II Hill sib g Silili mm II ■ if 111 3SS6T 111 Hill 1 I M 1 i tlillii Class / V '* _. Book. _____ GoppghtN .10 COPYRIGHT DEPOSIT. ARITHMETIC OF THE STEAM BOILEE THE POWER HANDBOOKS The best library for the engineer and the man who hopes to be one. This book is one of them. They are all good — and they cost $1.00 postpaid per volume. (English price 4/2 postpaid.) SOLD SEPARATELY OR IN SETS By PROF. AUGUSTUS H. GILL OF THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY ENGINE ROOM CHEMISTRY By HUBERT E. COLLINS BOILERS KNOCKS AND KINKS SHAFT GOVERNORS PUMPS ERECTING WORK SHAFTING, PULLEYS AND PIPES AND PIPING BELTING By CHARLES J. MASON ARITHMETIC OF THE STEAM BOILER McGRAW-HILL BOOK COMPANY, Inc. 239* WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E. C. THE POWER HANDBOOKS 5 ARITHMETIC OF THE STEAM BOILER A REFERENCE BOOK SHOWING THE VARIOUS APPLICATIONS OF ARITHMETIC TO STEAM BOILERS BY CHARLES J. MASON First Edition McGRAW-HILL BOOK COMPANY, Inc. 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E. C. 1914 TJ ■X16 .Hi. Copyright 1913, by the McGraw-Hill Book Company, Inc. THE.MAPLE.PRESS.YORK-PA JAN -9 1914 ©CI.A361529 c THIS BOOK IS RESPECTFULLY DEDICATED TO MR. FRED R. LOW, EDITOR OF POWER, WHO HAS EVER TAKEN A DEEP INTEREST IN THE AFFAIRS OF THE ENGINEERING FRATERNITY PREFACE This book is a compilation of arithmetical rules and formulas applicable to steam boilers of various types. The author claims no originality in the preparation of the material, excepting only the arrangement and manner of presentation. It is intended as a book of reference for those who may require rules and formulas directly related to steam boilers, and its aim is concentration and logical order in the arrangement and treatment of the various features introduced. Most of the material was gathered during the author's career as a steam and marine engineer, covering a period of twenty-five years. It is not intended to teach the elements and principles of arithmetic in this book, as might perhaps be inferred from the title, but only the application of arithmetic to steam-boiler calculations. It is presumed that those who may use it already understand arithmetic but desire to have a compact set of rules and formulas con- veniently ready for use, without having to look through several books for a certain one when required. Those who are preparing for examinations for engineer's certifi- cates and licenses will find the work of great assistance to them. The author desires to thank all those who have in any way contributed to the production of this work, vii viii PREFACE particularly the publishers and editor of Power, from which paper various extracts have been taken, kind permission to use the same having been granted, and Mr. William Kent, M. E., author of Kent's Mechanical Engineer's Pocket-book. Charles J. Mason. SCRANTON, PENNA, December, 1913. CONTENTS Page Preface vii PART I CHAPTER I The Sphere, Stress per Square Inch Section; Safe Pressure i Calculations pertaining to the sphere — the cylinder — riveted joints — efficiencies — bursting and safe pressures. CHAPTER II Boiler Heads. — Unstayed Heads 30 Boiler heads — unstayed bolts — convex and concave — flat unstayed heads — stays and staybolts — diagonal stays — segments to be braced — girder bars. CHAPTER III Manhole Reinforcing Rings 56 Reinforcing rings — heating and grate surface — corrugated furnaces — horse-power of boilers — ratio of heating to grate surface — equivalent evaporation — boiler efficiency — boiler trials. PART II Miscellaneous .Applications 83 Bursting pressure of pipe — cost of evaporating 1000 lb. of water — safe pressure of flat cast-iron heads — equivalent boiler performance — efficiency of diagonal seam — collap- sing strength of cone-shaped flue — strength of cone seam — safety valves — Roper's rules — tapered levers — chimneys — size of feed pipes ix x CONTENTS PART III Page Appendix 133 Abstracts from rules, United States Board of Supervising Inspectors of Steam Vessels — abstracts from Massachu- setts' Boiler Rules. Tables 191 Diameters, areas and circumferences of circles — decimal equivalents — squares, cubes, cube roots and square roots — factors of evaporation — standard boiler tubes — Kent's table of chimneys — Mark's and Davis' steam tables. Index 221 ILLUSTRATIONS Figure Page i Diagram of stresses in a sphere 2 2 The cylinder 6 3 Types of riveted joints 12 4 Data sheet, double butt-strapped joint 22 5 Quadruple, double butt-strapped joint 25 6 Bumped heads 30 7 Diagram to find radius of a bumped head 31 8 Arrangement of direct and diagonal stays 37 9 Diagonal stays 40 10 Segment of head to be braced 48 11 Approximate method of finding area of segment 51 1 2 Girder bars 53 13 Direction of stress in reinforcing rings 60 14 Lower part of Manning boiler 86 15 Diagonal steam 91 16 Cone-shaped flue ! 93 17 Strength of seam in cone 94 18 Diagram of fire box 97 19 Diagram of safety valve dimensions for calculations 100 20 Diagram showing tapered safety valve lever for calculation. 1 13 21 Diagram of locomotive boiler 130 XI PART I BOILER CALCULATIONS BOILER CALCULATIONS CHAPTER I The Sphere; Stress Per Square Inch Section; Safe Pressure The sphere is the strongest form in which a steam boiler could be made, but because of mechanical and commercial reasons, that form is not used. In order to understand the stresses, due to pressure, endured by steam boilers, it is well to start with the spherical form, for that is the simplest, and it forms a basis for calcula- tions on the prevailing forms in which boilers are made. Given a spherical vessel made of metal of a certain thickness and known diameter, it is desired to find what stress per square inch section the metal is subjected to, due to a known pressure per square inch contained within the sphere. The pressure would tend to separate the sphere in halves, through a diametral plane. Actually, pressure in a closed vessel of any form radiates from the center outward. But for convenience in calculations the radiating forces may be resolved in two, and acting perpendicular to any diametral plane. The total pres- sure or force tending to burst the sphere asunder will be the product of the known pressure per square inch, 2 ARITHMETIC OF THE STEAM BOILER existing in the sphere, and the area in square inches of a diametral plane. For illustration, assume a pressure of ioo lb. per square inch; a sphere whose internal diameter is 30 in., and made Diametrical Plane Arrows show Direction of Resolved Forces Shaded Rings shows Thickness of Sphere Hemisphere Fig. i.— Diagram of stresses in a sphere. of 1/2-in. steel plate, no joints, seams, nor rivets. The area of the diametral plane is: 30 2 X. 7854 = 706.860 sq. in. As the pressure is 100 lb. on each square inch, then: 100X706.860 = 70686.00 lb. This is the total force tending to separate the sphere in two parts. This force is resisted by the area of metal at the cir- THE SPHERE 3 cumference of the plane upon which the pressure is as- sumed to act. This is actually a ring whose internal diameter is 30 in. and whose outside diameter is 31 in. This is called sectional area, and it is the difference be- tween the area of the inner and outer circles of which the ring of metal is formed. The area of a circle whose diameter is 31 in. is: 3 i 2 X. 7854 = 754.769 sq. in. and, 30 2 X. 7854 = 706.860 sq. in. Difference = 47.909 sq. in. That is, 47.909 sq. in. cross-sectional area of steel plate is enduring a total stress of 70,686 lb. or, 70686 47.909 = 1475.42 lb. per square inch section. Assuming the tensile strength of the plate to be 60,000 lb. per square inch section, the foregoing shows that considerably more than 100 lb. per square inch could safely be carried in the sphere considered, for with a factor of safety of 6, the stress would be 10,000 lb. and this would be obtained by having the pressure at 677.77 lb. per square inch, as a trial in calculation will show. Ordinarily, in practical work, the difference between the outside and inside diameters is not taken. For example, the rule that would be employed to find the stress per square inch in a sphere whose inside diameter is 30 in., and whose thickness of plate is .5 in., is: Multiply the area of the dia- metral plane by the pressure per square inch, and divide that product by the product of the inside diameter, the constant 3. 141 6, 4 ARITHMETIC OF THE STEAM BOILER and the thickness of the plate in inches. This written in formula is: d*X-7SS4XP . d X3.i4i6X*~ StreSS - But by simple cancellation the formula reduces to pXd —ttt = stress. 4Xt The value of the letters used is: d= inside diameter, inches; p= pressure, pounds per square inch; t= thickness of plate, inches; s= safe stress in pounds per square inch; .7854= a constant; 3.1416= a constant. In order to find the pressure per square inch that may safely be carried, when the safe stress, the diameter of sphere, and the thickness of plate are given, it is simply a matter of changing the rule and formula to suit the purpose, thus: Multiply four times the thickness of plate by the given safe stress per square inch section, and divide by the internal diameter in inches. Written in formula it is : 4/ X stress -j--=p. Applying this to the example chosen, it becomes: 4X .5X10000 3° = 666.66 lb. THE SPHERE 5 which for practical purposes is 667 lb. per square inch. Here, the internal diameter only has been taken. In the first method shown the mean diameter was taken, which gave a safe pressure of 677.77 lb. per square inch which for practical purposes is 678 lb. The difference is 11.1 lb. or 1.6+ per cent, difference in favor of the usual method, which, though not absolutely correct, errs on the side of safety as shown. If the sphere in the example were .25 in. thick, or 60 in. internal diameter, then the difference in the methods explained would be less than that given. Grouping all the rules and formulas pertaining to the sphere, so that any term may be found having the remain- ing terms given: To find the stress per square inch section endured by the plates, multiply together the pressure per square inch and the internal diameter in inches. Divide the product by four times the thickness of the plate in inches. Written as a formula this rule becomes: pXd 4X7 = * (I) To find the safe pressure per square inch that may be carried, multiply four times the thickness of the plate by one-sixth of the ultimate tensile strength of the material of which the plates are made, and divide the product by the internal diameter in inches. 4XtXs To find the internal diameter, multiply four times the thickness of the plate by one-sixth of the ultimate tensile strength of the 6 ARITHMETIC OF THE STEAM BOILER material of which the plates are made, and divide by the safe pres- sure per square inch. ^ = <*. (3) P To find the thickness of plate, multiply the internal diameter by the safe pressure per square inch, and divide the product by one- sixth the ultimate tensile strength of the material of which the plates are made; divide the quotient by 4. JXs~ L (4) The Cylinder Next to the sphere, the cylinder is the form best suited for steam boilers, and excepting the spherical form, the (a) ' W Fig. 2. — Stresses in cylinders, (a) Stress acting at right angles to the longitudinal plane abed. Tendency to separate the cylin- der at a d and b c. (b) Stress acting parallel to the longitudinal plane abed. Tendency to separate the cylinder at ef. cylindrical is the strongest. This is not hard to under- stand, when it is considered that pressure inside any vessel tends to make that vessel assume a spherical form, as illustrated in the inflated toy rubber balloon. For this reason flat surfaces in boilers must be braced, but cylindrical surfaces do not require any bracing. THE CYLINDER 7 For the sake of convenience in calculations, the force due to the pressure in a cylindrical vessel may be consid- ered as acting in two directions. One acts in the direc- tion tending to blow off the head, and the other at right angles to the diametral longitudinal plane. In the former, the calculations are exactly the same as has been explained in connection with the sphere. In the latter, the pressure acting against the imaginary plane tends to separate the cylinder into two parts, and that which resists the tendency is the area of metal along both sides of the cylinder. The stress per square inch section of the plate, due to any given pressure, may be found from the follow- ing rule: Multiply together the pressure per square inch, the diameter of the cylinder in inches, and the length in inches; divide the product by two times the thickness of the plate in inches multiplied by the length in inches. This, written as a formula, is: pXdXl 2XtXl (S) As the factor / appears both above and below the line, it may be cancelled, and the formula becomes, From the formula just given it can be seen that the stress increases as the diameter or pressure increases, and it also can be seen that as the thickness of the plate increases the stress on the plates decreases. By comparing formula (1) with formula (6) it is seen 8 ARITHMETIC OF THE STEAM BOILER that in the latter the stress is twice as great as it is in the former. A careful study of these rules and formulas will show why the stress in one case is just twice that of the other. For that reason, the longitudinal seams in a. boiler must be made stronger than the circumferential seams, and this is accomplished by having more than one row of rivets in the longitudinal seams. Circumferential seams are frequently double riveted to make a mechanically tight job. The strength of riv- eted joints will be treated further on. So far, seamless vessels have been considered, so as to introduce the sub- ject of stress and resistance to stress in the most simple form. The following rules relate to the cylindrical form, and belong in the group under consideration. To find the stress on each inch in the circumference tending to blow off the head in a longitudinal direction (not the stress per square inch section), the following rule is applicable : Multiply together the area in square inches of the end of the cylinder and the pressure per square inch, and divide the product by the circumference of the cylinder in inches. <* 2 X. 7854x1 JX3.1416 ""*■ {7) To find the total stress caused by the pressure in a cylinder, multiply together the diameter in inches, the length in inches, and the pressure in pounds per square inch. d XI Xp = totals (8) / = length in inches. To find the total pressure on the entire shell of a cylinder, multi- THE CYLINDER 9 ply together the circumference in inches, the length in inches, and the pressure per square inch. cXlXp = total pressure. (9) To find the bursting pressure of a cylinder, multiply together the thickness of the plate in inches and the tensile strength of the material of which the plate is made, and divide the product by the radius of the cylinder in inches. ZX T = bursting pressure per square inch. (ro) In this formula, t = thickness of plate in inches. T = tensile strength in pounds per square inch section of the material, and r = the radius in inches. To find the safe working pressure of a cylinder, multiply together the thickness in inches and the tensile strength of the material of which the plate is made, and divide that product by the radius in inches multiplied by whatever factor of safety may be desired. tX T ,. = safe working pressure per square inch. (n) rX/ Here follows examples showing the application of the foregoing rules and formulas from (1) to (n) inclusive. For the sake of clearness and convenience the same values will be used in all. Numbers easy to operate have been chosen, for no matter what numbers may be contained in any example which may come up in practice, the method of operation will be exactly the same. In formula (1) assume the following values: pressure (p) =100 lb. per square inch; diameter (d) = 60 in. ; thickness (t) =1/2 or .5 in. Then to find the stress (s) the statement becomes: 100X60 . — — = 3000 lb. per square inch section. 4X -5 10 ARITHMETIC OF THE STEAM BOILER To find the safe pressure (p) which may be carried, formula (2), the statement becomes: 4X -5X3000 „ . . = 100 lb. per square inch. To find the internal diameter (d), formula (3), the statement becomes: 4X .5X300 , . - = 60 in. 100 To find the thickness of plate (/), formula (4), the statement becomes: 100X60 = .5 in. 4X3000 Formula (5) reduces to that given in (6), and to find the stress per square inch section in this case, the state- ment becomes: 100X60 . n — — = 6000 lb. 2X.5 per square inch section. To find the stress (s) on each inch in the circumference, formula (7), the statement becomes: 6o 2 X. 7854X100 — z~r^ 7— = 1500 lb. 60X3. 1416 J To find the total stress on the entire shell, formula (8), assume a length of 144 in. with the other values remaining the same, the statement becomes: 60X144X100 = 864,000 lb. THE CYLINDER II To find the total pressure on the entire shell, formula (9), the statement becomes: 60X3.1416X144X100=2,714,342.4 lb. To find the bursting pressure, formula (10), the state- ment becomes: assume a tensile strength of 50,000 lb. per square inch section. .5X50000 n . ,. . . — — = 833.33+ lb. per square inch. To find the safe working pressure, formula (n), and assume a factor of safety of 5, the statement becomes: s X ^0000 ^^b = 166.66+ lb. 30X5 In practice 170 lb. would be allowed with factor of safety 5 per square inch. Or a simpler method is to take one-fifth of the bursting pressure thus: 83 ^=i66.66 1b. 5 and as the bursting pressure is five times the safe pressure, 5X166. 66+ = 8 33 . 33+ lb. In actual practice, 6 is used as a factor of safety more frequently than 5. However, the method of operation is the same for any and all factors of safety that may be used; so that if the method is understood, it matters not as to the values that may be substituted in the various formulas treated of. 12 ARITHMETIC OF THE STEAM BOILER Riveted Joints In the previous sections, spheres and cylinders without joints or seams were assumed in order to simplify explana- Pitch Pitch Pitch A © ©!© © © ©^ >©©©©_© © © © © ©i © © © ©!© ~^- w Fig. 3. — Types of riveted joints. (A) Single-riveted lap joint; one rivet in single shear. (B) Double-riveted lap joint; two rivets in single shear. (C) Triple- riveted, double butt-strapped joint; four rivets in double shear, one rivet in single shear. (D) Quad- ruple-riveted, double butt-strapped joint; eight rivets in double shear, three in single shear. tions and calculations. In actual practice, however, steam boilers are constructed with both longitudinal RIVETED JOINTS 13 and circumferential joints, or seams. The seams are secured by rivets. There are several kinds of riveted joints known to engineers and others who have to do with boilers. The strength of a riveted joint depends upon how the joint is made, as to the size and pitch or spacing of the rivets, and the number of rows of rivets, and also as to whether the joint is what is known as a lap or butt-strapped one. A riveted joint of any kind is not theoretically as strong as the solid part of the plate, although in practice it has been known for boilers to tear apart at some place other than at the riveted joint. This probably was due to a flaw or weakness in the metal. If a cylindrical vessel were made of plates uniform in structure and thickness throughout, and if tested to destruction, it would likely break or pull apart at the riveted longitudinal joint. In making calculations it is presumed that a break would occur at the joint, rather than at any other place. Strength of Riveted Joints The strength of a riveted joint is compared with that of the solid plate, the latter being valued at 100 per cent. Nominally, the strength of joints varies from 56 per cent, to 94 per cent., the former value representing single- riveted lap joints, and the latter, quadruple, double butt-strapped joints. Between single-riveted lap joints and quadruple-riveted, double butt-strapped joints, there are: double-riveted lap joints, triple-riveted lap joints, quadruple-riveted lap joints, single-riveted butt joints, double-riveted butt joints, triple-riveted butt joints. The foregoing joints may be either chain riveted, 14 ARITHMETIC OF THE STEAM BOILER or what is termed zig-zag riveted. In triple-riveted joints and in quadruple-riveted joints it is customary to omit every alternate rivet in the outer rows; this admits of a stronger joint. The strength of a riveted joint depends upon the size and pitch of the rivets, the number of rows of rivets, type of joint as to lap or butt, the tensile strength of the plates, and the shearing stress of the material of which the rivets are made. In rinding the strength of a joint two things are to be considered, the strength of the plate section and the strength of the rivet section. The lesser value, as found from an analysis, is taken as the strength of the joint as a w T hole. Theoretically, riveted lap joints and those butt joints with one cover plate should be designed so that the rivet and plate sections are equal — or as nearly equal as possi- ble — in strength. But in practice it is usually considered desirable to so design a joint that the plate section is a little stronger than the rivet section; this particularly relates to joints having all the rivets either in single or double shear, for the reason that the plates become thin from w T ear, with a consequent reduction in strength, while the rivets suffer little if any from wear. But in joints having some of the rivets in single shear and some in double shear, the greatest strength usually obtains when the rivet section exceeds the strength of the net section of plate. Efficiency of Riveted Joints To determine the efficiency of a riveted joint, its re- sistance must be calculated for each of the different ways RIVETED JOINTS 15 in which it may fail, and then the lowest efficiency so found in relation to the solid plate, will be the one by which the joint is known. A riveted joint may fail in the following ways: (1) The plate may break asunder along the rivet holes, at the net section. (2) The rivets may shear off, leaving the plates intact. (3) The plate may shear out in front of the rivets. (4) The plate may crush in front of the rivets. (5) In joints having zig-zag rivets the plate may break diagonally between the rivet holes. (6) The joint may fail by a combination of the fore- going. With joints as usually proportioned, the liability to failure in the ways referred to in (3), (4), (5) and (6) is reduced to a minimum; this by having the distance from the edge of the plates to center of rivet holes one and one- half times the diameter of the rivet holes. It is custom- ary (except in special cases which will be referred to) to consider only (1) and (2) as possible ways of failure, and base all calculations upon those two ways. Therefore, calculate the efficiency of the net section of the plate part of the joint as compared with the solid plate, and then find the efficiency of the rivet section of the joint as compared with the solid plate; the lesser of the values found is to be taken as the final efficiency of the joint as a whole. Single-riveted Lap Joints To find the efficiency of a single-riveted lap joint. (The distance from the edge of the plate to the center 1 6 ARITHMETIC OF THE STEAM BOILER line of rivet holes must be not less than one and one-half times the diameter of the rivet hole, for all joints.) First, find the strength of a unit of length of the solid plate. PXtX S = strength of solid plate. In which, P = pitch of rivets in inches, from center to center. / = thickness of plate in inches. 5 = tensile strength of plate, in pounds per square inch section. d = diameter in inches of rivet holes. The next step is to find the strength of the net section of plate between the rivet holes: (P—d)XtXS = strength of plate between the holes. Next, find the shearing strength of one rivet in single shear: nXsXa = shearing strength of one rivet. n = number of rivets in single shear; 5 = shearing strength of rivet; a = cross-section area of rivet, after driving. Take the lesser of these two results and divide it by the value found for the strength of the solid strip; the quotient will be the efficiency of the joint decimally expressed. Example. — Single-riveted lap joint of the following dimensions: •S =55,ooo lb. P =1.625 in. t =.25 in. d = .6875 in. 5 =42,000 lb. a = .3712 sq. in. RIVETED JOINTS 1 7 The strength of the solid strip will be: i. 625 X. 25X55*000 = 22,343 lb. The strength of the net section of plate between the rivet holes will be: (i. 625-. 6875) X. 25X55,000 = 12,890 lb. The strength of the rivet in single shear will be: 1 X42,oooX . 3712 = 15,590 lb. As the net section of plate in this example is weaker than the rivet, its value must be used. Then: 12800 . = .576, or 57.6 per cent, efficiency. Shearing Strength or Rivets The shearing strength of rivets may be taken from the following table (from the Massachusetts Board of Boiler Rules) : Iron rivets in single shear, 38,000 lb. Iron rivets in double shear, 70,000 lb. Steel rivets in single shear, 42,000 lb. Steel rivets in double shear, 78,000 lb. These values are on the safe side, as they are lower than some others that are in use. Double-riveted Lap Joints To find the efficiency of double-riveted lap joints, the method of procedure is the same as that for single- riveted joints, with the exception that there are two rivets in single shear instead of one as in single-riveted joints. 18 ARITHMETIC OF THE STEAM BOILER Example. — A double-riveted lap joint has the following dimensions: 5 =55,000 lb. t = .3125 (5/16) in. P =2.875 (2 7/8) in. d = -75 (3/4) in. a = .4418 sq. in. 5 =42,000 lb. The strength of the solid plate is: 2. 875X. 3125X55,000=49,4*4 lb. The strength of the net section of plate is: (2. 875-. 75) X. 3125X55,000 = 36 ? 5 2 3 lb. The strength of the two rivets in single shear is : 2X42, 000X .4418 = 37,111 lb. Here again the plate section is the weaker, so the value for that must be used: — = . 739, or 73 . 9 per cent, efficiency of joint. 49414 Triple- and Quadruple-riveted Lap Joints In triple-riveted and quadruple-riveted lap joints (sometimes used in marine boilers) there are three and four rivets, respectively, in single shear. With this exception, the method of finding the efficiency of such joints is the same as for single and double, as just illustrated. Lap Joints Lap joints for the longitudinal seams are now consid- ered not safe for steam boilers of more than 36 in. in RIVETED JOINTS 19 diameter, and for pressures higher than 100 lb. per square inch; and probably as time goes on they will not be used at all; but it is important to know how to calculate the strength of such joints, hence the reference to them in this book. Butt Joints Butt joints with double cover plates are the strongest and safest joints in use. The minimum thickness of cover plates, or butt straps as otherwise called, is as follows : PRESCRIBED BY THE MASSACHUSETTS BOARD OF BOILER RULES Thickness of shell Minimum thickness of plates butt straps 1/4 in. 1/4 in. 5/16 in. 1/4 in. 3/8 in. 5/16 in. 7/16 in. 3/8 in. 1/2 in. 7/16 in. 9/16 in. 7/16 in. 5/8 in. 1/2 in. 3/4 in. 1/2 in. 7/8 in. 5/8 in. 1 in. 3/4 in. 1 1/8 in. 3/4 in. 1 1/4 in. 7/8 in. Single Butt Straps Single butt straps should never be thinner than the plates of the shell. In some instances (British Board of Trade and Canadian Rules) the minimum thickness must be not less than one and one-eighth the thickness of the shell plates. Double butt straps must be at least 3 20 ARITHMETIC OF THE STEAM BOILER five-eighths, and preferably the thickness of the shell plates. If the shell plate is light, say 7/16 in. or less, the outside strap should be as heavy as the plate, to admit of a tightly calked joint. When single butt straps are used, the method of finding the efficiency of the joint is the same as that for lap joints, for the rivets are all in single shear, and the pitch of the rivets is the same in each row. Number of Rivets Considered In lap joints, all the rivets in a given pitch strip of plate are taken into account when figuring for the effi- ciency of joint, while in butt joints only those rivets on one side of the center line of the joint are considered. A little thought on the part of the reader will make clear the reason. Number oe Rows of Rivets In double butt-strapped joints, three or four rows of rivets on each side of the center line are generally used. In the former (triple riveted) the pitch of the outer row of rivets on each side of the center line is twice the pitch distance of the two inner rows of rivets on each side of the center line. In the latter (quadruple-riveted joints) the outer row of rivets on each side of the center line of joint is four times the pitch distance of the two inner rows on each side of the center line; in the rows next to the outer rows, the rivets are pitched twice the distance of those in the two inner rows. (See Fig. 3.) RIVETED JOINTS 21 Number of Rivets in Double and Single Shear In triple-riveted butt joints, there are four rivets in double shear, and one rivet in single shear, in a given pitch strip. In quadruple-riveted butt joints, there are eight rivets in double shear, and three rivets in single shear, in a given pitch strip. When calculating the efficiency of triple- and quadruple- riveted joints, the strength of the net section of plate is taken at the outer row of rivets, where the pitch is the greatest. The reason for this will be explained presently. High Joint Efficiencies Due to Wide Spacing of Rivets at the Outer Rows It is because of the wide spacing in the outer rows of rivets that such high efficiencies can be obtained with those types of joints as compared with those joints in which the pitch of the rivets is the same for all the rows. In order to explain why the net section of the plate at the outer row of rivets is taken, an illustrative example of a triple-riveted, double butt-strapped joint will be used; the same principles may be applied to a quadruple joint of the same kind. Thickness of shell plates 3/8 in., tensile strength 50,000 lb. per square inch section, rivet holes 13/16 in., rivets 3/4 in. diameter; shearing stress of the rivets taken as 38,000 lb. per square inch section. The pitch of rivets in the two inner rows is 3 1/4 in., and in the outer row, 61/2 in. 22 ARITHMETIC OF THE STEAM BOILER The width of strip to be considered in this case is 6 1/2 in. The sectional area is .375X6.5 = 2.4375 sq. in. 2.4375X50,000=121,875 lb. strength of the solid strip, with which the joint is to be compared. Next find the strength of the net section of the plate at the outer row of rivets. As there is but 1 rivet in the 6 1/2 in. strip under con- Inside Cover Plate Outside Cover Plate Bottom Fig. 4. — Diagram of triple-riveted double butt-strapped joint. sideration, at the outer row, then, 6. 5— .8125 = 5.6875 in. width of net section of plate; and, 5. 6875 X. 375 X 50,000=106,640 lb. strength of net section of plate between the rivet holes. If the plate should break or pull asunder at the line of the outer row of rivets, the resistance to breaking is the metal in the net section of the plate as shown at c in Fig. 4. But should the plate break along the net RIVETED JOINTS 23 section at the inner row of rivets, the resistance offered is the strength of the sections of plate E D and F, and as well as that, the resistance to shearing offered by one-half of each of the rivets in the outer row, which is equivalent to one rivet in calculation. First, find what the plate resistance is. By measure- ment it is 6.5 — (2X .8i25)=4.875 in. Here there are two rivet holes to subtract from the width of the strip. The sectional area will be 4. 875 X .375 = 1 .828 sq. in.; and 1.828X50,000 = 91,400 lb. resistance. To this must be added the resistance offered by the two half rivets in the outer row. The area of a 13/16-in. driven rivet is .5185 sq. in., and as the shearing stress is 38,000 lb. per square inch, .5185X38,000=19,703 lb. This added to 91,400 lb. before obtained for the plate gives 111,103 lb. total resistance to the plate breaking at the inner row of rivets, as against the 106,640 lb. found for the net section of plate at the outer row of rivets. Therefore the latter is the weaker, and there the plate w T ill probably break, if at all. Continuing, there are four rivets in double shear, and one rivet in single shear, in the strip. The resistance to shearing of one rivet was shown to be 19,703 lb. The resistance to shearing offered by each rivet in double shear is: 38,oooX .85=32,300 and 38,000+32,300 = 70,300 lb. per square inch section. (The factor .85 is a value used for rivets in double shear as double shear does not necessarily mean twice that of single shear.) The area of each rivet is .5185 24 ARITHMETIC OF THE STEAM BOILER sq. in., and there are four to consider, therefore, .5185 X4X7o,3oo= 145,802 lb. and to this add the value for the rivet in single shear, 19,703 lb., which gives a total of 165,505 lb. shearing strength of all the rivets in the strip. The net section of plate at the outer row of rivets is the weakest part of the joint as a w T hole, and its value is to be compared w T ith the strength of the solid strip in order to find the efficiency. The net section of plate is 106,640.625 lb. and the solid strip is 121,875 lb.;theeffi- ciency is: 106640.62=; —^ = 87. s per cent. 121875 ' J ^ Quadruple-riveted, Double Butt-strapped Joint To analyze a quadruple-riveted joint and find the efficiency, proceed as follows: Fig. 5 shows the construction and arrangement of rivets. The data is given in this manner. A strip of the joint marked P is taken. The value of the letters is: P = pitch of rivets in inches. / = thickness of plate in inches. S = tensile strength of plates. d = diameter of the driven rivets, in inches. N = number of rivets in double shear. n = number of rivets in single shear. a = area of cross-section of rivets, in square inches. Strength of the solid strip of plate considered = PXtXS, represented by letter A. Strength of plate between the rivet holes at the outer row of rivets = (P — d)X S, represented by the letter B. RIVETED JOINTS 25 The shearing strength of 8 rivets in double shear, plus the shearing strength of 3 rivets in single shear = Na+na, represented by the letter C. The strength of the plate between the rivet holes in the second row plus the shearing strength of 1 rivet in single shear in the outer row = (P — 2d)XtX S+ na, represented by the letter D. Next, divide B, C, or D, whichever is the least in value, <■ 4> ±V Q O O- $- e o e r — ^ — tx> e- -e- ■& e 0- ^? f 1 L e ^> & -e O-O e -0- e- e ^ e- _ P — Fig. 5. — Quadruple-riveted double butt-strapped joint. by the value of A, and the quotient w T ill be the efficiency of the joint. The numerical values to be used are: 5 =55,000 lb. / = 1/2 in. or .5. P =15 in. d =15/16 in. or .9375 in. a =.6903 sq. in. N =8 rivets, double shear value of 78,000 lb. n =3 rivets, single shear value of 42,000 lb. 26 ARITHMETIC OF THE STEAM BOILER The final values are: A =isX. 5X55,000 = 412,500. B =(i5-.9375)X. 5X55,000 = 386,718. C = 8X78,oooX. 6903+3 X4-2,oooX. 6903 = 517,725. D =(i5-2X.937S)X.5X55 J oo°+ I X 42,000 X .6903 = 389,930. The value of B is found to be the least, therefore the strength of the joint depends upon the weakest part, and the efficiency is, 386718 412500 .937 or 93. 7 per cent. The foregoing example shows how to calculate the ef- ficiency of the various parts of the joint where possible failure may occur. The same line of reasoning, and simi- lar methods of operation, may be employed for any kind or type of riveted joint that may be used in a steam boiler. Size of Rivets and Pitch There are no absolute rules for determining the size of rivets to be used in any given case, for with different size rivets, the same efficiency of joint may be obtained. The size to be chosen depends upon several factors and varies with any one of them. The things to be considered are: the shearing strength of the material of which the rivets are made, the tensile strength of the plates to be riveted to- gether, the pitch of the rivets, and the type of joint to be made. The required efficiency of joint determines the type to be used and the greatest pitch of rivets allowable depends upon the thickness of plate to be used, the object sought being steam-tight joints. RIVETED JOINTS 27 For single-riveted lap joints, the United States Super- vising Inspectors of Steam Vessels recommend a rivet diameter equal to the plate thickness plus 7/16 in. using steel plates and steel rivets. For double-riveted lap joints, a rivet diameter equal to the plate thickness plus $/& in. Some authorities make the rivet diameter range from plate thickness plus 3/8 in. to plate thickness plus 1/2 in., with plates from 1/4 in. to 1/2 in. thickness. For triple-riveted lap joints rivet diameters range from plate thickness plus 3/8 in. to plate thickness plus 7/16 in. with plates from 1/4 to 1/2 in. thickness. For double-riveted butt joints, triple-riveted butt joints, and for quadruple-riveted butt joints, rivet diameters range from plate thickness plus 5/16 in. to plate thickness plus 7/16 in. The foregoing is intended to give a general idea only of rivet sizes that may be chosen to be some- what in proportion to the joint as a whole. In the end it is a matter of choosing that size rivet and a certain pitch which w r ill give the highest efficiency of joint, consistent with steam-tight work, type of joint, strength of materials and all other considerations. It is a matter of "cut and try" until the best is arrived at. Distance Between Adjacent Rows of Rivets The distance between adjacent rows of rivets, center to center, is sometimes called transverse pitch. When the rivets are subjected to the same kind of shear, this dis- tance should not be less than twice the diameter of the rivets, nor more than two and one-half times the diameter of the rivets used. If the distance between the rows of 28 ARITHMETIC OF THE STEAM BOILER rivets is too small, the plate is likely to fracture along a diagonal line, or diagonal pitch as it is termed. If the transverse pitch is at least equal to twice the diameter of rivets, failure of the plate will not occur along the diagonal line, but rather in the net section of plate along the line of rivets. This makes the calculation of joint efficiency somewhat simpler. In cases where the outer butt strap is not as wide as the inner strap, the dis- tance between the line of rivets in double shear and the line in single shear should be two and three-quarters or three times the diameter of rivets, in order to have a properly formed rivet head, and also room to calk the outer strap. Safe Working Pressure of Cylinders with Riveted Joints It w r ill be remembered that formula n, page 9, gives the safe working pressure of a cylinder without any visible joint. But cylinders having riveted joints must be calculated with the efficiency of longitudinal joint taken into consideration. The rule will be the same as that expressed in formula 11, with the additional factor of joint efficiency expressed as a decimal value. Let e represent the efficiency of the longitudinal seam or joint; the efficiency of the girth seam is not required for reasons explained at the beginning of this chapter. The formula now becomes: /X Ty< e — r-77 — = safe working pressure per square inch. RIVETED JOINTS 29 Using the same example illustrating formula 11 , page 11, and assuming a joint efficiency of say 85 per cent. the statement becomes: .5X50000X .85 — — — — =141.66 lb. safe pressure per square inch. No matter what the efficiency of the joint may be, nor by what method it may be found, it is always to be applied as shown in the example just given. Of course there are other ways of arranging and simplifying the fac- tors in the formula, but no matter what the arrangement or how simplified, the result will always come out the same if the work is correctly done. Abbreviated formulas and rules are convenient to those who know of their derivation, but they are not satisfying to those who do not know just how each simplified factor was obtained. For this reason, no attempt is made in this work to ab- breviate anything that will detract from the value of any problem presented, as far as underlying elements and principles are concerned. Any one who understands how to calculate the efficiency of riveted joints, and how to find the safe working pressure of spherical or cylin- drical vessels as given in this work, will be able to work out similar problems, no matter in what form they may be given, or what rule it may be desired to apply to them. So far, the shell and its riveted joints only have been considered. The rules given apply to the cylindrical parts of all boilers of whatever make. There are other rules relating to shells of special design, and these rules will be given further on. The next in order, at present, is the bracing or staying of boiler heads and flat surfaces in boilers. CHAPTER II Boiler Heads — Unstayed Heads Bumped heads may be either convex or concave accord- ing as to how placed in a shell. Fig. 6 shows the applica- tion of the two forms, (a) being a concave bumped head while (b) is a convex head. The arrows show the direc- tion of pressure acting against the heads. Bumped heads do not require bracing, particularly the convex (b) form Fig. 6. — Bumped heads, (a) Concave head, (b) Convex head. Arrows show the direction of pressure. as it is already in the form that internal pressure would tend to make it assume. In the concave (a) form, the tendency of internal pressure is to collapse the head, and allowance is made for this in the rule which will be given presently for the safe working pressure allowed. Bumped 30 BOILER HEADS— UNSTAYED HEADS 31 heads may be either single or double riveted to the shell. It is necessary to know the radius to which a head is bumped when making calculations for safe working pressure. A bumped head is presumed to be virtually part of the surface of a sphere. To find the radius to which a head is bumped, take half the diameter of the head where it fits into the shell, and multiply that value by itself, and divide the product by the height of the d= 4 Inches Radius =4 Inches h =.55 Inches do, Horizontal Centex Line ef, Vertical Center Line d= Diameter of Head, Inches /j=Height of Bump, Inches O— Center from which the Bump is Struck OC ^Radius to which Bump ie Struck Fig. 7. — Diagram showing how to find radius of bumped head. bump. To the quotient add the height of the bump and divide the sum by 2. It is usual to take the dimensions in inches. To find the radius to which a boiler head is bumped the following formula may be used: Referring to Fig. (7), h -= radius. 32 ARITHMETIC OF THE STEAM BOILER All dimensions to be taken in inches. To find the safe working pressure of a bumped head like that in (b) and when it is single riveted to the shell, the following formula may be used: * ixs ( \ P= ^Xr W When the head is double riveted to the shell, then the formula becomes: P = ^Xr (2) When the head is concaved like that in (a) and single riveted, then the formula becomes: txs P = 5Xr (3) When a concaved head is double-riveted the formula b ecomes: txs (a1 * 4iXr In these formulas the values of the letters are: p —safe working pressure in pounds per square inch. t = thickness of metal in the head, expressed in inches. r = radius in inches, to which the head is formed. S = tensile strength of the material of which the head is made, expressed in pounds per square inch section. BOILER HEADS— UNSTAYED HEADS 33 As the foregoing formulas are similar in construction, one example will serve to illustrate the operation of all. A boiler head is bumped to a radius of 60 in. made of plate .5 in. thick, with a tensile strength of 50,000 lb. and double riveted to the shell. What working pressure will be allowed? The operation is as follows: p = — — zn — = 166.66 lb. per square inch. 2.5X60 Unstayed Flat Heads When a boiler head is flat and not stayed, the following formula may be used: txs P = •54X.4 The letters have the same values as for the preceding formulas, and A equals the area of the head expressed in square inches. Example. — A flat head is 30 in. in diameter, . 75 in. thickness of plate having a tensile strength of 55,000 lb. per square inch, what pressure will be allowed? Operation: p = — — — — — — o — = 108 lb. per square inch. .54X30X30X.7854 F H All the rules for boiler heads of the kind just described are those prescribed by the Board of Supervising In- spectors of Steam Vessels in the United States. If it is desired to find what thickness a bumped head should be, having the tensile strength, the radius to 34 ARITHMETIC OF THE STEAM BOILER which the head is bumped, and the pressure in pounds per square inch to be carried, it is a matter of transpos- ing the terms of the proper formula in the group just treated of. Suppose in the last illustrative example given that it is desired to know what thickness of head should be em- ployed. The formula transposed will be: S ~ ~ l Applying this to the example, the statement becomes: 166.66X2.5X60 50000 -=.5 in. thickness. Thickness of Boiler Heads, Massachusetts Rules In actual practice, however, the thickness of boiler heads is not derived mathematically but empirically. The rules in the state of Massachusetts require the thick- ness to be as follows: Boilers up to and including 42 in. diameter, heads must be 3/8 in. From 42 in. to 54 in. diameter, heads must be 7/16 in. From 54 in. to 72 in. diameter, heads must be 1/2 in. Over 72 in. diameter, heads must be 9/16 in. Thickness of Boiler Heads, Ohio Rules The rules formulated for bumped heads by the Board of Boiler Rules in the State of Ohio differ slightly from those given. BOILER HEADS— UNSTAYED HEADS 35 The minimum thickness of a convex head shall be de- termined by this formula: RXF.S.XP T.S. ~ l The minimum thickness of a concave head shall be de- termined by this formula: RXF.S.XP .6(7.5.) In these two formulas the values are as follows : R = one-half the radius to which the head is bumped. F.S. =5= factor of safety. P = working pressure, in pounds per square inch, for which the boiler is designed. T.S. = tensile strength, in pounds per square inch, stamped on the head by the manufacturer. t = thickness of head in inches. The radius of head shall not exceed the diameter of the shell. When a convex or concave head has a manhole open- ing, the thickness as found by the formulas just given, shall be increased by not less than 1/8 in. The minimum thickness of plates in stayed flat surface construction shall be 5/16 in. The minimum thickness of tube sheets shall be as follows: " When the diameter of tube sheet is 42 in. or less, the thickness is 3/8 in.; over 42 in. to 54 in. inclusive, 7/16 4 36 ARITHMETIC OF THE STEAM BOILER in.; over 54 in. to 72 in. inclusive, 1/2 in.; over 72 in.. 9/16 in. Stays and Stay Bolts The maximum allowable stress per square inch net cross-sectional areas of stays and stay bolts as denned in the Massachusetts rules, is as follows: Weldless, mild steel head to head or through stays, 8000 lb. for sizes up to and including 1 1/4 in. diameter, or equivalent area, and 9000 lb. for sizes over 1 1/4 in. diameter or equivalent area. Fig. 8 (a) illustrates direct or through stay arrangement. Weldless, mild steel diagonal or crow-foot stays, 7500 lb. for sizes up to and including 1 1/4 in. diameter, or equivalent area and 8000 lb. for sizes over 1 1/4 in. diameter or equivalent area. Weldless, wrought-iron, head to head or through stays, 7000 lb. for sizes up to and including 1 1/4 in. in diameter or equivalent area, and 7500 lb. for sizes over 1 1/4 in. diameter or equivalent area. Weldless, wrought-iron, diagonal or crow-foot stays, 6500 lb. for sizes up to and including 1 1/4 in. or equiva- lent area, and 7000 lb. for sizes over 1 1/4 in. diameter or equivalent area. Welded mild steel or wrought-iron stays, 6000 lb. Mild steel or wrought-iron stay bolts 6500 lb. for sizes up to and including 1 1/4 in. diameter or equivalent area, and 7000 lb. for sizes over 1 1/4 in. diameter or equivalent area When a greater allowable stress per square inch on BOILER HEADS— STAYED HEADS 37 stays and stay bolts is required than those just given, the material shall conform to the following physical qualities: The tensile strength shall not exceed 62,000 lb. per square inch. The yield point shall not be less than one-half the tensile strength. The elongation per cent, in 8 in. shall not be less than 28. Direct Stays To find the safe working pressure per square inch that may be carried by stays of a given size, the following formula may be applied: Fig. 8. — Arrangement of stays in boiler heads, (a) Direct through stays in horizontal rows, seven stays, (b) Diagonal stays in concentric rows, seventeen stays. aXS (1) 38 ARITHMETIC OF THE STEAM BOILER To find the diameter of stays required, A -¥=°- - 41k' d (2) To find the area supported by one stay, and the dis- tance between stays, ^^=A, and ylA=D. (3) P X To find the required tensile stress that may be en- dured by stays: ^ = S (4) a The value of the letters in this group of rules is as follows : A = area in square inches, supported by one stay. a = area, cross-sectional, in square inches, of stays. p = pounds pressure per square inch. S = tensile strength of stays in pounds per square inch. d = diameter of stays in inches. D = distance between stays, in inches. Example, illustrating the foregoing rules. Assume stays of 1 in. diameter at the smallest part, with an allowable stress of 6000 lb. per square inch, and distanced 6 in. center to center. The area of each stay will be i 2 X . 7^54= • 7^54 sq. in. The area supported by each stay will be 6X6 = 36 sq. in. Applying these values to formula (1) the statement becomes: .^854X^000 = lb per square inch allowable pressure. 36 BOILER HEADS— STAYED HEADS 39 To formula (2), 36X130.9. 6000 = .7854 sq. in. area of each stay. and, \~~~i — =I * n * diameter of stays. To formula (3), .7854X6000 . 111 - J — = 36 sq. in. area supported by each stay. and, V 36 = 6 in. distance center to center of stays. To formula (4), z — —=6000 lb. tensile stress allowed on stays. •7854 In the foregoing no allowance has been made for the space occupied by the stays in the sheets supported. This is on the side of safety and is generally accepted. If in any case it is not accepted, it becomes a matter of subtracting the area occupied by the stays from the area as found from the center to center measurement. Diagonal Stays The size of a diagonal stay depends upon the angle it makes with the surface it is helping to support, when con- sidered in relation to a direct stay. The less the angle is, the larger in diameter must the stay be. The nearer a stay is to being at right angles to the surface it supports, the smaller in diameter it may be. The same principle 40 ARITHMETIC OF THE STEAM BOILER applies to any form of stay that may be used, other than a circular cross-section. In Fig. 9 is shown an ordinary diagonal stay attached to the boiler head at C and to the shell at E. The length of the stay is considered as CE; the distance CD also < enters the calculation as will be shown presently. Fig. 9. — Forms of attachment of diagonal stays, (a) Riveted at both ends, (b) Riveted at one end, nuts and washers at the other end. To find the area of a required diagonal stay, first find the area of a direct through stay, as has been explained. Call the length CE of the required diagonal stay, x and the distance, CD, y. BOILER HEADS— STAYED HEADS 41 Let a = the area of direct stay. Let A = the area of diagonal stay. Let S = tensile strength of stay. Then, aXx . , N = A (1) y AXy a AXy x aXx = x (2) =a (3) =y (4) x A ° Safe pressure per area supported by = sc l ua, ; e inch . ^ one stay ma ^ be carned - (5) Example, illustrating the foregoing group of rules. Assume that the area of a direct stay has been found to be .7854 sq. in. (due to 1 in. diameter as before taken). That the length of the required diagonal stay is to be 36 in. and that the distance between perpendiculars of points of attachment is 30 in. Then, applying these values to formula (1), 1 = .9425 sq. in., nearly, cross-sectional area of diagonal stay required, and the corresponding diameter is: V .9425 . .. — - — = 1 .095 in. diameter. •7^54 The nearest commercial size brace that would be used is 1 1/8 in. diameter. 42 ARITHMETIC OF THE STEAM BOILER To formula (2), jt = 36 in., length of the required diagonal stay, as measured on the line CE in Fig. 9. To formula (3), 7 = .7854 sq. in., area of direct or through stay. To formula (4), .7854X36 = 30 in., distance between perpendiculars as measured on the line CD in Fig. 9. To formula (5), .9425X30 X6000 = 96.1 lb. per square inch pressure allowed. Other formulas for obtaining the diameters of direct and diagonal stays. For direct stays, /SX.7854 For diagonal stays, , / AXp , v d Hsx^I (I) / xXAXp () \yXSX. 7854 K The values of the letters are the same as those used for the stay group of formulas. Examples, illustrating these formulas. Assume 36 sq. in. area supported by one stay (pitch, 6 in. center BOILER HEADS— STAYED HEADS 43 to center), pressure to be carried, 100 lb. per square inch, 6000 lb. allowed stress per square inch on stays, then to find the diameter of direct stay as stated in formula (1) the statement becomes: / 36XIOO , V for plates up to 7/16 in. thick. (1) V 120X/ 2 A= — ? for plates above 7/16 in. thick. (2) P 140 X^ 2 A= — - — ? for screw stay bolts and nuts. (3) 112 X^ 2 p = — -j — 7 for plates 7/16 in. and under. (4) 120X/ 2 p= — -j , for plates over 7/16 in. thick. (5) T AO X / 2 p = — -j — -> for screw stay bolts and nuts. (6) /112 X£ 2 S = \ > f° r plates 7/16 in. thick and less. (7) /l20X^ 2 S = \l f ioi plates above 7/16 in. thick. (8) S = \l— j for screw stay bolts and nuts. (9) -J*M (I0 ) \ 112 t If this formula gives more than 7/16 in. for the value of /, use the next formula with the factor 120 in it. =J^4 (II) \ I20 \ 12" [40 BOILER HEADS— STAYED HEADS 47 In the foregoing group of formulas, the values are as follows : A = area in square inches, supported by one stay. The pitch can be found by extracting the square root of A. t — thickness of plate, expressed in sixteenths of an inch. Example: if the plate were 7/16 in. thick, / in this case would be 7. In other words, t is the numerator of the fraction whose denominator is 16. p = pressure in pounds per square inch, allowed to be carried. # 5 = pitch of stays, center to center. 112, 120, 140 are constants used in the respective formulas. Examples, illustrating the application of the formulas in this group: Assume plate 7/16 in. thick; pressure to be carried, 100 lb. per square inch. Required, area that may be supported by one stay, from which the pitch distance may be found. Then, in formula (1), 112 X 7 2 A = = 54.88 square inches supported by one stay. In formula (4), 112 ^^ 1 p = «x— =100 lb. per square inch pressure allowed. In formula (7), /112 X 7 2 S = % — = 7.4081 in. center to center of stays. \ 100 4 8 ARITHMETIC OF THE STEAM BOILER In formula (10), t=\ — — = 7, numerator of the fraction 7/16, \ 112 thickness of plate in inches. Formulas (4), (5), and (6) in this group serve as a check on the design of stays, for by applying the proper one to any given case it can be determined as to whether or not the desired pressure may be carried. Segments to be Braced In Fig. 10 is shown that portion (shaded) of a boiler head that requires to be braced. The flange is presumed i/=Height of Segment in Inches R= Radius of Circle of which the Segment is a Part C = Center of the Circle of which the Segment is a Part Fig. 10. — Segment of head to be braced. to support the head for a distance of 3 in., and the tubes, a distance of 2 in. above as shown in the figure. These are the generally accepted figures by most designers of boilers and experience seems to have proved them to be safe. BOILER HEADS— STAYED HEADS 49 To find the height of the segment requiring bracing, subtract 5 in. from the distance from the tubes to the highest part of the shell. To find the diameter of the circle of which the segment is a part, subtract 6 in. from the diameter of the boiler. To find the area of a segment sufficiently close for all practical purposes, the following formula may be used (see Fig. 10): 4 4XH* 2XR , X a /— rr .608 = area. In which H = the height of segment, R = the radius of circle, and the numerals, constants. Suppose, for example, it is desired to know the area of a segment of a circle whose diameter is 50 in., the height of the segment being 20 in. The statement becomes: 4x^o! x i,_x*s_ 6o8= 3 \ 20 1600 co 3 \20 1600 XaJ 2 '5-- 6 ° 8 : *22 x Ji.892_ 3 x 1600 3 1600X1.37S b. per square inch. For a diameter of 16 in. the seam will fail at 7705.62 ^^| — = 974.45 lb. per square inch. Collapsing Pressure of Fire Box Fig. 18 illustrates a vertical fire-box boiler. It is required to figure the collapsing pressure of the fire box, assuming that 7/8-in. stay bolts are used and that the pressure is 125 lb. per square inch; it is also required to find what pitch to give the rivets in the vertical seam, using 11/16-in. rivets. In calculating the strength of cylindrical furnaces supported by stay bolts, it is customary to assume that the surface is flat; that is, the tendency of the form to lend strength to the construction is ignored. With a specified size of bolt the first step is to find. how MISCELLANEOUS APPLICATIONS 97 many square inches of surface one bolt will support at the given pressure of 125 lb. per square inch. Standard stay bolts up to 1 1/4 in. and larger are cut 12 threads per inch, and it is practically correct to assume that the depth of thread is the same as given by the United States Standard for that pitch, viz., .05425 in., . although this is not strictly correct, as the usual stay- bolt thread varies slightly from this standard. Using 7/8-in. bolts will give an effective area of .4614 sq. in. per bolt. The working stress allowed in stay bolts varies, according to different authorities, from 6000 to 10,000 lb. per square inch, the most generally ac- cepted figure being 7500, and on this basis a 7/8-in. stay would support 3460 lb. or at a pressure of 125 lb. per square inch it would be capa- ble of supporting 3460 r ^—2 hV 2 V — T.S. 56,000 Lbs. / I I 2 S = 27.6 Fig. 18. — Diagram from which to calculate the collaps- ing pressure of fire box. square inches of surface, or the bolts could be spaced 98 ARITHMETIC OF THE STEAM BOILER \/27.6, or about 5 1/4X5 1/4 in. Of course to be strictly correct, the area occupied by the bolt itself should be added to the 27.6 in. before extracting the square root to get the pitch, for the pressure does not act on this area. But in actual practice such allowance is not made. The surface to be supported in the problem is 2 ft. 6 1/8 in. in diameter, or 94.64 in. in circumference, and the approximate height or distance between the rivets on the leg ring and crown sheet is 24 in. and with the spacing given before it would be necessary to have 18 vertical rows of stays containing four stays each. There is another feature to be considered in the stay- ing of sheets that affects the strength, and that is the stresses in the sheets themselves. Unwin has devised a formula for the maximum stress in flat sheets supported at regular intervals, the supported points forming squares. The formula is: in which a is a side of a square, / the stress per square inch in the sheet, / the thickness of the sheet in inches, and p the pressure per square inch on the supported surface. Substituting present values in this formula, and assum- ing a maximum working stress in the sheet of 7500 lb. per square inch, the statement becomes: o= V 9x7500x.3125x.3125 2x125 5 3> MISCELLANEOUS APPLICATIONS 99 or 5 1/8 in., which is a little less than the pitch deter- mined by the strength of the stay bolt. The first lay- out should be modified, making 19 vertical rows of stays with four stays in each row, or the pitch would be about 5X4.8 in. In relation to pitching the rivets in the vertical seam, it is found in practice that the lapping together of the plate at this point stiffens the sheet so that strength of the joint is not a very important factor in the case, and this joint is generally designed to be least affected by the heat, i.e., single riveted, and of the best propor- tions to insure tightness, which w T ould be a pitch of about 2 1/4 in. if 5/16-in. plate and n/16-in. rivets be used. Relating to Safety-valve Rules Concerning the apparent discrepancy between the different rules relating to safety-valve calculations, particularly that in Reed's Engineer's Handbook (an English publication), which is used by candidates for British Board of Trade certificates of competency, and those used by a prominent educational institution the reader w T ill discover, after carefully reading the following statements, that both rules referred to will give the same results in any given problem when intelligently used. In other words, both rules are correct, although there is a possibility of misunderstanding them and so obtain- ing incorrect answers. If the safety-valve problem could be handled without taking into consideration the effective weight of the lever, then it is not probable that there would ever be any difficulty in candidates having any misunderstanding 8 IOO ARITHMETIC OF THE STEAM BOILER of the matter. But the effective weight of the lever must be taken into the calculation if accuracy is desired, and of course in such a matter accuracy should be desired. Consider what the effective weight of a safety- valve lever is and what relation it bears to the calculation. In the first place there is a difference between the effective weight of a lever and the effective moment of a lever. Either may be used in the safety-valve problem, but it must be clearly understood how and where each is to be used. Probably the whole difficulty that candi- , L =20 10 " F = 2' H ..4" Fva| Center of Gravity of Lever f W? Dia. 4 1/ V \]P= 1256.64 lb. Fig. 19. — Diagram for safety-valve calculations. dates have with the safety-valve problem lies just in this point. The effective weight of a lever is found by multiplying its weight in pounds by the distance its center of gravity is from that point called the fulcrum, and by dividing the product by the distance that the center of the valve upon which the lever acts is from the ful- crum. The effective moment of a lever is found by multi- plying its weight in pounds by the distance its center of gravity is from the fulcrum. The distances referred to are measured in inches. Referring to the following example: Required, the weight to be placed at the end of the safety-valve lever shown in the diagram with the given data. MISCELLANEOUS APPLICATIONS ioi Diameter of valve, 4 in. Area of valve, 12.5664 sq. in. Steam pressure per square inch, 100 lb. Weight of lever, 20 lb. Weight of valve and stem, 10 lb. Total upward pressure on the valve is 1256.64 lb. The letter c represents the sum of the effective moments of the valve and stem, and also the lever. The effective moment of the lever is 10X 20= 200; the effective moment of the valve is 2X10=20; and the sum is 200+20 = 220. xt FP-c . (2X1256.64)^-220 Now, — y — = weight, or -=114.66410., the weight to place at the end of the lever. (This method employs effective moments of valve and lever.) Now try the same example by using the other rule, which employs the effective weight of the lever, and the actual weight of the valve. The rule is stated thus: "To find the weight which must act on a lever at a given dis- tance from the fulcrum so that the valve is about to blow off at a given pressure, subtract the downward force due to the weight of the valve, stem and lever from the prod- uct of the area and the steam pressure. Multiply the remainder by the distance from the fulcrum to the center line of the valve, and divide this product by the distance from the fulcrum at which the weight is to act." Expressed in a formula it appears like this: w J Ap -^ D W L W = the required weight in pounds. A = the area of the valve in square inches. 102 ARITHMETIC OF THE STEAM BOILER P = the pressure per square inch. D = distance in inches from the fulcrum to the center line of valve. L = distance from fulcrum at which the weight is to be placed. w = the weight of the valve and stem in pounds plus the effective weight of the lever. Weight of valve and stem = 10 lb. Effective weight of lever = — = ioo lb. ° 2 Total downward force due to both valve and lever = no lb.; that is, w in the formula = no lb. ( 1 2 . 5 664 X i oo — 1 1 o) X 2 — = 114.6^4 lb., 20 J the same answer as obtained by the first method employed. The effective weight of the lever will be = 100 lb.; that is, this lever, because of its weight and its point of contact with the valve, is equivalent to 100 lb. being placed directly on top of the valve. The effective moment of the same lever will be: 20 X 10= 200. Now, either the effective weight of the lever or the effective moment may be used in the calculation as has been shown, but each has its own place in the operation, as will be seen a little further on. The equality of moments in the safety-valve problem are stated like this: WXD+W'XD'+wXd=pXAXd MISCELLANEOUS APPLICATIONS 103 In which W = weight of the weight in pounds. D = distance in inches that weight is placed from fulcrum. W' = weight of the lever in pounds. D f = distance of center of gravity of lever from fulcrum. w = weight of the valve and stem in pounds. d = distance center of valve is from the fulcrum. p = pressure in pounds per square inch. A = area of valve in square inches. By applying the formula to the problem it can be seen that the answer obtained is correct. Showing the let- ters, and substituting the numerals of the example, the statement becomes: WXD+W'XD'+wXd=pXAXd, 114.664X20+20X10+10X2 = 100 X 12.5664X2, 229.328+200+20 = 2513. 28, 2513.28 = 2513.28. Any safety-valve rule or formula that can successfully stand the above test is correct and may be safely employed. In order to arrive at the point aimed at, and also to make it as clear as possible, it may be said that from the foregoing formula of moments a series of formulas can be derived that will handle the safety-valve problem in all its phases. Here are the formulas: (1) FX S+ L = W. (2) FX S + W=L. 104 ARITHMETIC OF THE STEAM BOILER (3) LXW+ S=F. (4) LXW+ F = S. Where F = the force acting upward against the valve; this equals the pressure per square inch times the area of the valve; from the product must be subtracted the weight of the valve and stem. 5 = the distance from the fulcrum to the center of the valve. L = distance from the fulcrum to point where the w T eight is hung. (This may or may not be at the extreme end of lever.) W =the weight in pounds of the weight to be hung on lever. In cases (1) and (2) the effective weight of the lever must be subtracted just before multiplication occurs. If effective moment of lever is to be employed, then the effective moment is to be subtracted just before division occurs. In cases (3) and (4) the effective weight of lever is to be added just before multiplication occurs. If the effective moment of lever is to be employed, then it will be added just before division occurs in the formula. It is very important that a distinction be made between the two phases of the four cases, and then no trouble will be experienced. Bearing in mind what has just been stated and being ready to refer to it again, now try the problem by using Reed's rule, which is stated thus: (1) Find the area of the valve and multiply it by the pressure per square inch. MISCELLANEOUS APPLICATIONS 105 (2) From the product take the weight of the valve (which of course includes the stem). (3) Multiply the remainder by the distance from the fulcrum to the valve, then subtract the moment of the lever, and divide by the distance from the fulcrum to the weight. In Reed's Engineers' Handbook the effective moment of the lever is defined the same as that which appears in the earlier part of this subject. Using the figures in the example and following Reed's rule, 12.5664 sq. in. (area of valve) X 100 (pounds pressure per square inch) = 1256.64 lb. (total upward pressure) — 10 lb. (weight of valve and stem) = 1246.64 lb.X2 (inches distance from fulcrum to valve) = 2493.28 lb. — 200 (moment of lever) = 2293.28 lb.-^2o (inches, distance fulcrum to weight) = 114.664 lb. weight. Ans. Thus it is seen that exactly the same answer is obtained as found before. Reed's rule is therefore correct if intelligently used. Now, suppose that it is desired to use Reed's rule, but instead of having the effective moment of lever given, the effective weight is given instead. The effective weight of the same lever would be (as before explained) = 100 lb. The operation will be like this: 12.5664 sq. in. (area of valve) X 100 (pounds pressure per square inch) = 1256.64 lb. (total upward pressure against valve) — 10 lb. (weight of valve and stem) = 1246.64 lb. (balance) — 100 lb. (effective weight of lever) = 1146.64 lb. X2 (inches, distance, fulcrum to valve) = 2 293. 28 4- 20 (inches, distance, fulcrum to weight) = 114.664 lb. weight 106 ARITHMETIC OF THE STEAM BOILER answer, giving exactly the same answer as in the other two cases. Again is attention directed to the difference between the tw r o latter cases, and particularly the place in each w T here the subtraction of the lever factor takes place. Suppose the effective weight of the lever were subtracted after multiplication had occurred instead of before, then the following statement would occur: 12.5664X100 = 1256.64—10=1 246.64 X 2 = 2493 .28 — 100 (effective weight of lever) = 2393.28-^20=119.664 lb. w r eight. 119.664—114.664 = 5 lb. difference, too much weight. In other words, by so doing the valve would be over- weighted 5 lb. Of course, the percentage of difference is small, but it is further aggravated by the frictional resistances, and it is not on the safe side of the calcu- lation. Suppose on the other hand the effective moment of the lever were subtracted before multiplication occurs; the statement would be: 12.5664X100=1256.64—10 = 1246.64—200 (effective moment of lever) = 1046.64-^20 = 104.664 lb. weight. 114.664—104.664=10 lb. difference, too small. Let the interested reader study this matter out for himself and try the arithmetical operations in the different cases. In this way he cannot but come to a correct and complete understanding of all that is em- braced in the problem, as far as an operating engineer is concerned, when standing an examination for a certificate. Any rule relating to the safety valve, which a candidate is not sure of should be tested by the application of the equality of moments formula before referred to. MISCELLANEOUS APPLICATIONS 107 Roper's Safety-valve Rules Examiners of engineers in the United States Steam- boat Inspection Service sometimes prefer to have candi- dates for American marine engineers' license use what are known as Roper's Rules for safety-valve problems. Therefore it is well to have a knowledge of these rules in case they are required. In the following formulas let A = Area of valve in square inches, or diameter 2 X.7854. D = distance from center of valve to the fulcrum, in inches. L = distance of the weight from the fulcrum, in inches. P = steam pressure in pounds per square inch. W = weight of the ball in pounds to hang on the lever. V = w r eight of the valve and stem in pounds. w = w eight of the lever in pounds. / = distance of the fulcrum from the center of gravity of the lever, in inches. n (WXL) + (wXl) + (VXD) AXD (1) TT7 AXPXD-{wXl+VXD) , x W = — £ — (2) ■ AXPXD-(wXl+VXD) . , L= — w~ (3) The following examples illustrate the application of the formulas given. 108 ARITHMETIC OF THE STEAM BOILER Example i. — At what pressure will a safety valve blow off, hav- ing a diameter of 4 in., weight of valve and stem 12 lb., weight of lever 22 lb., weight of ball 125 lb.; the overall length of lever is 46 in., and it is straight and parallel; the weight is hung at 42 in. from the fulcrum, and the distance from center of valve to fulcrum is 4 in. Using formula (1) the statement becomes: P = (125X42)+ ( 22 X^) +(12X4) 4 2 X. 7854X4 = 115.466+ lb. per square inch, answer. Example 2. — Using the foregoing example (1), it is required to find the necessary weight to hang on the lever. Using formula (2) the statement becomes: W 4 2 X. 7854X115. 466X4- (^2Xy + I2 X4) 42 = 125 lb., answer. Example 3. — Using the same example (1), it is required to find at what distance the weight is to be hung. Using formula (3) the statement becomes: 12. 5664X115. 466X4- ( 22X^+12X4) 125 =42 in., answer. Safety-valve Capacity Extract from paper read before the A. S. M. E., February 23, 1909, by Philip G. Darling, Mechanical Engineer. MISCELLANEOUS APPLICATIONS 109 CAPACITY FORMULA FOR 45° SEATS 1. E=iosXlXPXD 2 . D = .oo 95 j£p Modified Forms for Special Applications For Locomotives H For Cylindrical Multitubular, Vertical and Water Tube Stationary Boilers For Water Tube Marine and Scotch Marine Boilers 5. D = .o 9 5j^p E = pounds of steam relieved per hour. / = vertical lift of valve in inches. P = steam pressure (absolute) pounds per square inch. D = nominal diameter of valve (inlet) in inches. H = total boiler heating surface in square feet. For flat-seated valves the constants in these formulae are as follows: i — 149.; 2 — .0067; 3 — .052; 4 — .065; 5— .090. no ARITHMETIC OF THE STEAM BOILER UNITED STATES STEAMBOAT INSPECTION SERVICE SAFETY- VALVE RULE The rule is that the areas shall be found by the formula: WG A = .2o 74 x 1 r' in which .4 = the area of the safety valve in square inches. W= pounds of water evaporated per square foot of grate surface per hour. P = the absolute pressure per square inch. G = grate area in square feet. In the case of spring-loaded valves, the effective area must be equal to that derived from the formula, and a lever must be provided which will raise the valve one- eighth its diameter from its seat. All seats to have an angle of 45 to the axis of the valves. Derivation of the United States Board of Super- vising Inspectors' Rule for Areas of Safety Valves Napier's rule for flow of steam through orifices: _. . . . Absolute pressure X area Flow in pounds per second = (This corroborated by Peabody's experiments.) P = absolute pressure = gage pressure+ 1 5 . W = pounds discharged per hour. A =area of valve opening or orifice. MISCELLANEOUS APPLICATIONS in Hence _ PXA * e , c 'T 36oX^XP W= X6oX6o= — — 70 7 For safety-valve practice, cut this amount down 25 per cent., leaving 75 per cent. Thus Restrict the lift of valve to 1/32 of its diameter = d_ then 32 d ~Xd 2 A — X^Xd = lift X circumference = — 3 2 3 2 Substituting this value for A = area of orifices 270 „ 7ld 2 w=^~xpx — 7 32 In a valve of diameter d the area = Tld 2 =a 4 To get W in terms of area of valve, substitute for d 2 its value in terms of a, W =- 7 -X P X— X 4 - =4-821 XPa 7 32 t In safety-valve practice this will represent the pounds 112 ARITHMETIC OF THE STEAM BOILER of steam that must escape per hour, which must be equal to the pounds of water that the boiler can evaporate per hour. To reduce this to a working basis, consider these quanti- ties per square foot of grate surface per hour. W — pounds of water evaporated per square foot of grate surface per hour. P = absolute pressure per square inch. A =area of safety valve per square foot of grate surface. Hence W TF=4.82iXPX^, anda = .2074X -p From which a table of areas required per square foot of grate surface may be found by assuming the different values of W and P. Finding the Center of Gravity of Tapered Safety- valve Levers In questions relating to the lever safety valve it is necessary to know how to find the center of gravity of the lever, in order to calculate the effective weight of the lever; how may the center of gravity be found w T hen the lever is tapered, and of uniform thickness throughout its entire length? Answer. — There are three ways that the center of gravity of tapered safety-valve levers is found: By taking the lever off and actually balancing it on a knife edge; diagrammatically, and mathematically. The last two methods only require an explanation. MISCELLANEOUS APPLICATIONS 113 First consider the diagrammatic method of finding the center of gravity. To be brief and to the point, assume a lever 20 in. long, 2 in. wide at one end, and 1 in. wide at the other end and of uniform thickness throughout. Do not consider the projection which is usually at the small end of the lever for preventing the weight slipping off, nor the holes at the other end through which the lever is attached to the fulcrum, and by which the pin is attached which bears upon the valve. These would Fig. 20. — Diagram for finding center of gravity of tapered lever. make but little difference in the result obtained, and would tend to complicate the calculations. An inspection and study of figure 20 should make clear how the center of gravity may be determined. A scale of quarter size may be chosen as a matter of convenience only. Any other scale w T ill do. The center line AB is ' drawn first. The length of A B is 5 in., or one-quarter as long as the lever assumed. At A and B respectively draw the lines CD and EF 1/2 in. and 1/4 in., represent- ing the actual 2-in. and i-in. dimensions of the lever. Next draw the diagonal line DE, and locate the point 1 exactly midway between points D and E. Join the 114 ARITHMETIC OF THE STEAM BOILER points C and i by the line Ci. From the point i locate the point 2 on the line Ci at a distance equal to one- third of its length. Now, from the point F draw the line Fi, and from the point 1 locate the point 3 on the line Fi, one-third of its length from the point 1. Join points 2 and 3; where the line 2-3 intersects the center line AB will be the point 4 at w T hich the center of gravity is. By carefully and accurately drawing the figure the measurements show that the center of gravity is about 1 1. 1 in. from the small end of the lever, and is about 8.9 in. from the large end of the lever. To find the center of gravity of a tapered lever mathe- matically, the following formula may be used: 2A+B T . ,. p 2A+B X== 3 A+$B X ; ° r m m? x= siA + B) X Where x = distance center of gravity is from the small end of lever. A= the width in inches of lever at the fulcrum end. B = width in inches of lever at the weight end. L = entire length of lever in inches. Applying this formula to our example we have 2X2 + 1 5 100 -7 : rX20=~X20= "=11.11 111., 3(2 + 1) 9 9 distance center of gravity from small end of lever. Thus it will be seen that the two methods which have just been explained verify each other, and no doubt the first method (that of balancing the lever on a knife edge) referred to would verify the others, for, as before MISCELLANEOUS APPLICATIONS 115 stated, the difference due to the projection at one end and the holes at the other end would hardly be noticeable. At an examination where time is limited, it would be better to use the formula in such questions. It does not require much study and only a little practice to become familiar with it. Furthermore, it will be found easy to remember after it has been used a few times. Marine engineers cannot afford to ignore this subject. Chimneys The " proportions of chimneys" vary very much according to the requirements. Every chimney should be large enough in cross-section to carry off the gases and high enough to produce sufficient draft to cause a rapid combustion. The object of a chimney being to carry off the waste gases, it naturally determines the amount of fuel that can be burnt per hour, and it is advisable to have always a good draft, as it can always be regulated by a damper. Draft pressure is caused by the difference in weight between a column of hot gases in the chimney and a column of air of equal height and area outside the chimney. Formula for finding the force of draft in inches of water of any given chimney: 7-64 7-95^ ^tj(7M 7-95\ Where F = force of draft in inches of water. H = height of chimney in feet. 9 Il6 ARITHMETIC OF THE STEAM BOILER T\ = absolute temperature of chimney gases (H-460). T 2 = absolute temperature of the external air (/1+460). t = temperature of chimney gases. ti = temperature of external air. Formula for finding the height of a chimney in feet for a given force of draft: F 11 = /7.6 4 _7. 9 5\ \T 2 Tj To find the maximum force of draft for any given chimney, the external air being 6o° F., and the heated column being 6oo° F., multiply the height above the grate in feet by .0073, and the product is the force of draft expressed in inches of water. 1. The draught power of the chimney varies as the square root of the height. 2. The retarding of the ascending gases by friction may be considered as equivalent to a diminution of the area of the chimney, or to a lining of the chimney by a layer of gas which has no velocity. The thickness of this lining is assumed to be 2 in. for all chimneys, or the diminution of area equal to the perimeter X 2 in. (neglecting the overlapping of the corners of the lining). Let D = diameter in feet, A = area, and E = effective area in square feet. 2 8D 2 — For square chimneys, E = D — =A — \/ A For round chimneys, £=- yD 2 — — j =A — .591 \A4 MISCELLANEOUS APPLICATIONS 117 For simplifying calculations, the coefficient of \/ A may be taken as .6 for both square and round chimneys, and the formula becomes 3. The power varies directly as this effective area E. 4. A chimney should be proportioned so as to be capable of giving sufficient draught to cause the boiler to develop much more than its rated power, in case of emergencies, or to cause the combustion of 5 lb. of fuel per rated horse-power of boiler per hour. 5. The power of the chimney varying directly as the effective area E, and as the square root of the height H, the formula for horse-power of boiler for a given size of chimney will take the form horse-power = CE\/e[, in which C is a constant, the average value of which, ob- tained by plotting the results obtained from numerous examples in practice, the author finds to be 3.33. The formula for horse-power then is horse-power = 3. 3sE\/jj y or horse-power = 3.33 (A — aVaWh If the horse-power of boiler is given, to find the size of chimney, the height being assumed, .3 H.P. ,- For round chimneys, diameter of chimney = diam. of £+4 in. For square chimneys, side of chimney = \/£+4 in. Ii8 ARITHMETIC OF THE STEAM BOILER If effective area E is taken in square feet, the diameter in inches is d= i^.S4\/E+4. in., and the side of a square chimney in inches is s= i2\/E+4 in. If horse-power is given and area assumed the height In proportioning chimneys the height is generally first assumed, with due consideration to the heights of surrounding buildings or hills near to the proposed chimney, the length of horizontal flues, the character of coal to be used, etc., and then the diameter required for the assumed height and horse-power is calculated by the formula or taken from the table. Size of Boiler Feed Pipe What size feed pipe should be installed to supply three ioo-H.P. boilers? At ordinary commercial rating the water required per horse-power-hour, is taken as 30 lb. Each boiler there- fore will need 3000 lb. of water per hour or 50 lb. per minute, which at 62.5 lb. per cubic foot would call for .8 of a cubic foot each minute. To provide for emergencies, twice the actual quantity of water required should be figured. The velocity of flow is usually taken as 100 ft. per minute. On this basis, the quantity to be taken care of will be 2X.8 = 1.6 cu. ft. per minute, which at 100 ft. per minute velocity would require an area of pipe of -=.016 sq. ft., or 100 2.3 sq. in., found like this: .016X144=2.304 sq. in. and A / '^ I *7 m - diam. \-7 8 54 MISCELLANEOUS APPLICATIONS 1 19 The nearest commercial size of pipe is 1 1/2 in. diameter, which is the size to connect to each of the boilers. For the main pipe, to supply all three boilers, a pipe of equivalent area to the three boiler pipes would be 6.9 square inches, and this would mean a 3-in. diameter pipe is required. Should not all three boilers require to be fed at the rate before referred to at any one time, then a 2 1/2-in. main feed pipe will be ample. Effect of Stiffness of Head on Braces Away back in 1876, Samuel Nichols, a practical boiler maker in charge of a large English works, wrote a book for boiler makers. In it he recounts some experiments made under the supervision of Robert Nelson, author of " A Treatise on Steam Boilers/' upon boilers built by himself for the purpose, with regard to unstayed heads in cylindrical shells. A boiler 30 in. in diameter with flat heads 3/8 in. thick, of plate having a "tenacity" of 21.2 tons per square inch with an elongation of 7.9 per cent., flanged on a radius of 1 in. inside the plate, was subjected to hydrostatic pressure. At 10 lb. the head had bulged 1/16, at 120 lb. 11/16 and at 150 lb. 13/16 in. Permanent set occurred somewhere between 50 and 65 lb. at which latter figure the deflection was 3/8 in. Rupture took place at about 300 lb. From the results of his tests, Mr. Nichols deduced a formula which is printed in his book for determining the bursting pressure of cylinders with unstayed heads, although he strongly disclaimed any advocacy of ex- posing an unstayed surface to high pressures. "On the 120 ARITHMETIC OF THE STEAM BOILER contrary," he writes of himself, "he is more convinced, now that he has witnessed these experiments, that a fiat unstayed surface is very weak indeed, and that they still require a larger amount of care and judgment on the part of boiler engineers than any portion of the boiler/' Evidently under the erroneous notion that if the head can take care of the amount of pressure indicated by this formula on its own account it will need correspondingly less bracing, the Board of Boiler Rules of Ohio, after instructing inspectors to determine the working pressure of boilers with respect to the bracing in the usual way but with an allowable stress of 8000 lb. per square inch irrespective of size, tells them, "To the above pressure may be added the Nichols formula w T ith a factor of safety of not less than 8." If this means, and it can seem to mean nothing else, that to the pressure which can safely be taken care of by the bracing may be added the pressure which, by the Nichols formula, would be allowed upon the unbraced head, it is wrong. Mr. Nichols shows that the head may be bulged considerably without straining the sheet beyond the elastic limit. But a brace is supposed to be tight before the head commences to bulge. Just as soon as the head starts to move it commences to stretch the brace. Under the allowable stress of 8000 lb. per square inch of section a brace wall extend only about 1/3750 of its length, or a 6-ft. brace would extend less than .02 in. The pres- sure which would produce the movement in the unbraced head is inconsiderable — a pressure of 10 lb. produced a movement of three times as much — and yet this is MISCELLANEOUS APPLICATIONS 121 all the help that the stiffness of the head would be to the brace. As an example, assume a 72-in. boiler, height of seg- ment to be braced 24 in., area of segment to be braced 814 sq. in., pressure 100 lb., thickness of head 1/2 in., tensile strength 60,000 lb. The Nichols formula with a factor of safety of 8 would allow tXT Xio 0.5X6000 0X10 ^4X8 "814X8 -46 1b. If this may be added to the pressure which the bracing is capable of carrying, it would be necessary to brace against only 100 — 46 = 54 lb. per square inch, which at 8000 lb. per square inch w^ould require seven i-in. braces. Practice calls for at least twice as many. Bracing Flat Surfaces in Steam Boilers There is considerable variation as to the load allowed per square inch of net section on diagonal braces, rod braces and stay bolts by the authorities who have laid down rules on this subject. The United States Government rules allow 6oco lb. on welded iron stays below 1 1/4 in., 7500 on 1 1/4 in. and above, and from 7000 to 9000 lb. on weldless steel stays. Chicago has a flat scale of 6000 lb. on all stays or braces, Philadelphia has a limit of 7500 lb.; the Massachusetts rules allow from 6500 to 9000 lb. per square inch net section, varying as the braces are w r elded or weldless and with the size, the latter for the reason that with a given waste of 122 ARITHMETIC OF THE STEAM BOILER material the percentage of reduction is greater with the smaller sizes. The above applies to flat surfaces and refers to flat heads, such as dome heads, segments of heads, etc. The United States Government has a rule to find the pressure on flat heads not exceeding 20 in. in diameter as follows: CXT 2 A where P = pressure. C=ii2 (7/16 or under) and 120 over 7/16. A = 1/2 the area. T = thickness in sixteenths. With a 3/4-in. head 20 in. in diameter, no lb. would be allowed by these rules. A short time ago the Board of Boiler Rules for the State of Ohio issued instructions to the inspectors holding certificates of competency that the following formula could be used in flat surface of heads: rxr.s.xio ax's w T here T = thickness. T.S. = tensile strength. A = area. In addition, a limitation allowance of 8000 lb. per square inch irrespective of size of brace is granted. This applies to boilers now in use, but not to boilers to be installed after July 1, 19 12. This ruling is far more liberal than any other authority has heretofore MISCELLANEOUS APPLICATIONS 123 allowed as a comparison will show. Assume a 72-in. boiler, height of segment 24 in., thickness of head 1/2 in., tensile strength 60,000 lb. The total area of the segment = 1 186.4 S Q- in., while the area requiring bracing = 814 sq. in. Hence .5X60000X10 r „ 814x8 " =461b - allowed without braces. Let the pressure required be equal to 100, then 100 — 46 = 54 lb. to be braced, and 54X814 = 43,956 lb. Assuming the proposed brace to be of .79 in. area, then .79X8000 = 6320 per brace, and 43,956^6320 = 7 braces of practically i-in. diameter. It may be said that flat surfaces subjected to internal pressure will spring and proportionally to the unsup- ported area. Samuel Nichols, in his tests of circular flat heads, showed the springing began with very low pressures, even at 20 lb. on 28-in. heads and increased as the pressure was raised. Applying this fact then to the Ohio ruling, it seems the head would so spring that at 100 lb. pressure the total load on the braces would be .79X7 = 5.53 into the toal load, 81,400 lb., or 14,718 per square inch net section instead of 8000 lb. Applying this ruling to a flat dome head 36 in. diameter, 1/2 in. thick, 60,000 lb. tensile strength, area to be braced 707 in., gives 53 lb. without bracing. The results of allowing a flat head unbraced 1 o spring and return times without number would be final failure due to such action. Reverting to the segment as in a horizontal tubular boiler, it may be said other authorities have been careful 124 ARITHMETIC OF THE STEAM BOILER to avoid allowing excessive stresses on the chord of the segment which is supported by the tubes inasmuch as the latter are not a constant in strength as is the flange of the head in the arc of the segment, and this view has been approved by most students as the tubes are subjected to more or less rapid wear and reduction in thickness. Further, such calculations apply to boilers now in use irrespective of age. Indeed, Ohio has no limitation as to age as respects pressure to be determined with a factor of safety of 4 together with this exceedingly liberal allow- ance on braces. Comparing this with the Massachusetts, Chicago, Philadelphia and Detroit rules, w T hat results may be expected? Three Boiler Questions In an examination, three out of live engineers failed to answer the following questions, which are given for the benefit of those w T ho may be called upon to make similar calculations. 1. A horizontal tubular boiler is 72 in. in diameter and 18 ft. long; thickness of plate .437 in.; efficiency of longitudinal joints 77 per cent., and steam pressure no lb. What should be the tensile strength of the plate, allowing a factor of safety of 5 ? 2. If the tensile strength had been 56,000 lb. and the efficiency of the joint 70 per cent., what thickness of plate should be employed? 3. If this boiler had been intended for 125 lb. of steam, what would the efficiency of the joint have been, using the data in the first question (except the pressure and efficiency) ? MISCELLANEOUS APPLICATIONS 125 The required tensile strength of plate is found by the rule pressure X diam. X factor of safety efficiency of joint X thickness of plate X 2 Substituting the figures given, instead of the words in the rule, we have Tensile strength = 1 -°~^ 2 —^ = 58,7 7 5 lb. .77X437X2 ° ,//0 The rule for thickness of plate is pressure X diam. X factor of safety tensile strength X efficiency of joint X 2 Again substituting the figures we have rrw 1 110X72X5 inickness =—- — — — •= . c m. 56000 X. 70X2 J Efficiency of joint is given by the rule . _ pressure X diam. X factor of safety ^ tensile strength X thickness X 2 This figured out gives 12^ X 72 X ^ Efficiency = —z — — — = .87^ or 87. < per cent. 58775X.437X2 /0 ' 0F To Find Pitch of Rivets How can the pitch of the rivets be determined for a double-riveted butt and double-strap joint which is to have 7/8-in. rivets and a strength of plate between the rivet holes on the outer row which will be 82 per cent, of the strength of the solid plate? Let P = pitch. t = thickness of plate. TS = tensile strength of plate. d = diameter of rivets. 126 ARITHMETIC OF THE STEAM BOILER Then PXtXTS = strength of solid plate and (P — d) XtXTS = strength of plate between the rivet holes on the outer row. The conditions require (P-d)tXTS _ PXtXTS W By canceling out t and TS from numerator and de- nominator of the first member of (i) we obtain, P-d P and as ^ = .875, the equation becomes ^-•875 P from which it is found that = .82 (2) ecomes =•82 ( 3 ) P=--^= 4 .86+ (4) or practically 4 7/8 in. Collapsing Pressure of Lap-welded Bessemer Steel Tubes of from 3 to 10 In. Diameter, and of Different Wall Thicknesses Formulas: and t P-- = 86,670-7- / -1386 / p\ p= - IOOO ( I — y J 1 " 1 6oo j2 ) p= = 50,210,000 (t)» MISCELLANEOUS APPLICATIONS 127 where p = collapsing pressure in pounds per square inch. d = outside diameter of tube in inches. / = wall thickness in inch measure. The first formula is applicable to cases where -j is greater than .023 and the others to the case of thin-walled tubes where the quotient is less than that value. Safe Working Pressure Calculations as Applied to the Shell of Climax, Hazelton and Porcupine Types of Steam Boilers Example. — Shell plate 5/8 in. thick, diameter 30 in., tensile strength 60,000 lb. per square inch, tubes 4 in. diameter. See Figs. 1 and 2 of this section. Consider the ring of the shell 3 27/32 in. wide (Fig. 2) included between any two transverse rows of holes. For each pound per square inch of pressure, any longi- tudinal section of this ring will be subjected to a stress of = 57.65625 lb. The net section of the ring on the axis ab of a longitudinal row of holes is 5/8 X (3 27/32 — 2) = i. 15+sq. in. The unit stress on this section for a pressure of 1 lb. per square inch is, therefore, 57.656-^1.15 = 50 lb. per square inch, nearly. The section on the line ac through two adjacent holes in a diagonal row is subjected to the stress of 57.656 lb., which acts in the direction of the line ef, perpendicular to ab. This stress may be resolved into two components, one of which, eg, acts perpendicular to ac and tends to pull the plate apart through that section, while the other, eh, acts along the line ac and tends to shear the plate through the 128 ARITHMETIC OF THE STEAM BOILER same section. Of these two components, eg is equal to the stress on a longitudinal section of the ring multiplied by the cosine of the angle bac, and eh is equal to the stress 'on the longitudinal section multiplied by the sine of the angle bac. The sine of bac is 3.5-^5 • 2 = .673, and the cosine 3 27/32-7-5.2 = • 739? which correspond to an angle of 47 40', nearly. The tensile stress on the section ac resulting from the stress of 57.656 lb. acting perpen- dicular to the section ab is, therefore, 57.656 X .739 = Dia. of Shell 30" Thickness of Plate % Fig. i. — Calculations relating Fig. 2. — Calculations relating to to porcupine type of boilers. porcupine type of boilers. 42.61 lb., nearly, and the shearing stress is 57. 656X^73 = 38.8 lb. In addition to the stresses due to the force that tends to break the ring through a longitudinal section, the section on ac is subjected to a stress from the action of the force that tends to rupture the shell along a trans- verse section. For a pressure of 1 lb. per square inch, MISCELLANEOUS APPLICATIONS 129 this force is equal to the area of the head multiplied by 1 ; that is, to 3o 2 X-7854X 1 = 706.86 lb. The number of sections among which this force is divided is equal to the circumference of a 30-in. circle divided by 3.5; that is, to ^ — — =27. The force on each section is, there- 3-5 fore, 706.86-^27 = 26.18 lb. This force acts on the section ac in the direction of the line el, parallel to the line ab. It may be resolved into two components, one, em, perpendicular to ac, which is equal to 26.18 multi- plied by the sine of the angle bac; the other, en, in the direction of ac, equal to 26.18 multiplied by the cosine of bac. Of these two components, the first acts in the same direction as the component eg; its value, 26.i8X.673 = 17.62 lb., nearly, is, therefore, to be added to the value represented by eg, thus giving us a total tensile stress in the section ac of 42.61 + 17.62 = 60.23 lb. The compo- nent en, whose value is 26. 18X. 739 = 19. 347 lb., acts in the opposite direction to eh; they therefore partly neu- tralize each other, and the resulting shearing stress, 38.8 — 19.347 = 18.453, is so much less than the tensile stress of 60.23 lb. that it is evident the section would fail by tension and not by shear. The area of the net section of the plate, which resists the tensile stress of 60.23 lb., is 1.2X5/8 = 75 sq. in.; the unit stress in this section for a pressure of 1 lb. per square inch is, therefore, 60.23-^.75 = 80.3 lb. per square inch. Since the stress on the section ab was but 50 lb. per square inch, it is evident that the plate will fail along the section ac. If we assume the safe working stress of the 60,000-lb. steel plate to be 10,000 lb. per square inch, the safe working 130 ARITHMETIC OF THE STEAM BOILER pressure will be 10,000^80.3 = 124.5 lb. per square inch. Figuring the Safe Working Pressure op the Shell of a Locomotive Boiler How to determine the safe working pressure of the shell of a locomotive boiler with several courses of vary- ing diameter, like that shown in Fig. 21. With the locomotive boiler, like Fig. 21, the safe work- ing pressure can only be ascertained by considering the Fig. 21. — Diagram from which to calculate the safe pressure in a locomotive boiler. diameters A, B and C. Also, the thickness of the plates and the efficiencies of the longitudinal seams of the re- spective courses must be considered. ! The inside diameter A is 72 in. and plate 3/4 in. in thickness, the inside diameter B 61 in. and plate 9/16 in. in thickness, and the inside diameter C 60 in. and plate 1/2 in. in thickness. The efficiency of the riveted joints for the respective courses is A, 82 per cent., course B 84 per cent., and course C 86 per cent. It may be asked, what is the difference in efficiency in MISCELLANEOUS APPLICATIONS 131 the respective courses? This is because the over-all distances, D, E and F, are such that the same maximum pitch could not be obtained in the respective courses, and a change of pitch, large or small, will make a dif- ference in the efficiency of the net section of plate, maxi- mum pitch of rivets, which point is made the weaker of the several parts of a riveted joint. Assuming the factor of safety to be 5, and the plate to ha\e a tensile strength of 60,000 lb. then the working pressure of the boiler, as far as the shell is concerned, may be determined by the following formula: Where T — thickness of the plate in inches. D = diameter of the boiler in inches. T s = tensile strength of the plate in pounds. F = factor of safety. E = efficiency of the longitudinal seam. P = pressure in pounds per square inch. TXT S X E_ DXF The safe working pressure for course A then will be (2X3/4)X6ooooX.82 — — = 205 lb. 72X5 D The safe working pressure for course B then will be (2X9/i6)X6ooooX.8 4 y-^~ = 186 lb. 61X5 The safe working pressure for course C then would be (2Xi/2)X6ooooX.86 , ., — = 172 lb. 60X5 ' 132 ARITHMETIC OF THE STEAM BOILER The calculations thus show that the course C, the course with the least diameter and the longitudinal seam with the greatest efficiency, to be the weaker of the three courses A, B and C. Therefore, the pressure for the boiler, considering the shell only, would be 172 lb. Had the designer in the first instance made the course C 9/16 plate and the course B 5/8-in. plate, the boiler shell in question could have been allowed a greater pressure than 172 lb. To a boiler designer these calcu- lations would suggest several things. First would be that if no change was to be made in course C, then the thickness of course A could be reduced, perhaps, 1/16 in. in thickness. This would make a saving in the cost of the boiler. Second, if need be the efficiencies of the longitudinal seams of courses A and B could be less. This is, of course, on the assumption that course A will undergo no changes in regard to thickness of plate, di- ameter, efficiency of longitudinal seam and tensile strength of plate. PART III APPENDIX EXTRACTS FROM UNITED STATES RULES- MARINE— AND FROM THE BOARD OF BOILER RULES STATE OF MASSACHU- SETTS—TABLES APPENDIX EXTRACTS FROM RULES OF THE UNITED STATES BOARD OF SUPERVISING INSPEC- TORS, STEAMBOAT INSPECTION SERVICE United States Rules Pertaining to Riveted Joints The following formulas, equivalent to those of the British Board of Trade, are given for the determination of the pitch, distance between rows of rivets, diagonal pitch, maximum pitch, and distance from centers of rivets to edge of lap of single- and double-riveted lap joints, for both iron and steel boilers. Let p = greatest pitch of rivets in inches. n = number of rivets in one pitch. pd = diagonal pitch in inches. d = diameter of rivets in inches. T — thickness of plate in inches. V = distance between rows of rivets in inches. E = distance from edge of plate to center of rivet in inches. TO DETERMINE THE PITCH Iron plates and iron rivets: C7854>0 ^ 1/2 For steel plates and steel rivets: 23Xd 2 X. 7854X7* . , P 28XT Example for single-riveted joint: Given, thickness of plate = 1/2 in., diameter of rivet 15/16. In this case, n — \. p ., , 2 3 Xd5/i6) 2 X. 7854X1 . Pitch = 28X1/2 1-15/16 =2.071 in. Example for double-riveted joint: Given, thickness of plate = 1/2 in., diameter of rivet = 7/8 in. 71 — 2. Then — p-*i. 2 3X(7/8) 2 X. 7854X2 Pitch = 28Xl/2 "+7/8 =2.85 in. FOR DISTANCE FROM CENTER OF RIVET TO EDGE OF LAP E= sXd 2 Example. — Given, diameter of rivet (d) = 7/8 in., required the distance from center of rivet to edge of plate. E= — =1.312 in., for single- or double-riveted lap joint. APPENDIX 137 FOR DISTANCE BETWEEN ROWS OF RIVETS The distance between lines of centers of rows of rivets for double, chain-riveted joints (V) should not be less than twice the diameter of rivet, but it is more desirable 4J+1 that V should not be less than * 2 Example under latter formula: Given, diameter of rivet = 7/8 in.; then — ( 4 X7/8) + i V= =2.25 in. 2 ° For ordinary, double, zigzag riveted joints: v _\/ (iip+4d)(p+4d) 10 Example. — Given, pitch = 2.85 in. and diameter of rivet = 7/8 in.; then — * T/-V / (nX2.85+4X7/8) (2.85+4X7/8) . . V— =1.487 m. DIAGONAL PITCH For double, zigz&g riveted lap joint. Iron and steel: . 6p+ 4 d pd= 10 Example. — Given, pitch = 2.85 in., and d = j/& in.; then — A (6X2.85)+ (4X7/8) pd= = 2.00 in. r 10 1 Extract the square root of the expression above the line only, then divide by 10. 138 ARITHMETIC OF THE STEAM BOILER Maximum Pitches for Riveted Lap Joints For single-riveted lap joints: Maximum pitch =(1.31 XT) + 1 s/8.J For double-riveted lap joints: Maximum pitch = (2.62 XT) + 1 5/8. Example. — Given, a thickness of plate = 1/2 in., required the maxi- mum pitch alio: -able. For single-riveted lap joint: Maximum pitch = (1.31X1/2)4-1 5/8 = 2.28 in. For double-riveted lap joint: Maximum pitch= (2.62X1/2) + ! 5/8 = 2.935 in. TO DETERMINE THE AREAS OF DIAGONAL STAYS Multiply the area of a direct stay required to support the surface by the slant or diagonal length of the stay; divide this product by the length of a line drawn at right angles to surface supported to center of palm of diagonal stay. The quotient will be the required area of the diagonal stay. aXL A = I Where A = sectional area of diagonal stay. fa = sectional area of direct stay. L — length of diagonal stay. 1 = length of line drawn at right angles to boiler head or surface supported to center of palm of diagonal stay. APPENDIX 139 Given diameter of direct stay = i in., a = .7845, L = 6o in., 1 = 48 in., substituting and solving, .7854X60 ^4 = q — - = .981 sectional area. 45 Diameter — 1 . 1 1 in. = 1 1/8 in. The sectional area of gusset stays, when constructed of triangular right-angled web plates secured to single or double angle bars along the two sides at right angles, shall be determined by formula for diagonal stays, and shall not be less than 10 per cent, greater than would be necessary for a diagonal bolt stay. STAYS The maximum stress in pounds allowable per square inch of cross-sectional area for stays used in the con- struction of marine boilers, when same are accurately fitted and properly secured, shall be ascertained by the following formula: p= AXC a Where P = working pressure in pounds. A= least cross-sectional area of stay in inches. a = area of surface supported by one stay in inches. C = a constant. C = 9ooo for tested steel stays 1 1/4 in. and upward in diameter, when such stays are not forged or welded. The ends may be upset to a 140 ARITHMETIC OF THE STEAM BOILER sufficient diameter to allow for the depth of the thread, provided it is the least diameter of the stay. All such stays after being upset shall be thoroughly annealed. C = 8ooo for a tested Huston or similar type of brace, the cross-sectional area of which ex- ceeds 5 sq. in. C = 7ooo for such tested braces when the cross- sectional area is not less than 1.227 and not more than 5 sq. in., provided such braces are prepared at one heat from a solid piece of plate without welds. C = 75oo for wrought iron through stays 1 1/4 in. diameter and upward. When made of the best quality of refined iron, they may be welded. C = 6ooo for welded crowfoot stays when made of the best quality of refined wrought iron, and for all stays not otherwise provided for when made of the best quality of refined iron or steel without welds. Furnaces The tensile strength of steel used in the construction of corrugated or ribbed furnaces shall not exceed 67,000 and be not less than 54,000 lb.; and in all other furnaces the minimum tensile strength shall not be less than 58,000 and the maximum not more than 67,000 lb. The minimum elongation in 8 in. shall be 20 per cent. All corrugated furnaces having plain parts at the ends APPENDIX 141 not exceeding 9 in. in length (except flues especially pro- vided for) when new, and made to practically true circles, shall be allowed a steam pressure in accordance with the following formula: CXT P = D LEEDS SUSPENSION BULB FURNACE CXT P = D Where P = pressure in pounds. T = thickness in inches, not less than 5/16 in. Z) = mean diameter in inches. C = a constant, 17,300, determined from an actual destructive test under the supervision of the Board, when corrugations are not more than 8 in. from center to center, and not less than 2 1/4 in. deep. MORISON CORRUGATED TYPE CXT p = D Where P = pressure in pounds. T = thickness in inches, not less than 5/16 in. Z) = mean diameter in inches. C = 15,600, a constant, determined from an actual destructive test under the supervision of the Board of Supervising Inspectors, 142 ARITHMETIC OF THE STEAM BOILER when corrugations are not more than 8 in. from center to center and the radius of the outer corrugations is not more than one-half of the suspension curve. [In calculating the mean diameter of the Morison furnace, the least inside diameter plus 2 in. may be taken as the mean diameter, thus — Mean diameter = least inside diameter +2 in.] FOX TYPE CXT F = D Where P = pressure in pounds. T = thickness in inches, not less than 5/16 in. Z) = mean diameter in inches. C — 14,000, a constant, when corrugations are not more than 8 in. from center to center and not less than 1 1/2 in. deep. PURVES TYPE CXT P = D Where P = pressure in pounds. T = thickness in inches, not less than 7/16 in. Where D== least outside diameter in inches. C= 14,000, a constant, when rib projections are not more than 9 in. from center to center and not less than 1 3/8 in. deep. APPENDIX 143 BROWN TYPE p= CXT D Where P = pressure in pounds. T = thickness in inches, not less than 5/16. D = least outside diameter in inches. C= 14,000, a constant (ascertained by an actual destruction test under the supervision of this Board), when corrugations are not more than 9 in. from center to center and not less than 1 5/8 in. deep. The thickness of corrugated and ribbed furnaces shall be ascertained by actual measurement. The manufac- turer shall have said furnace drilled for a 1/4-in. pipe tap and fitted with a screw plug that can be removed by the inspector when taking this measurement. For the Brown and Purves furnaces the holes shall be in the center of the second flat; for the M orison, Fox, and other similar types in the center of the top corrugation, at least as far in as the fourth corrugation from the end of the furnace. TYPE HAVING SECTIONS 1 8 IN. LONG p _CXT D Where P = pressure in pounds. T — thickness in inches, not less than 7/16. D = mean diameter in inches. 144 ARITHMETIC OF THE STEAM BOILER C= 10,000, a constant, when corrugated by sec- tions not more than 18 in. from center to center and not less than 21/2 in. deep, measur- ing from the least inside to the greatest outside diameter of the corrugations, and hav- ing the ends fitted one into the other and substantially riveted together, provided that the plain parts at the ends do not exceed 12 in. in length. TOPS OF COMBUSTION CHAMBERS AND BACK CONNECTIONS Formula for girders over back connection and other flat surfaces: Working pressure = (w _ F) ^p^Z Where W = width of combustion box in inches. P = pitch of supporting bolts in inches. D = distance between girders from center to cen- ter in inches. L = length of girder in feet. d = depth of girder in inches. T = thickness of girder in inches . C = 5So when the girder is fitted with one sup- porting bolt. C = 825 when the girder is fitted with two or three supporting bolts. C = 9i7 when the girder is fitted with four or five supporting bolts. C = 963 when six or seven supporting bolts are used. APPENDIX 145 C = ggo when eight or more supporting bolts are used. EXAMPLE Given W = S4 in., -P=7-5 i n -> D=y.7S in., L = 2.927 ft., ^=7.5 in., T=2 in., C = 82$, then, substituting in formula, w , • 825X7-5X7-5X2 _ \\ orkmg pressure = 7 ;— — 154-3 lb. F (34-7-5)X7-75X2. 9 27 FLAT SURFACES The maximum stress allowable on flat plates sup- ported by stays shall be determined by the following formula : All stayed surfaces formed to a curve the radius of which is over 21 in., excepting surfaces otherwise pro- vided for, shall be deemed flat surfaces. W T CXr2 Working pressure = — p 2 — Where T = thickness of plates in sixteenths of an inch. P = greatest pitch of stays in inches. C=ii2 for screw stays with riveted heads, plates 7/16 in. thick and under. C=i2o for screw stays with riveted heads, plates above 7/16 in. thick. C=i2o for screw stays with nuts, plates 7/16 in. thick and under. C=i25 for screw stays with nuts, plates above 7/16 in. thick and under 9/16 in. C=i35 for screw stays with nuts, plates 9/16 . in. thick and above. 146 ARITHMETIC OF THE STEAM BOILER C=i75 for stays with double nuts having one nut on the inside and one nut on the out- side of plate, without washers or doubling plates. C= 1 60 for stays fitted with washers or doubling strips which have a thickness of at least .5 of the thickness of the plate and a diameter of at least .5 of the greatest pitch of the stay, riveted to the outside of the plates, and stays having one nut inside of the plate, and one nut outside of the washer or doubling strip. For T take 72 per cent, of the combined thickness of the plate and washer or plate and doubling strip. C = 200 for stays fitted with doubling plates which have a thickness equal to at least .5 of the thickness of the plate reenforced, and covering the full area braced (up to the curvature of the flange if any) riveted to either the inside or outside of the plate, and stays having one nut outside and one inside of the plates. Washers or doubling plates to be substantially riveted. For T take 72 per cent, of the combined thickness of the two plates. C=2oo for stays with plates stiffened with tees or angle bars having a thickness of at least two-thirds the thickness of plate and depth of webs at least one-fourth of the greatest pitch of the stays, and substan- tially riveted on the inside of the plates, APPENDIX 147 and stays having one nut inside bearing on washers fitted to the edges of the webs that are at right angles to the plate. For T take 72 per cent, of the combined thickness of web and plate. No such flat plates or surfaces shall be unsupported at a greater distance than 18 in. REQUIREMENTS FOR HEADS All plates used as heads, when new and made to practically true circles, and as described below, shall be allowed a steam pressure in accordance with the following formula: CONVEX HEADS TXS P = R Where P = steam pressure allowable in pounds. T = thickness of plate in inches. 5 = one-fifth of the tensile strength. R = one-half of the radius to which the head is bumped. CONCAVE HEADS For concave heads the pressure allowable will be .8 times the pressure allowable for convex heads. Note. — To find the radius of a sphere cf which the bumped head forms a part, square the radius of head, divide this by the height of bump required; to the result add height of bump, which will equal diameter of sphere, one-half of which will be the required radius, n 148 ARITHMETIC OF THE STEAM BOILER Example. — Required the working pressure of a convex head of a 54-in. radius, material 60,000 lb. tensile strength and 1/2 in. thick. Substituting values, d -5X12000 P = = 222 lb. 27 The pressure allowable on a concave head of the same dimensions would be 222 X. 8 = 177 lb. ANGLE STIFFENERS FOR CURVED SURFACES Where rounded bottoms of combustion chambers are stiffened with single angle-iron stiffeners, such angles shall have a thickness of leaf eight-tenths that of the plate and a depth of at least one-half pitch. Where stiffened with double angle irons or tee bars, such angles or tee bars shall have a thickness of leaf at least two- thirds that of plate and a depth of at least one-fourth of pitch. Said angles or tee bars shall be substantially riveted to the plate supported. Where rounded tops of combustion chambers are stiffened with single or double angle-iron stiffeners, or tee bars, such angles or tee bars shall be of thickness and depth of leaf not less than specified for rounded bottoms of combustion chambers. Said angles or tee bars shall be supported on thimbles and riveted through with rivets not less than 1 in. in diameter, and spaced not to exceed 6 in. between centers. Working pressure allowed on rounded surfaces sup- ported by angle irons or tee bars shall be determined by the following formula: W T CXT2 Working pressure = p jz APPENDIX 149 Where T = thickness of plate in sixteenths of an inch. P = pitch of angle or tee stiffeners in inches. D = diameter of curve to which plate is bent, in inches. C = 9oo, a constant. Example. — Given T = g/i6 in. P = 7 in. D = s i in- Substituting values in formula and solving, ^ 1 • 900X81 . \\ orkmg pressure = — — =20410. per square men PRESSURE PERMISSIBLE ON ROUNDED BOTTOM OF COM- BUSTION CHAMBERS, ANGLES BEING OMITTED 5o(3oor- 2Z, ) ' D Where P = working pressure in pounds. T= thickness of bottom plate of combustion chamber in inches. L = extreme length of plate forming bottom of combustion chamber in inches. D = twice outside radius of bottom of combustion chamber in inches. Example. — Required the working pressure on the bottom plate of a combustion chamber, angles being omitted: Thickness of plate, .82 in., extreme length of plate, 33 in., twice the radius of bottom of combustion chamber, 50 in. Substituting: p _ 5oX(3QoX.82-2X33) = l8o i b . 50 T _ PXD+icoL 15000 ISC ARITHMETIC OF THE STEAM BOILER Pressure allowable on tube sheets where combustion chambers are not suspended from the shell of the boiler shall be determined by the following formula: (D-d)XTX2 7 ooo WXD Where P = working pressure in pounds. D = least horizontal distance between tube centers, in inches. d = inside diameter of tubes in inches. T = thickness of tube plates in inches. W = extreme width of combustion chamber in inches. The compressive stress on tube plates, as determined by the following formula, must not exceed 13,500 lb. per square inch, when pressure on top of combustion chamber is supported by vertical plates of such chamber. PXDXW L 2(D-d)T Where C = stress on tube sheet. P = working pressure in pounds. D = least horizontal distance between tube cen- ters in inches. d = inside diameter of tubes in inches. W = extreme width of combustion chamber in inches. T = thickness of tube sheet in inches. APPENDIX 151 Safety Valves The areas of safety valves shall be determined in accordance with the following formula and table: W # = .2074Xp- Where a = area of safety valve, in square inches, per square foot of grate surface. PF = pounds of water evaporated per square foot of grate surface per hour. P = absolute pressure per square inch = working gage pressure+15. From which formula the areas required per square foot of grate surface in the following table are found by assuming the different values of W and P. The figures (a) in table multiplied by square feet of grate surface give the area of safety valve or valves required. When this calculation results in an odd size of safety valve, use next larger standard size. Examples. — Boiler pressure = 75 lb. per square inch (gage). 2 furnaces: Grate surface = 2 (No.)X5 ft. 6 in. (long)X3 ft. (wide) =33 sq. ft. Water evaporated per pound of coal = 8 lb. Coal burned per square foot grate surface per hour =12 1/2 lb. Evaporation per square foot grate surface per hour = 8X12 1/2 = 100 lb. Hence W = 100 and gage pressure = 75 lb. From table the corresponding value of a is .230 sq. in. Therefore area of safety valve = 33 X. 23 = 7.59 sq. in. For which the diameter is 3 1/8 in. nearly. Boiler pressure = 215 lb. 152 ARITHMETIC OF THE STEAM BOILER 3 o ft w o < ft Oh p> Is Pa. -V H H ft 02 ft O <^ ft rt v- CD > O ^ 0 ^l" oo w OoO t>0 io ^t" oo N w OoO 00 r- l>0 OvO ir < ioiOiO't^^t'tT}-^-T)-r«5fOfOfOrorOfOfO -OoCroOc^Tl-OONI>'^-rorot^Q\ , ^i-ii>.iorONi-ii-iro rfo OC 00 lO Tj" 00 0 Tj-fOM m O O OOO t^ CO l> I>0 O lOlOlOlOTJ-TftrtTtrOfOrOfOf^fOfOrOfOWN 00 CN 03 'tNrooO iflrf N w m m n "3- J> O N u^ O i- 1 O (nooiom CiMO^h o oo O 10 tj- oo oq m o Ooo r-O O 10 i> i>0 OiO'fliO't^t'^'t^TOfOrofOfOrOfOrOfON M w o* cq N 5 mojooOOOOO'^ ^O Oi^ti-i t^iourt'^ioOoo O -^-t>0 "*t g i>0 Oioiom T t , t , t , *fO( v )fOfOfOrOfOroN 0* CS (N N N N N N o — — ! — : — : — : — : — - — - — : — '- — '- — - — - — - — - — - — '- — '- — '- — : — - — - — - — - — - — - — - 3 oOMMOONOOi-'^NOJOHO t~~MD r- t-00 On w rto O -^OO ro -i ^oiflHooorOHa r-\0 rtnH o O\oo r^\0 lOin^roootN n m o _J ' '. ! ! 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(long) X 3 ft. 4 in. (wide) = 110 sq. ft. Water evaporated per pound coal = 10 lb. Coal burned per square foot grate surface per hour = 30 lb. Evaporation per square foot grate surface per hour = 10X30 = 300 lb. Hence IF = 300, gage pressure = 215, and a = .2 70 (from table). Therefore area of safety valve = noX.2 7o = 29.7 sq. in., which is too large for one valve. Use two. 20. 7 = 14.85 sq. in. Diameter = 4 3/8 in. To determine the area cf a safety valve for boiler using oil as fuel or for boilers designed for any evaporation per hour, Divide the total number of pounds of water evaporated per hour by any number of pounds of water evaporated per square foot of grate surface per hour (W) taken from, and within the limits of, the table. This will give the equivalent number of square feet of grate surface for boiler; for estimating the area of valve, then apply the table as in previous examples. Example. — Required the area of a safety valve for a boiler using oil as fuel, designed to evaporate 8000 lb. of water per hour, at 175-lb. gage pressure. Make W = 200. 8000 . — =40, the equivalent 200 ^ ' n grate surface in square feet. For gage pressure = 175 lb. and W = 2oo, from table, a = .218 sq. in.; .218X40 = 8.72 sq. in., the total area required for this boiler, for which the diameter is 3 5/16 in. closely. WATER TUBE AND COIL BOILERS The working pressure allowable on cylindrical shells of water tube or coil boilers, when such shells have a row APPENDIX 155 or rows of pipes or tubes inserted therein, shall be deter- mined by the following formula: (D-d )XTXS DXR Where P = working pressure allowable in pounds. D = distance in inches between the tube or pipe centers in a line from head to head. d = diameter of hole in inches. T = thickness of plate in inches. S = one-sixth of the tensile strength of the plate. R = radius of shell in inches. n = number of tube holes in a pitch. When tubes on any one row are pitched unequally, nd must be substituted in the formula for d; where rows of tubes are pitched diagonally, each diagonal ligament shall not be less than three-fifths of each longitudinal ligament. Example. — Required the working pressure of a cylindrical shell having holes i in. in diameter, spaced 2 in. from center to center, in a line from head to head; material, 1/2 in. thick; diameter of shell, 20 in.; tensile strength of plate, 60,000 lb. Substituting values, we have (2 -i)X. 5X10000 P=— — — = 250 lb. 2X10 ° PORCUPINE-TYPE BOILERS The formula for determining pressure on boilers of the so-called Porcupine and similar types shall be as follows: Multiply the vertical distance between the centers of the horizontal rows of tubes in inches by one-half the diameter of shell of boiler in inches, which gives the area 156 ARITHMETIC OF THE STEAM BOILER upon which the pressure is exerted to break a diagonal ligament, then find the sectional area of the ligament at its smallest part and multiply by one-sixth the tensile strength of the material. This result, divided by the area upon which the strain is exerted, gives the working EFT pressure per square inch, which is as follows =W, the working pressure, in which E equals wddth of liga- ment in inches, F thickness of material in inches, T one- sixth of the tensile strength, C distance between vertical centers, and D one-half the inside diameter of the shell or central column. For the boiler proposed, 30 in. diameter, 5/8 in. thick, tensile strength 60,000 lb., 1.219 in. would be width of ligament, .625 thickness of plate, 10,000 one-sixth of tensile strength, 3 11/16 = 3.6875 in., distance of vertical centers; 15 in., one-half the diam- eter of shell, would be as follows: 1.219 multiplied by .625, this product multiplied by one-sixth the tensile strength, 10,000, equals 7618.75. This product, divided by the product of 3.6875, distance between vertical centers, multiplied by 15, one-half the diameter, equals 55.3125, gives 137.7 as pressure allowed. EXTRACTS FROM BOARD OF BOILER RULES, STATE OF MASSACHUSETTS Maximum Pressure on Boilers 1. The maximum pressure allowed on any steam boiler constructed wholly of cast-iron shall not be greater than twenty-five (25) pounds to the square inch. APPENDIX 157 2. The maximum pressures allowed on any steam boiler, the tubes of which are secured to cast-iron headers, shall not be greater than one hundred and sixty (160) pounds to the square inch. 3. The maximum pressure allowed on any steam boiler constructed of iron or steel shells or drums shall be calculated from the inside diameter of the outside course, the percentage of strength of the longitudinal joint and the minimum thickness of the shell plates; the tensile strength of shell plates to be taken as fifty-five thousand (55,000) pounds per square inch for steel and forty-five thousand (45,000) pounds per square inch for iron when the tensile strength is not known. SHEARING STRENGTH OF RIVETS 4. The maximum shearing strength of rivets per square inch of cross- section of area to be taken as follows: Pounds Iron rivets in single shear 38,000 Iron rivets in double shear 70,000 Steel rivets in single shear 42,000 Steel rivets in double shear 78,000 Factors of Safety 5. The lowest factors of safety used for steam boilers, the shells or drums of which are directly exposed to the products of combustion and the longitudinal joints of which are of lap-riveted construction, shall be as follows: (a) Five (5) for boilers not over ten years old. (b) Five and five-tenths (5.5) for boilers over ten and not over fifteen years old. 158 ARITHMETIC OF THE STEAM BOILER (c) Five and seventy-five hundredths (5.75) for boilers over fifteen and not over twenty years old. (d) Six (6) for boilers over twenty years old. (e) Five (5) on steam boilers, the longitudinal joints of which are of lap-riveted construction, and the shells of drums of which are not directly exposed to the product s of combustion. (f) Four and five-tenths (4.5) on steam boilers, the longitudinal joints of which are of butt and strap construction. Fusible Plugs 1. Fusible plugs as required by Section 20, Chapter 465, Acts of 1907, shall be filled with pure tin. 2. The least diameter of fusible metal shall not be less than one-half inch, except for working pressure of over one hundred and seventy-five (175) pounds gage, or when it is necessary to place a fusible plug in a tube, in which cases the least diameter of fusible metal shall not be less than three-eighth (3/8) inch. 3. The location of fusible plugs shall be as follows: (a) In Horizontal Return Tubular Boilers — in the back head, not less than two (2) inches above the upper row of tubes, and projecting through the sheet not less than one (1) inch. (b) In Horizontal Flue Boilers — in the back head, on a line with the highest part of the boiler exposed to the production of combustion, and projecting through the sheet not less than one (1) inch. (c) In Locomotive Type or Star Water Tube Boilers — APPENDIX 159 in the highest part of the crown sheet, and projecting through the sheet not less than one (1) inch. (d) In Vertical Fire Tube Boilers — in an outside tube, placed not less than one-third (1/3) the length of the tube above the lower tube sheet. (e) In Vertical Submerged Tube Boilers — in the upper tube sheet. (f) In Water Tube Boilers, Horizontal Drums, Babcock & Wilcox Type — in the upper drum, not less than six (6) inches above the bottom of the drum, and over the first pass of the products of combustion, projecting through the sheet not less than one (1) inch. (g) In Stirling Boilers, Standard Type — in the front side of the middle drum, not less than six (6) inches above the bottom of the drum, and projecting through the sheet not less than one (1) inch. (h) In Stirling Boilers, Superheater Type — in the front drum, not less than six (6) inches abo\e the bottom of the drum, and exposed to the products of combus- tion, projecting through the sheet not less than one (1) inch. (i) In Water Tube Boilers, Heine Type — in the front course of the drum, not less than six (6) inches above the bottom of the drum, and projecting through the sheet not less than one (1) inch. (j) In Robb-Mumford Boilers, Standard Type — in the bottom of the steam and water drum, twenty-four (24) inches from the center of the rear neck, and pro- jecting through the sheet not less than one (1) inch. (k) In Water Tube Boilers, Almy Type — in the tube directly exposed to the products of combustion. 160 ARITHMETIC OF THE STEAM BOILER (1) In Vertical Boilers, Climax or Hazelton Type — in a tube or center drum not less than one-half (1/2) the height of the shell, measuring from the lowest cir- cumferential seam. (m) In Cahall Vertical Water Tube Boilers — in the inner sheet of the top drum, not less than six (6) inches above the upper tube sheet. (n) In Scotch Marine Type Boilers — in combustion chamber top, and projecting through the sheet not less than one (1) inch. (o) In Dry Back Scotch Type Boilers — in rear head, not less than two (2) inches above the top row of tubes, and projecting through the sheet not less than one (1) inch. (p) In Economic Type Boilers — in the rear head above the upper row of tubes. (q) In Cast-iron Sectional Heating Boilers — in a section over and in direct contact with the products of combustion in the primary combustion chamber. (r) For other types and new designs, fusible plugs shall be placed at the lowest permissible water level, in the direct path of the products of combustion, as near the primary combustion chamber as possible. Size of Rivets 1. When the size of the rivets in the longitudinal joints of a boiler is not known, the diameter and cross- sectional area of rivet, after driving, shall be taken as follows : APPENDIX 161 Thickness of plate. I7/16 in.i7/i6 in.l 15/32 in.l 1/2 in. |q/i6 in. | 5/8 in. 7/8 in. 15/16 15/16 in. 15/16 1 1/16 I 1/16 Diameter of rivet up to in. over in. in. in. after driving. 2 1/4 in. pitch 2 1/4 in. pitch Cross-sectional area of .6013 .6903 .6903 .6903 .8866 .8866 rivet after driving. sq. in. sq. m. sq. in. sq. in. sq. in. sq. in. Thickness of plate. 1/4 in. 9/32 in. 5/i6 in. 11/32 in. 3/8 in. 3/8 in. 13/32 in. 1 1/16 1 1/16 3/4 3/4 3/4 in. 13/16 in. 13/16 Diameter of rivet in. in. in. in. up to over in. after driving. 2 in. pitch 2 in. pitch Cross-sectional • 3712 .3712 .4418 .4418 .4418 .5185 .5185 area of rivet after sq. in. sq. in. sq. m. sq. m. sq. 111. sq. in. sq. in. driving. Allowable Strain on Stays 1. The maximum allowable strain per square inch net cross-section for weldless mild steel shall be as follows: Size up to and in- Type eluding 1 1/2 in. diam- eter or equivalent Size over 1 1/2 in. diameter or equiva- lent Head to head or through stays. 8,000 lb. 9,000 lb. Diagonal or crowfoot stays. . . . 7,500 lb. 8,000 lb. Screwed stays (stay bolts) 7,000 lb. 7,000 lb. 2. For welded stays the strain allowed per square inch net cross-section shall not exceed six thousand (6000) pounds. 3. For wrought-iron stays or stay bolts the strain allowed per square inch net cross-section shall not exceed six thousand (6000) pounds. l62 ARITHMETIC OF THE STEAM BOILER ^ r Bs ^ IJ II ^ ^ ^ (X, ■ ^ &< ^ IT) cs ^ V* n > LO o i> l- ro oo ro 10 oo Tf C\OC\VOO\000 ro 10 i> rf 03 O O lO 00 -t t<) lfl O Tj- O O vO 00 rf 00 "i" O O f) fO N ^ O io a roo o Tfo o (Nro^ti>Mi>T}-MOO H H 0« fO "* ^ -(-> £ a3 o .d XJ o o +J a aj ft > >. 3 w 0) a 0) .£ bo 03 £ o Ifl > > H 00 CO M 00 N O "t O O M fO ^t t* 0\ 000 + I X42,oooX .6013 = 89,708. E= (4-875- 2X.87s).37sX5S»oo°+-875X.3i2S X95 ? ooo = 9o,429. F = 2X.87sX.37SX9S ? 000 +- 8 75X.3i2SX9S ? 000 = 88,320. G=2X.87sX.37SX9S,ooo+iX4 2 ) 00 °X.6oi3 = 87,599. 82500(^8) rr • r • • 4 — - — V-v = .82o, emciency of joint. 100547^) (See Fig. 3 in this group.) Butt, Triple-riveted, butt and double-strap joint A =strength of solid plate =PXtXT.S. Fig. 4. — Triple- riveted, double butt-strapped joint. B = strength of plate between rivet holes in the outer row= (P-d)tX T.S. C = shearing strength of four (4) rivets in double shear, plus the shearing strength of one (1) rivet in single shear = NXSXa+nXsXa. APPENDIX 175 D = strength of plate between the rivet holes in the second row, plus the shearing strength of one (1) rivet in single shear in the outer row = (P-2d)tXT.S.+nXsXa. E = strength of plate between rivet holes in the second row, plus the crushing strength of butt strap in front of one (1) rivet in the outer row = (P-2d)tXT.S.+dXbXc. F = crushing strength of plate in front of four (4 J rivets, plus the crushing strength of butt strap in front of one (1) rivet = XXdXtXc+nXdX bXc G = crushing strength of plate in front of four (4) rivets, plus the shearing strength of one (1) rivet in single shear = XXdXtXc+nXsXa. Divide B, C, D, E, F, or G (whichever is the least) by A j and the quotient will be the efficiency of a butt and double-strap joint, triple riveted. (See Fig. 4 of this group.) T.S. = 55,000 lb. # = .5185 sq. in. t = $/8 in. = .375 in. 5 = 42,000 lb. 6 = 5/16 in. = .3125 in. 5 = 78,000 lb. F = 6 1/2 in. = 6.5 in. £ = 95,000 lb. d= 13/16 in. = .8125 Number of rivets in single shear in a unit of length of joint = 1. Number of rivets in double shear in a unit of length of joint = 4. -4=6.5X.375X55> 000=I 34,o62. ^ = .( 6 -5-- 8l2 5)-37SX 55,000 =117,304. 176 ARITHMETIC OF THE STEAM BOILER C = 4 X 78,000 X. 5185 + 1 X 42,000 X-5i85 = 183,549- £ = (6.5-2 X.8i25). 3 75X55>°oo+iX 42,000 X .5185 = 122,323. £ = (6.5-2 X.8i25). 3 75 X 55,000+. 8125 X.3125 X95,ooo= 124,667. £ = 4X.8i25X.375X95>°oo+iX.8i25X.3i25X 95,000=139,902. G = 4 X.8125 X.375X95,ooo+iX 42,000 X. 5185 = i37 ; SS8. i34o?fe) == - 87S ' efficienc y of i° int - (See Fig. 4 in this group.) Butt, Quadruple-riveted. butt and double-strap joint, quadruple -riveted A = Strength of solid plate =PXtXT.S. B = Strength of plate between rivet holes in the outer row =(P-d)tXT.S. C = shearing strength of eight (8) rivets in double shear, plus the shearing strength of three (3) rivets in single shear = NXSX a-\-nXsXa. D = strength of plate between rivet holes in the second row, plus the shearing strength of one (1) rivet in single shear in the outer row = (P-2d)tXT.S.+nXsXa. E = strength of plate between rivet holes in the third row, plus the shearing strength of two APPENDIX 177 (2) rivets in the second row in single shear and one (1) rivet in single shear in the outer row=(P^4d)tXT.S.+nXsXa. F = strength of plate between rivet holes in the second row, plus the crushing strength of butt strap in front of one (1) rivet in the outer row=(P-2d)tXT.S.+dXbXc G = strength of plate between rivet holes in Fig. 5. — Quadruple-riveted, double butt-strapped joint. the third row, plus the crushing strength of butt strap in front of two (2) rivets in the second row and one (1) rivet in the outer row=(P-4d)tXT.S.+nXdXbXc. H = crushing strength of plate in front of eight (8) rivets, plus the crushing strength of butt strap in front of three (3) rivets = NXdXtXc+nXdXbXc 1 = crushing strength of plate in front of eight (8) rivets, plus the shearing strength of two (2) rivets in the second row and one 178 ARITHMETIC OF THE STEAM BOILER (1) rivet in the outer row, in single shear = NXdXtXc+nXsXa. Divide B, C, D, E, F, G, H, or / (whichever is the least) by A, and the quotient will be the efficiency of a butt and double-strap joint, quadruple-riveted. T.S. = 55,000 lb. t= 1/2 in. = .5 in. 6 = 7/16 in. = .4375 P=iS in. d= 15/16 in. = .9375 in. a = .6903 sq. in. 5 = 42,000 lb. 5 = 78,000 lb. c = 95,000 lb. Number of rivets in single shear in unit of length of joint = 3. Number of rivets in double shear in unit of length of joint = 8. ^ = i5X.5X55> 000 = 4i2,5oo. £ = (i5--9375)-5X55>°°o = 386,7i8. C = 8 X 78,000 X.6903 + 3 X 42,000 X.6903 = 5i7,723- #=(15-2 X.937S)-S X 55,ooo+ 1 X 42,000 X •69°3 = 389>93°- £=(15-4 X .9375)-5 X 55,000 + 3 X 42,000 X .6903 = 396,353. ^ = (15-2 X.9375)-5X 55>°°°+-9375 X-4375 X 95,000 = 399,902. APPENDIX 179 G=(i5-4 X .937S)-S X 55>°oo + 3 X -9375 X •4375X95> 000 = 426,269. H = 8X -937S X.5 X 95,000 + 3 X-937SX.437SX 95,000 = 473,145. 7 = 8 X -9375 X .5 X 95>°°° + 3 X 42,000 X .6903 = 443,229. 386718(5) ffi . , . . ■ 4i25oo04r- 937 ' efikienCy ° f JOmt * (See Fig. 5 of this group.) Bumped Heads The minimum thickness of a convex head shall be determined by the following formula: RXF.S.XP T.S. The minimum thickness of a concave head shall be determined by the following formula: RXF.S.XP .6(T.S.) R = one-half the radius to which the head is bumped. F.S. = 5 = factor of safety. P = working pressure, in pounds per square inch, for which the boiler is designed. T.S. = tensile strength, in pounds per square inch, stamped on the head by the manufacturer. t = thickness of the head in inches. When a convex or a concave head has a manhole 13 180 ARITHMETIC OF THE STEAM BOILER opening, the thickness as found by the preceding for- mulas shall be increased by not less than one-eighth (1/8) inch. FORMULA TO FIND AREA OF SEGMENT OF CIRCLE TO BE BRACED ~ a / jj— .608 = area in square inches. H = distance from tubes to shell, minus five (5) inches. R = radius of boiler, minus three (3) inches. FORMULA FOR DIAMETER OF STAY BOLTS AT BOTTOM OF THREAD D — (PX 1.732) =d, or for 12 threads per inch, £>-(.o8333Xi.73 2 )=^ then D = diameter of stay bolt over the threads. P = pitch of threads = 1/ 1 2 = .08333. d = diameter of stay bolt at bottom of threads. 1.732 = a constant. When U. S. threads are used, the formula becomes: Z) — (PX1.732X.75W- FORMULA FOR CAST-IRON NOZZLES The minimum thickness of cast-iron nozzles shall be determined by the following formula: Pdf , APPENDIX 181 P = working pressure in pounds per square inch. d = inside diameter of nozzle in inches. /= factor of safety = 12. 5 = ultimate tensile strength of cast iron, not less than eighteen thousand (18,000) pounds per square inch. .5 = a constant. t = thickness of nozzle in inches. Maximum Pressure on Boiler Shells The maximum pressure to be allowed on a steel or wrought-iron shell or drum shall be determined from the minimum thickness of the shell plates, the lowest tensile strength stamped on the plates by the plate manufacturer, the efficiency of the longitudinal joint or ligament between the tube holes, whichever is the least, the inside diameter of the outside course, and a factor of safety of not less than five (5), the formula being: T.S. X^X % 7?v7? y — = maximum allowable working pressure per square inch in pounds. T. S. = tensile strength of shell plates in pounds. t = minimum thickness of shell plates in inches. % = efficiency of longitudinal joint or ligament be- tween tube holes, whichever is the least. R = radius = one-half (1/2) the inside diameter of the outside course of the shell or drum. F.5. = 5, the lowest factor of safety allowed on boilers installed after May 1, 1908. 182 ARITHMETIC OF THE STEAM BOILER The method of determining the efficiency of the longitudinal joint has already been explained and illus- trated. To find the efficiency of ligaments, the following formulas are to be employed. Efficiency of Ligaments When a shell or drum is drilled for tube holes in a line parallel to the axis of the shell or drum, the efficiency of the ligament between the tube holes shall be deter- -(£-&- Q Ct) (fr d> <-5 1 4-> <— &£->*— VA-* ^ ) Q Cp Cp Cp Cp C p-^)- Longitudinal Line >. Fig. i. — Diagram for calculating the efficiency of ligament. mined as follows: (a) when the pitch of the tube holes on every row is equal the formula is: p d — - — = efficiency of ligament. p = pitch of tube holes in inches. d = diameter of tube holes in inches. Example. — Pitch of tube holes in the drum of a water-tube boiler = 5 1/4 in. = 5.25 in. Diameter of tube holes = 3 1/4 in. = 3.25 in. P~d 5-25-3-25, : .38, efficiency of ligament. 5-25 (See Fig. 1 of this group.) APPENDIX 183 (b) when the pitch of the tube holes on any one row is unequal, the formula is: P-na = efficiency of ligament. ■e e-©-e ^._ 5 4-/W--6^-^-54i^--64f , ->U--544^[<---6^-^--54- h— -1-2^— h Longitudina l Line Fig. 2. — Diagram for calculating the efficiency of ligament. P = unit length of ligament in inches. n = number of tube holes in length, P. d = diameter of tube holes in inches. -29 H- J Longitudinal Line Fig. 3. — Diagram for calculating the efficiency of ligament. Example. — P — nd _i2 — 2X3. 25 = .458, efficiency of ligament. (See Fig. 2 of this group.) Example. — P—nd _ 2g. 25-5X3.25 .444, efficiency of ligament. (See Fig. 3 of this group.) 1 84 ARITHMETIC OF THE STEAM BOILER When a shell or drum is drilled for tube holes in a line diagonal to the axis of the shell or drum, the efficiency of the ligament between the tube holes shall be determined as follows: P-d — = efficiency of ligament. P P = diagonal pitch of tube holes in inches. d = diameter of tube holes in inches. p = distance between rows of tubes longitudinally. { p'cfo ($> OF CIRCLES FROM 1 TO 100 (C ontinued) Dia. Area Circum. Dia. Area Circum. Dia. Area Circum. 64 3217.0 201.062 70 3848.5 219.911 7°~ 4536.5 238.761 1 8 3229.6 201.455 1 8 3862.2 220.304 1 8 45514 239-I54 i 3242.2 201.847 1 4 3876.0 220.697 1 4 4566.4 239.546 1 3 2 54.8 202.240 3 8 3889.8 221.090 1 4581.3 239.939 4 3 26 7.5 202.633 1 2 3903-6 221.482 4 4596.3 240332 § 3280.1 203.025 5 8 3917-5 221.875 f 461 1. 4 240.725 I 3292.8 203.418 3 4 3931-4 222.268 3 4 4626.4 241. 117 i 33°5- 6 203.811 7 8 3945-3 222.660 7 8 4641.5 241.510 65 33i 8 -3 204.204 71 3959-2 223.053 77 4656.6 241.903 I 333 1 - 1 204.596 1 3973-1 223.446 1 8 4671.8 242.295 1 4 3343-9 204.989 1 4 3987.1 223.838 1 4 4686.9 242.688 1 335 6 -7 205.382 3 8 4001. 1 224.231 3 5 4702.1 243.081 1 5 3369.6 205.774 4 4015.2 224.624 1 47I7.3 243-473 1 3382.4 206.167 5 8 4029.2 225.017 5 8 4732.5 243.866 f 3395-3 206.560 3 4 4043-3 225.409 3 4 4747-8 244.259 1 3408.2 206.952 7 8 4057-4 225.802 7 8 4763.1 244.652 66 3421.2 207.345 72 4071.5 226.195 78 4778.4 245.044 1 8 3434-3 207.738 1 8 4085.7 226.587 i 4793-7 245-437 1 4 3447-2 208.131 1 4 4099 .8 226.930 1 4 4809.0 245.830 1 3460.2 208.523 3 8 4114.0 227.373 3 8 4824.4 246.222 4 3473-2 208.916 4 4128.2 227.765 1 4839.8 246.615 1 3486.3 209.309 5 8 4142.5 228.158 1 4855.2 247.008 1 3499-4 209.701 3 4 4156.8 228.551 3 4 4870.7 247.400 I 35 I2 -5 210.094 i 4171.1 228.944 i 4886.2 247-793 67 3525-7 210.487 13 4185.4 229.336 79 4901.7 248.186 J 3538.8 210.879 1 8 4199-7 229.729 1 8 4917.2 248.579 1 3552.o 211.272 1 4 4214. 1 230.122 i 4932.7 248.971 1 3565-2 211.665 3 8 4228.5 230.514 3 8 4948.3 249.364 J 3578.5 212.058 1 2 4242.9 230.907 1 2 4963.9 249-757 f 359L7 212.450 5 8 42574 231.300 5 8 4979-5 250.149 a 4 3605.0 212.843 3 4 4271.8 231.692 3 4 4995.2 250.542 1 3618.3 213.236 I 4286.3 232.085 7 8 5010.9 25 -935 68 3631.7 213.628 74 4300.8 232.478 80 5026.5 251-327 1 8 3645 -° 214.021 1 8 43I5-4 232.871 i 5042.3 251.720 i 3658.4 214.414 1 4 4329.9 233.263 1 • 4 5058.0 252.113 | 3671.8 214.806 | 4344-5 233-656 f 5073-8 252.506 4 3685.3 215.199 1 2 4359-2 234.049 i 5089.6 252.898 5 8 3698.7 2I5-592 5 8 4373-8 234.441 5 8 5I054 253.291 3 4 3712.2 215.984 3 4 4388.5 234-334 3 4 5121.2 253.684 1 37257 216.337 8 4403.1 235.227 i 5i37.i 254.076 69 3739-3 216.770 75 4417.9 235.619 81 5 J 53-o 254.469 i 3752.8 217.163 i 4432.6 236.012 i 5168.9 254.862 i 3766.4 217-555 1 4 4447.4 236.405 1 4 5184.9 255.254 i 3780.0 217.948 § 4462.2 236.798 3 8 5200.8 255.647 i 3793-7 218.341 •4 4477.0 237.190 i 5216.8 256.040 f 3807.3 218.733 5 8 4491.8 237.583 5 8 5232.8 256433 i 3821.0 219.126 3 4 4506.7 237.976 3 4 5248.9 256.825 I 3834.7 219.519 7 8 4521.5 238.368 i 5264.9 257.218 14 195 TABLE I.— AREAS AND CIRCUMFERENCES OF CIRCLES FROM I TO ioo (Continued) Did. Area 82 5281.O 1 8 5 2 97-l 1 4 53I3-3 i 1 2 1 3 4 53294 5345-6 5361.8 5378.1 7 8 5394.3 83 1 8 5410.6 5426.9 i f 5443-3 5459-6 5476.0 5 8 I 5492.4 5508.8 * 5525-3 84 1 8 1 4 5541.8 5558.3 5574-8 i i I 3_ 4 1 5591-4 5607.9 5624.5 5641.2 5657.8 85 1 5 5674.5 5691.2 i 5707.9 i 5724-7 i f 574L5 5758.3 4 5775-1 8 579L9 86 i I 1 8 5808.8 5825.7 5842.6 5859.6 5876.5 5893.5 3_ 4 5910.6 8 5927.6 87 i 1 4 1 i a 4 5944-7 5961.8 5978.9 5996.0 6013.2 6030.4 6047.6 6064.9 Circum. JDia. Area Circum. |Dia. Area 257.611 88 6082.1 276.460 94 6939.8 258.OO3 1 8 6099.4 276.853 i 6958.2 258.396 i 6256.7 277.846 i 6976.7 258.789 1 6134.1 277.638 3 8 6995.3 259.181 J 61514 278.031 i 7013.8 259.574 5 8 6168.8 278.424 5 8 7032.4 259.967 i 6186.2 278.816 I 7051.0 260.359 i 6203.7 279.209 I 7069.6 260.752 89 6221. 1 279.602 95 7088.2 261.145 i 6238.6 279.994 i 7106.9 261.538 1 4 6256.1 280.387 1 4 7125.6 261.930 f 62737 280.780 1 7M4.3 262.323 i 6291.2 281.173 1 2 7163.0 262.716 f 6308.8 281.565 5 8 7181.8 263.103 3 4 6326.4 281.958 3 4 7200.6 263.501 7 8 6344.1 282.351 i 7219.4 263.894 90 6361.7 282.743 96 7238.2 264.286 i 63794 283.136 i 7 2 57.i 264.679 1 4 6307.1 283.529 1 4 7276.0 265.072 3 8 6414.9 283.921 3 8 7294.9 265.465 1 6432.6 284.314 1 7313-8 265.857 5 8 6450.4 284.707 I 7332.8 266.250 3 4 6468.2 285.100 3 4 7351-8 266.643 i 6486.O 285.492 7 8 7370.8 267.035 91 6503.9 285.885 97 7389.8 267.428 i 6521.8 286.278 i 7408.9 267.821 1 4 65397 286.670 i 7428.0 268.213 1 6557.6 287.063 3 8 7447-1 268.606 i 6575.5 287.456 1 2 7466.2 268.999 5 8 6593.5 -287.848 5 8 7485.3 269.392 3 4 66 1 1. 5 288.241 3 4 7504.5 269.784 i 6629.6 288.634 8 75237 270.177 92 6647.6 289.027 98 7543.0 270.570 i 666s. 7 289.419 1 8 7562.2 270.962 1 4 6683.8 289.812 1 4 7581.5 27L355 3 s 6701.9 290.205 3 8 7600.8 271.748 1 6720.1 290.597 1 2" 762O. T 272.140 5 8 6738.2 290.990 5 8 7639.5 272.533 3 4 6756.4 291.383 3 4 7658.9 272.926 1 6774.7 291.775 I 8 7678.3 2 73-3 I 9 93 6792.9 292.168 99 7697.7 273.711 i 681 1. 2 292.561 i 77I7.I 274.104 i 6829.5 292.954 i 7736.6 274.497 3 8 6847.8 293.346 3 8 7756.1 274.889 1 2 6866.1 293-739 I 7775-6 275.282 8 6884.5 294.132 5 8 7795-2 275.675 3 4 6902.9 294.524 3 4 7814.8 276.067 7 8 6921.3 294.917 7 8 78344 196 TABLE II.— DECIMAL EQUIVALENTS OF FRACTIONS OF AN INCH. (ADVANCING BY 8THS, 16THS, 32NDS AND 64THS.) 8ths 3- >nds 64ths 64ths I = - T2 5 1 _ .32 — •°3 I2 5 A = .015625 If = .515625 i = -250 3 32 — •°9375 A = .046875 ff = .546875 1 = -375 32 = .15625 A = .°7 8 i25 H - -578125 1 = .500 3 2 = .21875 A = - io 9375 If = .609375 f - .625 9 _ 3 2 — .28125 A = -140625 tt = .640625 i = -75o 11 3 2 — •34375 *i - -171875 H = .671875 i - .875 13 "3"2 — .40625 « = -203125 It = .7°3 I2 5 15 32 — 46875 a = .234375 H = .734375 i6ths. A = -0625 17 _ 3 2 — 03125 a = .265625 f| = .765625 A - .1875 « = •59375 H = .296875 « = .796875 A = -3I25 2 1 — 32 — .65625 *i - -328125 II - .828125 A = -4375 2 3 32 — •71875 If = .359375 H = .859375 A = 0625 25 _ 32 — .78125 If = .390625 II = .890625 tt - -6875 2 7 32 — .84375 U = .421875 If = .921875 if - .8125 29 3 2 — .90625 II = -453 I2 5 ft = .953125 x# = -9375 31 3 2 — .96875 tt = 484375 If = .984375 197 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 {Continued) No. Square Cube Sq. Root Cube Root Circle Circum. Area 41 1681 68921 6.4031 3.4482 128.81 1320.25 42 1764 74088 6.4807 3.4760 I3I-95 138544 43 1849 795°7 6-5574 3-5°34 I35.09 1452.20 44 1936 85184 6.6332 3-53°3 138.23 1520.53 45 2025 91125 6.7082 3o569 141.37 I590.43 46 2116 9733 6 6.7823 3-5830 144-5 l 1661.90 47 2209 103823 6.8557 3.6088 147.65 1734-94 48 2304 1 10592 6.9282 3-6342 150.80 1809.56 49 2401 1 1 7649 7.0000 3.6-93 153.94 1885.74 5° 2500 125000 7.0711 3.6840 157.08 1963.50 51 2601 132651 7.1414 3.7084 160.22 2042.82 5 2 2704 140608 7.2111 3-7325 163.36 2123.72 53 280Q " 148877 7.2801 3-7563 166.50 2206.18 54 2916 157464 7-3485 3-7798 169.65 2290.22 55 3° 2 5 166375 7.4162 3.8030 172.79 2375.83 56 3 T 3 6 175616 74833 3-8259 175.93 2463.OI 57 3 2 49 185193 7-5498 3.8485 179.07 2551-76 58 3364 195112 7.6158 3.8709 182.21 2642.08 59 348i 205379 7.6811 3-893° 185.35 2733.97 60 3600 2 1 6000 7.7460 3-9!49 188.50 2827.43 61 3721 226981 7.8102 3-9365 191.64 2922.47 62 3 8 44 238328 7.8740 3-9579 194.78 3019.07 63 3969 250047 7-9373 3-9791 197.92 3II7.25 64 4096 262144 8.0000 4.0000 201.06 3216.99 65 4225 274625 8.0623 4.0207 204.20 33^-3* 66 435 6 287496 8.1240 4.0412 207.35 3421.19 67 4489 300763 8.1854 4.0615 210.49 3525-65 68 4624 3 J 443 2 8.2462 4.0817 213.63 3631.68 69 4761 328509 8.3066 4.1016 216.77 3739.28 70 4900 343000 8.3666 4-1213 219.91 384845 7i 5°4i 3579" 8.4261 4.1408 223.05 3059. T 9 72 5184 373248 8.4853 4.1602 226.19 4071.50 73 53 2 9 389017 8.5440 4.1793 229.34 4185.39 74 5476 405224 8.6023 4.1983 232.48 4300.84 75 5625 421875 8.6603 4.2172 235.62 4417.86 76 5776 438976 8.7178 4.2358 238.76 453646 77 5929 456533 8.775o 4-2543 241.90 4656.63 78 6084 474552 8.8318 4.2727 245.04 4778.36 79 6241 493°39 8.8882 4.2908 248.19 4001.67 80 6400 512000 8.9443 4.3089 251.33 5o 2 6.55 199 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERE.XCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) ClR^T T No. Square Cube Sq. Root Cube Root Circum. Area 8i 6561 53 T 44i 9 .OOOO 4-3267 254-47 5 J 53-oo 82 6724 551368 9-0554 4-3445 257.61 5281.02 83 6889 571787 9.1104 4.3621 260.75 5410.61 84 7056 592704 9.1652 4-3795 263.89 5541-77 85 7225 614125 9-2195 4.3968 267.04 5674.50 86 7396 636056 9.2736 4.4140 270.18 5808.80 87 7569 658503 9-3274 4.4310 273-32 5944-68 88 7744 681472 9.3808 4.4480 276.46 6082.12 89 7921 704969 9.4340 4.4647 279.60 6221. 14 90 8100 729000 9.4868 44S14 282.74 6361.73 9i 8281 753571 9-5394 4-4979 285.88 6503.88 92 8464 778688 9-5917 4-5*44 289.03 6647.61 93 8649 8o4357 9-6437 4.5307 292.17 6792.91 94 8836 830584 9.6954 4.5468 295.3I 6939.78 95 9025 857375 9.7468 4.5629 298.45 7088.22 96 Q2l6 884736 9.7980 4.57 8 9 301.59 7238.23 97 9409 912673 9.8489 4-5947 304.73 7389.81 98 9604 941192 9.8995 4.6104 307.88 7542.96 99 9801 970299 9.9499 4.6261 311.02 7697.69 100 I OOOO I 000000 10.0000 4.6416 314.16 7853.98 IOI I020I 1030301 10.0499 4.6570 3*7-30 8011.85 102 IO404 1061208 10.0995 4-6723 320.44 8171.28 103 I0609 1092727 10.1489 4.6875 3 2 3-58 8332.29 104 I0816 1 1 24864 10.1980 4.7027 3 26 -73 8494.87 105 IIO25 II57625 10.2470 4.7177 329.87 8659.OI 106 II236 1191016 10.2956 4.7326 333-0 1 8824.73 107 1 1 449 1225043 10.3441 4-7475 336.15 8992.02 108 1 1 664 1259712 10.3923 4.7622 339-29 9160.88 109 11881 1295029 10.4403 4.7769 342.43 933I-32 no 12100 1331000 10.4881 4.7914 345-58 9503-32 III 12321 1367631 io.5357 4.8059 348.72 9676.89 112 12544 1404928 10.5830 4.8203 351.86 9852.03 113 12769 1442897 10.6301 4.8346 35 5 -oo IOO28.7 TI4 12996 I 48 1 544 10.6771 4.8488 358.14 IO207.O "5 13225 1520875 10.7238 4.8629 361.28 IO386.9 Il6 I345 6 1560896 10.7703 4.8770 364.42 IO568.3 117 13689 1601613 10.8167 4.8910 367o7 I075I-3 Il8 13924 1643032 10.8628 4.9049 37o.7i 10935-9 119 14161 1 685 1 59 10.0087 4.Q187 373-85 III22.0 I20 14400 1728000 10.9545 4.9324 376.99 I I309.7 200 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENXES AXD CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) ClRC'i -V. No. Square Cube Sq. Root Cube Root Circum. Area 121 14641 1771561 1 1 .OOOO 4.9461 380.13 1 1 499.O 122 14884 1815848 11.0454 4-9597 383.27 1 1 689 .9 123 15129 1860867 II.0905 4.9732 386.42 11882.3 124 15376 1906624 H-I355 4.9866 38906 12076.3 125 15625 i953 I2 5 11. 1803 5 .OOOO 392.70 12271.8 126 15876 2000376 11.2250 5-°i33 395.84 12469.O 127 16129 2048383 11.2694 5.0265 398.98 12667.7 128 16384 2097152 "•3137 5-°397 402.12 12868.O 129 1 664 1 2146689 11.3578 5.0528 405.27 13069.8 13° 16900 2197000 1 1. 4018 5.0658 408.41 13273.2 131 17161 2248091 n-4455 5.0788 411-55 13478.2 132 17424 2299968 1 1. 489 1 5.0916 414.69 13684.8 133 17689 235 26 37 11.5326 5.1045 417.83 13892.9 134 I795 6 2406104 11.5758 5-1172 420.97 14102.6 135 18225 2460375 11. 6190 5.1299 424.12 I43I3.9 136 18496 2515456 1 1. 6619 5.1426 427.26 14526.7 137 18769 2571353 11.7047 5-I55 1 430.40 14741.I 138 19044 2628072 11-7473 5.1676 433-54 I4957-I 139 19321 2685619 11.7898 5.1801 436.68 15*74-7 140 19600 2744000 11.8322 5- I 9 2 5 439.82 *5393& 141 19881 2803221 11.8743 5.2048 442.96 15614-5 142 20164 2863288 1 1. 9 1 64 5.2171 446.II 15836.8 143 20449 2924207 n.9583 5.2293 449.2 5 16060.6 144 20736 2985984 12.0000 5.2415 452.39 16286.0 J 45 21025 3048625 12.0416 5-2536 455-53 1 6513.0 146 21316 3112136 12.0830 5.2656 458.67 16741.5 147 21609 3176523 12.1244 5.2776 461.81 16971.7 148 21904 3241792 12.1655 5.2896 464.96 17203.4 149 22201 3307949 12.2066 5.3015 468.IO 17436.6 15° 22500 3375°°° 12.2474 5.3133 471.24 I767I-5 151 22801 344295 1 12.2882 5.3251 474.38 17907.9 152 23104 35 1 1808 12.3288 5.3368 477.52 18145.8 153 23409 358i577 12.3693 5.3485 480.66 18385.4 154 23716 3652264 12.4097 5.3601 483.81 18626.5 155 24025 372387=: 12.4499 5.3717 486.95 18869.2 156 24336 3796416 12.4900 5.3832 490.09 191134 157 24649 3869893 12.5300 5-3947 493-23 19359.3 158 24964 3944312 12.5698 5.4061 496.37 19606.7 159 25281 4019679 I2.0OQ5 5.4175 499 -5 1 19855.7 160 25600 4096000 I2.649I 54288 502.65 20106.2 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) No. Square Cube Sq. Root Cube Root Circle Circum. Area 161 25921 4173281 12.6886 5.4401 505.80 20358.3 162 26244 4251528 12.7279 5-45 J 4 508.94 20612.0 163 26^69 433°747 12.7671 5.4626 512 .08 20867.2 164 26896 4410944 12.8062 5-4737 515.22 21124.1 165 27225 4492125 I2.S452 5.4848 518.36 213S2.5 166 27556 4574296 12.8841 5-4959 521.50 21642.4 167 27889 4657463 12.9228 5-5069 524.65 21904.0 168 2G224 4741632 12.9615 5.5178 527-79 22167. 1 169 28561 4826809 1 3 .0000 5.5288 530.93 22431.8 170 28900 4913000 13.0384 5-5397 534.07 22698.O 171 29241 500021 T 13.0767 5-5505 537-21 22965.8 172 29584 5088448 13.1149 5-56i3 540.35 23235.2 173 29929 5!777 J 7 *3-*S*9 5-5721 543-5° 23506.2 174 30276 5268024 13.1909 5.5828 546.64 23778.7 175 30625 5359375 13.2288 5-5934 549-78 24052.8 176 30976 5451776 13.2665 5.6041 552.92 24328.5 177 3*3 2 9 5545233 13-3041 5.6i47 556.o6 24605.7 178 31684 5639752 I3-34I7 5.6252 559-20 24884.6 179 32041 5735339 ^3-3791 5.6357 562.35 25164.9 180 32400 5832000 13.4164 5.6462 56549 25446.9 181 32761 5929741 I3-4536 5.6567 568.63 25730.4 182 33 I2 4 6028568 13.4907 5.6671 57L77 26015.5 183 33489 6128487 I3-5277 5.6774 574.91 26302.2 184 33856 6229504 I3-5647 5.6877 578.05 26^90.4 185 34225 6331625 13.6015 5.6980 581.19 26880.3 186 34596 6434856 13.6382 5.7083 584.34 27171.6 187 34969 6539203 13.6748 57185 587.48 27464.6 188 35344 6644672 i3-7ii3 5.7287 590.62 27759.1 189 35721 6751269 13-7477 57388 59376 28055.2 190 36100 6859000 15.7840 5-7489 596.90 28352.9 191 36481 6967871 13.8203 5.7590 600.04 28652.I 192 36864 7077888 13.8564 5.7690 603.19 28952.9 193 37 2 49 7189057 13.8924 5.7790 606.33 29255.3 194 37636 73 OI 384 13.9284 5.7890 609.47 29559.2 195 38025 7414875 13.9642 57989 612.61 29864.8 196 38416 7529536 14.0000 5.8088 6i5.75 30171.9 197 38809 7645373 14.0357 5.8186 618.89 30480.5 198 39204 7762392 14.0712 5.8285 622.04 30790.7 199 39601 7880599 14.1067 5-8383 625.18 31102.6 200 40000 8000000 14.1421 5.8480 628.32 3I4I5.9 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) ClI?^T TT No. Square Cube Sq. Root Cube Root Circum. Area 201 40401 8120601 14.1774 5.8578 631.46 31730.9 202 40804 8242408 14.2127 5.8675 634.60 320474 203 41209 8365427 14.2478 5-877I 637.74 32365.5 204 41616 8489664 14.2829 5.8868 640.89 32685.1 205 42025 8615125 14.3178 5.8964 644.03 33006.4 206 42436 8741816 I4.3527 5-9059 647.17 33329.2 207 42849 8869743 14.3875 5-9*55 650.31 33653-5 208 43 2 64 8998912 14.4222 5-9250 653-45 33979-5 209 43681 9129329 14.4568 5-9345 656.59 34307.0 210 44100 9261000 14.4914 5-9439 659-73 34636.1 211 44521 939393 1 14.5258 5-9533 662.88 34966.7 212 44944 9528128 14.5602 5.9627 666.02 35298.9 213 45369 9663597 14.5945 5.9721 669.16 35632.7 214 45796 9800344 14.6287 5.9814 672.30 35968.1 215 46225 9938375 14.6629 5.9907 675-44 36305.O 216 46656 10077696 14.6969 6.0000 678.58 36643.5 217 47089 10218313 14.7309 6.0092 681.73 36983.6 218 47524 10360232 14.7648 6.0185 684.87 37325-3 219 47961 io 5°3459 14.7986 6.0277 688.01 37668.5 220 48400 10648000 14.8324 6.0368 691.15 38013.3 221 48841 10793861 14.8661 6.0459 694.29 38359.6 222 49284 1 094 1 048 14.8997 6.0550 69743 38707.6 223 49729 11089567 14.9332 6.0641 700.58 39057.1 224 50176 11239424 14.9666 6.0732 703.72 39408.I 225 50625 1 1390625 15.OOOO 6.0822 706.86 39760.8 226 51076 11543*76 *5-°333 6.0912 710.OO 40115.O 227 5^29 1 1 697083 15.0665 6.1002 7 I 3-*4 40470.8 228 5!9 8 4 11852352 15.0997 6.1091 716.28 40828.I 229 52441 12008989 I 5-i3 2 7 6.1180 719.42 41187.1 230 52900 1 2 167000 15.1658 6.1269 722.57 41547.6 231 5336i 12326391 15.1987 6.1358 725-7I 41909.6 232 53824 12487168 i5- 2 3 I 5 6.1446 728.85 42273.3 233 54289 12649337 15.2643 6.1534 731-99 42638.5 234 54756 12812904 15.2971 6.1622 735-13 43005.3 235 55225 12977875 I5-3297 6.1710 738.27 43373-6 236 55696 13144256 15-3623 6.1797 741.42 43743-5 237 56169 I 33 I2 o53 I5-3948 6.1885 744.56 44115.O 238 56644 13481272 15.4272 6.1972 747.70 44488.I 239 57i2i 13651919 I5-4596 6.2058 750.84 44862.7 240 57600 13824000 15.4919 6.2145 753-98 45238.9 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 {Continued) No. 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 Square 58564 59049 59536 60025 60516 61009 61504 62001 62500 63001 635 4 64009 64516 65025 65536 66049 66564 67081 67600 68l 2 1 68644 69169 69696 70225 70756 71289 71824 72361 729OO 73441 73984 745 2 9 75076 75625 76176 76729 77284 7784I 784OO Cube 3997521 4172488 4348907 4526784 4706125 4886936 5069223 5252992 543 82 49 5625000 5813251 6003008 6194277 6387064 6581375 6777216 6974593 7I735I2 7373979 7576000 777958i 7984728 8191447 8399744 8609625 8821096 9034163 9248832 9465109 9683000 19902511 20123648 20346417 20570824 20796875 21024576 21253933 21484952 2 1 717639 21952000 Sq. Root Cube Root 15.5242 6.2231 I 5-55 6 3 6.2317 15-5885 6.2403 15.6205 6.2488 15.6525 6.2573 15.6844 6.2658 15.7162 6.2743 15.7480 6.2828 15.7797 6.2912 15,8114 6.2996 15.8430 6.3080 15.8745 6.3164 15.9060 6.3247 15.9374 ^•333° 15.9687 6.3413 16.0000 6.3496 16.0312 6.3579 16.0624 6.3661 16.0935 6-3743 16.1245 6.3825 16.1555 6.3907 16.1864 6.3988 16.2173 6.4070 16.2481 6.4151 16.2788 6.4232 16.3095 6.4312 16.3401 6.4393 16.3707 6.4473 16.4012 6-4553 16.4317 6.4633 16.4621 6.47*3 16.4924 6.4792 16.5227 6.4872 16.5529 6.4951 16.5831 6.5030 16.6132 6.5108 16.6433 6.5187 16.6733 6.5265 16.7033 6-5343 16.7332 6.5421 Circle Circum. I Area 757-12 760.27 76341 766.55 769.69 772.83 775-97 779.12 782.26 785.40 788.54 791.68 794.82 797.96 801. 11 804.25 807.39 810.53 813.67 816.81 819.96 823.10 826.24 829.38 832.52 835.66 838.81 841.95 845.09 848.23 851.37 854.51 857.66 860.80 863.94 867.08 870.22 873-36 876.50 879.65 45616.7 45996.1 46377.0 46759.5 47M3.5 -47529.2 47916.4 48305.1 48695.5 49087.4 49480.9 49875.9 50272.6 50670.7 51070.5 5I47L9 51874.8 52279.2 52685.3 53092.9 53502.1 53912.9 54325-2 54739-1 55J54.6 55571.6 55990.3 56410.4 56832.2 57255.5 57680.4 58106.9 58534.9 58964.6 59395-7 59828.5 60262.8 60698.7 61 136.2 6i575. 2 204 .TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF XOS. FROM i TO 520 (Continued) No. Square Cube Sq. Root Cube Root Circle Circum. Area 281 78961 22188041 16.7631 6.5499 882.79 62015.8 282 795 2 4 22425768 16.7929 6-5577 885.93 62458.0 283 80089 22665187 16.8226 6.5654 889.07 62901.8 284 80656 22906304 16.8523 6.5731 892.21 63347.1 285 81225 23149125 16.8819 6.5808 895-35 63794.O 286 81796 23393656 16.9115 6.5885 898.50 64242.4 287 82369 23639903 16.9411 6.5962 901.64 64692.5 288 82944 23887872 16.9706 6.6039 904.78 65144. 1 289 83521 24137569 17.0000 6.6115 907.92 65597.2 290 84100 24389000 17.0294 6.6191 911.06 66052.O 291 84681 24642171 17.0587 6.6267 914.20 66508.3 292 85264 24897088 17.0880 6.6343 9^7-35 66966.2 293 85849 25153757 17.1172 6.6419 920.49 67425.6 294 86436 25412184 17.1464 6.6494 923.63 67886.7 295 87025 25672375 17.1756 6.6569 926.77 68349-3 296 87616 25934336 17.2047 6.6644 929.91 68813.5 297 88209 26198073 i7- 2 337 6.6719 933-05 69279.2 298 88804 26463592 17.2627 6.6794 936.19 69746.5 299 89401 26730899 17.2916 6.6869 939-34 70215.4 300 90000 27000000 17-3205 6.6943 942.48 70685.8 301 90601 27270901 17-3494 6.7018 945.62 7II57-9 302 91204 27543608 17.3781 6.7092 948.76 71631.5 3°3 91809 27818127 17.4069 6.7166 951.90 72106.6 3°4 92416 28094464 I7-4356 6.7240 955-°4 725834 3°5 93° 2 5 28372625 17.4642 6.73*3 958.19 73061.7 306 93636 28652616 17.4929 6.7387 •961.33 7354L5 307 94249 28934443 17-5214 6.7460 964.47 74023.O 308 94864 29218112 17-5499 6.7533 967.61 74506.O 3°9 9548i 29503629 I7-5784 6.7606 97o.75 74990.6 310 96100 29791000 17.6068 6.7679 973-89 75476.8 3 11 96721 30080231 17-6352 6.7752 977.04 75964.5 312 97344 3 37I328 17.6635 6.7824 980.18 76453.8 3*3 97969 30664297 17.6918 6.7897 983-32 769447 3M 98596 30959144 17.7200 6.7969 986.46 77437-1 3^5 99225 31255875 17.7482 6.8041 989.60 77931. 1 316 99856 31554496 17.7764 6.8113 992.74 78426.7 3 X 7 100489 31855013 17.8045 6.8185 995.88 78923.9 318 101124 32157432 17.8^26 6.8256 999.03 79422.6 3 X 9 101761 32461759 17.8606 6.8328 1002.20 79922.9 320 102400 32768000 17.8885 6.8399 1005.30 80424.8 205 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) No. Square Cube Sq. Root Cube Root Circle Circum. Area 321 IO3041 33076161 17.9165 6.8470 IO08.5 80928.2 322 103684 33386248 17.9444 6.8541 IOU.6 81433.2 3 2 3 IO4329 33698267 17.9722 6.8612 IOI4.7 81939.8 3 2 4 104976 34012224 18.0000 6.8683 1017.9 82448.O 3 2 S 105625 3432S125 18.0278 6.8753 1 02I.0 829577 326 106276 34645976 18.0555 6.8824 1024.2 83469.O 3 2 7 106929 349657 8 3 18.0831 6.8894 1027.3 83981.8 328 107584 35287552 18.1108 6.8964 m IO30.4 84496.3 3 2 9 108241 35611289 18.1384 6.9034 ' 1033.6 85012.3 33° 108900 35937000 18.1659 6.9104 1036.7 85529.9 33 1 109561 36264691 18.1934 6.9174 1039.9 86049.O 33 2 IIO224 36594368 18.2209 6.9244 IO43.O 86569.7 333 I 10889 36926037 18.2483 6.9313 1046.2 87092.O 334 III556 37259704 18.2757 6.9382 1049.3 87615.9 335 112225 37595375 18.3030 6.9451 1052.4 88141.3 336 112896 37933 56 18.3303 6.9521 I055.6 88668.3 337 II3569 38272753 18.3576 6.9^89 IO58.7 89196.9 33^ I 14244 38614472 18.3848 6.9658 1 06 1. 9 89727.O 339 114921 38958219 18.4120 6.9727 106^.0 90258.7 340 I I 5600 39304000 18.4391 6.9795 1068. 1 90792.O 34i 116281 39651821 18.4662 6.9864 1071.3 91326.9 342 1 1 6964 40001688 18.4932 6.9932 1074.4 91863.3 343 1 1 7649 40353 6o 7 18.5203 7 .0000 1077.6 92401.3 344 118336 40707584 18.5472 7.0068 1080.7 92940.9 345 II9025 41063625 18.5742 7.0136 1083.8 93482.O 346 119716 41421736 18.6011 7.0203 I087.O 940247 347 120409 41781923 18.6279 7.0271 1090. 1 94569.0 348 121 IO4 42144192 18.6S48 7.0338 I093.3 95114.9 349 I2l8oi 42508549 18.6815 7.0406 1096.4 95662.3 35o I225OO 42875000 18.7083 7.0473 1099.6 9621 1.3 35i I232OI 43243551 I8.7350 7.0540 IIO2.7 96761.8 352 I23904 43614208 18.7617 7.0607 1105,8 97314.O 353 I24609 43986977 18.7883 7.0674 1 109.0 97867.7 354 I25316 44361864 18.8149 7.0740 III2.I 98423.O 355 I20025 44738875 18.8414 7.0807 III5.3 98979.8 356 I26736 45118016 18.8680 7-0873 II18.4 99538.2 357 I27449 45499293 18.8944 7.0940 II2I.5 IOOO98 358 128164 45882712 18.9209 7.1006 II24.7 IO0660 359 I2888l 46268279 18.9473 7.1072 1 127.8 IOI223 360 I2960O 46656000 18.9737 7.1138 II3I.O 101788 206 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) No. Square Cube Sq. Root Cube Root Circle Circum. Area 361 130321 47045881 19.OOOO 7.1204 II34.I IO2354 362 131044 47437928 19.0263 7.1269 H37-3 IO2922 3 6 3 131769 47832147 19.0^26 7-1335 1 140.4 IO3491 364 132496 48228544 19.0788 7.1400 II43.5 IO4062 365 133225 48627125 19.1050 7.1466 1146.7 IO4635 366 133956 49027896 19.1311 7-I53I 1149.8 IO5209 3 6 7 134689 49430863 19.1572 7-I596 H53-0 105785 368 135424 49836032 19-1833 7.1661 1156.1 106362 369 136161 50243409 19.2094 7.1726 1159.2 106941 37° 136900 50653000 I9-2354 7.1791 1 1 62 .4 IO7521 37i 137641 51064811 19.2614 7.1855 H65.5 108103 372 138384 51478848 19.2873 7.1920 1168.7 108687 373 139129 51895117 19.3132 7.1984 1171.8 IO9272 374 139876 52313624 19-3391 7.2048 I175.O IO9858 375 140625 52734375 19.3649 7.2112 II78.I I IO447 376 I4I376 53157376 19.3907 7.2177 Il8l.2 IIIO36 377 142129 53582633 19.4165 7.2240 1184.4 IH628 378 142884 54010152 19.4422 7.2304 1187.5 II222I 379 143641 54439939 19.4679 7.2368 H90.7 II28I5 380 144400 54872000 19.4936 7.2432 1193.8 II34II 381 145161 553 634i 19.5192 7.2495 II96.9 I I 4OO9 382 145924 55742968 19.5448 7.2558 1200. 1 I I4608 3^3 146689 56181887 19.5704 7.2622 1203.2 II5209 384 147456 56623104 19-5959 7.2685 1206.4 I I 581 2 385 148225 57066625 19.6214 7.2 748 1209.5 H6416 386 148996 57512456 19.6469 7.2811 1212.7 II702I 3^7 149769 57960603 19.6723 7.2874 1215.8 1 1 7628 388 I5°544 58411072 19.6977 7.2936 1218.9 H8237 389 1513 21 58863869 19.7231 7.2999 1222. 1 I 18847 390 152100 59319000 19.7484 7.3061 1225.2 I 19459 39i 152881 59776471 19-7737 7.3124 1228.4 120072 392 153664 60236288 19.7990 7.3186 I23L5 I20687 393 154449 60698457 19.8242 7.3248 I234.6 I21304 394 155236 61162984 19.8494 7-33 10 I237.8 I2I922 395 156025 61629875 19.8746 7-337 2 I24O.9 122542 396 156816 62099136 19.8997 7-3434 I244.I I23163 397 157609 62570773 19.9249 7.3496 1247.2 I23786 398 158404 63044792 19.9499 7.3558 I25O.4 I244IO 399 159201 63521199 19.9750 7.3619 I253v5 I25036 400 1 60000 64000000 20.0000 7.3684 I256.6 I25664 207 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES, AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) No. Square Cube Sq. Root Cube Root Circle Circum. Area 401 I 6080 I 64481201 20.0250 7-3742 1259.8 126293 402 161604 64964808 20.0499 7-38o3 1262.9 126923 403 162409 65450827 20.0749 7.3864 1266. 1 127556 404 163216 65939264 20.0998 7.392 5 1269.2 128190 405 164025 66430125 20.1246 7.3986 1272.3 128825 406 164836 66923416 20.1494 74047 1275.5 129462 407 165649 67419143 20.1742 7.4108 1278.6 130100 408 166464 67917312 20.1990 7.4169 1281.8 130741 409 167281 68417929 20.2237 7.4229 1284.9 13*382 410 168100 68921000 20.2485 7.4290 1288. 1 132025 411 I 6892 I 69426531 20.2731 7.4350 1 291.2 132670 412 169744 69934528 20.2978 7.4410 1294.3 I333I7 413 170569 70444997 20.3224 7.4470 l297o 133965 414 171396 70957944 20.3470 74530 1300.6 134614 415 172225 7*473375 2o.37 J 5 74590 I303.8 135265 416 173056 71991296 20.3961 7.4650 1306.9 I359i8 417 173889 725H7I3 20.4206 7.4710 1310.0 136572 418 174724 73034632 20.4450 74770 I3I3-2 137228 419 I7556I 73560059 20.4695 7.4829 i3 l6 -3 137885 420 176400 74088000 20.4939 7.4889 13*9-5 138544 421 177241 74618461 20.5183 7.4948 1322.6 139205 422 178084 75151448 20.5426 7*5007 1325.8 139867 423 178929 75686967 20.5670 7.5o67 1328.9 14053 1 424 179776 76225024 20.5913 7.5126 1332.0 141196 425 180625 76765625 20.6155 7.5185 1335.2 141863 426 181476 77308776 20.6398 7.5 2 44 1338.3 14253* 42 7 182329 77854483 20.6640 7o30 2 1 341-5 143201 428 183184 78402752 20.6882 7o36i 1344.6 143872 429 184041 78953589 20.7123 7.5420 13477 144545 43° 184900 79507000 20.7364 7.5478 1350.9 145220 43i 185761 80062991 20.7605 7-5537 i354.o 145896 43 2 186624 80621568 20.7846 7-5595 1357.2 146574 433 187489 81182737 20.8087 7-5654 1360.3 147254 434 188356 81746504 20.8327 7-5712 1363.5 147934 435 189225 82312875 20.8567 7-577o 1366.6 148617 436 190096 82881856 20.8806 7.5828 1369.7 149301 437 190969 83453453 20.9045 7.5886 1372.9 149987 438 191844 84027672 20.9284 7-5944 1376.0 150674 439 192721 84604519 20.9523 7.6001 1379.2 i5!3 6 3 440 193600 85184000 20.9762 7.6059 1382.3 152053 208 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) ClRrr.v. No. Square Cube Sq. Root Cube Root Circum. Area 441 19448 I 85766121 2 I .OOOO 7.6117 I385-4 J 5 2 745 442 195364 86350888 21.O238 7.6174 1388.6 153439 443 196249 86938307 2I.O476 7.6232 I39I-7 154134 444 197136 87528384 2I.O713 7.6289 1394.9 154830 445 198025 88121125 2I.O950 7.6346 1398.O 155528 446 198916 88716536 2I.H87 7.6403 I40I.2 156228 447 199809 89314623 2I.I424 7.6460 1404.3 156930 448 200704 899I539 2 2I.l66o 7-65I7 1407.4 157633 449 201601 90518849 2I.1896 7.6574 1410.6 *5%337 45° 202500 91125000 21.2132 7.6631 I4I3-7 159043 45i 203401 9I73385I 21.2368 7.6688 I416.9 i5975i 45 2 204304 92345408 2I.2603 7.6744 1420.O 1 60460 453 205209 92959677 2I.2838 7.6801 I423.I 161171 454 206116 93576664 2I.3073 7.6857 1426.3 161883 455 207025 94196375 2I.3307 7.6914 1429.4 162597 45 6 207936 94818816 21.3542 7.6970 1432.6 163313 457 208849 95443993 2I.3776 7.7026 1435-7 164030 458 209764 96071912 2 1 .4OO9 7.7082 1438.9 164748 459 210681 96702579 2I.4243 7-7I38 1442.0 165468 460 2 1 1600 97336000 2I.4476 7-7 J 94 I445- 1 166190 461 212521 97972181 2I.4709 7.7250 1448.3 1 66^ 14 462 2 13444 98611128 21.4942 7.7306 145 1 -4 167639 463 214369 99252847 21.5174 7.7362 1454.6 168365 464 215296 99897344 2I.5407 7.7418 1457-7 169093 465 216225 100544625 2I.5639 7-7473 1460.8 169823 466 217156 101 194696 2I.587O 7-7529 1464.0 170554 467 218089 101847563 2I.6I02 7-7584 1467. 1 171287 468 219024 102503232 21.6333 7-7639 i47o-3 172021 469 219961 103161709 2I.6564 7-7695 1473-4 172757 470 220900 103823000 2I.6795 7-775o 1476.5 173494 47i '22 1 841 104487111 2I.7025 7-7805 1479-7 174234 472 222784 105 1 54048 2I.7256 7.7860 1482.8 174974 473 223729 105823817 2I.7486 7-7915 1486.0 i757i6 474 224676 106496424 2I.7715 7.7970 1489. 1 176460 475 225625 107171875 21-7945 7.8025 1492.3 177205 476 226576 107850176 21.8174 7.8079 1495-4 177952 477 227529 Io8 53i333 21.8403 7-8x34 1498.5 178701 478 228484 109215352 21.8632 7.8188 1501.7 1 7945 1 479 229441 109902239 21.8861 7-8243 1504.8 180203 480 230400 1 10592000 2 1 .9089 7.8297 1508.0 180956 209 TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, CIRCUMFERENCES AND CIRCULAR AREAS OF NOS. FROM i TO 520 (Continued) No. Square Cube Sq. Root Cube Root Circle Circum. Area 481 231361 111284641 21.9317 7.8352 ISII.I 181711 482 232324 111980168 21-9545 7.8406 I5I4.3 182467 483 233289 112678587 21.9773 7.8460 I5I74 183225 484 234256 1 13379904 2 2 .OOOO 7.8514 1520.5 183984 485 235225 114084125 22.0227 7.8568 J523.7 184745 486 236196 114791256 22.0454 7.8622 1526.8 185508 487 237169 H550I303 2 2.o68l 7.8676 1530.0 186272 488 238144 116214272 22.O907 7.8730 I533-I 187038 489 239121 1 1 6930 1 69 22.1133 7.8784 1536.2 187805 490 240100 1 1 7649000 22.1359 7.8837 15394 188574 491 241081 118370771 22.1585 7.8891 1542.5 189345 492 242064 1 19095488 22.l8ll 7.8944 1545-7 190117 493 243049 119823157 22.2036 7.8998 1548.8 190890 494 244036 120553784 22.226l 7.905I I55I-9 191665 495 245025 121287375 22.2486 7.9I05 I 555-i 192442 496 246016 122023936 22.2711 7.9I58 1558.2 193221 497 247009 122763473 22.2935 7.92 1 1 1561.4 194000 498 248004 T2 35 599 2 22.3159 7.9264 1564.5 194782 499 249001 124251499 22.3383 7.9317 1567-7 195565 500 250000 125000000 22.3607 7.9370 1570.8 196350 5oi 251001 125751501 22.383O 7-9423 1573-9 197136 502 252004 126506008 22.4054 7.9476 I577.I 197923 503 253009 127263527 22.4277 7-9528 1580.2 I987I3 5°4 254016 128024064 22.4499 7.958l 1583-4 199504 5o5 255 25 128787625 22.4722 7-9634 1586.5 200296 506 256036 i295542i6 22.4944 7.9686 1589-7 201090 507 257049 130323843 22.5167 7-9739 1592.8 201886 508 258064 131096512 22.5389 7-9791 1595-9 202683 5°9 259081 131872229 22.56lO 7-9843 i599.i 203482 5io 260100 1 3 265 1 000 22.5832 7.9896 1602.2 204282 5ii 261121 13343283 1 22.6053 7.9948 1605.4 205084 512 262144 134217728 22.6274 8.0000 1608.5 205887 5i3 263 1 69 135005697 22.6495 8.0052 1611.6 206692 5i4 264196 135796744 22.6716 8.0104 1614.8 207499 5i5 265225 136590875 22.6936 8.0156 1617.9 208307 5i6 266256 137388096 22.7156 8.0208 1621.1 2091 1 7 5i7 267289 138188413 22.7376 8.0260 1624.2 209928 5i8 268324 1 3899 1 83 2 22.7596 8.0311 1627.3 210741 5i9 269361 139798359 22.7816 8.0363 1630.5 2II556 520 270400 140608000 22.8035 8.0415 1633.6 212372 APPENDIX 211 TABLE IV —FACTORS OF EVAPORATION Temper- ature of Boiler gage pressures in pounds per square inch above the atmosphere feed- water 5 10 15 20 25 30 35 40 45 ° Fahr. 32 1. 187 1 .192 1. 195 1. 199 I. 201 1.204 1 .206 1 .209 1 .211 1. 212 35 1. 184 1. 189 1 . 192 1. 196 I. 198 1.20T 1.203 1 .206 1.208 1 .209 40 I.T79 1. 184 1. 187 1 .191 I. 193 1 .196 1. 198 1 .201 1.203 1.204 45 1. 173 1. 178 1. 181 1. 185 I. I87 1 .190 1 .192 1. 195 1. 197 1. 198 50 1. 168 1. 173 1. 177 1. 180 I. 182 1. 185 1. 187 1 . 190 1. 192 1. 193 55 1. 163 1. 168 1 .171 1. 175 I. 177 1. 180 1. 182 1. 185 1. 187 1. 188 6o 1. 158 1. 163 1. 166 1 . 170 I .172 1. 175 1. 177 1. 180 1. 182 1. 183 65 I -153 1. 158 1 . 161 1. 165 I. 167 1 .170 1. 172 1. 175 1. 177 1. 178 70 1. 148 1. 153 1. 156 1. 160 I .162 1. 165 1 .167 1. 170 1. 172 1. 173 75 1. 143 1. 148 1. 151 1. 155 I. 157 1. 160 1. 162 1. 165 1. 167 1. 168 8o 1. 137 1. 143 1 .146 1. 149 I .151 1. 154 1. 156 '1. 159 1 . 161 1. 162 85 . 1. 132 1. 137 1 . 140 1. 144 I .146 1. 149 1. 151 1. 154 1. 156 1. 157 90 1 .127 1 .132 1. 135 1. 139 I .141 1. 144 1 .146 1. 149 1. 151 1. 152 95 1 . 122 1 .127 1. 130 1. 134 I .136 1. 139 1 .141 1. 144 1 .146 1. 147 100 1 . 117 1 . 122 1 .125 1 .129 I. 131 1. 134 1. 136 1. 139 1 .141 1 .142 105 1 . in 1 . 117 1 .120 1 .123 I .125 1. 128 1 .130 1. 133 1. 135 1. 136 no 1 .106 1 . in 1 .114 1. 118 I .120 1 .123 1. 125 1. 128 1. 130 1. 131 ii5 1 . 101 1 .106 1. 109 1. 113 I. 115 1. 118 1 . 120 r .123 1. 125 1 .126 120 1 .096 1 . 101 1 .104 1. 108 i . no 1. 113 1. 115 1. 118 1 .120 1 .121 125 1 .091 1 .096 1.099 1. 103 1 .105 1. 108 1 . no 1 .113 1. 115 1 . 116 130 1.085 1. 091 1.094 1.097 1.099 1 . 102 1 .104 1 .107 1. 109 1 .110 135 1 .080 1.085 1.088 1 .092 1.094 1.097 1.099 1 .102 1 .104 1. 105 140 1.075 1 .080 1.083 1.087 1 .089 1 .092 1.094 1.097 1.099 1 .100 145 1 .070 1.075 1.078 1 .082 1.084 1.087 1 .089 1.092 1.094 1.095 150 1.065 1 .070 1-073 1.077 1.079 1.082 1 .084 1.087 1.089 1 .090 155 1.059 1.065 1.068 1 .071 1.073 1.076 1.078 1. 081 1.083 1 .084 l60 1.054 1.059 1 .062 1 .066 1.068 1 .071 1.073 1 .076 1.078 1.079 165 1.049 1.054 1.057 1. 061 1.063 1.066 1.068 1 .071 1.073 1.074 170 1.044 1.049 1.052 1.056 1.058 1. 061 1.063 1 .066 1.068 1 .069 175 1.039 1.044 1.047 1. 051 1.053 1.056 1.058 1 .061 1.063 1 .064 180 1.033 1.039 1 .042 1.045 1.047 1.050 1.052 1.055 1.057 1.058 185 1.028 1.033 1.036 1 .040 1.042 1.045 1.047 1.050 1.052 1.053 190 1.023 1.028 1. 03 1 1.035 1.037 1.040 1 .042 1.045 1.047 1 .048 195 1. 018 1.023 1.025 1.030 1.032 1.035 1.037 1 .040 1 .042 1.043 200 1 .013 1. 018 1. 021 1.025 1 .027 1.030 1 .032 1.03s 1.037 1.038 205 1 .008 1. 013 1. 015 1.020 1.022 1.025 1.027 1.030 1.032 1.033 210 1.008 1.008 I .Oil I. 015 1. 017 1 .020 1.022 1.025 1.027 1.028 212 1.002 1.002 15 212 ARITHMETIC OF THE STEAM BOILER TABLE IV— FACTORS OF EVAPORATION {Continued) Temper- ature of Boiler gage pressures in pounds per square inch above the atmosphere feed- water 50 60 70 80 90 100 120 140 160 ] 80 ° Fahr. 32 1. 214 1 .217 1 .219 1 .222 1 . 224 1 . 227 1 .231 1 .234 1.237 1 239 35 1 .211 1 .214 1 .216 1 .219 1 .221 1 .224 1 .228 1 .231 1.234 1 236 40 1 . 206 1 . 209 1 . 211 1 . 214 1. 216 1. 219 1. 2 23 1. 226 1 .229 1 231 45 1 .200 1 .203 1 .205 1 .208 1 .210 1 . 213 1 .217 1 .220 1.223 1 225 50 1. 195 1. 198 1 .200 1 .203 1 . 205 1 . 208 1 .212 1 . 215 1. 218 1 220 55 1 . 190 1. 193 1. 195 1. 198 1.200 1.203 1.207 1. 210 1. 213 1 215 60 1. 185 1 . 188 1 . 190 1 . 193 1. 195 1 . 198 1 .202 1 .205 1.208 1 210 65 1. 180 1. 183 1. 185 T.188 1 . 190 1 . 193 1 . 197 1 .200 1 .203 1 205 70 1. 175 1 . 178 1 . 180 1 . 183 1. 185 1 . 188 r . 192 1 . 195 1. 198 1 200 75 1 . 170 1. 173 1. 175 1. 178 1. 180 1 . 183 1 . 187 1 . 190 1. 193 1 195 80 1 . 164 1 . 167 1 . 169 1 . 172 1. 174 1. 177 1 . 181 1 . 184 1. 187 1 189 85 1 -159 1 . 162 1 . 164 1 . 167 1 . 169 1 . 172 1 . 176 1 . 179 1. 182 1 184 90 1 -154 1. 157 1 . 159 1 . 162 1 . 164 1 . 167 1 . 171 1 • 174 1. 177 1 179 95 1. 149 1 • 152 1. 154 1. 157 1. 159 1 . 162 1 . 166 1 . 169 1 . 172 1 174 100 1. 144 1. 147 1 . 149 1 . 152 1. 154 1. 157 1 . 161 1 . 164 1. 167 1 169 105 1. 138 1 . 141 1 . 143 1 . 146 1. 148 1.151 1. 155 1. 158 1. 161 1 163 no 1. 133 1. 136 1 . 138 1 . 141 1. 143 1 .146 1. 150 1. 153 1. 156 1 158 115 1. 128 I. 131 1. 133 1-136 1. 138 1 . 141 1. 145 1. 148 1. 151 1 153 120 1. 123 1 . 126 1 . 128 1 . 131 1. 133 1. 136 1. 140 1. 143 1 . 146 1 148 125 1. 118 1 . 121 1 . 123 1 . 126 1. 128 1. 131 1. 135 1. 138 1. 141 1 143 130 1 . 112 1. US 1 . 117; 1 . 120 1 . 122 1. 125 1 .129 1. 132 1. 135 1 137 135 1 . 107 1 . no I . 112 I . 115 1. 117 1 . 120 1 .124 1 . 127 1. 130 1 132 140 1 . 102 1 .105 i . 107 i . no 1 . 112 1. US 1 . 119 1 . 122 1. 125 I 127 145 1.097 1 . 100 1 . 102 1. 105 1. 107 1 .110 1 . 114 1 .117 I . 120 I 122 150 1 .092 1.095 1.097 1 . 100 1 . 102 1. 105 1 .109 1 .112 1.1.31 117 155 1.086 1 .089 1 .091 1.094 1 .096 1.099 1 . 103 1 . 106 I . 109 I in 160 1. 081 1.084 1.086 1 .089 1 .091 1.094 1 .098 1 . 101 I . 104 I 106 165 1 .076 1.079 1. 081 1 .084 1.086 1 .089 1 .093 1 -096 1.099 I ior 170 1 .071 1.074 1 .076 1.079 1. 081 1 .084 1 .088 1 .091 1.094 1 096 175 1.066 r .069 1 .071 1.074 r .076 1.079 1.083 1.086 I.O89 I 091 180 1 .060 1.063 1.065 1.068 1 .070 1.073 1.077 1 .080 I.O83 I 085 185 1.055 1.058 1 .060 1 .063 1.065 1.068 1 .072 1.075 1.078 I 080 190 1.050 1.053 1.055 1.058 i.o6oji .063 r .067 1 .070 1.073 I 075 195 1.045 1.048 1.050 1-053 1.055 1-058 1 .062 1.065 1.068 I 070 200 1 .040 1.043 1.045 1.048 1.050 1.053 1057 1 .060 I.063 I 065 205 1.035 1.038 1 .040 1.043 1.045 1.048 1.052 1.055 1.058 I 060 210 1 .030 1033 1 .035 1 .038 r . 040 1 . 043 1 . 047 1 . 050 1-053 1 055 212 APPENDIX 213 How to Interpolate the Table of Factors of Evaporation It sometimes happens when it is desired to use the table of factors of evaporation that the given figure for any case falls be- tween two certain figures in the table, and therefore the correct result cannot at once be found without resorting to what is called " interpolation." Suppose, for example, that the average steam pressure in the boiler is 64 lb. per square inch gage, and that the average tempera- ture of the feed water is 13 2 F.; what is the factor of evaporation? By referring to the table, there are no columns with heading or side heading corresponding to these figures, and unless there is some definite method of obtaining exact figures, it would be necessary to strike an average between two sets of figures in the table, nearest to those given in the example. While for ordinary purposes this would be close enough, yet because of the ease with which the real figures may be found it is worth while to learn what to do. The factor for 60 lb. gage and 130 feed water is 1.115; the factor for 70 lb. gage and 130 feed water is 1.118; the factor for 64 lb. gage amd 130 feed water is therefore, 1. 118 — 1. 115 1. 115+ X4= 1.1162 In the same manner, the factor of evaporation for 64 lb. gage pressure and 140 feed water is found to be , 1. 107 — LICK 1.105 + '— ^X4= 1. 1058 There is now the factor for 64 lb. gage and 130 feed water, and 64 lb. gage and 140 feed water, and it only remains to inter- polate between these values to get the factor for 64 lb. gage and 13 2 feed water. This is done as follows: 1.1162 — 1.1058 1.1162 X 2 = 1.1141 10 which is the factor of evaporation corresponding to 64 lb. gage pressure and 13 2 F. feed water, as given in the example. The foregoing method may be applied to any figures within the range of the table. 214 ARITHMETIC OF THE STEAM BOILER m w cq P H rr Ph d W J fl O F3 pq ;..; U c^ '■j < P ■a Z - < M H M-l c/> o | o > & d a H j pq < H Ui c3 £1 X Tf Tt CO CO CM ri O o O X vO X CM X X 2 « X i> m m 0\ CO H IC o Ov PC X 0! X CM vO M M CM n 0) CO «* Tf m lO vO vO X Ov M rj- 43 'tOO M r^ ro tc cm vO 0\ M M M CO C O CO H/ 4 tJ- oo co VO lO vO X CO 00 ^ X O X Eh vO CM X h g & 00 M M M cs tt M ~~- CO -t 4 IO LT, vO I- Ov CM 1$ a "8 o si tJ- ro N |_| a lO r^ o vC CM X o X CM Ov CJ o H C/2 ^ +j sur- foot e , cm i> X 00 "* a r» CO 00 X CO On t» X M Tf i— i M I> Tf o CO o vO CM X I/-. 1-- X X U 1 C cm cm X -^ Tf io lO vO i- *> X o O *~[ CM "?f rea of se per of tut M M M ' CM t^ CM X CO On Tf O LO M vO 01 r- X Ov w O cm O U1 (N ^r LO ^i X IO y -T r^ O l> -s 5 cr d cm x x H/ LC LO O t> l.- X Ol Oi C H X X < «£ w - w 1 x cm oo M P0 CO o LO o\ o r- vO Ov o Ov O co 0) ^ H 00 't r- ro o PO ^t t^ X vO r- x o 4 d X 00 CO a LT. co c o o H PO *o Ov c X CM CD W a « W »*H H M cq co n- VO vO 1^ y, a O -t f- X > n3 w w r^ — 0\ o a 01 \C O X o vO CM H W '- 1 H H M r^ CO ^1- 1/J i^ X o\ M 01 X Ov 00 H H H M CM « J lfl Ifl Ifl in lil »o a a 0\ o o "fr H/ 00 lO Ov CN a 0\ C^ o Ol CM CM CO X t vO *tf «g o o o o c c ■g.9 C/3 +3 H >- o 00 CO 00 vC CO X o t- £ rt d 00 O CO LT; 00 M LT. t- CM LO t> 01 t^ vO In