point of contrary flexure; that is, the point where the line ceases to be convex and begins to be concave. 628 A Library of Steam Engineering. THE EXPANSION CURVE. The expansion curve BC (Figs. 406 and 407) begins at the point of cut off B and terminates at the point of exhaust C. The latter, like the former, is presumably at the point of contrary flexure, except when the load is so light that the return pressure falls below the atmos- phere. Such a case is represented by the dotted expansion curve B" C" (Fig. 406), which is about what would be produced if all load were thrown off; in other words, it would produce a friction diagram. THE EXHAUST LINE. The exhaust line C D represents the change of pressure which results from the opening of the exhaust. This change may, in auto- matic cut-off engines, be a fall or rise, or no change at all, when termi- nal pressure is the same as back pressure, as shown by curve B' D (Fig. 406), in which case the exhaust line is absent. THE BACK PRESSURE LINE The back pressure line D E represents the pressure acting against the piston during the return stroke. With non-condensing engines it should coincide with atmospheric pressure line, or not materially exceed it-exact coincidence is seldom attained-and with condensing engines it should not materially exceed the pressure shown by the vacuum gauge. With the high pressure cylinders of compound engines it should show sufficient pressure above atmosphere to permit the next cylinder to do its proper share of the work. THE POINT OF EXHAUST CLOSURE. The point of exhaust closure E, being anticipated by some rise of pressure, and there being no change in the direction of curvature, can not certainly be located, but it can be approximated closely enough for all practical purposes, when a clear understanding is had of the laws governing the formation of expansion and compression curves. THE COMPRESSION CURVE. The compression curve E F indicates the rise of pressure due to the compression of the steam remaining in the cylinder after the clos- ure of the exhaust. It begins at the point of the exhaust closure and ends where steam begins to be admitted, a point far more difficult to locate in such diagrams, as shown in Figs. 406 and 408, than any of the other points named, as the engines from which they were taken- Buckeye-are provided with a graduated lead, so adapted to existing conditions as to cooperate with compression in reducing to the utmost The Buckeye Engine. 629 possible degree the tendency to pound when passing the dead centers; and it is found that when such running qualities are best attained the terminal point of the compression curve is most obscure. THE LEAD OR ADMISSION LINE. The lead or admission line F A, when one exists, completes the dia- gram. Its beginning F, as previously explained, is mostly, and should be, very uncertain, but its termination at the beginning of the steam or initial pressure line is the most clearly defined point of all, except as it may seem obscured by vibrations of the pencil lever of the indi- cator, which are always present at high speeds and with a free working instrument. The line is absent when compression begins so early that FI F" F B B B" ale 50 G E" E E' V C D V Fig. 408. the resulting pressure equals the initial pressure, as is the case with the inner or smallest diagram A B" C" D E" F", in the group shown in Fig. 408, which is from what is called a single valve automatic engine, in which a single valve, possessing the capacities and limitations of a common slide valve, has its travel so varied in time and amplitude extent by the governor as to maintain substantially constant lead with variable cut off, but at the expense of considerable variation in the points of exhaust and exhaust closure, as is clearly shown, when such engines are run empty, the exhaust closure takes place near mid stroke, hence the need for large clearance, 7 to 10 per cent. being about the in practice. Most of the high speed engines are of this class. range THE ATMOSPHERIC LINE The atmospheric line G is drawn while the indicator cock is so placed as to admit the atmosphere below the piston. It may be drawn by applying the pencil while the drum is in motion, but it is preferable 630 A Library of Steam Engineering. to make it larger than the intended diagram, and that is done by pull- ing the cord by hand. The foregoing list includes all that the indicator itself can pro- duce, but for further analysis two more lines must be considered. THE VACUUM LINE. The vacuum line V V is located below and parallel with the atmos- pheric line, at a distance from it corresponding with the known or estimated pressure of the atmosphere at the time and place of taking it. In the absence of such knowledge, however, it is customary to assume the sea level mean, 14.7 pounds per square inch, as has been done with the diagrams shown herewith; but since the pressure dimin- ishes over one-half of a pound, .533, for each 1000 feet of elevation above sea level, there are many cities and large areas of country of sufficient altitude to require material allowance for it. For instance, Denver, Col., being situated at 5267 feet altitude, the mean atmos- pheric pressure is but 11.9 pounds; at the altitude of the lakes, except Ontario, it is 14.4 pounds; and at Altoona and Johnstown, Pa., and a large portion of Ohio, it is 14 pounds. A barometer which has been adjusted for weather readings at a particular locality, is of no use for determining atmospheric pressure, but if made or adjusted to read according to altitude, whether in terms of inches of mercury, feet altitude, or pounds atmospheric press- ure, it is exactly what is required, and such can be had. THE CLEARANCE LINE. The clearance line V A (Fig. 409) is drawn perpendicular to the atmospheric line, at a distance from the initial end of the diagram and bears the same relation to its length that the volume of clear- ance space does to the piston displacement. When the construction permits, the clearance is most readily and accurately measured by fill- ing the space with water. The volume so ascertained in cubic inches is multiplied by 100, and divided by the cubic inches of the piston displacement; the quotient will give the clearance as usually expressed, namely: the per cent. of the displacement. When clearance can not be ascertained, and it becomes necessary to estimate it, the following facts should be borne in mind: The same volume of clearance makes a larger per cent. of the dis placement when the stroke is short than when it is long. THE CLEARANCE OF BUCKEYE ENGINES. The clearance of Buckeye engines is somewhat less than that of any other build, there being but two ports, and these short and direct. The Buckeye Engine. 631 They are built in three styles: Style "A," having the stroke a little less than twice the bore; style "B," stroke slightly shorter in propor- tion, and style "C," stroke one and one-third times the bore and less. Their clearances are about 2, 3 and 5 to 6 per cent., respectively, the latter figure being that of one or two sizes whose strokes are but little more than the bore. The large port area required for high speed also raises the percentage in the latter class. FOUR-PORT ENGINES. Four-port engines, of which the Corliss is a type, have 3 to 4 per cent. clearance. The smallness of the figures. The smallness of the figures being due mainly to their relatively long strokes. A .74 .86 .98 1.15 ck 1.38 1.72 2.3 90 V 301 80=V 335 3" 70 =V 380 -60=V 438 -50=V 521 -40-V 643 -30-V 842 a V. 1081 V 23 lbs Fig. 409. SINGLE VALVE HIGH-SPEED ENGINES. Single valve high-speed engines have, as before stated, about 7 to 10 per cent. clearance, so as to accommodate the excessive compression which accompanies early cut off, moreover, with steam ports of the length required by the central position of the valve, and of the capac- ity required for high speed, such clearance is unavoidable. THEORETICAL EXPANSION CURVES. In the construction of theoretical expansion and compression curves by which to test the correctness of actual ones, it is customary to assume the unqualified correctness of the Mariotte law, that the volume of elastic fluids is inversely as the pressure-that is to say, if steam at a certain pressure be allowed to expand to double its original vol- ume its pressure will fall one-half, if expanded to three times its orig- 632 A Library of Steam Engineering. inal volume its pressure will fall to one-third, and so on for all degrees of expansion-so that if the different volumes be multiplied by their resultant pressures a constant product will be obtained. But the law is true only for a constant temperature at all pressures, whereas, giv- ing no transmission of heat to or from the fluid during the expansion, the sensible temperature will fall with the pressure, causing the latter to be materially less after expansion than the Mariotte law calls for. Thus, if steam at 90 pounds total pressure, above vacuum, be expanded 6 times, the pressure should fall, according to the Mariotte law, to 15 pounds, whereas, owing to the accompanying fall of temperature from 320° Fahrenheit to 213° Fahrenheit, the pressure will fall to a small fraction over 13 pounds, or nearly 2 pounds more in consequence of the fall of temperature. A curve constructed in accordance with the Mariotte law, as above defined, is called the isothermal curve, the term signifying same temper- ature. A curve in which the effect of temperature is correctly allowed for, is called the adiabatic curve, the term signifying no transmission, since if no heat be transmitted to or from the fluid while undergoing a change of pressure its temperature will change directly with the pressure. Since the expansion of steam in an engine cylinder takes place too quickly to allow any considerable amount of heat to be transmitted to it from the walls of the cylinder, it would seem that the expansion curve should be more nearly adiabatic than isothermal, and under favorable conditions, such is sometimes the case. The conditions required are: engine not too small, not less than 150 horse power; tight-fitting valves and piston, and cylinder and steam pipe well protected with non-conducting covering, or otherwise; steam sufficiently superheated to reach the cylinder without entire loss of superheat. Generally, and particularly with small engines, it will be found that actual expansion curves are still more incorrect than the isother- mal curve in the same direction, namely: undue terminal pressure; so that the isothermal curve is sufficiently correct for all practical pur- poses, and when an actual curve agrees closely with it, the curve is considered practically perfect, but since more perfect curves are some- times met with, it is well for the student of steam engineering to understand the nature and extent of the inaccuracy of the isothermal curve. Such an understanding will result from a study of the follow- ing processes for their delineation : THE MATHEMATICAL METHOD OF DRAWING THE ISOTHERMAL CURVE. Having drawn the vacuum and clearance lines as correctly as pos- sible, select a point in the expansion curve at which it is desired that it shall coincide with the theoretical curve. The point may be near The Buckeye Engine. 633 the point of cut off or exhaust, or anywhere between the two. If near the cut off it will show what should be the resulting terminal, if near the exhaust it will show where the point of cut off should be, and if later than actual it will show how much more work should have been done with existing consumption of steam; or a point may be found somewhere in the curve which will give a theoretical diagram showing equal work, in which case, if theoretical terminal is lower than actual, it will show how much less the consumption should have been for the the same work. In the present case we will select a point near the terminal at a (Fig. 409), where the total pressure is 23 pounds (40 scale), and the distance from the clearance line c (which represents the volume due to that pressure) is 3 inches. Multiplying these together we have 69, which on our present assumption will be the constant product of pressure and volume throughout the curve, therefore, we may divide it by any number of other volumes-distance from clear- ance line to obtain resultant pressures, or other pressures to obtain resultant volumes. Choosing the latter plan, we draw horizontal lines at 30, 40, 50 up to 90 pounds, or as high as the diagram requires. The constant 69, divided by each of these pressures, gives the volume num- ber marked on each line, which is the distance in inches from the clear- ance line to the desired point in the curve. The points so obtained may be connected by a continuous curve, as shown in Fig. 409, or if sufficiently numerous they will serve the purpose without connecting. THE GEOMETRIC METHOD FOR DETERMINING THE THEORETICAL CURVE. The geometric method for determining the theoretical curve requires the card to be pinned to a board accurately squared, and a small draughting tee square and a hard, finely-pointed pencil. The same diagram as shown in Fig. 409 is shown in Fig. 410, and the point of coincidence at C is the same as before. From C a vertical line is drawn upward to B, where it joins the horizontal line A B, the exact height of which above the diagram is immaterial. It may represent the boiler pressure when known, other- wise it may be drawn as most convenient. From the intersection of these lines at B, draw the diagonal line B V, to the intersection of the vacuum line and clearance lines. From C draw line to D parallel with the atmospheric line cutting BV at D, and from their point of inter- section draw vertical line D E, which locates the theoretic point of cut off at E, assuming A E to be the theoretical initial pressure line. Then from points F G H I J K draw perpendicular lines downward far enough to pass the probable path of the desired curve, and from same points draw diagonal lines F V, GV, H V, etc., cutting the line ED at f 634 A Library of Steam Engineering. gh, etc., from which points draw horizontal lines to intersect the ver- tical lines under F G H, etc., which points of intersection will be in the desired curve. Points F G H I, etc., may be located at random, but should be closer together near E than they need be near B. All vertical lines should be parallel to the clearance line A V, and all horizontal lines parallel to the atmospheric line L M. The two curves agree substantially and would do so exactly with exact calculations and measurements in the first case (Fig. 409) and equally exact location of lines and intersections in the second case (Fig. 410.) They show the expansion curve a little more correct than the isothermal curve. A f h E F G H D L Fig. 410. K B C M THE ADIABATIC CURVE. The most that can be done on this subject is to give the student a tolerable correct idea of the difference between this curve and the other, since the basis of all rules for tracing it must be the volumes of saturated steam of different pressures experimentally determined, something that has not yet been accomplished with absolute cer- tainty. Rankin's formula that "the pressure varies nearly as the reciprocal of the tenth power of the ninth root of the space occupied," is not correct unless the assumed volumes on which it is based are correct. In view of these considerations we prefer to give a process which is correct in itself, the correctness of the result obtained depend- The Buckeye Engine. 635 ing upon that of the assumed volumes, which being merely factors may be changed for others which may be known or believed to be more correct. TO TRACE THE ADIABATIC CURVE. The process that will be here given is like that first given for trac- ing the isothermal curve, except that instead of treating the different volumes inversely to the pressures, they are obtained from the tables of properties of steam given elsewhere in this book. Referring to the diagram, as shown in Fig. 409, suppose the volume- of steam at 23 pounds pressure, as at the selected starting point a, to be 1081 times that of water of same weight (this volume is repre- sented as before by the distance of the clearance line 3"), then, as 1081 is to 3", so is the volume of any other pressure to the length of the line representing it, measuring from the clearance line, hence the following rule: RULE.-Divide the length of the line 3" representing the volume of the pressure at the starting point by that volume, 1081, and multiply the quotient by the volumes of any number of other pressures, and the products will be the lengths of the lines representing such volumes. 1 Thus, beginning at the highest pressure, the quotient as above begins with .002775, and the volume of 90 pounds being 301 the length of that line will be .002775 times 301 equal .835275", or .83" instead of .74", as per the isothermal curve. In like manner the other lines are found to be as follows: For 80 pounds .93" instead of 86". For 70 pounds 1.06" instead of 98". For 60 pounds 1.21" instead of 115″. For 50 pounds 1.44" instead of 1.38". For 40 pounds 1.78" instead of 1.72". For 30 pounds 2.33" instead of 2.3". Setting off these measurements on the pressure lines the points: in the desired curve are obtained through which is drawn the dotted curve. The test shows that the actual curve is between the two, which is a very good showing for so small an engine 12" x 20". But in all such cases it should be borne in mind that a certain amount of piston leak- age would cause a deceptive appearance of correct expansion. Whether any such leakage existed in the present case is not known with certainty, but the correctness of the compression curve negatives the supposition. 636 A Library of Steam Engineering. ANOTHER METHOD. Move the clearance line toward the diagram, a distance equal to about one-fortieth of the length of the diagram, about in contact in the present case, and construct an isothermal curve based on such a changed clearance. The result will not be exactly the same as the foregoing, but it will be about the same at the two extremes, and not noticeably different throughout. THEORETICAL CLEARANCE. Select two points in the expansion or compression curve, as A B or a b, as shown in Fig. 411, from which points draw horizontal and vertical lines, forming the parallelogram A CD B, or a cd b. Through angles C D or c d draw a diagonal line, continuing it until it intersects. 06 V a d A D B b Fig. 411. the vacuum line at V, from which point draw a vertical line which will be the theoretical clearance line. Sometimes the clearance thus indicated will be less than the actual, and sometimes more. When the conditions are such that an adiabatic curve should be produced, it should show about 2 per cent. less than actual. HORSE POWER CALCULATIONS. To find the horse power of an engine it will be found most con- venient to first find its horse power constant, that is the horse power due to one revolution per minute and one pound mean effective press- The Buckeye Engine. 637 ure. This constant number multiplied by the number of revolutions per minute will give the horse power per pound mean effective pressure, and the product multiplied by the mean effective pressure will give the horse power for that pressure. RULE.-Multiply the area of the piston in square inches by twice the length of the stroke in feet, and divide the product by 33,000; the quotient will give the constant. Example.-Let 22 inches equal diameter of cylinder. Let .7854 equal a constant. Let 2.75 feet equal length of stroke. Let 2 equal the number of strokes for each revolution. Let 33,000 equal a constant. Then we have: 22X22X.7854×2.75X2 33,000 =.0633+ A constant. Entire accuracy, however, requires allowance for the area of the pis- ton rod, which, assuming it to be one-sixth the diameter of the piston, or one thirty-sixth of the area of the piston, and that it affects one stroke only of the two, the area of piston may be diminished by of itself making it 374.8, instead of 380.13 and giving .0624 as a constant instead of .0633. THE MEAN EFFECTIVE PRESSURE. The mean effective pressure is most conveniently measured with the planimeter, but as complete instructions in the use of this instrument will be found in a preceding chapter, we will here confine ourselves. to the method for obtaining the mean effective pressure by ordinates. It is customary to draw ten vertical lines, called ordinates, across the diagram, as shown in Fig. 412, and measuring and recording the press- ure shown at each, adding the pressures and dividing the total by ten to obtain the mean effective pressure. If the spaces are all equal the measurements for pressure should be taken in the centers of the spaces, but it is better and more accurate to begin and end with spaces one-half the width of the rest, so that measurements taken on the ordinates will stand for the centers of equal spaces. But the following method of measuring the ordinates is pre- ferable to the one described, and, if done with reasonable care, will be more accurate. Cut a narrow strip of paper and apply it to each ordinate in suc- cession, setting off its length accurately with a sharp hard pencil, thus 638 A Library of Steam Engineering. getting the aggregate length of all of them. This length 9 inches in diagram, as shown in Fig. 412, is then multiplied by the scale and divided by 10. Thus: 9.03 X 40 10 -=36.12 pounds. Mean effective pressure. When negative ordinates are encountered, as Nos. 6, 7, 8, 9 and 10, in the friction diagram, shown by the dotted curve in Fig. 412, they must be measured in the opposite direction on the paper from the rest, so as to diminish the aggregate length of the positive ordinates, for the reason that they represent a partial vacuum behind the piston. a -e- B 2 3 M. e. p. 36.1 5 1 C sat-198- Fig. 412. 8 9 10 b d f SPACING ORDINATES. A parallel ruler is made for the purpose, and which is very con- venient, but it can not be compared with the planimeter. In the absence, however, of any such instrument, the spaces may be laid off with div- iders; or, having drawn perpendiculars touching the ends, a rule, grad- uated to sixteenths, twelfths, tenths and eighths may be applied in a more or less diagonal position, varying the angle until a distance between the verticals is formed which contains a number of one of the above subdivisions which is divisible by 10, 20, or 30, when nine full spaces with half spaces at the ends can be pointed off with a pencil and the ordinates drawn through the points. Thus line ab (Fig. 412) 34 inches or fg long; hence of an inch at each end and of an inch for each of the rest gives the desired spacing. 16 The Buckeye Engine. 639 A QUICK ESTIMATE OF MEAN EFFECTIVE PRESSURE. Draw line c d, as shown in Fig. 412, touching the exhaust end and cutting off an area A equal to B and C, as nearly as can be estimated by inspection, when the pressure at the middle (ordinate No. 5) will be the mean effective pressure, since the middle width of the wedge cdef is its width. This method is not applicable, with any degree of accuracy, to diagrams representing the extreme and sudden changes due to light load and high pressure; but under favorable conditions, a little prac- tice, with results tested by other methods, will reduce the margin of error to within one or two pounds. First attempts will generally give too high a result, since area A looks larger than B and C when they are equal. COMPOUND ENGINES. The entire calculations of each cylinder may be made separately, and the results in indicated horse power (I. H. P.) added together; but for certain purposes, it is best to add to the mean effective pressure (M. E. P.) of one cylinder the equivalent of the other or others, thus: Suppose there are two cylinders-one high and the other low press- ure--and it is desired to reduce the mean effective pressure of the low pressure cylinder to an equivalent pressure for the high pressure cylinder. RULE.-Multiply the square of the diameter of the low pressure cylin- der by the mean effective pressure of that cylinder, then divide the product by the square of the diameter of the high pressure cylinder, and the quotient will give the desired equivalent; which, if added to the mean effective pressure of the high pressure cylinder, the sum will give a pressure that, if acting in the high pressure cylinder alone, would equal the energy of both high and low pressure cylinders; and the same process may be applied to the other cylinders of triple and quadruple expansion engines, and any one of the cylinders may be selected as the exponent of all. Example. Let 21 inches equal diameter of low pressure cylinder. Let 20 pounds equal mean effective pressure of low pressure cylinder. Let 12 inches equal diameter of high pressure cylinder. Let 40 pounds equal mean effective pressure of high pressure cylinder. Then we have: 212X20 40+ -101.25 lbs. 122 640 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 21 21 21 42 441 20 12X12= 144) 8820 (61.25 lbs. Equivalent to 61.25 pounds 864 180 144 360 288 mean effective pressure in the high pressure cyl- inder. 720 720 Adding the equivalent mean effective pressure to the actual mean effective pressure of the high pressure cylinder, we have: 40 lbs. M. E. P. high pressure cylinder. 61.25 lbs. Equivalent M. E. P. 101.25 lbs. M. E. P. of both cylinders combined in one. THEORETICAL WATER CONSUMPTION. The theoretical or actual economy of an engine is best expressed in terms of the number of pounds of water used per indicated horse power per hour with dry steam. In all thorough tests of engine per- formance both theoretical and actual consumption are tabulated-the former as water accounted for by the indicator", and the latter as the “remainder found by subtracting from the water actually fed to the boiler" the water found to be present in the steam, as spray or super- saturation, by the calorimeter test. The water accounted for by the indicator will seldom exceed 90 per cent. and may be considerably below that, according to the condi- tions as to load, pressure, protection against radiation, leakages, etc. With simple non-condensing automatic engines the best theoret- ical economy is shown with the highest pressure, and such a point of cut off as will give a terminal pressure of one or two pounds above the atmosphere. But since the unindicated loss from leakage and condensation-is much more nearly a constant quantity than a constant percentage of steam used, it follows that increase of load diminishes. the percentage of unindicated loss, and for that reason improves the The Buckeye Engine. 641 actual economy up to the point where the increased consumption due to the increased terminal pressure begins to overbalance the gain from diminished percentage of unindicated loss. Hence, the propriety of rating engines on the basis of a mean effective pressure in excess of that which would show the best theoretical economy, at about one- fourth cut off instead of one-fifth or one-sixth. TO CALCULATE THE THEORETICAL CONSUMPTION OF WATER. C b- a Volumes 310 - 1.03 -87 lbs: +1.85 19.42 2.77: + 3.5 in. 541 -48 lbs. 19.99 B M. e. p. 42 lbs. 792 32 lbs 20.44 Fig. 413. The horse power being 33,000 foot pounds per minute is 33000X 60×12" 23,760,000 inch pounds per hour, which would be the num- ber of cubic inches of water per horse power per hour that would be consumed by an engine of one horse power driven by water instead of steam at one pound pressure. The number of cubic inches of water per pound is 27.648, hence 23,760,000-27.648-859,375 pounds of water indicated horse power per hour such an engine would consume. per With more than one pound pressure the consumption of water would be proportionately less; and with steam instead of water it would be as much less as the steam at the pressure used is lighter than water, hence the following process will give as correct results as any other: Select one or more points in the expansion curve at which it is desired to ascertain the number of pounds of water per indicated horse power per hour, accounted for by the indicator, as shown in the diagram of Fig. 413. 41 642 A Library of Steam Engineering. Three points have been selected in the diagram: A, at or near the exhaust; B, near the middle of the curve, and C, at or near the cut off. If one only be selected, it should be A; if two, A and C, which are enough for all practical purposes. The following table contains the data: MEAN EFFECTIVE PRESSURE, 42 POUNDS. ABC Total Pressure. Volume. Length of Consumption Line. 228585 32 pounds. 48 66 792 times water. 2.77 inches. 541 66 (C 1.85 87 66 . 310 66 1.03 66 C By the term "consumption line" is meant a line the length of which represents the amount of steam consumed at a given pressure as compared with the length of the diagram, thus: Line A a, as com- pared with the whole length of the diagram, represents the amount of steam consumed at 32 pounds; B b, the amount at 48 pounds; and C, the amount at 87 pounds. The lines begin at the point of given pressure in the curve and end at the compression curve, running pår- allel with the atmospheric line. The reason for stopping at the com- pression curve is that the pressure at the selected point in the expansion curve, after further expansion and release, is restored by compression at the point in the return stroke corresponding with the point in the compression curve where the consumption line meets it; so that the line being shortened by compression in proportion to its amount, the process correctly credits the gain due to compression, while the loss of power from compression is debited in the calculation of the mean effective pressure. The greater the volume of clearance, the more gradual the rise of compression pressure and the longer the consump- tion line; and it will be seen that the process also debits the loss due to exhausting from the clearance space. RULE.-Divide the constant obtained according to the rule previously given by the mean effective pressure, then divide the quotient by the length of the diagram, in inches; then multiply the last quotient by the length of each of the consumption lines, and divide the product by their respective volumes. Thus: 859375 42 =20461.3095+ Dividing the quotient by the length of the diagram (3.5 inches), we have: 20461.3095 -5846.08+ 3.5 The Buckeye Engine. 643 This quotient serves for all of the selected points. Omitting the decimals, we have the indicated rate: 5846×2.77 For point A, -=20.44+ 792 5846×1.85 For point B, -=19.99+ 541 5846×1.03 For point C, =19.42+ 310 66 It is evident that with perfectly adiabatic expansion and com- pression curves, according to the table of volumes in the chapter on Evaporative Tests," the above calculations, if based on the same table, would give the same result for all points in the curve. Such discrepancies in the results, as shown, will be found in practice, though the degree of discrepancy will sometimes be greater, and sometimes the indicated rate will be less, in the middle of the curve than anywhere else; particularly so with small engines, owing to considerable con- densation of steam by the cylinder walls soon after cut off, and the re-evaporation of the resulting water toward the end of the stroke. This process of condensation and re-evaporation will always go on, to a greater or less extent, unless prevented by the use of super- heated steam or steam-jacketing the cylinder, since the temperature of the cylinder walls will be lower than that of the initial steam, but higher than that of the terminal steam. The process of condensation frequently gets credit for more curve distortion than is really caused by it, for the reason that cut-off valve leakage is frequently responsible for most of it, especially when the terminal pressure is considerably in excess of that due to the cut off, and the condition of steam genera- tion and transmission are such as to give fairly dry steam. In such cases there will be little or no water to re-evaporate, except that result- ing from condensation in the cylinder, and that alone could only raise the terminal pressure to what it would have been without condensa- tion; so that the presence of terminal pressure in excess indicates either wet steam or leaky valves, or both. SLIDE VALVE PERFORMANCE. Engineers should understand that the several functions of the com- mon slide valve can not all be equalized, for the reason that there is unequal piston velocity during the two halves of the piston's travel, caused by the virtual shortening of the connecting rod by being deflected from the center line, as it is at all times except when the crank is on its dead centers. When the crank pin is at the middle of 644 A Library of Steam Engineering. its path between centers the piston is not at the middle of its travel, but it is nearer the shaft by an amount which may be found by sub- tracting the square root of the difference between the squares of the length of the crank and connection from the length of the connection. Thus, suppose the crank to be 12 inches and the connection 60 inches long, then we have: ✓ (60×60)—(12×12)=58.78+ inches. The distance from center of shaft to center of cross head wrist when the crank is midway of its travel between centers. Hence, 60-58.78-1.22 inches. The distance the piston is nearer the shaft than to the middle of its stroke. But in practice there is no necessity for making this calculation; all that need be known in regard to the possibilities and limitations of the slide valve and its relation to the piston movement distortion can be shown geometrically by the process here given. TO DETERMINE THE POINTS OF CUT OFF AND EXHAUST CLOSURE. Draw a circle of any convenient diameter, which for large engines may be equal to the valve travel, and for small engines may be larger- in some cases equal to the piston travel. The larger the circle the more accurate the results. This circle represents both valve and piston travel as occasion requires. In the illustration (Fig. 414) it represents an assumed stroke of 24 inches, so that when it represents piston travel it is on a scale of 1½ inches to the foot, and inch represents 1 inch. When the two travels are not so conveniently related to each other a scale may be constructed by which to make measurements for stroke events, as shown in Fig. 414, and also for valve measurements, when, as may sometimes be most convenient, it is drawn larger than the valve travel. Through the circle draw diameter line A B, and extend it beyond the circle far enough for subsequent operations. Now suppose a valve with § inch steam lap and inch exhaust lap, to determine the points of cut off and exhaust closure due to 3 inch travel and inch lead at each end. At the intersections of the circle with line A B, draw small circles, with given leads (1 inch) for radius, and from center C draw two cir- cles, one with given exhaust lap for radius, and one with given steam lap for radius. Draw lines ab and a' b', tangent to the lead circles and the steam lap circle, and parallel to each other. From points of inter- The Buckeye Engine. 645 section bb' of these lines, with the circle draw ares be and b'c' with radius equal to the length of the connection, according to the scale by which the scale equals the stroke of the piston, and with the center on the prolongation of the line A B. In the illustration the connec- tion is assumed to be five times the length of the crank, or 7 inches by the scale. The intersections cc' of these arcs, with the lines A B locates the points of cut off-A c 20 inches, and B c' about 18 inches. A a d' g'f' e F D b e d c' a' B C b E G Scale of inches for piston travel. Fig. 414. Also draw lines d e and e' d', tangent to the exhaust lap circle and parallel to a b and a' b'. Draw arcs efef' and d' g' with radius and center on line A B, locating points of exhaust closure ff' and exhaust These events are also unequal, compression A f' being 2 inches and Bf 12 inches. 9 g'. Line D E is drawn at right angles to a b and the others, and FG at right angles to A B, hence, the angle D F or E G is the angular advance of the eccentric. 646 A Library of Steam Engineering. TO DETERMINE THE CONSEQUENCES TO OTHER EVENTS OF SETTING THE SAME VALVE FOR EQUAL CUT OFF. It becomes evident by inspection that cut-off equalization must be effected at the expense of considerable lead inequality, and that the greatest lead must be at B, which corresponds with the crank end, so called, of the cylinder. We have now no use for lead and lap circle, as these must be unequal and subject to the requirements in hand. A d g' f h b C' D F h E' e d a a B f g G Fig. 415. E The valve and its travel being assumed to be unchangeable, we can not locate the cuts at will, but they must be determined as follows: The sum of the laps, equal or unequal, remains unchanged. This sum, of steam lap, being 14 inches, with that amount as radius draw an arc h h, as shown in Fig. 415, from A as the center, and tangent to which draw a short portion of line b' a, placing the ruler above B a distance equal to the estimated lead that will be required with none at A to accomplish the requirement, but make the line long enough only to cut the circle at b', from which point draw the arc b' c', locating one point of cut off at c'. Set off A c equal to B c', or B c equal to A c'; draw arc cb and line b A. Complete line b' a, making it parallel to A b, The Buckeye Engine. 647 but tangent to arc h h, giving the lead required at a' B. If this lead has been considerably misestimated when establishing point b', that point will be slightly changed, and strict accuracy will require a cor- responding correction of the resulting points of cut off; but the points of tangency with arc h h and of intersection with the circle are so close together that the error may be neglected in practice. A I' C d' e' b D' C F E' b e d aa B fg G Fig. 416. E TO LOCATE EXHAUST CLOSURE. To locate exhaust closure draw D E (Fig. 415) through center and at right angle to the other lines, as shown in Fig. 414. Bisect D' E', obtaining point C", from which set off each way on line D E, the former exhaust laps (1 inch), or draw a 3 inch circle, as shown; tangent to which draw lines d' e and e' d parallel to A b; then draw arcs d' g' e' f' d g and ef, locating exhausts at g' and g, and exhaust closures at f and f'. The latter will be unequal, being about 25 inches at end A, and 211 inches at B. Exhaust inequality is of no great consequence. It is now evident that it is not advisable to entirely equalize cuts at the cost of such great lead inequality, although if clearance, load and steam pressure were such that compression pressure would just about 648 A Library of Steam Engineering. 16 reach the initial pressure, the excessive lead at crank end would cause no loss of good running qualities, and the compression might be equal- ized by changes of exhaust laps, now about and inch; but first let it be determined what the result will be by equalizing the com- pression by adjustment alone, without change in the proportions of the valve. The compressions obtained in Fig. 415, being slightly less than 21 inches at end A, and slightly more at B, set off at f' and ƒ equal compressions of 23 inches, and draw arcs ƒ e and f' e', as shown in Fig. 416. Next, set off exhaust g slightly less than corresponding one in Fig. 414, and draw arc g d and line de', obtaining the angle. Then line e d parallel to de' and arc d' g' locates the other exhaust at g'. Obtain point C", as before, and set off steam laps C" D' and C" E' on line DE. Through points D' and E' draw lines parallel to the others, and arc b c and b' c', locating the cut offs at c and c', 19 inches and very nearly 19 inches respectively. Owing to the use of a little guess work in locating exhaust at g the sum of the exhaust laps slightly exceeds inch, which, if entire accuracy is sought, would call for a slightly earlier exhaust point at g, which would change the angle of the parallel lines and bring the exhaust lap lines d'e and de' closer together. The inaccuracy is left uncorrected, however, to show that the "cut and try" process is to some extent required when lead equality is departed from and fixed valve proportions are adhered to. TO FIND THE VALVE LAPS REQUIRED FOR GIVEN RESULTS. Having learned that entire equality of all events is impossible, it now remains to find the best possible compromise among the several distortions shown in the illustrations. The excessive lead inequality required to equalize the cuts is objectionable, if for no other reason than that many engineers would consider it a mistake or accidental mal-adjustment that should be cor- rected. Moreover, since unequal cuts are neither detrimental to the economy or running qualities of the engine, we may safely take the illustration shown in Fig. 414, so far as the steam laps and leads are concerned, and change the exhaust laps so as to give equal compression of chosen amount, determining the amount with reference to the known volume of clearance, so as to compress from three to five volumes into one, according as the speed may be high or low. The construction of Fig. 417, which will now be sufficiently under- stood without further explanation, shows equal leads and compres- sions-the former of an inch, as shown in Fig. 414, and the latter The Buckeye Engine. 649 12 2 inches. The cuts are at 18 inches and 20 inches. The exhaust laps are inch and inch, the smallest determining the compres- sion at the end of the cylinder farthest from the shaft, where the piston moves faster than it does at the opposite end. The steam laps are both ğ of an inch, as with Fig. 414. Angular advance of eccentric 28 degrees, as shown in Fig. 417. 28° Fig. 417. TO DETERMINE SLIDE-VALVE FUNCTIONS- CONDENSED FORM. Having presented the subject of slide-valve functions in detail for the benefit of the student, it will now be given in condensed form. The circle A C B D, as shown in Fig. 418, may be of any convenient diameter; but the larger it is the more accurate will be the results obtained. It represents the strokes of both the valve and piston, and if not drawn to agree with either, scales may be constructed, by one of which its diameter will equal the stroke of the piston, and by the other the stroke of the valve. In the present case it will be assumed to equal the valve travel, so that only a piston travel scale will be needed; and assuming a piston travel of 24 inches, a scale rule of 1½ inches per foot will answer. 650 A Library of Steam Engineering. Line A B is drawn through the center and extended to the length required outside of the circle. Lines a b, c d, a' b', c' d' and ee are drawn parallel to each other, and e e', shown broken, through the center. Ares bf', dg, c'h', b'f, d' g and c h are each drawn with a radius. equal to the connecting rod of the engine according to the scale by which the diameter of the circle equals the piston travel, and from C E k b 288 d A g f O B k a f' i' b F Fig. 418. centers located on the prolongation of the line A B. They are drawn from the intersections of lines a b, c d, etc., with the circle to the line A B. By the scale according to which the circle represents the valve travel-the scale in the present case-the radii of the small circles at A and B are the leads, the centers of the circles being at the intersec- tions of the line A B with the main circle, and lines a b and a' b' being drawn tangent to them. Distance Cand i D are the maximum port openings. Distance i O and i O are the steam laps, O being the center of the circle. The Buckeye Engine. 651 K O and K'O are the exhaust laps, the smallest K' O being at the crankward end of the cylinder to equalize compression. The points of cut off are ƒ and f'. They are unequal in conse- quence of the slower movement of the piston during the crank ward half of its stroke than during the other half, due to the virtual short- ening of the connecting rod when deflected from the center line. The points of the exhaust closure g g' are located at 2 inches from the ends of the stroke, in the present case. The points of the exhaust fall at h and h' as a result of all the other conditions. Angle C E is the angular advance of the eccentric, line C D being at right angles with the parallel lines, a b, c d, etc., and E F at right angles with line A B. 9 It is obvious that many variations may be made on the above lay- out, for example: by making more lead at B and less at A the cut offs ff' may be made equal. If less compression be desired distances A and g' B may be diminished accordingly, from which new points arcs gd and g' d' and lines dc and d'c' are redrawn, showing the required diminution of exhaust laps K O and K' O. Different angles for CE may also be chosen, showing effects on the several events, all of which may be done correctly by the student by following these instructions. COMPOUND ENGINES. RATIO OF CYLINDERS. 10 10 The best ratio of capacity of cylinders depends on the range of pressures, that is, on the ratio of initial pressure to pressure at release. For example, suppose the initial to be 115 pounds, and the steam is to be expanded until its total pressure is only 11 pounds; which would be an expansion of steam from one volume to about 10 volumes, and the ratio of expansion would therefore be 101. If the steam be cut off in the small cylinder at of the stroke, then the total capacity of the large cylinder would be 10 times of the small cylinder, for of the small cylinder represents the amount of steam admitted at 115 pounds pressure, which finally fills the large cylinder at 11 pounds pressure. But 10 times is about 31; therefore the ratio of cylinders in this case would be 31 to 1. This is about the ratio used in most of the Buck- eye Compound Condensing Engines, and while the figures employed are approximated the result obtained is accurate enough for all prac- tical purposes. A larger ratio of cylinders and a greater expansion would produce greater economy in case the load was carefully adjusted to the engine; but as that is not practical except where engines are specially designed to meet certain conditions, a ratio of cylinders of 3 to 1, for 10 to 10 expansions of steam will be found good practice. 10 652 O O O O e @ CLEGG GOESEC (0. Fig. 419. BUCKEYE TANDEM COMPOUND ENGINE-REAR VIEW. The Buckeye Engine. 653 போ Fig. 420. SECTIONAL PLANS OF CYLINDERS AND VALVE GEAR. CLEGG EDESER CO CIN D 654 A Library of Steam Engineering. O O O Fig. 421. PLANS OF CROSS COMPOUND ENGINE. O O O O The Buckeye Engine. 655 O O O O O O O O Fig. 422. END ELEVATION OF CROSS COMPOUND ENGINE. 656 A Library of Steam Engineering. CHAPTER XXIII. THE PORTER-ALLEN ENGINE. This engine presents features in its valve motion distinctly pecu- liar to itself, and differs materially from others herein presented. Its central feature is a link, actuated by a single eccentric, from which sep- arate and independent movements are given to the admission and exhaust valve O O e C O B O จ O Ο O O Fig. 423. THE ECCENTRIC. The eccentric is placed on the shaft in the same position with the crank, and can not be altered from that position. The lead of the valves is adjusted by other means. The first requirement in this system is that the crank and the eccentric shall have coincident movements, and so shall arrive on their dead points, or lines of centers, simulta- neously. To insure the permanence of the eccentric in its correct posi- tion, it is formed in one piece with the shaft, and its low side is brought down to the surface of it, as shown in Figs. 425 and 426. THE LINK. The construction of the link is also shown in Figs. 425 and 426. It is of the form known as the stationary link, and consists of a curved The Porter-Allen Engine. 657 arm, partly slotted, formed in one piece with the eccentric strap, and pivoted at its middle point on trunnions, which vibrate in an arc whose chord is equal to the throw of the eccentric, about a sustaining: pin secured rigidly to the bed. The radius of the link is equal to the length of the first rod, by which its motion is communicated to the admission valves. Only the upper end of the link is here shown, being the portion used in engines which are run in the forward direction; backward- running engines require the lower end, and in reversing engines, both ends are employed. The terms "forward" and "backward" are employed here in the reverse sense of that employed in locomotive practice; that is, the forward end of a locomotive cylinder is the rear end of a stationary engine cylinder. Fig 424. In the slot is fitted a block from which the admission valves only receive their motion. This block is moved by the action of the gov- ernor, which thus varies the point of cut off. If the center of the block is brought to the center of the trunnions the port is not opened. at all, except by the lead given to the valves, and this opening is closed before the piston has advanced a sensible amount. If, on the other hand, the block is brought to the end of the slot, as here represented, the steam is not cut off until the piston has reached the half stroke, which is the usual limit of the admission; although where great power is occasionally required, the steam is admitted through five-eighths of the stroke. The exhaust valves are driven from a fixed point on the link, and have, therefore, an invariable motion. The movements of the link, at this point, are suited to this function, causing the steam, wherever may have been cut off, to be held until near the end of the stroke, it 42 658 A Library of Steam Engineering. when it receives a free and ample release, and is confined again at an earlier point of the return stroke. Fig. 425. ECCENTRIC AND LINK-SIDE AND FRONT ELEVATION. Fig. 426. THE LINK MOTION. An explanation of this motion is here given, reference being had to Figs. 427 and 428. O is the center of the shaft; A B is the path of the crank; CHDI is the path of the center of the eccentric; and EK F is the arc in which the trunnions of the link vibrate, about the center G of the sustaining pin. This arc is here divided into twelve equal parts, the terminations of which are indicated by numbers, continued through both vibra- tions, from one to twenty-four. The same numbers indicate corre- sponding points in the path of the center of the eccentric. The curved lines show the positions of the center line of the link corresponding with the positions of the eccentric and trunnions, and the progression of the numbers indicates the direction of the motion. K H and K I represent the line connecting the center of the trunnions with the center of the eccentric, at each extreme of its vibration. This link, it will be seen, is a right angle lever, of which the above line represents one arm, which is termed its driven arm, and the curved center line of the link represents the other arm, which is termed its driving arm, the whole turning on a vibrating fulcrum. The Porter-Allen Engine. 659 We will at present confine our attention to the adjustment shown in Fig. 427, to show the manner in which the link imparts the move- ments to the admission valves. When the engine is on either dead. center, the center of the trunnions of the link stand at E or F; the line connecting this center with the center of the eccentric coincides with the line of centers D F; the center line of the link stands on E M or F L, which are called the lead lines, being the arcs whose cen- ter is the joint at the other end of the valve rod; and the block can B M 12/D O H 8 20- 19 18- 17 16- -3 15- -2 14 13 K F 13 23 DIAGRAM SHOWING MOVEMENTS OF THE LINK IN INCORRECT ADJUSTMENT.. 6 Fig. 427. be moved from end to end of slot, without imparting any motion to the valves. The piston is then at the commencement of a stroke, and the valve has opened the port to the extent of the lead given to it. This lead is, therefore, the same, whatever may be the position of the block in the link, or, in other words, the engine has a constant lead for all points of cut off. The peculiar motion of the link is given to it by a combination. of the horizontal and vertical throws of the eccentric. The horizontal throw alone only moves the link from one to the other of the lead lines, E M, and F L, which motion only draws off the lap of the valves. The opening movement is produced by the tipping of the link alter- nately in the opposite directions beyond the lead lines, and these tip- 660 A Library of Steam Engineering. ping motions are given by the vertical throws of the eccentric. Its upward throw, from the point C, tips the link in the direction from the shaft, and opens the port at the further end of the cylinder; and its downward throw, from the point D, tips the link toward the shaft, and opens the port at the crank end of the cylinder. At the same time, its horizontal throw is drawing the valve back, and when, in this return movement, that point in the link at which the block stands crosses the lead line, the steam is cut off. DISTINGUISHING FEATURE OF THE VALVE MOTION. There is another distinguishing feature about the valve motion which will now be described. The angular vibration of the connecting rod causes a considerable difference in the motion of the piston in the opposite ends of the cyl- inder, retarding it in the end nearest to the crank, and accelerating it in the end furthest from it. When the length of the connecting rod equals six cranks, as it does in this engine, this difference in velocity averages 20 per cent., and at the commencement and termination of the strokes reaches 40 per cent. Now the driven arm of the link is also equal to six eccentric cranks, and its angular vibrations, to K H, and K I, coincide, in degree, as well as in time, with those of the con- necting rod, and so the trunnions of the link receive a motion coinci- dent with that of the piston, and the link gives to the valves, in open- ing and closing their ports, different velocities, accelerated at one end of the cylinder, and retarded at the other, corresponding to the differ- ence in the velocity of the piston. In these diagrams, therefore, the divisions of the arc represent also corresponding divisions of the stroke, and the movements of the link are correctly shown in their relation to the motion of the piston. This difference in the piston velocity, at the opposite ends of the stroke is plainly to be seen in any engine, and in these engines the same difference is to be observed in the admission valves. Fig. 428 is a modification of the adjustment shown in Fig. 427. The adjustment of the link, shown in Fig. 427, by which the vibrations of the trunnions terminate on the line of centers, causes the link to tip beyond the lead line, further in the direction towards the shaft than in the other direction, giving a wider opening and later cut off at the crank end of the cylinder. Fig. 428 shows the adjustment by which this inequality is corrected, and equal openings are made, and the steam is cut off at identical points, on the opposite strokes, from the commencement to the middle of the stroke. The tipping of the link in the direction from the shaft is produced by the upward movement of the point C, at the extremity of the line The Porter-Allen Engine 661 CF. If the point F were at rest, the point C would describe an arc, about F as a center, opposite to the arc C H. But the point C is com- pelled to move in the arc C H, and thus draws after it the point F, through a distance equal to the interval between these rapidly diverg- ing arcs, limiting the opening movement, and causing an early closing of the valve. On the other hand, the tipping of the link toward the shaft is produced by the downward movement of the point D at the extremity of the line D F. If the point E were at rest, the point D B N 20 19 -7 181 17- 5 16- 6 -8 S 15- 3 144 2 R 13- H IX VI III DO 12 24 -K- A 13 23 T DIAGRAM SHOWING MOVEMENTS OF THE LINK IN CORRECT ADJUSTMENT. G Fig. 428. would describe an arc, about E as a center, and this arc would not be opposite to the arc D I, but would coincide with it in direction, only being drawn with a longer radius. The point D, moving in the are D I, draws after it, therefore, the point E more slowly, permitting a wider opening and a later closing of the port. It is the same angular vibration of the driven arm of the link, which causes the motion of the valves to differ in velocity in a degree corresponding to the difference in the velocity of the piston, that also limits the opening, and hastens the closing of that valve whose motion it accelerates, and enlarges the opening, and delays the closing, of the valve whose motion it retards. 662 A Library of Steam Engineering. The action of the link would be mathematically more correct if the trunnions, instead of vibrating in an arc, moved in a direct line EF (Fig. 427) along the line of centers. Then the vibration of its driven arm would be equal to the sine of angle made by it, and also by the eccentric, with the line of the centers. The lead would, as already described, be constant, and at the mid stroke the opening and cut off on the opposite strokes would be equal, since they are deter- mined by the amount of this vibration at the point of cut off, and sine H K O equal IKO (Fig. 428). But these vibrations would be equal at the mid stroke only; as we pass backward in the stroke the vibrations for corresponding points of the opposite strokes grow un- equal, and before we reach the commencement the difference in the diagrams becomes considerable. Although by changing the line of motion of the trunnions of the link from an arc terminating on the line of centers, to this line itself, the more serious inequalities of the opposite valve movements are corrected, it still remains to remove them in the earlier part of the strokes. That is done in the following manner: If, with a radius equal to one and a half the driven arm of the link, an arc be drawn below the line of centers, so that this line is tan- gent to it at O, it will be found that ordinates drawn to this arc, from points in the path of the eccentric corresponding to equal advance of the piston on its opposite strokes, as from 1 and 13, 2 and 14, and so on, are practically equal in length. If the sustaining pin G be lowered the short distance necessary to cause the trunnions to vibrate in the arc (Fig. 428) tangent to the line of centers at K, equal valve action will be obtained on the oppo- site strokes, at every point of cut off, from the commencement to the middle of the stroke. The steam is cut off at the latter point when the driven and driving arms of the link are of equal length. The link will now, however, arrive on its lead line F L, when the crank is at A (Fig. 428), before it reaches the line of centers for its for- ward stroke, while, on the return stroke, it will not arrive on its lead line E M, until the crank has reached a position corresponding to A, after passing the line of centers; the driven arm of the link standing then on a line below and parallel with the line of centers, as shown in Fig. 428. This adjustment serves another purpose. It gives a difference of lead, to the amount above shown, in favor of the center where the motion of the piston is most rapid. This is an important feature, since, to obtain at high speed an equally good admission at the further end of the cylinder, it is necessary, not only that the motion of the valve shall be accelerated in the same degree with that of the piston, but also that the area of opening shall be proportionately enlarged, The Porter-Allen Engine. 663 and this enlarged area is, at the commencement of the stroke, where the excess of the piston velocity is the greatest, completely furnished by the greater lead. It will be interesting to the student to observe that the movements of the link are the same as those that would be imparted by a series of eccentrics of increasing throw and diminishing angular advance. To show this: At the point K (Fig. 428) erect K P, perpendicular to the line of centers, and through the center O draw the perpendicular S T, and draw CR tangent to the arc CH. Then, from the center O, draw a diagonal line, terminating on any point on the perpendicular K P. A portion of such a diagonal line will form a secant to the tangent C R. Four illustrations of these diagonals are given in Fig. 428, forming the secants, O I, O III, O VI and O IX. The identity of the link with the eccentric is then exhibited, as follows: First. The length of the secant is equal to one-half the throw of the link at the point at which the diagonal terminates. Second. The circle being equal to the distance between the lead lines, it follows that the section of the secant beyond the circle repre- sents the opening movement of the link at that point. Third. The movement of that point of the link, and the open- ing given by it, are therefore the same that would be derived from an eccentric whose throw was equal to twice the secant, and whose advance was equal to the angle made by the secant with the perpen- dicular S T. Fourth. The intersection of the secant with the circle shows the point in the revolution of the eccentric at which the full opening is found. Fifth. This point bisects the portion of the arc that is included between the commencement of the opening and the point of cut off. The diagonal represents, therefore, the line of such larger eccentric. The eccentric movement here exhibited assumes a connection unaffected by angular vibration. The motion of the eccentric imparting its movement to the link, at right angles, is in direct opposition to that of eccentrics imparting. similar movements directly. STEAM AND EXHAUST VALVES The valves are four in number, two admission and two exhaust valves, as shown in Fig. 429. The valves to the right in the illustration are the admission valves, and those to the left are the exhaust valves. There is one admission and one exhaust valve at each end of the cyl- 664 A Library of Steam Engineering. inder. They stand in a vertical position, and are perfectly balanced. Each of these valves opens simultaneously four passages, two on each face, for the admission and release of the steam. The exhaust valves open and close their ports as the center line of the link crosses the line K N. The movements at the point of the link at which connection is made with the valves are shown at top of illustration. It will be seen that the opening is made while the valve is moving swiftly, and that one-half of the opening move- ment has been accomplished when the piston arrives at the end of its stroke. The valves are so constructed that this portion of the move- ment opens the full area of the port, which does not begin to be con- tracted again until the center line of the link has recrossed the lead lines on its return. The speed of the piston is then diminishing also, and the exhaust is not throttled at all until the port is just about to be closed. DIFFERENTIAL VALVE MOVEMENT. A form of the Corliss wrist motion is introduced into the connec- tions of the admission valves, which is shown in the following plan and elevation (Figs. 431 and 432), and which effects a modification of the movements of these valves. In this movement an arm, which is connected by a rod with the block in the link, communicates, through a rock shaft, motion to two other arms, causing them to vibrate in the same vertical plane in which the valves move. Each of these arms, alternately, rises nearly to the vertical position, while the other, at the same time, descends to and beyond its dead point. Each, by a separate connection, imparts motion to one of the admission valves, and at the top of its vibration causes it to open and close its port swiftly, and then, descending to its idle arc, reduces the motion of the valve to an interval practically of rest. These movements can be followed in the illustration, where the upper arm is about to move in its arc to the left, and thus, through the lower connection, to open the port at the further end of the cyl- inder, while the lower arm will be scarcely moving its valve at all. In this manner the width of opening is largely increased, chiefly by a difference in the length of the levers, while, at the same time, fully one-half of the lap, or the useless motion of the valve after it has covered its port, is gotten rid of, so that smaller valves and narrower seats are employed, and, notwithstanding the greater opening move- ment, the total motion of the valves is very much reduced. The valves work between opposite parallel seats, the exhaust valves are nearly and the admission valves are wholly in equilibrium. Means are provided for keeping them perfectly steam tight, the exhaust valves The Porter-Allen Engine. 665 Fig. 429. 666 A Library of Steam Engineering. Fig. 430. Fig. 431. Fig. 432. The Porter-Allen Engine. 667 automatically and the admission valves by hand. The latter is accom- plished by the employment of what is termed ADJUSTABLE PRESSURE PLATES. The construction of these pressure plates, and the method of adjust- ing them, are fully represented in Figs. 429 and 430. On the right-hand side of the perpendicular section (Fig. 429) both admission valves are shown, working between their opposite parallel seats, one of which is formed on the cylinder, and the other on the pressure plates, the latter having cavities opposite the ports. The valve at the upper end of the cylinder is at the extremity of its lap, while the one at the crank end has commenced to open the four passages for admission of the steam. The vertical cross section (Fig. 430) passes through the middle of one pressure plate, and shows its form, and the means employed for its adjustment, it is made hollow, and most of the steam supplied to two of the openings passes through it. It is arched to resist the pressure of steam without deflection. It rests on two inclined supports, one above and the other below the valve. These inclines are steep, so that the plate will move freely down them under steam pressure, and also that it may be closed up to the valve with only a small vertical movement. It is prevented from moving down these inclines by a screw pass- ing through the bottom of the steam chest, the point of which, and the plug against which it bears, being of hardened steel. The pressure plate is held in its correct position by projections. in the chest on one side, and tongues projecting from the cover on the other, which bear against it near each end, as shown. Between these guides, it is capable of motion up and down its inclined supports, and also directly back and forth between the valve and the cover, from which it is separated by a space varying, in the different sizes, from to of an inch. The pressure of steam is always on this plate, and tends to force it down the inclines, to rest on the valve. By means of the screw, it is forced, against the steam pressure, up the inclines, and away from the valve. This adjustment is capable of great precision, so that the valve works with entire freedom between its opposite seats, and still is steam tight. These plates also act as relief valves. Whenever the pressure in the cylinder exceeds that in the chest, the pressure plate is instantly moved back to contact with the cover; thus affording an ample pas- sage for the discharge of water before it can exert a dangerous strain. To enable the student to get a still clearer idea of the construction and adjustment of the pressure plates, his attention is directed to the sectional views presented in Figs. 433, 434, 435 and 436. 668 A Library of Steam Engineering. C B A Fig. 433. C OF B A Fig. 434. d O Fig. 435. A C B O The Porter-Allen Engine. 669 Figs. 433 and 434 are horizontal sections, through the steam chest. at one end of the cylinder, showing the four openings in the valve B in Fig. 433, and also showing commencement of valve opening; and in Fig. 434 showing the valve B at the extreme point of its lap. Figs. 435 and 436 are vertical sections showing the pressure plate C. In Fig. 435 the bolt d has been turned forward and forced the plate C against the inclines cc and away from the valve B, producing C B O A Fig. 436. a leak. In Fig. 436 the bolt d has been turned back and the valve B has been let down to its proper working position. In the illustrations, A represents the steam port. TO SET THE VALVES. To set the admission valves place the engine on one of its dead centers; then raise the governor, bringing the center of the block between the centers of the trunnions of the link. 16 With the governor remaining up, set the valve that is about to open, 3 and give it a lead of from to of an inch, according to the size of the engine and the speed at which the engine is to run. High speed engines require considerable lead. Repeat this same operation for the admission valve on the other end of the cylinder, using the opposite dead center of the crank. On letting the governor down, the crank remaining on the dead center, it will be seen that the valve is moved a short distance. This motion of the valve, produced by moving the block from the trun- nions to the extremity of the link while the crank stands on the cen- ter, is the same in amount on either center, and takes place in the same direction; namely, toward the crank. Its effect is, therefore, to cover the port nearest to the crank, and to enlarge the opening of the 670 A Library of Steam Engineering. port furthest from it; so that the lead, which is equal at the earliest point of cut off, is at the crank end of the cylinder gradually dimin- ished, and at the back end increased in the same degree as the steam follows further. The effect of this is to equalize the opening and cut-off move- ments, so that, on setting the governor at any elevation whatever, and turning the engine over, the openings made and the points of cut off will be found to be identical on the opposite strokes, from the com- mencement up to the maximum admission. This difference in the lead is also adapted to the difference in the piston velocity at the two ends of the cylinder. In case the indicator shows that the lead of either admission valve requires to be changed, the change can be made without opening the chest, by lengthening or shortening the stem at the socket of its guide, bearing in mind that each valve moves toward the middle of the cylinder to open its port. The exhaust valves have an invariable motion, and they are set so as to open before the end of the stroke, enough to give ample lead, and close again when the piston is on the return stroke, early enough to effect the required compression. All the valves are held between pairs of brass nuts, of which the inner one is flanged. These nuts must be securely locked, and should be so set upon the valve that it is free to adjust itself between the nuts while yet sufficiently tight that no lost motion exists. To avoid the consequences of a mistake, care should be taken, before closing the valve chests, to turn the engine slowly through an entire revolution, while the movements of the valves are carefully watched, so as to be sure that they have not been set so as to bring the valves or their nuts into contact with the ends of the chest at the extreme of their move- ments. TO SET THE ENGINE. The foundation should be made of hard bricks laid in cement. The bricks should be wet, and the cement washed into every course. Time should be allowed for the cement to set before any weight is put upon it. A week, at least, is required for this purpose; four weeks would be better still. The bolts should have some play in the masonry, and the best way of insuring this is to surround each bolt with a wooden box of inch material, about 16 inches long, which is drawn up as the courses are added, and removed entirely before the engine is placed on the foundation, so the bolt holes may be poured full of cement after the setting is completed. The Porter-Allen Engine. 671 THE BED PLATE. This is lined in the usual manner by a line through the cylinder, which is bolted to the end of the bed, in line with the guides. In case the cylinder is not yet in place, it is represented by the bore in the head of the bed, and the line is to be continued mid way between the side rails of the lower guide bars. The guides lie in one plane, and are to be used for leveling the bed in both directions. The base of the bed is not brought in contact with the founda- tion. Thin parallel packing pieces are to be placed on each side of each bolt, and under each end of the main bearing, and the bed must bear equally on all these, when the guides are level in all directions, before any strain is put on the bolts. After these have been tight- ened and the guides are found to be level, the broad flange of the bed is brought to a general bearing on the foundation by running with sulphur, or by calking with iron borings, wet with water, made only slightly acid with sal ammoniac. THE SHAFT. In placing the shaft in position three requirements must be observed: First. That it be placed at a right angle with the axis of the cylinder. Second. That it be level. Third. That it lies fairly in its bearings. The shaft is readily squared. The crank disc is finished on the shaft centers, after the pin has been set, so that if its rim on the oppo- site side is equally distant from its center line, the shaft is square. It is leveled by plumbing the crank disc. When thus set it will lie fairly in the main bearing; and if the outer bearing has been correctly set, it will lie fairly in that also. This is tested by rotating the shaft entirely dry. Brightened rings will show what part of the journals have found bearings, and on lifting the shaft bright spots on the Babbitt metal will show where these bearings were. The boxes are slightly larger than the journals, and so the latter should bear along the center of the lower box and not on the sides. The journals of the shaft, if set as here directed, will, with ordin- ary lubrication, run cold from the start. Should the shaft get out of line, it may be squared by gauging between the rim of the crank disc and bases provided on the bed. 672 A Library of Steam Engineering. TO LINE THE ENGINE WITH THE SHAFT PLACED AT A HIGHER OR LOWER LEVEL. We will suppose the shaft not yet in place, but to be represented by a line tightly drawn. From two points, as far apart as practicable, drop plumb lines nearly but not quite touching this line. Then by these stretch another line parallel with the first, and at the same level as the center line of the engine, and at right angles with this, stretch another, representing this center line, and extend both each way to permanent walls, on which their terminations, when finally located, should be carefully marked, so they can at any time be reset. A G C- No. 1 E No. 2 F -D Fig. 437. The problem is to get the latter line exactly at right angles with the former. Everything depends on the accuracy with which this right angle is determined. It is done by the method of right-angled trian- gles. There are two ways of applying this method: In the first, one end of a measuring line is attached to some point of line No. 1, and its other end is taken successively to points on line No. 2 on opposite sides of their intersection, as illustrated in Fig. 437, in which A B is a portion of line No. 1, and CD of line No. 2, the direction of which is to be determined. B F and B G are the same measuring lines fixed at B, and applied to the line C D successively at the points F and G. The distance B F and BG being therefore the same, when E F is equal to EG the lines A B and C D are at right angles with each other. In the second, application is made of the law that the square of the hypothenuse of a right-angle triangle is equal to the sum of the squares of the other two sides. Example.-Let 5 feet equal the length of the hypothenuse of a right-angle triangle. Let 3 feet equal the length of the base. Let 4 feet equal the height. som The Porter-Allen Engine. 673 Then we have: 5X5-25. 3x3=9. 4X4 16. Then: 9+16=25. Square of the length of hypothenuse. Square of the length of the base. Square of the height. The sum of the squares of the other two angles. So if the above figure E F=3, E B=4 and B F=5, the angle at E is a right angle. Any unit of measure may be used a foot is gener- ally the convenient one-so any multiple of these numbers may be taken, as for example, 6, 8 and 10. Respecting the comparative advantages of these two ways, the sit- uation will often determine which is to be preferred. In the former, the diagonal being the same line, fixed at B and brought successively to the points F and G, its length is immaterial, though generally the longer the better; and the only point to be determined is the equal- ity of E F and E G, which may be compared with each other by marks on a rod. In the latter the proportionate lengths, 3, 4 and 5, or their multiples, must be exactly measured. It is better adapted to places where a floor is laid, and the measurement can be transferred by tram- mels. The result should be verified by repeating the operation on the opposite side of the intersection at E, and when so verified we have, in fact, the first process, without the additional and unnecessary trouble of determining the relative length of the lines. Care should be taken when a measuring line is used to avoid errors from elasticity. On this account a rod is often employed. Points on the lines are best marked by tying on a white thread. TO LINE THE ENGINE WITH A SHAFT COUPLED DIRECT. In this case it is supposed that the engine bed and the bearings for the shaft are already approximately in position. They are leveled by a parallel straight edge and a spirit level. To line them horizontally a line must be run through the whole series of bearings, and continued to a permanent wall at each end, and its terminating points, when deter- mined, carefully marked, as already directed. A piece of wood is tightly set in each end of each bearing, and the surface of these are painted white or chalked. Then, the middle of each piece being found by compasses, two fine lines are drawn across it, equally distant from the middle, and having a space between them a little wider than the thickness of the line. This then being strained, nearly touching those blocks, or if long, having its sag supported by them, the two marks on each block must be seen, one on each side of the line, with the line of white between them. 43 674 A Library of Steam Engineer ing. O .. EE 0.0 O Fig. 438. FRONT ELEVATION. The Porter-Allen Engine. 675 C Fig. 439. PLAN. A- B- O O οι E 676 A Library of Steam Engineering. EET இ Fig. 440. REAR ELEVATION. The Porter-Allen Engine. 677 MW!! 117 ACTION OF RECIPROCATING PARTS OF STEAM ENGINES. These parts are the piston and rod, the cross head and the con- necting rod. The student, or the engineer, who is unacquainted with the forces developed in the running of these parts, and the important function which they perform, can have, at best, but a very imperfect idea of the action of high speed engines. wpf Fig. 441. CRANK END ELEVATION-PULLEY PARTLY IN SECTION. The great desideratum is perfect smoothness in the running of all steam engines, and to accomplish that, in high speed engines, mechan- ical ingenuity has encountered an obstacle which has been hard to overcome. Nevertheless it has been overcome, and this engine pre- sents a splendid example of the triumph of mechanical ingenuity over what seemed an insurmountable obstacle for many years. The secret of it was the inertia of the reciprocating parts between the steam and 678 A Library of Steam Engineering. the crank. On every center they are at rest. At the middle of each stroke they have a velocity equal to that of the crank. This velocity is all imparted to them in the first half of each stroke, and taken from them in the last half. Now it is plain to be seen that these parts offer resistance to being put in motion, and when put in motion they offer resistance to being put at rest; and this resistance to changes in the amount and direction of its motion is termed its inertia, and this inertia throughout the universe of matter is a constant thing; and a certain force acting through a certain space is required to impart or to Fig. 442. VIEW FROM CYLINDER END. O arrest a certain motion in a given body; and the force exerted varies as the square of the motion imparted or arrested within such space. Therefore, if a double velocity is imparted, it must be imparted in one-half the time, and hence requires the exertion of four times the force. The inertia of the reciprocating parts of an engine culminates on the dead centers, those points at which their motion in one direction. is finally arrested, and that in the opposite direction begins to be imparted; and there it is equal to the centrifugal force which these parts would exert if they were revolving in the path of the crank, and at every other point in their stroke it is equal to the horizontal com- The Porter-Allen Engine. 679 ponent of this centrifugal force, at the corresponding point in the revolution of the crank. That is, the centrifugal force is merely the inertia of a revolving body, or its resistance to a continual change in the direction of its motion. THE UNIT OF CENTRIFUGAL FORCE. The unit of centrifugal force, or the centrifugal force of one pound, making one revolution per minute, in a circle of one foot radius, or 2 feet in diameter, is .000341 of a pound; and this force varies directly as the weight revolving, and as the length of the crank, and as the square of the number of revolutions per minute. So the sim- plest computation enables us, in any case, to find what this final setarding and initial accelerating force is. b a Fig. 443. For example, in an engine of 2 feet length of crank, making 50 revolutions per minute, it is 1.705 times the weight of the reciprocat- ing parts; for 50X50X2X.000341-1.705. That is, they are being put in motion 1.705 times as rapidly as gravity would do. We then mul- tiply the weight of these parts by 1.705, and we have the total force; and by dividing this by the number of square inches on the piston, we have the pressure of steam per square inch that will furnish it. At the above speed it will be insignificant, probably 7 or 8 pounds per square inch; this depends, of course, upon the size of the piston. But it increases as the square of the speed; at 100 revolutions per 680 A Library of Steam Engineering. O HHHH Fig. 444. PLAN VIEW OF TANDEM COMPOUND CONDENSING ENGINE. תס ח ט םים The Porter-Allen Engine. 681 minute it is four times as much; for 100 X 100 X 2 X.000341 = 6.82. At 200 revolutions per minute it is sixteen times as much; for 200 X 200 X 2 X.000341=27.28. The first is between 7 and 8 pounds per square inch, which does not amount to very much; the second is 30 pounds per square inch, something to be considered; the third is 120 pounds per square inch, something to be avoided. The force on the square inch having thus been found for the dead centers, we have only to measure, at the commencement of the dia- gram, a height representing it, as, for example, a b in Fig. 443, and from the point b to draw a diagonal, crossing the line of counter pressure at the middle of the diagram, and continue to the end; and we will have enclosed two triangles, the first representing, at every point, the force exerted to give these parts their velocity which they must have to keep up with the crank; and the second the precisely opposite force exerted to stop them. At the mid stroke, acceleration passes insensibly into retardation. The curved end of the lower triangle in the figure shows the extent to which the motion of these parts is arrested by the compression on the opposite side of the piston, and not by the crank. COMPOUND CONDENSING ENGINES. Fig. 444 represents a plan of a Porter-Allen Compound Condens- ing Engine, and Fig. 445 represents a side elevation. COMPOUND EXPANSION. Economy in the use of steam is the real object of compounding, but to insure economy in a compound expansion engine provision must be made for preventing too great a condensation of the steam between the point of admission into the first, or high pressure cylinder, and the point of its discharge into the second, or low pressure cylinder. Or, if the engine is of the triple or quadruple expansion type, provi- sion should be made to prevent too great a condensation between the point of admission of steam into the first and the point of its discharge into the last cylinder. If this is not provided against com- pounding can result in little, if any, benefit so far as economy is con- cerned. In the construction of this engine, however, this important mat- ter has received due attention, and all of the conditions necessary to high economy are fully met. But, in order that the student may have a clear conception of what changes take place in a steam engine cyl- inder, we will first consider the subject of conversion of heat into work. 682 2 A Library of Steam Engineering. 100 0 Fig. 445. SIDE ELEVATION OF TANDEM COMPOUND ENGINE. The Porter-Allen Engine. 683 CONVERSION OF HEAT INTO WORK. When the steam enters the cylinder it is as invisible as in the gauge glass, but as the piston advances it grows clouded, and when it is exhausted, it bears through it minute particles of water. This is because the latent heat of the steam has been partly converted into work, as its atomic has been changed into dynamic energy. For each one horse power exerted 42.746114 units of heat must disappear per minute, since 33,000 units of work per minute make one horse power, and one unit of heat is the equivalent of 772 units of work. The unit of heat, or heat unit, is fully explained in another part of this work, but it will bear repetition here. The heat unit is the amount of heat required to raise one pound of water one degree of the Fahren- heit scale from a temperature of 32°. It has also been observed that the heat units increase at the rate of .305 for every degree of temper- ature above 212°. TO DETERMINE THE NUMBER OF HEAT UNITS EXPENDED PER HORSE POWER. RULE.-Divide the number of units of work (33,000) in one horse power by the units of work (772) performed by one unit of heat. Then we have: 33000 772 -42.746114- Units of heat expended in one horse power exerted. At the pressure of 80 pounds per square inch by steam gauge, one pound of steam generated at that pressure requires an expenditure of 1212.6 units of heat to raise the water from zero to 80 pounds pressure and evaporate it at that pressure, and one pound of water at that press- ure has required the expenditure of 326.6 units of heat to raise it to. that pressure from zero. So that 886 units of heat are required to evaporate one pound of water from and at a pressure of 80 pounds. per square inch. There must, therefore, be found by condensation in the cylinder, nearly 3 pounds of water per hour for each one horse power exerted. 42.746114X60 886 -=2.89476+ Pounds of water condensed per horse power per hour. If this steam is expanded to and exhausted at the atmospheric pressure, .34 of a pound of water will be re-evaporated by the heat set free during the expansion, by the fall of the boiling point, which has fallen from 326.6 heat units to 212.9 heat units, showing a loss of 113.7 heat units. 684 A Library of Steam Engineering. At the atmospheric pressure one pound of water contains only 212.9 units of heat, and 965.7 units are required to evaporate it at that pressure. Then we have: 326.6 212.9 113.7 Heat units expended. Then, multiplying the heat units expended by the weight of water condensed per horse power per hour, and dividing the product by the heat units required to evaporate one pound of water at a temperature of 212° Fahrenheit, or 212.9 heat units, we have: 113.7×2.895 965.7 -=.34+ Decimals of a pound of water re-evaporated. This leaves 2.555 pounds of water which must be exhausted in the form of water unaccounted for by the indicator. If the expansion is carried down to 4 pounds absolute pressure then one pound of water would contain only 153.396 units of heat, but it would require 1007.229 units to evaporate it; so that .5 of a pound would be re-evaporated, leaving 2.395 pounds unaccounted for by the indicator at the termination of the expansion. If this steam were exhausted at only 2 pounds pressure absolute, a slight further re-evaporation would take place, but there must still remain in the exhaust 2.33 pounds of water per horse power per hour. The exhaust must therefore always be wet. Why is not this disappearance of the steam amounting, in ordin- ary good practice, to 10 per cent. of the whole quantity, shown by the indicator? The answer is twofold: First, that which takes place before the steam is cut off the indicator can not show; second, the steam which thus disappears during the expansion is restored-(1) by the re-evaporation of a small part of the water formed by the work done on the expansion itself; and, (2) by the partial re-evaporation of the water present in the cylinder at the point of cut off. This latter embraces the water formed by work done before the cut off, that which was formed by condensation of the entering steam, and that which the entering steam brought with it, either from the boiler or condensed in the pipes and chest. In the case above supposed of 80 pounds ini- tial pressure, exhausting to the atmosphere, not one-twentieth of the water formed during the expansion is re-evaporated, because the aver- age fall of the boiling point from the successive points in the expan- sion, is less than one-half its extreme fall; but of the water in the steam at the point of cut off, between one-eighth and one-ninth will reappear as steam, as the boiling point falls. (965.7-113.7-8.5-.) The Porter-Allen Engine. 685 This re-evaporation, where there is no steam jacket or super- heating, always restores the whole loss of pressure from work done on the expansion, and generally adds to it, sometimes considerably. In fact, it requires a good deal of additional heat to be imparted to the steam, in one or both the above ways, to bring the expansion curve below the isothermal line, or that theoretical curve that is obtained when the product of the volume into the pressure is a con- stant quantity. Although the pressure at the end of the expansion thus always accounts for all the steam in the cylinder at the point of cut off, and often considerably more, still the water supplied to the boiler is not accounted for. The fact, always observed, that the higher the terminal pressure above the isothermal line, the smaller the portion of the water that is accounted for, is at least partly understood, when we remember that this excess is produced by re-evaporation of water in the cylinder, which, so far as the heat of vaporization is supplied from the water itself, by fall of the boiling point, will represent from eight to nine times as much water remaining unevaporated. In addition to this, however, an uncertain quantity of heat is imparted during the expansion by the hotter surfaces. Having thus described this phenomenon of the conversion of heat into work in the cylinder, and explained some of the indications of the diagram, we will now consider the obtainable limit of economy. ATTAINABLE LIMIT OF ECONOMY. The heat that must be lost by conversion into work is a fixed quantity, namely, 2565 units per horse power per hour. 33000X60 772 2564.76683+ The practical problem is, how little heat can be put into the water and supply this quantity for conversion into work? At 100 pounds pressure each pound of steam contains 1217 units of heat, of which, for a condensing engine, about 1100 units need to be supplied in the boiler. The above number, 2565 is .0933 of heat thus added to 25 lbs. of water. (6 (6 2565" .1166 66 66 20 (( (6 66 66 2565" .1555 (( 15 (( 66 66 2565" 2000 (( (C 11.66" (C An economy better than the third one is attained by this engine. Theoretically, this economy is to be attained by twenty-fold expan- 686 A Library of Steam Engineering. sion, in two cylinders. Three-fourths of all the work is then done by the expansion of the steam, and, with a vacuum in the cylinder of 26 inches of mercury, the steam accounted for by the indicator will not exceed 10 pounds per horse power per hour. lows: But why not expand to the same extent in a single cylinder? There are three important reasons. The principal one is as fol- Heat is lost in cylinders by being imparted to their interior sur- faces by the entering steam, which condenses on them, and during the exhaust is re-evaporated, taking up again the heat it had parted with, and conveying it to the atmosphere or the condenser. Water brought over with the steam does not add to this action; so also water formed in doing work passes out unevaporated, except by the fall of its own boiling point; since no more water can be evaporated from the surfaces than has been condensed upon them. But this alternate condensa- tion and re-evaporation must always go on to the fullest extent possible under the conditions, which, with a metal of given conducting power, are the time, the surface, and the difference between the initial and the exhaust temperatures. The cylinder in this case, however, must be large and the surface extensive. The difference of temperature between steam of 100 pounds gauge pressure and a vacuum of 26 inches, is 337°-126°-211°, or 31° more than the difference between boiling water and ice. Under these con- ditions, of the small amount of heat entering the cylinder at the com- mencement of each stroke, by far the greater part would be lost; being locked up in the metal while the work was being done. Steam jacketing and superheating are the means employed to pre- vent or diminish this action; but this case is quite beyond their power to remedy, unless the superheating is carried to an extent that would be destructive. On account of this loss of heat, it has been practically settled that, even with the use of the steam jacket and fairly super- heated steam, six expansions are about as many as will give any gain in a single cylinder. But these give, theoretically, only 60 per cent. of the gain that is to be derived from a twenty-fold expansion. The second reason why a twenty-fold expansion in a single cylin- der is not of any advantage is, that the waste room is then about equal to the piston displacement, up to the point of cut off. It is obvious that, on this account also, six expansions are as many as can give any gain. The third reason is, that such extreme variations of pressure, at the opposite ends of the stroke, are undesirable from a mechan- ical standpoint. By the use of the two cylinders, according to the plan here pre- The Porter-Allen Engine. 687 sented, the steam is cut off in each one at from one-fourth to one-fifth of the stroke, and we gain in the following ways: 6 In the first, or high pressure cylinder, in round numbers, the fall of temperature is one-half as great, and the extent of surface is one- quarter as great, giving only one-eighth the amount of alternate con- densation and re-evaporation. Now this amount can be almost entirely prevented by steam jacketing and superheating. If no water has been condensed on these surfaces they can be cooled during the exhaust only by evaporation of the small portion of the water from work done diffused through the steam that comes in contact with them, and the heat thus lost may be restored by conduction from the jacket, and by loss of the superheat, without causing any condensation of the enter- ing steam. If any heat should be carried away it is not carried direct to the condenser, but is utilized in the second cylinder. Again, the waste room bears but one-fourth or one-fifth the pro- portion to the piston displacement up to the point of cut off that it did in the single cylinder, and the compression, commencing at the same pressure at which the expansion terminates, or higher, is carried up to the initial steam pressure, or nearly so, thus avoiding loss from the waste room in this cylinder almost entirely. So much for the first cylinder. We now come to the point where compounding frequently fails. In theory the work done in the second cylinder is but a continuation of the expansion, as if the first cylinder were five times as long, and the expansion, instead of stopping at 25 pounds, would continue down to 5 pounds pressure. This being impractical, both for mechanical and refrigerative reasons, the problem is to transfer the steam, without loss, to another cylinder four times as large, in which the expansion can be completed. But here is the difficulty. Usually a large part of the steam disappears in the transfer. This disappearance is caused partly by the filling of passages and clearance, but chiefly by conden- sation of the steam as it enters the low pressure cylinder. There being no boiler to furnish an unlimited supply, but the supply being only the quantity discharged from the first cylinder, the loss from condensation of the entering steam becomes plainly apparent. The fall of pressure is hardly ever less than 8 pounds, while it frequently reaches 25 pounds in some engines, and consequently a corresponding loss of heat. How to prevent this loss in the second cylinder without superheating has been the question. The exhaust from the high pressure cylinder is already charged with water from the work done, and which must escape mostly unevaporated. If the steam in this state must go into the low pressure cylinder, there can be no hope of first-rate economy. 688 A Library of Steam Engineering. STEAM JACKETED INTERMEDIATE RECEIVER. Mechanical ingenuity has devised a number of contrivances to prevent any undue loss of heat in the steam in passing from the first to the second cylinder; an intermediate receiver, provided with a steam jacket, is employed. The pressure maintained in the receiver is that to which the steam is expanded in the high pressure cylinder, or above this; there is no fall of pressure suffered in exhausting from this cylinder. The low pressure cylinder takes its steam from the receiver as from a boiler, and cuts it off and works it expansively, as a separate engine would do. This action is shown in the following diagrams, taken from one of these compound engines, 12 inch and 21 inch by 24 inch stroke, at 180 revolutions per minute. Fig. 446. Fig. 447. The steam exhausted from the high pressure cylinder is wholly accounted for in the low pressure cylinder. There is a slight fall of pressure sufficient to produce the current through the ports, but this only requires the cut off to take place a little later. The product of the volume into the density is everywhere the same, giving the same weight of steam from the cut off in the high pressure cylinder to the exhaust in the low pressure cylinder. To effect this the receiver is formed to serve first as a separator; part of the water contained in the The Porter-Allen Engine. 689 steam is caught and drawn off by a trap. This is let run to waste, being greasy and little of it. The steam thus partly freed from water is dried and sufficiently superheated by the jacket, so that it enters the second cylinder under very nearly the same advantages that it had at first. Both cylinders are steam jacketed, body and heads, and the spe- cial advantage in applying the jacket to the low pressure cylinder, is in the greater difference of temperature, combined with the more extended surfaces, which increases its efficiency to a very great extent. The object here is, by means of superheat retained, and by radiant heat from the walls, to supply, as it is required, the heat to be converted into work, so as to obtain in this cylinder, and to maintain to the end of the stroke, all the pressure possible from the steam supplied by the first cylinder, without re-evaporation except from fall of the boiling point. This cylinder is kept dry and the temperature low, therefore, by proper lubrication, no injury can result to it. There are no passages between the cylinders to be filled anew at each stroke, the receiver extends from valve to valve, and only about. 5 per cent. waste room has to be suffered. The water condensed in the jackets is returned to the boilers by a separate pump placed below the receiver. The diagrams shown in Figs. 446 and 447, were taken from one of these compound engines whose cylinders give a ratio of 1 to 3 by 24 inch stroke. The better ratio between the cylinders, especially for lar- ger engines, would be 1 to 4. In this engine the full indicated horse power was to be 200, with 100 pounds initial pressure; the work, how- ever, was not all on, the average load being 105 horse power. These diagrams were among the largest taken, and represent 115 horse power. The cut off in the high pressure cylinder was impared by a long- pipe, with several elbows; that in the low pressure cylinder is given by corresponding port openings. The waste room in clearance and. port is excessive, being 6.5 per cent. in the high pressure cylinder, and 8.33 per cent. in the low pressure cylinder. The steam was expanded about sixteen times, and the same weight of steam was accounted for on the diagrams, from the cut off in the first cylinder to the final release in the second. The water drained from the receiver was 50 pounds per hour, and that from the jackets 210 pounds per hour, making together about 2 pounds per horse power per hour. This was thrown away, as the means of returning it to the boilers were not then provided. The steam was not superheated, and in fact was. not dry, as no means for moderate superheating had at that time been found, the want of which seriously affected the result. 44 690 A Library of Steam Engineering. Fig. 448. MARKE The Porter-Allen Engine. 691 The steam accounted for by the indicator averaged 11.75 pounds per horse power per hour, and the quantity drained, as before described, brought this up to 14.25 pounds. The quantity of water pumped into the boiler was 18.5 pounds, showing 4.25 pounds still unaccounted for. This result shows just what the system is capable of and what was required to perfect it. INDEPENDENT AIR PUMP AND CONDENSER. Fig. 448 is a perspective view of independent air pump and con- denser, used in connection with these engines. INDICATOR DIAGRAMS, The following cards have been taken from various sizes of these engines and under different conditions: Fig. 449. Fig. 450. Steam pressure, 80 pounds. Diameter of cylinder, 6 inches. Stroke, 12 inches. Revolutions per minute, 305. Piston speed, 710 feet. Horse power, front end, 6.28. Horse power, back end, 6.45. Total horse power, 12.73 Water consumption per horse power per hour, by card, 22.13 pounds. 692 A Library of Steam Engineering. noo bas quug Fig. 451. Fig. 452. Steam pressure, 55 pounds. Diameter of cylinder, 12 inches. Stroke, 20 inches. Revolutions per minute, 230. Piston speed, 766 feet. Horse power, front end, 24.15. Horse power, back end, 23.65. Total horse power, 47.80. Water consumption per horse power per hour, by card, 22.10 pounds. The Porter-Allen Engine. 693 Fig. 453. Fig. 454. Steam pressure, 68 pounds. Diameter of cylinder, 16 inches. Stroke, 30 inches. Revolutions per minute, 125. Piston speed, 625 feet. Horse power, front end, 36.48. 0 Horse power, back end, 37.05. Total horse power, 73.53. Water consumption per horse power per hour, by card, 18.72 pounds. 694 A Library of Steam Engineering. Fig. 455. asde Fig. 456. Steam pressure, 63 pounds. Diameter of cylinder, 54 inches. Stroke, 66 inches. Revolutions per minute, 82. Piston speed, 902 feet. Horse power, front end, 695.1. Horse power, back end, 687.3. Total horse power, 1382.4. Water consumption per horse power per hour, by card, 19.26 pounds. The Porter-Allen Engine. 695 Fig. 457. Fig. 458. TANDEM COMPOUND ENGINE. Initial steam pressure, 90 pounds. Diameter of high pressure cylinder, 12 inches. Diameter of low pressure cylinder, 22 inches. Stroke, 20 inches. Revolutions per minute, 160. Piston speed, 533 feet. Horse power high pressure cylinder, 45.5. Horse power low pressure cylinder, 56.73. Total horse power, 102.23. Water consumption per horse power per hour, by card, 14.85 pounds. 696 A Library of Steam Engineering. Fig. 459. Fig. 460. Initial steam pressure, 120 pounds. Diameter of high pressure cylinder, 24 inches. Diameter of low pressure cylinder, 46 inches. Stroke, 42 inches. Revolutions per minute, 116. Piston speed, 812 feet. Horse power high pressure cylinder, 444.2. Horse power low pressure cylinder, 424.1. Total horse power, 868.3. Water consumption per horse power per hour, by card, 12.87 27 pounds. The Porter-Allen Engine. 697 TABLE OF PORTER-ALLEN HIGH PRESSURE ENGINES. Horse Power Con- stant for One Revolution. One Pound M. E. P. DIMENSIONS OF CYLINDER. SPEED. Revo- Piston Bore Stroke lutions in of per Speed in Feet per Inches. Piston. Minute. Minute. CAPACITY IN HORSE POWER. 85 POUNDS INITIAL PRESSURE. Rated Horse Pow'r at approxi- mately Economical Range, Maxi- mum Horse to Cut Off. Power 4 Cut Off. 2 Cut Off. 10 11 10 12 BORIO2 8 16 12 9 16 12 6262 280 746 .0040 45 39 300 600 .0044 53 44 3868 56 72 66 84 280 746 .0051 60 50 71 92 300 600 .0057 64 52 80 102 20 230 766 .0079 75 63 90 108 12 300 600 .0069 80 68 95 122 12 14 290 677 .00788 91 74 114 148 12 20 230 766 .0114 107 91 131 170 13 14 290 677 .00927 107 87 134 174 14 14 290 677 .01077 124 101 156 202 13 24 200 800 .0166 130 112 160 208 14 16 275 733 .01226 134 109 168 218 15 16 275 733 .01410 154 125 194 251 14 24 200 800 .0200 160 140 200 260 16 16 275 733 .01607 175 142 221 286 16 18 255 765 .0182 180 145 225 290 16 20 230 767 .0200 180 145 225 290 17 18 255 765 .0206 200 175 250 325 16 30 165 825 .0304 200 175 250 325 18 18 255 765 .0231 225 185 272 348 18 20 230 766 .0257 235 206 295 384 18 30 165 825 .0385 250 222 317 412 20 20 230 767 .0314 285 230 360 455 20 36 140 840 .0571 320 279 399 519 22 24 200 800 .0460 350 295 450 575 22 36 140 840 .0690 400 338 483 627 24 24 200 800 .0545 410 345 509 652 24 42 125 875 .0958 480 419 598 778 26 42 125 875 .1126 560 492 703 914 28 42 125 875 .1304 645 570 815 1,059 28 48 112 900 .1490 670 586 838 1,089 30 42 125 875 .1482 740 656 910 1,289 32 48 1123 900 .1949 870 767 1,096 1,425 36 48 112 900 .2467 1,100 971 1,387 1,803 40 48 112 900 .3046 1,360 1,199 1,713 2,227 44 48 112 900 .3686 1,600 1,451 2.073 2,695 44 66 82 902 .5066 1,650 1,493 2,146 2,750 48 66 82 902 .6030 2,000 1,730 2.472 3,213 48 72 75 900 .6580 2,000 1,730 2,472 3,213 54 66 82 902 .7634 2,500 2,180 3,072 4,000 698 A Library of Steam Engineering. CHAPTER XXIV. THE WEIR-HARDEN COMPOUND ENGINE. This is a recent invention, and bids fair to become a most formi- dable competitor for first place among the leading American steam engines. It is a separate and distinct type, and presents many new and novel features worthy of careful study on the part of engineers and students of steam engineering. It is therefore presented in detail, with full instructions in regard to construction and arrangement of valve gear, shaft, governor, method of adjusting and regulating the same, and setting and adjusting the valves and valve gear. This engine is provided with two valves which regulate and con- trol the flow of steam to and from the cylinders, as shown in Figs. 461, 465 and 466. Fig. 461 is a front elevation of vertical engine partly in section. Fig. 462 is an end elevation of vertical engine. Fig. 463 is the rear pillow block. Fig. 464 is a cross section through the cylinder and steam chest. Figs. 465 is a longitudinal section of steam and exhaust chambers of steam chest, showing the arrangement of the valves, steam passages, balancing and other features of the valves. Fig. 466 is a longitudinal section through the center of the cylin- der, piston, valves and steam chest. A (Fig. 465) is the cut-off valve; B the main valve; C the relief valve; DD DD, openings in the side of the steam chest for the admis- sion of live steam; 1, balance recess of cut-off valve; 22, steam pas- sages through cut off valve connecting auxiliary recesses 5 5, with bal- ance recess 1; 33, openings into cut-off valve for main steam supply; 4, main admission port from supply openings 3 3 for admission of steam alternately into ports 9 9, in center partition plate A', which plate separates the live steam part of the chest from the exhaust steam part of the chest, the whole being cast in one casting; 5 5, auxiliary recesses in the under side of the cut-off valve; these communicate alternately with the outer edges of the valve seat of partition A' and recess 10, and admit steam to ports 9 9 from the live steam chest, and to ports 99 from port 4, the whole forming a multiple steam open- ing equal to four times the amount of valve movement, so that when The Weir-Harden Compound Engine. 699 00 Fig. 461. FRONT ELEVATION, PARTLY IN SECTION. O O Q Felger Cin 700 A Library of Steam Engineering. க 60 அ Fig. 462. END ELEVATION. Folger Gn. The Weir-Harden Compound Engine. 701 Fig. 463. END ELEVATION OF PILLOW BLOCK. Fig. 464. CROSS SECTION THROUGH CYLINDER AND STEAM CHEST. 702 A Library of Steam Engineering. the valve has traveled one-eighth inch after commencement of opening the combined opening will be equal to one-half inch; 6 6, cross packing strips in the upper surface of the cut-off valve; there are also two side strips, not shown in the engraving, the whole forming a steam tight enclosure of proper area for balancing the valve; 7, a plate cast on the under side of the steam chest cover, which forms a surface for the move- ment of the packing strips; 8, space cored between 7 and the steam chest cover for the reception of lubricants, which, for locomotives, are generally admitted through the steam chest cover; 99, main ports leading from the cut-off valve through partition plate A' forming port 11; 10, auxiliary recess in the seat of the steam chest partition, com- municating alternately with recesses 5 5 for the transmission of steam to ports 9 9; 11, port formed at the junction of 9 9 for the admission of steam into the main valve; 12 12 12 12, ports through the main valve, connecting with spaces 13 13 13 13, forming passageway for the steam to ports 16 16 from port 11; lower spaces 13 13 form a passage- way for the steam from port 11 into the high pressure part of the cyl- inder through ports 16 16 alternately, and from port 16 of the high pressure to port 17 of the low pressure part of the cylinder-the upper spaces 13 13 are employed for counterbalancing the pressure in lower 13 13, and for conveying steam into ports 12 12; 14, a small passageway through the center bridge of the main valve B to recess 18, to counterbalance the pressure from port 11; 15 15 15 15, cross packing strips in the upper side of the main valve-there are also two longitudinal strips, one on each side of the valve, forming a bear- ing against the end of the cross strips, the whole enclosing an area sufficient to balance the valve-the side strips are not shown in the illustration; 16 16, ports leading into the high pressure part of the cyl- inder; these ports answer the double purpose of a passageway for the high pressure steam into the high pressure part of the cylinder and exhausting it after expansion into the low pressure part of the cylinder through recesses 13 13 and ports 17 17; 17 17, receive the exhaust steam from high pressure ports 16 16 for low pressure expan- sion, after which they serve as passageways for the exhaust steam from the low pressure ends of the cylinder to the exhaust steam chest 19 19; 18, recess in valve seat for the main valve, for counterbalancing pressure on top of main valve from port 11; 19 19, exhaust steam chest, this has two openings, one at each end of the cylinder, leading into a longitudinal passageway to an outlet into a heater or to the atmosphere; or, as in the case of a locomotive, through the saddle to the exhaust nozzle; a the valve stem yoke of the cut-off valve; a' the valve stem of the cut-off valve; b the valve stem yoke of the main valve; b' the valve stem of the main valve. 4 The Weir-Harden Compound Engine. 703 NKA A AA AA A 9 / A A A X DJ J a D 1 a 13 b' B- B 13- 19 17 D 17 Fig 465 LONGITUDINAL SECTION THROUGH CENTER OF STEAM CHEST. 19 Folger Cin. 704 A Library of Steam Engineering. E (Fig. 466) the cylinder, is a continuous casting; d d, partition ring cast in the center of its bore. This ring is of such depth as to give the differential area between the high and the low pressure piston surfaces, which surfaces being made in a ratio of about 3 to 1; e, steam packing ring in partition ring d, forms a bearing on the outer surface of the shell f of the piston F. This piston is made in three parts, two followers g g, and a cylindrical shell ƒ interposed between them. It will therefore be seen that there are four piston faces in one cylinder, two annular surfaces h h of a smaller area for high pressure steam, and two h'h' of a larger area for low pressure steam. There are two steam packing rings i i, one in each follower. All of the steam packing rings are of the Harris type, made in sections, with a spiral spring at each joint, set into a socket just large enough to hold and guide it. In the partition ring e, the springs are placed in a reverse position from those in the followers, so as to close the ring around the barrel, or cylindrical shell, of the piston. The piston rod j passes through both followers and the cylindrical shell, and holds all of the parts of the piston firmly together by means of a steel collar and shoulder against one of the followers, and a nut on the other to tighten up against it. The ports are designed to allow a large flow of steam, and being long and narrow, give a quick opening and equally quick closure. The high pressure ports are very short from the cut-off valve to the cylin- der, and much longer than the diameter of the high pressure part of the cylinder. The low pressure ports have an easy bend and are pro- portioned to suit the area of the low pressure part of the cylinder. The clearance in the high pressure ports is larger during expansion than during compression. This holds up the terminal pressure and gives a higher initial pressure for the low pressure part of the cylinder, while the clearance during compression is small; therefore, the waste of steam in filling the ports is materially reduced, and the necessity for early compression is, to a great extent, obviated. The steam is expanded directly from the high pressure side of the piston to the opposite or low pressure end of the piston; consequently, the fall in pressure from the terminal of the high to the initial of the low is very small without compressing the low-pressure steam. THE MAIN VALVE. The main valve is a plain skeleton slide valve, and is perfectly bal- anced by means of its own steam pressure. It gives the high pressure port a large opening before the piston starts on its return stroke, and when the piston has traveled a short distance the port is wide open, and remains so until the piston has traveled from one-half to five- eighths of its stroke. The compression can be varied without affecting The Weir-Harden Compound Engine. 705 ㅂ ​PYATTOODUSS h E F g 60 j h h g h' Fig. 466. LONGITUDINAL SECTION THROUGH CENTER OF CYLINDER. Folger: Cin 45 706 A Library of Steam Engineering. the cut off, and the travel of the valve is less than that in general prac- tice. The exhaust ports from the low pressure end of the cylinder are open one-half the width of port when the engine is on its center, hence, giving a free exhaust of steam on account of the size of the ports. This valve performs all the functions for controlling the admission of steam to the high pressure part of the piston, its passage from the high to the low pressure part of the piston, and the exhaust from the low pressure ends of the piston to the exhaust steam chest. BALANCING FEATURE OF THE MAIN VALVE. The main valve B is balanced by means of the ports 12 12 12 12, spaces 13 13 13 13, recess 18, and passage 14 through the center of the valve, as shown in Fig. 465. When the valve is in the center of its travel, and the central bridge closes port 11, and subjects the central portion of the valve to pressure from above, steam has access to recess 18 in the valve seat below through passage 14, and counterbalances from below the pressure on top of the valve, as the area of the recess is nearly equal to that enclosed by packing strips 15 15 in the central bridge of the valve. When port 11 is open and the steam has access to port 16 on that side of the valve, the valve is balanced on that side by the steam pressure being equal in spaces 13 13 on the same side. The ports 16 16 are always wide open, so that when the valve is admit- ting steam to the cylinder on one side, the expansion pressure of steam from the cylinder on the opposite side has access to upper space 13, through passage 12 12, and equalizes the pressure in lower space 13, and thus balances the valve on that side. The area on top of the valve subjected to steam pressure is just enough in excess of that at the lower face subjected to pressure to insure the valve being kept down on its seat just hard enough to keep it tight and free from leakage. The steam exhausting from the large or low pressure part of the cylinder has very little effect on the valves, as it exhausts into the exhaust steam chest at the outer edges of the valves. 10 THE CUT-OFF VALVE. The cut-off valve can be operated to cut off from the initial line to any point up to 1% of the engine stroke without any wire drawing of steam, or dropping of the steam line at high piston speed, as shown in diagrams 467, 468, 469, and 470, taken from this engine: Diameter of high pressure cylinder, 10 inches. Diameter of low pressure cylinder, 20 inches. Ratio of cylinders, 1 to 4 Stroke, 20 inches. Scale, 60. Engine exhausting through heater. The Weir-Harden Compound Engine. 707 Boiler pressure, 83 lbs. Revolutions, 204. Speed, 680 feet per minute. Fig. 467. 1 cut off. Boiler pressure, 90 lbs. Revolutions, 208. Speed, 693 feet per minute. Fig. 468. cut off. Boiler pressure, 77 lbs. Revolutions, 208. Speed, 693 feet per minute. Fig. 469. 1/4 cut off. Boiler pressure, 70 lbs. Revolutions, 206. Speed, 6862% feet per minute. Fig. 470. cut off. 708 A Library of Steam Engineering. The steam line is carried very nearly parallel to the atmospheric line to the point of cut off, which is sharp and well defined. The lead of the valve commences when the crank is very nearly on the center, and when it is on the center the valve has very nearly a full port opening, which opening is obtained by the number of ports and the opening of the valve on each side of the ports simultaneously and in uniform proportion, so that the valve has nearly a full port opening when the piston starts on its return stroke, and does not con- tract the opening until the point of cut off is nearly reached, when the valve closes quickly, just as it opened quickly. As the auxiliary port in front of each main port closes, the port behind it opens in the same proportion, until a point is reached where both forward and backward openings are equal to each other, and also equal to the full port opening, when they are decreased until the ports are closed, which is done in a very short movement of the valve. It will, therefore, be seen that the operation described produces a sudden opening, an almost constant full port opening during the movement of the valve, and a sudden closure. The steam flows through the center of the valve from the upper side of both ends down through the main ports and under each end directly, and also through auxiliary passages, one-half going through the ports in the valve and the other half going in at and also under the ends of the valve through recesses into ports in the partition plate, making a multiple action, as shown in the following illustrations: 1 10 9 9 8 11 Fig. 471. Fig. 471 shows the valve A, traveling in the direction shown by the arrow, and just at the point of opening. The Weir-Harden Compound Engine. 709 13 1 8 10 1/1 Fig. 472. 92 Fig. 472 shows the valve traveling in the direction shown by the arrow, and steam ports full opening. The ports open on one side as fast as they close on the other, and as the area of the ports at the valve face is twice as great as at the port entrance, the valve is required to travel only from the position shown in Fig. 471 to that shown in Fig. 472 to give a full port opening. 1 8 3 124 A 9 Fig. 473. Fig. 473 shows the valve in its extreme position, in the direction shown by the arrow, and is full port open. 710 A Library of Steam Engineering. 121 1 5 5 9 9 11 8 Fig. 474. Fig. 474 shows the valve in the center of its travel, and ports closed. BALANCING FEATURE OF THE CUT-OFF VALVE. This valve A is so constructed as to be perfectly balanced during every part of its travel. The upper surface of the cut-off valve, as shown in Fig. 474, is provided with a recess 1, so that when the valve is subjected to steam pressure under it, the steam has access to recess 1 through passages 2 2, and the area enclosed by packing strips 6 6 is equal to the area of the lower face of the valve subjected to pressure; this balancing is performed when the valve is open and the steam pressure gets under it. When the valve is closed, and no pressure is operating on the lower surface, there is no pressure on the top of the valve on the space enclosed by the packing strips 6 6. If any pressure should come under the valve, whether from the cylinder or from leak- age, it will find its way to the top of the valve through passages 2 2, and thus counterbalance the pressure on the valve below. It will therefore be seen that no matter what position the valve may be in, it will always be perfectly balanced. THE GOVERNOR. The cut-off valve is controlled by a shaft governor of peculiar con- struction, a description of which will be here given. Fig. 477 is a plan view of the governor, showing the position of the parts at approximately full stroke. Fig. 478 is a similar view, showing the position of the parts at the short or reduced stroke. Fig. 479 is a section on line x x (Fig. 477). The Weir-Harden Compound Engine. 711 Farger Guin Fig. 475. SIDE ELEVATION. 712 A Library of Steam Engineering. O Fig. 476. PLAN. Folger Cin The Weir-Harden Compound Engine. 713 Fig. 480 is a plan view of the oscillating disc. Fig. 481 is a cross section of the same. Fig. 482 is a plan view of the eccentric ring. Fig. 483 is a plan view of the pendulous lever. Fig. 484 is a modification of Fig. 477. Fig. 485 is a vertical section on line x x (Fig. 484). Fig. 486 is a sectional elevation on line y y (Fig. 484). In Fig. 479, 1 represents the hub of the governor wheel; 2 the rim; 3 the main shaft of the engine; 4 the arms connecting the hub and rim together. In Fig. 483, A represents the pendulous lever which supports the eccentric ring shown in Fig. 480. The pendulum is provided with an oblong opening 5, through which passes the main shaft 3. 6 repre- sents the center on which the pendulum is hung, and it is secured to boss 7 of one of the arms by the axial bolt passing through lug h and boss 7, as shown in Fig. 479. In Fig. 482, B represents the eccentric ring. It is grooved around its periphery, in which groove the eccentric strap is journaled. It is provided with slotted openings 8 8. 9 represents bosses on the pen- dulum A, upon which the eccentric rests and is secured by means of bolts passing through the oblong openings of the eccentrics and tap- ping into the bosses 9 (Fig. 483). This eccentric can therefore be adjusted radially upon the lever A by loosening the bolts and moving it on the bosses 9, to lengthen or shorten the valve travel by increasing or decreasing the throw of the eccentric. In Fig. 480, C represents the eccentric-moving lever plate. Its movement is rotary and journaled upon the shaft 3 (Fig. 479). It is provided with cam or eccentric guides D D'. When the device is to be used for a right-hand governor, only one such guide is employed. If it is desired to be used either right or left, a secondary cam or guide D' is provided. In Figs. 477 and 478, E represents a slide. It may be either of the pin or other suitable form to be held in the guide D, in which it jour- nals; a represents an ear to which the pendulous lever A is bolted. The opposite ear on the lever is employed only when the governor is required to run in the reverse direction. F represents centrifugal or weighted arms counterpoised by springs II. Two arms are used in the preferred form of construction, but only one may be employed, as shown in Fig. 484. These arms are each journaled upon a center, and one end of the arms is hinged to the eccentric-moving plate, and the other arm is connected to the spring which opposes the action of the weighted arms. When the speed of the governor wheel is increased, the weighted arms move outwardly and compress the spring and oscil- 714 A Library of Steam Engineering. 2 4 E A Ꮻ O I HAAAAAAAAA] 4 a E D 13 13 F 4 13 F 13 AAAAAAAAN 4 F O 4 Fig. 477 2 77 2 4 4- F h6 Fig. 479 D 13 A D' 4 C 13 C [M] F 4 Fig. 478 D Fig. 480 Fig. 481 10 10 B C The Weir-Harden Compound Engine. 715. Fig. 482 8 B A -9 Fig. 483 5 H E D1 2 4 LEHD e X + H 11 a E Db БААААЕР Fig. 486 11 A' B 3 y- C Fig. 485 4 B 1-8 B F A -F e B 3 -y Fig. 484 4 716 A Library of Steam Engineering. late or move the plate C, which moves the slide E forward in the slot D, swinging the pendulum across the shaft and changing the position of the parts from that shown in Fig. 477 to that shown in Fig. 478. When the speed decreases the springs will cause the opposite move- ment of said parts. Fig. 484 is a modification of Fig. 479, as previously stated, and shows a governor constructed with but one weighted arm F. The lever plate C' is mounted in a guideway supported upon the enlarged hub or plate H; bb represent the guide in which the lever plate C' moves. The eccentric supporting plate A', instead of a pendulous lever, is supported by and moves in slides e e. The spring 11, which counterpoises the weighted arm in this instance, is attached to the opposite ends of the lever plate C'. These parts, A' and C', are the equivalents of the parts A and C shown in Figs. 477 and 478, except that they move in straight lines instead of curved. The weighted arm F, and spring 11, are the equivalents of the weighted arms and springs shown in Figs. 477 and 478, but are placed in different relation to act in substantially the same manner. It will be observed that the slide E moves a considerable distance in the guideway D to a compar- atively short movement of the plate A', owing to the slight eccentric- ity of the guideway; consequently the slide and guideway form a lock or a resistance sufficient to prevent the friction of the valve movement from affecting the operation of the governor. It will be observed that the movement of the lever A (Figs. 477 and 478), or plate A' (Fig. 484), is nearly at right lines to the path of the eccentric guideway D and the slide E traveling therein (Figs. 476, 479, and 484); consequently it makes it impossible for the movement of the valve to affect the governor, and yet the latter operates positively to move the plate A, or A', and lengthen or shorten the valve travel. The plate C, when it is moved forward to arrest the increase of speed, trav- els in the same direction as the governor wheel and ahead of its speed. In order to prevent the bumping of the slide E, in case of extreme throw, elastic cushions 13 13 (Figs. 477 and 480) are provided for the ends of the guideways, against which the slide E will strike; but under ordinary circumstances the cushions are not required. When the governor is to be changed, say from right to left, the slide E is detached from the guideway D, and placed in the guideway D', and the levers and springs are reversed and occupy opposite quad- rants of the wheel. The main object of this construction is to provide a positive means for regulating the throw of the valve, and consequently the speed of the engine, which will readily move under slight variations of load and yet positively stop at the desired speed, avoiding oscillations or The Weir-Harden Compound Engine. 717 tremor in changing under varying speed. This is accomplished by means of the traveling plate carrying a pin or slide engaging with the cam or eccentric guide for shifting the eccentric, which travels nor- mally in right lines to the movement of the shifting plate, thereby locking the rapidly moving eccentric against the strain of the valve rod. TO SET THE VALVES. These valves, being slide valves, are set in a manner similar to that of ordinary slide valves, but as this engine contains marks pecul- iar to itself for setting the valves to avoid the necessity for removing the steam chest cover, an explanation will be necessary. TO SET THE MAIN VALVE. On the main valve stem will be found five marks, the center mark being 8 inches from the end of the steam chest; the two extreme marks represent the extreme travel of the valve, the two inner marks represent the lead or opening of the main valve port when the crank is on the dead center. This is not exactly the lead, as that is governed by the cut-off valve, and as this port is merely the passageway from the cut-off valve, it should have a large opening when the engine is on the center, so that the steam from the cut-off valve will have a free entrance. 16 For catching the marks on the valve stems, a trammel should be made of or inch square or round steel, with one end bent at right. angle, about 2 or 2 inches long, with end sharpened to a point, and the other end to be offset 1 to 1 inches, parallel to the body of the trammel-the offset to be not over 2 inches in length, with end sharp- ened to a point. One end is placed in the center punch mark in the steam chest face, and the other is employed to catch the marks on the valve stems. The crank is then placed on the dead center-if on the center toward the cylinder use the marks on the valve stem toward crank, and if on the outward center use the marks on stem nearest the cyl- inder. Then turn the eccentric in the opposite direction in which the engine is to run to a line inside of the 90 degree line, and tighten the set screw; and then turn the engine over on the other center and see if the opposite mark coincides with the trammel; if it does, the valve is correctly set; if it does not, equalize it by lengthening or shortening the eccentric rod until both are correct. A line will be found on the shaft 90 degrees behind the crank, representing the normal line; and another will be found at an angular advance toward the crank. The center line of the eccentric has also a line marked on it, which, if placed to coincide with the angular 718 A Library of Steam Engineering. advance line, will be found to give the best general results, and also will coincide with the lines on the valve stem. Lines will also be found in just the reverse position for running in the opposite direc- tion. If the compression should have to be changed, move the eccentric forward or backward from this angular advance line, as may be required; this will change the marks, as they are made according to the compression for the average condition, power, and size of the engine, as engines are often subjected to conditions that require them to be changed accordingly. TO SET CUT-OFF VALVES. On the cut-off valve stem will be found five marks, the center mark being 8 inches from the face of stuffing box, similar to those of the main valve, the two outer marks represent the extreme travel of the valve like those on the main valve stem, the exceptions being that the two inner marks represent the valve line to line over the ports without any lead. This valve is set exactly like the main valve according to the marks; the only difference being that the inner marks represent the valve line to line instead of the opening of the valve. This is done for the reason that the opening or lead in the main valve is constant, while the opening or lead in the cut-off valve may require to be varied to suit the condition of the work. The lead is then obtained after the valve has been adjusted and equalized, by simply moving the eccentric forward or backward for more or less lead. TO ADJUST THE GOVERNOR. In placing the governor in position the pendulum is set squarely opposite, or 180 degrees to the crank. The pin on which the pendulum swings is to be in the opposite position to that of the crank pin, and to get the required lead the governor is moved around the shaft, the pendulum pin following behind the crank the same as the main eccen- tric shaft in the direction in which the engine is to run, lessening the angle of 180 degrees between the crank and the pendulum as may be required. The lead can also be varied by sliding the eccentric forward or backward on the pendulum, giving a longer or shorter stroke to the cut-off valve. Any adjustment, such as lead, compression, cut off, etc., can be varied independently, and all adjustments of the valve can be made without taking off the steam chest cover, by simply using the marks on the valve stem. The Weir-Harden Compound Engine. 719 The speed of the engine can be varied about 10 per cent. each way, by tightening the spring for increase of speed, or slackening the spring for decrease of speed. For any greater variations of speed it is preferable to change the weights and springs. The governor can be set to run the engine in either direction. If it is desired to reverse the motion of the engine, it can be accomplished by taking off the weight arms and reversing them on their pins, and coupling up the connecting rods to the cam plate, and the spring rods in the opposite position, and then taking out the sliding shoe in the cam slot that connects the pendulum lug and placing it in the oppo- site slot, and setting the pin in the corresponding lug. The weight arms should be placed so that the weights will follow behind the pivot pin of the arm in whichever direction the engine is to run. 720 A Library of Steam Engineering. CHAPTER XXV. SLIDE-VALVE ENGINES. The main feature of all steam engines is the valve gear, and in this consists the chief difference in the various types of engines. It is therefore of the utmost importance that engineers should carefully study the anatomy of the various types of steam engines to enable them to acquire a thorough knowledge of the construction and opera- tion of their valve gear; for without that knowledge no engineer can lay claim to proficiency in his profession, any more than a physician can lay claim to proficiency in his profession when he does not thor- oughly understand the anatomy of the human body. In order, then, to enable engineers to acquire the knowledge so essential to a successful prosecution of their profession, it has been the purpose and aim of the author to lay down a course of instruction upon the construction and operation of the valve gear of the various distinctly different types of steam engines, in the simplest form, and yet embody all the information necessary to enable engineers to ac- quire a thoroughly scientific, as well as practical, knowledge of the valve motion of steam engines generally. Therefore, to complete the course of instruction in regard to stationary engines, the slide-valve engine will receive consideration here. The slide-valve engine is the pioneer of modern steam engines, and no her type of engine has ever gone into such universal use. Its construction and operation are therefore better understood among engineers, generally, than the construction and operation of any other type of steam engines. But as no work on steam engineering would be complete without embracing the slide-valve engine, this chapter will be devoted to an explanation and illustration of the main fea- tures of this particular engine, taking for such explanation and illus- tration two high-class and well-known makes, beginning with THE SINKER-DAVIS SLIDE-VALVE ENGINE. This engine is provided with a balanced valve, as shov in Fig. 487, which is a longitudinal elevation partly in section. Fig. 488 is a plan view of the engine. Slide-Valve Engines. 721 עד 0x0 Fig. 487. Fig. 488, Ο o O Ο O 46 722 A Library of Steam Engineering. VALVE AND CRANK MOVEMENT. Fig. 489 represents the position of the valve at the beginning of the forward stroke, and shows the lead opening. Fig. 490 shows the position of the crank when the steam port P¹ is full open. Fig. 491 shows the forward position of the crank when the steam port P¹ is closed, and also the point of the stroke at which expansion commences. Fig. 492 shows the position of the crank at which the valve closes the exhaust, and the point of the stroke at which compression com- mences on the opposite side of the cylinder. In this position port P2 is closed. Fig. 493 shows the port P¹ open to exhaust, and the to lead. open port P2 Fig. 494 shows port P1 full open to the backward or return move- ment of the crank pin. Fig. 495 is a diagram showing the various positions of the crank pin corresponding to the various positions shown in Figs. 489, 490, 491, 492, 493 and 494. TO SET THE VALVES. First. Put the engine on the dead center. Second. Place the valve central over the ports. Third. Place the rock shaft in a perpendicular position and con- nect valve rod without moving valve or rock shaft. Fourth. Place the throw of the eccentric down on the shaft, and adjust the nuts on the eccentric rod until the cam hook will drop onto the rock shaft pin. Fifth. Turn the eccentric around on the main shaft, and adjust the nuts on the eccentric rod until the valve travels an equal distance on the outer side of each of the steam ports. Sixth. Turn the eccentric in the direction toward the crank pin in which the engine is to run, until the valve has 2 inch opening, and then tighten the set screws in the eccentric. Seventh. Revolve the main shaft and put the engine on the other dead center, and note whether or not the lead is the same as it was on the original dead center, if so, the valve gear is correctly adjusted. If there is a difference, continue putting the engine from one dead center to the other and adjusting the eccentric rod until the lead is equal on both sides. The eccentric in this engine follows the crank. When putting the engine on the dead center revolve the main shaft in the direction oppo- Slide-Valve Engines. 723 P 1. P 2. P.1. P 2. 1 Fig. 489. Fig 490. 724 A Library of Steam Engineering. P 1. P 2. P1 ci Fig. 491. Fig. 492. Slide-Valve Engines. 725 ci %d T O I - P 2 Fig. 493. Fig. 494. 726 A Library of Steam Engineering. site to that in which the engine is to run, and never allow the crank pin to pass the dead center. If the crank pin should pass the dead cen- ter, turn the main shaft back about one-quarter turn and come ahead again until the crank pin comes fairly to the dead center, that will take up any lost motion that may exist in any of the working parts and bring the valve to its proper position. Fig. 490. Fig. 491. Fig. 492. Fig. 493. Fig. 489. Fig. 494. Fig. 495. THROTTLING SLIDE-VALVE NON-CONDENSING COMPOUND ENGINES. The slide valve has long been recognized as the simplest instru- ment for distributing steam to the cylinder of a steam engine. It has never been considered as economical as an automatic cut-off gear when applied to single cylinders, chiefly because it can not be made to satis- factorily cut off steam much earlier than one-half stroke, thus lim- iting the number of expansions of steam to about two, whereas the expansions found most economical in non-condensing simple engines range from three to five, according as stroke is short or long. Slide- valve engines have not until recently been constructed to carry high mean effective pressures, the poor grade of workmanship, design, and. material common to many of them being such that anything over 25 pounds mean effective pressure would seriously strain the machine. On the other hand, with stronger parts and better material and work- manship, so that 40 pounds or 50 pounds mean effective pressure can be maintained, a better economy has been secured, although not as good as that of the best long stroke four-valve automatics. Slide-Valve Engines. 727 ADVANTAGES OF THE SLIDE VALVE. The slide valve of itself possesses two prominent advantages-it gives easy relief to water, especially if the valve is located at the side of the cylinder; and it can be made and kept tight very easily, thus preventing the serious loss of leakage common to many other styles of valve gear. With the advent of the compound and triple expansion engines, the chief limitation to the usefulness of the slide valve has disap- peared, for the number of expansions does not now depend altogether upon the cut off, but chiefly upon the ratio of the cylinder areas. For instance, if two cylinders are employed, one two and a half times the volume of the other, and steam be cut off on the smaller at one-half stroke, we obtain by this means about five expansions, and the result can be accomplished by means of slide valves, one for each cylinder. This principle lies at the base of the modern marine engine design where cut offs of five-eighths and one-half are used, and yet with triple and quadruple expansion engines fourteen and twenty expansions are obtained. ADVANTAGES OF SLIDE-VALVE NON-CONDENSING COMPOUND ENGINES. When applied to non-condensing compound engines the slide valve, in combination with a throttling governor, possesses-first, one important advantage over simple automatic engines; and second, another important advantage over the automatic compounds. The first advantage consists in the fact that it can work under light loads-that is, with small mean effective pressures-more econom- ically than any simple automatic engine, especially when its high pressure cylinder is steam jacketed. There are four reasons for this: First-A smaller mean effective pressure can be obtained by means of the expansion due to a fixed cut off of a throttling compound, without expanding below atmospheric pressure, than by means of automatic gears in which small mean effective pressures are secured by early cut offs, and the expansion extends below the atmosphere, accompanied by a great loss of power and an expenditure of steam. Second-Small mean effective pressures can be obtained by moderately late fixed cut off and a throttling governor without an increase in the relative amount of cylinder condensation. With automatic cut offs the small mean effective pressures developed by early cut offs are attended by increased percentages of cylinder condensation. Third-A steam jacket on the high pressure cylinder of a throttling compound is most effective in checking cylinder condensation with small mean effective pressures, as at that time there is the greatest difference in pressure and tempera- ture between steam inside of cylinder and the steam at boiler pressure 728 A Library of Steam Engineering. in the jacket. Fourth-Clearance losses are less with the throttling compound than with automatic cut offs under light load; for with the former the ratio of clearance to volume of cylinder at cut off is constant at all loads, and the clearances are filled with steam at the throttle press- ure only and from a back pressure equal to receiver pressure; with the latter the ratio of clearance to volume of cylinder at cut off becomes larger the earlier the cut off, and the clearances are filled with steam at boiler pressure and from a back pressure equal to the atmospheric pressure. The second important advantage possessed by throttling com- pound over automatic compound non-condensing engines, lies in the fact that the power can be almost evenly divided between two cylin- ders, and the total maximum strain upon pistons is approximately proportional to the power developed. There are some automatic com- pounds which have the valve gears of both cylinders under control of the governor, so that the work of the cylinders is quite evenly divided, but the strains thrown upon the journals and reciprocating parts are as great, and often greater, under light loads than under heavy loads. The even division of load obtained by a throttle that regulates the initial and receiver pressures in proportion, and that produces small mean effective pressures without expansion below the atmosphere, per- mits the employment of the cross compound type of engine, which has been found impracticable with automatic cut off compounds, in which, with light loads, the low pressure cylinder is converted into a pump, and becomes noisy and difficult to operate. THE HOUSTON, STANWOOD & GAMBLE CROSS COMPOUND ENGINES. ADVANTAGES OF CROSS COMPOUND ENGINES. The advantages possessed by a cross compound engine with cranks at right angles, where each engine performs its share of the work, are important They chiefly consist in great steadiness of motion, as the fly wheel does not have to carry load when crank pins pass their centers; small fluctuation of speed under sudden changes of load, as nearly the entire inertia of fly wheel is available to resist rapid external disturbances; strains are divided between two engines and are not concentrated in a single machine; speed can be reduced as much as 50 per cent. on account of high efficiency of fly wheel, so that large engines may be installed at low speeds for temporary light loads; in case of repair of either high or low pressure engine, the one or the other may be operated singly under control of governor at half load; and finally pistons are accessible and are easily removed, which is not the case with the tandem engines. Slide-Valve Engines. 729 O O GOREV-LO. USG Fig. 496. HOUSTON, STANWOOD & GAMBLE CROSS COMPOUND ENGINE. 730 A Library of Steam Engineering. Figs. 497 and 498 are diagrams taken from the high and low press- ure cylinders of a 200 horse power engine of this type; they show how evenly the power is divided between both cylinders. These diagrams when combined into a single block diagram, as shown in the chapter on Indicator Practice, will show the action of steam, supposing the expan- sions to have all taken place in the low pressure cylinder only. This arrangement of diagrams indicates the losses that occur by the passage of steam from one cylinder to the other. It also shows the small amount of power required to fill both high and low pressure clearance spaces with steam by compression, as compared with a single cylinder. 111 Lbs. 110 Lbs. 25 lbs. 77 H. P. ד 32 lbs.. 34 lbs. Fig. 497. HIGH PRESSURE CYLINDER-60 SPRING. 73 H. P. 25 lbs: Fig. 498. LOW PRESSURE CYLINDER-30 SPRING. IT 31½ lbs. TO FIND THE HORSE POWER. To find the horse power of these engines, indicator diagrams have to be taken from each cylinder; the horse power of each cylinder is calculated as if each cylinder were a simple engine, and the sum of horse powers is the total power of the engine. Thus, multiply the mean effective pressure found by diagram of the high pressure cylinder, by its area in square inches, by its piston speed in feet per minute, and divide the product thus found by 33,000, the quotient equals horse power developed by high pressure engine; then multiply mean effective pressure found by diagram of the low pressure cylinder, by its area in Slide-Valve Engines. 731 square inches, by its piston speed in feet per minute, and divide the product thus found by 33,000, the quotient equals horse power devel- oped by low pressure engine. In Figs. 497 and 498 these results are- 77 for high pressure engine and 73 for low pressure engine, or total horse power equals 77+73-150 indicated horse power, or the load under which engine was working when diagrams were taken. To determine the maximum horse power that engines of this type develop, the builders have arranged a series of numbers which have been chosen for each of the thirty-three sizes they build, so that if the number representing the size of a given engine be multiplied by the boiler pressure, and by the number of revolutions and the product divided by 10,000, the quotient will give the maximum indicated horse power of the engine. Example.-What horse power will engine No. 49 develop with 80 pounds of steam at 170 revolutions? 49X80 3920; 3920X170=666400 666400 10000 66.64 Horse power. TO SET THE VALVES. The valves are set in each engine as if each was a plain slide-valve engine. Houston, Stanwood & Gamble have marks on their engines which will assist the engineer in this work. The shaft and eccen- trics are marked so that eccentrics can be located correctly for engines to run either "over or under." The valves have from inch to inch lead, according to the size of the engine. To ascertain if valve has proper opening, there are tram marks on the crank disc by which the crank can be placed exactly on center. To do this, select a piece of thick telegraph wire, about 18 inches long, sharpen ends to a point, and bend 2 inches of same square at both ends to make a trammel 14 inches from point to point. Place the engine with any crank nearly on center by the eye, then search on corner lug of outside foundation bolt for a trammel point; from it set the other point of trammel in the point easily found on edge of crank disc, at a distance of 14 inches by the wire tram. When this is done the engine is exactly on its center. This can be done for all the centers. When cranks are thus placed the lead can be examined and equalized, so that the opening is the same for all dead center positions of cranks. TO SET THE VALVES OF SLIDE-VALVE ENGINES GENERALLY. First. The steam chest cover being removed, and the valve stem and eccentric rod being disconnected, place the valve V exactly in the 732 A Library of Steam Engineering. center of its travel, as shown in Fig. 499; that is, place it so that the edge of the valve at one end will extend over the outer edge of the steam port S at that end, exactly the same distance that the edge of the valve at the other end does over the outer edge of that steam port. Second. Place the rocker arm R straight up and down, and adjust the valve stem V S so that it will connect with the valve and rocker arm without moving either from the positions in which they have been placed, and then connect the valve stem to the valve and rocker arm without moving either of them. S E V. S. R A E' E. R. Fig. 499. Third. Turn the eccentric E' loosely around the main shaft until its throw will be at a right angle with the eccentric rod when connected to the rocker arm, as shown by line A A, and then adjust the eccentric rod so that it will connect with the rocker arm without moving the rocker arm or the eccentric. Fourth. With the eccentric rod and valve stem connected, as shown in Fig. 499, turn the eccentric around the main shaft and note the distance the valve at each end has uncovered the steam ports, and if the travel of the valve from the outer edge of each steam port toward the exhaust port is equal, the eccentric rod is properly adjusted. If the travel of the valve, as described, is not equal at both ends, lengthen or shorten the eccentric rod, as may be required, and again turn the eccentric around the shaft, and if there still appears to be a difference in the travel of the valve at each end, lengthen or shorten the eccentric rod, as may be required, and continue the operation of revolving the eccentric and adjusting the eccentric rod until the valve uncovers each steam port exactly the same width. Fifth. If the travel of the valve is too great lower the eccentric- rod pin of the rocker arm; if the travel of the valve is too short raise the eccentric-rod pin of the rocker arm, and continue these adjustments until the travel of the valve is correct, Slide-Valve Engines. 733 Sixth. Place the engine on either dead center. But for simple and convenient description we will begin by placing the engine on the dead center toward the cylinder, and if the engine is to run under on the outward stroke, turn the eccentric around the shaft in the direc- tion in which the engine is to run, until the eccentric stands up at right angle with the crank; then move the eccentric toward the cyl- inder and downward toward the crank pin until the valve at the pis- ton end of the cylinder has the desired lead, and then fasten the eccen- tric securely in position. Seventh. If the engine is to run over on the outward stroke, and the engine is on the dead center toward the cylinder, turn the eccen- tric around the main shaft, in the direction in which the engine is to run, until the eccentric hangs down at right angle with the crank; then turn the eccentric toward the cylinder and upward toward the crank pin until the valve at the piston end of the cylinder has the desired lead, and then fasten the eccentric securely in position. Eighth. Place the engine on the other dead center, and if the lead of the valve is the same as it was on the first dead center, the adjustment is properly made. If the lead is not the same move the eccentric one-half the difference in the required direction, and adjust the eccentric rod the other half of the difference, and continue putting the engine from one dead center to the other, and making adjustments by moving the eccentric and adjusting the eccentric rod until the lead of the valve is equal at both ends. If the engine is put exactly on the dead center in the first instance, there will be no difficulty in setting the valve properly if the eccentric rod was properly adjusted, and the valve and rocker arm properly placed to begin with. Care must also be taken during subsequent operations to get the engine exactly on the dead center, as any failure to get the engine on the dead center will make it impossible to set the valve correctly. 734 A Library of Steam Engineering. CHAPTER XXVI. LOCOMOTIVES. The science of steam engineering applies to locomotives as well as it does to stationary or marine engines. The same knowledge of that science requisite to make a stationary or a marine engineer proficient in his calling is just as indispensable to the completion of the educa- tion of a locomotive engineer as it is to those who follow the other two branches of steam engineering. It is, therefore, unnecessary to repeat in this chapter anything which has preceded it. There are some things, however, which belong peculiarly to the profession of locomo- tive engineering, not met with in either of the other branches of steam engineering science. Therefore, to complete the course of instruction laid down in this work, and to make it a library of steam engineering in fact as well as in name, that part of the science applicable chiefly to locomotive engineering will be treated in this chapter. The chief feature of this branch of steam engineering, and the one in which locomotive engineers are, as a general rule, mostly interested, is that of valve setting. In order, then, that the instructions here given may be easily and readily understood, they will be given in their simplest form possible. But to secure accuracy in performing the work, the student is informed that the greatest care must be exer- cised in putting the engine on the dead center, for the reason that unless that is accurately done, all subsequent movements in the set- ting of the valves, or in the adjustment of any other part of the valve gear, will be incorrectly performed. There are various methods employed for putting locomotives on the dead center, but few of which insure the accuracy so indispensable to correct valve setting. There- fore, the instructions here contained will embody the most accurate methods known to the profession. HOW TO SET THE VALVES OF A LOCOMOTIVE. PUTTING THE ENGINE ON THE DEAD CENTER. The right or left-hand engine may be the first to be put on the dead center, and they may be put on either the forward or backward centers; but for the sake of simplicity and convenient description the right-hand engine will be put on the forward dead center to begin Locomotives. 735 with. But before proceeding to do that disconnect both parallel rods. unless there is a perfectly level track on which to perform the opera- tion; and then set up the forward driving box wedges just tight enough to prevent any play of the driving boxes between the jaws of the frame. Then take up all the lost motion in the back and forward ends of the main rods, and the engine will be ready to put on the dead center. The crank pins of locomotives are set at right angles to each other, that is, the crank pins on one side are set at right angles to the crank pins on the other side. In some locomotives the left-hand crank pins lead the right-hand crank pins, but in this case we will assume that the right-hand crank pins lead the left-hand crank pins. We will now proceed to show how to put the locomotive on the dead center. First. To put the right-hand engine on its dead center, revolve the forward driving wheels forward until the cross head has traveled to within a short distance of the forward end of the stroke on the right- hand side; then put a straight mark on the cross head and continue it onto one of the guides. Second. Before moving the driving wheels again put a similar mark on the cross head and guide of the left-hand engine. Third. Return to the right-hand side of the locomotive and resume turning the driving wheels forward until the crank pin has passed over the dead center and moved the cross head forward to the end of its travel and back until the mark on the cross head and the mark on the guide of the right-hand engine come exactly line and line. Fourth. Before moving the driving wheels again, and while the marks on the cross head and guide of the right-hand engine are line and line, proceed to the left-hand side of the locomotive and put a mark on the guide of the left-hand engine, so that it will be in line with the mark already on the cross head. Then find the center between the two marks on the guide on that side and place a center mark there similar to the two already on the guide. Fifth. Remain on the left-hand side of the locomotive and turn the driving wheels backward until the mark on the cross head of the left-hand engine comes squarely up to the center mark on the guide on that side, and the right-hand engine will then be practically on the dead center-the difference in the angularity of the rod when the cross head is at either end marks and the center mark, alone prevents abso- lute accuracy. The angularity of the rod when the first mark is made on the cross head and guide is exactly the same as it is when the sec- ond mark was made; but as the wrist pin of the driving wheel moves in a circle, there is a very slight difference in the angularity of the rod when the mark on the cross head is placed midway between the two 736 A Library of Steam Engineering. outside marks; but the length of main rods is generally such that the difference between the results obtained by this method and absolute accuracy is hardly perceptible; while the result generally obtained by the old method with marks on the rim of the driving wheel tire and wheel cover is uncertain; because the locomotive, being on springs, the wheel cover is liable to move up or down with the boiler every time the driving wheels are turned, and hence the mark on the wheel cover rarely, if ever, settles to the identical position occupied before the wheels were revolved. PUTTING THE ENGINE ON THE DEAD CENTER-ANOTHER METHOD. Another method, which if performed according to instructions here given, will insure perfect accuracy in putting the engine on the dead center, is as follows: E F F D CBA F F E' Fig. 500. First. Disconnect both ends of the main and parallel rods, and move the cross head of each engine to the forward striking point, so that they will be out of the way for the operations described further on. Second. Set up the wedges of the forward driving boxes just tight enough to prevent any play between the boxes and the jaws of the frames. Third. Procure a planed board for each engine, of any conve- nient thickness, 3 or 4 inches in width, about 2 inches longer than half the length of the stroke of the engine, and having at least one straight edge to work from. Draw a line square across each end of the board, at such distance apart that the distance from the line E to the line E', on the board B, as shown in Fig. 500, will be exactly equal to one-half the length of the stroke of the engine. Fourth. Fasten the board B, with its face against the back of the board A, with the face of the board A against the inner edges of the outside guides, as shown in Fig. 500, or in any other convenient man- Locomotives.ALL 737 ner; but so that the face of the board B will be in line with the mid- dle of the forward crank pin. Which position, it will be observed, is determined by the thickness of the board A, or by a number of boards. To determine the thickness of material to be placed between the face of the board B and the inner edges of the outer guides, in the space occupied by the board A, measure from the middle of the for- ward crank pin to the middle of the main rod journal; then measure the same distance from the middle of the cross head wrist pin toward the inner edge of the outer guides, and the distance between the end of that measurement and the inner edge of the outer guides will give the thickness of material required to be placed between the face of the board B and the inner edges of the outer guides. For example: If the length of the forward crank pin from the face of the hub of the driv- F BA E E' Fig. 501 ing wheel to the end of the pin is 10 inches, the middle of the pin will be 5 inches. If the distance from the middle of the main rod journal to the middle of the pin is, say 2 inches, then the distance from the middle of the cross head wrist pin toward the inner edge of the outer guides will be 2 inches, and the distance between those 2 inches and the inner edge of the upper outside guide will give the thickness of material required to be placed between the face of the board B and the inner edges of the outer guides; so that if the distance from the mid- dle of the cross head wrist pin should be, say 4 inches, the thickness of material required to fill the space occupied by the board A (Fig. 500) will be just 2 inches. After determining the required thickness of the board A, place it against the inner edges ef the outer guides F F, and as near the cross head as possible, and wedge it firmly in place by the aid of the wedge C and block D, against the inner edge of the inner guides F' F', as shown in Fig. 500, or by clamping the boards against the inner edges 47 738 A Library of Steam Engineering. of the outer guides F F, by a bolt passing through the boards A B C, as shown by the position of the boards in Fig. 501, or in any other convenient manner. After fastening this board B, as shown in Figs. 500 and 501, with the face of the board in line with the middle of the forward crank pin, adjust the board so that it will be square with the guides, and that the lines E E'.will be equidistant from the center of the piston rod. Or, in other words, so that the line E will be the same distance above the top of the piston rod as the line E' is below the bottom of the piston rod, and the board will be in its proper position. Should the board B be a quarter of an inch or so out of line with the middle of the forward crank pin, or if the lines E E' be not exactly equidistant from the cen- ter of the piston rod, it will be near enough correct for practical pur- poses. But exactness should be the aim of all engineers. Fifth. If a trammel can not be procured of the length here required, procure a board thick enough and wide enough to prevent much sagging at its center, and long enough to reach about one inch beyond the center of the board B, on the lines E E', and one inch beyond the center of the forward driving axle. Then, if the centers in the forward driving axle and crank pins are large, plug them with lead, and with the aid of compasses, locate a small center mark at the center of each end of the axle, and at the center of each forward crank pin. Then take the long board, previously procured, and construct a temporary trammel by driving a round wire nail through each end, and at such distance apart as to reach from the center of the forward driving axle to somewhere near the center of the board B, on the lines E E' (Figs. 500 and 501). Then put the point of the nail of one end of the trammel into the center of the forward driving axle, and press the point of the nail at the other end of the trammel into the board B at line E at the top, and then press the same nail into the board B at line E' at the bottom with the point of the nail in the back end of the board in the center of the driving axle. Then perform the same operation on the other side of the engine. We are now ready to put either side on the forward or back dead center, but for the sake of simplicity and convenient description, we will begin by putting the right-hand crank pin on the forward dead center; and to accomplish that proceed on the left-hand side of the locomotive, as follows: First. Place the point of one of the nails in one end of the tram- mel into the small hole previously made in the board B at line E, and bring the point of the nail in the other end of the trammel to the cen- ter of the axle, to see if there has been any change. If there has been no change, and the point of the nail at the axle comes fairly into the center of the axle, raise that end of the trammel to where the center Locomotives. 739 of the crank pin in revolving will meet the point of the nail in the trammel, keeping the nail at the other end in the hole in the board B at line E, and cause the forward driving wheels to be revolved so that the crank pin will move upward until its center comes fairly in line with the point of the nail in the trammel at that end, and the crank pin on the right-hand side of the forward driving wheel will be exactly on the dead center. Second. To put the left-hand forward crank pin on the dead center take the trammel to the right-hand side of the locomotive, and put the point of the nail at one end into the hole or center mark in the board B at line E' at the lower end of the board below the guides, and bring the point of the nail at the other end into the center of the forward driving axle, to see if there has been any change. If there has been no change, and the point of the nail in the trammel at the axle end comes fairly into the center of the axle, lower that end of the trammel down to where the center of the crank pin in revolv- ing will meet the point of the nail in the trammel, keeping the nail at the other end in the hole or center mark in the board B at line E' at the lower end of the board, and cause the driving wheel to be turned forward until the crank pin center comes fairly to the point of the nail in the trammel at that end, and the crank pin of the forward driving wheel on the left-hand side of the locomotive will be exactly on the forward dead center. Third. Return to the left-hand side with the trammel and place the point of one of the nails in the trammel into the hole or center previously made in the board B, at the line E', below the guides, and bring the point of the nail at the other end of the trammel to the cen- ter of the forward driving axle to see if any change has taken place. If there has been any change, simply put the point of the nail at the axle to the center of the axle, and make another center in the board B, at the line E', below the guides, with the nail at that end of the trammel; then keep the nail in that position, lower the other end of the trammel down to where the center of the crank pin in revolving will come in line with the point of the nail, and cause the forward driving wheels to be revolved forward until the center of the crank pin comes fairly to the point of the nail in the trammel at that end, and the crank pin of the forward driving wheel on the other side of the locomotive will be exactly and squarely on the back dead center. Fourth Return to the right-hand side of the locomotive with the trammel and place the point of one of the nails in the trammel in the hole or center mark of the board B at the line E previously made above the guides, and bring the point of the nail at the other end of the trammel to the center of the forward driving axle, to see if any change has taken place. If there has been no change, and the point 740 A Library of Steam Engineering. of the nail at the axle end of the trammel comes fairly into the center of the axle, raise that end of the trammel so that the center of the crank pin in revolving will meet the point of the nail in the trammel, keeping the nail at the other end in the hole or center in the board B, at the line E, above the guides, and cause the driving wheels to be turned forward until the center of the crank pin comes fairly to the point of the nail in the trammel at that end, and the crank pin of the forward driving wheel on the left-hand side of the locomotive will be exactly and squarely on the back dead center. is This concludes the operation of putting the engine on dead cen- ters; but in the foregoing description it has been assumed that the right-hand crank pin leads the left-hand crank pin. Still, it does not make a particle of difference in the operation of placing crank pins on the dead center. In cases where the left-hand crank pin leads the right-hand crank pin, when the right-hand crank pin is up and at right angle with the line of travel of the piston, the left-hand crank pin is on the forward dead center. In cases where the right-hand crank pin leads the left-hand crank pin, and the right-hand crank pin up and at right angle with the line of travel of the piston, the left- hand crank pin is on the back dead center. It is, therefore, obvious that where one crank pin leads the other, as is always the case in loco- motives, the other is always one-quarter of a revolution behind the one that leads, so that whenever the crank pin that leads is on the for- ward dead center the other crank pin is up and at right angle with the line of the piston travel; if the crank pin that leads is on the back dead center, the crank pir on the other side is down and at right angle with the line of the piston travel. If the crank pin that follows is on the forward dead center, the crank pin on the other side is down and at right angle with the line of the piston travel; if the crank pin that follows is on the back dead center the crank pin on the other side is up and at right angle with the travel of the piston. We are now prepared to take the first step in locomotive valve setting. TO ADJUST LOCOMOTIVE VALVE GEAR. While operations may be begun on either side of the locomotive, yet, for the sake of simplicity and convenient description, we will begin each of the series of operations on the right-hand side, after dis- connecting all of the driving rods, eccentric blades and valve stems and removing steam chest covers, and setting up the wedges for forward driving boxes so as to take up all lost motion in driving boxes; taking care to get them just tight enough to prevent any forward or back- ward movement of the driving boxes between the jaws of the frames. Locomotives. 741 This being done, we are prepared to begin adjustment of the valve gear. First. Place the center of the right-hand valve exactly in the middle of its travel, that is with the center between the two steam ports, having first placed the reverse lever in the center notch of the quadrant. Second. Observe whether or not the rocker arm is exactly at right angle with the guides; if it is not, make adjustments so that it will be exactly at right angle with the guides, no matter whether the guides are horizontal or whether they incline little or much down- ward toward the center of the driving axle. Third. After having placed the rocker arm at right angle with the guides, connect the valve stem with the rocker arm without mov- ing the valve from its central position, or without moving the rocker arm from its central position. Fourth. Place the link in its neutral position at right angles with a line drawn through the center of the axle and center of the lower rocker-arm pin, and fasten the link in that position. Fifth. Turn the forward motion eccentric loosely around the axle, in the direction the engine is to run, until the eccentric has been turned upward and stands at right angle with the eccentric blade when con- nected to the link; fasten the eccentric in that position temporarily; then adjust the eccentric blade and connect it with the link without moving the link or eccentric. Sixth. Turn the backward motion eccentric loosely around the axle, in the direction the engine is to run, until the eccentric has been turned downward and stands at right angle with the eccentric blade when connected with the link; fasten the eccentric in that position temporarily; then adjust the eccentric blade and connect it with the link without moving the link or eccentric. Seventh. Place the reverse lever in the forward notch of the quad- rant, and loosen the forward eccentric and turn it completely around the axle, and note the forward and backward extreme travel of the valve. If the valve has traveled an equal distance on both sides of the outer edge of each outer steam port, the adjustment is correct. If the valve has traveled too far forward, lengthen the eccentric blade, if the valve has traveled too far backward shorten the eccentric blade. For example, if the valve has traveled half an inch too far forward, lengthen the eccentric blade half an inch; if the valve has traveled half an inch too far backward, shorten the eccentric blade half an inch. Keep up these adjustments until the eccentric, in being revolved loosely around the axle, will cause the valve to travel an equal distance on each side of the ports. 742 A Library of Steam Engineering. When these adjustments have been completed for the forward motion eccentric, place the reverse lever in the back notch of the quad- rant and repeat, with the backward motion eccentric, the operation described for the forward motion eccentric, until the travel of the valve on both sides of the ports has been equalized for the motion of the backward eccentric. Then, to equalize the valve travel on the left- hand side of the locomotive, repeat the above described operation for both left-hand eccentrics. Eighth. Put the crank pin of the forward right-hand driving wheel on the forward dead center; place the reverse lever in the for- ward notch of the quadrant. Place the forward eccentric straight up and the backward eccentric straight down, then turn the upper part of the forward eccentric forward and downward toward the crank pin until the valve has the desired lead for going head, and fasten the eccentric in that position. Ninth. Place the reverse lever in the back notch of the quad- rant, and turn the lower part of the backward eccentric forward and upward toward the crank pin on the forward center until the valve has the desired lead for backing, and fasten the eccentric in position. Tenth. Put the crank pin of the forward left-hand driving wheel on the forward dead center; place the reverse lever in the forward notch of the quadrant; place the forward eccentric straight up and the backward eccentric straight down; then turn the upper part of the forward eccentric forward and downward toward the crank pin until the valve has the desired lead for engine going ahead, then fasten the eccentric in that position. Eleventh. Place the reverse lever in the back notch of the quad- rant, and turn the lower part of the backward eccentric forward and upward toward the crank pin on the forward center until the valve has the desired lead for engine backing, then fasten the eccentric in that position. Twelfth. Put the crank pin of the forward right-hand driving wheel on the back dead center, and place the reverse lever in the for- ward notch of the quadrant, and notice if the lead of the valve is the same as it was when the crank pin was on the forward dead center; if it is, the adjustment is right; if it is not, divide the difference by lengthening or shortening, as may be required, the blade of the for- ward eccentric one-half of the difference and move the eccentric the other half of the difference, and fasten the eccentric in that position. Thirteenth. Place the reverse lever in the back notch of the quadrant, and notice if the lead of the valve is the same as it was when the crank pin was on the forward dead center; if it is, the adjust- ment is right; if it is not, divide the difference by lengthening or Locomotives. 743 shortening, as may be required, the blade of the backward eccentric one-half the difference, and then moving the backing eccentric the other half of the difference. Fourteenth. Put the crank pin of the forward left-hand driving wheel on the back dead center, and place the reverse lever in the for- ward notch of the quadrant, and notice if the lead of the valve is the same as it was on the forward dead center; if it is, the adjustment is right; if it is not, adjust the eccentric blade and move the eccentric according to directions given in twelve and thirteen. Fifteenth. Place the reverse lever in the back notch of the quad- rant, and notice if the lead of the valves is the same as it was on the forward dead center; if it is, the adjustment is right; if it is not, adjust the eccentric blade and move the eccentric according to direc- tions given in twelve and thirteen. If the valve gear was readjusted on either side of the locomotive, when the crank pin was on the back dead center, that side should again be put on the forward dead center, to see if the lead is right; if it is not, the adjustments must continue until it has been made right; that is, the engine must be put from one dead center to the other until the valve lead has been equalized. If the valve lead when engines were on back centers was the same as when they were on the forward centers, it will not be necessary to again put them on the forward dead centers; but all parts of the valve gear should be properly secured to place, so that there will be no likelihood of any of the parts getting loose when the locomotive is under way. This completes the valve setting. The steam chest cover can now be replaced and the main rods connected. But before the parallel rods are connected the back wedges should be set up, and the axles and crank pins properly trammed; after which the parallel rods can be connected, and the locomotive will be ready for service. TO ADJUST ECCENTRICS BEFORE DRIVING WHEELS ARE PLACED UNDER A LOCOMOTIVE. Adjusting the eccentrics when the driving wheels are under the locomotive, and the valve gear is properly connected, and one engine is on the dead center, has already been thoroughly explained; it will therefore be the purpose here to explain how to adjust locomotive eccentrics to their proper positions before the driving wheels are put under the locomotives. The first thing to be done is to find the diameter of the inside col- lars of the crank pin of the driving wheel attached to the axle on which 744 A Library of Steam Engineering. the eccentrics are to be placed; then scribe a circle on the end of the axle, making the diameter of the circle the same as the diameter of the inside collar of the crank pin. Then revolve the driving wheels until the inside collar of the crank pin and the circle on the axle come in line horizontally, using a spirit level to get them in line accurately. The spirit level is placed on top of the collar of the crank pin, and on top of the circle on the end of the axle, and the wheels are adjusted until the spirit level in that position comes to a perfect level, as shown in Fig. 502, when the wheels are blocked in place for the adjustment of the eccentrics on that end of the axle. B A' Fig. 502. A' represents the inside collar of the crank pin; B the circle on the end of of the axle, corresponding to the diameter of the crank pin collar; and C a spirit level resting on top of crank pin collar in line with top of circle B. In adjusting the eccentrics according to this method, the driving wheels should have their crank pins put on the forward center, as shown in Fig. 502, and the eccentrics on that end of the axle should be adjusted with the driving wheels in that position, after which the crank on the other or left-hand side should be put on the forward center. The next thing to be done is to procure a board E, as shown in Figs. 503 and 504, with parallel edges, thick enough to set on edge without side support; and one of such edges to be true. Before putting the board in position, as shown in the illustration, lay off the diam- eter of the driving axle and the diameter of the eccentrics on the Locomotives. 745 board near its middle. Make a center mark in the board at A, and from that lay off one-half of the diameter of the axle on each side of that center, and one-half of the diameter of the eccentric on each side of the center, so that the two inner marks at the ends of the arrows A will represent the diameter of the axle, and the marks at the ends of the arrows B will represent the diameter of the eccentric. If there is any difference in the diameters of the eccentrics, the diameter of each should appear on the board; but eccentrics should always be made equal in diameter and in throw. F H A G A- C E B Fig. 503. After laying off the diameter of the axle and the diameter of the eccentric on the board, ascertain the lap of the valve and add the lead required to the length of the lap, and lay off that distance in front of the forward mark on the board representing the diameter of the eccen- trics, as shown by the mark C in Figs. 503 and 504. For example: If the lap of the valve is found to be, say & inch, and it is desired to give the valve inch lead, lay off the forward mark C 3 inch from the forward mark B, and that distance will represent the lap and lead of the valve. Next, place the board under the eccentric, with the true edge of board up, in position as shown in the illustration, low enough to clear the eccentric when down. Level the board with a spirit level, move the eccentric to one side out of the way for the time being, and place a square on top of the board E, with the corner of the square at the forward mark A, and move the board toward the axle until the upper part of the square touches the axle, then the marks on the board, rep- resenting the diameter of the axle, will be squared with the axle. 746 A Library of Steam Engineering. If the board is under the forward eccentric turn the eccentric up; if it is under the backing eccentric, turn that eccentric down, and keep it there temporarily. H represents the axle, F the go-ahead eccentric, G the backing eccentric, A the crank pin, and D the square. Now place the square D, as shown in Fig. 504, in line with the lap and lead mark C. If the board is under the backing eccentric, move the eccentric forward and upward until it touches the square, then put a mark on the axle and eccentric with the eccentric in that position, to mark the proper position of the eccentric; then fasten the eccentric securely in that position, taking care to see that the eccentric is the proper distance from the inside of the hub of the driving wheel. H F D E C B Fig. 504. Next, move the board over to and under the forward or "go ahead eccentric, and move the eccentric out of the way for the time being. Level the board and get the marks on the board, representing the diam- eter of the axle squarely in line with the sides of the axle. Bring the eccentric to its proper place on the axle and turn it into an upright position. Place the square at the lap and lead mark on the board and move the eccentric forward and downward toward the crank pin until the eccentric touches the square; fasten the eccentric in that position, and put a mark on the axle and eccentric to mark the proper position of the eccentric. Repeat the above operation for the opposite side and the eccentrics will be properly adjusted. Locomotives. A 747 TO ADJUST LOCOMOTIVE VALVE GEAR IN CASE OF DERANGEMENT ON THE ROAD. In case the valve gear of a locomotive becomes deranged while the engine is out on the road, it is of the utmost importance that the engi- neer should understand how to determine the cause, how to locate it, and how to apply the remedy quickly. It will, therefore, be the pur- pose of these instructions to impart the information necessary in such cases. As soon as any derangement occurs in the valve gear when the locomotive is under way, it will be indicated by irregular exhaust, and the engine should be brought to a stop at once; and when brought to a stop the reverse lever should be thrown into the back notch of the quadrant, and the throttle valve opened and the engine backed. If the engine stops after backing, or refuses to back, it is a pretty sure sign that something about the valve gear is broken. If the engine backs and stops on the center on either side, or refuses to start to back- ing from the center on either side, the cause is pretty sure to exist on the opposite side, and an examination may show a broken arm or a broken valve stem. In which event the valve on that side should be placed squarely over the ports, and if the rocker arm is broken, the eccentric straps, rods, and link should be removed, and the main rod should be disconnected. But in no case should the parallel rod be dis- connected unless the opposite parallel rod, or one of its straps, is bro- ken; in which event, both parallel rods should be disconnected. If the valve stem is broken, disconnect it from the rocker arm; place the valve squarely over the ports; disconnect the main rod on that side; pinch the engine over the center, and go ahead. If the engine backs all right, start the engine ahead, and if it stops on the center on one side, it will indicate that the trouble is on the other side, and that the forward eccentric has slipped. Pinch the engine, on the side on which the eccentric has slipped, onto the for- ward center, and throw the reverse lever into the back notch in the quadrant, and put a mark on the valve stem just outside of the gland of the stuffing box. This mark will indicate that the valve has uncov- ered the forward steam port just to the extent of the lead of the valve. Now throw the reverse lever into the forward notch of the quadrant and the mark made on the valve stem will disappear in the stuffing box gland, or move outward from its face. Now move the forward eccentric loosely around the axle in the direction the engine is set to run, until the mark on the valve stem appears at the face of the stuff- ing box gland, and fasten the eccentric firmly in that position; the valve will now be properly set, and the engine will be in condition to go ahead. 748 A Library of Steam Engineering. COMPOUND LOCOMOTIVES. Figs. 505, 506, 507, 508 and 509, represent the Weir-Harden com- pound locomotive. The only difference between this and ordinary locomotives is the cylinder, valve, and steam chest; all of which can be attached to any ordinary locomotive without changing or altering any other part of the valve gear. The cylinder is a continuous casting, as shown in Fig. 509. This cylinder is cast with an annular partition ring or bridge wall at the center, of sufficient thickness to withstand the steam pressure, and to allow a packing ring to be inserted therein. The partition ring is made of such depth as to give the differential area between the high and low pressure piston surfaces, the latter hav- ing an area of about three to that of one for the high pressure surface. The piston is made in three parts-two followers and a cylindrical shell or piston barrel interposed between them. It will thus be seen that there are four piston faces in one cylinder-two of a small annu- lar area for high pressure steam, and two of a larger area for low press- ure steam. There are three packing rings in each follower and one in the center partition ring or bridge wall. These rings are of the Harris type of two thin rings riveted together, and then returned and cut, with a lap joint, into six or eight sections, each of which is provided with a special spring set into a pocket just large enough to hold and guide it at each joint. A small brass casting is set into each spring to give it a bearing at the joint of the ring. In the partition ring or bridge wall the springs are placed in the reverse position from that of the follower packing rings, so as to close the rings around the piston barrel. The piston rod passes through both followers and the piston barrel, and is provided with a shoulder and steel collar at the rear and two nuts at the forward follower, which constitute the means for tight- ening up the piston parts and locking them securely in place. It will thus be seen that this presents greater simplicity of arrangement than that presented in the ordinary piston. Beside, the followers being some distance apart, together with the partition ring or bridge wall, form three bearings, and thus providing a most excellent guide for the piston, giving it a smooth running effect, and preventing the plunging or gouging on the cylinder surfaces, especially in the case of large pis- tons, thus obviating the necessity of passing the piston rod through the forward cylinder head to avoid the difficulties encountered in many large engines. The piston barrel is subjected only to external pressure, and may be made very light on that account. The followers can also be made very light on account of their peculiar construction, so that the whole is but a trifle heavier than the ordinary piston. Another important feature in this style of engines is the absence of unequal strain to Locomotives. 749 00 O BO Fig. 505. Folger Gin 750 A Library of Steam Engineering. OL OO Fig. 506. REAR ELEVATION. SECTION SHOWING POSITION OF LEVERS. LOOKING FORWARD. Fig. 507. SECTION BACK OF ROCK SHAFT. SECTION FRONT OF ROCK SHAFT. LOOKING FORWARD. Folger Cin Fig. 508. FRONT ELEVATION. FRONT ELEVATION, PILOT REMOVED. LOOKING BACKWARD. Locomotives. 751 which the pistons of ordinary compound engines are subjected, espec- ially where they are connected by separate connecting rods to their respective cranks. This style of engine has also the advantage of having but one stuffing box, instead of from three to four as in some types of compound engines. Again, the oil for lubricating the interior of the cylinder is admitted with the high pressure steam, and lubricates the high and low pressure cylinders at the same time, as the high and low pressure cylinders are one and the same in this engine. ՌՈ Induction Port Low Pressure Low Pressure High Pressure Cylinder Low Pressure Cylinder Cylinder Cock Rod Converting Rod Fig. 509. High Pressure Cylinder O Another feature of this engine is that the low pressure steam per- forms its work in the same space that the high pressure steam performs. its work; hence it is not subjected to the evils of condensation that the steam in the ordinary type of low pressure engines is subjected to, because the walls of the cylinder are kept at a higher temperature. Another feature of this engine is the shortness of the distance the steam has to travel from the high to the low pressure surface of the piston. The ports are very large in area, and are proportioned accord- ing to the area of the respective piston surfaces and the maximum speed at which the engine is to run, yet, they being exceedingly short, the clearance is kept at a minimum. The induction port is located in the sweep of the cylinder where the exhaust port is commonly located in simple engines. The exhaust ports are also located in the sweep of the cylinder, one at each end of the seat, or where the induction port is located in ordinary slide-valve 752 A Library of Steam Engineering. engines. The exhaust steam leads to a large exhaust opening in the saddle which gives it a free exit. The steam chest contains only the exhaust steam, as the live steam is admitted through the induction port under the valve. The steam chest forms a passage for the exhaust steam to the exhaust ports in the saddle. The steam chest cover has a plate cast to its under side, which forms the balance plate or upper seat of the slide valve. This plate has recesses cast in it opposite the ports for counterbalancing. These recesses have ribs cast across them flush with the seat to keep the packing strips in the valve from springing in. The valve is of the slide-valve type, made in skeleton form, having the same area on top as on the bottom, except that the area of the upper portion of the valve to which the steam has access is slightly in excess of the lower portion subjected to steam pressure; the difference being provided for the purpose of keeping the valve to its seat; and yet the valve is so constructed as to present a very small surface either at its top or bottom. Each port controlling section is double or pierced by an auxiliary port, so that when any port is uncovered or open or inch the top of the valve opens that same amount, thereby doub- ling the amount of opening as compared with the actual travel of the valve. This action is performed on the steam induction port for the admission to the high pressure piston, and also for the release from the high and admission to the low pressure ports, as well as the exhaust. from the low pressure ports, thereby giving large openings. This same action also balances the valve, by letting steam into the recesses in the steam chest cover. When the valve is cut back to its finest cut off all the ports are wide open on center except the steam inlet, which has less than a full port opening, but double that of the ordinary valve operated by link motion. The high pressure steam is admitted through the induction port under the valve alternately to each recess connecting with the high pressure port, thence transferring it to the low pressure port, and from there exhausting over the under side of the outer edge of the valve into the chest. The upper face of the valve contains packing strips, the outer edge of each is opposite to and corresponds with the port edge on the lower seat, and they are held up in position by light springs; in addi- tion thereto live steam is admitted under them by a series of holes leading from the center auxiliary port. The valve is arranged loosely between the upper and lower seat so as to lift if necessary; but that feature is of no consequence in revers- ing the engine, as the valve is so constructed in connection with the ports that the steam can pass from one port to another or out into the chest, thus rendering it impossible to cause any damage. Locomotives. 753 CONVERTING VALVES. The converting mechanism consists of a short U pipe, as shown in the cross section of cylinder (Fig. 510) each end of which connects with one of the high pressure ends of the cylinder at the center partition. Each of these pipes is provided with a quick opening gate valve, the whole occupying but a very small space, under and back of the cylinder, so as to be protected against being hit when the locomotive is under way; at the same time they are so situated as to be easily acces- sible. The mechanism for operating these valves consists of a small lever situated in the cab in front of the seat box, convenient to the AM S T E EXH A Folger, Cin Converting Valve Ո S T Fig. 510. engineer, and connected by rods and arms to the converting valves. By the use of this lever the engine can be converted from a simple to a compound engine, and vice versa, at the pleasure of the engineer. The method of converting the engine from a compound to a sim- ple engine is as follows: When the converting valve is opened the high pressure steam is admitted from the side of the piston which is using it through the converting valve and into the other high pressure part of the cylinder; and, as the main valve is in communication with the high and low pressure ports of that end of the cylinder, as it would be if working compound and using low pressure steam, the high pressure steam is acting on the large or low pressure part of the piston at that end; and that part of the piston having an area three times as great as that of 48 754 A Library of Steam Engineering. the high pressure part of the piston, the engine is given greatly increased power. The opening through the converting valve is of suf- ficient area to maintain boiler pressure in the low pressure part of the cylinder at slow speed. VALVE GEAR. The shifting link motion, as shown in Fig. 511, or any other valve gear employed on simple or ordinary locomotives, can be used without change except that the eccentrics are required to be reversed, with the position of crank pin on forward center, as shown at A (Fig. 511), or, in other words, set as if it were a direct-acting engine without the inter- vention of any rocker arm. With this single exception the valves and valve gear are adjusted and set in the same manner precisely as those of an ordinary locomotive. Hence the instructions relating to valve setting given in the forepart of this chapter are as applicable to this locomotive as they are to ordinary locomotives. THE MAIN VALVE. Fig. 512 is a longitudinal section through the center of the chest, valve and upper and lower seats. A A represents the valve seat of the cylinder, showing location of ports; B B, steam chest; C, steam chest cover, with plate D cast thereon, forming the upper or balance seat; E E, main valve; F, valve yoke around the valve; G, vacuum valve to relieve the cylinder while rolling without steam; H, valve stem packing case; IIIIII, packing strips in the upper face of the valve. There is also one strip running lengthwise on each side of the valve, not shown in the illustration, into which the ends of the cross strips are set, and forming a steam tight joint on top of the valve. J, oil cup for valve stem; K, oil plug through which the oil is supplied to the valve from the lubricator in the cab; L, valve stem metallic pack- ing; M, valve stem; N, induction port in the saddle of the cylinder, through which the cylinder is supplied with live steam through the valve, as shown by the arrows; O O, steam ports leading to the high pressure parts of the piston; P P, steam ports leading to the low press- ure parts of the piston; Q Q, passages in slide valve through which the steam passes from the induction port N to the high pressure ports O O, and from ports O O to low pressure ports P P; R, auxiliary port for live steam, to give the double supply over the top of the valve through recess T, for passages Q Q; S S, auxiliary ports for giving the double supply of high and low pressure steam over the top of the valve through U U. The auxiliary ports R S S and recesses T U U are also for balancing the valve, by constantly keeping the steam over the valve on the same area as that which is exposed to the action of steam underneath. The recesses T U U have ribs level with the surface of the seat, to prevent the packing strips from jumping Locomotives. 755 Forward Eccentric Reverse Lever Eccentric Backward Main Reach Rod Forward Backward Convertin Lever, Compound Tumbling Shaft Crank Valve Rocker Fig. 511. Simple Converting Reach Rod Valve Rod Folger Cin Converting Rocker 756 A Library of Steam Engineering. t 21 8 9 1 11 31 M L H LO B F E Y K N P W Fig. 512, E Y B Folger Cin Locomotives. 757 into them. V V, blocks cast in the valve, with a lightning core through each; these blocks are made large for the purpose of taking up space to reduce the amount of clearance; W, bridge wall or partition ring in the high pressure part of the cylinder; X, bridge wall packing which surrounds the barrel of the piston; Y Y, cavity formed in the cylinder and sweeping around the same, forming a passage for the exhaust steam to the saddle and into the exhaust pipe; Z Z, holes drilled from the live steam auxiliary port R to the under side of the packing strips, to admit live steam to the strips to hold them tight against the upper seat. MARKS ON VALVE STEM FOR VALVE SETTING. The marks 1 2 2 3 3 on the valve stem represent the usual marks put on by which to set the valve; which is done by an L-shaped tram- mel, the long end of which is placed against the point shown by the arrow at 5. The trammel is made to reach from 5 to 1 when the valve is in its central position over the ports. 2 2 represent the lap and lead of the valve for each direction from 1. 3 3 represent the extreme travel of the valve. PISTON SPEED OF LOCOMOTIVES. In determining the proper area for steam and exhaust ports of locomotives it is necessary to ascertain the maximum speed of piston in feet per minute, which may be done according to the following rule: RULE.-First, multiply the number of feet in one mile (5280) by the maximum number of miles the locomotive is to run per hour, and divide the product by 60 (the number of minutes in one hour), and the quotient will give the number of feet the engine travels per minute. Second, determine the circumference of the driving wheel by multiplying the constant 3.1416 by the diameter of the driving wheel in feet, and then divide the speed of the locomotive in feet per minute by the circumference of the driving wheel, and the quotient will give the number of revolutions of driving wheel per minute. Third, multiply the number of revolutions of the driving wheel per minute by twice the length of the stroke of the piston in feet, and the product will give the piston speed in feet per minute. Example.-Let 5280 equal number of feet in one mile. Let 72 miles equal speed of locomotive per hour. Let 60 equal number of minutes in one hour. Let 3.1416 equal a constant. Let 6 feet equal diameter of driving wheel. Let 2 feet equal length of stroke of piston. Let 2 equal number of strokes of piston for each re70- lution of driving wheel. 758 A Library of Steam Engineering. Then we have: (5280×72)÷60 3.1416×6 -X2X2=1345.72+ feet. Piston speed per minute. Performing the operation in the ordinary way, we have: 5280 Number of feet in one mile. 72 Number of miles locomotive travels per hour. 10560 36960 60) 380160 (6336 Number of feet locomotive travels por minute. 360 201 180 216 180 360 360 Next, determining circumference of driving wheel, we have: 3.1416 A constant. 6 feet. Diameter of driving wheel. 18.8496 feet. Circumference of driving wheel. Next, dividing the number of feet the locomotive travels per min- ute, by the circumference of the driving wheel in feet, we have: 18.8496) 6336.0000 (336.13+ Revolutions of driving wheels 5654 88 681 120 per minute. 565 488 115 6320 113 0976 253440 1 88496 649440 565488 Finally, multiplying the number of revolutions of driving wheels per minute by twice the length of the piston stroke in feet, we have: 336.13 1344.52 feet. Piston speed per minute. 2X2= 4 Steam Boiler Injectors. 759 CHAPTER XXVII. STEAM BOILER INJECTORS. The knowledge of the capacity of a moving jet of steam, or other fluid, to produce a vacuum in properly formed ducts, for the purpose. of raising air, water or other fluids, and conveying them from one place to another, may be traced back to the time of Venturi, Nicholson and others. Nevertheless, it must be admitted that the eminent French, engineer, H. J. GIFFARD, was the first to conceive the idea that the kin- etic energy of a moving mass of fluid could be utilized to overcome the static energy of a mass of water under boiler pressure, so that the first mass of fluid would enter such boiler against the resistance of the second. The terms "kinetic energy" and "static energy" may not be abso- lutely correct in a mechanical sense in this connection, but they may be used here for the purpose of designating the inherent difference between the conditions of the two masses of fluid under consideration, which will receive notice further on. THE ESSENTIALLY ACTIVE PARTS OF AN INJECTOR. First. A steam nozzle through which the operating steam from the boiler enters the injector. Second. A combining and condensing nozzle, in which the steam and feed water meet, and in which the steam condenses and transmits its dynamic force to the water. Third. A delivery nozzle in which the maximum velocity of the combined mixture of steam and water is attained. If these three parts, which are to be found in every injector, are looked upon as the essential components of an injector, the Marquis MANNAURY D'ECTOT must be considered the inventor of the injector. In 1818 he was granted a French patent for a steam jet apparatus, which was capable of raising water from a tank and delivering it into a second tank. After d'Ectot, the French engineers, Pelleton and Bourdon (of monemeter fame), published various inventions of a sim- ilar character. On March 8, 1858, H. J. GIFFARD was granted his first patent for an injector to be used as a boiler feeder. The difference between his appa- ratus and those of former inventors, consisted mainly of a certain dis- 760 A Library of Steam Engineering. tance between the outlet orifice from the condensing nozzle and the inlet orifice to the delivery nozzle, which is called the overflow space, and which communicates with the outer atmosphere. In starting the injector more water, as a rule, enters the apparatus than the injector is capable of delivering against the back pressure of the boiler, and if there were no communications with the atmosphere the injector would "break," that is, refuse to work, and the steam blow back into the tank from which the water is taken; because the steam would naturally follow the line of the least resistance. By providing for an overflow space between the condensing and delivery nozzles, the surplus water is given an opportunity to escape, until the jet of com- bined steam and water has attained a sufficient amount of velocity and over pressure to open the boiler check valve and deliver the jet into the boiler. After the apparatus has been started and is in opera- tion, the waste of water may be avoided by cutting down the supply until no more water is seen at the overflow. Giffard's invention, therefore, consisted in the discovery that in order to deliver water through an injector into a boiler, by means of steam taken from the same boiler, the jet must first overflow into the atmosphere. Injectors made prior to Giffard's were, in fact, based upon the same principle as that of the Giffard injector, but were unable to deliver against pressure. The advantages of this new method of boiler feeding, the simplicity and efficiency of the apparatus, and the comparatively small expense of installation and maintenance, were soon appreciated by steam users, and to-day the injector is among the most popular boiler feeding appa- ratus in use. Soon after Giffard's injector had been placed upon the market, the manufacture of injectors became a most important and extensive indus- try in the United States, as well as in Europe. Hundreds of thousands of injectors have been manufactured for various purposes, and to-day there is hardly a locomotive engine running anywhere which is not provided with at least one injector, though by far the greater major- ity of them are equipped with two. Nearly every steam vessel in the United States and many stationary plants on land are equipped with injectors as boiler feeders. Hence it is of the utmost importance that every engineer, stationary, marine and locomotive, should thoroughly familiarize himself with the theory and principles which underlie the operation, management and construction of injectors. Giffard seems to have so thoroughly mastered the laws relating to the action of a moving fluid mass, that the curves and tapers of the nozzles, as constructed by him, have been adopted by all his followers, and are used to-day with scarcely any perceptible change. Whatever Steam Boiler Injectors. 761 changes or modifications have been made by the numerous manufac- turers of boiler-feeding injectors, relate chiefly to constructive details of the bodies of the instruments, and of the different valve mechan- isms which control the flow of the steam and of the water. To attempt to describe the numerous types of injectors now in service would be largely a matter of reiteration, and for this reason it will answer the purpose of this chapter to select a few representative examples of mod- ern efficient injectors, and describe their construction and operation, and in this way convey to the minds of engineers and students of steam engineering a clear idea of the construction and operation of boiler feeding injectors generally. We will therefore begin with the Monitor injector, as shown in Figs. 513 and 514. S represents handle of steam spindle; J, handle of lifting jet; W, handle of water valve; H, handle of heater cock; O, overflow; L, line check (Fig. 514). P represents plug for oiler (Fig. 513). Fig. 515 is a longitudinal sectional view, showing the interior of the injector: 1, body (back part); 2, body (front part); 3, body screw; 4, yoke; 5, yoke gland; 6, yoke packing nut; 7, yoke lock nut; 8, steam-valve disc and nut; 9, steam-valve spindle; 10, steam- valve handle; 11, steam-valve rubber handle; 12, steam-valve top nut; 13, jet-valve disc and nut; 14, jet-valve spindle; 15, jet-valve bonnet and nut; 16, jet-valve gland; 17, jet-valve lever handle; 18, jet-valve top nut; 18a, jet tube; 186, lifting nozzle; 19, water valve; 19a, eccentric spindle; 20, water-valve bonnet; 23, water-valve lever handle; 25, steam nozzle; 26, intermediate nozzle; 27, condensing nozzle; 28, delivery nozzle; 30, line check; 31, line-check valve; 32, stop ring; 33, overflow nozzle; 33a, overflow chamber with nut; 34, heater-cock check; 35, heater-cock bonnet and nut; 36, heater- cock spindle; 37, heater-cock T handle; 38, coupling nut (steam end); 39, coupling nut (water end); 40, coupling nut (delivery end); 38a, tail piece (steam end); 39a, tail piece (water end); 40a, tail piece (delivery end). This injector is provided with the usual inlet openings for steam and feed water, and an outlet opening for the delivery, which are des- ignated in the illustration (Fig. 515) by arrows, and by the words ""water," "delivery." "steam," The body of the injector consists of two parts, 1 and 2, bolted together, as shown in Figs. 513, 514 and 515. This construction facil- itates a thorough inspection and cleaning of the interior of the body whenever such should be deemed necessary or desirable. Part 8, Fig. 515, is the steam valve which controls the flow of steam from the boiler into the nozzles, and is operated by means of the handle, parts 9, 10, 11 and 12. The gland or follower 5 has two arms cast on it, 762 A Library of Steam Engineering. S C W P Fig. 513. EXTERIOR VIEW Fig. 514. INTERIOR VIEW. O H Steam Boiler Injectors. 763 passing by the threaded neck of the yoke 4, engaging a circular groove in the packing nut 6. Part 7 represents a lock nut to fix the position of nut 6. By these means the adjusting of the packing is accomplished, uniformly, by a single central nut, in the same manner as in an ordin- ary stuffing box, without exposing the thread on the spindle to the action of the steam within the chamber. Parts 186 and 33 form an independent lifting mechanism, receiv- ing the operating steam directly from the boiler through tube 18a, the inlet to which is controlled by valve 13 and handle 17, and connec- tions 14, 15, 16 and 18. 12 10 6 9 88 STEAM +38 a 37 +38 36 35 2 3 25 26 27 28 +32 31 30 40 a 40 15 13 11 16. 18 14 18 a 18b 19 19 33 17 20 23 WATER 39a +39 33a Fig. 515. Part 19 represents the valve controlling the water inlet by means of the handle 23 and parts 19a and 20. This handle 23 is a double one, and in its upper part is provided with a steel pin, which engages notches milled into the round flange of bonnet 20. In this manner, handle 23 may be placed and retained in any position, and by working the notch, the engineer is enabled to place the handle into any certain position which he may have found most suitable for a certain degree of pressure or amount of delivery. Part 34 is a check valve, working automatically, and preventing the drawing in of air after the injector has been started. Part 31 is a check valve, which, in addition to an ordinary main boiler check valve, placed between the injector and the boiler, prevents the back flow of water from the boiler, while the injector is not in operation. DELIVERY 764 A Library of Steam Engineering. Parts 25, 26, 27 and 28 are the steam, intermediate, condensing, and delivery nozzles, respectively. The operation of the apparatus may be described as follows: وو Connection having been made with the boiler at the points marked "steam and "delivery," and with the water supply at the point marked “water," jet valve, part 13, is opened first. This will cause steam to rush through the jet tube 18a, lifting nozzle 186, and overflow noz- zle 33 into the atmosphere, and will produce a vacuum in the overflow chamber 33a. It will follow, as a natural result, that the atmospheric pressure will force the water up the water tube, through intermediate and condensing nozzles 26 and 27, lifting the heater-cock check 34, and throwing water through overflow nozzle 33. Water appearing at this overflow, steam valve 8 is opened, and then jet valve 13 is closed. The boiler steam passing through steam nozzle 25 will mix with the water in nozzles 26 and 27, gradually accelerating the velocity of the mixture, until it becomes sufficiently great to pass through delivery nozzle 28, and to open both check valve 31 and the boiler check valve. The quantity of water to be allowed may be regulated by means of valve 19. At high degrees of steam pressure, no regulation will be necessary, because the injector will then readily take up and deliver all the water supplied to it. At low pressures, however, more water will reach the injector than it is capable of delivering, and in such cases the supply must be regulated by the water valve 19. In locomotive practice it becomes occasionally necessary to warm the feed water in the tank. This can be done with the aid of the injector, by screwing down heater-cock spindle 36 and 37. The open- ing of heater-cock check 34 will thus be prevented, and if steam valve 8 is kept open steam will flow back through the water passages into the tank. AUTOMATIC INJECTORS. Fig. 516 is an exterior view, and Fig. 517 is an interior view of an automatic injector. It is automatic in that no regulation of the water supply is required. It will lift the water and force it into the boiler with a single steam jet. This class of injectors is largely used as feed- water feeders for stationary boilers, and for which they are peculiarly adapted. The Monitor, the injector previously described, however, has the advantage of being able to force water against a greater press- ure than the type of injector shown in Figs. 516 and 517. The prin- ciple upon which the latter works does not differ materially from other injectors. Therefore, when an engineer has familiarized himself with the construction and manner of operating any of the standard inject- ors he will experience little or no difficulty in operating any injector. Steam Boiler Injectors. 765 Fig. 516. STEAM WATER DELIVERY Fig 517. OVERFLOW DOUBLE JET INJECTORS. Fig. 518 is an exterior view, and Fig. 519 an interior view of the World injector, showing all of its working parts. This is a double jet instrument and works somewhat similar to the Monitor, giving about the same results as to over pressure and long and short lift. When the valves C and D are closed the valve C comes in contact with the valve seat at the bottom of the counterbore in steam nozzle K, so that before the valve can present a steam opening into the steam nozzle K it will have to be withdrawn from the counterbore. This construction is for the purpose of allowing the valve D to be with- drawn from its seat in the steam nozzle J, and lifting water to space 1 and forcing it through the nozzle F into spaces 2 and 3, and overflow- ing through the overflow nozzle 8, before the steam valve C is opened. When both steam valves are closed the handle L is up, as shown in Fig. 518; when in operation the handle is down, as shown in Fig. 519. To start the injector the throttle valve is opened and the steam let into the passageway G H H and I. The handle L is turned toward the left about a quarter of a turn, and the valve D is withdrawn from its seat in steam nozzle J, which forms a vacuum in the passageway 1, and the atmospheric pressure forces the water up and fills the space 1, from there it is syphoned in and through the nozzle F into spaces 2 and 3. The valve B is open and the water overflows into 7 and 8 from 766 A Library of Steam Engineering. TO BOILER ARCH TH STEAM WATER B Fig. 518. STEAM G TO 9 BOILER 8 B 2 3 E WATER L Fig. 519. Steam Boiler Injectors. 767 2 direct, and from 3 through the nozzle 5 into space 6, and thence into 7 and 8 and out into the atmosphere. The valve B fits loosely on the end of the central stem E, similar to a globe valve. The small end of the valve is plug shaped, and made to fit the opening from 2 into 7 and 8, so that the overflow from 2 may be closed, while that from 6 into 7 and 8 is still open. The valve B is bored out at its small end to fit the central stem at B, and counterbored to fit a collar on the end of the WATER STEAM OVERFLOW Fig. 520. TO BOILER central stem. The valve is slipped onto the central stem from the handle end before the stem is put into position. A plug is screwed into the valve close to the collar on the end of the central stem, so as to allow the stem to work loosely in the valve, and yet not have play enough to allow the valve to tip downward or get out of line. When water appears at the overflow the handle L may be turned far enough to admit steam into the steam nozzle K, when the water 768 A Library of Steam Engineering. will be drawn from 3 and forced through the nozzle 5 into 6, and from there into 7 and 8. Then by moving the handle L as far as it will go, the overflow valve B will be closed, as shown in Fig. 519, and the water will flow by the check valve M into discharge pipe 9, and thence into the boiler. In the construction of this injector the utmost simplicity has been strictly observed; at the same time it presents a most ingenious arrange- ment of all the parts so as to produce the highest standard of effi- ciency in steam boiler feeding. The injector, shown in Figs. 520 and 521, is a single jet device for feeding boilers. It is automatic in its operation; that is, it will start itself when the steam is turned on without any adjustment of valves or other parts. If, from any cause the water should be cut off, the injector will start again the moment water returns to it. With this kind of device not quite so much over pressure can be attained as with the double jet injector already described. With the double jet injector the discharge water can be forced, without difficulty, against a pressure of 225 pounds per square inch, with a steam press- ure of 100 pounds per square inch; while with the single jet injectors water can be forced against a pressure of 140 pounds per square inch, with a steam pressure of 80 pounds per square inch. This injector will work on a lift of 3 feet, with from 15 to 145 pounds of steam; with a longer lift it will not do quite as well, although with 80 pounds of steam pressure it will lift water about 20 feet. When the throttle valve is opened the steam passes through the steam nozzle 1 (Fig. 521) into nozzle 2, syphoning the water from inlet A and forcing it through 2 into 3, at the top of which it overflows into space B until the injector is in full operation, when all of the water flowing to it is taken up and discharged through 2 3 4 into 5, thence into 6 to 7 and thence into the boiler. The water overflowing into B passes down to C, thence under the valve 8 into D, and down around 6 into E, and thence into the atmosphere; which overflow, however, will cease as soon as the injector is in full operation, as it will then take up all the water flowing through the intermediate water nozzle 2. This automatic injector, like that shown in Figs. 516 and 517, is used largely in feeding stationary boilers, and boilers where excessive high steam pressure is not carried. CAUSES WHICH PREVENT INJECTORS WORKING. One of the main causes which prevents an injector working is air in the suction pipe, caused by leakages in the pipe or improper pack- ing, or lack of proper packing in stem of valve in suction pipe. When injectors are used on steamers, faulty construction of supply tank fre- Steam Boiler Injectors. 769 DELIVERY 7 Fig. 522. STEAM NOZZLE. 6 E 8 C STEAM 2 3 OVERFLOW 5 Fig. 523. INTERMEDIATE NOZZLE. Fig. 521. C O O B Fig. 524. CONDENSING NOZZLE. A WATER F 49 770 A Library of Steam Engineering. quently interrupts the working of the injector. The lower end of the injector suction pipe should always be surrounded and submerged in a solid body of water and entirely free from air. Fig. 525 is a perspective view of a properly constructed supply tank for injectors. A A' represent the supply pipes leading into the tank T, and D D' represent short pieces of pipe extending upward from and attached to the inner ends of the supply pipes; allowing the sup- ply water to overflow from the upper ends of the short pipes. In case any air should pass into the tank with the water it will escape through the air pipe C, in the top and at the opposite end of the tank. B resents the suction pipe leading to the injector; the lower end extends B rep- A' B Τ C Fig 525. SUPPLY TANK FOR STEAMBOATS. down into the water, to within a short distance from the bottom of the tank. This tank and connections present a very simple arrangement, and when constructed, as shown in Fig. 525, removes entirely the main cause which prevents injectors working properly. Injectors work best with dry steam; for that reason the steam sup- ply pipe should be attached to the highest point in the steam drum or dome. Sediment and scale in nozzles also are frequent causes of injectors failing to work properly. The nozzles should be removed occasionally and thoroughly cleaned. When the injector fails to get water, the supply may be cut off by absence of water at the source; strainer clogged up; the supply pipe, hose or valve stopped up; supply pipe or water too hot; lack of suffi- cient pressure for the amount of lift; leak in the supply pipe, its con- nections or its valve-the latter three are the most frequent causes. When the injector gets water and fails to force it into the boiler, it may be that there is too much or too little water, dirt in the Steam Boiler Injectors. 771 delivery tube, faulty check valve, obstruction between injectors and boiler check valve. When the injector starts but "breaks," the supply of water may not be properly regulated; the supply pipe may be admitting air to the injector; there may be some obstruction to the free delivery of the water on account of dirt in the delivery tube, faulty check valve, or obstruction between injector and boiler check valve. Sometimes a globe valve used on the supply connection has a loose disc, and after starting the disc is drawn down, partially closing the valve, which is equivalent to giving the injector too little water. This only occurs, however, when starting with low steam. To remedy this take the valve off and reverse it. Having described the practical operation of injectors and the man- ner of their construction, as well as the causes which tend to obstruct their working, the next subject for consideration will be the theory of the injector. SPECULATIVE THEORY OF THE INJECTOR. WHY AN INJECTOR WORKS. The reason why and how an injector works may be mathemat- ically explained as follows: For the purpose of making deductions as simple as possible, it will be assumed with Combes, that steam flows into the atmosphere with the same density and under the same pressure which it had within the boiler. Under such a condition the velocity of steam issuing from a certain orifice, and not taking into account the influence of the form or shape of the orifice, may be computed from the following formula: V= 29 P-p q In which V equals the velocity of steam in feet per second. 2 equals a constant. g equals a co-efficient of acceleration (32.16 feet). P equals steam pressure per square foot in the boiler. p equals atmospheric pressure per square foot. q equals specific gravity of steam (weight of one cubic foot). To dwell upon a concrete example, we will assume the boiler pressure to be 70 pounds per square inch, and that P in the equation 772 A Library of Steam Engineering. represents the boiler pressure per square foot; p the atmospheric press- ure per square foot, and q the weight of a cubic foot of steam at 70 pounds pressure per square inch equals 0.168 pounds, therefore, P-p (144x70)-(144x14.7) q 0.168 -47400 And V=√2X32.16X47400-1746+ feet per second. As the student may not be familiar with algebra, the formula. P-P V= 2 g q will be explained and reduced to simple arithmetic. The operation to be performed, as indicated by the formula, is that the value of p is to be subtracted from the value of P, and the remainder divided by the value of q, and the quotient to be employed to multiply the product of 2 multiplied by the value of 9, and finally that the square root is to be extracted of the last product, and the answer will equal V; which represents the velocity of steam in feet per second. The value of g is 32.16, and that represents the velocity a falling body attains in feet at the end of the first second, when falling in a vacuum. The distance a body will fall in one second in vacuum is 16.083 feet; but when it has reached that distance its velocity will be at the rate of 32.16 feet per second; in other words, if it would con- tinue at the rate of speed it had attained at the end of its first second. of fall, without any variation in speed, it would travel a distance of 32.16 feet per second. The velocity of 32.16 feet is therefore taken as a unit or measure of force of gravity, and is denoted by the letter g, and this in the formula is multiplied by 2, which equals 64.32. P in the formula represents the steam pressure per square foot, no matter what that pressure may be. In the case under consideration it is assumed to be 70 pounds per square inch, so that in the equation the steam pressure per square foot would be 144 X 70-10080 pounds. In the formula p represents the atmospheric pressure per square foot. In the case under consideration it is assumed to be 14.7 pounds per square inch, so that in the equation the atmospheric pressure per square foot would be 144 X 14.7-2116.8 pounds. q, like the steam pressure, is a variable quantity. It represents the weight, in pounds, of a cubic foot of steam at any given pressure. As the weight of a cubic foot of steam varies with the pressure, recourse must therefore be had to the steam tables to ascertain the weight of a cubic foot of steam for any given pressure per square inch. In the present case the Steam Boiler Injectors. 773 steam pressure is given at 70 pounds per square inch, and by reference to the table of "Properties of Saturated Steam," it will be found that the weight of a cubic foot of steam for that pressure is 0.168 pounds, and this weight, whatever it may be, is represented by the letter q. We are now prepared to demonstrate the formula arithmetically. VELOCITY OF STEAM FLOWING INTO THE ATMOSPHERE RULE.-First, multiply the number of square inches in a square foot (144) by the given steam pressure per square inch, in pounds, and call the product "Product No. 1." Second, multiply the number of square inches in a square foot (144) by the atmospheric pressure per square inch (14.7 pounds), and call the product "Product No. 2." Third, subtract "Product No. 2" from "Product No. 1," and call the answer "The Remainder." Fourth, ascertain the weight of a cubic foot of steam due to the given steam pressure, as found in the table of "Properties of Saturated Steam," and divide "The Remainder" by the weight of a cubic foot of steam, and call the answer "The Quotient." Fifth, multiply the velocity per second a body falling in vacuum has attained at the end of one second of time (32.16 feet) by the constant 2, and call the product "Product No. 3." Sixth, multiply "The Quotient" by "Product No. 3," and then extract the square root of the last product, and the root thus found will give the veloc- ity of steam per second flowing into the atmosphere. Example. Let 144 inches equal a square foot. Let 70 pounds equal given steam pressure per square inch. Let 14.7 pounds equal atmospheric pressure per square inch. Let .168 pounds equal weight of a cubic foot of steam at 70 lbs. pressure. Let 32.16 feet equal velocity per second a body fall- ing in vacuum has attained at the end of the first second. Let 2 equal a constant. Then we have: 144 X 70-144 X 14.7 .168 X32.16X2=1746+ feet. Velocity of steam per second. 774 A Library of Steam Engineering. Performing the operation in the ordinary way, we have: 144 Number of square inches in a square foot. 70 Pounds of steam pressure per square inch. 10080 "Product No. 1." 144 Number of square inches in a square foot. 14.7 Pounds of atmospheric pressure per square inch. 100 8 576 144 2116.8 "Product No. 2." Next, subtracting "Product No. 2" from "Product No. 1," we have: 10080.0 "Product No. 1." 2116.8 "Product No. 2." 7963.2 "The Remainder." Next, dividing "The Remainder" by the weight of a cubic foot of steam due to 70 lbs. steam pressure per square inch, we have: .168) 7963.200 (47400 "The Quotient " 672 1243 1176 67 2 67 2 00 Next, multiplying the velocity a body falling in vacuum has attained at the end of the first second, by the constant 2, we have: 32.16 2 64.32 "Product No. 3." Next, multiplying "The Quotient" by "Product No. 3," we have: 47400 "The Quotient." 64.32 "Product No. 3." 948 00 14220 0 189600 284400 3048768.00 The last product. Steam Boiler Injectors. 775 Finally, extracting the square root of the last product, we have: 3048768 (1746+ feet. Velocity of steam flowing into 1 the atmosphere per second. 27) 204 189 344) 1587 1376 3486) 21168 20916 PROPORTION OF WATER TO THAT OF STEAM. Before this steam can issue from the orifice it comes in contact with water, by which it is condensed, and with which it forms a mixed jet of steam and water. The velocity of this water which mixes with the steam is very slow, as compared with that of the steam itself, and may therefore be entirely ignored, without materially affecting the results in view, so that we arrive at this equation: (m+M)v=mV in which M equals mass of water mixing with the steam. m equals mass of steam issuing per time unit. V equals velocity of steam figured above. v equals velocity of the mixed jet. v=V m m+M (1) The first thing is to determine the value of the different letters in the equation. This is done by ascertaining the temperature of the water flowing to the injector, the total heat units of the steam due its boiler pressure per square inch, and the temperature of the mixture; that is the temperature of the water discharged into the boiler by the injector. In other words, the temperature of the water after the steam is mixed with it. As the boiler pressure in the case under consideration has been taken at 70 lbs. per square inch, by reference to the steam table it will be found that the total heat units in one pound of steam at that press- ure is 1206, and as the temperature, we will say, of the water after being mixed with the steam has been raised from 75 degrees to 160 degrees, we deduct the degrees of temperature to which the water has been raised from the heat units in the steam due the pressure, which in this case is 70 lbs. per square inch, and the heat units due that 776 A Library of Steam Engineering. pressure will be found by reference to the steam table to be 1206, and deducting the degrees of temperature of the mixture (160), we have: 1206 160 1046 "Remainder No. 1.' In the next place we deduct the degrees of temperature of the water flowing to the injector from the degrees of temperature of the mixture, or water flowing from the injector, which in this case is 75 degrees for the former and 160 degrees for the latter, and we have: 160 75 85 "Remainder No. 2." We next divide "Remainder No. 1" by "Remainder No. 2" and the quotient will give the amount or mass of water discharged by the injector as compared with that of the steam, and we have: 85) 1046 (12+ This represents the proportion of the 85 water to 1 of steam. 196 170 And hence 1 represents the value of m and 12 represents the value of M in equation No. 1. We are now prepared to demonstrate equation No. 1 arithmetically. TO DETERMINE THE VALUE OF M AND M IN EQUATION No. 1. RULE.-First, subtract the number of degrees of temperature of the water discharged by the injector from the total heat units in one pound of steam due the given pressure, and call the remainder "Remainder No. 1." Second, subtract the number of degrees of the water flowing to the injector from the number of degrees of temperature of the water discharged by the injector, and call the remainder "Remainder No. 2." Third, divide "Remainder No. 1" by "Remainder No. 2," and the quotient will give the proportion of the water to 1 of steam contained in the mixture, and consequently the value of m and M in equation No. 1. Example.-Let 1206 units of heat equal the number in one pound of steam at 70 pounds pressure per square inch. Let 160 degrees equal temperature of mass of water dis- charged by the injector. Let 75 degrees equal temperature of water flowing to the injector. Steam Boiler Injectors. 777 have Then we have: 1206-160 160-75 =12+ Proportion of water to 1 of steam. Performing the operation in the ordinary way, we have. 1206 160 1046 "Remainder No. 1." Next we have: 160 75 85 "Remainder No. 2." Finally, dividing "Remainder No. 1" by "Remainder No. 2," we 85)1046 (12+ Proportion of water to 1 of steam. And 85 196 170 12 1 12 and hence, 12 represents the value of M and 1 represents the value of m in equation No. 1. TO DETERMINE THE VELOCITY OF THE MIXTURE DISCHARGED BY THE INJECTOR. Before laying down the rule for determining the velocity of the mixture, it will be well to remind the student that this velocity does not represent the velocity with which the mixture flows into the boiler, but it represents the velocity with which it would flow into the atmos- phere. RULE.-Add the proportion of steam to the proportion of water con- tained in the mixture, and divide the proportion of steam by the sum, and multiply the quotient by the velocity of the steam per second; the product will give the velocity of the mixture in feet per second. Example. Let 1 represent the proportion of steam to water con- tained in the mixture. Let 12 equal the proportion of water contained in the mixture. Let 1746 feet equal velocity of steam flowing into the atmosphere per second at 70 pounds pressure per square inch. 778 A Library of Steam Engineering. Then we have: 1 1+12 X1746-134.3+ feet per second. Velocity with which the moving jet from the in- jector will flow into the atmosphere. Performing the operation in the ordinary way, we have: 12 Proportion of water to steam. 1 Proportion of steam to water. 13 The sum. Next, dividing the proportion of steam by the sum of the steam and the water, we have: 13) 1.00 (0.07692+ The quotient. 91 90 78 120 117 30 26 Finally, multiplying the quotient by the velocity of steam per second flowing into the atmosphere under a pressure of 70 pounds per square inch, we have: .07692 1746 46152 3 0768 53 844 76.92 134.30232 feet per second. Velocity with which the moving jet from the injec- tor will flow into the atmos- phere. If this velocity is greater than that with which the water would flow from the boiler, under the same pressure of 70 pounds per square inch, it becomes evident that the moving fluid, connected to the boiler by suitable piping, will enter it. The velocity with which the water would issue from the boiler may be calculated by the following formula: V'=√2 g h, in which g is again 32.16, and h equals the height of a water column corresponding to the pressure of 70 pounds per square inch (161 feet, in round numbers) from which it follows that V'=101.76+ feet per second. Again we will reduce the formula to simple arithmetic and give the rule for performing the operation. Steam Boiler Injectors. 779 TO DETERMINE THE VELOCITY OF WATER FLOWING FROM THE BOILER UNDER A GIVEN PRESSURE. First determine the height of a column of water in feet correspond- ing to the given pressure per square inch. RULE.-Divide the number of cubic inches in a cubic foot of water (1728) by 12, and divide the quotient by the weight of a cubic foot of water (62.5 pounds); then multiply the last quotient by the given steam pressure (70 pounds in this case), and the product will give the height of a column of water in feet corresponding to the given pressure per square inch. Example.-Let 1728 equal number of cubic inches in a cubic foot. of water. Let 12 equal a constant. Let 62.5 pounds equal weight of a cubic foot of water. Let 70 pounds equal given steam pressure per square inch. Then we have: 1728 12 62.5X70-161.280 feet. Height of a column of water cor- responding to a pressure of 70 pounds per square inch. Performing the operation in the ordinary way, we have: 12) 1728 (144 The quotient. 12 52 48 48 48 Next, dividing the quotient by the weight of a cubic foot of water, we have: 62.5) 144.0 (2.304 The last quotient. 125 0 19.00 1875 2500 2500 Finally, multiplying the last quotient by the given steam press- ure per square inch, we have: 2.304 70 161.280 feet. Height of a column of water cor- responding to a pressure of 70 pounds per square inch. 780 A Library of Steam Engineering. Then, to determine the velocity of water flowing from a boiler under a given pressure, we proceed as follows: RULE.-Multiply the velocity in feet that a body falling in vacuum will attain at the end of the first second of time (32.16 feet per second) by 2, and multiply the product by the height in feet of a column of water corresponding to the given steam pressure per square inch (70 pounds in this case), and then extract the square root of the product, and the answer will give the velocity, in feet per second, the water will flow from the boiler: Example. Let 32.16 feet equal velocity per second a body falling in vacuum will attain at the end of the first second. Let 2 equal a constant. Let 161 feet equal height of a column of water corre- sponding to a pressure of 70 lbs. per square inch. Then we have: ✓32.16X2X161-101.76+ feet. Velocity per second. Performing the operation in the ordinary way, we have: 32.16 Velocity of falling body in feet. 64.32 2 A constant. 161 feet. Height of a column of water corresponding to a pressure of 70 lbs. per square inch. 64 32 3859 2 6432 10355.52 The product. Finally, extracting the square root of the product, we have: 10355.52 (101.76+ feet. 1 201) 0355 201 Velocity per second with which water will issue from a boiler with a pressure of 70 lbs. per square inch. 2027) 154 52 141 89 20346) 12 6300 12 2076 Therefore, as the water flowing from the injector, under a steam pressure of 70 lbs. per square inch, has a velocity of 134.3 feet per sec- ond, and the water flowing from the boiler under a pressure of 70 lbs. Steam Boiler Injectors. 781 per square inch has a velocity of 101.76 feet per second, the difference between the two velocities clearly explains the action of the injector; while it also shows that, theoretically, the quantity of water delivered by an injector may be twelve times the quantity of steam consumed for the purpose. The actual quantity delivered will, however, be materially less. The friction of the steam and of the mixture of steam and water, within the nozzles, and within the pipe conduits, will reduce the velocity of the jet, and this, taken in connection with imperfect condensation, and with more or less air finding its way into the instru- ment, will tend toward reducing its delivering capacity. It would probably lead too far for the purpose of an elementary work of this kind, to enter minutely into the details of the very intri- cate item of injector theory, as treated by various authorities; espe- cially in view of the fact that there is no general or universally applic- able theory or formula by which the proportions of injector nozzles, as to the areas of orifices, tapers and relative positions could be computed. The conditions upon which the proper function of an injector depends are so manifold that it is impossible to combine them in one compre- hensive formula. DELIVERING CAPACITY OF INJECTORS. The German scientist, Dr. Grashof, who wrote extensively on this. subject, and who is regarded as high authority, gives the following formula for the delivering capacity, C, of an injector, in litres or kilo- grams: C=A d²√p-1 in which d equals the smallest diameter of the delivery nozzle, p equals steam pressure in atmospheres, A is a co-efficient dependent upon vari- ous conditions, but independent of the size of the apparatus. The C quotient is a measure for the capacity of the injector per square d2 millimeter of the smallest delivery opening, and may therefore be used for the purpose of ascertaining the comparative capacities of dif- ferent makes of injectors. Dr. Grashof also gives the size of the steam nozzle, and the area of the water inlet around the steam nozzle, in the following formula: A G2 G' y' v' x y'v A' G2 y v2 in which A equals area of steam nozzle. $782 A Library of Steam Engineering. A' equals area of water inlet opening around the steam nozzle. G' equals weight of steam issuing per second. v' equals its velocity. G2 equals weight of water entering per second. equals specific gravity of the mixture of steam and water. x equals the proportion G2 G' y equals the specific gravity of the water. v² equals the velocity with which the water passes by the steam nozzle. It is evident that these formulæ are not directly applicable, as they depend chiefly upon the velocity and density of the steam, which naturally varies according to pressure. But even if they were directly applicable, they give no information as to taper, length and relative position of the nozzles, which items are fully as important as the areas themselves, and they are, and are likely to remain, largely matters of experimental research. The efficiency of the injector as a boiler feeder has been fully demonstrated without the aid of any of the advanced theories. Engi- neers have long been skeptical regarding the reliability of injectors and their ability to force water successfully against a pressure per square inch equal to that of the pressure of the steam supplied to the injector. Experiments made, by the author of this work, with an injector constructed upon the principle shown in Figs. 513, 514, 518 and 519 with a steam pressure of 80 pounds per square inch in the boiler which supplied the injector, the hydrostatic pressure in a battery of three boilers was forced up to 180 pounds to the square inch. And with a steam pressure of 115 pounds per square of 115 pounds per square inch the hydrostatic pressure in the battery of three boilers was forced up to 225 pounds per square inch. The final test was made with 150 pounds steam pressure, which was much greater than was necessary, and the hydrostatic pressure in the battery of three boilers was forced up to 265 pounds to the square inch without any break in the jet, thus proving conclusively that the injector may be implicitly relied upon as a thoroughly reliable and efficient boiler feeder. INDEX. PAGE. To handle an Cycloidal engines, description of .. Foundation plan and elevation Governor, description of the.. Materials for foundation, table of.. Valves, description of.... Valves of the engine, how to set... Adiabatic curve.. How to trace the. Armature, to connect an To wind an... Atlas Automatic Engine-Description and illustration of the... Engine foundation template. 634 635 303 . 302 301 570 587 573 572 577 573 Specifications and dimensions of foundations.. Ball or sphere, to determine the weight of a cast-iron. Battle Ship, United States 574 576 584 47 471 Boilers 117 Bumped heads 355 Bursting pressure, simple rule for determining Bursting pressure of double-riveted. 124 ..124, 339 Bursting pressure of double-riveted-simple rule. Bursting pressure of single-riveted..... 340 ..117, 338 Bursting pressure for single-riveted-simple rule. Data for double-riveted.. 338 222 Data for single-riveted.. 221 Diameter-how determined. 218 Diameter for double-riveted, when the working pressure, thickness and tensile strength of material are given.. Diameter for single-riveted, when the working pressure, thickness and ....126, 470 tensile strength of material are given. Flat surfaces, construction of . Horse power of... ..... 119 373 218 Hydrostatic pressure, for double-riveted... 361 Hydrostatic pressure, for double-riveted-simple rule 362 Hydrostatic pressure, for marine 360 Hydrostatic pressure, for single-riveted... 360 Hydrostatic pressure, for single-riveted-simple rule. 361 Hydrostatic pressure, for single and double riveted. . 362 Longitudinal seams, strain on ... .127, 216 Longitudinal strain on-how determined 128 783 784 INDEX. Boilers-Continued. Longitudinal strain on, simple rule for determining Plans and specifications for tubular Reduction of working pressure of Riveted joints PAGE. 129 188 363 132 Safe-working pressure, based on ductility of plate and tensile strength.. 351 Safe-working pressure, based upon data for any given marine Safe-working pressure for marine..... Safe-working pressure of double-riveted.. Safe-working pressure of double-riveted-simple rule. Safe-working pressure of double-riveted-another rule Safe-working pressure of double-riveted-a simpler rule Safe-working pressure of double-riveted, holes drilled Safe-working pressure of double-riveted, holes drilled-simple rule Safe-working pressure of single-riveted Steam, weight of, discharged under a given pressure... Steam discharged, to determine weight of Steam drum legs, dimensions of..... Steam pressure allowable, table of... Steam pressure based upon hydrostatic pressure. Strain produced on plate by pressure in. Strength of ..... Tensile strength of plate, for double-riveted.. Tensile strength of plate, for single-riveted Thickness of plate, for double-riveted.. Thickness of plate, for single-riveted.. Boiler Flues. ... ..... 349 340 ..120, 345 .121, 346 122 123 345 146 ...117, 341 129 129 436 365 364 344 338 .126, 221 .120, 220 ..125, 219 .118, 218, 363 Boiler Plate Coupon, attached to sheet Coupon, form of, for testing iron. Ductility Ductility of-how determined... Ductility of, percentage-how determined. Ductility of, percentage required Elongation of special test piece, to determine percentage of.. Elongation of test piece Maximum thickness of Strength of, actual. Tensile strength and safe-working pressure. Tensile strength of, for double riveting... Tensile strength of, for single riveting.. Tensile strength of-how determined Test piece, iron, form of.. Test piece, steel, form of.. Test piece, steel, length of reduced section.. Test piece, steel, special form of.... Test piece, steel, special test of . Thickness required for single riveting. Thickness required for double riveting.. Thickness required to stand a given strain. Breeching, proportion and arrangement of. ..167, 439 112 113 113 347 ..115, 348 ... 351 116 354 353 222 342 351 126 120 .114, 346 350 353 353 354 354 118 125 344 186 INDEX. 785 PAGE. .... Brown Engine-Description and illustration of valve gear of the.... Cross head of the Exhaust ports, area of Exhaust ports, area of simple rule.. Exhaust ports, width of . Main bearings of the... Performance of the 591 597 602 .. 602 .. 605 . 599 599 Plan and elevation of the Steam and exhaust ports... 599 600 Steam and exhaust ports, dimensions of. Steam and exhaust ports, engines having several... Steam ports, area of.... Steam ports, area of-rule in detail.. 603 604 600 601 Steam ports, width of.. Trips, to set the... 605 .. 596 Compound engine, ratio of cylinders for Valves, to set the..... Brush holders, to insulate the Brushes for a dynamo, to make the Buckeye Engine. ... Adjustments by indicator cards.. Compression, adjustment for equalizing. Compression, amount of... Compression, to adjust the amount of. Cut off equalization.... . Diagram showing displacement of main valve. 596 304 304 606 614 . 651 612 612 612 612 616 Showing fast main eccentric.. 616 Showing perfect engine performance 615 Showing slow main eccentric... 615 Showing unequal cut off .. 617 Test diagrams 615 Distortion of diagrams from elasticity of the cord 614 Eccentric, to set the..... 611 Equalization, adjustment for ... 613 Equalization, choice of methods 613 Governor. 617 Containing wheel, to set the. 612 Friction, centripetal.. Frictional disturbance Motion, reversal of direction of Regulation, close... Regulation, to secure close Speed, to change... Speed, to decrease Speed increase, minor changes of ... Independent of marks, adjustments, how made Indicator diagrams, rudiments and analyses of .. Adiabatic curve... Adiabatic curve, to trace the Atmospheric line 620 620 621 619 619 620 621 621 608 626 634 635 . 629 50 786 INDEX. Buckeye Engine (Indicator diagrams, etc.)—Continued. Back pressure line... Clearance line. Clearance, theoretical .... PAGE. 628 630 636 Clearance of Buckeye engines 630 Clearance of four-port engines Clearance of single valve high-speed engines. Compound engines, to determine horse power of Compression curve.... Cut off, the point of.. Exhaust closure, the point of.. Exhaust line Expansion curve.. Expansion curve, theoretical. Horse power calculations.... Isothermal curve, mathematical method Lead or admission line. 631 631 639 628 627 • 628 628 628 631 636 632 629 Mean effective pressure-how determined.. 637 Mean effective pressure, a quick estimate of... 639 Ordinates, spacing 638 Steam line 626 Theoretical curve, the geometric method for determining the. 633 Vacuum line 630 Putting the engine together 606 Eccentric rods... 608 Eccentric, the main .... 606 Governor containing wheel, the 606 Shop marks.... 606 Slide-valve performance... 643 Cut off and exhaust closure, to determine the points of.. 644 Exhaust closure, how to locate the 647 Functions, to determine 649 Laps, for given results, to find the 648 Setting valve for equal cut off, to determine consequence of other events of 646 Test diagrams..... 615 Valve arrangement 622 Valve arrangement, girder bed and compound rocker arm. 626 Valve closure, to determine point of cut off.... 613 Valve, main.. .... 608 Valve, to set the main Water consumption, table of data Water consumption, theoretical Water consumption, to calculate the theoretical. Bumped heads for boilers 611 642 640 641 355 Calorimeter Tests.. 234 Experiments 265 Moisture in steam after leaving drip box. Moisture in steam intercepted by drip box 263 262 Steam discharged per hour by orifice of heat gauge Steam, how to determine percentage of water in 260 237 INDEX. 787 Calorimeter Tests-Continued. Tables of.. Tests of coil boilers.. Tests of land boilers Water, table of properties of Cams... Movement of full-stroke To lay off cut-off.... To lay off full-stroke. To set cut-off... To set full-stroke.. Centrifugal force, the unit of. Chimneys, force of draught of.. Circle, area, to find the... Circumference, to determine the. Diameter, to determine the..... To reduce area to a perfect square.. Ciphers, annexation of, in division Clearance in four-port engines.... Clearance in single-valve high-speed engines. Coal, combustible matter in-how determined. Moisture in-how determined... Coil boilers, evaporative and calorimeter tests of. Commutator of a dynamo, to make and connect a Compound engines. .... Horse power of-how determined.... Ratio of cylinders Compound expansion. Compression curve Condensers.. Constant .7854-how obtained. Constants .7854 and 3.1416, use of. ... Construction of flat surfaces of steam boilers. Conversion of heat into work Copper steam pipes, thickness of material-how determined... Corliss Engine.. Columbian Corliss engine. Compound engine Condensers Connecting rods Cross head, to adjust the.... Cylinder. Cylinder heads Dash pot. Dash pot, to adjust the..... Dead center, to put an engine on the .. Diagrams of slide valve and Corliss engines Engine performance, first-class Engine performance, table of first-class Factors of evaporation, table of Feed-water heaters, table of percentage of saving in fuel by using Feed-water heating ... PAGE. .251, 257 .243, 250 259 240 507 513 509 508 510 ..508, 516 679 186 39 35 35 36 41 631 631 227 227 243 302 .548, 651 639 651 681 628 549 ..37, 38 35 373 683 457 528 534 548 549 542 541 536 537 532 . 546 560 • 554 549 552 558 559 ...... . 556 788 INDEX. Corliss Engine-Continued. Frames Friction. Governor... Governor, care of Governor, to adjust... Lagging for cylinders of the.. Limitations of engine speed.... Mechanism for operating the valves of the Pillow blocks of... Piston packing Slides of the.... Steam and exhaust chests of. Steam and exhaust valves of... Steam, economy in the generation and use of Valves and valve rods of Valve motion of the .. Valves of, how to set the Reynolds Corliss engine... Dash-pot rods, to adjust the length of. Valves, to set the ..... Valves, table for setting. Cosines PAGE. 537 533 543 566 565 536 532 530 547 540 .. 538 536 • 536 553 542 544 564 566 569 566 567 100 Coupons for testing boiler plate ...113, 350, 353, 354 Cut off and exhaust closure, to determine the points of, of slide valves...... 644 Cycloidal engines . 587 Dash pot, to adjust a . 546 Dead center, to put engines on the 560 Dead center, to put locomotives on the. ..734, 736 Decimal fractions .... 35 Diagrams, rudiments and analyses of ... 626 Adiabatic curve. 634 Atmospheric line 629 Back pressure line.. 628 Clearance line ... 630 Clearance line, to determine theoretically 636 Compression curve. 628 Cut off, point of.... 627 Exhaust closure, point of . 628 Exhaust line 628 Expansion curve... 628 Expansion curve, to construct the theoretical 631 Horse power calculations from 636 Isothermal curve, mathematical method of drawing the 632 Lead or admission line ... 629 Mean effective pressure ... 637 Mean effective pressure, a quick estimate of. 639 Theoretical curve, geometric method for determining the. Mean effective pressure of compound engines-how determined Ordinates, how to space them.. • Throttling engine.. Showing fast eccentric ... 639 638 . 627 633 616 INDEX. 789 Diagrams, rudiments and analyses of-Continued. Showing first-class engine performance Showing main valve displacement. PAGE. 615 616 Showing slow eccentric.. 615 Showing unequal cut off 617 Steam line.... 626 Vacuum line..... 630 Water consumption, to calculate the theoretical... 641 Draught of chimneys, force required.. 186 Ductility of boiler plate 347 Amount required 116 How determined.... 348 How to determine the percentage of 351 Dynamo, how to build a 299 Names of different parts of a.. 299 Economy, attainable limit of, in steam engine.. 685 Electrical resistance.. 299 Electrical units, definition of terms of.. 298 Electricity.... 298 Armature, to connect the.. 303 Armature, to handle the ... 302 Armature, to wind the .. 301 Brushes, to make 304 Brush holders, to insulate... 304 Commutator, to connect the 302 Dynamo, names of different parts of a 299 Dynamo, to build a.... 299 Electrical resistance 299 Electrical units, definitions of terms of... 298 Electro-motive force 299 Field magnets, to wind the.. 303 Machine, to operate the... 304 Multiple system of coupling... 306 Volts, to wind machine for a given number of.. 305 Electro-motive force. 299 Engine performance, first-class ... 549 Engine performance, first-class, table of. 552 Engine speed, limitation of..... 532 Engineers, qualifications required of, by United States laws 337 Evaporation, economic ...... 266 Evaporation as from and at 212° Fahrenheit-how determined.. 226 Evaporation, rate of... 266 Evaporation, reduction of tests to standard of from and at 212° Fahrenheit.. 229 Evaporative Tests of Steam Boilers... 223 Capacity results-how determined. 265 Combustible matter in coal. 227 Commercial horse power-how determined 266 Economic evaporation-how determined. 266 Evaporation from and at 212° Fahrenheit. 226 Evaporation, rate of-how determined. Factor of evaporation-how determined.. 266 224 790 INDEX. Evaporative Tests of Steam Boilers-Continued. Factors of evaporation, table of.. Heat units, explanation and definition of Heat units in steam-how determined Moisture in coal-how determined.. PAGE. 558 223 224 227 Observed data and computed results.. Saturated steam, table of properties of Temperature of a furnace-how determined... Tests... 265 230 275 248 Tests-how reduced to standard of from and at 212° Fahrenheit.. 229 Tests of coil boilers. Tests of land boilers. Tests, tables of ... 243 259 .249, 255 Water, table of properties of ... 240 Examination questions and answers for engineers Exhaust ports, area of .. Exhaust ports, dimensions of Exhaust ports, engines having several 458 602 603 604 Factors of evaporation, table of.. Exhaust ports, width of.... Expansion curve, theoretical Factor of evaporation-how determined. .... Feed pumps, United States regulations... Feed pumps, to compute the required dimensions.. Feed water heaters, table of percentage of saving in fuel by using Feed water heating. Fire pumps for steam vessels. 605 631 224 558 176 177 559 .. 556 405 Flat surfaces of steam boilers. 370 Flues... ..167, 439 Adamson and angle-iron rings 441 Collapsing pressure of ..... 173 Diameter and thickness of material, table of.. 456 Diameter required for furnace flue of given length, given thickness of material and given steam pressure.. 444 Diameter required for given thickness of material and given pressure.... 171 Diameter required when supported by half-round iron strengthening rings. 451 Distance between half-round iron strengthening rings. Experiments of Sir WILLIAM FAIRBAIRN 450 167 Furnaces, length required, given thickness of material, diameter and pressure.. 445 • Furnaces, plain cylindrical, pressure allowable 441 Furnaces, thickness of material required for given diameter, length and pressure.. 443 Furnaces, when supported with Adamson or angle-iron rings 441 ... Lap-welded and riveted, exceeding 6 and not exceeding 40 inches in diameter 457 Lap-welded, exceeding 6 and not exceeding 16 inches in diameter and not exceeding 18 feet in length 457 Lap-welded, not exceeding 6 inches in diameter 456 Length of 172 Pressure allowable, for increase in length of sections. 455 INDEX 791 Flues-Continued. Riveted and lap-welded, made in sections PAGE. 452 Safe-working pressure of..... 168 Safe-working pressure allowable for ribbed and corrugated .... 439 Safe-working pressure-how determined for flues over 10 inches in diameter ... 455 Strengthening rings, half-round iron.. 446 Strengthening rings, required diameter of half-round iron .... 446 Strengthening rings, required diameter of half-round iron-simple rule.. 447 Strengthening rings, half-round iron, steam pressure allowable .. 448 Table of pressures, thickness of material and length of sections. Thickness of material required for certain flues and tubes .. Thickness of material required for corrugated. 453 456 440 Thickness of material required for flues supported with half-round iron strengthening rings... .. 449 Thickness of material required for given pressure, diameter and length.. 169 Fractions, common, how reduced to decimals... 42 Friction... 533 Furnace, Hawley down draught 268 Furnaces, temperature of-how determined.. 275 Gearing, speed of-how determined 292 Heaters, percentage of saving by use of. Horse power of boilers... Governor of a Corliss engine, how to adjust the. Grate bars, percentage of air space required. Heat, conversion of, into work.. Heat units. Heat units expended per horse power Heat units in steam.. Horse power calculations. 565 186 683 223 683 224 559 636 218 Description of the indicator. Showing slow eccentric. Horse power of boilers, to determine the commercial Horse power of compound engines how determined. engines-how Horse power of marine engines-how determined.. Horse power required to elevate water... Hydrostatic pressure for marine boilers Indicator, Steam Engine. Indicator Practice.. Admission and exhaust of steam, pipes and passages for. Condensation, percentage of loss by cylinder, tables of.. Derangement of valve gear-how revealed.... Diagram, a well-formed.... Essential features of the Showing fast eccentric Showing first-class engine performance Showing main valve displacement. Showing unequal cut off. Diagrams, rudiments and analyses of .. Adiabatic curve.. Adiabatic curve, to trace the. Atmospheric line.. 266 730 406 175 360 307 307 321 323 321 308 320 319 616 615 616 ... 615 617 626 634 .635, 636 629 792 INDEX. Indicator Practice (Diagrams, rudiments and analyses of)-Continued. Back pressure line ..... Clearance, to determine theoretical Clearance line. Combining the, from compound engines Compression curve... Cut-off point. Exhaust closure, the point of.. Exhaust line Expansion curve PAGE. 628 636 630 325 628 627 628 628 628 Expansion curve, to construct the theoretical... 631 Expansion line, non-condensing engines carrying light load 330 From throttling engines.... 627 Isothermal curve, mathematical method 632 • Lead or admission line 629 Mean effective pressure. Ordinates, how to space. Steam line... Test diagrams ... Theoretical curve, geometric method. Vacuum line. .328, 637 638 626 615 633 630 Volume of high pressure cylinder 327 How to attach the indicator Volume of low pressure cylinder Engines long in use, examination of valves of . Horse power, how computed from diagrams How to indicate a steam engine Lazy tongs Management and care of the indicator Manner of taking diagrams Pantagraph. Planimeters . Reducing lever... Regulating the exhaust Springs, table of maximum pressures Usefulness of the indicator. Injectors 328 322 ..328, 636 314 314 316 312 318 317 330 315 321 310 322 759 Active parts of an injector 759 Automatic injectors.... 764 Causes which prevent injectors from working.. 768 Delivering capacity of injectors. 781 Double jet injectors.. 765 Height of a column of water corresponding to a given pressure. 779 Speculative theory of the injector 771 Supply tanks for steamboats... 770 Velocity of steam flowing into the atmosphere 773 Velocity of the mixture discharged by the injector.. 777 Velocity of water flowing from a boiler under a given pressure.. 779 Water, proportion of, to that of steam ... 775 Why an injector works... 771 Inspector of boilers, qualifications required of United States 336 Isothermal curve 632 INDEX. 793 Land boilers, evaporative and calorimeter tests of Lining an engine shaft Link motion. Locomotives... Compound Converting valves Cylinder Dead center, putting the engine on the Eccentrics, to set the, before placing wheels under locomotive Main valve Piston speed of locomotives-how determined... Steam chest Valve gear Valve gear, to adjust the . Valve gear, to adjust the, in case of derangement on the road………. Valves, to set the Magnets of a dynamo, to wind the field... Marine engines.. Mathematical terms. Mathematics of Steam Engineering. Addition ... Ball or sphere, to determine the weight of a cast-iron Circle, area of-how determined... Area of, how reduced to a perfect square Circumference of-how determined.. Diameter of how determined.. Constant .7854-how obtained.. Constants 3.1416 and .7854, use of. Decimal fractions .. Decimal fractions, multiplication of. Decimal parts of a square inch.. Division, annexation of ciphers. Division, long . Fractions, common, to reduce to decimals. Mathematical terms. . Product Quotient Remainder Sum Safety-valve, to determine weight of mathematically. Lever, to determine weight of mathematically Spindle, to determine weight of mathematically Square inch, decimal parts of a Valve, the correct diameter of a · Valve circumference, to determine the.... PAGE. 259 672 658 734 748 753 748 ..734, 736 743 754 757 752 754 740 747 734 .. 303 .471, 502, 506, 511 34 33 34 47 39 36 35 35 .37, 38 35 35 39 36 41 34 42 34 34 34 34 34 46 44 45 36 39 41 40 41 44 43 ..406, 637 639 227, 264 Valve of given area, to determine diameter of a . Valve of given circumference, to find the diameter required of a Weight, cast-iron, to determine number of pounds in... Weight, cylindrical, to determine cubic inches in... Mean effective pressure-how determined ...... Mean effective pressure, a quick method for estimating the.. Moisture in coal-how determined. 794 INDEX. PAGE. Piston speed of locomotives-how determined..... Plans and specifications for tubular boilers, how to draught...... Plans for non-condensing engine for Western river side-wheel steamer Porter-Allen Engine... Air pump and condenser. 757 215 496 656 691 Bed plate 671 Centrifugal force, unit of. 679 Compound condensing engines. 681 Compound expansion.. 681 Conversion of heat into work. 683 Eccentric... 656 Economy, attainable limit of ..... 685 Engine, to line the, with shaft at higher or lower level.... 672 Engine, to line the, with shaft coupled direct. 673 Engine, to set the.. 670 ... Exhaust valves... Heat units, number expended per horse power Indicator diagrams.. Link.. 663 683 691 658 Link motion.. .... 658 Pressure plates, adjustable Reciprocating parts of steam engine, action of. Shaft.. Steam jacketing intermediate receiver Valve motion, distinguishing features of the. Valve movement, differential. 667 677 671 688 660 664 Valves, to set.... Ports, steam and exhaust.. Area of steam-how determined 669 600 Pumps. Area of exhaust-how determined. Dimensions of steam and exhaust. Width of exhaust-how determined Width of steam-how determined. Pulleys, speed of-how determined... Feed-simple method for computing dimensions . 602 600, 601 603 605 604 277 175 179 Feed, to compute the dimensions of 177 Feed, United States regulations.... 176 Fire, size of, required under United States regulations-how determined. 405 Horse power required to elevate water... 175 Reciprocating parts of steam engine, action of 677 Riveted Joints.. 132 Butt-strap joints.. 154 Diameter of rivet holes for double-riveted lap joints-iron plates and iron rivets. .... 135 .... Diameter of rivet holes for double-riveted lap joints-steel plates and steel rivets Diameter of rivet holes for single-riveted lap joints-iron plates and iron rivets 136 ... 135 Diameter of rivet holes for single-riveted lap joints-steel plates and steel rivets 135 INDEX. 795 Riveted Joints-Continued. Distance between centers of rows of rivet holes for chain riveting Distance between centers of rows of rivet holes for zig-zag riveting. Distance of rivet holes from edge of plate .... PAGE. 132 133 ..... 132 156. 155 158 Pitch of rivet holes for double-riveted double butt-strap joints. Pitch of rivet holes for single-riveted double butt-strap joints. Pitch of rivet holes for triple-riveted double butt-strap joints. Pitch of rivet holes for double-riveted lap joints-iron plates and iron rivets ..... 138 Pitch of rivet holes for single-riveted lap joints-iron plates and iron rivets .. 136. Pitch of rivet holes for triple-1iveted lap joints-iron plates and iron rivets. 139 Pitch of rivet holes for double-riveted lap joints-steel plates and steel rivets 141 Pitch of rivet holes for single-riveted lap joints-steel plates and steel rivets 140 Pitch of rivet holes for triple-riveted lap joints-steel plates and steel rivets... 142 Shearing strength of rivets. 132 Single butt-strap joints 154 Strength of double-riveted double butt-strap joints . 162 .... Strength of single-riveted double butt-strap joints Strength of triple-riveted double butt-strap joints Strength of double-riveted lap joints-iron plates and iron rivets Strength of double-riveted lap joints-steel plates and steel rivets Strength of single-riveted lap joints-iron plates and iron rivets.. Strength of single-riveted lap joints-steel plates and steel rivets Strength of triple-riveted lap joints-iron plates and iron rivets.. Strength of triple-riveted lap joints-steel plates and steel rivets. Table of riveted joints. Safety-valves--Lever weighted Adjustment of simple method 160 164 145 150 143 148 147 .. 152 166 48, 410 87 Area-how determined 55 Area of, to determine the proper 92 Area required-how determined. 411 Center of gravity of lever.... 48. Diameter of, to determine the proper 92 Diameter required--how determined. 411 Diameter of, simple rule for determining the proper.. 92 Diameter for valve to rise at given pressure 75 Diameter for valve to rise at given pressure-rule in condensed form.... 78 Distance from fulcrum to place given weight for given pressure 416 Fulcrum 48 Lever, center of gravity of ... 48 Lever, how to graduate a. 91 Lever, names of different parts of a 48 Lever, pressure produced on valve by weight of 49 Lever, pressure produced on valve by weight of-how determined .... 90 Lever, short arm of, to determine the length of.. 71 Lever weighted.. 410 Practical adjustment of.. 414 Pressure required to raise ..50, 414 796 INDEX. Safety-valves-Lever weighted Continued. Pressure required to raise a given Pressure required to raise, how determined without weighing any of its parts Pressure required to raise, rules for determining the. Pressure required to raise, to determine the. Pressure required to raise, with conic weight on lever... Pressure required to raise, with spherical weight on lever.. Pressure required to raise, with two or more weights on lever... Size of cubic cast-iron weight required to be placed a given distance from fulcrum for given pressure Spindle, pressure produced on valve by weight of United States laws relating to.... United States lever valve Specifications for..... ... 51 80 52 .56, 60 427 424 431 • 421 90 409 433 433 .....66, 418 70 Weight, amount to be placed a given distance from the fulcrum for given pressure.. Weight, amount to be placed a given distance from the fulcrum for given pressure-rule in condensed form.. Weight, distance from fulcrum to place given weight for given pressure..59, 61 Weight, distance from fulcrum to place given weight for given pressure— rule in condensed form... Weight, pressure produced on valve by weight of Safety-valves Spring loaded Area of how determined .. Diameter of-how determined... Saturated steam, tables of properties of.. Sines Slide-valve and Corliss engines diagrams Slide-valve Engines.. Advantages of .... • Compound throttling non-condensing Compound, advantages of non-condensing Cross compound. Cross compound, advantages of.. • 65 90 .412, 434 412 413 230 100 554 720 727 .. 726 727 728 728 Horse power of compound, to find the... Houston, Stanwood & Gamble cross compound Sinker-Davis 730 728 720 Valve and crank movement 722 Valves of cross compound, to set the. 731 Valves of Sinker-Davis, to set the 722 Valves of slide-valve, to set the 731 Slide-valve performance. 643 Consequence to other events of setting valve for equal cut off, how to determine the 646 Cut off and exhaust closure, to determine the points of 644 Exhaust closure, how to locate... 647 Functions, condensed form-how determined.... 649 Laps required for given results, how to find ... 648 Position of piston when crank is at middle of its travel Smoke Stack Proportions Air space required for grate bars ..643, 644 184 186 INDEX. 797 Smoke Stack Proportions-Continued. Area of stack. Breeching, proportions and arrangement of Diameter of round stack Diameter of square stack... Draught, force of, required... Grate bars, percentage of air space required. Height of stack for bituminous lump coal Height of stack for bituminous nut coal.. Height of stack for bituminous slack coal. Specifications and plans for Western river steamboat boilers and attach- ments... ... PAGE.. 184 186 185 184 186 186. 186 187 187' 517 Specifications for engines and boilers for Western river stern-wheel steamer.. 483 Specifications for tubular boilers... Speed of engines, limitation of ... 188 532 Speed of Pulleys and Gearing 277 Diameters of driving and driven, on intermediate shafts.. Diameters of-how determined 290 277 Diameters of, in a series-how determined 282 Diameters of intermediate driven-how determined.... 286 Diameters of intermediate driving-how determined .. 284 Diameters of intermediate driving and driven.... 288 Diameters of intermediate-general rule... 288 Speed of driven, in a series-how determined.. Speed of general rule .. .278, 281 288 Speed of-how determined 278 Speed of gearing.. 292 Diameter of driven wheel-how determined 294 Diameter of driven wheel-how determined, simple method. 297 Diameter of driving wheel-how determined.... 293 Diameter of driving wheel-how determined, simple method ....... 296 Diameter of gear wheels-simple method.... 296 Spring-loaded safety-valves.... 434 Square inch, decimal parts of a. 36 Stay bolts... 370 Steam, economy in the generation and use of.. 553 Percentage of water in-how determined. 237 Tables of properties of saturated.. 230 Velocity of, flowing into the atmosphere. 773 Weight discharged under given pressure through given orifice 129 Steamboat Practice, Western River 483 Steam Boilers... 117 Steam drum legs.... 436 Steam Engineering, Mathematics of.. 33 Steam engines, action of reciprocating parts of. 677 Steam jacketing intermediate receiver. 688. Steam pipes, copper, to determine thickness of material. 457 Steam ports, area of-how determined .. 600, 601 Steam ports, dimensions of-how determined.. 603 Steam ports, engines having several.. 604 Steam ports, width of..... 605 Steam pressure, based on hydrostatic pressure 364 $798 INDEX. Steam pressure, reduction of working.... Steam pressure, table of Temperature of boiler furnaces-how determined Tensile strength of boiler plate-how determined Test diagrams. Test piece, form of iron.. Test piece, form of special Test piece, form of steel.. Theoretical curve, the geometric method for determining Theoretical expansion curves... Tubular Boilers, specifications and plans for. Boiler plate, actual strength of... Diameter required-how determined. Eclipse man hole and cover .... .. Horse power, heating surface required for. How to draught plans and specifications for. Longitudinal seams, data for double-riveted.. Longitudinal seams, data for single-riveted Longitudinal seams, strain on Material for setting, table of ... ... Plan and elevation of left-hand engine connected to boilers Plan and elevation of right-hand engine connected to boilers.... PAGE. 363 365 275 .114, 346 615 350 354 353 633 631 188 ..217, 342 218 192 218 215 222 221 216 214 212 213 Plans and dimensions for single setting, full front..........194, 195, 196, 197 .200, 201 .202, 203 Plans and dimensions for single setting, half arch front.....198, 199, 200, 201 Plans and specifications for single setting, half arch front... Plans and specifications for single setting, full front... Plans and specifications for double setting, half arch front Plans and specifications for double setting, full front.... Plans and specifications for triple setting, half arch front.. Plans and specifications for triple setting, full front. Plate, maximum thickness of..... Plate, tensile strength for double-riveted Plate, tensile strength for single-riveted.. Plate, thickness of, for double-riveted.. Plate, thickness of, for single-riveted. Water column, safety United States Battle Ship. Air pumps and condensers Boilers, arrangement of ... Grate surface of.... ... ..204, 205 ...206, 207 .208, 209 .210, 211 222 221 220 219 218 192 471 476 482 482 482 476 482 471 482 Engines 476 Horse power of ... 471 Evaporators 482 Feed water, how supplied to boilers.. 482 Speed of vessel. 471 Triple screws 471 Heating surface of .. Main and auxiliary Type of Clutch coupling.. Draught, how supplied to boilers INDEX. 799 PAGE. United States lever safety-valve United States Marine Boiler Inspection and Engineers' License Laws... Boiler plate 433 335 342 Ductility, percentage of, required.. Actual strength of Ductility of.... Ductility of-how determined Steel, special test of.. Tensile strength of-how determined Tensile strength and safe-working pressure Test piece, iron, form of..... Test piece, length of reduced section of steel. Test piece, steel, elongation of... Test piece, steel, form of 312 347 348 351 354 346 351 350 353 353 353 Test piece, special test of steel, percentage of elongation. 354 Thickness for marine boilers limited.. Thickness required ... Boilers, bursting pressure for double-riveted.. 363 344 339 Bursting pressure for double-riveted-simple rule Bursting pressure for single-riveted Bursting pressure for single-riveted-simple rule.. Diameter of double-riveted-how determined Hydrostatic pressure for double-riveted 340 338 338 470 361 ... Hydrostatic pressure for double-riveted-simple rule Hydrostatic pressure for marine. 362 360 Hydrostatic pressure for single-riveted. 360 Hydrostatic pressure for single-riveted-simple rule 361 Hydrostatic pressure for single and double riveted. 362 Reduction of working steam pressure of.. 363 Safe-working pressure based upon data for any given marine .... Safe-working pressure based upon ductility and tensile strength of 349 plate.. 351 Safe-working pressure for double-riveted 345 Safe-working pressure for double-riveted-simple rule 346 Safe-working pressure for marine. 340 Safe-working pressure for single-riveted. 341 Steam drum legs for, dimensions of ... 436 Steam drum legs for, how to determine dimensions of... 436 Steam pressure based on hydrostatic pressure. 364 Steam pressures allowable, table of .... 365 Strain produced on plates by pressure in. 344 Thickness of material for single-riveted. 363 Bumped heads for .. 355 Radius to which heads are required to be bumped.. Thickness of material... 359 357 Thickness of material-simple rule.. 358 Working steam pressure.. 355 Working steam pressure-simple rule 356 Examination questions and answers Fire pumps To determine size required.. 458 405 405 800 INDEX. United States Marine Boiler Inspection, etc.-Continued. Flat surfaces Area of surface stayed by bolt .... Engineers required to calculate strains on stays and braces... Plates not over of an inch thick Diameter of stay-bolt-how determined.... Pitch of bolts based on diameter of bolts and steam pressure Pitch of bolts based on thickness of plate and steam pressure. Steam pressure based on diameter and pitch of bolts PAGE, 370 370 370 373 373 375 376 378 Steam pressure based on thickness of plate and pitch of bolts... 379 Steam pressure, United States Government rule 380 Thickness of plate based on pitch of bolts and steam pressure 376 Thickness of plate based on pitch of bolts and steam pressure- simple rule... Plates over of an inch thick Diameter of stay bolt-how determined ..... ... 377 381 ... ... 381 386 Pitch of bolts based on diameter of bolts and steam pressure 382 Pitch of bolts based on thickness of plate and steam pressure 384 Steam pressure based on diameter and pitch of bolts Steam pressure based on thickness of plate and pitch of bolts... 386 Thickness of plate based on pitch of bolts and steam pressure 385 Plates over of an inch thick, for other than furnaces, etc.. Diameter of stay bolt-how determined.... 387 387 389 ... Pitch of bolts based on diameter of bolts and steam pressure Pitch of bolts based on thickness of plate and steam pressure... 390 Steam pressure based on diameter and pitch of bolts 392 Steam pressure based on thickness of plate and pitch of bolts... 392 Thickness of plate based on pitch of bolts and steam pressure... 391 Plates over of an inch thick, for other than furnaces, etc., stay bolts with double nuts, etc...... Diameter of stay bolt-how determined... 394 394 395 Pitch of bolts based on diameter of bolts and steam pressure Pitch of bolts based on thickness of plate and steam pressure... 396 Steam pressure based on diameter and pitch of bolts 398 Steam pressure based on thickness of plate and pitch of bolts... 399 Thickness of plate based on pitch of bolts and steam pressure 397 Steel stay bolts exceeding 14 inches and not exceeding 2 inches in diameter .... Diameter of stay bolt-how determined.. 400 400 401 402 402 404 Steam pressure based on diameter and pitch of bolts Steel stay bolts exceeding 2 inches in diameter Diameter of stay bolt-how determined.. Steam pressure based on diameter and pitch of bolts... Flues, Boiler-Corrugated and ribbed, safe-working pressure allowable... 439 Corrugated and ribbed, thickness of material required Flues, Furnace-Half-round iron rings for . Plain cylindrical, pressure allowable.. Required diameter of half-round iron rings Required diameter of half-round iron rings-simple rule With Adamson and angle-iron rings 440 446 441 446 447 441 INDEX. 801 United States Marine Boiler Inspection, etc.-Continued. ... PAGE. 444 445 442 With Adamson and angle-iron rings, diameter required. With Adamson and angle-iron rings, distance between rings. With Adamson and angle-iron rings, pressure allowable... With Adamson and angle-iron rings, thickness of material.. With half-round iron rings, diameter of-how determined With half-round iron rings, how to determine distance between rings. 450 With half-round iron rings, pressure allowable With half-round iron rings, thickness of material required. Flues, Lap-welded .... 443 451 448. 449 456 Lap-welded and riveted, over 6 and not over 16 inches in diameter and not over 18 feet in length. 457 Made in sections, increase length of sections-how allowable. Not over 6 inches in diameter Made in sections over 10 inches in diameter, pressure of-how deter- mined.... Lap-welded and riveted, over 6 and not over 40 inches in diameter, re- quired to carry pressure exceeding that prescribed in table (p. 453). 457 Lap-welded and riveted-made in sections 452 455 .. 455 .. 456 Over 6 and not over 16 inches in diameter and not over 18 feet in length. 457 Table of diameters and thickness for tubes and flues not otherwise provided for... 456 Area required-how determined.... Table of dimensions and pressures Horse power of marine engines-how determined. Mean effective pressure-how determined... Inspectors of boilers, qualifications of..... Safety-valves, United States laws relating to Safety-valves, lever weighted.... Diameter required-how determined Distance from fulcrum to place given weight for given pressure..... 416 Practical adjustment of... 453 406 406 336 409 · .. 410 411 411 414 Pressure required to raise Pressure required to raise, with conic weight on lever. Pressure required to raise, with spherical weight on lever ...50, 414 427 424 pressure.. United States lever valve ... Pressure required to raise, with two or more weights on lever..... Size of cubic weight required given distance from fulcrum for given United States lever valve, specifications for .... Weight, amount to be placed a given distance from fulcrum for given pressure Safety-valves, spring-loaded... Area of how determined Diameter of-how determined 431 421 433. 433. .. 418 .412, 434 412 413 Steam drum legs, dimensions of 436. Steam drum legs, diameter of-how determined 437 Steam pipes, copper 457 Steam pipes, copper, thickness of material-how determined. United States marine examination questions and answers 457 458. 51 802 INDEX. Valve, circumference of a-how determined. Correct diameter of a..... Diameter of, for given area-how determined .... Diameter of, for given circumference-how determined Valves of Atlas automatic engine, how to set the... Brown engine, how to set the ………. Buckeye engine, how to set the..... Compound locomotives, how to set the Corliss engine, how to set the ……….. Frisbie marine engine, how to set the.. Houston, Stanwood & Gamble engine, how to set the... Ordinary locomotives, how to set the... Porter-Allen engine, how to set the... Rees marine engine, how to set the Sinker-Davis engine, how to set the Slide-valve engine, how to set the... Sweeney marine engine, how to set the Weir-Harden engine, how to set the Valves Generally Bevel-seated, area of opening produced by lift. .... PAGE. 41 39 40 41 584 596 611 754 .564, 566 508 731 734 669 516 722 731 504 ... 717 95 96 98 97 ... 102 110 109 .... 100 95 Area of opening produced by lift when lift is above the seat... Area of opening produced by lift when lift is not above the seat.. Area of opening produced by lift when seat has an angle of 45° Diameter of required valve of given lift to equal area of opening produced by given lift of given valve.... Lift of valve, to determine when area of opening required and diam- eter are given….. Sines and cosines Flat-seated, area of opening produced by lift.... Diameter of required valve of given lift to produce area of opening equal to area of opening produced by given lift of given valve.. 109 Lift of required valve to produce area of opening equal to that pro- duced by lift of given valve.. 105 Lift of required valve to produce area of opening equal to that pro- duced by lift of given valve-simple method.... 106 Lift of required valve to produce area of opening equal to area of a given valve or cross section of a given pipe... 106 Lift of required valve to produce area of opening equal to area of a given valve or cross section of a given pipe-simple method .... 108 Round hole, size of, to determine from area of opening produced by lift.. 95 .... Square hole, size of, to determine from area of opening produced by lift..... 95 Valve proportions 105 175 .. Water-Horse power required to elevate.. Column of, how to determine height corresponding to a given pressure.. 779 Consumption, how to calculate the theoretical Consumption, table of data for calculating. Consumption, theoretical.. ... 641 642 640 240 ... 779 Table of properties of, from 32° to 212° Fahrenheit Velocity of, flowing from a boiler under given pressure INDEX. 803 Weight of cast-iron-To determine the Ball or sphere, to determine the Conic weight, to determine the.... Cylindrical weight, to determine the. Weight of steam discharged under given pressure through orifice of given diameter Weir-Harden Compound Engine... Cut-off valve Cut-off valve, balancing feature of the. Cut-off valve, to set the Diagrams showing cut-off valve performance. Governor.... Governor, to adjust the... Main valve, balancing feature of the Main valve, to set the..... Western River Steamboat Practice Boilers and attachments, specifications and plans for Cams Cut-off, how to lay off...... Frisbie marine engine. Full-stroke, how to lay off. Balance puppet valves of.. Cut-off cam, to set a... Full-stroke cams.. Full-stroke cam, to set a.... Plans for non-condensing engine for side-wheel steamer Rees marine engine.. Adjustable cut-off of the Full-stroke cam, to set the.. Full-stroke cam, movements of the Steam and exhaust valves of the PAGE. 44 ..47, 424 427 43 129 698 706 710 718 707 710 718 . 706 717 483 517 507 509 . 508 506 506- 510 507 508 496 511 . 511 516 513 513 Specifications for engines, boilers and attachments for stern-wheel steamer. 483. Specifications and plans for boilers and attachments Sweeney marine engine. Valve gear of the .... Valve gear of the, to adjust the .... 517 502 502 504 ADVERTISEMENTS. Nathan Manufacturing Co. 92 & 94 Liberty St., NEW YORK, N. Y. INJECTORS, EJECTORS, BOILER TESTING and FILLING APPARATUS ...OIL CUPS AND CYLINDER LUBRICATORS... For Marine, Stationary, Portable and Locomotive Engines. MONITOR "Monitor" INJECTORS ...FOR... Feeding High Pressure Land and Marine Boilers on Lake and River Steamers. The attention of river and lake navigation companies, and the proprietors and builders of steamboats is called to the su- periority of "MONITOR" IN- JECTORS. These injectors combine within themselves the double capacity of Boiler Feeders and Boiler Testers. They are constructed upon the same principle upon which the celebrated Loco- motive "Monitor" Injectors are constructed, and which are in universal use on locomotives. The Superiority of "Monitor" Injectors over pumps as boiler feeders has been fully demonstrated by the rapidity with which they have displaced the latter on river steamers. They are not only superior to pumps as boiler feeders, but they have the advantage of occupying much less space, besides having much less weight, which is a great desideratum in river steamers, where lightness is one of the principal objects. In addition to weighing considerably less, and taking up much less room than the "doctor" pump, the "Monitor" Injectors have no movable parts to get out of order, and are, therefore, more reliable in service than the "doctor," with its many valves and stuffing boxes, which are a source of con- stant trouble, expense and delay. The "Monitor" Injectors feed hot water to the boilers at all times when in operation, thus avoiding the injurious effects upon boiler plates, tubes, heads, etc., produced by the injection of cold water. This Company also manufactures a great variety of other Injectors and Ejectors; also Boiler Testing and Filling Apparatus, Oil Cups and Cylinder Lubricators for Marine, Stationary, Portable and Locomotive Engines. WRITE FOR CATALOGUE NATHAN MANUFACTURING CO. 92 & 94 Liberty St. NEW YORK, N. Y. ADVERTISEMENTS. Steam Boilers... TANKS Heavy and Light Sheet Iron Work POINT BOILER WORKS R.MUNROE & SON PLTSBURGH PA AUGUST 1892 FOR ECONOMY, DURABILITY, AND LOW PRICE, MUNROE'S PATENT WATER TUBE WATER VILER BOILERS HAVE NO EQUAL. ..Manufacturers of.. Munroe's Patent Water Tube Boilers AND. Eclipse Manholes. West Point Boiler Works- R. Munroe & Son, 23d Street, PITTSBURGH, PA. ADVERTISEMENTS SPECIFY Pittsburgh Safety Water Columns Write for Prices. THE BEST! THE SIMPLEST! Because.. THEY SAVE FUEL by the carrying of a uniform gauge of water in the boiler at all times. Moreover, THEY AFFORD ABSOLUTE SAFETY in running a low gauge of water to save fuel and gain more steam space-more power. THEY PREVENT MISTAKES when new men are employed; when the watch- man is in charge; when the pump stops, or when it furnishes too much water. NO EXPLOSIONS: no leaky tubes; no burned crown sheets; no burst engine cylinders; no aggravating or ex- pensive stoppages. Altogether THE RELIEF from anxiety is well worth their slight cost. Pittsburgh Rotating Gauge Cocks. An Improvement in Self-Grinding. No Packing. No Leaking. The Best! Gauge Cocks. Metal Seat. Metal Valve. The Latest! The Pittsburgh Rotating Gauge Cocks remain closed with or without pressure on the boiler. The Pittsburgh Gauge Co. 201 Market St. PITTSBURGH, PA. ADVERTISEMENTS. The I. & E. Greenwald Co. Engineers, Founders AND Machinists 1.& CINCINN HEADQUARTERS FOR Mill Gearing of all kinds, Shafting, Hangers, Pulleys, Sprocket Wheels, Chains, etc. Mill Gearing a specialty. Steam Power Plants complete. Automatic Cut-off and Slide-valve Steam Engines. 720 EAST PEARL ST. ..SEND FOR. SEE CATALOGUE OF PATTERNS, ETC. PAGE 592. || CINCINNATI, O. ADVERTISEMENTS. THE LANE & BODLEY CO. CINCINNATI, OHIO. Corliss Engines SIMPLE and COMPOUND. UNRIVALED FOR DURABILITY, PERFORMANCE AND ECONOMY. SLIDE VALVE ENGINE DEPARTMENT. SIMPLE ENGINES (15 to 100 Horse Power), HORIZONTAL and VERTICAL CROSS COMPOUND, FIXED CUT= OFF ENGINES. SHAFTING, HANGERS, PULLEYS, FREIGHT ELEVATORS. SEE PAGES 528-550. FULTON OIL WORKS, HIGH GRADE C. P. WAGNER & CO. -OILS- Cylinder, Engine and Dynamo. Lubricating, Lard and Cooking. Elevator, Journal and Belt Greases. No. 116 BROADWAY, TELEPHONE 1556. CINCINNATI, OHIO. THE.. 'Scientific'.Feed Water Purifier ..IS.. Guaranteed to Prevent .the. REMOVES The Scale-Forming Properties from the Feed Water while It is COLD, Accumulation of Scale. and BEFORE IT GOES into the HEATER and BOILERS. THE WE-FU=GO COMPANY, Smith and Augusta Streets, CINCINNATI, OHIO. ADVERTISEMENTS. For Estimates ON BOILERS, TANKS, ETC. WRITE TO THE MCILVAIN & SPIEGEL BOILER AND TANK CO. CINCINNATI, O. JAMES REES & SONS Co. IRON, STEEL, AND WOODEN STEAMBOATS BY CONTRACT. MARINE AND LAND ENGINES AND BOILERS. Locomotives, Tanks, Stills, and Iron Work Generally, PITTSBURG HIGH SPEED AUTOMATIC .. AND.. HIGH SPEED ROLLING MILL ENGINES. Duquesne Way and Fourth Street, PITTSBURG, PA. SEE PAGES Nos. 483, 498, 512, 513 AND 515. non SEP 27 1957 UNIVERSITY OF MICHIGAN 3 9015 07321 8870