FIVE YEARS QUESTIONS AND ANSWERS NATIONAL ASSOCIATION OF STATIONARY ENGINEERS NATIONAL ASSOCIATION/ STATIONARY ENGINEERS Five Years Questions and Answers As originally published ^^^^^^^^ in ^^^^~ The National Engineer VOLUMES ONE TO FIVE INCLUSIVE Marsh & Grant Co Printers and Engravers Chicago Copyright lyoz by GEORGE D. B. V AN TASSEL, Sec'y P R E F A C E / T A HE Questions and Answers found herein were first published in the National Engineer in a competitive educational course, designed to stimulate technical work and theory study among the members of the National Association of Stationary Engineers. This volume becomes the fourth bound edition of the same, which represents the work of separate committees acting through a period of five years. These committees were as follows : F. W. ENLOY . . OTTO LUHR . . CHAS. W. NAYLOR E. J. STODDARD E. G. JACQUES E. P. GlLROY CHAS. H. Fox . ARTHUR O. HALL S. J. GRAIN . . M. M. CHILDS . THOS. P. BURKE . HY. C. HOFFMAN M. M. CHILDS . GEO. F. HAVEN . 1896-1897 . Illinois No. 29, Chicago. . Illinois No. 38, Chicago. . Illinois No. 28, Chicago. 1897-1898 . Michigan No. i, Detroit. . Michigan No. i, Detroit. . Michigan No. I, Detroit. 1898-1899 . Ohio No. 36, Cincinnati. . Ohio No. 15, Cincinnati. . Ohio No. 2, Cincinnati. 1899-1900 . Rhode Island No. I, Providence. . Rhode Island No. 2, Pawtucket. . Rhode Island No. 5, Providence. 1900- 1901 . Rhode Island No. I, Providence. . . New York No. 48, Brooklyn. The successful associations and individuals for the several years were : For 1896-7: Ohio No. 36, Cincinnati; Iowa No. 8, Sioux City, and Illinois No. 22, Rockford. For 1897-8: Ohio No. 15, Cincinnati; Illinois No. 22, Rockford, and Iowa No. 8, Sioux City. For 1898-9: Iowa No. 8, Sioux City; Louisiana No. i, New Orleans, and Massachusetts No. 17, Lowell; and also J. S. Gillespie, Pennsylvania No. 12, Philadelphia, and H. H. Carman, Ohio, No. 28, Akron. For 1899-1900: New York No. 48, Brooklyn; Massachu- setts No. 17, Lowell, and Ohio No. 45, Canton. Also for elemen- tary work, New York No. 48, Brooklyn; Ohio No. 37, Dayton, and Michigan, No. 24, Detroit. For 1900-01: Massachusetts No. 17, Lowell, and Massa- chusetts No. 1 6, Waltham. Volumes I, II and III, containing the problems for the years 1896-7, 1897-8 and 1898-9, were edited by Chas. Desmond, at that time editor of the National Engineer. This volume is presented by the undersigned committee in the role of editors, who ask your kind indulgence for their effort, expressing their thanks to the several past committees for assistance in handling the proofs, and hopeful that the book may attain its object, which is to be of service as a guide and reference to the practical student-engineer. The sequence of the questions and answers by years has not been respected in the present arrangement, which attempts to clas- sify the several problems under their logical headings. The result has not, however, been entirely satisfactory because the questions were originally presented without any particular thought of their ever becoming a part of an harmonious whole. A large demand for the book will, if it comes, assure the committee that its work has been fairly well done. CHAS. W. NAYLOR Illinois No. 28, Chicago. F. ELMO SIMPSON Illinois No. jj, Chicago. GEORGE NOWARD Illinois No. 4.0, Chicago. N ATIONAL ASSOCIATION OF STATIONARY ENGINEERS QUESTIONS w ANSWERS 1896-1901 Boilers, Furnaces, Fuel, Combustion, Chimneys, Etc. Q. 1. (1896-7.) What causes worst form of external corrosion of steam boilers? Ans. 1. External corrosion is frequently formed by exposure of the shell to the cold air. Is more likely to occur when cold than when under pressure and moisture condensing on it when cold aids in forma- tion of rust. It is likely to occur along line of brick work in externally fired boilers. Leaky joints are another source of corrosion, either from riveted seams, man or hand holes, or from imperfectly fitted attachments. Certain coals are rich in sulphur, the products of whose combustion contain sulphurous and sulphuric acid this has a bad rusting effect on boilers. [Wet ashes or soot, when permitted to remain in contact with the boiler plate, cause corrosion.] Q. 2. (1896-7.) Give safe strain for braces in steam boilers. Ans. 2. For iron 6,000 Ibs., for steel 7,000 Ibs. per square inch cross-section. Q. 3. (1896-7.) Does heat or pressure cause greatest strain in boil- ers? Ans. 3. Heat, on account of unequal expansion. Q. 4. (1896-7.) How to construct a tubular boiler to insure free circulation? Ans. 4. To insure free circulation in horizontal tubular boilers of the ordinary type the tubes must not be staggered but disposed in straight vertical lines. The spacings between tubes should be ample and a clear space of 4" to 6" is proper between the outer row of tubes and the sides of the boiler shell, together with a space from 14" to 16" at the bottom. The upper row of tubes to be well covered when the water line is carried at a point a little above 2/3 of the shell diameter. Some designers prefer omitting the center row of tubes or modifying the space to allow more freedom of the circulating currents at that point. Q. 6. (1896-7.) Name the cheapest and most commonly used scale preventer. Ans. 6. Taking the country as a whole, sal-soda is the cheapest and most commonly used. The Eastern section is, however, using kerosene oil to a great extent. Q. 7. (1896-7.) For best results in tubular boiler with natural draft, give ratio between grate and heating surface. Ans. 7. This would depend on the amount of draft and the kind of coal used. Slack coal requires a larger grate surface than lump coal. Also when the draft is strong the grate area does not need to be so great as when draft is poor. All chemists appear to agree that the perfect combustion of one pound of coal gives exactly the same number of heat units, as the perfect combustion of another pound of the same coal. The question would then appear to resolve itself into the question, whether we can get nearer the perfect combustion with strong draft and with small grate surface, or with a moderate draft and large grate sur- face. Ratios vary from 30:1 to 60:1. Probably 45:1 is the best ratio. Q. 11. (1896-7.) What is pitting? How caused? Ans. 11. Pitting is conical or spherical depressions which are filled with a yellowish brown deposit, consisting mainly of peroxide of iron. Pitting is caused by the action of oxygen which has been held in solu- tion by the water, its action is hastened by the presence of carbonic acid gas which is liberated when the temperature of the water is increased. Boilers that are kept lukewarm, and in which the circulation is poor, are most likely to be affected by pitting. Q. 14. (1896-7.) Is a "mysterious gas" formed in boilers at times of explosion? Ans. 14. Its existence has not yet been proved, although some be- lieve there is such a thing. Q. 17. (1896-7.) Give relative values of solid boiler plate, double- riveted and single-riveted joints; who established them? Ans. 17. Solid plate 100 %, double riveted 70 %, single riveted ra Wm ' Fairbairn - [ These values are not often found in Q. 20. (1896-7.) What advantages do water tube boilers possess over other types? b6 boilers nave the followin S advantages over 1. Thin heating surface in boilers. 2. Joints removed from the fire. 3. Complete combustion. 4. Large draft area. 5. Thorough absorption of heat. 6. Efficient circulation of water. 7. Quick steaming. 8. Freedom of expansion. 9. Safety from explosions. 10. Ease of transportation. 11. Ease with which they can be repaired. [Some of these statements are questionable.] Q. 21. (1896-7.) If a boiler evaporates 3,000 Ibs. of water per hour, what should be size of safety valve? (Pop or lever.) Ans. 21. Rankine says the constant .006 X the water evaporated per hour = the square inch area of the required safety valve. Then, .006 X 3,000 = 18" area and 4.8" for the diameter of valve needed. The Commission of United States Supervisors say the constant shall be .005. Then, .005 X 3,000 = 15" area or 4.375" diameter of valve. Taking it that one square foot grate surface will burn 12 Ibs. of coal per hour and that each pound of coal will evaporate 8% Ibs. of water, then 12X8% = 102 Ibs. of water will be evaporated per hour per square foot grate surface and 3,000 -f- 102 = 29.4 sq. ft. grate surface required. With the ratio of one square inch of safety valve opening (pop valve) to every 3 square feet of grate surface we have 29.4^ 3 = 9.8 sq. in. required in valve; or a diameter of 3.53". With a ratio of one square inch of valve opening (for lever valve) to each 2 sq. ft. of grate surface we have 29.4 -j- 2 = 14.7 sq. in. of area in valve or a diameter of 4.32". If the boiler pressure had been given the calculation would be made on the basis that the number of pounds of steam that will flow through an orifice ot one square inch area in one second may be found by divid- ing the absolute pressure by 70. The following formula is also some- times used: % water evaporated per hr. Area valve Steam pressure plus 10. Working on this formula with assumed pressures as below: 1500 50 Ibs. B. P. = = 25 sq. in. area = 5.65" diameter. 50+10 1500 100 Ibs. B. P. = = 13.6 sq. in. area = 5.17" diameter. 100+10 1500 150 Ibs. B. P. = = 9.37 sq. in area = 3.46" diameter. 150+10 Q. 22. (1896-7.) Should horizontal externally fired boilers be set ievel or with a pitch? If with a pitch, which end should be the lower, and why? Ans. 22. First: with a pitch. Second: back end. Third: to facili- tate the draining of the boiler and the blowing out of mud, also have more heating 1 surface directly exposed to fire, and as a safeguard to the water column by keeping more water at the back than is indicated in the front end of boiler. Q. 23. (1896-7.) Name in order of general merit the four most com- monly used boiler metals. Ans. 23. Steel, wrought iron, copper, cast iron. Q 25 (1896-7.) What should he the temperature of gases in an uptake for the best economy, with natural draft and a gage pressure of 60 Ibs.? Ans 25 Theoretically same as temperature of the steam; or 307 P.; in practice from 50 to 75 or sometimes 100 greater than this, influenced somewhat by draft. Q. 29. (1896-7.) How many tons of air are needed to burn one ton of average coal? \ns 29 Theoretically about 12 tons, but in practice 18 to 24 tons of air. (One pound of coal requires 12 to 24 Ibs. air to burn it.) Q. 32. (1896-7.) Can smoke once formed be burned. Ans 32 Yes and no. As a chemical process it can be done, but practically, in a boiler furnace, we think not. [Carbon from a hydro- carbon is the sole source of smoke; heated to 800 or upwards, it will combine with the oxygen of the air, if present in sufficient quantity, and smoke will be prevented in the furnace or consumed if such condi- tions can be obtained in the combustion chamber of a boiler setting.] Q. 40. (1896-7.) What should be the pitch of rivets to give the best possible percentage of strength, in a staggered double-riveted lap joint; thickness of plate, 5/16"; diameter of rivet, 11/16" tensile strength of plate, 55,000 Ibs.; shearing strength of rivet, 38,000 Ibs.? Ans. 40. In a riveted joint of the character indicated the strain transmitted by the load carried on that part of the structure repre- sented by a distance equal to the "pitch" is resisted by the shearing strength of two rivets. The cross section of a single rivet 11-16" diameter is .3712 square inches; hence the strength of the joint, as far as the riveting is con- cerned, is found by doubling the area and multiplying by the assumed shearing strength or value of the metal per square inch of section, viz. : 2 X .3712 = .7424 X 38,000 = 28,211.2. The best possible percentage of strength in any riveted joint is realized when the resistance of the unimpaired plate section between the rivets is equal to the strength of the rivets. To obtain this result the centers of the rivets must be proportioned so that the thickness of the metal, taken in inches, its tensile strength per square inch and the length of the effective section remaining between the rivets, when taken together as factors, produce a product representing a sectional value for the plate equivalent to that found for the rivets. The decimal equivalent of 5-16", the thickness given, is .3125, which, multiplied by 55,000, its tensile strength, gives 17,187.5. The value of the third factor is unknown; therefore, considering the conditions in hand in the form of an equation, we have: The resistance of the rivets, as previously found, 28,211.2, giving 28,211.2 = 17,187.5 X D, D representing the unknown factor, from which we deduce by transposing: D = 28,211.2 -=- 17,187.5 = 1.641 inches. Since D, or 1.641, represents the distance between rivet holes, the actual pitch is one diameter of the rivet more, or 1.641 + .687, or 2.328 inches. Q. 41. (1896-7.) If a boiler evaporates 5,000 Ibs. water per hour from and at 212, how many HP. will it be rated? Ans. 41. Horse-power is strictly a measure of work; the term is conventionally used to express the capacity of boilers. Since the actual power that may be developed by any given volume of steam depends upon conditions foreign to the generator producing it, it is essential that we assume a "base" or arbitrary standard in connection with the expression by which the relative capacities of dif- ferent boilers may be intelligently known, without reference to the duty actually performed thereby. An evaporation equivalent to 30 Ibs. of water per hour taken from a feed temperature of 100 F. into steam at 70 Ibs. gage pressure, as fixed by a board of experts during the Centennial Exposition in 1876, is generally accepted as the unit expressing the capacity of steam boilers in horse-power. The conversion of one pound of water into steam, under the con- ditions specified, is effected by the absorption of 1,110 thermal units of heat, which is equivalent to the evaporation of 1110 -^ 966 = 1.149 Ibs. at a temperature of 212 as required. Using 1.149 as a factor and multiplying the same by 30 gives 34.47, which represents the weight of water that may be changed to steam, from and at 212, by the same number of heat units required to effect the evaporation of 30 Ibs. of water at 100 into steam at 70 Ibs. gage pressure. The quantity 34.37 is usually assumed at 34.5 Ibs., hence on the basis we have noted the direct answer to the question may be found by dividing the total amount evaporated by the boiler by the number which represents the evaporation required for one horse-power, giving: 5000-^-34.5 = 144.92 HP. [At the meeting of the A.S.M.B. a boiler HP. was defined as 34.5 Ibs. water evaporated per hour into steam, from and at 212, equivalent to the transfer of 34.5 X 965.7 = 33,317 B.T.U. per hour. See part of Ans. 65 (1896-7).] Q. 42. (1896-7.) Where should the feed water enter the boiler and how be distributed, in the best practice, and for the best results? Ans. 42. A prominent authority recommends the introduction of the feed pipe at the front head, just above the upper row of tubes, and ex- tending along the side of the boiler nearly to the back head. It then crosses to the opposite side of the shell, and, turning downward, dis- charges between the shell and the tubes. [If pipe is too large it will fill with scale until force of water is sufficient to prevent the further formation of scale.] There are strong arguments in favor of introducing the feed water into the steam space in the form of a spray, this preventing in a great measure the formation of a hard scale in the lower parts of the boiler. In any case the comparatively cool feed water should not be allowed to come into contact with the hot part of the shell that is over the fire. It is calculated that if a plate be cooled 200 a longitudinal strain of 8,000 Ibs. to 10,000 Ibs. per square inch is produced. This, in addi- tion to the normal strain of the steam pressure, may tax the seams beyond their elastic limit. By the same authority it is given out that girth seams develop leaks and cracks in 99 cases out of every 100 in which the feed dis- charges directly upon the fire sheets. Feeding througn the blow-off or the mud-drum is not considered good practice. Q. 43. (1896-7.) What part or parts of a horizontal externally fired boiler are subjected to greatest strains in working? Ans. 43. The furnace sheets and seams over or near the fire or bridge wall and the back head or tube sheet, brought about through unequal expansion or contraction. Frequent and unnecessary opening of flue doors will effect front head and flooding the boiler with cold feed water may start local strains. Q. 46. (1896-7.) Sketch to a scale of % size, the head of a 48" tubular boiler, showing the thirty 4" tubes, and a 4" X 6" handhole properly placed. Ans. 46. See cut on following page. Q. 47. (1896-7.) If one Ib. of coal will evaporate 10 Ibs. water from and at 212 F., how many pounds will it evaporate 'from 80 into steam at 310? Ans. 47. The total heat of steam is, at all temperatures, separable into two parts latent and sensible heat. The sensible heat is that indicated by the thermometer, and it varies as its pressure. The latent heat absorbed in converting water into steam is by far the greater portion of the total heat. Thus, steam at a temperature of 212 F. has a total heat of 1178.6 units, so that in evaporating one pound of water at 212 into steam of 212 we have added 1178.6 212 = 966.6 units; to evaporate 10 Ibs. we have 10 X 966.6 = 9666 units. The total heat of steam at 310 is 1208.5, from which we subtract the temperature of the water, and the remainder will be the units of heat required to evaporate one pound of water from 80 into steam at 310. 1208.580 = 1128.5, and as we have 9666 units to apply, then 9666 -f- 1128.5 = 8.56 + pounds of water. Q. 50. (1896-7.) A ten-pound piece of iron is left in the gases of a chimney until thoroughly heated and then inserted in 100 Ibs. of water at 50 F. The resulting temperature of the water and submerged iron is 55. Required: The temperature of the chimney gases. Ans. 50. One hundred pounds of water has been raised 5 in tem- perature; this will require 500 units of heat, which must be taken from the ten Ibs. of iron in reducing its temperature from that of the chimney gases to 55 F. The specific heat of wrought iron is .1138, which is the amount of heat it will be necessary to put into 1 Ib. of iron to raise its temperature 1, or the amount of heat that lib. of iron will give up in falling 1 in temperature. Therefore, 10 Ibs. of iron v/ill give up 1.138 heat units for each degree drop in temperature, then 500 -f- 1.138 = 439 fall of temperature, and 439 + 55 = 494 F. as the temperature of the chimney gases in the supposed case. Q. 63. (1896-7.) Name five things or conditions upon which the efficiency of boiler heating surfaces depend. Ans. 63. 1. Difference in temperature between two sides of the plate L. Thickness of plate. 3. Thermal conductivity of plate. 4. Cleanliness of surfaces inside and out. Hn t Ci . ula tion of water in boiler. [Position of the surface in rela- surface ] S UrCe f ^^ Angle at WMch the heat rays Strike the setifn,t Sh uld always be P ut n the inside of the fire causing the LtJhit 6 ff^J 6 " makes a place for sediment to gather! mg the patch itself to be injured by the action of the fire. Interna SECTION OF BOILER HEAD SHOWING ARRANGEMENT OF TUBES AND BRACES pressure will hold the patch in place and not have a tendency to blow It off the boiler, as would be the case were it external. The calking area is reduced by putting the patch on the inside In case the patch comes at the edge of the plate it will be necessary to let it come on the outside of the adjacent plate on account of the difficulty in making a tight joint if it was on the inside of both plates. Q 65 (1896-7.) Which is the more accurate way of measuring the HP of a boiler, by amount of water it will evaporate in a given time, or the number of sq. ft. of heating surface it contains? Give reasons. Ans 65 In measuring the horse-power of a boiler it is advisable to assume a set of practically attainable results in average good practice, and to take the power so obtainable as a measure of the power of the boiler in commercial and engineering transactions. The unit generally assumed has been the weight of steam demanded per horse-power per hour by a fairly good engine. The magnitude has gradually been decreasing since the early times of the steam engine. in the time of Watt, one cubic foot of water per hour was thought fair; at the middle of the present century, 10 Ibs. of coal was a good figure, and 5 Ibs., commonly equivalent to about 40 Ibs. evaporation, was allowed for the engine. After the introduction of the modern form of the engine the last figure was reduced 25%, and the most recent developments have still lowered this consumption of fuel and steam. By general consent the unit has now become 30 Ibs. of dry steam per hour per horse-power, which represents the performance of a good non- condensing engine. Large engines, with condensers and compounded cylinders, will do better. A committee of the A.S.M.E. recommended 30 Ibs. of water as a unit of boiler power, and this is now generally accepted. They advised that the commercial rating be taken as an evaporation of 30 Ibs. of water per hour from a feed temperature of 100 F. into steam of 70 Ibs. gage pressure, which may be considered to be equal to 34.5 units' of evapora- tion; that is 34.5 Ibs. of water from a feed temperature of 212 into steam at same temperature. This standard is equal to 33,305 B.T.U. per hour. A boiler may have a large heating surface and not be as efficient as one with less. Some parts of heating surfaces are more effective than others, and the horse-power per square foot of heating surface will vary largely. Therefore we think the proper way of rating a boiler is by the amount of water it will evaporate in a given time. Square feet of heating surface is no criterion by which to judge different styles of boilers, but when an average rate of evaporation per square foot for any boiler has been fixed upon by experiment this becomes a more convenient way of rating the power of other boilers of the same style. Q. 69. (1896-7.) Is scale, or deposit, in any quantity, a good thing in a steam boiler? Why? Ans. 69. A light film of scale is believed to be desirable in a steam loiler. It is well known that a free contract of pure water is destruc- tive to boiler plate. The same may be said of the acids and other :hemicals commonly introduced with the feed and liberated by the LCtion of heat. The action of such corrosive influences is believed to be JSSST^K ch ^ c ^ ed b y the interposition of a thin film of scale. Some : that a light scale prevents, in a measure, the leakage that would occur at seams of a poorly built boiler. Q. 71. (1896-7.) If a boiler is to carry a working pressure of 105 Ibs. with double riveted longitudinal seams of 70 per cent strength, and using a factor of safety of 5, having 60,000 Ibs. tensile strength of plate, with inside diameter of shell 60", what should be the thickness of the plates? Ans. 71. To find the thickness of boiler plates we are given the following formula: TX SX .70 = PXRXF. When T = thickness of plate in inches. S = tensile strength of plate. .70 = percentage of strength of joint. p = pressure in pounds per square inch. R = radius of boiler shell in inches. F = factor of safety. Transposing above formula we have: P X R X F 105 X 30 X 5 15750 T = = = =.375 = %" S X .70 60000 X .70 4200 or thickness of plate required. Q. 84. (1896-7.) How many feet of rod needed to put three guys on a smoke stack, if fastened 55 ft. above the ground and inclined at an angle of 45? Ans'. 84. At a height of 55 ft. above a base line a line is drawn to an angle of 45 from the vertical line and a line drawn from this point at this height at this angle would pass through a point 55 ft. on the base line from its intersection with the vertical. We have a right angle triangle with 2 equal sides. To find the hypothenuse. This equals the square root of the sum of the squares of the two sides or the V55 2 + 55 2 3= V3025 + 3025 = V6050 =77.78 +ft, which is the hypothenuse or slant side. There are three guys, then 3 X 77.78 = 233.34 ft. = the length of the guy rod needed, making no allowance for eyes, links, or fasteners. Q. 89. (1896-7.) Is the use of the steam jet as an auxiliary to fur- nace combustion economical? Ans. 89. A committee of experts in St. Louis in 1891, after 39 tests by various methods, report that in one case fuel consumption was in- creased 12% for the same work. Prof. Landreth then said "Steam jets to draw air in or inject air into the furnace above the grate and also to mix the air and combustible gases together form an efficient smoke preventer, but one liable to be wasteful of fuel." From above and also from our own experience we incline to the opinion that, generally speaking, in ordinary practice, a jet of steam above the grates will not increase the efficiency of a boiler as a steam generator. Q. 90. (1896-7.) Given a lever safety valve: Weight of ball 90 Ibs., distance from center of weight to fulcrum 40.41", weight of lever and valve 11% Ibs., distance from fulcrum to center of gravity of lever 18", from fulcrum to center of valve 4", and diameter of valve 3%"; at what pressure will it blow off? Explain. Ans. 90. The length of lever being 40.41" and multiplying this by 90 Ibs., or the weight of the ball, will give 3639.9, or the moment of force or leverage of the ball; to this add the weight of the valve-stem and valve, or the weight of the valve-stem, lever and valve, by its ful- crum distance, or 11% X 18=211.5, this sum +3639.9=3848.4, gives the to- tal moment tending to hold the valve down, against which we have a cer- tain pressure trying to force valve up. As the valve is 3V 2 diam t. its area is 9621 sq. in , which multiplied by 4", its leverage, gives 38.484 as its leverage moment. This divided into the total downward force 3848 4 gives 100 Ibs. as the pressure needed to balance the valve. Prac- tically a little more would be needed to make the valve blow off. Q 93 (1896-7.) A tubular boiler has an inside diameter of 60", and is to carry a working pressure of 105 Ibs. by the gage; longitudinal seams are double-riveted, 70% strength of joints, plates %" thick, factor of safety being 5, what should be the tensile strength of plates? How found? Ans. 93. Formula: R X P X 5 _ T X S Where R = radius in inches. P = pressure in Ibs. per sq. in. 5 = factor of safety. T = thickness of sheets in inches. S = strength (efficiency) of joint. ts = tensile strength. 30 X. 105X5 15750 .-. = =60000 ts. .375X70% .26250 NOTE: These lectures to be used in explanation of Questions 1 and 20 inclusive of 1897-8 series. BOILER AND FURNACE. The efficiency of the furnace and boiler is always directly dependent upon the care, skill and knowledge of the engineer. If he properly handles the coal and regulates: the supply of air, the coal produces the maximum amount of heat of which it is capable and the boiler absorbs the greater part of the heat produced and uses it in the production of steam ; if he does not properly regulate the supply of coal and air, either the maximum amount of heat is not produced, or, if produced, the greater part is carried up the chimney with the gases and is radiated, or both these sources of loss occur at the same time. A good quality of coal should produce about 14,000 heat units per pound, when properly burned, and may generate as little as 3,500, when improperly burned (Ans. 1 and 2). Of this heat the boiler should absorb from 60 to 80 per cent, and use it for making steam. (Ans. 10.) Suppose one had 50 pounds of gases passing up the chimney per pound of coal burned, and these gases were at a temperature of 500 above the temperature of the engine room. Then, as raising one pound of the gases one degree takes .24 of a heat unit, raising it 500 would take .24 X 500, or 120 heat units. Raising 50 Ibs. to that temperature would take 50 X 120, or 6,000 heat units. Thus he is sending 6,000 of his 14,000 available heat units, or 43 %, up the chimney, and he could not get more than about 50 % into the boiler, as about 7 % would be radiated and lost in other minor ways. Twenty-five pounds of chimney gases per pound of coal is sufficient (practical trials with specially skillful firemen generally show MS) and the average temperature of the chimney gases above the out- side air in the 137 tests recorded by Mr. Geo. H. Barrus is not over JO . Taking these values, only about 17 % of the heat would be car- ried up the chimney in this way. An understanding of the operation of the furnace and boiler requires that we should know the following facts: 14 (1) The heat value of the fuel. (2) The proportion of this heat used in making steam. (a) By incomplete combustion. (b) By hot gases passing up the chimney. (c) By radiation and dropping of fuel through the grate. (3) The proportion of this heat lost. The second and third added together should equal the first. Results may be checked in this way: First: In the answer to the first question we have tried to collect data that shall enable us to ascertain the approximate number of heat units (quantity of heat) that should be produced per pound of fuel burned. This has been collected from papers received from all over the country in reply to the first twelve questions. In said answer the first column contains the name of the coal, the second the number of heat units that will be produced by the complete combustion of one pound, the third column the number of pounds of water, from and at 212, that would be evaporated, provided 60 % were used for this pur- pose; the fourth column the same as the third, except that 80 % instead of 60 % is; here assumed; the fifth column gives the pounds of water, from and at 212, that have been evaporated by one pound of some of the coals mentioned, in actual practical tests; the sixth column gives the per cent efficiency, corresponding to the actual evaporations named in column five. Second: The amount of water evaporated during a given time may be estimated as indicated in question No. 21 or in various other ways. This amount in pounds divided by the number of pounds of coal burned during the same time will give the evaporation per pound of coal. This multiplied by the number of heat units required to evaporate one pound will give the amount of heat utilized in making steam. Or the evaporation may be reduced to an equivalent evaporation from and at 212 by multiplying it by the proper number (factor of evaporation) contained in the following table abridged from "Kinealy on Engines and Boilers." This product multiplied by 966 will give the quantity of heat used in forming steam. Dividing the heat value of the coal by this last product will give the efficiency of the boiler and furnace. Temperature Pressure of Steam by Gauge in Ibs. per Square Inch, of Feed Water. Factors of Evaporation. O.P. 40P. 50P. 60P. 70P. SOP. 90P. 100P. HOP. 120P. 40 1.179 1.203 1.206 1.209 1.212 1.214 1.217 1.219 1.221 1.222 60 1.158 1.182 1.185 1.188 1.191 1.193 1.196 1.198 1.200 1.201 70 1.148 1.172 1.175 1.178 1.181 1.183 1.186 1.188 1.190 1.191 90 1.127 1.151 1.154 1.157 1.160 1.162 1.165 1.167 1.169 1.170 110 1.106 1.130 1.133 1.136 1.139 1.141 1.144 1.146 1.148 1.149 130 1.085 1.109 1.112 1.115 1.118 1.120 1.123 1.125 1.127 1.128 150 1.065 1.089 1.092 1.095 1.098 1.100 1.103 1.105 1.107 1.108 170 1.044 1.068 1.071 1.074 1.077 1.079 1.082 1.084 1.086 1.087 190 1.023 1.047 1.050 1.053 1.056 1.058 1.061 1.063 1.065 1.066 TEST OF BOILER AND FURNACE. A complete test of a boiler and furnace would involve ascertaining the following facts: First. The quantity of heat generated by the combustion of the coal. Second. The heat used in making steam. Third. The proportion of the heat lost by (a) Incomplete combustion. (b) Hot gases passing up the chimney. (c) Radiation and dropping of fuel through the grate. The second divided by the first is the efficioncy of the furnace and boiler. 15 flso ea P sy to fstimatewhetheror not the loss from incomplete combus- U "/thTfuenfnot evenly distributed on the grate one may have too much air supplied to a part of the furnace and not enough to the re- mainder, so that excessive quantities of free oxygen (0.) and of carbon monoxide (CO) in the chimney gases would indicate both an excessive amount of air and also incomplete combustion. If the remaining sources of loss are very considerable it is owing to inexcusable mismanagement or imperfect apparatus. One author remarks that the remedy is to get a new fireman. The following gives a tabulated result of an 1897 practical test: Coal Used Pocahontas (run of mine) ; Heat Value, 14,289. Per cent of Heat units. total heat. Useful evaporation 11374 ' C ! 4 t Loss by chimney gases 1900.437 16.6 Unconsumed carbonic oxide 271.491 1.9 Loss by radiation, unconsumed hydrocarbons, evaporation of moisture in coal (2.3%) and unaccounted for 743.028 5.2 Total.. 14289 100 The Quantity of Heat Generated. Mr. R. S. Hale, in his paper on "Fuel Gas Analysis in Boiler Tests," read before the American Society of Mechanical Engineers in 1897, remarks that it is not necessary to have an analysis of the coal, inas- much as the data of the text books can be relied upon within very narrow limits. In answer to Q. 1, your committee has endeavored to supply reliable data; of course it is open to correction. To find the quantity of heat that should have been generated in the furnace, one would look in the table, Ans. 1, for the kind of coal he was using, and opposite that he would find the number of heat units gen- erated per pound of coal burnt. This multiplied by the number of pounds of coal burned in a given time would be the total heat that should have been produced during that time. The coal is usually weighed in the barrow upon platform scales. The Proportion of Heat Used in Making Steam. In the report of the Committee on Code, at this year's meeting of the A.S.M.E., occur these words: "The elaborate directions and multiplicity of details provided for in the foregoing code should' not divert the mind from the fact that the principal elements to be ascertained in a boiler test are the weight of water evaporated and the weight of the fuel required to produce such evaporation." The water fed to the boiler is best weighed in a separate tank, wnich is emptied into a tank from which the boiler draws its supply. The amount may be approximately estimated as indicated in Ans. it would perhaps pay to actually measure the water necessary to draw from the boiler to make the level drop from one mark to another the water glass when the boiler was not in use. It might also be measured by a water-meter. th^^Sf g ? fc , t . he , water evaporated, this amount would be divided by J number indicating the weight of coal used and the result reduced equivalent evaporation from and at 212 by multiplying by the 16 factor of evaporation taken from one of the numerous tables published. The above last result would then be multiplied by 966, and the product divided by the heat value of the coal, to get the efficiency of the furnace and boiler, which should be from 60 % to 80 %. ANALYSIS OF FLUE GASES. As a check upon the above results, also a reliable indication of the performance of the furnace, we wish to know two things: First. How many pounds of gas are going up the chimney per pound of coal burned, together with its temperature? Second. Whether or not a considerable amount of combustible gag is going up the chimney? The first may be determined approximately by the rules given in Ans. 15. A rough but practical estimate of the second may easily be made when we recollect that air is about 20 % oxygen by volume and 80 % nitrogen; that the nitrogen goes through the furnace unchanged; that as much of the oxygen as is properly used is changed to an equal vol- ume of CO* (carbonic acid). Therefore if all the oxygen of the air is either unchanged or is changed to carbonic acid, the volume of the carbonic acid and oxygen taken together must still be 20 %, so that if we measure the volume of carbonic acid and oxygen, find what per cent they are of the whole, taken together, subtract this result from 20, we shall have as a remainder a per cent which is roughly propor- tional to the combustible gases in the chimney. If this remainder is greater than 2 (two) it would generally indicate that insufficient air is being supplied to the furnace. In all ordinary cases the second is neglibly small. For the volumetric analysis of the chimney gases the Orsat appa- ratus is generally recommended as accurate, convenient and reliable. Its cost is somewhere from $16 to $30. The form using pinch-cocks instead of glass-cocks is recommended by Prof. Gill as simpler, cheaper and more practical. Of this apparatus the Committee on Code of the A.S.M.E. says: "For the past year the writer has made extensive use of the Orsat apparatus in his boiler testing, and has found the work not only inter- esting but exceedingly instructive and valuable. Its chief value lies in the guide which it affords in determining what kind of firing is most advantageous where the fuel is bituminous; coal. That the instru- ment is reliable and useful for the purpose noted is quickly ascertained and without any very extended practice. When the thickening up of the fire is invariably attended with an increase in the percentage of carbonic oxide and a reduction in the percentage of oxygen, as the writer has found, he feels at once assured that the instrument is' not a plaything or something that is influenced in unexplained ways by whim or caprice, but rather that it is an important adjunct to the engineer's outfit." HOME-MADE ANALYZING TUBE. The device described below, which has been constructed and used by your committee, is upon the same principle as the Orsat apparatus, but is, of course, not so elaborately constructed and graduated. This apparatus is illustrated in Figs. 1 and 2. Anyone can make it, and the apparatus itself ought to cost about 50 cents. Fig. 1. A is a stick held in a vertical position by a vise. B C is a small cup. D is a graduated measuring glass; it is immaterial into what units the glass is graduated, so that the unit is small. E is a glass; tube %" or more in diameter and about 15" long, drawn down in a gas flame at the ends, so that a small rubber (black preferred) tube, as F, may be slipped on and make air-tight joints. G is a pinch- cock by which the tube F may be closed air-tight. H is a funnel; in this case it is the part of the tube E that was separated from the rest when the ends were drawn down. I is a rubber band, which is placed around the stick A and funnel H to hold the apparatus in a vertical position. J are rubber bands upon the tube E. These may be moved to different positions on the tube, to serve as marks. If a graduated tube is used the glass D may be dispensed with. METHOD OF USING. The pinch-cock G is opened and the tube E filled with water, which is then allowed to run out slowly into the graduate and measurer. The measurement of this water determines the volumetric contents of the tube once for all. Suppose, for convenience, it is 100 cubic centimeters (though it may be anything; the larger, the more accurate will be the measurements) . It is generally thought best to lead a W or %" lead pipe from the "breaching" or a part of the flue near the up-take to a convenient place for filling the tube. We have used a small rubber hose. It is said that if rubber gets too hot it will generate gases that will effect the results. To get the gases into the tube (see Fig. 3) one simply connects the ; C, in line with the pipe A, to the breeching X, and also to con- t said tube to the inlet port of a hand air-pump B, and draws a charges of the pump through the tube. This will clear out the Lir that was in the tube and replace it with flue gases. Perhaps it might be better to take a larger number of strokes of the pump. A sort of ejector, like that shown in Fig. 4, may be used instead of a pump. By the aid of a small jet of steam a steady stream of flue may be drawn through the tube. A barrel, keg, or large can, filled with water, may be used for this purpose, as indicated in Fig. 5, in which C is the tube connected with the top of the barrel, and with the pipe A (Figs. 3 and 5). Letting the water run out of the bottom of the barrel draws the flue gases through the tube at the top. Care should be taken not to let the water get low enough so that air will be drawn back by the draft. The tube E is connected, by means of rubber tubes upon its ends, with a pipe or tube leading to the chimney, and the flue gases are drawn through by means of an air pump or other convenient device until all the air is removed and the tube is full of chimney gases. Having got the gases into the tube, the pinch-cock G is closed, the other end of the tube B is closed, say, by the finger (Fig. 2). The cup C is about two-thirds full of caustic potash solution. The finger and end of the tube E are now put into the cup and the finger removed from the end of the tube while both are under the surface of the fluid. The apparatus is now in the position shown in Fig. 1, though the fun- nel H need not be on. The pinch-cock G is now opened for an instant until some of the fluid in the cup has run into the tube, and a rubber band, J, moved down to mark the surface of the fluid in the tube. The finger is again placed over the end of the tube (under the surface) and the tube turned up so that the fluid will run along the walls of the tube and expose a good deal of surface to the contained gas, as shown in Fig. 2. Upon placing the tube E back into the cup in the position of Fig. 1 it will be noticed that the fluid is drawn up into the tube. Repeat the above operation three or four times at intervals of about a minute, or until the fluid no longer rises. Push another of the bands J down to mark the surface of the fluid in the tube. The distance between the two rubber bands measures the volume of carbonic acid gas (CO 2 ) that was in the tube; this volume divided by the total volume of the tube above the first band is the per cent of carbonic acid gas in the flue gases. Now leaving everything in position, put the funnel H in the end of the rubber tube F, as shown. Then fill the glass graduate or any convenient vessel about two-thirds full of water and put into it all the snow-like pyrogallic acid that it will absorb. Fill the funnel with this pyrogallic acid solution and carefully open the pinch-cock G a very little, allowing a little of the contents of the funnel to run down. Be careful and not overdo this; enough to fill the tube *4" to W is suf- ficient. On repeating the operation indicated by and described in refer- ence to Fig. 2 it will be noticed that the fluid in the tube begins to turn a dark orange color and the level to rise. This indicates the absorption of oxygen. When the level no longer rises, move another rubber band down to mark its position. The distance between the second and third rubber bands measures the free oxygen that was in the tube and has been absorbed; this volume divided by the volume of the tube above the first band is the per cent of free oxygen in the flue gases. As long as the rubber bands, Jl, J2, J3, are not moved, one may measure the volume of the tube between them as often as he pleases by drawing water up into the tube and allowing it to flow out into the graduate until it falls from one band to the other. If one wants very great accuracy he should calculate the effect of the column of fluid drawn up into the tube, and should be careful to let the tube drain long enough before measuring, and also consider the pressures of the water vapor. The temperature of the tube and contents must not vary during the operation. The finger must never be removed from the end of the tube E, except when under the surface of the fluid in the cup C. Unless one has particularly thin skin, or unless the operation is unnecessarily protracted, the caustic will do no harm. Running water, olive oil and muriatic acid will neutralize or remove the potash. The writer has protected his finger by placing over it a thin rubber bag taken from the toy technically called a "cry-baby," sold at most of the shops for a nickel. Our brethren in New York and Chicago may be able to obtain something better for the purpose something specially made to resist the action of corrosive fluids. [Rubber finger-stalls can be purchased from dealers in rubber goods. Ed.] COST OF APPARATUS. Glass tube $0.10 1 ft. black rubber hose 0.10 Glass graduate 0.25 Pinch-cock 0.10 1 Ib. caustic potash 0.10 1 can pyrogallic acid... .. 0.35 Total $1.00 ESTIMATING THE WEIGHT OF THE CHIMNEY GASES. For this purpose it would only be necessary to ascertain the per cent by volume of the carbon dioxide. Ordinarily the weight of the gases per pound of coal would run about as follows for the corresponding per 20 Weight of Per cent Flue Gases carbon dioxide. per Ib. coal. 3. 61 4. 50 5. 40 6. 33.3 7. 28.6 8. 25 9. 22.2 10. 20 11. 18.2 12. 16.7 13. 15.4 14. 14.3 15. 13.3 16. 12.5 Under the conditions at each end of the column there would prob- ably be considerable inflammable gases in the chimney gases. (See Ans. 16 and remarks under the heading "Analysis of flue gases.") That would have to be estimated as previously described. For the pur- pose of finding the volume of CO 2 only, the tube would not haye to be manipulated, nor would it be necessary to put the hand nor any part of it into the alkali. A little of the potash solution would be put in the cup, the funnel would be filled with it, and a little let down occa- sionally by opening the pinch-cock so as to keep the walls of the tube covered with the solution. The correction for the weight of the col- umn in the analyzing tube is made as above described in referring to the oil column except that the constant would be .0063. AN ILLUSTRATIVE TEST. Duration of test, 10 hours. Kind of coal used, Pocahontas. Heat value, 14,290. Weight of coal consumed, 6,507 Ibs. Water fed to boiler, 64,877 Ibs. Water per pound of coal, 64,877 -^ 6,507 = 9.97 Ibs. Equivalent evaporation from and at 212 9.97 X 1.19 = 11.80 Ibs. Heat used in making steam per pound of coal used, 11.8 X 966 = 11,398 B. T. U. Efficiency of boiler and furnace, 11,398 4- 14,290 = .80. Gas analysis per cent by volume carbonic acid 15.1, oxygen 4. Leaving 20 19.1 =.9 %, indicating a small amount of combustible gas in chimney. Weight of chimney gases per pound of coal, 11 X (15.1 + 4) -r- 15.1 = 13.92. Temperature of flue gases above outside air, 529. Sensible heat in flue gases, 529 X 13.92 X .24 = 1,767 heat units. Per cent of heat in flue gases, 1,767 -f- 14,290 = 12.3. Heat Units. Per Cent. Steam making 11,398 80 Hot gases 1,767 12.3 Balance: radiation, unconsumed gases, dropping coal through grate, etc 1,125 7.7 Total 14,290 100 FIG. 6. FIG. 4. FIG. 5- FIG. 7. For practice in analyzing gases one may find the proportion of carbonic acid and oxygen in his breath. It is better to hold the breath as long as convenient before blowing it through a tube, and to blow through the tube enough times to be sure that all the air originally in the tube is expelled. TEMPERATURE OF CHIMNEY GASES. We have felt that the methods of ascertaining the temperature of chimney gases were not very available and practical for our purposes. Below will be found described a device for that purpose, which we be- lieve accurate, simple and convenient. The drawings are the work of one of the committee, and are in- tended simply to be illustrative, and no great accuracy nor proportion in sizes have been striven for. We have experimented, successfully, we believe, with the following described apparatus: In Fig. 6, A represents the front wall of the breeching; B is a glass tube, say 1 /4" outside diameter, and as long as can conveniently be se- cured, drawn together at one end, b, in a gas flame; the other end, which is open, or slightly contracted, stuck through a cork or plug, C. The cork, C, is then forced into the hole in the breeching, so that the glass tube extends into the hot flue gases, as shown, and left there until it has the same temperature as the gases. It Is then taken out and placed with its open end under the surface of some high boiling point (flashing point) oil (we use engine oil) as shown in Fig. 7, and left there to cool. When it has cooled down to the temperature of the room, the oil will have risen up into the tube a certain distance, which is proportional to the amount the col- umn of air in the tube has contracted in cooling, and therefore pro- portional to the change in the temperature of the tube. The column of air, b, e, Fig. 7, is now measured. Inasmuch as the volume of air is proportional to its absolute temperature and the volume of air in the tube, b, e, is at the temperature of the room and equal to the length of the tube at the temperature of the flue gases, therefore, b, e, is to the length of the tube as the temperature (absolute) of the room is to the absolute temperature of the flue gases. The tube we used was 15.5" long. The temperature of the room was 90; that is 460 + 90 = 550 absolute. The oil was drawn up 7", leav- ing 15.5 7 = 8.5" as the length of the column, b, e, above the oil. Then 8.5 : 15.5 : : 550 : the absolute temperature of the flue gases. Therefore the absolute temperature of the flue gases was 550 X 15.5 -f- 8.5 = 1003 or in ordinary measurement 1003 460 = 543. This would be 543 90 = 453 above the outside air. The weight of the column of oil has a slight stretching out effect upon the air above it, because of its weight. To find how much should be added to the height of the column for this effect, one multiplies the height of the column of oil by the length of the tube above the oil and this product by .004; the result is the fraction of an inch that should be added to the height of the oil column. In the above example the correction would be as follows: 7 X 8.5 X .004 = .238. Therefore to the measured height of the oil column should be added about .24 of an inch. One might use a metal tube, for instance a piece of gas pipe, with one end closed air tight and the other nearly closed, instead of the glass tube, if he prefers; but in this case he must have a measuring glass (or a pair of scales) to find the volume of the air in the tube origin- ally,, and the volume after contraction. For example: Suppose we had a tube of 2" inside diameter, and 5" long. We will fill it with water, then pour the water out and measure it, and find that it is 15.5 cu. ins. Then we put the tube into the breeching and let it get as hot 23 In Fig. 8 we have drawn our idea of this last form: A is a hook to hang it by in the flue or chimney. Of course the tube must be perfect- ly dry when placed in the flue and no steam in it. We do not think the evaporation of the water will effect the result. When metal is used it would be well to have pretty thick walls so that the interior would not cool off before one could get it into the water. We would remark that the method of placing the tube in the chimney is a matter of personal choice and judgment. COMPRESSED AEB. The usual formula in the book is PVi-*=C in which P is the pressure, V the volume, and C a constant which has to be calculated for each compressor. This formula is somewhat difficult to use because it has a decimal fraction for the exponent of V. Moreover, it assumes that no heat is lost by the air which, owing chiefly to the relatively cool cylinder walls, but to some extent to the moisture in the air is never the case. If, however, we assume an exponent of 1 1/3 or 4/3 we shall have a fundamental formula, that may be solved by the more familiar opera- tions of squaring, cubing and extracting the square and cube roots. We shall also have a formula that will give results approximating closer to practice than the usual formula, because it allows for a small loss of heat. Absolute pressures and temperatures are always used. The fundamental formula then is (1) This may be put in the form, (2) which is a more convenient form to use. In words formula No. 2 says: "The pressure multiplied by the volume and this product by the cube root of the volume is always the same, that is, is equal to a constant " Prom equation No. 1 the following formulas' and rules may be de- 24 FIRST. To calculate the pressure, the volume being known: Rule 1. Divide the constant by the volume multiplied by its own cube root. C That is (3) p = We generally know the volume at the beginning of compression, and we also know that the pressure then is one atmosphere, say 14.7 Ibs. If we wish to know the pressure at some other volume, we may use the following rule, as well as Rule 1. Rule 2. Divide the greater volume by the lesser, and multiply this quotient by its own cube root. Multiply this last product by the pres- sure at the larger volume and we have the pressure required. That is v /y~ (4) P=P 1 ^ Vi For example, suppose we have a compressor that has a 12" stroke and 10 sq. ins. area of piston, compressing to 74.7 Ibs. (60 Ibs. gage). The figure shows the indicator diagram. For convenience we measure the volume in inches of cylinder length. We first find the constant, C, by formula 2. At the commencement of the stroke the pressure, P, is one atmosphere (14.7 Ibs.) and the volume V is 12" of cylinder length. Therefore, by formla 2 14.7 X 12 X f/r2 = C. 14.7 X 12 X 2.29 = C = 404. Now suppose we want to know what the pressure is after the piston has traveled 6 inches. The volume remaining, or the distance the piston has to travel yet before reaching the end of its stroke, is six inches. Therefore by Rule 1 we divide the constant, 404, by the vol- ume, 6", multiplied by its own cube root. 404 -H 6 X 1^6 =404 -+- (6 X 1.817) = 404 -f- 10.9 = 37 Ibs. By Rule 2, we first divide the larger volume by the lesser, 12 -f- 6 = 2. We then multiply this quotient by its own cube root, 2 X 1^2 = 2 X 1.26 = 2.52 We then multiply this product, 2.52, by the pressure at the larger volume, 14.7, thus 14.7 X 2.52 = 37 Ibs. Rule 2 is the simpler, and the form that will be generally used. We notice that the diagram is still inaccurate at this point. I I I I I I I I I I TT I I I i I i I -T-T ill J\Q \9 \f \r \6 Iff U la ie If SECOND. To find the volume, the pressure being known. Rule 1 Divide the constant by the pressure, and multiply this quotient by its own square root, then extract the square root of the product, that is (5) V= Rule 2 Divide the lesser pressure by the greater, and multiply the quotient by its own square root, extract the square root of this product, and multiply the larger volume by this square root Thus, if we want to know the volume at the end of compression in the above example, when we know the pressure is 74.7 Ibs. By Rule 1. We divide the constant by the pressure 404 -^ 74.7 = 5.408. We then multiply this quotient (5.408) by its own square root, 5.408 X 1/O08 = 5.408 X 2.326 = 12.58. We then extract the square root of this product (12.58) \/ 12.58 = 3.537 or about 3.54 inches of cylinder length. By Rule 2. We first divide the lesser pressure (14.7 Ibs.) by the greater 14.7 -=- 74.7 = .1968. We then multiply this quotient (.1968) by its own square root, .1968 X ]/. 1968 = .1968 X .444 = .0874 and extract the square root of this product ( N/M74) v/T08T=.2956. We then multiply the greater volume, 12", by this root (.2956), 12 X .2956 = 3.54" inches of cylinder length. Which is the result re- quired. To Find, the Net Work per Stroke of the Compressor. Rule. First multiply the stroke (in feet) by the area of the piston (in square inches) and this product by 58, and call this result No. 1. Second. Divide the greater by the less pressure, extract the fourth root (the square root of the square root) of this quotient, diminish it by unity (1) and call the remainder, result No. 2. Third. Multiply result No. 1 by result No. 2 and the product is the work per stroke in foot pounds (of work = 59 APV Thus, if we wish to know the work per stroke in the above example, by Rule 1, we First. Multiply the stroke in feet (1) by the area of the piston in square inches (10) and this result by 58, so that we have 1 X 10 X 58 = 590. Result No. 1. Second. We divide the greater (74.7) by the less pressure (14.7) and extract the fourth root of the quotient and diminish it by one 74.7 + 14.7 = 5.082; 4/57082 = ^ /I/TOM = 1/2^55 = 1.502 Diminishing this by one, 1.502 1 = .502. Result No. 2. Multiplying result No. 1 by result No. 2 we have 590 X .502 = 296 ft. Ibs. per stroke. 26 The above rules, and formulas, are also applicable to the ammonia compressor. To Find the Compression Temperature. Rule 1. Multiply the initial temperature (absolute) by the fourth root of the quotient obtained by dividing the greater by the less pres- sure. The result is the final temperature absolute. Thus in the example taken, assuming that the initial temperature is 60 460 + 60 or 520 absolute, we have ^=520 4/5.082=520 X 1.50 = 781 absolute or 781 .460 = 320 by the thermometer. Compression temperature may also be found by drawing the iso- thermal line (steam expansion line) under the actual diagram. The initial absolute temperature is to the final absolute temperature as the lengths of the horizontal lines drawn to the points of the curves where the temperature is to be measured. Thus referring to the figure or diagram: T :Tj ::ab : ac ac = 3.54 14.7 ab = - X 12 = 2.36 74.7 .-. 2.36 : 3.51 :: 520 : Tj .-. ^ = 780+ MISCELLANEOUS RULES. The work, in foot pounds, per pound of air compressed may be ob- tained by multiplying the rise in temperature by the constant 213. The weight of one cubic foot of air in pounds at any temperature and pressure may be found by multiplying its pressure by the constant 2.707 and dividing by its absolute temperature. Weight per cubic foot in Ibs. = Under constant pressure the volume of air is proportional to its absolute temperature. Under constant volume the pressure is proportional to the absolute temperature. The specific heat of air is: Under constant volume, .1688; under constant pressure, .2377. In two stage compression the intermediate pressure is equal to the square root of the initial and final pressure. Question 1. (1897-8.) How much heat is produced by the complete combustion of (a) one pound of anthracite coal, (b) one pound of bituminous coal of average quality? Express answer in heat units and also in foot pounds. Name kind of coal (See table on the following page) 27 Ans. 1. KIND OF COAL. Heat Evapo from a 21 ration ndat 2 ||| O N ." "0 lit! Value 60$ m% 5P s A PENNSYLVANIA ANTHRACITE. 12,000 to 14,900 Average 13,916 7.45 9.245 8.63 9.93 12.32 11.51 10.07 70 j 14,098 8.74 11.52 13,551 8.4 11. 2L 13,800 8.56 11.42 11.17 78 13,104 8.13 10.84 14,500 9. 12. 11.13 74 i_/acKdw 12 200 7.57 1009 Honeybrook Lehigh Massachusetts Anthracite 15,200 9.44 12.59 10.75 BITUMINOUS, SEMI-BITUMINOUS, ETC. R' 14,000 8.7 11.6 T ^ ^Bl ' 'k 14000 8.7 11.6 14,400 8.94 11.93 Pittsburgh Coking 14,400 14 265 8.94 8.87 11.93 11.82 Pocahontas, run of mine 14,289* 13,614 8.87 8.46 11.82 11.27 11.53 8.74 78 62 Pennsylvania, semi-bituminous 13,368 13 000 8.30 807 11.07 10.76 Bureau Co 111 . 13,025 8.07 10.76 11,083* 6.88 9.18 7.78 68 12,139* 7.54 10.05 8.25 66 12,240* 7.60 10.13 8.98 71 Big Muddy Jackson Co 111 12 600 7.83 10.43 Bituminous 1896 tests heat value for ten mines scattered over Eastern anc Southern Ohio (data furnished by Akron, Ohio, No. 28) Wood (assumed as .4 of coal_used) . . . Lignite 13,100 10,300 8.14 3.48 6.40 10.85 4.68 8.50 Crude Petroleum 19,200 11.93 15.90 *1897 Q. 2. (1897-8.) How much heat will be produced by the incomplete combustion of one pound of coal of the kind assumed in Q. 1 (a) ? Ans. 2. Assuming 90% carbon, the heat produced will be 4,400 X .90 = 3,960. Q. 3. (1897-8.) What are the products of combustion under the conditions; of Q. 1 (a) and what are their approximate proportions? Ans. 3. Carbonic acid gas (COa), also called carbon dioxide, a colorless gas consisting of 3/11 carbon and 8/11 oxygen by weight; formed by the union of the oxygen of the air with the carbon of the coal and having the same volume as the oxygen from which it was formed: steam (H 2 O) a colorless gas formed by the union of the oxygen of the air with the free hydrogen of the coal; consisting of one part by weight of hydrogen and eight parts of oxygen. The proportion of steam in the chimney gases is usually negligibly small. 28 Q. 4. (1897-8.) What are the products of combustion under the con- ditions of Q. 2, and what are their approximate proportions? Ans. 4. Carbon monoxide (CO), a colorless gas consisting of 3/7 carbon and 4/7 oxygen by weight, formed by the union of the oxygen of the air with the carbon of the coal, and having twice the volume of the oxygen from which it was formed. Q. 5. (1897-8.) What gas' in large proportions in the chimney gases indicates that an insufficient quantity of air is being supplied to the whole or a part of the fuel? Ans. 5. Carbon monoxide (CO). Q. 6. (1897-8.) What gas in large proportions in the chimney gases indicates that too much air is being supplied to the furnace? What proportion of this gas is allowable in good practice, with natural draft? Ans. 6. Oxygen (O 2 ). About 10% by volume. Q. 7. (1897-8.) What would be the weight of chimney gases per pound of coal burnt, in good practice, with natural draft? Ans. 7. About 25 Ibs. Q. 8. (1897-8.) What is the approximate specific heat of chimney gases? What do you mean by this answer? Ans. 8. Twenty-four one-hundredths (.24). Each pound of the chimney gases raised 1 F. in temperature, absorbs twenty-four one- hundredths (.24) of a unit of heat. Q. 9. (1897-8.) With the facilities of the average engine and boiler- room how can the temperature of the chimney gases be approximately determined? Ans. 9. (1) By a thermometer, properly protected, placed in the chimney gases. (2) By exposing a certain weight of iron to the gases until it acquires their temperature and then estimating the temperature of the iron [see answer to question 50 (1896-7).] (3) By the melting point of alloys or metals exposed to the gases. Q. 10. (1897-8.) What proportion of the heat produced by the burn- ing of the coal should go to the formation of steam, in good practice and with a good boiler plant? Ans. 10. From 60% to 80% (.60 to .80). Q. 11. (1897-8.) How can the amount of heat produced by the fur- nace be approximately estimated and how can the proportion of said 29 number of pounds of the same bu rned evaporated by the By multiplying the numb f * J^" The evaporation is usually heat required to eyaporat e one poun^ d 1^ e y ^ ^ reduced ^ an equivalent evapora^on from ^ ^ ^^ ^ Znd'of waYer'at'^S' to^e'pound of steam at the same temperature. -See answers to last year's questions. n 12 ( 1897-8 ) What would be the weight of the chimney gases per Q pound ( of coai^urnt, in good practice, with a forced draft? Ans. 12. About 19 Ibs. (CO,, from a mixture (b) For the absorption of free oxygen (O) from a mixture of Ans 13 (a) A solution of caustic potash (commercial) in water from the hydrant. Proportions about 1 lb. of potash to 2 Ibs. of water. ^fSffiSSySSto acid (pyrogallol) in water mixed with the above solution of caustic potash. Prof. Thurston used 5% pyro- gallic acid. Phosphorus may also be used. Q. 14. (1897-8.) Describe a cheap and simple apparatus by which the proportions, by volume, of carbonic acid gas (CO 2 ) and ot tree oxygen (O) in the chimney gases may be determined. Describe meth- od of use. Ans 14. Ga. No. 1 sends the following description: "Fill a graduated test-tube with chimney gases. Close the mouth, invert the tube and put the mouth under water. Arrange the tube so that the level of the water inside and out is the same. Now intro- duce a piece of caustic potash fastened to the end of a wire; allow it to stand about 45 mins.; withdraw the potash. It will be found that the volume (of the gas) has diminished, which represents the per cent of C0 2 in gases. Rearrange the tube So the water level is the same inside and out; introduce a piece of phosphorus and allow to stand 24 hours. Withdraw the phosphorus. It will be found the volume has again diminished, which represents the per cent of oxygen." The device which your committee has constructed and used is de- scribed and illustrated in a separate chapter. Q. 15. (1897-8.) Assuming that a sufficient amount of air is sup- plied to the fuel and that the per cent by volume of carbonic acid (CO 2 ) and of free oxygen (O) in the chimney gases is known, how can the weight of the chimney gases per pound of coal be approximately deter- mined? Ans. 15. Add the per cent by volume of carbonic acid (C0 2 ) to the per cent by volume of oxygen (0 2 ) ; divide this result by the per cent by volume of carbonic acid, and multiply the quotient by the constant number ten and one-half (10.5). 39 2. Divide the constant number two hundred and ten (210) by the number indicating the per cent by volume of the carbonic acid (C0 2 ). We remark that the above rules are for finding the weight of gases per pound of coal, and not per pound of carbon or combustible in the coal. It is therefore necessary to assume an arbitrary constant which will give the average value. For the combustible the above constants would be about 12 and 240; for a particularly good coal they would be about 11 and 220. Q. 16. (1897-8.) Knowing the per cent by volume of carbonic acid gas (CO-) and of free oxygen (O) in the chimney gases, how can the per cent by volume of carbon monoxide (CO) be calculated, neglecting water vapor and hydrocarbons? Ans. 16. Add the per cent by volume of carbonic acid (COa) and of oxygen (Oa), multiply the result of 5, subtract the product from 100 and divide the remainder by 3. Q. 17. (1897-8.) If coal, having a heat value of 14,000 units per Ib. is used, the feed-water is at 100 degrees, the pressure is 70 Ibs. gage, it is found that 30 Ibs. of water is evaporated by 4 Ibs. of coal burned: What per cent of the heat produced by the combustion of the fuel is used in making steam? Ans. 17. The heat produced by the coal should be 14,000 X 4 = 56,000. The factor of evaporation for feed-water at 100 and pressure 70 Ibs. is 1.149. Therefore the equivalent evaporation from and at 212 is 30 X 1.149, which is 34.47, 34.47 X 966 = 33,298, which is the number of heat units used in making steam. Therefore the per cent of the heat used in making steam is 33,298 -=- 56,000 = 59.46. Q. 18. (1897-8.) If under the conditions of Q. 17 it were found that the temperature of the chimney gases was 450 F. above the air in the boiler room, and that 40 Ibs. of air was being supplied per pound of coal burned, what per cent of the heat produced by the combustion of the coal would be going out of the chimney with the gases? Ans. 18. If 40 Ibs. of air was supplied per pound of coal the weight of the chimney gases would be about 41 Ibs. per pound of coal; the heat carried up the chimney by this would be about 41 X 450 X .24 = 4,428, and the per cent is 4,428 ^ 14,000 = 31.6 (about). Q. 19. (1897-8.) What is the relative approximate permeability to heat of clean boiler plate, %" thick, and the same plate covered with %" of hard sulphate scale? Ans. 19. From two to one (2 to 1), to two to one and a half (2 to Q. 20. (1897-8.) How is the condition of the boiler indicated by a change in the temperature of the chimney gases, and why? Ans. 20. A rise in the temperature of the chimney gases would indicate a dirty boiler, as scale on the inside or soot on the outside would reduce the conductivity of the metal, and a greater quantity of heat would pass out with the gas'es. Q. 21. (1897-8.) A horizontal boiler is 18 ft. long, 72" diameter, pressure 70 Ibs. gage. The lower end of a twelve inch water glass is 48" from the lowest point of the boiler. At 9 a. m. the water is 4" as a of steam is the boiler supplying per hour.' Ans 21 The water has fallen 3" or one-quarter of a foot. The average width of the surface while falling multiplied by the length of the boiler (18 ft.) and by the distance through which the level has fallen, gives the volume of the water used. This expressed in cubic feet and multiplied by 56.8, the weight of water at the temperature corresponding to 70 Ibs. gage (316), will give the weight of water- evaporated in the given time three-quarters of an hour. This result divided by .75 will give the weight of water per hour that the boiler is evaporating. To find the average width of the surface one might take the width at the higher and at the lower levels, add the two together and divide The width of the surface (a d Fig. 1) at the higher level may be found in subtracting the square of its height (be) above the center of the boiler from the square of the radius (36"), extracting the square root of the result and multiplying by 2. Thus: Radius equals 36", square of radius = 1296. Height from center of boiler 20", square of height 400; 1296 400 = 896. The square root of 896 is 29.933, and twice this is 59.866, which is the width of the surface at the higher level. The width of the surface at the lower level may be found in the same way to be 63.46; 59.866 + 63.46 -T- 2 = 61.66", or about 5.14 ft, as the average width. It would perhaps be better to draw a diagram like Fig. 1 to as large a scale as practicable, say one-half, then draw a line f g half-way be- tween the lines a d and h e, which indicate the higher and lower levels. Then carefully measure the line g f. With the scale suggested, the line will be found to be about 30 13/16 long, which, multiplied by 2, because of the reduced scale, will give about 61.63". This illustrates the simplicity and practical accuracy of graphical methods of calculation. Taking the average width as 5.14 ft, the length as 18 ft. and depth as .25 ft., the volume of the water evaporated would be 5.14 X 18 X .25 = 23.13 cubic feet, and its weight is 23.13 X 56.8 = about 1313 Ibs. This multiplied by 4 and divided by 3 gives about 1751 Ibs. of water per hour. 32 Q. 25. (1897-8.) The gage pressure being 120 pounds, the feed water 110 and the engine using 15 Ibs. of steam per horse-power-hour, what part of the heat that went to the making of steam is transformed into useful work by the engine? What per cent is this of the heat generated in the furnace, the efficiency of the furnace and boiler being .70? Ans. 25. (a) The total heat in steam above 32 due to 120 Ibs. gage pressure is 1188 H.U. and the heat already in the water is 110 32 = 78 H.U. Then, the heat required to change 1 Ib. water at 110 into steam at 120 Ibs. gage pressure is 1188 78 = 1110 H.U. The engine in question uses 15 Ibs. of steam per horse-power, which is equal to 15 X 1110 = 16,650 H.U. As 1,980,000 is equal to one horse-power per hour, and 1 H.U. equals 778 ft. Ibs. of work. 1,980,000 = 2545. 778 the H.U. changed into work per HP. per hour. Then the per cent of heat that went to make steam and was changed into work by the en- gine is 2545 X 100 = 15.22%. 16650 (b) If the furnace and boiler has an efficiency of 70% and the boiler uses 16,650 H.U. for 15 Ibs. of steam, the heat of the furnace would have to produce for each 15 Ibs. of steam 16650 =23785.7 H.U. .70 and the per cent of this heat that went to do useful work is 2545 X 100 = 10.7. 23785.7 Q. 42. (1897-8.) Give three ways of calculating the area of the end of the boiler above the tubes, which is supported by braces? Ans. 42. Taking the segment as less than a half circle and of the size that it is usually necessary to brace on the end of a boiler: FIRST METHOD. Take one-half the area of the entire circle of which the segment is a part. Call this result No. 1. Multiply the diameter of said circle by the height of the base of the segment above the center of the circle. Call this result No. 2. Subtract result No. 2 from result No. 1 and the remainder is the area of the segment very nearly. The result is a little too small. SECOND METHOD. Subtract twice the distance from the base of the segment to the cen- ter of the circle from half the circumference of the circle and multiply this result by one-half the radius of the circle. Call this result No. 1. Multiply the radius of the circle by the height of the base of the seg- ment above the center. Call this result No. 2. Subtract result No. 2 from result No. 1 and the remainder is the area of the segment very nearly. The result is a little too small. THIRD METHOD. Draw the segment to as large a scale as convenient and measure its area by the planimeter. Inasmuch as two or more methods of solving a problem are some- times useful as a check upon each other, and as some will prefer to use 33 one way and some another, the graphical methods of solving the seg- ment problem might be entertaining and useful. We believe that these methods as given are sufficiently accurate for practical purposes. The second method is quite accurate if one-half the base of the segment is taken instead ot the radius of the circle, in calculating the triangular area. AN EXAMPLE. The circle is 53" in diameter. The height of the base of the segment above the center of the circle is 7.5". (See Fig. 6.) First Method: The area of the circle, a b c f, 53" diameter, is 2206.18 sq. ins. The area of the semi-circle, g b h, is one-half of 2206.18 or 1103.09 sq. ins. Which is result No. 1. If you multiply the diameter g h, equal 53", by the height e d, equal 7.5", you get 397.5 sq. ins., which is a little greater than the area g a c h. If you take the area g a c h from the area g a b c h, you have left the area of the segment, a b c e. Therefore 1103.09 397.5 = 705.59 sq. ing., the approximate area of the segment. Second Method: If you subtract twice the height e d, from the semi- circumference g a b c h, you have approximately the length of the line a b c. If you multiply the line a b c by one-half the radius, a d, you have the area d a b c. If you multiply the radius by the height e d, you have approximately (a little large) the area of the triangle dace. If you take the area dace from the area d a b c, you have the area a b c e remaining, which is the area of the segment. FIG. 6 Referring again to Fig. 6: The circumference of a circle having a diameter of 53" is 166.5". One-half this (the line g b h), is 83.25". If from this you subtract twice the height d e, that is twice 7.5" or 15", you have 83.25 15 = 68.25", which is approximately the length of the "l ie _ a b c - [ f you multiply this by one-half the radius, you have J.25 = 904.31 sq. ins., or the area d a b c. This is result No. 1. 7 K J multiply the height d e, by the radius of the circle, you have 198.75 sq. ins., which is nearly the area of the triangle a c d. A little large. This is result No 2 7 ^ ubt J actin S result No ' 2 from result No - x y u have 904.31198.75 56 sq. ink, for the area of the segment, which is a little small because the area of the triangle was a little large. prmula of the answer to last year's Q. No. 15, and by accur- 710 l iS" ng SeC nd method ' tne area wil1 be found to b e about 34 Q. 43. (1897-8.) If a boiler brace is attached to the shell 4 ft. from the end of the boiler and to the end of the boiler 2 ft. from the shell, what would be the strain upon the brace as compared to an end-to-end brace supporting the same area? Ans. 43. The strain upon the brace attached to the shell would be to the strain upon the end-to-end brace as the length of the first men- tioned brace is to the distance of its point of attachment to the shell from the end of the boiler. In this instance 1.118. Q. 48. (1897-8.) What are the different ways in which a riveted joint may give way? Ans. 48. The plate may tear across along th'e line of least cross- section a, b, Fig. 7, (a). 2. The plate and rivet may be crushed, as shown in (b). 3. The plate may break across in front of the rivet, (c). 4. The rivet may shear across, (d). Q. 49. (1897-8.) What is the general rule for selecting the diameter of rivets for a given thickness of boiler plate? What should be consid- ered in determining the diameter of rivets? Ans. 49. If the rivet holes are to be punched, the punch should have a diameter as great as the thickness of the plate, otherwise it is liable to be broken. Drilled holes are not made less in diameter than the thickness of the plate. As the shearing strength of the rivet increases with the square of its diameter, and the crushing strength of the plate in front of the rivet increases only as the first power of the diameter, there will evidently soon come a time, as the rivet is increased in diameter, when the shearing strength of the rivet is greater than the crushing strength of the plate. A correctly designed joint should be equally strong in all its parts. The rivets should be close enough together to hold the joint tight. The rule given by Unwin is to make the diameter of the rivet 1.2 times the square root of the thickness of the plate d = 1.2 Vt. -0- (b) (c) Q. 50. (1897-8.) What effect' does the length of a tube have on its collapsing strength? Ans. 50. No. 12 of Boston, Mass., says: "The longer the tube the lower will be its collapsing pressure. First, because a short tube will retain its circular form better than a long one owing to the tendency of the long tube to sag in the middle and 35 get out of shape and, again, the short tube has the advantage of the support of the heads to which it is attached. This support would also be given to the long tubes, but would not have the same effect at the middle of its length. Large flues, like those in marine boilers are strengthened by rings placed at regular intervals to hold the flue in shape. Another method is to make the flue of short lengths, joining the ends by riveting them to the rings. Another method is to make the flues corrugated. All of which goes to show that a short tube will stand a greater external pressure than a long one." While the above seems about right to the committee, still it is said that for the purpose of calculating the thickness of the tube, the length beyond ten or twelve diameters may be neglected. Q. 58. (1897-8.) If a boiler weighing, with its contained water, 10 tons, is supported by four symmetrically-placed round wrought iron rods, what should be the diameter of the rods? Ans. 58. Each rod would have to support % of the entire weight (10 tons), or 5000 Ibs. Allowing 10000 Ibs. as the safe working strain per square inch of cross-section each rod would require .5 of a square inch cross-section, which corresponds to a diameter of .8". To allow for the change of temperature and corrosive action to which the rods would be subjected, the rods would probably best be made 1" or 1.125" in diameter. BOOKS REFERRED TO BY THE COMMITTEE OF 1897-8. Gas and Fuel Analysis, for Engineers Gill published by John Wiley & Son. Hempel's Gas Analysis translated by Dennis published by Mac- millan & Co. Boiler Tests by Geo. H. Barrus published by the author. We would refer any one wishing to read up on the subject of the graphical representation of the forces of inertia in steam engines, to Holmes on "The Steam Engine," published in Appleton's "Text Books of bcience series. The subject is treated, however, in nearly all books on mechanics or the steam engine. Upon the subject of lubricating oils. Thurston's work on "Lost Work ^Machinery," etc., published by Wiley, or G. H. Hurst on "Lubricating Gill on Oil Testing. *"* standard and reliabl e works on the strength of Strains in Framed Structures, by Bindon B. Stoney Unwm's Machine Design. Rauleaux'g Constructor, translated by Supplee Weisbach's Mechanics Vol. 1, translated by Cox !f M Q the T Fk Silvanus R Thompson, "Elements of Elec- erv" * M v gn S V "Electromagnets and Electromagnetic Machin- thls subject 1C Machiner y>" are ^out as good as any on iroi rZ k rn? t i tled " M , ec * ai ca l Draft," published as an advertisement, the best wort re f rd ^ flrst hundr ed or hundred and fifty pages as st work on the subject of combustion they have met. The Educational Committee. INTRODUCTION TO QUESTIONS 83 TO 112 (1898-9). Tnt^d loV^f th ^ S been V* a * with care. The questions 36 ulty of as much value as the more refined calculations which the same points are also susceptible of neither is superfluous and both should be cultivated energetically. Q. 83. (1898-9.) At what standard of temperature is the evapora- tion of steam boilers usually expressed, irrespective of the actual tem- perature of the feed water or the pressure at which the steam is taken off? Ans. 83. For comparison it is usual to reduce the results of boiler tests to a known standard. This standard is technically known as the "equivalent evaporation from and at 212." This means that the evap- oration is considered to have taken place at mean atmospheric pressure and at a temperature due to that pressure and the feed water also being assumed as supplied at that temperature. One pound of water evaporated "from and at 212" is equivalent to 965.7 B. T. units. Q. 84. (1898-9.) Explain the difference between a pound of coal and a similar weight of "combustible." Ans. 84. The distinction between equal weights of coal and com- bustible is as follows: A pound of coal always contains a "per cent" of moisture and non-combustible matter in the form of ash and clinker. After deducting this the remainder is useful for combustion. A pound of combustible consists of such substances as can be completely burned or consumed leaving no refuse. Q. 85. (1898-9.) How many pounds of coal burned each hour, per square foot of grate, is considered a fair and economical rate of com- bustion? Ans. 85. When economy is a consideration the rate of combustion must evidently not exceed the absorbing capacity of the boiler. With grate and heating surfaces properly proportioned the best economy is claimed for a rapid combustion of the fuel. From 15 to 20 Ibs. of coal per hour burned on each square foot of grate is considered good practice for factory boilers', the latter rate requires an exceptionally strong draft and is well up to the limit unless mechanical means are resorted to for hastening the consumption of the fuel. Q. 86. (1898-9.) What is a good "evaporation" per pound of coal for boilers of horizontal tubular type, under fair average conditions? Ans. 86. The evaporation from boilers of the horizontal tubular type, under fair conditions, may range from 7% to 9* Ibs. The lower figure is probably more common and the latter may be slightly exceeded. Much depends upon the kind of coal and the methods and manner in which the plant is conducted. Q. 87. (1898-9.) What "check" or test is necessary to obviate decep- tive results or magnify the evaporative efficiency of a steam boiler? Ans. 87. A calorimeter test should always be made where accurate results are wanted to determine the quality of steam and per cent of primage. This percentage must be deducted from the apparent evap- oration to arrive at a true result; that is, actual evaporation. Q. 88. (1898-9.) As an example, name a rate of evaporation, per pound of combustible, that you would regard with suspicion. Ans. 88. Efficiency of the furnace and the boiler is determined by a comparison of the heat units accounted for in the steam produced and the theoretical value of the fuel consumed. Combustion is always more 37 or less incomplete; fuel is lost by dropping through the grate; heat is carried away by the hot gases and lost by radiation, hence when the difference between the theoretical and realized values seems too small to cover these losses, there is good cause for caution in accepting the accuracy of figures which represent an abnormal rate of evaporation. - Naming 12 Ibs. of water evaporated from and at 212 Fahr. as a rate open to suspicion we may reckon 966 X 12 = 11592 B.T.U. accounted for in steam. This is about 82% of 14000, the latter figure being the assumed value of the combustible. Experience shows that 18% will not cover the losses attending common practice; 40% is not an unusual loss and less than 20% is 1 seldom, if ever, attained. Q. 89. (1898-9.) For a safe working pressure of 100 Ibs. gage pres- sure give the maximum diameter and also the greatest length desirable for horizontal tubular boilers. Ans. 89. A well-proportioned horizontal externally fired tubular boiler working under 100 Ibs. pressure should not be larger than 72" diameter and 18 feet long. The Hartford Steam Boiler Inspection and Insurance Company, at the request of this committee, kindly furnished the following on this question : "This company does not approve of the employment of plate thicker than y 2 " in the construction of externally fired boilers. Keeping within this limit of thickness and assuming the tensile strength of plate at 60000 and that all the usual requirements as to ductility and elongation are complied with. The diameters for a plate thickness of 15/32" are as follows, the sizes varying with the efficiency of the joint used in the longitudinal seams: Safe working pressure 100 Ibs. Factor of safety five. (a) For double-riveted lap-joint having 70% efficiency, diameter 78". Proof: 60000 X 0.46875 X .70 19687.5 = = 101 Ibs. (Radius) 39 X 5 195 (b) For triple-riveted lap-joint having 75% efficiency the diameter may be increased to 84 inches. Proof: 60000 X. 0.46875 X 75 21094 = = 100 Ibs. (Radius) 42 X 5 210 (c) For triple-riveted butt joint with double straps and securing an efficiency of 86%, the corresponding diameter is 96 inches. Proof: 60000 X 0.46875 X .86 24187.5 = 100% Ibs. (Radius) 48 X 5 240 Length of boiler having 3V 2 " or 4" tubes should not exceed 18 feet. The proper length should be determined by the ratio of heating surface to grate area, and the area of tube opening to grate area. By our tests we have found the former should be about 46 to 1 and the latter when bituminous coal is used should have an area of from 1/6 to 1/7 of the grate surface. All specifications issued by this company for this type of boiler have a factor of safety of five and is considered none too much The efficiency of the three joints mentioned, that is, 70, 75 and 86% %?* P^te when* properly proportioned, can be realized. The latter form of joint is illustrated herewith " W fL&t 9 ,?h ( 1 898 ; 9-) T, ^ hat is enerall y claimed for all boilers of the water-tube class? Define also the advantages and the disadvantages ANSWER No. 89. 39 attributed to those constructed with straight, and with curved, water tubss? An's 90 It is generally claimed for water-tube boilers: (a) They are safe from destructive explosion at extremely high pressures^ ^ forced far be y 0n d their normal capacity with impunity. (c) Occupy small space and weigh less per unit of power developed. (d) Economy in the use of fuel. (e) Are easily cared for, etc., etc. Some of these claims, in some instances at least, have no founda- tion in fact. The advantages' of straight water tubes are, that they can be easily inspected and cleaned and a standard tube of the required length can be used for renewals. One alleged disadvantage is, that owing to their straight form they are liable, on account of expansion and con- traction, to severely strain the headers they are expanded into. This, however, is thought to be more imaginary than real and the thousands of boilers of this type in successful daily use bear out this assumption. The advantages claimed for a "curved" tube are, that neither the tube or its connections are injured by unequal expansion or contraction. Such tubes are, however, difficult to properly inspect and clean; they are hard to repair, and this is especially true of boilers containing different lengths and bends. There are unusual conditions where boilers with bent tubes are a necessity and are also desirable, but we do not believe that these conditions obtain in stationary practice. There are many of this type and most of them are a delusion and a snare in bad water. Q. 91. (1898-9.) How many square feet of heating surface per horse-power is a customary allowance in horizontal tubular boilers? Give also a similar approximation for water-tube boilers of average efficiency? Ans. 91. Fifteen square feet of heating surface per horse-power is a common allowance for externally fired boilers of the horizontal tubular type, and ordinarily this is sufficient when such boilers are used to sup- ply simple throttling engines. For "Corliss" class of engines the allow- ance is more usually restricted to 12 square feet. Water tube boilers are rated at 7 to 11 sq. feet. All such ratings should be regarded only as an approximation; a better method is to take into consideration the various condition's which apply to any par- ticular case. Q. 92. (1898-9.) Give an average ratio of grate and heating surfaces in horizontal tubular boilers which give good results and note such conditions as may prompt a change in the proportions given? Ans. 92. For horizontal tubular boilers a ratio of heating surface approaching 35 to 46 to one square foot of grate is usually productive of good results. High furnace temperature and rapid combustion yield economy; the rate of combustion is always determined by the "draft" and the more rapid the consumption of fuel on the grate, the greater should be the extent of the heating surface. Draft in turn, however, is ordinarily dependent on the temperature of the escaping gases, hence the increased ratio referred to must be limited, of course, to a point where the draft remains equal to the requirements of the furnace. Q. 93. (1898-9.) For a plant requiring 200 HP, working steam pressure 100 Ibs;., and which is to be operated reasonably free from interruption, due to cleaning and repairs, give general dimensions and the number of horizontal tubular boilers you would install. State also how such should be set, i. e., singly or in pairs, to fill the requirements at a minimum cost. Ans. 93. (a) Assuming that the 200 HP plant referred to in the question is to be used for manufacturing purposes and is not to run on Sundays or holidays the boilers required would be two 66" dia. 16 ft. horizontal tubular type 54-4" tubes. To be of standard design and material, first-class workmanship and capable of safely carrying 100 Ibs. pressure per square inch. Boilers to be set over separate furnaces and properly connected so they could be run independently if necessary. (b) Another answer provides a third boiler which would be neces- sary, of course, for a plant designed for continuous operation. This answer is expressed as follows: We would recommend three boilers of the horizontal tubular type set singly; that is, with partition walls between adjacent boilers each to be provided with separate grate, steam and water connections, enabling any one of the three to be operated independently. Bach smoke connection to stack to have damper and each boiler to be pro- vided with an independent gage in addition to the steam gage con- nected to the main header. Each boiler to develop 100 HP; that is, must be capable of evaporat- ing 3450 Ibs. of water from and at 212 per hour. Allowing 12 square feet of heating surface per HP would therefore require 1200 square feet in each, boiler. A boiler 66" diameter 18 feet long, fitted with 54-4" tubes fulfills this requirement. Q. 94. (1898-9.) Describe briefly, without detail, the method of supporting horizontal tubular boilers which you regard with favor? Ans. 94. Boilers are best supported by being hung up on steel fram- ing in a way that the brick work is relieved from all weight. This makes it much easier on their "settings" and greatly facilitates work in case of needed repairs. Properly designed, such an arrangement allows freedom for expansion, without cracking brickwork. Q. 95. (1898-9.) Which is preferable in horizontal tubular boilers, the dome, steam-drum, or dry-pipe? Ans. 95. The dome or steam drums are expensive adjuncts to hori- zontal tubular boilers and are useless on boilers which are well designed and proportioned with a view of dispensing with either. A properly designed "dry-pipe" is an equivalent, answering every requirement when all other things are equal. Q. 96. (1898-9.) Do you consider the advantages claimed for mud- drums sufficient to outweigh such objections as their use entails? Ans. 96. The trend of modern "ways" is to dispense with mud drums on horizontal tubular boilers. They deteriorate" rapidly and are there- fore considered costly and not ordinarily necessary, although in Missis- sippi River practice and other places where only muddy unfiltered water is available, the mud-drum is still regarded with some favor. Q. 97. (1898-9.) What style of cock or valve is best adapted for lower "blow-off" of boilers? Ans. 97. For "blow-off" use a valve with passages' calculated to pass freely all particles likely to get into the blow-off pipe. A globe valve is out of place, because it does not meet the above requirement. Valves "handle" easier than plug cocks and are less liable to "jar" and thereby injure the pipe and connections. Asbestos-packed cocks are claimed to be more satisfactory than the ordinary kind. Q 5,8. (1898-9.) Give the proper position for a "surface blow"; state how its details should be arranged and explain the purpose it serves. Ans 98 A "surface blow-off," as applied to a horizontal tubular boiler consists of a flat funnel-shaped appendage inside of the shell, preferably located near the rear end with the wide mouth of the funnel placed at the water line and directed toward the front of the boiler. This should be connected to a 1%" pipe leading out of the back head of boiler and a valve should be located in an accessible position so it can be used frequently while the boiler is in operation. The purpose served is the removal of the sludge and scum that is always more or less upon the surface of the water when boilers are being worked. When properly installed and attended to, a surface blow-off is very helpful and greatly relieves certain conditions which tend to cause "foaming." Q. 99. (1898-9.) What form of connection do you approve for a water-column and how should gage-cocks and water-glass be placed with reference to the tubes in horizontal tubular boilers? Ans. 99. A water-column should be connected to boiler with pipe not smaller than l 1 ^". (The Hartford Steam Boiler Inspection and Insur- ance Company advocates seamless brass tubing, iron pipe size and brass ground-joint unions.) The fittings for lower or water end should be crosses or tees with brass plugs for ready removal in cleaning and inspection. Valves should not be put on the pipes between the boiler and col- umns, as these are a possible source of danger. By making the "blow- off" pipe to column %" or 1" there is no need of any valves at this dangerous point. The lower gage cock and bottom of water glass should be placed on a line 3" above top row of tubes; that is, water should appear in the glass when tubes are covered to a depth of 3". Q. 100. (1898-9.) Give size and specify the number and style of safety valves necessary for the boilers referred to in Q. 93 pop or lever pattern? Ans. 100. Good 'pop" safety valves: of standard make are preferred; each boiler to have its own valve; every valve to be in direct communi- cation with the boiler which it serves and placed with no intervening means of shutting off same. 'ihe grate area to be provided for either of the boilers described in Ans. 93 will hardly be less than 30, or more than 38, square feet. Figuring on the maximum grate area and allowing one square inch of valve area for uiree square feet of grate we should provide a valve having 38-^3 = 122/3 square inches, which is nominally the area corresponding to a 4" -diameter. (Ed. Com.) The above is the U. S. government allowance and is only one of the many rules on this subject. There are other modes, more logical, perhaps; see Kent's Pocket Book, 1898, page 723. While it is not a common practice to provide two smaller valves instead of one larger one, yet there seems much in this idea that is 1 commendable, especially when the valves are set to blow off at slightly different pressures. Q. 101. (1898-9.) What are fusible plugs made of and where should same be placed in horizontal tubular boilers? Ans. 101 Fusible plugs are usually % or 1" pipe size with a taper 5 through them which is filled with Banca tin. The tin is calculated 42 to fuse when not covered with water, but certainty of action depends on proper exposure; that is, scale should not be allowed to accumulate on the face of the plug. The plugs should be placed in the back head, in rear combustion chamber, well exposed at highest point on the fire line, which should be slightly above top row of tubes. Q. 106. (1898-9.) Give size of "stack" adapted for 200 HP, i. e., for boilers proposed in Q. 93, height of chimney assumed at 100 ft. Ans. 106. For 200 HP boilers referred to in Q. No. 93, the diameter for chimney corresponding to a height of 100 ft. should be 38 inches. This assumes the consumption of 5 Ibs. good coal per HP per hour and therefore provides for overload and other contingencies as to the quality of the fuel. (Kent's Formula.) Q. 107. (1898-9.) What is claimed for the modern steel chimney, as compared with well-constructed brick chimneys? Ans. 107. The advantages claimed for the modern steel chimney over similar brick structures are: Greater strength; less space required; foundations smaller; the cost is less, and furthermore, the draft is not impaired by infiltration of air. Steel chimneys are smaller, lighter and offer less area to wind pressure and are also better adapted to stand the strains, due to unequal temperatures, etc. Q. 108. (1898-9.) Explain the distinction between "forced" and "induced" draught; state briefly the advantages of both systems when acquired "mechanically," i. e., by fan blowers. Ans. 108. Under "forced" draft the air is forced through the fire, the pressure above the atmosphere being maintained either in the closed ash-pit or in the closed fire-room. The latter arrangement is only practical in marine service. Under the "induced" or suction method the products of combustion are exhausted from the furnace, a partial vacuum is produced therein and air thereby caused to flow through the fuel. Both systems of "mechanical draft" operate to produce a pressure difference between the ash-pit and the combustion chamber, and other things being equal, there is no conclusive evidence that one method is superior to the other; the one to be adopted must depend upon the conditions and the advantages as compared with "chimney draft" are common to either. In operation, the primary advantage of a mechanical system lies in its intensity and its controllability. Intense draft such as may be read- ily produced by means of a fan makes possible the burning of the cheapest fuel, which is always mostly in fine particles and packs closely on the grate. It is also conducive to higher rates of combustion and the carrying of deeper fires. A deep fire gives better opportunity for contact between the air and the fuel so that the air supply can be reduced and the efficiency of the plant increased. Ample draft for the prevention of smoke can be best produced by mechanical means, which at the same time prevents the formation of carbonic oxide. With a chimney the temperature of the gases must be high to insure sufficient draft, whereas with the fan the gases may be cooled to the lowest possible degree and the heat transferred to the feed water or the air supply, thereby greatly reducing the loss usually resulting from this cause. It is claimed also that the exhaust steam from the fan engines may be utilized and the expense of producing draft by the heat savings noted, is reduced to practically nothing. 43 Mechanical draft can be automatically regulated to instantly change both the intensity of the draft and the volume of the air. It is thus entirely independent of the conditions of weather and is capable of responding to sudden demands. By its use a given boiler plant may be instantly and greatly increased in capacity to meet such demands as are common in electric railway work. Reserve capacity can thus be stored in light and comparatively cheap fans rather than in ponderous and expensive boiiers. When the plant is properly designed less boilers are required than with the ordinary chimney, for a higher com- bustion rate and greater evaporative capacity can be secured. Q. 109. (1898-9.) What difficulties attending "hand firing" are over- come by mechanical stokers? Ans. 109. Mechanical "stokers" feed the fuel with constant regular- ity; they obviate the frequent opening of fire doors with the usual inrush of cold air; the fires are practically self-cleaning, and by properly arranged coal and ash handling machinery the cost of steam production may be much reduced. Stokers have to be judiciously selected, maintained and properly run or they may be a source of annoyance and expense. Q. 110. (1898-9.) Name what considerations should govern the dis- tance between grate and boiler. Ans. 110. In furnace construction the distance from grate to boiler shell should be governed by the kind of coal to be burned. Anthracite coal requires but 20" to 24 for the best results; for semi-bituminous coal the distance should be between 26" to 30" and more in proportion for rich bituminous coals. Q. 111. (1898-9.) What portions of a horizontal tubular boiler should be exposed to the fire and heated gases? Ans. 111. Only such parts of a horizontal tubular boiler as are pro- tected by water should be exposed to the fire or gases of combustion. This ordinarily includes the lower 3/5 of the shell and such surfaces as are concerned therewith. The return of gases over top of shell does not meet with general favor. Q. 112. (1898-9.) Explain the meaning of "ultimate tensile strength ' and "elastic limit" and state what proportions the latter should bear to the former, in good steel boiler plate. Ans. 112. "Ultimate tensile strength" is the maximum stress that a piece will endure before being torn asunder. This strength depends somewhat upon the mode of testing; the more rapidly the testing pro- ceeds, the higher will be the apparent strength. Elastic limit defines the limiting strength, i. e., a load greater than expressed by the elastic limit will produce a permanent elongation or "set" after the load is removed. The elastic limit for the best grades of boiler steel should be 50% of e tensile strength; the "elongation" varies with the length and thick- ness of the specimen, but is usually fixed at 25% for plate %" thickness and thereabouts. (1899 " 1900 - ) Wha ^ is the effec t of increasing the height of a How is the difference in pressure measured? ' the C0mmon draft of a chimney as measured in inches of 44 What is the effect of an increase in the difference of pressure or in- tensity of draft? What relation would the increase of pressure or intensity of draft have upon the velocity of the gases which flow through the chimney? Ans. 77. Increasing the height of a chimney increases the intensity of the draft, because it makes a greater difference per unit of base between the weight of the inside column of gases and a column of equal height of the outside air. The difference in pressure is measured by the height of the column of water it will support; for such a test, a U-shaped tube has one leg connected with the inside of the chimney and the other open to the air; the greater pressure of the atmosphere pushes the water toward the chimney until the difference of the heights of the two columns in the glass tube is sufficient to balance the difference in pressure between the flue and atmosphere. The pressure is measured by taking the dif- ference between the heights of the two columns. The common draft of a chimney is about Vz in. for ordinary heights and temperatures. The effect of an increase in pressure, or intensity of draft, is to increase the velocity with which the gases flow through the chimney. The velocity increases as the square root of the pressure; in order to double the velocity, the pressure would have to be increased 4 times; to get 3 times the velocity, the pressure would have to be increased 9 times. Q. 78. (1899-1900.) What relation has the height of a chimney to its capacity? The area of a chimney to its capacity? How does the area increase? Give general statement for the capacity of chimney. What effect has temperature upon the capacity? Ans. 78. The capacity of a chimney varies as the square root of the height. In a chimney 200 ft. in height, the velocity will be, theoretically, twice as much as a chimney 50 ft. high, so that twice as much gas would be discharged in a given time. The capacity varies directly as the area that is, a chimney of twice the cross-section will discharge twice the amount of gas in a given time. The area increases as the square of the diameter. The capacity varies as the square root of the height and as the square of the diam- eter. The effect of temperature on the capacity is that the greater the inside temperature, the greater the difference in pressure and velocity of the gases; but, as the density of the gases decreases with the tem- perature, there is a point where as much is lost in weight of the gas passed, by the lightness of the gas, as is gained by the increased ve- locity. Q. 79. (1899-1900.) How would you find the area required for a chimney of a given height and the number of sq. ft. of grate surface connected? How could the area for a given H.P. be determined? Ans. 79. Multiply the number of sq. ft. of grate surface by 120, and divide the product by the square root of the height and the quotient will be the required cross-section in square inches. Divide the H.P. by 3 1-3 times the square root of the height, and the quotient will be the required effective area in square feet. To the diameter, or length of the required side to give this area, add 4 in., to compensate for friction. 45 O 80 (1899-1900.) What would be the area of a chimney for 55 sq. ft. of grate surface, height being 125 ft., allowing for burning 5 Ibs. f No^e-Th^'allowancTof 5 Ibs. of coal per H.P. per hour is usually made for difference in condition in the atmosphere; also for an in- crease in demand for extra power, and is the usual base for formulat- ing. Ans. 80. 125 ft. = height of the chimney. 55 sq. ft. = grate surface. 55 X 120 = 595 sq. in. = area of cross-section. 1/125 / 595 Diameter = ^ / = 27.4 in. ; 595 "\ .7854 Adding 4" for friction, diameter 27.444 = 31.4 in. Should a square chimney be called for, then the square root of the area (595 in.) would equal the length of the required side, adding the 4 in. as before. Should a parallelogram be required in the cross-section of the chim- ney, having the area (595 sq. in.) given, divide the area by the estab- lished length of side, for the side adjacent, adding the 4 in., as before, to both sides. Q. 91. (1899-1900.) What pressure (draft) will a chimney 135 ft. high produce? Outside temperature 62 F., temperature of flue gases 580 F. Ans. 91. The province of a chimney is to serve a double purpose of creating a movement of air, or as commonly called, creating a draft, and conducting away the products of combustion or obnoxious gases. The draft depends upon the gases having a higher temperature, con- sequently lighter than an equal column of outside air. The air flowing or moving from the place of the higher pressure to that of the lower. The intensity of the draft depends upon the height of the stack and the difference in temperatures of the outside air and the hot gases in the stack, called the pressure difference. As an illustration: Take a stack 100 ft. high, outside air at a temp, of 62 F., the gases at a temp, of 570 F. A pound of burned gas at 62 F. has a volume of about 12.5 cu. ft, while the same gas at 570 F. will have a volume of 22.27 cu. ft. 12.5 (570 + 460) = 22.27 cu. ft. 62 + 460 The 460 representing absolute temperature (some works use 461 as absolute temperature). A column of gas 100 ft. high with a base of one square foot equals a pressure of 100 -=- 22.27 = 4.49 pounds per square foot. A pound of air at 62 F. has a volume of 13.14 cu. ft., and a column 100 ft. high and one foot square in section will weigh 100 -~ 13.14 = 7.61 pounds. The pressure difference equals 7.61 4.49 = 3.12 pounds per sq. foot. Equal in oz. pressure to 3.12 X 16 144 1 oz. = 1.7295 in. of water. .35 oz. X 1.7295 = .605 the draft pressure in inches of water. It will be seen that the higher the stack the lower will be the tem- erature of the gases in order to obtain the same pressure difference, or the same draft. 46 The pressure (draft) of a chimney 135 ft. high; outside temp. 62 C F. ; gases temp. 580 F., may be found by the following formula: 7.6 7.9 P = H( )= .945 inches of water. (7.6 7.9 \ 1 = .945 inches of T Ta/ Substituting / 7.6 7.9 \ P _ 135! \- 135 (.0146 .0076 = 135 X .0070 = .945 inches of V 522 1040 / water, the pressure or draft of the chimney. P = draft in inches of water. H = height of chimney. T = absolute temp, outside air (460 + 62) =522 F. Ta = absolute temp, chimney gases (460 + 580) ) =1040 F. 7.6 and 7.9 constants obtained by experiment. Q. 92. (1899-1900.) A chimney 120 ft. high, temperature flue gases 660 F. It is desirable to add an economizer, which will reduce the temperature of the gases to 390 F. To what height should the chim- ney be raised in order that the draft will remain the same as at first? Outside temperature in both cases 60 F. Ans. 92. H = height chimney. H' = height chimney after economizer is installed. T = absolute temp, outside air, 460 -+- 60 = 520 F. Ta = absolute temp, gases, 460 + 660 = 1120 F. Tb = absolute temp, gases after economizer is installed, 460 + 390 = 850 F. P = pressure in inches of water. 7.6 and 7.9 are constant numbers found in part, by experiment and mathematical process. (7.6 7.9 \ I = P before economizer was installed. T Ta/ (7.6 7.9 \ I = P after economizer is installed. T Tb/ H' == 7 . 6 7.9 the height of chimney after economizer is installed. T Tb Substituting / 7.6 7.9 \ P = 120 ( - ) = 120 (.0146 .0071) = 120 X .0075 = .90, the V 520 1120/ pressure in inches of water. .90 .90 .90 H'= 7.6 7.9 = = = 169.81 ft. .0146 .0093 .0053 520 850 the height of the chimney after the economizer has been installed. By the use of another formula we obtain an answer of 168.35 ft.; second, 168.88 ft; third, 170.52. A slight variation in the constant numbers probably in disuse of decimals make the slight variation in the different answers. 47 Q 94 (1899-1900.) A boiler evaporates 12,468 Ibs. of water per hour' (feed-water 154 F.) to a steam pressure of 145 Ibs. (absolute) per square inch, find the equivalent evaporation? Ans 94 The total heat of steam, divided by the latent heat of evaporation of water, at 212 F., gives a multiplier, by which the weight of water actually evaporated is to be multiplied ( by, to reduce it to the equivalent evaporation of, "from and at 212 F." This total heat of steam is modified somewhat by circumstances. From the total heat of steam, subtract the heat units in the feed water, and divide the remainder by the latent heat of evaporation at 212 F. Solution 1190.4 B. T. U. equals the number of heat units in steam at 145 pounds pressure absolute. 122.33 B. T. U. equals the number of heat units in water at 154 F. 1190.4 122.33 = 1068.07, the total heat. 1068.07 -4- 966 = 1.1056, the factor of evaporation. 12468 Ibs. steam X 1.1056 = 13784.62 Ibs., "from and at 212 F.," the equivalent evaporation. Q. 5. (1900-1901.) Assume you are operating a plant furnishing steam for purposes other than supplying steam for your engine, how would you arrange your coal report so as to charge each department with the proper amount of fuel consumed? Ans. 5. To answer this question other than as an approximation is beyond the limit of a work of this character. There are plants of this kind where the operator generally takes such methods as he may be possessed of for forming approximate ideas relative to the coal and water consumption. The indicated power is, of course, first taken care of and the balance is subdivided for different purposes. The first thing requisite is good judgment, and a careful scrutiny of the per cent loss not returned in condensation, etc. By measuring the returning condensa- tion water and steam for each system (by meter or otherwise). If the returns are returned to the boiler, then the difference be- tween the heat units in the steam and the heat units in the returns will equal the heat units used; these multiplied by the number of pounds of returned condensation equal the total number of heat units used; divide this product by the effective heat units in one pound of coal consumed and the quotient will equal the number of pounds of coal used. Or, the number of pounds of returned condensation divided by the number of pounds of water evaporated will also give the number of pounds of coal used, providing proper allowance is made for the heat that is returned to the boiler. Q. 6. (1900-1901.) Give opinion relative to the several advantages and disadvantages of the two general types of boilers, water tube and fire tube. Ans. 6. The water tube boiler, while not rated at as high efficiency as the return tubular boiler, has an element of safety superior to the return tubular boiler. It is the tendency at the present day with the multi-cylinder engines to carry extreme high pressures, especially in the marine' service. ucn boilers must be carefully designed for strength. It is likewise necessary to reduce its weight and size to the lowest possible limit, e best of material should be used. For carrying high pressures, is clear to see, for the sake of safety, the Scotch type or any tubular Her must be made extremely heavy and bulky, and much 48 attention is now being paid to the devising of a new type, which, while retaining the good features of the Scotch type, will be lighter, smaller and cheaper for the same power. Some of the advantages claimed for water-tube boilers are as follows: 1. That the portions of the boiler which contain the water are so small in diameter that the material used in the construction can be made comparatively light without impairing the strength. Conse- quently the heat is transmitted to the water more readily and the danger of burning the iron where it is exposed to the fire is greatly diminished, 2. There are no riveted joints, and consequently no double thick- ness of metal exposed to the fire. 3. That the draught area, being much larger than in the fire-tube boilers, gives ample time for the absorption of the heat of the gases before their exit to the chimney. 4. That the gases being thoroughly mingled in their passage be- tween the staggered tubes, the combustion is more complete. 5. That the gases impinging against the heating surface perpen- dicularly instead of gliding along the same longitudinally, the absorp- tion of heat is more thorough. It is claimed that experiments have proven that a gain of 30 per cent in the efficiency of the heating sur- face is effected. 6. That the circulation of water is rapid and all in the same direc- tion, there being no conflicting currents. As a result the temperature of the boiler in all its parts is practically the same and the tendency to deposit scale is materially lessened. 7. That the water being divided into many small streams in thin envelopes, steam may be raised rapidly. 8. That the large area of disengaging surface in the drums, together with the fact that the steam is delivered at one end and taken out at the other end, insures dry steam without the aid of any superheating device. 9. That the water level may readily be kept steady. 10. That the whole structure is so flexible that the parts may expand and contract without producing strains. 11. That the division of water into small masses avoids destructive explosions. 12. That the space occupied by this type of boiler for a given power is much less than in fire-tube boilers. 13. That by a suitable arrangement of hand and man holes every part of the boiler is accessible for cleaning and repairs. 14. That the loss of effect from dust collecting on the top of the tubes is far less than in fire-tube boilers, where it collects on the interior surface. In the latter case there is no limit to the amount of dust which may collect, while in the former only a limited amount is retained. 15. That since no part of the boiler above the water level is ex- posed to the fire, and because of the absence of deteriorating strains and thick plates and joints in the fire, it is much more durable. 16. That, being made in sections, it is less cumbrous and much more easily transported and erected. 17. That a new tube may be easily inserted or most any other repair be made by an ordinary mechanic with ordinary tools. Q. 7. (1900-1901.) Give rules for selecting material for cylindrical shells, also shell plate formula. Also rules for flat plates. Give furnace formula. Give rules for selecting stays, also rules for allowable safe load on stays. Material for cylindrical shells subject to internal pressure. 49 Ans. 7. Board of Trade Rules. Tensile strength between 27 and 32 tons. In the normal condition the elongation is not less than 18 per cent in 10 inches, but should be about 25 per cent; if annealed not less than 20 per cent. Strips 2 inches wide should stand bending until the sides are parallel at a distance from each other of not more than three times the thickness of the material. Lloyd's. Tensile strength between the limits of 26 and 30 tons per sq. in. Elongation not less than 20 per cent in 8 inches. Test strips heated to a low cherry-red and plunged into water at 82 F. Must stand bending to a curve the inner radius of which is not greater than 1% times the plates' thickness. U. S. Statutes. Plates of %-inch thickness and under shall show a contraction of area of not less than 50 per cent; when over % inch and up to % inch not less than 45 per cent; when over % inch not less than 40 per cent. From a paper on boiler construction by Nelson Foley, we find the following comments: "The Board of Trade rules seem to indicate a steel of too high a tensile strength, when a lower and more ductile one can be got; the lower tensile strength limit should be reduced, and the bending test might, with advantage, be made after tempering, and made to a smaller radius. Lloyd's rule for quality seems more satisfactory, but the temper test is not severe. The U. S. statutes are not sufficiently stringent to insure an entirely satisfactory material." Mr. Foley suggests a material which would meet the following: 25 tons the lowest limit in tension; 25 per cent in 8 inches minimum elongation; radius for bending after tempering equals the plate's thickness. Shell plate formula: Board of Trade: TxBxtx2 2 (TBt) P = orP = . DxFxlOO 100 (DF) D = diameter of boiler in inches; P = working pressure in Ibs. per sq. in.; t = thickness in inches; B = percentage of strength of joint compared to solid plate; T = tensile strength allowed for material in Ibs. for sq. in.; F = factor of safety; being 4.5, with certain additions depending upon method of construction. C (t 2) B Lloyd's: P = - D t= thickness of plate in sixteenths. B and D = same as in previous formula. C = constant depending on the kind of joint. When longitudinal seams have double butt straps, C 20. When longitudinal seams have double butt straps of unequal width, only covering on one side the reduced section of plate at the outer line of rivets, C = 19.5. When longitudinal seams are lap-jointed C = 18 5 2tT U. S. statutes: P = f or single riveting; add 20 per cent for 6D S6 th6 Sam6 notation as in Board of Trade rules or criticises the u - S. statutes as follows: "The rule ignores tf e ?n that * distln guishes between single and double, the latter 20 per cent advantage; the circumferential riveting of X S ^ mS 1S alt sether ignored. The rule takes no account ^ +L P r meth d adopted of construction of joints. The -sixth simply covers the actual nominal factor of safety, 5 as well as the loss of strength at the joint, no matter what the per- centage; we may therefore dismiss it as unsatisfactory-" Plat Plates The Board of Trade rules for flat surfaces, being based on actual experiment, are especially worthy of respect; sound judgment appears also to have been used in framing them, and will be the only ones given in this work. Board of Trade: C (t + 1) 2 P = S 6 p = working pressure in Ibs. per sq. in. S = surface supported in sq. in. t = thickness in sixteenths of an in. C = constant. C = 125 for plates not exposed to heat or flame, the stays fitted with nuts and washers, the latter three times the diameter of the stay and two-thirds the thickness of the plate. C = 187.5 for the same conditions, but the washers two-thirds the pitch of the stays, and thickness the same as the plate. C = 200 for the same conditions, but doubling plates in place of washers, the width of which is two-thirds the pitch and the thickness the same as the plate. = 112.5 for the same condition, but the stays with nuts only. C = 75 when exposed to impact heat or flame and steam in contact with the plates, and the stays fitted with nuts and washers three times the diameter of the stay and two-thirds the plate's thickness. = 67.5 when the same condition exists, but the stays fitted with nuts only. = 100 when exposed to heat or flame and water in contact with the plates, and stays screwed into the plates and fitted with nuts. C = 66 for the same condition, but stays with riveted heads. Furnace Formula: Board of Trade. Long Furnaces: Ct 2 P = , but not where L is shorter than (ll.St 1), at (L + l) D which length the rule for short furnaces comes into play. P = working pressure in Ibs. per sq. in.; t = thickness in inches; D = outside diameter in inches; L = length of furnace up to 10 ft.; = constant as per following table for drilled holes: = 99000 for welded or butt-jointed with single straps, double riveted; C = 88000 for butts with single straps single riveted; = 99000 for butts with double straps single riveted. Provided, however, that the pressure so found does not exceed that given by the following formula, which applies also to short furnaces: Ct P = for all patent furnaces named; D Ct / 12IA P = 15 I when with Adamson rings. 3D\ 67.5t/ = 8800 for plain furnaces; = 14000 for "Fox"; minimum thickness 5/16 inch, greatest 5/8 inch, plain part not to exceed 6 inches in length; = 13500 for "Morrison;" minimum thickness 5/16 inch, greatest 5/8 inch, plain part not to exceed 6 inches in length; = 14000 for "Purves-Brown;" limit of thickness 7/16 and 5/8 of flange next nre 1* inches. U. S. Statutes. Short Furnaces: Plain and Patent. When less than 8 ft. in length: 89600t* LD tc P = - when C = 14000 for Fox corrugations where D equals mean diameter. C = 14000 for Purves-Brown where D equals the diameter of the 6 = 5677 for plain flues over 16 inches in diameter and less than 40 inches, when not over 3 ft. in length. Material for stays: Board of Trade. The tensile strength to lie within the limit of 27 and 32 tons per square inch, and to have an elongation of not less than 20 per cent in 10 inches. Steel stays, which have been welded or worked in the fire, should not be used. Lloyd's. 26 to 30 tons, steel, with elongation not less than 20 per cent in 8 inches. U. S. Statutes. The only condition is that the reduction in area must not be less than 40 per cent if the test bar is over % inch in diameter. Safe loads on stays: Board of Trade. 9000 Ibs. per square inch is allowed on the net section, provided the tensile strength range from 27 to 32 tons. Steel stays are not to be welded or worked in the fire. Lloyd's. For screwed and other stays not exceeding IV 2 inches in diameter, effective, 8000 per square inch is allowed; for stays above 1% inches 9000 Ibs. No stays are to be welded. U. S. Statutes Braces and stays shall not be subjected to a greater stress than 6000 Ibs. per square inch. Q. 8. (1900-01.) ^ive rules for tube plates, finding thickness and distance between tubes; that is, how much material should properly be left between the tubes? Give rules for establishing the value of material for boiler tubes. Would you allow for the holding power of boiler tubes due to fric- tion between the outer surface of the tube and the surface of the hole in the tube sheet? What do you consider as the best form of hole through the tube What would be the relation in efficiency between a tube simply ex- panded in and one expanded in and beaded? That is, as regards the holding power of tube. Ans. 8. Tube Plates: Board of Trade rule: t (D d) X 20000 WD D = least horizontal distance between centers of tubes in inches; d = inside diameter of ordinary tubes; t = thickness of tube plate in inches; W = extreme width of combustion-box in inches from front tube- plate to back of fire box, or, distance between combustion box tube plates when boiler is double-ended and the box common to both ends; P = pitch of tubes; 52 The thickness of tube plates is generally one-eighth of an inch in excess of the sheet forming the shell of the boiler. Material for boiler tubes: If of iron, the quality to be such as to give at least 22 tons per sq. in. as the minimum tensile strength, with an elongation of not less than 15 per cent in 8 inches. If of steel, the elongation to be not less than 26 per cent in 8 inches for the material before being rolled into strips; and after tempering, the test bar to stand completely closing together. Provided, the steel welds well, there does not seem to be any objection in providing tensile limits. The ends should be annealed after manufacture and stay-tube ends should be annealed before screwing. The holding power of tubes: Experiments made in the Washington Navy Yard show that, with 2%-inch brass tubes, in no case was the holding power less than 6000 Ibs., while the average was upwards of 20000 Ibs. It was further shown that with these tubes, nuts were superfluous, quite as good results being obtained with tubes simply expanded into the tube-sheet and fitted with a ferrule. In five experiments with steel tubes 2 inches and 2 1 / 4 inches in diameter, the first five tubes gave way on an average of 23740 Ibs., which would appear to be about 2/3 the ultimate strength of the tubes themselves. In all these cases the hole through the tube-plate was parallel with a sharp edge to it, and a ferrule was driven into the tube. Another test of five steel tubes made under the same conditions as the first five, with the exception that the ferrule was omitted, the tubes simply being expanded into the plate. The mean pull required 15270 Ibs., or considerably less than half the ultimate strength of the tubes. The effect of beading the tubes, the holes through the plate being parallel and ferrules omitted. The mean of the first three, which are tubes of the same kind, gives 26876 Ibs. for the holding power, under these conditions, as compared with 23740 Ibs. for tubes fitted with ferrules only. This high figure is, however, due mainly to an excep- tional case when the holding power is greater than the average strength of the tubes themselves. It is disadvantageous to cone the hole through the tube-plate unless the sharp edge is removed, as the results are much worse than those obtained with a parallel hole. In experiments on tubes expanded into tapered holes and simply beaded over, better results were obtained than with ferrules; in these cases, however, the sharp edge of the hole was rounded off, which appears in general to have a good effect. Experiments by Yarrow & Co.: In fifteen experiments on 4 and 5-inch steel tubes, the strain ranged from 20720 to 68040 Ibs. Beading the tubes does not necessarily give increased resistance, as some of the lower figures were obtained from beaded tubes. Q. 9. (1900-01.) What controls the diameter of rivets, and what is the extreme limit of their pitch? What should be the thickness of double butt straps (each) and what should be the thickness of single butt straps? What should be the distance from edge of plate to the center of rivet holes? What should be the distance between the rows of rivets when chain riveted and when zig-zag riveted? Ans. 9. For Single-Riveted Plates. Lap Joint: The diameter of the hole should be two and one-third (21-3) times the thickness of the plate, and the pitch of the rivet two and three- 53 eighths times the diameter of the hole, making the mean plate area 71 per cent of the rivet area. For double-riveted lap-joints: The ratio of diameter to thickness remains the same as in single- riveted lap-joints; while the ratio of pitch to diameter of hole should be 3.64 for 30-ton plates, and 22 and 24 ton rivets, and 3.82 for 28-ton plates with the same rivets. The distance from the edge of plate to center of rivet hole should be 1% diameters. The thickness of double butt straps should be at least % of the thickness of the plate, and for single-butt straps the thickness should be 1% times the thickness of the plate. "Kent" gives the following formula for the pitch: Single riveted plates P = .571 h a d 1 Double riveted plates r = 1.142 f- d t P = pitch of the rivets; d== diameter of hole; t = thickness of plate; The co-efficients .571 and 1.142 agree closely with those given in the report of the committee of the Institution of M. B. Distance between rows of rivets in chain riveting = [(diam. X 4) +1] D = 2 X diam. of rivet, or 2 Zigzag = ; D= V [(pitch X 11) + (diam. X 4)] X (pitch + diam. X 4) (pitch X 6 + diam. X 4) Diagonal pitch = 10 Note. The subject of riveting, in its many forms of joints and con- ditions, is exhaustless, and to do proper justice to the subject is not within the province of this work, and to those seeking a thorough knowledge on this subject, the standard authorities should be consulted. Q. 10. (1900-01.) What is your opinion of iron versus steel boiler tubes? Give rule for finding allowable. pressure on bumped heads of boil- ers. Ans. 10. A good grade of charcoal iron makes the best boiler tube. Mild homogenous steel is used to a great extent and makes a very good tube. If wrought iron is used, it should have a tensile strength of not less than 45000 pounds per sq. in., and an elongation of 15 per cent in 8 inches, and after tempering the test bar should stand completely closing together. Experiments seem to indicate that, so far as leakage is concerned, iron is preferable because it is not subject to the same degree of ex- pansion and contraction as steel. Bumped Heads: In the construction of bumped heads for boilers, in order that the head should have the same strength as the shell, the head should be bumped; that is, the spherical part of the head should be curved to a radius equal to the diameter of the boiler. Should a larger radius be used the tendency would be to weaken the boiler The nearer hemispherical the head is the stronger it is. 54 Rules for allowable pressure: Multiply the thickness of the plate by 1/6 of the tensile strength and divide this product by 6/10 of the radius, to which the head is bumped, which will give the pressure per sq. in. allowable. Q. 11. (1900-01.) What should be the tensile strength (T. S.) elon- gation and contraction of area of the materials for rivets? What shearing resistance per square inch, and what factor of safety should be used for steel rivets? What difference should be allowed in the calculations between rivets in single and double shear? Ans. 11. Rules Connected with Riveting: Board of Trade. The shearing resistance of the rivet steel to be taken at 23 tons per sq. in., 5 to be used as a factor of safety independ- ently of any addition to this factor for plating. Rivets in double shear to have 1.75 times the single section taken in the calculation instead of 2. The diameter must not be less than the thickness of the plate and the pitch never greater than 8% inches. Lloyd's. The shearing strength of rivet steel to be taken at 85 per cent of the tensile strength. The tensile strength of rivet bars between 26 and 30 tons, with an elongation in 10 inches of not less than 25 per cent and a contraction in area not less than 50 per cent. Q. 12. (1900-01.) Give rules for proportioning the areas of flues, tubes and other gas passages for both anthracite and bituminous coals. For air passages in grate bars. Ans. 12. For anthracite coal 1/9 to 1/10 of the grate surface. For bituminous coal 1/6 to 1/7 of the grate surface. The tube or flue area should be of sufficient area so as not to impede the passage of the gases and not impair the efficiency of the boiler. If too large, there is a tendency for the gases to select passages of the least resistance, escaping without wholly giving up their heat, also impairing the efficiency of the boiler. The grate bars are usually constructed with an allowance of 45 to 55 per cent air space. Some forms of construction require more air space than others. Q. 13. (1900-01.) How is the working pressure of a boiler calcu- lated from the pressure of the usual hydrostatic test? What are the several rules used in establishing the nominal factor of safety in boiler construction? Ans. 13. The hydrostatic test, as applied to boilers, is usually 1% times the working pressure of the boiler. Nelson Foley proposes that the proof pressure should be 1% times the working pressure plus one atmosphere (15 Ibs.). The rules for finding factor of safety are somewhat conflicting. The factor of safety equals the bursting pressure of the boiler as figured by the rules before given, divided by the working pressure for which the boiler is built. The factor is usually considered in connection with the tensile strength of the material, the character of the joint and the workman- ship, and range from 3% per cent upward, according to conditions. Q. 14. (1900-01.) The term "horse-power" as applied to steam boil- ers, means the capacity of a boiler to evaporate 30 pounds of water from and at a temperature of 100 F. into steam of 84.7 pounds abso- lute pressure per square inch. 55 What should be the average proportion for maximum economy, hand firing, good anthracite coal, of Heating surface per horse power? Grate surface per horse-power? Ratio of heating to grate surface? Water evaporated from and at 212 F. per square foot of heating surface per hour? Combustible burned per horse-power per hour? Combustible burned per square foot of grate surface per hour? Coal with 16 2/3 % refuse in pounds per hour? Coal with 16 2/3 % refuse in pounds per hour per square foot of grate surface? Water evaporated from and at 212 F. per pound ot combustible? Water evaporated from and at 212 F. per pound of coal, 16 2/3 % refuse? Ans. 14. Steam Boiler Proportions (as per conditions of Question 15). Heating surface per horse-power, 11.5 sq. ft. Grate surface per horse-power, 1/3 sq. ft. Ratio of heating to grate surface, 34.5 to 1. Water evp'd from and at 212 per sq. ft., H. S. per hour, 3 Ibs. Combustible burned per H. P., per hour, 3 Ibs. Coal with 16 2/3 % refuse, Ibs. per hour, 3.6 Ibs. Per sq. ft. grate surface, 10.8 Ibs. Combustible burned, per sq. ft. grate surface, 9 Ibs. Water evaporated from and at 212 per Ib. comb., 11.5 Ibs. Water evaporated from and at 212 per Ib. coal, 9.6 Ibs. Q. 15. (190U-01.) Specify the thickness of sheets or plates, style of seams, braces, size of rivets, pitch, etc., together with the diameter and length of shell, the diameter and number of tubes, for a horizontal tubular boiler, builders rating 150 horse-power, maximum working pressure to be 120 pounds per square inch, gauge pressure. Which is advisable to use, a drum or nozzle? Ans. 15. Under conditions of the questions: A boiler 72 inches in diameter, 17 feet long, 132 3-inch tubes; heat- ing surface 1800 sq. ft. The sheets % inch in thickness of "open hearth fire-box steel," with tested tensile strength of 60000 Ibs. per sq. in. of section. Horizontal seams "triple-riveted double butt-joint;" holes drilled 1 inch in diameter; rivets 15/16 inch in diameter; pitch of rivets 3% inches by 7% inches. The efficiency of joints, 86.6 per cent. Through bracing from head to head. Heating surfaces: 6 X 3.1416 X 17 One-half shell = = 160.21 sq. ft. 2 Area of heads = 37.7 sq. ft. Area of tube ends in both heads equals 6.48 X 2 = 12.96 square feet. Deducted from head area equals: 37.68 12.96 = 24.72 sq. ft. heating surface on heads. The inner diameter of 3-inch tube equals 2.78 inches. Heating surface per ft. in length equals 2.78 X 3.1416 -^ 144 = .7283 sq. ft. per foot length. Sq. ft. heating surface in each tube equals .7283 X 17 = 12.38 sq. ft. Total heating surface in tubes equals 12.38 X 132 = 1634.16 sq. ft Total heating surface of the boiler: Shell, 160.21 sq. ft; heads, 24.72 sq. ft; tubes, 1634.16 sq. ft total 1819.09 sq. ft. of heating surface. The nozzle is used in preference to the dome. The steam dome has 56 a tendency to weaken the shell of the boiler and has not proved of any real value in providing dry steam for the engine. Q. 16. (1900-01.) Give a description of what you consider a first- class typical boiler setting for a horizontal tubular boiler of given dimensions, with total area and height of chimney included (for natural draft). Ans. 16. Boiler Setting. Return Tubular: In a boiler setting, three things are to be obtained: First A firm support for the boiler shell; this of course includes foundations, walls and all necessary supports. Second A properly proportioned and arranged ash-pit, furnace and combustion chamber. Third A protected covering for the boiler, which shall, as far as possible, prevent the loss of heat by radiation. The Hartford Boiler Insurance Company describes the setting of a 60-inch horizontal return-tubular boiler as follows: "The foundation is heavy stone work laid to the depth of three to four feet below the surface. Upon this the brick wbrk is laid. "The side and rear walls are double, with a two-inch air space between the inner and outer walls; there are projecting brick laid in these walls, simply to steady the brick work, but not to interfere with the expansion of the inner wall, caused by the extreme furnace heat. "The inside wall next to the furnace, and hot gas passage is faced with fire brick. "The bridge wall in some instances is built wholly of fire brick; in other cases faced. "The boiler is supported by cast-iron lugs, riveted to the shell. These lugs rest upon iron plates placed upon the tops of the side walls. "The front lugs rest directly upon the plates, while the back lugs rest upon rollers of one inch round iron allowing free expansion and contraction of the boiler. "The rear wall is 24 inches from the rear head of the boiler, allow- ing ample space for the gases to enter the tubes; above the tubes, how- ever, the wall is built in to meet the head, and forms a roof for the chamber. "The rear wall is provided with a door, to remove the dirt and soot that collects back of the bridge, and also to provide means for inspection. "For anthracite coal the grate is placed 24 inches below the shell. For bituminous 28 to 30 inches. "The grate has a fall of three inches from front to rear, so that the fuel is thicker near the back end of the fire; this is believed to lead to more even combustion, since the air has naturally a greater ten- dency to pass through the fire nearest the bridge, and upon meeting a thick bed of coals its passage is somewhat retarded. "The end of the boiler which contains the man-hole or hand-hole should be set one inch lower than the other end; this aids the flow of the water and sediment toward the man-hole through which it can be removed. "The brick-work is closed into contact with the shell at the level of the center of the upper row of tubes; this prevents the gases coming in contact with the plates above the water-line. "A safe rule is, Never expose to fire or gases of combustion any part of the shell not completely covered with water. "The brick work is strengthened by buck-staves held together by tie-rods. The buck-staves are of wrought iron, channel or angle irons. "In the matter of covering: the tops of boilers and other portions 57 of the surface not in contact with furnace gases should be covered with some non-conducting substance to prevent the radiation of heat. "A chimney 30 inches in diameter, or its equivalent area, if square (V06.9 sq. in.), with a height of 90 feet, will be ample for a boiler of "The location of a chimney oftentimes should govern the height. A chimney should have height enough so that the draught should not be impaired by surrounding objects." Q. 17. (1900-01.) Give full detailed description of how you would conduct a boiler test, accompanied by a report, either genuine or fic- titious, hand firing. Give rule for finding the factor of evaporation. Ans. 17. Boiler Tests First In preparing or conducting trials of steam boilers the specific object of the proposed trial should be clearly defined and steadily kept in view. Second Measure and record the dimensions, positions, etc., of grate and heating surfaces, flues, chimneys, proportion of air space in the grate surface, kind of draught, natural or forced. Third Put the boiler in good condition, have heating surface clean inside and out, grate-bars and sides of furnace free from clinkers, ashes and dust removed from back connections, all leaks in masonry stopped and all obstructions to draught removed. That the damper will open to full extent, and that it will close when it is desired. Fourth Have an understanding in regard to the character of coal to be used. The coal must be dry; if wet a sample must be dried carefully and the per cent of moisture be obtained, to correct finally the results of the test. Fifth In all important tests a sample of coal should be selected for chemical analysis. Sixth Establish the correctness of all apparatus used in testing, weighing or measuring. These are: (1) scales for weighing; (2) tanks, or water meters (water meters as a rule should be used only as a check on other measurement) ; for accurate work the water should be weighed or measured in a tank; (3) thermometers and pyrometers for taking temperatures of air and steam, feed water, waste gases, etc.; (4) pressure gauges, draught gauges, etc. Seventh Before beginning the test, the boiler and chimney should be thoroughly heated to their usual working temperature. If the boiler is new, it should be in continuous use at least one week before testing, so as to dry the mortar thoroughly and heat the walls. Eighth Before beginning a test all superfluous pipes and connec- tions should be disconnected (including the blow-off), or stopped with a blank flange, unauthorized opening of valves should be guarded against. If an injector is used, it must receive the steam directly from the boiler under test. See that the steam pipe is so arranged that the water from condensation cannot return to the boiler. If necessary, it must be trapped. Starting and Stopping a Test: A test should last at least ten hours of continuous running and longer if practicable. The conditions of boiler and furnace in all re- spects should be, as nearly as possible, the same as at the beginning of the test. The steam pressure and the water-level kept the same, the fires should be kept as near as possible the same; in fact, all con- ditions as enumerated kept the same. Standard Method: Is to pull the fires, the steam being raised to the proper pressure, and the water-level at the proper point. Close damper and clean ash- pit, etc. Rebuild with weighed wood and coal, noting the time and 58 all conditions of water and steam. At the end of the test remove the fire and ashes as at the beginning of the test, and make notes of all conditions, etc. An Alternate Method: Clean the fires, and note all conditions as in the standard method, and at the end of the test the fires should be burned low, as at the beginning of the test and all other conditions observed. During the test: "A." Keep all conditions uniform; the boiler should be run con- tinuously, without stopping for meal-times or for rise or fall of pres- sure of steam due to the change of demand for steam. The draught being adjusted to the rate of evaporation or combustion desired before the test is begun, it should be retained constant during the test by means of the damper. If the boiler is not connected to the same steam pipes with other boilers, an extra outlet for steam with valve in same, should be pro- vided, so that in case the pressure should rise to that at which the safety-valve is set, it may be reduced to the desired point by opening the extra outlet without checking the fires. If the boiler is connected to the same steam pipe with other boilers, the safety-valve on the boiler being tested should be set a few pounds higher than those on the other boilers, so that in case of a rise in pressure the other boilers will blow off, and the pressure be reduced by closing their dampers, allowing the damper on the boiler to remain open. Conditions must be kept uniform. Should (owing to the character of the coal) the fires need cleaning during the test, the time and all conditions should be noted. Keeping the Records: The coal should be weighed and delivered to the fireman in equal portions, each sufficient for one hour's run, and a fresh portion should not be delivered until the previous one has all been fired, noting the time required to consume each portion. It is desirable that at the same time that the amount of water fed into the boiler should be accurately noted and recorded, the height of the water, and steam pressure, and temperature of the feed water (average). By thus recording the amount of water evaporated by each suc- cessive portions of coal, the record of the test may be subdivided into divisions, if desired, and at the end of the test to discover the degree of uniformity of combustion, etc., at the different stages of the test. Priming Tests. Calorimeter tests should be made to ascertain the percentage of moisture in the steam or of the degree of superheating. At least ten such tests should be made, and the greatest care should be taken in the measurements of weights and temperature. Analysis of Gases. Measurements of air supply. For commercial purposes are not necessary, only in cases of scientific research. These are the measurements of the air supply, the determination of its contained moisture, the measurement and analysis of the flue gases, the determination of the amount of heat lost by radiation, of the amount of infiltration of air through the setting, the direct deter- mination by calorimeter experiments of the absolute heating value of the fuel and (by condensation of all steam made in the boiler) of the total heat imparted to the water. The analysis of the flue gases is an especially valuable method of determining the relative value of different methods of firing or of different kinds of furnaces. The final results should be recorded upon a properly prepared blank and should include all the items as are adapted for the specific object for which the test is made. 59 REPORT OP ACTUAL BOILER TEST. Credited to "Mass. No. 17, Lowell," under A. S. M. E. Code, Stand- and Wilcox boiler, with B. & W. regular setting plain used, "Georges Creek," Cumberland, of good quality. Grate surface, 67.6 sq. ft. Water heating surface, 3,195 sq. ft. Builders' rating, 319.5 H. P. TOTAL QUANTITIES. 1. Date of trial, May 30, 1899. 2. Duration of trial, 12 h. 3. Weight of coal fired, including equivalent wood, 14,966 pounds. 4. Percentage of moisture in coal, 3.1 per cent. 5. Total weight dry coal consumed, 14,537 pounds. 6. Total ash and refuse, 1,165 pounds. 7. Percentage of ash and refuse, 8.1 per cent. 8. Total weight of teed water, 137,266 pounds. 9. Water actually evaporated, corrected for moisture in steam, 136,580 pounds. 9a. Factor of evaporation, 1.1913. 10. Equivalent water evaporated into dry steam from and at 212 F., 162,708 pounds. HOURLY QUANTITIES. 11. Dry coal consumed per hour, 1,211 pounds. 12. Dry coal per sq. ft. grate sur., per hour, 17.9 pounds. 13. Water evaporated per hour, corrected for quality of steam, 11,382 pounds. 14. Equivalent evaporation per hour from and at 212 F., 13,565 pounds. 15. Equivalent evaporation per hour from and at 212 F., per sq. ft. water heat surface, 4.25 pounds. AVERAGE PRESSURE, TEMPERATURE, ETC. 16. Steam pressure by gage, 91.9 pounds. 17. Temperature of feed water entering boiler, 66 F. 18. Temp, of escaping gases from boiler, 386 F. 19. Force of draught between boiler and damper, .7 inch. 20. Percentage of moisture in steam, 5 per cent. HORSE-POWER. 21. Horse-power developed, 393 H. P. 22. Builders' rating H. P., 319.5 H. P. 23. Percentage of builders' rating developed, excess, 23 per cent. ECONOMIC RESULTS. 24. Water apparently evaporated under actual conditions per pound of coal as fired, 9.15 pounds. 25. Equivalent evaporation from and at 212 F., per pound of coal as fired, 10.85 pounds. 26. Equivalent evaporation from and at 212 F., per pound of dry coal, 11.19 pounds. 27. Equivalent evaporation from and at 212 F., per pound of combustible, 12.17 pounds. EFFICIENCY. 28 (assumed). Calorific value of dry coal, per pound, 14,000 B. T. 29 (assumed). Calorific value of the combustible per pound 15134 B. T. U. 60 30. Efficiency of boiler (based on combustible), 77.6 per cent. 31. Efficiency of the boiler, including grate (based on dry coal), 77.1 per cent. COST OF EVAPORATION. 32. Cost of coal per ton (2,240 pounds), $3.40. 33. Cost of coal required to evaporate 1,000 pounds of water from and at 212 F., $0.136. The Factor of Evaporation is the difference between total heat of the steam at the observed pressure and the total heat of the feed water at the observed temperature, divided by 965.7. H-h Formula. = Factor of evaporation. 965.7 H = the total heat of steam at the observed pressure, h = the total heat of feed water at the observed temperature. 965. 7 = Latent H. U. in steam, atmospheric pressure. Q. 18. (1900-01.) If your test should show that you were evapo- rating fourteen pounds of water per pound of coal, or six pounds of water per pound of coal, would you regard these results with sus- picion, and if so, why? Ans. 18. In the common forms of horizontal tubular land boilers and water-tube boilers, with ample horizontal drums, supplied with water free from substance likely to cause foaming, the moisture in the steam does not usually exceed 2 per cent, unless the boiler is over- driven or the water level carried too high. If, in an evaporative test, it was found that fourteen pounds of water was being evaporated per pound of coal, the result would be deemed erroneous, and another test should be made, subject to a cal- orimetric correction for the unprecedented amount of moisture in the steam. There are not sufficient number of H. U. in coal to evaporate 14 pounds of water per pound of coal. On the other hand, there might be losses enough to reduce the evaporation down to 6 Ibs. per pound of coal, which should be sought out and corrected at once. Q. 19. (1900-01.) What is the difference between incrustation and corrosion? Define, and give causes for each; also good practice with respect to prevention of each. Ans. 19. Incrustation is due to the presence of salts in the feed water (carbonates and sulphates of lime and magnesia for the most part), which are precipitated when the water is heated, and form hard deposits upon the boiler plates. Corrosion (internal): In marine boilers corrosion is generally due to the combined action of salt water and air when under steam, and when not under steam to the combined action of air and moisture upon the unprotected surfaces of the metal. Other deleterious influences are at work, such as the corrosive action of fatty acids, the galvanic action of copper and brass, and the inequalities of temperature; these latter are considered of minor importance. There is a condition where a boiler is badly scaled, the high tem- perature required to form steam predisposes the material forming the boiler to oxidize rapidly. The higher the temperature at which iron or steel is kept, the more rapidly it oxidizes, and at a heat of 600 F. it soon becomes gran- ular and brittle, and is liable to bulge, crack, or otherwise give way from the internal pressure. As a preventative from scale formation, no general rule can be given; it is only through a careful analysis of the feed water that the proper antidote is found. There are also mechanical appliances known as "live steam purifiers," by the use of which the impurities are pre- cipitated and blown off before reaching the boiler proper. When the quantity of these salts is not very large (say 12 grains per gallon) scale preventatives may be found effective. The chemical preventa- tives either form with the salts, other salts soluble in hot water, or precipitate them in the form of soft mud, which does not adhere to the plates, and can be blown and washed out at times. The selection of the chemical must depend upon the analysis and be introduced regularly with the feed. The preventation of corrosion in a boiler is a subject upon which there is a diversity of opinion. A boiler in use, especially where re- turns from heating systems and otherwise, or where pure water is used which has a tendency toward oxidization, should be mixed with a certain quantity of the regular service water; so_ne propose that a boiler after cleaning should be coated with some compound as a wash of Portland cement, to form a thin hard scale to protect the iron from corrosive action, etc., etc. Boiler laid up in ordinary. There are also different opinions as to which is the proper mode of procedure. If a boiler could be thoroughly cleansed and dried and kept dry, it would probably be as good a way as any, but in connection with other boilers, there is more or less vapor from leaky stop valves and it is impossible to keep all parts of the boiler in a dry condition; in a case of this kind I would suggest the following mode of procedure: The boiler being cleaned, fill with water and attach a ^-inch steam pipe at the highest point and keep the boiler under pressure at all times. Q. 20. (1900-01.) Would you regard a boiler compound offered for universal use as safe to use? What may be considered as a proper preventative under any and all possible conditions? Ans. 20. A boiler compound offered for universal use should be condemned. As a proper preventative to be used at all times and under all circumstances; again there is a divergence of opinion. Generally the usual substance in water can be retained in soluble form, or be precipitated as mud by using caustic soda or lime. This is especially desirable when the boilers have small interior spaces. Some of the products of petroleum are highly advocated by some, others condemn it. Kerosene (refined) contains more or less of the acids used in the refining process, yet, still it is recommended. Crude petroleum of the grade known as rock oil is probably the best product of the petroleum series; the grade known as surface oil should be barred, as it contains explosive gases, and its tarry con- stituents are apt to form a spongy incrustation. Care should be taken on opening up a boiler or its connections at any time when these oils have been used, to guard against explosions of gas. Lights, cigars and pipes should be left in the background and the boiler have ample time to air out. Q. 24. (1900-01.) Assume a coal, either anthracite or bituminous, having 5 per cent moisture, 15 per cent refuse and the balance com- bustible capable of being transformed into C 2 at the rate of 14,544 B. T. U. per pound. Assume, also, that one horse-power requires 2545 heat units per hour, how many pounds of said coal will be required r 1200 horse-power per day of twelve hours? No allowance being made for banking fires and waste due to cleaning, etc. 62 Ans. 24. 100 (15+5) =80% the thermal efficiency of the coal. 14544 X 80 = 11635.2 H. U. in 1 pound of the coal. 2545 X 1200 X 12 = 36,648,000 heat units required under conditions of question for 1200 H. P. for a run of 12 hours. 36,648,000-^11635.2 = 3149.7525 pounds of coal required per day of It hours for 1200 H. P. This amount of coal consumed of course looks ridiculous, and the only assumption would be that this consumption of coal was based upon a theoretically perfect boiler, and an engine working to the absolute zero with no condensation or radiation losses. Should we assume that this engine working 1200 H. P. for day of 12 hours had a thermal efficiency of 9 %; Then will 36,648.000 H. U. equal 9 % of the total heat-units in the entire amount of coal consumed; Then will 100 %, or total consumption of coal equal 36,648,000 X 100 =34,997 pounds =17.49 tons of coal of 2,000 Ibs. per 9 ton, required for 1200 H. P. for a day of 12 hours. Q. 27. (1900-01.) What is the common cause of smoke and what is smoke called when deposited on solid bodies? Ans. 27. The ingredients of every kind of fuel commonly used may be classed: First Fixed or free carbon, which is left in the form of charcoal or coke after the volatile ingredients of the fuel have been distilled away. These ingredients burn either wholly in the solid states (CtoCOa), or part in the solid state and part in the gaseous state (CO + O = COs), the latter part being first dissolved by previously formed car- bonic acid by the reaction, C O 2 + C = 2 C O. Carbonic oxide, C O, is produced when the supply of air to the fire is insufficient. Second Hydro-carbons, such as oleflant gas, pitch, tar, naphtha, etc., all of which must pass into the gaseous state before being burned. If mixed on their first issuing from amongst the burning carbon with a large quantity of hot air, these inflammable gases are com- pletely burned with a transparent blue flame, producing carbonic acid and steam. When mixed with cold air they are apt to be chilled and pass off unburned. When raised to a red heat, or thereabouts, before being mixed with a sufficient quantity of air for perfect combustion, they disengage carbon in fine powder, and pass to the condition partly of marsh gas and partly of free hydrogen, and the higher the tem- perature the greater is the proportion of carbon thus disengaged. If the disengaged carbon is cooled below the temperature of igni- tion before coming in contact with oxygen, it constitutes, while float- ing in the gas, "smoke," and when deposited on solid bodies "soot." Q. 28. (1900-01.) What occurs when the disengaged carbon is maintained at the temperature of ignition and supplied with sufficient oxygen for its combustion? Ans. 28. If the disengaged carbon is maintained at the tempera- ture of ignition and supplied with oxygen sufficient for its combustion, it burns while floating in the inflammable gas, and forms red, yellow or white flame. The flame from the fuel is the larger the more slowly its combustion is effected. The flame itself is apt to be chilled by radiation from the heating surface of a steam boiler, so that the com- bustion is not completed, and part of the gas and smoke pass off. Q 29 (1900-01.) Does the flame from the fuel come into contact with' the fire sheets of a horizontal tubular boiler, or the outer surface of the tubes of a water tube boiler, or does it penetrate into the tubes of any type of fire tube boilers? Ans 29. In the use of wood and similar class of fuels there is no doubt but what the flame comes into contact with the fire sheets of the boiler, also the tubes and sometimes the stack. In the use of coal (anthracite and semi-bituminous) the flame proper is hardly long enough to either strike the sheets or penetrate the tubes. In mixing the hydro-carbons with the necessary amount of air the product of perfect combustion takes place and the flame encircles the exposed sheets, penetrates the tubes, especially in case of hard driven toilers. Q. 30. (1900-01.) Can secondary combustion take place in the back connections or flues? Ans. 30. As we have been taught that combustion is the chemical combination of the constituents of the fuel, mostly carbon and hydro- gen, with the oxygen of the air. The nitrogen in the air remains inert and causes loss of useful effect to the extent of the heat it carries off through the chimney. The hydrogen combines with the proportional part of oxygen to form water which passes off as steam. The carbon combines with enough oxygen to form perfect combus- tion, or with only enough to form imperfect combustion. To insure perfect combustion the sufficient quantity of air having been admitted and properly mixed with the fuel solid and gaseous, and these, the air and combustible gases should be brought together and maintained at a sufficiently high temperature. The hotter the elements the greater is the facility of good combustion. If perfect combustion ensues there is no need of secondary combus- tion in the back connections and flues. If imperfect combustion takes place, then a portion of the mixed air and gases pass off unconsumed; now, unless a proportional amount of air can be admitted and mixed with this product, and the tempera- ture raised to and above the point of ignition secondary combustion will not take place. Secondary combustion is desirable, and there has been various devices and settings introduced at various times to accomplish this result, but the gain has not warranted their contin- uance or extended use. Q. 31. (1900-01.) What is the comparative value of coal and oil as fuel from the evaporation standpoint, also from the standpoint of economy in cost? Ans. 31. In 1892 there were reported to the Engineers' Club, Phila- delphia, some comparative figures from tests undertaken to ascertain the relative value of coal and oil as a fuel. 1 Ib. anthracite coal evaporated from and at 212 F., 9.70 Ibs. water. 1 Ib. bituminous coal evaporated from and at 212 F., 10.14 Ibs. water. 1 Ib. oil evaporated from and 212 F., 16.48 Ibs. water. Taking the efficiency of the bituminous coal as a basis the calorific energy of petroleum is more than 60 % greater than that of coal; whereas, theoretically, petroleum exceeds coal only 45 %, the one con- taining 14500 H. U. and the other 21000 H. U. As a result of tests made by the Twin City Rapid Transit Company of Minneapolis and St. Paul, showed that with the ordinary Lima oil weighing 6.6 Ibs. per gallon, and costing 2% cents per gallon, and coal that gave an evaporation of 7% Ibs. of water per pound of coal, the 64 two fuels were equally economical when the price of coal was $3.85 per ton of 2,000 Ibs. With the same coal at $2.00 per ton, the coal was 37 per cent more economical than the oil. With the coal at $4.85 per ton the coal was 20 % more expensive than the oil. These results include the difference in the cost of hand- ling the coal, ashes and oil. Q. 43. (1900-01.) Calculate what percentage of the energy contained in the fuel is utilized in a steam plant consisting of a boiler which evaporates eight pounds of water per pound of fuel, and an engine which consumes twenty-five pounds of steam per IHP. per hour. The fuel contains 10,000 B. T. U. per pound. Ans. 43. Pounds of coal consumed in evaporating 1 Ib. of water at 200 F. into steam at 80 Ibs. pressure = 1 -^ 8. = .125. Pounds of water consumed per H. P., 25 Ibs. Pounds of coal consumed in evaporating 25 Ibs. of water at 200 F. into steam at 80 Ibs. pressure = 25 X .125 = 3.125 Ibs. Pounds of coal consumed per hour for 100 H. P. = 100 X 3.125 = 312.5 Ibs. Heat units in pound of fuel, 10,000 H. U. Heat units contained in all the coal consumed, 10,000 X 312.5 = 3,125,000 H. U. Mechanical equivalent for one H. U., 778 foot pounds. Foot pounds of work stored in all of the coal consumed each hour = 778 X 3,125,000 = 2,431,250,000 ft. Ibs. Foot pounds of work done per hour by each I. H. P., 33,000 X 60 = 1,980,000 foot pounds. Foot pounds of work per hour by 100 I. H. P., 1,980,000 X 100 = 198,000,000 foot pounds. Losses in overcoming friction of engine, about 10 per cent, 10 % of 198,000,000 = 19,800,000 foot pounds. Total foot pounds of useful work at the engine shaft per hour, 198,- 000,000 19,800,000 = 172,000,000 foot pounds. Foot pounds lost per hour, 2,431,250,000 172,000,000 = 2,259,250,000 foot pounds. Percentage of useful work, about 92-10 per cent; 2,259,250000 2,431,250,000 = Percentage of lost work, about 90 8-10 per cent. Engines, Indictaors, Shafting, Belting, Etc. Q. 5. (1896-7.) Are liners between bearings good practice? Why? How properly adjusted? Ans. 5. Liners between bearing boxes are good practice. 1st, they will keep dust or grit from getting in the journals, prevent the caps from working loose and prevent rattling of the boxes. 2d. For large engines the weight of the top box and cap would otherwise increase the friction. Liners are adjusted as follows: After the boxes have been carefully fitted, lead wires of suitable thickness are placed between them. The cap is then screwed down hard, thereby flattening the leads. They are taken out and used as a gage to fit the liners, which should be a trifle thicker. Q. 18. (1896-7.) How much change made in valve travel by turning W off eccentric [all around]. Ans. 18. None; except that rod length will have to be altered to maintain position of valve. Q. 31. (1896-7.) What is the proper relation between cylinder and steam pipe areas for automatic cut-off engines? Ans. 31. Velocity of entering steam should not exceed 8,000 ft. per minute. Area of cyl. X piston speed Area pipe = Velocity of steam in pipe. A constant, 6,000 ft., is much used in later practice. Examples: For 8,000 ft., area pipe = area piston X .075. For 6,000 ft. area pipe area piston X .10. eccentric?^ 896 ?-) Wh&t IS meant by an & ular advance of an engine Ans. 34. If a valve has neither lap nor lead and delivers steam in te usual manner, that is, over or around the ends or outside, the eccentric will stand at an angle of 90 with the crank pin either in advance of the crank if direct connected, or behind the crank if indi- If a valve has lap and it is desired that the valve open the port for steam at the instant the crank pin moves from the dead center, or if the valve is to have lead, that is, to open for 2SS W H S 6 C v ank is on tne dead center > the eccentric must be dyanced on the shaft in the direction in which the engine is to run until the edge of the valve is either in line with the edge of the steam +1 1 th f V . alVe has given the required lead opening, the engine -^ad center. Therefore, the angle of advance or the of an eccentric is the angle (beyond ninety degrees) scentric is advanced or pushed around on the shaft to bring 66 the steam edge of the valve in line with the steam edge of the port, if the valve has lap. Or, if the valve has lead or both lap and lead, it is the angle, the eccentric is advanced (beyond 90) on the shaft to give an opening for steam, the engine in either case being on the dead center. Q. 37. (1896-7.) What is meant by angularity of the connecting rod of an engine? Ans. 37. When the crank pin of an engine stands at any point in its path of motion, except when on the dead center, the connecting rod forms an angle with the line of centers, this is termed the angularity of the connecting rod. The angle thus formed is greatest when the piston is in mid-position. By the line of centers is meant a straight line drawn from the center of the crank shaft through the center of the cylinder. The shorter the connecting rod is in proportion to the stroke of the piston, the greater the angularity of the rod. Or the angularity of the connecting rod may be described as the equivalent, but not actual, shortening of the rod. by its being deviated from the line of centers, which it is at all times except when the crank pin is on the dead centers. Q. 38. (1896-7.) Why should an engine be given compression? Name all the reasons. Ans. 38. Compression is given to an engine 1st. To furnish a cushion or gradually increasing resistance to bring the reciprocating parts of an engine to a stop at the end of the stroke and change the direction of the thrust upon them without shock, which would other- wise follow upon opening the steam valve. 2d. If obtained by early closing of the exhaust valves it fills the clearance spaces with steam that would otherwise go to waste by being blown into the condenser, or to the atmosphere. 3d. It insures quiet and smooth running of the engine, taking up the lost motion, if any, without jar. 4th. It reheats the clearance space in the cylinder [cylinder head and piston] promoting economy by thus lessening the amount of in- ternal condensation, a very important item. Q. 39. (1896-7.) What is the best practice and most accurate way for setting Corliss valves? Ans. 39. Take off caps of valve chambers; marks will be found; then set wrist plate so that central line on wrist plate coincides with line on stand ; set steam valves with 1-4" lap for 10" diameter, and 1-3" lap for 32" diameter; for intermediate diameters in proportion. Set exhaust valve 1-16" lap for 10" diameter, and 1-8" for 32" diameter. Proportion intermediate diameters on non-condensing engines; on con- densing engines give nearly double this. The rods connecting steam valve arms to dash pots should be ad- justed by turning wrist plate to extremes of travel and adjusting the rod so that when it is down as far as it will go the steel block will just clear the shoulder of the hook. Hook the engine in with the eccen- tric loose on shaft, and adjust the eccentric rod so that wrist plate will come to extremes of travel as indicated by lines. Place the engine on dead center, turn eccentric (in direction engine runs) until steam valve shows from 1-32" to 1-8" lead, owing to conditions; turn engine over to other dead center and note lead on valve; if not the same, adjust rod from wrist plate to valve. To adjust rods to cut off from governor: Have wrist plate at one extreme of travel, then adjust rod connecting the opposite cam of 67 steam valve so that cam will clear the steel tail of the hook about 1-32": turn wrist plate to other extreme and adjust the rod the same way. To equalize the cut-off, block the governor up, then turn engine over and note where cross-head is when valve is tripped; then turn on over until other valve is tripped; if not the same, adjust one or the other until they are the same. Next, start the engine and verify results obtained by an intelligent application of the indicator. Q. 44. (1896-7.) Is there a gain realized in practice from steam jacketed cylinders? If so, under what general conditions? Ans. 44. The fact that a gain may be realized by steam jacketing is as well established as statements to the contrary, the economy in any particular case depending upon design and conditions attending their application. The subject is quite too intricate to be treated satis- factorily in an abbreviated manner. Briefly stated, the function of the steam jacket is to diminish initial condension and more fully main- tain the temperature of the expanding steam through the stroke. For this purpose the jacket serves as a medium for conveying from the boiler a reinforcement of heat to the working steam, and the conditions to be observed, both in design and management, are such that will sub- serve this end. The margin of gain at the best is so small that but a slight defect in design or operation may set up conditions with a tendency just the reverse of those which the use of a jacket implies. The surplus heat stored in superheated steam performs to some extent the functions usually attributed to the jacket, consequently the benefit is less pronounced when such is used within the cylinder proper. The gain is materially greater when the heads as well as the body of the cylinder are steam jacketed. Efficiency is greater for long stroke and varies somewhat with the period of admission. For cylinders in series, compound, etc., the gain is greatest in the high pressure cylinders, since the absorption of heat during the exhaust periods is not lost as is the case of the simple engine. For the best results the design should insure a brisk circulation of "live" steam, taken directly from the boiler through all parts of the jacket, and ample provision must be made for freeing the jacket of air and the water of condensation, the latter to drain back into the boiler. Q. 51. (1896-7.) What should be the ratio between steam cylinder and air pump capacity for a condensing engine? Ans. 51. For a jet condensing engine, the single-acting pump should have a capacity of from 1-5 to 1-10 that of the cylinder; a double-acting pump would have 1-8 to 1-16 the cylinder capacity. For surface con- densing engine the single-acting pump would have a capacity of 1-10 to 1-18 and a double-acting from 1-15 to 1-25 that of the cylinder. In both cases the proportions depend upon the terminal pressure, increasing with same, and for a jet condenser varies with amount of injection used. These proportions are for pumps making same number of strokes 3 the engine. If the speed varies from that of main engine the size is so calculated as to give same relative displacement as shown above. When an engine is compounded the volume of the low pressure cylinder alone is considered. Q. 52. (1896-7.) Is brass or babbitt best for bearings? Elucidate. Ans. 52. Except in case of bearings where the pressure is excessive, where pounding and jarring is liable to occur, or unusually high speeds are attained, babbitt metal must be given the preference over brass as a material for bearings. Babbitt bearings can be the more easily kept cool and there is less liability of the journal cutting. These advantages 68 are due to its anti-friction properties and to the fact that on account of its softness it will accommodate itself to any slight irregularities in the surface of the journal, and, should the shaft get slightly out of alignment with the bearings, the metal will yield where the pressure is greatest until there is a uniform bearing surface. Another advantage is that the wear of the shaft will be less rapid, and when the bearings become worn they can be rebabbitted at a slight expense. To give best results, babbitt metal of the right formula or proportion for the work in view should be selected, as the babbitts are made of all degrees of hardness and consistency. It is our opinion that for all around usefulness there is no bearing metal that will approach babbitt. Q. 55. (1896-7.) How set the valves on a medium speed, simple, non-condensing, automatic, Buckeye engine? Ans. 55. Under the conditions of the question it is understood that the regular form of Buckeye engine is meant, and it will be necessary to bear in mind that the eccentric follows the crank instead of leading it, as in the ordinary engine, and that the steam chest is not a steam chest as commonly understood, but is an exhaust chamber or box; that is, the cylinder discharges its exhaust steam into the steam chest and takes its supply of steam through the main valve direct from the steam main. There are two valves to this engine, both of the slide valve type and both flat, and their functions are the same essentially as those of the common slide and riding cut-off. The riding cut-off in this case is placed on the inside of the main valve, which becomes a hollow box or chest through which the steam passes. The main valve has ports on its face nearest to the cylinder which alternately travel over the usual cylinder ports and then away from same far enough to allow the end of the valve to uncover the cylinder port and allow the steam to be exhausted from the cylinder into the steam chest, thus enveloping the valve and keeping it at all times in a bath of hot steam. There is a separate eccentric for each valve, one for the main valve fixed to the shaft and the other for the riding valve or cut-off, operated by a loose eccentric, which is thrown by the centrifugal force of a weight or weights around the shaft, or is, in other words, given a variable amount of angular advance. The main, or box valve, is first set independently of the cut-off by putting the eccentric 90 behind the crank, plumbing the rocker arm and equalizing the compressions just as would be done with the leads on a common slide valve, remembering that the ends of the valve are exhaust edges, and regulate the compression or exhaust and not the steam supply. The matter of equal leads in this engine is not of so much importance. The valves and cylinder faces are marked to designate the position of their respective ports, which are in themselves invisible from the outside, and the necessary lead is given by advancing the eccentric toward the crank or in the direction the engine will run, and equalized in the usual way. Thus far the engine is the same as a common slide valve type, except that the steam comes in through the valve and exhausts out of and around it. The ends being exhaust edges, the eccentric follows instead of leading the crank. Bearing this in mind, there should be no trouble in setting the valve. As to the cut-off, this is another common slide valve, much smaller and lighter than the main valve, and working entirely out of sight and inside of the main valve, being rendered visible only, and then with difficulty, by removing the back plates from the main steam chest and withdrawing the balance plates found therein. This valve is driven from a separate eccentric, fastened, not to the engine shaft, but by links to the governor wheel or frame, so that it may be revolved to the extent of about 90. Its normal position when at rest is coincident with the crank of the engine, and is held there by the tension of the springs in the governor. If the eccentric is set in the same direction as the main crank, that is all that is necessary. The governor is then set out about two-thirds of the throw of its arms and blocked, and the valve so adjusted by its rods and stems that, as shown through the sight holes, it is cutting off equally on each end. It has been assumed that the experimenter is skilled in handling and setting the ordinary slide valve. Adjustments in the case of each of the valves of this engine are made at the ends of the eccentric rods and valve stems, as usual, keeping in mind that the main valve acts just exactly opposite to the manner of an ordinary slide. Compression is equalized by moving the main valve toward the end of the cylinder showing greatest compression, and similarly the cut-off is equalized in the same manner by moving valve toward end of cylin- der showing greatest load. After the cut-off is adjusted and equalized, any adjustment for compression equalization on the main valve had better be made on main eccentric rod and not on valve stem. If made on the latter it will be again necessary to make adjustments for cut- off. It is surmised that the valves and parts have the usual shop marks for guidance and that the governor wheel and its parts are properly placed. The whole of the work should be finally proved and corrected by use of an indicator. Q. 57. (1896-7.) If we get an initial pressure of 70 Ibs. in the cylin- der and the compression runs up to 35 Ibs., both gage pressure will the clearance be half filled? Ans. 57. Yes, and more than half filled. Thus: 70 Ibs. -f 15 = 85. 35 Ibs. + 15 = 50. 85H-2 = 42y 2 Ibs. clearance, 50% filled, or 42.5-^85X100 = 50% or V 2 filled; hence, under the above conditions, we have 50 85 X 100 = 58.8%, which is more than half filled. Q. 59 (1896-7 ) Will the slide valve cut off the same at both ends of the stroke if it has equal lap and lead? hv ^ft 59 " A* ai \ engine has a connecting rod, no. If crank is turned by yoke and piston rod, yes. of ? pulley? 1896 ' 70 What are the CaUS6S f a belt runnin S to one si( le 1 ;* ^S? carrying P ullevs n t being parallel; pulleys not in . n ; ne 6dge f the belt bein S stretched more than nhr. n n S srece more an fdtrue orb^ n0t t Cr T ned correctl y: PiHeys not being bored or ed true, or belts not cut square before lacing or glueing. Ans 62 The valve must travel in each direction from its mid- travel an amount equal to the lap and width of the port, which in this case is the sum of 1 1-16" and 15-16", which is 2", thus making the valve travel 4". This may be found by the use of a valve model or any of the many valve diagrams in use. Q 68 (1896-7.) If an engine is to develop 300 H.P. with 35 Ibs. m.e.p. and 600 ft. piston speed per minute, what will be the cylinder diameter? Explain how found. Ans 68. Rule for finding the required diameter of cylinder for an engine of any given horse-power, the travel of piston and available pressure being given. Multiply 33,000 by the number of horse-power; multiply the travel of piston in feet per minute by the available pressure in the cylinder; divide the first product by the second; divide this quotient by the deci- mal .7854. The square root of this last quotient will be the required diameter of the cylinder. Example: 33000 X 300-^- (600 X 35) = 600. .7854 V 600 = 24.4949" diameter required. Q. 72. (1896-7.) What should be the terminal pressure in a steam cylinder of 40" stroke, 2^% clearance, cut-off at 1 A stroke, initial pres- sure 108 Ibs. by the gage, making no allowance for leakage, condensa- tion or wire drawing, and assuming 15 Ibs. atmospheric pressure? Ans. 72. The terminal pressure will be in the same proportion to the absolute initial pressure as the volume of the cylinder, plus the clearance at point of cut-off, is to the total cylinder volume plus the clearance. Stroke 40" + clearance (2%%) 1" = 41". Cut-off ( 1 A ) 10" + clearance 1" = 11". Total pressure (initial) 108 lbs.+ 15 lbs.= 123 Ibs. Thus: P : 123 : : 11 in. : 41 in. 123 X 11 = 33 Ibs. absolute; the answer. 41 Q. 73. (1896-7.) What is the distinction between indicated and actual steam consumption of an engine? Ans. 73. An indicator diagram does not show the actual amount of steam used by an engine for the following reasons: Leakage of steam often occurs and cylinder condensation is inevi- table, yet the extent of these losses is not revealed by any marked effect upon the lines of a diagram. The measurement of steam consumption by a diagram cannot, there- fore, be taken to show actual performance without allowance for these losses, but diagrams taken in connection with feed-water tests will reveal the extent of the losses produced by leakage and cylinder con- densation. These losses are represented by that part of the feed water consumption which remains after deducting the steam computed from the diagram, or "accounted for by the indicator," as it is termed. The loss by condensation is nearly constant for different engines working under similar conditions, and an allowance may therefore be made for its amount. The other leakage, depending upon the wearing surfaces of valves, pistons and cylinder, is variable in different cases. Q ZEUNER VAI aS resards centrifugal force is independent g H 8 ? 2 y S at 5 ft per min " an 18 ft - wheel m ay run . and at 6000 ft. per minute may run 105 R.P.M., but does not recommend such high rim speeds for cast iron wheels. norse abolf 10^ U HR atiC CUt " ff Single cylinder non-condensing engines of practice? ^ 16 expansion con densing engines of about 500 HP., in good Ans. 23. (a) About 25 to 30. (b) About 12 to 15. conden ^ng engine has a ter- 1 u recommend to'better adapt ^ T Us' wor'k? " f ^ ^^ W Uld Y U Ans. 24. Compound the engine by adding a low pressure cylinder. 80 Q. 28. (1897-8.) An engine has a twelve (12") inch stroke. Its con- necting rod is three (3) feet long. It is running at six hundred (600) revolutions per minute. The weight of its piston, piston-rod and cross- head is 250 Ibs. Draw to scale a diagram such that horizontal distances shall represent piston-positions and vertical distances, piston-velocities at the respective positions, neglecting the angularity of the connecting- rod. Ans. 28. Lay off horizontally, to a convenient scale, the larger the better, a line A, C, Fig. 2, to represent the stroke of the piston. Draw upon this line a semi-circle A, B, C. Then will points upon the horizon- tal line represent piston positions and the vertical distances above said points to the semi-circle will represent piston velocities at the corre- sponding positions of the piston. The radius B, D, of the semi-circle represents the velocity of the crank-pin, and the vertical lines at other points of the stroke velocities proportional to their lengths. FlG. 2 Q. 29. (1897-8.) In the engine of Q. 28 what is the approximate velocities of the piston at positions one and one-half (1-5") and two (2") inches from the commencement of its stroke? Explain how you find this by the diagram of Ans. 28. Ans. 29. In Fig. 2, the scale is V" to the inch. The radius of the semi-circle is therefore 1.5". The vertical line b, b, at the point which represents a position 1.5" from the commencement of the stroke, is .992" long. The velocity, therefore, at this point is found by the pro- portion 1.5 : .992 : : 31.4 : answer. By multiplying 31.4 by .992 and dividing by 1.5 we find the velocity at this point to be 20.77 ft. per second. The vertical line, c, c, at the position corresponding to 2" from the commencement of the stroke is 1.118" long. Calculating as above the proportion would be 1.5 : 1.118 : : 31.5 : 23.4 Therefore the velocity at 2" from the commencement of the stroke is 23.4 ft. per second. Q. 30. (1897-8.) With the engine of Q. 28 what is the approximate average force due to the inertia of the piston, piston-rod, and cross-head, while the piston is passing from a position' one and one-half (1.5") to a position two (2") inches from the commencement of its stroke? Solve by the use of the rule of Ans. 27 and the diagram of Ans. 28, and explain fully the method of calculation. Ans. 30. The weight of the parts is 250 Ibs. The velocity at 1.5" is 20.77 ft. per second. The velocity at 2" is 23.4 ft. per second. There- Si fore, according to the rule of Ans. 27, the work done upon the parts to change their velocity from one value to the other is 250 X [(23.4) 2 (20.77) 2 ] ~ by 64.4, that is 250 X (547.56 431.4) -=- 64.4 = 250 X 116.16 -=- 64.4 = 541 foot-pounds of work done upon the reciprocating parts to increase their velocity in W travel. Now this work divided by the distance will give the average force exerted. Thus 451 foot-pounds of work divided by one twenty-fourth of a foot equals 10,824 Ibs. as the average force due to inertia. The change of angularity of the connecting rod may be taken into account by multiplying the result by one plus the ratio between the crank and connecting rod, that is, in this instance, by one and one-sixth, [f the calculation had been made on the latter half of the stroke, the correction would have been made by multiplying by unity, minus the ratio of the crank to the connecting rod; in this instance by 7/6. Q. 34. (1897-8.) In the engine of Q. 28, if the steam is suddenly shut off while the engine is running at full speed and running over, what force would be necessary to keep the cross-head from pressing against the upper guide when the piston is one inch from the com- mencement of its forward stroke? Explain method of calculation. Ans. 34. As the steam is shut off we only have to deal with the forces of inertia. By referring to the diagram Fig. 3, we find that the force at this point, 1" from the commencement of the stroke, is 14,000 Ibs. Now draw a line a, b. Fig. 4, to represent the position of the con- necting rod. Let the line a, b, represent, by its length, 14,000 Ibs., which is the tension upon the rod. Complete the rectangular parallelo- gram a, b, c, d, then will the line a, c, represent by its length the upward effect of the tension upon the connecting rod. Thus the length of the line a, b, in the figure is 6", the length of the line a, c, is .5527". Therefore, 6 : .5527 : : 14,000 : 1.289. Therefore it would be necessary to exert a downward force of 1,289 Ibs. to keep the cross-head from pressing upon the upper guide at this point. Q. 35. (1897-8.) If the engine of Q. 28 has a cylinder 10" internal diameter and is running with a gage pressure of 70 Ibs., what is the greatest strain brought upon the connecting rod? The weight and angu- larity of the connecting rod, lead, back-pressure and compression being neglected. Ans. 35. We have the force of inertia and the steam pressure to take into account. By referring to Fig. 3 we find that there is at the commencement of the stroke a .tension of about 18,000 Ibs. on the con- ectmg rod due to the inertia of the parts; but the action of the steam to produce a compression upon the connecting rod. That is at the commencement of the stroke the steam pressure acts in the opposite direction to the force of inertia. Therefore the strain upon the con- necting rod at the commencement of the stroke is 18000 70 X 7854 (area of the 10" piston) = 18,000 5498 = 12,502 Ibs. tension upon the connecting rod. At the end of the stroke the force of inertia is 15,310 82 2552 = 12,758 a force of compression; to this should be added the final pressure of the steam upon the piston, if any. Therefore, the greatest strain brought upon the connecting rod is equal to about 12,758 Ibs. and steam pressure, if any, and is a force of compression at the end of the stroke. Q. 36. (1897-8.) Draw, to scale, a diagram of the engine of Q. 28 that shall show the distances traveled by the piston at each angular position of the crank. Ans. 36. With the compass set to a radius equal (to a convenient scale) to the connecting rod, draw the circle b, c, d, i, 1, Fig. 5. Upon the same diameter with the compass set to a radius equal to the length of the connecting rod plus the length of the crank, draw a circle b, e, f, h, k, touching the first circle at b. Draw radial lines a e, a f, a h, from the center a, to the larger circle. Then will the portion of said lines between the two circles be equal to the travel of the piston at the corresponding angular position of the crank. Thus, when the crank is in the position a, e, the travel of the piston is equal to c, e, measured by the scale adopted, and when the crank is in the position a, f, then d, f, is the travel of the piston from the commencement of the stroke. If the diagram of Ans. 36 is drawn around the same center as the Zeuner valve diagram, both the valve and piston travel will be shown for each angular position of the crank, by the same diagram. 83 Q 37 (1897-8.) What per cent of the steam is condensed by the cylinder walls in simple non-condensing, fast running engines, without steam jackets, of from 50 to 100 HP.? Ans. 37. From 25 to 50 per cent. The object of Q. No. 37 is to call attention to the great importance of cylinder condensation. The quantity of steam condensed being much too great to be accounted for by radiation from the outside of the cylin- der, and only to be accounted for upon the supposition that much of the heat goes to re-evaporating the condensed steam. Q. 41. (1897-8.) Does the fly-wheel increase the power of the engine? Ans. 41. It regulates but does not increase the power. Q. 51. (1897-8.) What should be the shape in cross-section of the connecting-rod of a fast running engine? Why? Ans. 51. The cross-section of the connecting rod in fast-running engines should be rectangular and greater in the plane of motion than perpendicular to said plane. In his paper read before the A. S. M. E. Mr. John H. Barr finds as the average of a large number of engines that the depth is 2.7 times the breadth. The depth is made greater in order to resist the bending action of the inertia of the rod itself. Q. 52. (1897-8.) Assuming a factor of safety of 20 how would you determine the diameter of a connecting rod having a circular cross- section? Ans. 52. To calculate the diameter of a circular connecting-rod may be used the following: Rule: Multiply the maximum pressure that may be brought upon the rod, in pounds, by the factor of safety, and extract the fourth root (i. e., the square root of the square root) of this product. Call this result No. 1. Extract the square root of the length of the connecting-rod expressed in inches. Call this result No. 2. Multiply result No. 1 by result No. 2 and divide by 61.75. This will give the maximum diameter of the rod in inches. This rule is expressed algebraically as follows: 1/PFVL 61.75 Suppose we take the following data for example: Area of piston 100 sq. ins. Weight of reciprocating parts 250 Ibs. Speed 250 turns per minute. Maximum steam pressure 50 Ibs. Stroke 1 ft. Connecting-rod 36" long; then the greatest pressure of the steam on the piston would be 100 X 50 = 5000 Ibs. The pressure due to inertia would be about 250 X 13 X 13 -=- 16 = 2641 Ibs. Therefore the maximum pressure to adapt the rod to is 5000 + 2641 = 7641 Ibs. Usirg a factor of safety of 20, we would have, pressure X factor of safety = 7641 X 20 = 152,820. The square root of 152,820 is about 391, and the square root of 391 is about 19.77. Therefore the fourth root of 152,820 is about 19.77. This is result No. 1. The square root of 36, the length of the connecting-rod is 6. Six times 19.77 is 118.62, and this divided by 61.75" is 1.92, say a 2" Circular connecting-rod. 84 In Mr. Barr's paper, above referred to, the average factor of safety in slow running engines with circular rods was found to have been 13. in the examples of Ans. 52 and 54, the pressure due to the inertia of the parts is taken into account. In some cases this would be mate- rial, in others it would have but little effect. It is in any case easily calculated. Cylindrical and square parts are so common in machine construction that the rules for calculating their strength are very useful. Such rules are not intended to give exact results, they are intended to tell us when a part is liable to be overloaded, and when it is safe. For this purpose they are reliable. Q. 53. (1897-8.) How are the dimensions of a connecting-rod having a rectangular cross-section determined? Ans. 53. The method of securing the ends of the connecting-rod would require that the height (H) of the cross-section should be twice its breadth (B). The inertia of the rod, its whipping action, would require a still greater relative height to breadth. Mr. Barr found the minimum H to be 2.2 B, the maximum H, 4 B, and the average H to be 2.7 B. Assuming this average relative value of H, the following may be taken as the RULE FOR CALCULATING THE DIMENSIONS OF A RECTANGULAR ROD. First. Multiply the maximum pressure, in pounas, that may be brought upon the connecting-rod, by the factor of safety, and extract the fourth root of the product. That is, get the square root of the square root of this product. Call this result No. 1. Second. Find the square root of the length of the connecting-rod (expressed in inches). Call this result No. 2. Third. Multiply result No. 1 by result No. 2 and divide this product by 127.8. This will give the breadth of the rod in inches. The height is 2.7 times the breadth. This is expressed algebraically as follows: 127 8 In which, as before, F is the factor of safety, P the maximum pres- sure that may be brought upon the rod, L the length of the rod in inches, 127.8 is a constant. Q. 54. (1897-8.) What would be the dimensions of a proper con- necting-rod for the engine of Q. 28 assuming a gage pressure of 70 Ibs.? Ans. 54. The pressure due to the inertia of the parts is 12758, the maximum pressure due to the steam would be 78.54 X 70 = 5498. There- fore the greatest pressure might be 12758 -=- 5498 18256. Assuming a factor of safety of 20 the calculation would be as follows: 20 X 18256 = 365,120. The square root of 365,120 is about 604 and the square root of 604 is about 24.58. Result No. 1. The length of the connecting-rod is 36"; the square root of this 6: 6 X 24.58 =147.48, and this divided by the constant, 127.8, gives 1.154 as the breadth of the connecting-rod; 2.7 X 1.154 = 3.1158 for the height of the cross-section. Q. 57. (1897-8.) If the pinion to the elevator should be broken, what data would you send to the factory to secure a wheel to replace it? Give specifications. Ans. 57. Diameter of wheel, diameter of shaft, diametrical pitch, width of face, width and depth of keyway. 85 PREFACE TO 1898-9 QUESTIONS AND ANSWERS. "FOUNDATIONS." The utility of good, firm and lasting foundations under a boiler, engine or other moving mechanism about a steam plant will hardly be questioned, and while some situations are attended with natural advan- tages favoring this end, it is not infrequently the case that considerable extra labor and forethought are entailed to bring about the desired Examples of poorly constructed and overloaded foundations are abundant and their general " effect too well known to require much comment; all that could be said on this subject, however, might be accepted as an apt simile of the difficulties encountered by engineers who endeavor to build '^'temples" of knowledge upon a substructure inadequate to receive it. The absurdity of placing a hundred horse-power engine on a ten horse-power foundation is striking, but there is no evident reason why this is not to be considered just as practicable as an effort to acquire the higher branches of engineering science without regard to the fundamental principles which are the natural approach to this greatly coveted goal. It is not believed that this slight of elementary principles is wholly intentional on the part of many who are seeking to place themselves on a higher plane, hence if a few pertinent suggestions will serve as a beacon to some of our workers the purpose of this article will be fully accomplished. Solutions of various engineering problems, by means of rules and formulas, do not afford much satisfaction to the engineer, who is not, in a general way at least, familiar with the process of reasoning upon which the final result depends. By a rigid application of the several operations indicated by the mathematical prescription arranged for him, a proper answer is obtained, but the "spice" of perfect confidence and assurance which should be felt in the work does not attend the result. To safely use even the most ordinary engineering formulae it is regarded as essential that a keen knowledge of ordinary arithmetic is at ready command, and further, that the application of its principles to the usual methods of measuring and computing surfaces and volumes has been mastered well. It is equally important that a precise knowledge of the points embraced in a course of natural philosophy should be well founded in an engineer's mind, and while engaged in crowding in knowledge at the "bung" it is well to see that none gets away at the "spigot." Should the conditions just noted be reversed, it is a foregone conclusion that the case is just about as hopeless as when the "cask" is claimed to be so full that it can hold no more. Among many other things that should be incorporated in an engi- neer's "foundation" of knowledge, no other is of more value than abso- lutely correct "notions" concerning the methods of loading a structure and thoroughly understanding the strains induced thereby. Well fixed ideas of forces and their different effects should be studied, for it is the effect of forces that the engineer has to do with, and if he fails at any time to recognize their existence, and does not succeed in bending them to his will or foretelling their possible action, it is frequently due to the fact that he may not know enough to properly locate them and predetermine results that are to be avoided. In its broadest sense engineering stands for the acquirement of the knowledge necessary to determine with refined accuracy the results or possible effects due to the action of forces; it makes mankind the master over matter and is a science in which theory, practice and reason are combined for the one grand purpose of furthering the ends of civiliza- tion. It is beyond the scope of this article to do more than merely outline some of the obstacles that seem to hinder the progress of our earnest 86 plodders, and of these there is probably no one other hindrance as great as the common fault of not grasping broadly the important prin- ciples upon which the whole fabric is supported. A perfect mastery of rudiments breeds a sort of intuition, that grows unconsciously and develops the man into the successful engineer. The elementary principles of mechanics and mathematics should be acquired in a way enabling the engineer to base his investigations on a knowledge of such principles as are actually involved in any given case; he must be able to detect these principles, no matter how "clothed" or disguised, and his reasoning is often more effective for a given purpose than either the eye, the ear or the other senses all of which, however, must be more or less acute to serve his ends. The foundation, therefore, need not be particularly ostentatious; rather see that the several courses are of the proper material, thor- oughly laid and well cemented. If the original proportions are insuffi- cient to carry the final structure, don't continue to build, but tear it down to the bottom and start again with a broader base and a deeper footing. C. H. F. INTRODUCTIONS TO QUESTIONS 43 TO 62 (1898-9). INTRODUCTORY. The questions under the heading "Practical Steam Engineering," are designed to provoke general discussion at association meetings; the aim has been to arrange them in a way, tending to draw out indi vidual opinions without the need of special preparation on the part of even the more advanced members. No special pains have been taken to evade points upon which engineers differ, and it is not the intention of the committee to allow their own particular preferences to interfere with able arguments advanced on either side of a vexed question. FIRST LIST OF QUESTIONS. Note. This embraces some of the principal practical points con- cerned in the installment of 100 HP engine in an ordinary manufactur- ing establishment. Answers based on the assumptions indicated in the first question. The answers selected are brief and to the point; the information conveyed will be useful to all engineers who may be new in matters pertaining to such installations. The committee has added the interlineations which are set out in separate paragraphs, the purpose being to further elucidate the topics, handled in the answers and to point out the line of reasoning followed in arriving at the results. The comment thus offered in connection with the answers in no way detracts from the worth of the original paper; it proves the figures to be correct and places the work far above the plane of a good guess. It is the "reasoning" that represents the "kernel in the nut." Q. 43. (1898-9.) A simple, non-condensing, automatic engine, with releasing valve-gear is to be chosen, the pressure at boiler is 100 Ibs. per square inch and the maximum load is estimated at 100 HP; give bore, stroke and speed of engine you would select to develop this power economically. Ans. 43. We would select a first-class Corliss engine of "world wide" reputation size 14" X 36", same to run at 80 r.p.m. (Ed. Com. Majority of answers favored Corliss engines of 36" stroke. Cylinder diameters in fractional inches were given by some and though such sizes would, under the assumed conditions, develop the required 100 HP yet, for ordinary purpose, there would be no 87 greater need for a IS 1 ^" dia. steam cylinder, than for a 4*4" dia. pipe; commercial sizes should be adhered to. Speeds given in the various: answers ranged from 75 to 90 revolu tions. Answers to other questions are based on engine dimensions given in Ans. No. 43.) Q. 44. 1898-9.) Thirty-five feet of overhead pipe, one bend and two elbows are required to convey steam from boiler to engine; 50 feet of exhaust pipe extends to the roof and discharges to atmosphere. Give diameters of both steam and exhaust pipes; state if steam pipe should drain to, or from, boiler and what arrangements are necessary for expansion and other incidental details, also how feed water heater should be provided for, and what is necessary if exhaust steam is to be utilized in heating building. Ans. 44. Diameter of steam pipe = 4". Diameter of exhaust pipe = 5". These sizes are taken from well known formulas for steam and exhaust pipes for Corliss engines running under 550 ft. piston speed per minute. Steam pipe should drain toward engine. As close to the throttle valve as possible should be placed a steam separator of ample capacity, correctly drained and trapped, and of approved design. Steam pipe should be well supported and so arranged as to not tear itself to pieces by expansion and contraction. We prefer all bends made of pipe bent to a long radius for first-class work in lieu of cast fittings of any description. All connections should be flanged. Exhaust pipe should be led to a heater of ample capacity (we ap- prove of an open heater for this section) and pipe properly drained in low points between engine and heater. The exhaust should be arranged with a proper by-pass, with the necessary valves so that heater can be shut off for cleaning and repairs during operation of engine. For exhaust heating, a suitable back pressure valve with ample area should be provided. (Ed. Com. It is generally conceded that velocity of steam should not exceed 6,000 ft. per minute through pipes 100 ft. long. The flow is always assumed to continue throughout the stroke, and with this as a basis we may analyze the given case as follows: Engine 14" X 36", 80 rev. 480 ft. piston speed per min. Sq. dia. of 14 (cylinder) 196 = 12.25. Sq. dia. of 4 (pipe) 16 The quotient represents the ratio of cylinder and steam pipe areas. Piston and velocity of steam will be in like proportion, hence piston speed 480 X 12.25 = 5880 ft. which is "inside" the limit prescribed. For exhaust pipes a velocity of 4000 ft. is much used; a comparison similar to the above gives for the 5" pipe specified in the answer 196X480 = 3763 feet. 25 Also within the given limit. Q. 45. (1898-9.) Give initial pressure and point of "cut off" assumed in connection with your answer to Q. No. 43; also the mean effective pressure corresponding thereto and the probable weight of steam required per hour for each horse-power. Ans. 45. The initial gage pressure assumed in the answer to Q. No. Ibs. per sq. in. upon piston. The point of cut-off for 100 HP at 80 rev. or 480 ft. of piston speed . 88 per minute will equal nearly 22% of the stroke. The mean effective pressure will equal 44 1/2 Ibs. for this steam pressure and point of cut- off. These calculations are based upon an assumed clearance of 4% and an absolute back pressure of 16 Ibs. (Ed. Com. Figures submitted in Ans. No. 45 may be subjected to the following line of reasoning: 100 HP = 3,300,000 ft. Ibs. The distance or space is represented by 480 ft. piston travel per min., hence 3,300,000 -~ 480 = 6875 is the corresponding "weight to be raised." This "weight" is equally distributed on the face of a 14" dia. piston; hence the quotient 6875 divided by the area of a 14" circle = 6875 -M53.9 = 44.6, which is the mean effective pressure necessary for a 100 HP under the specified conditions, and which agrees substantially with the answer. Point of "cut-off" given at 22% of stroke becomes 36 X. 22 = 7.92 inches. The clearance at 4% of the cylinder volume, that is 153.9 X 36 X .04 Area X stroke = = 1.44 in. of stroke. 153.9 Ratio of expansion. Stroke + clearance 36 + 1.44 = 37 . 44 A Cut-off + clearance 7.92 + 1.44= 9.36 Terminal pressure. Absolute initial 90 + 14.7 = 104.7. = 26.17. Ratio of expansion 4 Mean pressure. Hyperbolic log. ratio of expansion 4 = 1.3863, to which add 1, giving 2.3863. The terminal pressure 26.17 X 2.3863 = 62.43, which is the absolute mean pressure of piston. The answer assumes 16 Ibs. absolute back pressure; hence 62.43 16 = 44.43, which is the mean effective gage pressure. Fixing the initial pressure at 90 Ibs. with 100 Ibs. at boiler provides for "drop" in pressure between boiler and engine. Steam required per horse-power is not given. It is safe to assume 30 Ibs. per hour for average conditions, although greater economy is possible with such engines. Q. 46. (1898-9.) The transmission of 100 HP from engine to shaft is by belt; explain the advantages and disadvantages of having main driving pulley distinct from fly wheel. State whether this, or a band wheel, is preferable, provided either could be used. Ans. 46. This engine is to transmit its power by belt, consequently there can be no valid reason for the use of a separate fly wheel. There may be cases where it is desirable to use a separate band wheel, but this being a new installation it should be arranged without it. Its disadvantages in this particular case would be extra cost and a display of poor engineering judgment. Q. 47. (1898-9.) The full power of the 100 HP engine referred to, in Question No. 43, is to be belted to a shaft which is to run at 240 revo- lutions per minute. Give diameters, weight and face of band wheel you would approve of; diameter of first driver pulley and best and most convenient position of engine with reference to shaft. Ans. 47." A band fly wheel for this engine should be for 80 rev. per min., 10 ft. diam., with a weight of 9,000 Ibs. Face should be 20"; 89 engine running 80 rev., band wheel 10 ft. dia., to drive first driven pulley at 240 rev. we have 10 ft. = 120 ins.; hence 80 X 120 240 . fo'mul'a for "fly whee,. Cor.iss eng.nes running at 48fr ft. piston speed, is as follows: W = 700,000 X = 7717.5 D 2 X R 2 W = weight of wheel in pounds, d = dia. of cylinder in inches. W s = stroke in inches. D = dia. of wheel in feet. R = rev. per min. According to Thurston's formula the constant number 700,000 be- comes 785 400 and applied in that form to the case in hand we have 196 X 36 W = 785,400 X = 8659 lbs. 100 X 6400 The answer agreeing substantially with the round number given in repiy to the question. Q. 48. (1898-9.) What width, double leather belt, should be pro- vided to transmit satisfactorily 100 HP according t:> the condition not- ed in Q. 47? State what you consider the best method of joining the belt and at what angle should same be run for l/est results? Ans. 48. The belt should be oak-tanned stoc'j, short lapped and double; width should be 18 ins. The best and only Al method of joining a belt of this size is a care- fully lapped and cemented joint. For best conditions the belt should not be lea off from the main shaft at a greater angle than 45 from the horizontal. * * * (Ed. Com. When the size of a oelt is determined by one of the numerous rules on the subject, it is well to size up the result from practical points of view before finally deciding the question. Liberality in belt proportion is a good investment. The need of extremely tight belts should be avoided. According to some formulas on the subject, an 18" double belt is quite small for 100 HP; by another rule it is ample, and to choose safely between such extremes is a matter of judgment. Q. 49. (1898-9.) How wide and how deep should the trench for an engine foundation be excavated, in solid earth, and what modifications would be deemed necessary if the conditions were less favorable? Ans. 49. The length, depth and width of trench for this engine foundation would depend upon the character of the soil. The length for good conditions would be approximately 24 ft. Depth, 8 ft, with 12" of concrete before beginning brick work. The widtn opposite main bearing for outboard bearing will depend upon the dis- tance between main bearing and center of outboard bearing, as varying lengths of shafts require different measurements between bearings, and consequently different brickwork as to width at this place. The width of the "L" part of this trench should approximate 13 ft. In many places it is found desirable to drive piling and upon that place a proper grillage before ramming in concrete. This will satisfy all but extraordinary conditions. 90 Q. 50. (1898-9.) Which is the preferable material for an engine foundation brick or stone, and how should same be laid? Ans. 50. The best material for engine foundations, all things con- sidered, is good hard-burned brick laid in cement mortar and every course wet down and grouted. Q. 51. (1898-9.) How should the fly wheel pit be proportioned and built? Ans. 51. The pit should be proportioned for a reasonable amount of room. The walls should be washed and plastered with first-class ce- ment. Also the bottom should be treated likewise, and should, if possi- ble, be connected to the sewer or its equivalent. Q. 52. (1898-9.) What precautions are to be observed in making, handling and setting a foundation template? Ans. 52. In a foundation template the center lines of cylinder and main shaft should be square with each other and all bolt holes very carefully located as per foundation drawing. It should be well braced to preserve its alignment. It should be handled with care. It should be set at the proper height and carefully leveled, and also lined to exist- ing lines, if any have been run, and properly made secure to preserve its correct position. Q. 53. (1898-9.) Explain how foundation bolts should be set and walled up. Describe also the form of fastening preferred for lower ends of same. Ans. 53. Foundation bolts should be placed in template holes for their reception, the tops raised to their proper heights and the bolts plumbed vertically. Timbers should be slid over the bolts with room enough for plenty of lateral play. The lower ends are best made secure by nuts placed in "pocket plates." Recesses should be placed in bottom of foundation so that these can always be reached. The bottom threads should be long and the point of bolt well leveled, so in case of necessity they can easily be screwed in from top of foundation. Q. 54. (1898-9.) Would you use artificial or natural stone for coping or capstone? What advantage has one over the other in engine foundations? Ans. 54. There is but very little advantage of one over the other, providing the artificial stone is properly made. It usually resolves itself into a question of cost and facilities at hand for procuring a natural stone. Whichever are used, they should be always painted to exclude any oil. Q. 55. (1898-9.) Which are the best for leveling up preparatory to bedding iron or wood wedges? What precautions are to be observed in this operation? Ans. 55. Iron wedges are better than wooden ones for this work. It depends much more upon the ability of the erecting engineer than upon the material that the wedges are made of. Many men, erecting engines, lack judgment in driving wedges of all kinds. In driving wedges it is necessary that they be of the same uniform taper; that they are placed in correct positions under the several parts of the frame, and that they be most carefully driven so as to produce no distortion. Q 56 (1898-9.) Explain how to bed the 100 HP engine assumed m Q No 43 noting the materials you believe best adapted for the purpose. Ans 56 The engine must be placed upon the foundation, carefully lined up and leveled up on wedges. A space of about 5/16" or so should be left between foundation stones and bottom of frame. This space should be filled in solidly with a thin grouting of 2 parts best Portland cement to 1 part of sharp sand. If the engine has a girder bed with a center bearing under front end of girds, this should be left until the last before grouting. After joint has hardened thoroughly the wedges should be backed out, all joints pointed and trimmed and the joint well painted to exclude all oil. Q. 57. (1898-9.) How would you treat the foundation and the set- ting of the outboard bearing? Ans. 57. The outboard bearing should receive the same careful attention both as to material and care in bedding and should be square with center line of engine and lie in the same horizontal plane. Before starting the shaft should be rolled over in its bearings by hand and then raised and the babbit linings scraped to a good fair bearing. Q. 58. (1898-9.) What two tests can readily be applied to an ordi- nary spirit level to prove its fitness for use in lining up an engine? Ans. 58. A spirit level can be tested for truth by the usual method of reversing ends and noting position of bubble or by the rather unusual method of an accurate square and plumb bob. The following from "Instruction to Engineers," issued by the Buck- eye Engine Company, is of interest in this connection: "Good spirit levels will be needed for this work better ones than can usually be had at the stores one iron-bodied one in particular, 3" to 6" long, to be applied crosswise on slides and elsewhere. "The test of a good level is the distance its bubble travels for a given amount of departure from the true level. To test a level place it on a planed strip of board or plank two feet long (if the level itself is not a two-foot wooden one) ; support the ends of the strip on blocks close to ends, so adjusted that the bubble will be in center. Then lift one end and insert a piece of postal card or other card of about the same thick- ness (a postal card is full 1-100" thick) and note how much the bubble has moved. It should have moved quite perceptibly; if about 1/8", and same whichever end is raised, it is very good, if 1/16", fairly good, but requires very close observation." Q. 59. (1898-9.) What methods or devices would you use in lining up the crank-shaft? Ans. 59. For this work we would use a fine waterproof silk line for drawing through center of cylinder and girder and extend it beyond main bearing a convenient distance. Bring the crank-pin up to the line in nearly extreme forward and backward positions; when the distance from line to face of crank-pin hub measures alike when crank-pin is brought up to the line in the two different positions, the shaft is square laterally. A spirit level can be used to adjust shaft in the other direc- tion to the best advantage in this particular case. Q. 60. (1898-9.) Outline briefly the successive steps and precautions to be observed before a newly-placed engine is ready for steam? Ans. 60. Before connecting to engine permanently, a new steam pipe should be thoroughly blown out with steam of good pressure. The steam chamber, valve ports, valves, piston, etc., should be carefully cleaned of all foreign matter and liberally supplied with good cylinder oil; also a 92 generous application of flake graphite. The striking points of piston and also clearance of same should be determined and properly marked. Valve stems and piston should be properly packed. The valves should be properly set. Everything should be made secure and the bearings properly adjusted. The engine should be turned over by hand one complete revolution and moving parts carefully watcned for interference and fouling, one part with another. Gradually warm up cylinder and start cylinder lubricator and adjust all oil cups for a liberal supply of oil at first. After engine uas run long enough, say three or four days, always apply indicator and make final adjustments to valves. Q. 61. (1898-9.) Indicator cards taken prove the engine to under- loaded and likely to remain so for some time to come what course would you pursue? Ans. 61. First reduce steam pressure. Second, reduce speed if the character of the load as regards regulation will admit of it. Third, in extreme cases it may be necessary to reduce both. Q. 62. (1898-9.) Give an approximate, itemized estimate, covering the cost of the outfit outlined in the preceding questions, freights, cart- age or extraordinary conditions not to be considered. Ans. 62. The following estimate is based upon Al material and work: 14" X 36" Corliss engine complete $1,600 Foundation complete, cap and bottom stones 450 Piping complete, including separator, 4" 200 Total $2,250 (The lowest estimate received was $1,762. The highest was $3,539.75, and includes an indicator outfit and all incidentals. The average of all estimates is $2,233. The asking for estimates seemed an innovation, yet it is proper that men should have some sort of a notion of the actual money value of the apparatus in their care. A discussion of the "cost of things" is a topic worthy of consideration. Ed. Com.) Q. 67. (1898-9.) Steam pressure at boilers is 100 pounds; give bore, stroke and revolutions per minute for a 100 HP high-speed, simple and direct connected automatic engine, suitable for work of the char- acter indicated in Q. 65; the power to be developed within the economi- cal range of the engine, all general conditions assumed to be favorable. Ans. 67. For 100 Ibs. steam pressure at boilers the following pro- portions are deemed suitable for 100 HP engine of the "high-speed" type referred to in the question. Diameter of cylinder, 12"; stroke of piston, 12". Revolutions per minute, 300. Piston travels 600 ft. per minute. Average mean effective pressure for 100 indicated HP = 49 Ibs. Q. 68. (1898-9.) Why are "high-speed" engines of the type referred to in Q. 67 considered well adapted for electrical installations of the size noted in the preceding question? Ans. 68. High speed engines of the type referred to in Q. 67 require less floor space per horse-power and are better adapted for direct con- nection to dynamo; they afford better regulation and, by a higher speed of rotation, effect a saving and a resultant smaller size. Q. 69. (1898-9.) What are the relative advantages of horizontal and vertical engines and what reasons govern your selection, assuming "make-up" of both types to be about the same? 93 Ans 69 Assuming the "make-up" of horizontal and vertical engines to be about the same the only advantage the former have over the other is in the matter of head room or height. In favor of the vertical engine may be noted: Economy of floor space and the fact that piston and valves are less affected by friction and unequal wear and therefore require less lubrication. It is largely a question of location and the ground room at com- mand- better accessibility of the working parts is a point argued in favor 'of the horizontal engine. For very large engines the tendency of modern practice is decidedly towards vertical engines. 070 (1898-9.) Submit sketch, showing by indicator card, the con- ditions you expect to realize in the cylinder of the engine selected when driving the normal load indicated by Q. 67. Ans. 70. See drawing herewith: Card JVo.J - Sea Ze of Sp-ri77 b - Ut Under good P ractica l conditions and circumstances a belt 57 inches in width will be ample in all respects There is no doubt whatever that a belt 50 inches will do the re- quired work without trouble if properly cared for PPOT,nV h i%r mmItt ?' S opinion that a ple width in belting is more fn So *\ working too close to the width that will actually per- orm the work, both in care of and durability of the belt. i % ^ < 18 "-1900.) What is the effective pull of a 30 in two-ply 90 R.p r Ml ' transmittin * 30 HP > *i of drive pully 20 ft. dia," turning Ans. 90. The velocity in feet per minute equals 90 X 20 X 3.1416 5654.88 feet. Then 300 X 33000 = 1750.7 Ibs., the effective pull upon the belt under condi- 5654.86 tions stated in the question. The effective pull per 1 inch of width of the belt equal 1750.7 -=- 30 = 58.356 pounds. Rule: Reduce the horse-power to foot pounds and divide the num- ber of foot pounds by the velocity of the belt in feet per minute and the result will be the effective pull upon the belt in pounds. Q. 95. (1899-1900.) At what point in the stroke of an engine is the pressure the greatest on the guides? Ans. 95. With a uniform pressure upon the piston, the maximum thrust upon the guides will occur when the cranks are at right angles to the axis of the cylinder, or at an angle of 90. A drop in pressure, due to an early cut-off or other causes, would modify this statement somewhat, depending upon the point of cut-off, and the angularity of the rod. The maximum thrust upon the guides might occur at a point earlier in the stroke, or before the crank had reached its greatest angle. By referring to sketch, the triangle DEF represents the crank at an angle of 90. The indicator diagram, drawn to represent an initial pressure of 100 pounds absolute scale, 60 pounds per inch in height, with a 25 per cent of the length of the stroke, cut-off, shows that when the piston has reached the point E, representing 50 per cent or one-half stroke, and the crank at 90, the pressure has fallen to about 48 pounds abso- lute. Consequently the maximum pressure will take place at a point in the stroke before the crank has reached the angle of 90 to the axle of the cylinder. Expressed in proportion DE : DF : : horizontal thrust : the down- ward thrust on guides. Horizontal thrust equals area of piston multi- plied by steam pressure. The distance DE is found by the rules of finding the sides of a right angle triangle, hence DE = the sq. root of the difference of the squares of the height on the hypothenuse. Assuming DF to equal I, the ratio of the length of the crank to the connecting rod = EF. ' DE = 1/EF 2 DF 8 when the maximum thrust on the guide at E equals DF. X horizontal thrust DE. (See sketch.) Q. 96. (1899-1900.) What will be the greatest thrust on the gui'l-s (due to the steam pressure) of an engine 24"x60", having a connecting- rod 5y 2 times the length of the crank, with an unbalanced pressure of 50 Ibs. per square inch (gage) on the acting, or working, side of the piston? Ans. 96. Engine 24" X 60". Area of piston = 24 2 X .7854 = 452.39 sq. in. Horizontal thrust parallel with the axis of the cylinder = 452.30 X 50 = 22619.50 pounds. Length of crank = 60 -~ 2 = 30 inches, or 2.5 feet. 99 Lengrth of connecting rod equals 2.5 X 5.5 = 13.75 feet. Referring to sketch, question 95. The altitude DF of the triangle DEF equals 2.5 feet; the hypothe- nuse EF equals 13.75 feet. The base DE equals 1/EF 8 DF 2 =1/13. 75 a 6.25 s =1/182.8125= 13.52 + feet, or length of the base DE; therefore 2.5 X 22619.50 13.52 or downward thrust on the guides. = 4182.6 pounds Q. 97. (1899-1900.) With a guide 5y 2 " wide and shoe of cross-head 18" long, what is the greatest thrust on the shoe, per square inch? Taking the same conditions as stated in question 96. Ans. 97. The superficial area of shoe equals 18 inches X 5^ inches = 99 sq. in. As found in Question 96, the total thrust upon the guides is 4182.6 pounds. Then the total thrust per sq. inch equals 4182.6 -4- 99 = 42.2 pounds. Q. 98. (1899-1900.) In comparison, how do the lengths of eonnect- ings rods affect the thrust or pressure upon the guides? Ans. 98. In comparison the thrust, or pressure exerted upon the guides is affected inversely, as the length of the rods; that is, the longer the connecting rod, as compared with the length of the crank, the less will be the pj )ssure upon the guide. Q. 99. (1899-1900.) If the diameter of a piston is 18 inches, the pressure at the beginning of the stroke is 130 Ibs. (gage) .per square inch, what is the pressure on the crank-shaft at this point? When the crank has turned to an angle of 60, with the axis of the cylinder, the steam pressure has dropped to 115 Ibs. gage per square inch, what is the pressure on the crank-pin? described by the crank-pin? What is the force tangent to the circle, Ans. 99. The area of 18 inch piston equals 18 2 X .7854 = 254.47 square inches. Horizontal thrust at beginning of stroke equals 254.47 X 130 = 33081.10 pounds. The horizontal pressure on crank-pin when at an angle of 60 to the axis of cylinder with a pressure of 115 Ibs. per sq. inch equals 254.47 X 115 = 29264.05 pounds. When the crank is in a position, termed dead center, that is. when piston connecting rods and crank are all in the same line, the horizontal thrust, simply produces a pressure on the bearings of the crank-shaft. When the crank has turned to an angle of 90 or at right angle with the axis of the cylinder the pressure of the steam on the piston is all exerted in turning the shaft, and none on the bearings; this is because the pressure exerted is at right angles to the crank, or tangent to the crank circle. By referring to-the diagrams, when the crank is at 60 with the axis of the cylinder, the pressure is exerted in the direction of a c and a e. The diagram line a d b represents the horizontal thrust of 29264.05 pounds. The tangential thrust in the direction a c will be as a b : a f 29264.05 : X. Substituting. 2" : 1.73212 : : 29264.05 : X. X = 25343.545 pounds, equals the force tangential to crank-pin circle a c. SEE QUESTION 99. Note. The sine of the angle of 60 = .86606. Multiplying the length of the line a b (2") by the sine of the 60 (.86606) equals the length of the line a f equals 1.73212 inches. The force, or thrust, in direction of a e is in proportion as ab : ad : : 29264.05 : X Substituting 2" : 1 : : 29264.05 : X X= 14632.025 pounds, the pressure on the crank shaft when the crank is at an angle of 60 with the axis of the cylinder, or in the direction of the line a e. Note. The cosine of the angle of 60 equals .5 Multiplying the .length of the line a b (2") by the cosine of the angle of 60 (.5) equals the length of the line a d equals 1 inch. The horizontal thrust is represented by line a b. The thrust on crank shaft at 60 by line a e The thrust tangential at 60 by line a c. Q. 106. (1899-1900.) How does the Corliss type of engines differ from the slide-valve type of engines? excels 6 the m St important advantage in which the Corliss type release> by the me chanical mechanism, how are the valves Ans 106. The Corliss type of engine differs from the common slide valve type in that it has four valves, one at each end of the cylinder Qe admission of steam and one at each end of the cylinder for the exhaust or escape of the steam. 'he steam entering the cylinder at, or near, the boiler pressure for ?S!n? P - , he Str ke> at which P int the mechanism that opens >n - , > off thP^T V ^ lve ^ is . released and allows the valves to close, cutting remainder of the work to be performed type excels the slide valve type in that the steam is venien e t X1 i an: ely > c nse( l ue ntly, more economically, also more con- t in the adjustment of the valve gear, as each valve is inde- pendent of the others, less friction of the valve gear per unit of power developed, and a close regulation of speed. After the valve mechanism is released, the steam valves are closed by dash-pots, or, as sometimes called, vacuum pots, consisting of a cast iron cylinder fitted with an air-tight piston attached to a bell-crank on the valve stem. As the opening mechanism opens the valve it raises the piston in the dash pot, forming a partial vacuum on the under side of the piston; at the point of cut-off the opening mechanism is released the atmospheric pressure on the upper side of this piston closes the valve rapidly. To prevent pounding a little air is admitted for cushioning the piston before striking the bottom. Q. 107. (1899-1900.) An engine working at its maximum capacity, exhausting to the atmosphere, it is desired to increase the load 25 per cent, without an increase in steam pressure or the speed, how could this change be effected? Ans. 107. When an engine exhausts to the atmosphere the steam which fills the cylinder at the end of the stroke has to be forced out against the atmospheric pressure, usually reckoned 15 Ibs. per sq. in. The nature of steam being such that a part of the atmospheric pressure can be removed. One pound of steam at atmospheric pressure occupies about 1642 times as much space as it does in the form of water. . If this steam, when it is exhausted from the cylinder, is allowed to come in contact with cold water, or a cold surface within a vessel that is air-tight, it will be immediately condensed, occupying about 1/1600 of its original volume, creating a partial vacuum in the con- denser, the water and air being removed by the air pump, allowing the piston to make its return stroke relieved from a portion of the atmospheric pressure of 15 Ibs. per sq. in. This reduction of back pressure is equal to a corresponding in- crease of the M. E. P., increasing the mean efficiency of the engine from 25 to 30 per cent. Therefore, the adding of a condenser will increase the capacity of the engine the required amount. Q. 108. (1899-1900.) In good practice, how marv expansions are advisable in a single-cylinder engine? Why are more expansions objectionable? When more expansions are desired, how can they be obtained? When the engine is so constructed as to properly give more expan- sion, what is this type of engine called? Ans. 108. Usually in good practice it is desirable to have from 3 to 5 expansions in a single cylinder engine; an allowance of from 20 to 33 per cent cut-off, which is about the economical range of cut-off engines'. When more expansions are desired it is more practical to allow the expansions to take place in more than one cylinder. This type of engine is called a compound engine. Q. 109. (1899-1900.) When the expansion takes place in more than one cylinder, for example, a two-stage compound, how are the number of expansions found, and are they effected by the cut-off in the low pressure cylinder? Ans. 109. When the expansions take place in a two-stage compound the number of expansions are found by dividing the initial absolute pressure in the H. P. cylinder by the absolute terminal pressure in the L. P. cylinder, or it is the product of &, number of expansions in the H. P. cylinder by the number of expansions in the L. P. cylinder. 103 Formula the number of expansions of the two cylinders. t t 1 In which T = absolute initial pressure H. P. cyl. t absolute terminal pressure H. P. cyl. t*=i absolute terminal pressure L. P. cyl. The number of expansions are not affected by the point of cut-off in the low pressure cylinder. Q. 110. (1899-1900.) A compound engine, working with an initial pressure of 130 Ibs. absolute pressure expanded to a terminal pressure of 26 Ibs. absolute in the H. P. cylinder; then received and expanded to a terminal pressure of 8 Ibs. absolute in the L. P. cylinder; find the total number of expansions in the two cylinders. Ans. 110. 130 pounds absolute = initial pressure H. P. cyl. 26 pounds absolute = terminal pressure H. P. cyl. 8 pounds absolute = terminal pressure L. P. cyl. The number of expansions taking place in high press, cyl. = 130 -f- 26 = 5 expansions. Low pressure cyl. = 26 -=- 8 = 3.25 expansions. Whole number of exp. = 5 X 3.25 = 16.25 expansions. By formula T t = total number of expansions. t t 1 130 26 130 Substituting X = = 16 1 /4 total exp. 26 8 8 Q. 111. (1899-1900.) For what purpose is a receiver placed between the two cylinders of a compound engine? What should be its relative capacity, and to which cylinder of the engine does the relation exist? Does the disposition or arrangement of the cranks affect the size of the receiver? What conditions would be favorable for the non-use of a receiver, the steam passing through the exhaust pipe from the high to low pressure cylinder? Ans. 111. A receiver consists of an enclosed vessel or cylinder, gen- erally provided with a steam jacket or some means for reheating the steam and to provide against loss of heat from radiation, etc. They are sometimes called re-heaters. The H. P. cylinder exhausting into this receiver and the L. P. cylinder taking its supply from it. It provides for a volume of steam close at hand to supply the low pressure cylinder without a serious reduction or drop in the pressure, and at a low maximum velocity. The volume of the receiver is usually from 1 to 1.75 times the volume of the H. P. cylinder. The larger ratio is more favorable to a straight back pressure line of the H. P. cylinder and steam line of the L. P. cylinder indicator cards. The angle that the cranks are set to each other does affect the required volume of the receiver. The most favorable condition for the non-use of a receiver is when the cylinders are placed in line with each other and both pistons are attached to the same crank, as a tandem compound, or, when the cranks are set at an angle of 180, or nearly so, to each other. 104 Q. 112. (1899-1900.) Give the rule to find the receiver pressure necessary to balance the load on an engine? What receiver pressure is required to balance the load on a 28" X 52" X 72" cross-compound engine, the steam pressure 140 Ibs. absolute; vacuum maintained at 28 inches? Ans. 112. In order that the work of a compound engine may be divided equally, or nearly so, the receiver pressure should be propor- tional to the initial pressure of the H. P. cylinder, according to the ratio of the area of the piston of the H. P. cylinder and the area of the piston of the L. P. cylinder. Rule. To find the proper pressure to maintain in the receiver of a compound to balance the load. Form the proportion: As the area of the H. P. piston is to the area of the L. P. piston so is the required receiver pressure to the initial pressure less the receiver pressure (which acts as a back pressure). As the areas of circles vary directly as the square of their diameters, it is not necessary to find the areas of the pistons use the square of the diameters instead. Example Compound 28" X 52" X 72". Initial press. 140 Ibs. ab. X = the receiver pressure. Then . 28 2 : 52 2 : : x : (140 x) = 784 : 2704 : : x : (140 x) The product of the extremes = 784 (140 x) The product of the means = 2704 x or 2704 x = 784 (140 x) 784 (140 x) = 109760 784 x. Then 2704 x = 109760 784 x = 3488 x = 109760. 109760 X = =31.47 Ibs. ab. receiver pressure. 3488 28" vacuum = 28 X .491 = 13.7 pounds. 31.47 13.7 = 17.77 pounds receiver pressure, as shown by the gage. Q. 113. (1899-1900.) In good practice, how should the work be dis- tributed on a cross-compound engine? How is this distribution brought about? Ans. 113. The work should be so distributed that each cylinder will perform an equal share of the same; that the initial strains in each cylinder sb^"M be as near the same as possible. That the drop in pressure between the high pressure cylinder and the receiver should be as small as possible. This distribution is brought about by the adjustment of the cut-off in the L. P. cylinder, shortening up the cut-off increases the receiver pressure, allowing the low pressure cylinder to perform a greater pro- portion of the work. Lengthening out the cut-off has a vice versa effect. Q. 114. (1899-1900.) What are the advantages derived from the use of compound engines over the use of simple engines? What conditions in practice are necessary for the best economy in the use of multiple cylinder engines, condensing or non-condensing? Ans. 114. The advantages of dividing the expansion among two or more cylinders, or the use of multi-cylinder engines are the use of higher initial pressures, a wider range of expansion, with a minimum 105 of cylinder condensation, avoiding excessive strains on the metal by sudden expansion and contraction. Sizes of cylinders adapted to the work to be performed: A con- stant steam pressure and a constant load are required for economy. Q. 115. (1899-1900.) Upon what depends the ratio for capacity of the several cylinders of multiple cylinder engines? Explain how the ratio is found. Ans. 115. The ratio of capacity of the cylinders depends upon the range of temperature or pressure desired, or the ratio of the initial pressure to that of the desired terminal pressure in the following cylinder; this is the relation that the volume of the H. P. cylinder bears to the L. P. cylinder. Rule. Divide the absolute initial pressure of the first cylinder by the absolute terminal pressure of the last cylinder and multiply this quotient by the desired per cent of cut-off in first cylinder. Q. 116. (1899-1900.) What should be the ratio of capacity of the cylinders of an engine using steam at an initial pressure of 135 Ibs. absolute per square inch, expanding to a terminal pressure of 10 Ibs. absolute in the L. P. cylinder, the point of cut-off 30 per cent of the length of the stroke? Ans. 116. 135 Ibs. absolute = initial pressure. 10 Ibs. absolute = terminal desired in L. P. cyl. 30% cut-off. Then 135 -r- 10 = 13.5 expansions. 13.5 X 30 = 4.05 ratio of capacity. Or, -.1 : 4.05 = ratio of capacity of the H. P. cyl. : L. P. cyl. Q. 117. (1899-1900.) Give rule, or formula, for finding the approx- imate H. P. of a compound engine. Ans. 117. Rule for finding the approximate H. P. of a compound engine Assume that the entire work is to be done and the expansion all taking place in the L. P. cylinder. Then, 1 + hyp. log. of the total number of the expansions, multi- plied by the terminal pressure in the L. P. cylinder will equal the average mean effective pressure, due to the expansions; this product multiplied by the area of the L. P. cylinder's piston and by the piston speed in feet per minute; this product divided by 33,000 will give the approximate H. P. of the engine. Formula H. P. = horse-power of compound engine. H = * + hyp. log. of total number of expansions. terminal pressure, L. P. cylinder. = average M. B. P. due to number of expansions. b = piston speed in ft. per minute. A = area of piston. E = T. H. Then A.S.E. H. P. = 33,000 106 Q. 118. (1899-1900.) Required size of cylinders for a compound engine, condensing of 2,000 H. P., 5 feet stroke and 72 R. P. M., or 720 piston feet per minute. Boiler pressure 140 Ibs. absolute, an allowance of 5 Ibs. for drop be- tween the boiler and throttle. Terminal pressure L. P. cylinder 6 Ibs. per square inch. Back pres- sure 3 Ibs. per square inch. Prove the work by rule found in Question 117. Ans. 118. The theoretical diameter for a two-stage compound of 2,000 H. P., 60-inch stroke, 72 R. P. M. Boiler pressure 140 Ibs. absolute. Terminal pressure 6 Ibs. Back pressure 3 Ibs. Drop in pressure between boiler and throttle 5 Ibs. 140 5 = 135 Ibs. initial pressure absolute at throttle. 135 -j- 6 = 22.5 total number of expansions. The expansion in each cylinder is equal to V22.5=4.74. The terminal pressure in the H. P. cyl, is equal to 135 H- 4.74 = 28.48 Ibs. absolute. Average M. E. P. of H. P. cylinder is equal to (1 + hyp. log. of 4.74) 28.48 = 72.79 Ibs. absolute. 72.79 28.48 = 44.31 Ibs. = M. E. P. of H. P. cyl. Average M. E. P. of L. P. cylinder is equal to (1 + hyp. log. of 4.74) 6 = 15.34 Ibs. absolute. 15.34 3 =12.34 pounds. M. E. P. of L. P. cylinder. It is desirable to have the cylinders so proportioned that the work, with the above conditions, one-half of which, to wit, 1,000 H. P., be performed in each cylinder. Then, the area of the low pressure cylinder will equal 33,000 X H. P. piston speed in ft. X average M. E. P. Substituting 33,000 X 1,000 33,000,000 = =3,714.21 sq. in. 720 X 12.34 8,884.8 the area of L. P. cyl. V3.714.21 = 68.70 in., diam. of L. P. cyl. .7854 Area of the high pressure cyl. will equal 33,000 X 1,000 33,000,000 = =1,034,379 sq. in. 720 X 44.31 31,903.2 V1 ' 034 ' 3;9 =36.29 i n..dia.ofH.P.c y l. .7854 The size of the compound engine as per the conditions stated in question will be 36" X 69" X 60" at 72 R. P. M. with 140 Ibs. absolute boiler pressure, 22.5 expansions, or 4.74 expansions in each cylinder exhausting into condenser against 3 Ibs. back pressure, or into a vacuum of 14.7 3 = 11.7 Ibs. 11.7 Ibs. X 2.036 = 23.82 in. vacuum. 69 2 The ratio of cylinder = = 3.67. 36 2 The engine under the above conditions will develop 2,000 H. P. Proof by rule foun.d in Ans. 117. 107 Whole number of expansions = 135 - 6 = 22.5, (1 + hyp. log. of 22.5) =4.11353. 6 X 4.11353 = 24.68118 Ibs. effective pressure. Area L. P. piston = 3,729.38 sq. in. Speed of piston = 720 ft. per min. Average effective pressure = 24.68 Ibs. Tnen 3,729.38 X 720 X 24.68 J - = 2,012.05 H. P., 33,000 a difference only of 12.05 horse-power. Q 119. (1899-1900.) Describe the manner of attaching the indi- cator to an engine. What care should be taken in such preparation for the attachment? How to obtain the drum-motion and what care should be taken? Describe the manner of taking diagram, and note the important data usually taken. Ans. 119. The indicator is attached to the end of the cylinder, through holes drilled and tappec 1 tor y 2 " pipe, into the side or top, whichever is the most convenient, to connect with a reducing motion, attached to the crosshead. The holes must be drilled in such a position that they will com- municate with the clearance space and not become covered by the piston when it is at the extreme end of the cylinder; the indicator should be attached as direct as possible to the cylinder. In. most engines the holes are drilled and tapped in the shop before it is sent out. Some are not, however, and in those that are not care should be taken to locate the holes and to keep the chips from the drill from getting into the cylinder, especially the front end. As the length of the diagram represents the travel of the piston for % revolution, or one stroke of the engine, and the length of this dia- gram is determined by the rotation of the paper drum, this motion must be taken from some part of the engine which has a motion coin- ciding with that of the piston; the crosshead is the most convenient part for this purpose. This drum motion, or reducing motion, as it is called, is obtained by the Brumbo pulley, or the pantograph, reducing wheels, and nu- merous other devices, so long as they reduce the- motion proportion- ately to the stroke of the engine and parallel with the crosshead or piston; so that when the crosshead has made *4 of its stroke the pencil will have traveled % the length of the diagram, and so on. In taking diagrams from engines, the indicators should be thor- oughly warmed up by opening the steam cock between the indicator and the cylinder, the working parts should work very smooth, the spring selected to give the diagram should be one that will make a diagram from 1% to 2 inches high; dividing steam pressure per gage by about 1% gives a scale of spring that makes a very good height for diagram. The pencil should be sharp, smooth, and so adjusted that when the sssure is applied it will make a smooth, fine, distinct line. The cord leading from the crosshead or reducing motion to the indicator should be so adjusted that the paper drum in rotating will not bring up against the stops. After placing the paper around the drum taut and smooth start the paper drum in motion, open the shutoff cock, allowing steam on the under side of the piston. Press the pencil against thf pape? for one revolution or more, then shut off the steam to the indicator and 1 08 again press the pencil to the paper and draw the atmospheric line, which is drawn with the atmospheric pressure on both sides of the piston. As much of the following data as cu should be obtained and noted on the diagram: The date, the hour of raking the diagram, the type or make of the engine, which end of cylinder and which engine if one of a pair, the diameter of cylinder, and length of stroke, the diameter of the piston rod, the number of R. P. M., the steam pressure, by gage, at the boilers and at the engine, atmospheric pressure by barom- eter, the vacuum, by gage, in the condensers (if the engine is a con- densing type), temperature of feed water at boilers, the receiver pressure by gage if the engine is a compound, if not note the back pressure per gage, and the scale of spring used in taking the diagram; also note any other matter connected with the plant that may come to your notice. Q. 120. (1899-1900.) What is the value of the indicator diagrams for the successful operation of a steam plant? Describe the workings and use of a planimeter. Ans. 120. The indicator diagrams are the result of two motions thus: The horizontal of the paper in exact correspondence with that of the piston, it represents the stroke of the engine on a reduced scale, and the vertical movement of the pencil in exact ratio to the steam pressure, exerted in the cylinder; consequently, it represents by its length the stroke of the engine, by its height at any point, the pres- sure on the piston, at a corresponding point in the stroke. The shape of the diagram depends very much upon the manner in which the steam is admitted to and released from the cylinder of the engine. It shows the arrangement of the valves for admission of steam, cutoff, release and compression of the steam. The adequacy of the ports and passages for admission and exhaust, and when applied to the steam chest, show the adequacy of the steam pipe. The amount of power developed, the quantity lost in various ways by wire drawing, back pressure, premature release, or any other maladjustment of the valves. It has proved itself very useful to the designers of steam engines in figuring out the distribution of the horizontal pressure on the crank pin, the angular distribution of the tangential components of the hori- zontal pressure; in other words, the rotative effect around the path of the crank. Taken in combination with a measurement of feed water and coal used, the economy of the plant can be found. The degree of excellence to which the steam engine of to-day has been brought is due principally to the use of the indicator, as a careful study of the diagrams and different conditions, load, pressure, etc., is the only means of becoming familiar with the action of steam in the cylinder of an engine. The indicator furnishes many other items of importance when the economy of the generation and the use of steam is to be considered. The planimeter is an instrument designed to measure the areas of irregular figures. It is operated by moving a tracer, with which it is fitted, over the lines of the diagram or figure to be measured, and records the area upon a graduated wheel and Vernier scale. Some of the instruments are made to read the M. E. P. or the I. H. P. direct from the wheel without the use of any figures. In measuring the I. H. P. of cards much time and labor can be saved by the use of the planimeter. 109 Q 1 (1900-1901.) Give three ways of establishing the theoretical expansion curve on an indicator diagram, accompanying answer with a sketch of each method. Ans. 1. To establish the theoretical curve it is well to remember that the pressure of steam varies inversely as its volume; from this con- sideration we can readily locate any number of points in the curve. Again, all pressures must be measured from perfect vacuum, and clearances must also be taken into account. By referring to the sketches we find the vacuum line denoted by V, and the clearance line by C; the vacuum line laid off by the scale of the spring, 14.7 pounds below the atmospheric line, and the clearance line C the distance h from the beginning of the diagram, representing the added length of the cylinder that would represent or equal the capacity of the clearance space on one end of the cylinder. There are several ways of establishing the points in the theoretical curve. The curve may be drawn from any point in the real curve, the only restriction being that the point must be selected when the steam valve and the exhaust valve are closed. The following are several methods of establishing the theoretical Sketch ~/ Through the point of release b, draw any line as aB and make AB equal to ab. Then draw any other line as GD, and Gd, equal to AD, then will d be a point in the curve bA, as shown by the dotted lines above the real curve. By drawing a number of lines through the point A, and following the same method in regard to laying off the distances any number of points in the curve may be located. The following geometrical method for locating the points in the theoretical curve is perhaps the most used: Select any point as I in the actual curve and from this point draw a vertical line to J, on the line B; the line B representing the boiler pressure, or, it may be a line drawn at any convenient height. From the point J draw a line to the point K, K being the inter- section of the vacuum and clearance line; from I draw the line IL, parallel with the atmospheric line and at right angles to IJ. From L the point of intersection on the diagonal line JK and the horizontal line IL draw the vertical line LM. The point M is the point of theoretical cut-off. Fix upon any number of points as 1, 2, 3, 4, 5 on the line B and draw from these points lines to the point K; from the intersections of these lines with the line LM draw horizontal lines, and from points 1, 2, 3, etc., draw no vertical lines parallel to IJ. The intersection of these horizontal and vertical lines will establish the desired points in the theoretical curve. On the diagram draw vertical lines, commencing at the clearance line spacing equal distances apart and number them as shown in sketch. It is not necessary to pay any attention to coming out even with the end of the diagram, and the more numerous these vertical lines are the greater the accuracy in developing the curve. Selecting the point 14 on the line V, the pressure represented by the scale of the spring, we find to be 18 pounds absolute. The number of volumes is according to the units adopted and represented by the distance apart of the vertical line 14, hence to find the pressure for any other line: Multiply the pressure by the number of lines and divide by the number of lines upon which it is desired to determine the pressure. For example, 18 X 14 = 252. If we divide this by 7, for instance, it equals 36 pounds to be set off by the scale of the diagram on line 7. By following out the above method and connecting the points together the theoretical curve for the diagram will be established. Q. 2. (1900-1901.) Of what advantage is the theoretical curve and what does it indicate or represent? Also describe the adiabatic and isothermal curves. Which one is used in plotting the theoretical curve? Why? Ans. 2. The defects in engines as revealed by the indicator diagram are most clearly understood by comparing the diagram of the engine in question with a theoretically perfect diagram that is, one which would be obtained from an engine in which both the designs and the adjustments were perfect in every way. Adiabatic curve is defined as the curve of no transmission of heat. Isothermal curve is defined as the curve of equal temperature and is a common or rectangular hyperbola. The isothermal curve is usually used in plotting the theoretical The reason why this curve is usually used is that according to "Mariotte's" law the volume of a perfect gas, the temperature being kept constant, varies inversely as its pressure, and in calculations of the mean pressure of expanded steam in engines it is generally assumed that steam expands according to "Mariotte's" law, the curve of the expansion line being a hyperbola. Q. 3. (1900-1901.) Explain fully how you ascertain by card the number of pounds of water consumed per indicated horse power (I. H. P.) of an engine, either single high pressure or compound con- densing. Ans. 3. The indicator diagram enables us to find approximately the amount of steam consumed by the engine. The rule for calculating the number of pounds of steam consumed by an engine per horse-power is as follows: Rule- Take two points, one on the expansion line before release and one on compression line equally distant from the vacuum line. Find by the scale of the spring used the pressure of steam at these points, and from the steam table find the weight of a cubic foot of steam at that pressure. Multiply this weight by the distance between the two points and by the constant 13750. Divide the product by the M. E. P. and by the length of the dia- gram. The result will be the number of pounds of steam consumed per I. H. P. per hour, as shown by the diagram. This computed consumption of steam is for various reasons always less than the actual consumption. The difference between the theoretical water-consumption found by the rule and the actual consumption as found by test represents "water not accounted for by the indicator" due to cylinder condensa- tion, leakage through ports, radiation, etc. In relation to multi-cylinder engines the supposition is that the work is done in one cylinder. Q. 4. (1900-1901.) Explain the various methods of obtaining the mean effective pressure (M. E. P.) from indicator cards. Ans. 4. In general practice, the M. E. P. is found by the use of a planimeter or by computing by the use. of ordinates. It can also be found by plotting on cross-section paper. The planimeter, as has been heretofore stated, is an instrument by which the area of irregular surfaces may be accurately measured. Some instruments are so constructed as to read the M. E. P. direct while others give the area only. Having the area given the M E P' is found by multiplying the area in square inches by the scale of the spring and dividing this product by the length of the diagram. ihe computing by the means of ordinates is quite generally under- stood without much explanation; the total number of ordinates that is, their length divided by the total number equals their mean length m inches. This multiplied by the scale of the spring equals the Q. 32. (1900-01.) How would you determine if the steam pipe lead- ing to your engine was large enough, that is, of sufficient diameter? Ans. 32. The sufficient diameter of steam pipe for the work of the engine could be determined by the use of the indicator both on the engine cylinders and on the pipe proper. If in indicating an engine, the drop of initial pressure, the falling off of the steam line, would prove conclusively that there was too con- tracted area in the steam passages of the engine or that the pipe was too small, or both, to locate the trouble the indicator can be placed on the steam chest; also on the pipe and a diagram taken while the engine is working at its usual or rated load. The sudden falling off of the pressure would indicate the trouble. Q. 34. (1900-01.) Give in a general way your practice for using and handling of valves. Also the best practice as to kinds of valves to use and not to use. Ans. 34. A general rule for locating and handling valves is to avoid forming water pockets. Junction valves should never be placed close to the boiler if the main steam pipe is above the boiler, but put it on at the highest point of the junction pipe. Never let a junction pipe run into the main at the bottom, but into the top or side. Always use an angle valve when convenient, as there is more room in it. Ail gate valves of size 4 inches and upwards should be by-passed; this saves the gate and seat of valve from wear and is also convenient in equalizing the pressure before opening the large valve. In cutting in a boiler a -by-pass will oe found very convenient in equalizing the pressure before opening the main junction valve. Blow-off valves should be opened wide open to allow sediment to pass out and not lodge in the valve, thus ruining it when closed. Globe valves should not be used on indicator pipes. For water always use gate valves, or angle valves, or stop-cocks. Valves with renewable discs are recommended in some cases. It is also recommended that in case of inspection of a boiler connected to others working, as a precaution, that the junction pipe be disconnected and a blank flange, or a blank be inserted between the flanges, making the boiler surely safe for inspection or cleaning. Q. 36. (1900-01.) For what purpose is a separator placed on a steam pipe,- and for the greatest efficiency where should it be placed? Ans. 36. The function of a separator is to remove from vapors and gases the liquids and solids which they carry along with them. Thus a live steam separator is intended mainly to remove the en- trained water from steam, while the separator placed in the exhaust piping when the exhaust steam is to be condensed and again used in the boilers is intended to remove the grease which it has accumulated in its passage through the engine. The use of dry steam in an engine cylinder is very important, not only because an accumulation of water is a menace to safety, but also because entrained water involves a very considerable reduction in the economy of operation. The water carried into the steam cylinder not only carries away heat from the boiler which is incapable of doing any work in the engine, but it also materially increases the initial condensation, one of the most important losses of power in the engine. Hence there should be some device provided to ensure dry steam, The principles upon which the action of steam separators should be based are as follows: In the first place, they should be constructed in such a way that the momentum which has been acquired by the liquids and solids has been destroyed. This has been accomplished by baffle or deflecting plates, which alter or reverse the flow of the steam, or by allowing it to expand and give the heavier particles time to fall by the action of gravity. After this has been accomplished it is important to prevent 'the separated water from being again picked up and carried along by the purified gases. Finally, care must be taken that an ample and easy passage is af- forded to the current of steam, so that there will be no loss of energy from friction. A separator should be placed as close to the engine cylinder as pos- sible. There are several well known types of proved merit, made for both horizontal and vertical pipes, allowing in some cases to placing directly over the cylinder. Q. 37. (1900-01.) If a number of engines are receiving their steam supply from a common steam main, where would you locate the sepa- rator? Ans. 37. Should there be a number of engines connected to the steam main and only one separator used, it would be placed at the last engine. This answer might be modified in certain conditions in which the pipe was not in the same plane. The proper way if the engines were of fair size, to place a separator at each engine. The piping off to the engines would determine the position of the separator in the line of the main steam main. There might be instances where the separator would be placed near the first engine. If we take into account the difference in the specific gravity of water and steam, the momentum obtained by the entrained water as it is swept along with the steam we will see that the heavier particle will have a tendency to keep in a right line direction rather than take to a turn in direction, thus if the first engine was piped off from a side outlet, or taken from the top of the pipe there would be a tendency for the water to follow the straight pipe. Should the steam main take a turn at right angles at the first engine then the probability is that the separator should be placed near first engine. Q. 38. (1900-01.) In erecting a steam main of considerable length, what do you consider the best practice for the return of the conden- sation? Ans. 38. In erecting a steam main of considerable length a large drum or separator should be provided near the engine, for trapping the water condensed in the pipe. A steam drum 3 feet in diameter and 15 feet high has given good results in separating the water of condensation of a steam pipe 10 inches in diameter and 800 feet long. There are several different ways of conducting the water of conden- sation back to the boiler. The steam loop, the Holly system, pump and receiver system, the return trap system, etc. Where there are a number of returns of the same pressure coming back bring them to a receiver and return them automatically with pump. This system gives about as little trouble in handling as any. Still the other methods have their advocates. 114 Q. 39. (1900-01.) Is it safe practice to return the condensation from the steam cylinders of pumps or engine's cylinders to the boiler? Explain your reasons, "pro and con," setting forth dangers, etc.. and preventives for eliminating said danger. Ans. 39. In localities where a high tariff exists on water, it is deemed advisable to save and re-use all of the drips of condensation. In this case, care should be taken to eliminate from the exhaust steam as it passes from the steam cylinders all oils or greases. It is also advisable to use in such cases a high grade of mineral oil for cylinder lubrication. In discussing the reasons "pro and con," they could be made the subject of an extended paper, but while it is easily recognized that it is better not to return them to the boiler, economy says in this instance that we must. A discussion can easily arise as to the merits and demerits of open vs. closed heaters, etc. In which ever way it is done constant care should be taken that all apparatus is kept in proper condition and the water returned as free from oil and foreign elements as possible. It is good practice to feed in some service water at all times to make up what little deficiency is lost by evaporation or leaks. Q. 40. (1900-01.) In connecting drips from a number of engines into a common drip line, what safeguard should be applied? Ans. 40. Several drip lines (all under the same pressure) connected to a main drip line, should have a check valve to prevent the steam or condensation from backing up from any individual line. Q. 41. (1900-01.) What are the advantages of high piston speed? What kind of service requires high rotative speed? Why? Why should high speed engines run under high pressure? Ans. 41. The advantages of high piston-speed over slow-speed en- gines may be briefly stated as follows: .b'irst For a given steam pressure and cut-off the power of an engine varies directly as its speed. There are four factors which determine the power of an engine, viz: (a) The mean effective pressure on the piston. (b) The length of the stroke. (c) The area of the piston. (d) The speed. In other words, an engine of a given diameter and length of stroke, acting under a given mean effective pressure, will develop power in proportion to its speed, and if the speed is doubled its power will also be doubled, and so to obtain a given power under a given mean effec- tive pressure we need make an engine only half as large if we double its speed. This constitutes the first argument for high speeds, economy both in first cost and in space. Secondly Where power is transmitted to shafting or directly to machine running at a much higher speed, it can be performed more efficiently if the ratio of the speeds is not too great. xn many cases when the power is transmitted by belting the ratio of speed is too great to admit of transmission in a single step, since the arc of contact on the driven pulley would be too small to prevent slip- page of the belt. In such cases it is common to use an intermediate shaft which performs no other duty than to make the reduction more gradual, and thus insure a satisfactory running of the belt, "5 By increasing the speed of the engine this is done away with in many cases. In the case of dynamo machines it is now common practice to couple the shaft of the engines and dynamo direct without any belting what- ever. In spite of all that may be said against this practice, it cannot be denied that it saves a very considerable amount of valuable space. Thirdly It is claimed for high-speed steam engines that tue econ- omy in the use of steam far exceeds that of the older forms of slow- speed engines. There is no doubt a great deal of justice in this claim, because one of the main losses in the engine, viz., the cooling of the cylinder walls and passages during exhaust and re-evaporation, is greatly reduced. The disadvantage of admitting steam by the same passage through which it exhausts is clearly demonstrated in the increased economy of the Corliss type of engine, where this is not the case. As far as re- evaporation is concerned, it is obvious that the cylinder walls are chilled very considerably during this process, and the more steam passed through the engine in a given space of time the less will be the re- evaporation. Hence the increased economy of the high-speed engine. It is a fact that the uniformity and smoothness of running of the high-speed engines make them particularly adapted for dynamo run- ning. This uniformity of running is in part due to the fact that the influence of the fly-wheel is greatly enhanced and partly to the steady- ing influence of the reciprocating parts. It is a well-known fact that the steadying influence of a fly-wheel is proportional to the square of its speed. For instance, if one engine runs twice as fast as another of same design and the same weight of fly-wheel, it will run four times as steady. Now, in every steam engine the pressure is a maximum at the beginning of the stroke, decreasing as the stroke advances until the end, when it is a minimum; hence it is evident that the action of the reciprocating parts, which is to absorb or store a portion of the pressure during the first half of the stroke and restore it during the second half, has the. effect of tending to keep the pressure on the crank uniform during the entire stroke, or, in other words, to steady the running of tne engine. High-speed engines must be high pressure engines that is, the initial steam pressure must be sufficient always to set the reciprocating parts in motion, and the pressure should never be reduced by the means of the throttle-valve, otherwise the engine will be subjected to strains which will impair its life. The steam passages and parts should be of ample size, the higher the speed the larger the passages, otherwise the pressure will be reduced before entering the cylinder. The steam should be cut off early, and a moderate amount of com- pression to provide cushioning the reciprocating parts as they come to rest at the end of each stroke should be provided for by an early closing of the exhaust valve. Q. 42. (1900-01.) What are rotary engines? Why are they not more extensively used? What are the principal sources of loss in steam engines? What is the principal source of waste in the engine proper? How may it be reduced? Ans. 42. In the rotary engine there is no reciprocating motion what- ever, the force of the steam being used to produce at once a motion of rotation. Rotary steam engines other than steam turbines have been invented by the thousands, but no one has attained a commercial suc- 116 cess. The possible advantages, such as speed and the saving of space, to be gained by the rotary engine are overbalanced by its waste of In designing and constructing steam engines for additional econ- omy in the use of fuel, and yet with all that has been done in this direction since the time of James Watt, the steam engine is still a very wasteful machine, and only a small percentage of the energy con- tained in the fuel is actually converted by the steam engine into useful work. There are certain losses of energy which it has hitherto been im- possible to avoid, and so long as these must be incurred so long will the process of transforming the latent energy of the coal into mechani- cal work by the means of the steam engine be a wasteful one. It has been demonstrated that in the use of high-speed non-condensing type that of the total energy contained in the fuel that only from 5 per cent to 10 per cent is available at the shaft of the engine. A large part of this loss of energy occurs in the steam generator, due in part to the high temperature of the gases escaping into the chimney, which is necessary for the production of the draught, and partly to the excess of air over and above that necessary for the complete combustion of the fuel. Additionally there are losses due to radiation, to leakage and to the heat carried away in the ashes. The principal source of waste in the engine proper is cylinder con- densation and re-evaporation. The condensation of steam in the cyl- inder is greater, the greater the surface exposed for a given weight of steam passing through the engine, consequently it is not as great proportionately in large engines as in the smaller ones. Obviously, also, it is proportional to the range of temperature to which the cylinder is exposed, and it is for this reason that multiple expansion engines are more economical. As for the condensation in the steam passages, it has been found that this could be almost entirely eliminated by closing the exhaust before the end of the stroke, so as to compress the steam and thus raise the temperature of the walls and passages, or more effectually by providing separate passages for admission and exhaust. Losses in the engine proper by the presence of moisture in the steam cylinder, which is carried into the cylinder with the steam from the boilers or is produced by the condensation on the cylinder walls or in the steam passages, is re-evaporated during the expansion and exhaust periods, and re-evaporating, abstracts heat from the steam. Very little of this heat which is used in this way is returned to the engine as useful work, and consequently the presence of moisture may be said to rob the engine of an amount of useful energy proportional to the heat required for its evaporation. It is, therefore, a matter of economy to prevent the entrance of water to the cylinder or its formation by condensation on the walls of the cylinder and in the steam passages. Q. 44. (1900-01.) Which is the most economical type of engine, and why? What is the cause of cylinders wearing unevenly? How can it be avoided? What is meant by the term "clearance"? , . j Why is it a necessity anu how can it be determined in a given engine? Ans. 44. The four-valve and the Corliss type of engines are more economical than those in which the same passages are used for both admission aini exhaust. The losses of heat due to radiation may be effectually prevented by surrounding the cylinder with non-conducting material. 117 The Steam jacket is also frequently used and is effective in reducing the losses due to initial condensation. It is a general impression among engineers that the cylinders of very large horizontal engines are more liable to wear oblong than those of vertical engines of the same bore; but experience and obser- vation have proved this to be a mistaken idea. The trouble is frequently due to imperfect alignment, and it is dif- ficult to imagine why an engine piston should exert any pressure against the cylinder walls, other than that due to gravity, when all the reciprocating parts are perfectly true with the center line of the engine and with each other. Care should be taken in packing the rod that there should be no inequality in the packing, as any material inequality may throw the engine out of alignment. The wear which would occur on the bottom of large cylinders of large horizontal engines on account of the weight of the piston is frequently avoided by extending the rod through a stuffing-box in the outer cylinder head. The term clearance is understood to mean the unoccupied space be- tween the piston and cylinder-heads when the crank is on the dead center; but it also applies to the space between the cylinder and the face of the valves. The amount of clearance of any engine affects its economy; that is, if the clearance is small, the engine will be more economical than if large; for obvious reasons a certain amount of clearance is a necessity to prevent the piston from striking the heads of the cylinder, due to the wearing of boxes and pins or an unequal adjustment of the different reciprocating parts. The most accurate method of ascertaining the exact amount of clear- ance is to weigh up a certain amount of water, place the engine on the dead center, and then from this weighed or measured water fill the clearance space up to the face of the steam valve; reweighing the water remaining and subtracting from the first weight will give the number of pounds that are required to fill the clearance space, which can be reduced to cubic inches, and in comparison with the cubic con- tents of the cylinder, the percentage is easily arrived at. Q. 45. (1900-01.) What consideration determines the length of con- necting rods? How long are they usually made? What is the function of the cross-head? What are three different forms of cranks? Where would they be used? Explain the difference of the terms "throwing over" and "throwing under"? Which way is advisable? What are the disadvantages of center cranks? Of what material should they be constructed? Ans. 45. The length of stroke is one of the principal considerations that govern the length of the connecting rod. The connecting rod, like the piston rod, is subjected to both a tensile and compressive stress; undue stress, due to accidents, etc., must be taken into consideration. Long connecting rods have many advantages, but the longer they are the greater must be their thickness, and they are not as economical in the use of material as the short connecting rods. For long-stroke engines they are generally made from two to four times the length of H>6 stroke. The usual length for high-speed engines is five times the length of the crank. 118 The cross-head is that part of the engine which, moving between guides, preserves the rectilinear motion of the piston rod and at the same time supplies a bearing, called a wrist-pin, for the rocking motion of one end of the connecting rod. A crank is a simple lever, at one end of which acts the steam pres- sure which is transmitted to it by the reciprocating parts, while the other is secured to the shaft of the engine. It is the final link in the transformation of reciprocating into rotary motion. The three differ- ent forms of crank are the side crank, disc crank, and the center crank. The center cranks are used wnen the shaft extends on either side of the crank. They are usually forged in one piece with the shaft, but weaken it somewhat. The side and disc cranks can be used in any case where the connect- ing rod is on the side of the engine. The advantage of the disc over the crank is that it affords better facilities for balancing. An engine throws over when the crank-pin traverses the upper por- tion of its travel while the piston is moving towards the main shaft and throws under when the crank-pin traverses the lower portion of its travel while the piston is moving toward the main shaft. In the first case the stress on the guides is downward, which is preferable; in the second case the stress is upward. Engines are usu- ally built to throw over, and it is only advisable to throw under in such cases where the transmission is such that the tight side of the belt is on the top of the pulley, which is never advisable. Q. 46. (1900-01.) What is an eccentric? In what respect does it differ from a crank? When is it used in preference to a crank? Why? Explain the throw of the eccentric. Ans. 46. An eccentric is substantially a crank, with its pin enlarged in diameter so as to enclose the shaft on which it is placed within its periphery- It gives exactly the same motion that would be obtained from an ordinary crank of equal throw. Eccentrics are generally used for converting rotary into reciprocat- ing motions, while cranks are used for the opposite purpose, although the latter can accomplish both results. The principal reason why eccentrics are used instead of cranks to actuate the valve gear is because the motion must generally be taken off at some point near the middle of the shaft, hence a center-crank would be required, which would weaken the shaft. The distance between the center of the eccentric sheave and the center of the shaft is called the throw of the eccentric or eccentricity. Q. 47. (1900-01.) Calculate the diameter of a shaft that will safely transmit 1,000 HP. at 100 R. P. M., considering torsional stresses only. If the transverse stresses were taken into consideration, how would you proceed to find the true diameter? Ans. 47. Rule (a) If steel, multiply the H. P. to be transmitted by 75 and divide the product by the number of revolutions per minute. Extract the cube root of the results, which will be the required diameter of the shaft in inches. (b) If wrought iron, multiply the H. P. to be transmitted by 100, then proceed as above. Example Use formula for steel shaft. 119 1000 X 75 = S V 750 = 9.0856 inches = dia. of shaft. 100 The usual method to determine the proper diameter for the crank shaft is to calculate it according to the above rule, then consider it as a beam carrying a load equal to the total maximum pressure of the steam on the piston, and determine what diameter would be necessary to safely carry the load. The greater of the two results will be the proper diameter for the shaft. _____ Q. 48. (1900-01.) What are the functions of a "fly-wheel"? Where should the bulk of the metal be concentrated? Should it be evenly balanced? How would you proceed to find the safe weight of a fly-wheel for any size or style of engine? Suppose an engine is 12 " X 24 ", making 140 R.P.M., diameter of wheel 6 feet; what would be the proper weight of the wheel? Ans. 48. The function of a fly-wheel is to equalize the motion when- ever either the power communicated or the resistance to be overcome is variable. In one case where the fly-wheel is used to overcome a varia- ble resistance, it may be considered a conservator of power. In the other case the fly-wheel may be said to be a distributer of power. The fly-wheel, as before stated, is a regulator and a reservoir, and not a creator of motion. As regularity of motion is of much greater importance in some cases than in others, the weight and diameter of the fly-wheel must depend upon the work and character of the machinery it is intended to drive; so that in proportioning a fly-wheel to a given engine, attention must be paid to many particular circumstances rather than to any given rule. The effectiveness of the fly-wheel in steadying the motion of the engine depends upon the distance of the metal from the center. For this reason the material of which the fly-wheel is composed should be concentrated as much as possible in the rim. The steadying action also varies as the square of the speed of the rim. Hence within certain limits increasing the diameter saves weight. The speed of the rim is rarely above 80 feet per second, and if car- ried beyond 200 feet per second the strains produced by centrifugal force would probably be sufficient to rupture the wheel. Great care should be taken in erecting fly-wheels to see that they are perfectly balanced that is, that the center of gravity of the wheel coincides with the center of the shaft. Rule For finding the weight of a fly-wheel having the size of the cylinder, the diameter of the wheel and the revolutions per minute given , First Multiply area of the piston by the length of stroke in feet. Multiply this product by constant, 12,000,000. Second Square the number of revolutions. Multiply this by the square of the diameter of the wheel in feet. Divide the first result by the second and the quotient is the proper weight of the fly-wheel in pounds. Example As per question 12 in. dia. = 113 in. area. First 113 X 2 X 12,000,000 = 2,712,000,000. Second 140 2 X 6 2 = 705,600. 2,712,000,000 -^ 705,600 = 3,857 Ibs. The weight of the fly-wheel Q. 49. (1900-01.) Explain the terms: "-Valve gear" "releasing valve gear," "automatic cut-off," "positive cut-off," "riding cut-off" and "re- versing gear." Ans. 49. The term valve-gear embraces all intermediate connections between the eccentric on the driving-shaft and the valves, and is appli- cable to all mechanical arrangements employed for working the valves of steam engines. Releasing valve-gear is an arrangement in which the valve is lib- erated from the control of its moving agent, and allowed to close in obedience to the action of a spring, weight or other force independent of that which opened it. An automatic cut-off valve-gear is one in which the movement of the cut-off vaive is so controlled by the governor as to cut off the steam as early or as late in the stroke as may be required, to maintain the desired uniformity of speed, under variations of load and pressure. Positive cut-off is an arrangement of valve-gear by which the expan- sion of steam is effected by what is known as lap on the valve, the steam being cut off at the same point in each stroke, independent of load or pressure. Riding cut-off is a term applied to cut-off valves which ride on the back of the main steam valve. A "reversing" valve-gear is an arrangement employed for reversing the motion of engines. It is effected in different ways: in some cases with a single eccentric, while in others with two eccentrics, as in the case of the link; and in others still, by the means of a loose eccentric which revolves on the shaft, but is prevented from making a complete revolution by two stops so placed that one arrests it in the proper posi- tion for the forward, and the other for the backward motion. This last arrangement is peculiarly adapted to tug-boats and ferries, owing to the ease and quickness with which the engine can be reversed. Q. 50. (1900-01.) What are "relief valves," "balance valves," "ro- tary valves" and "gridiron valves"? Describe the plain slide valve and its action. Why are plain slide valves not used in large engines and high pres- sures? Ans. 50. Relief valves are used on the cylinders of large engines to prevent fracture of the cylinder-head and cylinder, in consequence of an accumulation of water in the latter. They are also used in many places for a relief from overpressure. Balance-valves are arrangements by which the weight on the back oi slide-valves, induced by the pressure of the steam, is relieved by the action of the steam in the steam-chest. Rotary-valves is a term applied to any valve that describes a revolu- tion in working. Semi rotary-valves is a term applied to all valves that have a vibra- tory or rocking motion, similar to the Corliss. Gridiron-valves are a modification of the slide-valve, containing a number of openings for the steam, by which means its travel and fric- tion are materially diminished. Multi-ported valves. The function of the common slide- valve is to admit steam to the piston at such times when its force can be usefully expended in pro- pelling it, and to release it when its pressure in the cylinder is. no longer required. Owing to the amount of power expended in moving the valve in large engines and high pressures, the plain slide-valve is but little used. Q. 51. (1900-01.) Define the terms "admission," "exhaust," "cut-off," "expansion," "compression," "angle of advance," "travel," "over travel," "inside lap," "outside lap," "steam lead," "exhaust lead," and "negative lead." T2I Ans. 51. Admission: The period during which the steam passages are open and steam is admitted behind the piston. Exhaust: The period during which the exhaust passages are open and the steam is exhausted from the cylinder. Cut-off: The point in the stroke at which the steam-valve closes. Expansion: The period during which the steam expands in the cylinder, beginning at the point of cut-off and continuing until the steam is released or the exhaust port is opened. Compression: The period during which the steam is compressed, which is from th time the exhaust port closes until the end of the Angular advance: The angle which would be formed by the eccen- tric when in its actual position with the position of the eccentric cor- responding to the central position of the valve, the crank being on its dead point. Travel of the valve: The total distance which the valve moves in one direction. If the valve rod was of infinite length, this would be equal to twice the throw of the eccentric. Overtravel: Is the distance traveled by the valve over and above tnat necessary to fully open the steam port Lap on the valve: The term lap on the valve denotes the amount the edges of the valve extend over the ports when the valve is in the center of its travel. The object of lap is to secure the benefit to be de- rived from the working of steam expansively. Lap on the steam side of the valve is termed outside lap, while lap on the exhaust side is termed inside lap. Steam lead: Is the amount the port is open at the beginning of thp stroke. The object of lead is to enable the steam to act as a cushion against the piston before it arrives at the end of the stroke, to cause it to reverse its motion easily and also to supply steam of full pressure to the piston the instant it passes the dead center. Generally the high- er the speed and the more irregular the work, the more lead will be required for any engine. Lead varies in different engines from 1-32 to 3-16 of an inch. Some valves have no lead at all, others less than none, or what is termed negative lead. Lead on the exhaust end means the amount of open- ing the valve has on the end from which the steam is escaping. The name applies alternately to each end of the cylinder. Q. 52. (1900-01.) Given the throw of the eccentric, angle of advance and inside and outside lap. show how, by the "Zeuner diagram," the distribution of steam may be studied. Ans. 52. Draw a line OX to represent the crank at the beginning of the stroke, and with this as a radius draw the crank circle XX 1 , X\ *. s , X 4 . Suppose the crank to turn in the direction of the arrow. Through the point O draw the line RR 1 , making the angle R'OY 1 equal to the angle of advance, and lay off the distances OR and OR 1 equal to the eccentricity or throw of the eccentric. On the lines OR and OR 1 as diameters draw the two circles ORCD and OER'F. With as a center and a radius OA equal to the outside or steam lap draw a circle ACD, and similarly with a radius OB equal to the inside or exhaust lap draw a circle BEF. Through the point O and the intersections C, D, E and F draw the lines OX 1; OX 2 , OXs and OX4. We are now able to take from the diagram all of the data necessary for a complete understanding of the distribution of steam in the cylin- der: OXi is the position of the crank when admission of steam takes place. OXa is thfe position of the crank when cut-off takes place. Hence X t OX 2 is the angle traversed by the crank during the period of ad- mission. OXs is the position of crank when exhaust opens. OXi is the position of crank when exhaust closes, hence XsOX4 is the angle traversed by the crank during the period of exhaust and X^OXi is the angle traversed by the crank during the period of com- pression. The distances from the intersection of the circles R and Ri with the lines OX 3 OX,, etc., representing the crank in its different positions to the center, represent the travel of the valve corresponding to those positions of the crank. The circle R represents the forward and the circle R 1 the return stroke, hence OK is the distance the valve has traveled from its central position at the beginning of the stroke. OK 1 , the same for return stroke. OA is the outside or steam lap, hence AK is the distance the steam port is open at the beginning of the stroke or steam lead. OR is the full travel of the valve. OB is the inside or exhaust lap, hence BK is the distance the ex- haust port is open at the beginning of the stroke or the exhaust lead. THE ZEUNER DIAGRAM. At the points C and D the travel of the valve is just equal to the outside lap; hence in these positions of the crank the steam port opens and closes respectively; similarly at the points E and F the travel is just equal to the exhaust lap; hence in these positions of the crank the exhaust port opens and closes respectively. If we lay down from the point A at a distance AH, equal to the width of the port, and with O as a center and a radius OH, draw an arc cutting the line OR, at J. JR is the distance the valve travels more than enough to fully open the port, or the over-travel. Similarly if we lay off from B the dis- 123 tance BL equal to the width of the port, and from the center O and a radius equal to OL draw an arc cutting OR at M, MR is the distance the valve travels more than enough to fully open the port for exhaust. It will thus be seen that by a careful study of the diagram all in- formation necessary for the proper design and setting Of the valve gear may readily be had. For example, in the above diagram the cut-off takes place a little later than % stroke. It is evident that if it is de- sired to have the cut-off take place earlier, say y 2 stroke, it will be necessary for the outside lap circle ACD to intersect the valve circle R in the line YY 1 . This may be accomplished by increasing the outside lap, by reducing the eccentricity, or by changing the angle of advance. However, any one of these changes would also effect the entire distri- bution, and it would probably be necessary to lay down several dia- grams before the most advantageous dimensions could be obtained. Q. 53. (1900-01.) Explain how the lap and lead may be determined without removing the cover to steam chest. Describe the piston valve and state its advantages and disadvantages as compared to a plain side valve. Ans. 53. Open the cylinder drain cocks and disconnect them from the drain-pipe, so that the steam may be seen and heard to issue from them. Or, open the holes made for the indicator, if there are any; then let in a little steam and turn the engine over by hand, and note the commencement and cessation of the flow of the steam, just when the steam is admitted and cut off. The point of cut-off can be most accurately ascertained by turning the engine backwards. The steam in this case will commence blowing at the same point of the stroke at which it would cease blowing when turning it forward; and, owing to the elasticity of the steam, the commencement of the issue is always more clearly defined than the cessation when the issuing orifice is small. For the same reason, the point of admission can be most accurately located by turning the engine forward. To determine the lead, having found the point of admission, make a mark on the valve-stem at a known distance from some fixed point, and another after the pin has reached the dead center; this will give the lead. If the admission forward takes place when the crank-pin is exactly on the dead center, there is no lead. Having obtained the lead and cut-off for both ends, the travel and length of the connection being known, a diagram may be constructed similar to the one in the previous question, which will give the lap and port openings. To obviate the extreme amount of friction of the common slide-valve, especially those of large size, where high pressure is used, the piston- valve has been designed to overcome this difficulty, it being a per- fectly balanced valve, and the only pressure on the seat is due simply to the weight of ihe valve. As usually constructed, the valve-chest is bushed, the bushing being accurately turned to form the valve-seat, and the valve is made tight by the use of piston rings, the same as steam pistons. In the Armington and Sims type it will be seen that the steam enters the cylinder around the inside edge of the valve, and also through additional passage cut in the valve. In this way the effective opening of the port for a given travel is greatly -increased. The ex- haust takes place around the outside of the valve, so that in reality the inside is the steam lap and the outside is the exhaust lap opposite as in the case of the ordinary D-slide valve. The principal objection to piston-valves is that the seat wears un- evenly that is; at the bottom only, and thus become leaky. However, the seat may be readily removed and replaced by a new one with little expense. 124 It is used in many of the best types of stationary high-speed engines, because it is simple, light and perfectly balanced. It is also much used in marine engines, especially for the high-pressure cylinders. Q. 54. (1900-01.) What is meant by a poppet valve? How is the lift determined? What is meant by a variable cut-off gear? Explain the Stephenson link motion by the means of a sketch fur- nished. Ans. 54. Poppet valves are those that open by rising from their seats. They are extensively used for the water distribution in pumps, and also in gas engines. In steam engines their use has been restricted to slow-running marine engines. The lift of such valves, if single, would be about one-quarter of their diameter; if double, about one-eighth; in either case would give an area equal to the steam port. Reversing the direction of motion of an engine is accomplished by a device, invented by Stephenson, by means of which the engine may be THE STEPHENSON LINK. not only reversed, but the cut-off raised to any desired extent in a very simple manner. This is called the link motion. It consists of two eccentrics, with straps and rods. The eccentrics are so placed that when one is in the right position for the engine to move forward, the other is in the position to run backwards. By rais- ing or lowering the link, motion will b^ communicated to the valve and the engine will move forward or backward as the case may be. The result of this combination is that the link receives a reciprocat- ing motion in its center, since when one eccentric is moving the end of the link in one direction the other eccentric is moving the other end of the link in the other direction; so that the link will have nearly the same motion communicated to it as if it were suspended from a pivot at its center. The horizontal motion communicated to the link by the joint action of the eccentrics is a minimum at the center of its length, which is equal to twice the linear advance and it increases towards its extremi- ties, being nearly equal at either extremity to the motion which would be imparted to it by the eccentric at that extremity alone without re- gard to the other. The valve rod is attached to a block which slides in the link, and the position of this block is varied by the means of a combination of rods and levers, attached in some cases to the block and in others to the link itself. In either case the link is suspended by a rod at some point, and the length of the rod, as well as the location of the point on the link to which it is attached, have an important influence on the motion of the link. The travel of the valve depends upon the distance of the block from the center of the link. By moving the link up or down on the block or the block up or down in the link, the travel of the valve may be increased or diminished. The central position corresponds to no mo- tion whatever, while the nearer the block is to either eccentric the more its motion will be under the control of that eccentric. Now, since the travel of the valve, other things being equal, determines the point of cut-off, it follows that the degree of expansion raises with the position of the block relative to the link. In the sketch, for example, suppose the front eccentric is set for forward motion and the back eccentric for reverse motion. The link is suspended by a rod (not shown), and in the position represented in the sketch the block is at the top, and it is therefore entirely under control of the forward eccentric. Consequently the engine is going forward and the steam is cut off at the latest possible moment, the exact point of the stroke when the cut-off takes place depending on the lap and other dimensions of the valve gear. Now, as the link is raised the travel of the valve decreases and cut-off takes place earlier, until the block is in the center, when there is not enough travel to un- cover the ports, and the engine comes to rest. It is obvious without further explanation that if the block is placed in the several positions controlled by the back eccentric that the effect will be the same, only in the reverse or opposite direction. The term "full-gear forward" means that the link is dropped to its full extent; while "full-gear backward" means that the link is lifted to its full extent When the link-block stands directly under the saddle-plate both parts are closed, and neither admission nor exhaust can take place. The distance between the block and the end of the link when in full gear is termed the clearance. The radius of the link is the distance from the center of the driving- axle or the shaft upon which the eccentric is located, to the center of the link, while the link itself is segment of the circle of that diameter. The length may be greater or less, but any variation from these pro- portions will give more lead at one end than at the other while working steam expansively, but the radius may be several inches shorter or longer without materially affecting the motion. The vital point in designing a valve-link motion is the point of suspension of the link. If suspended from the center it will invaria- bly cut off steam sooner in the forward stroke than in the backward stroke, while working expansively. It is customary to suspend the link at a point which is most used in the running of the engine. The ease and facility with which the link may be handled is a very important feature in its favor. Q. 55. (1900-01.) What is the function of a governor? Explain the action of a single throttling governor. What are its principal defects? In high speed machine why is the shaft governor used so exten- sively? Explain the action of a single form of a fly-wheel governor? Ans. 55. The function of a governor is to regulate or govern the speed of the engine, admitting the requisite amount of steam to do the work, sustaining the speed uniform or within a small percentage of it. s subject of regulating the speed of steam engines has of late years 126 received no little attention from engineers and practical inventors, and as a result various kinds of governors have been introduced. Governors when attached to throttle-valves, work under circum- stances that necessitate the use of openings for the passage of the steam that are too small in area, so much so that the useful effects of the steam are considerably diminished. On this depends the ill repute of throttling engines as compared with those which regulate with gov- ernor-controlled valve motions or variable cut-off. If the valve of a governor has too large openings it will, owing to the unsteady action of the governor, admit too large a quantity of steam and cause a jumping of the engine; then in trying to shut off the extra amount it shuts it all off; in fact, the governor cannot fix it exactly right, being incapable of delicate changes. This difficulty is best met by making the openings in the valve of peculiar shape, so that they open and close in a ratio different from that of the governor. The principle of centrifugal force, as embodied in the old fly-ball governor of Watt, has been more resorted to than any other; but, aside from this, the governor has been so improved, altered, and reconstruct- ed since his time as to be almost unrecognizable; but still the old prin- ciple remains, and also the prominent defects which so materially inter- fere with its efficiency. The principal defects are three (3) in number. First is the friction which arises from the joints; second, defect is due to the fact that the balls as they assume different positions in keeping with the speed with which they revolve are obliged to rise and fall. This is necessary in order that the resistance which the weights offer to centrifugal force should constantly increase. If it did not so increase, the weights, when once started from their position of rest, would instantly go to the extreme limit of motion. The rising of the balls shortens the distance which they are allowed to move for a given variation by bringing the centers of the balls and arms on which they swing into a straight line, so that a variation which moves the balls a given distance upward, if it occurs again, will not move them nearly so far in the same direc- tion. Again, the same force that would support the balls in any plane would not raise them to that plane from a lower one. So between fric- tion, which destroys the delicate power that the balls assume under a slight change, and the necessity for a large change to overcome their inertia, it is almost impossible to obtain a degree of regulation which would be equal to all requirements. The third defect in governors on throttling engines is that the valve stem has of necessity to pass through steam-tight packing boxes. There is also friction on the governor valve necessary to overcome the power required to move the valve-stem through all its bearings, guides, stuffing-boxes, etc., under the pressure of steam. Were it possible to construct a governor for throttling engines which would approach in practice what theory would demonstrate the fly-ball or centrifugal governor would be a perfect regulator; but this appears under mechanical laws impossible. By the use of insochronous governors, which would not admit of any variation of speed, but would be in equilibrium at any speed, whether the balls were up or down or in any other position, the defects of the common governor were supposed to be obviated; but it was found by experience that power and stability were necessary, and isochronism in its strict sense unattainable. In the fly-wheel governor these defects have been partially elimin- ated, and it has been found that much closer regulation is attainable with the aid of governors of this class and also that there is less variation of speed during each single stroke. For these reasons nearly all modern steam engines, excepting the roughest type, are equipped with governors which act upon the cut-off. Jt will be readily understood that the tension of the springs and the 127 weights may be so adjusted as to maintain a position of equilibrium and thus produce an approximately uniform speed of rotation. With the aid of a well-designed governor of this type it is possible to regu- late the engine to within 1 per cent of its rated speed from full load to no load or no load to full load. The action of the fly-wheel governor, which is the type used on most high-speed engines, may be described as follows: The arms carrying weights are pivoted to the fly-wheel and springs are attached to these arms and the rim of the wheel, the links con- necting the arms and collar, which is placed loosely on the main shaft, which carries the eccentric. When the fly-wheel revolves there is a tendency for the weights to travel outward that is, away from the shaft and the centrifugal force which actuates to move in that direc- tion is counteracted by the tension of the springs. If the speed in- creases this tension is partially overcome and the weights move out- ward, and in so doing shift the eccentric and changing its eccentricity, angle of advance, or both, and thus produce an earlier cut-off in the cylinder, or, as the case may be, this operation may be reversed and a later cut-off take place. Governors should be kept perfectly clean and free from accumula- tions induced by the use of inferior oils, as such gummy substances have a tendency to interfere with the easy movement of the different parts; parts working through packings should be frequently packed, that the packing may be kept soft. Q. 56. (1900-01.) What is the effect on the steam distribution if the cut-off is varied by altering the angular advance only or the eccentricity only? In what form of valve gear is it not necessary to vary both the angular advance and the throw of the eccentric? Ans. 56. In the case of a single valve operated by a shaft governor, if the angle of advance only is altered, the cut-off may be varied to suit the conditions of load, but in that case the lead will also vary in such a way that it would increase as the cut-off decreased. If, on the other hand, the regulation is performed by varying the eccentricity alone, the reverse would take place, and hence for shaft regulation in connection with a single valve it becomes necessary to vary the eccentricity and angle of advance simultaneously. This is not the case where separate distribution and expansion valves are used, as in the "Buckeye Engine," because the admission and ex- haust closure are affected by the distribution valve, and the governor acts only on the cut-off. Q. 57. (1900-01.) How would you calculate the diameter of gover- nor pulleys? What is the maximum variation in speed allowable with a good shaft governor? Explain the action of the governor in the Corliss type of engines. Why can Corliss engines not operate at a high rotative speed? Ans. 57. To find the diameter of the governor shaft-pulley Multiply the number of the revolutions of the engine by the diameter of the engine shaft pulley and divide the product by the number of revo- lutions of the governor. To find the diameter of the engine shaft pulley Multiply the number of revolutions of the governor by the diameter of the governor shaft pulley and divide the product by the number of revolutions of the engine. 128 The maximum variation in speed in the modern first-class high-speed engines should not exceed more than 2 per cent. Many engines are guaranteed that from full load to no load the variation will not exceed 1 per cent. In Corliss engines the economy in the use of steam and the close regulation which has been attained certainly makes this class of engines very desirable for a great many purposes. Owing to the method of effecting the cut-off they are, however, lim- ited as to speed of rotation, and it is consequently customary to secure the advantages which are to be derived from high-piston speeds by making the stroke very long. Of course, this renders them unfit for direct coupling to electrical and other machinery which needs to run at a comparatively great number of revolutions per minute, but there are many other uses to which the Corliss can be put to advantage. Q. 58. (1900-01.) What are the characteristics of high speed auto- matic cut-off engines? For what class of service are they especially adapted? Why? What is the steam consumption per horse-power in this class of en- gine? Ans. 58. The class of engines known as the high-speed automatic cut-off type, which now comprise a large variety of designs, owes its development largely to the unusual growth of electric lighting and power, isolated plants and electric railway service. Engines of this class, while applicable under various conditions, are designed primarily to meet the requirements which the nature of this service imposes. An engine used for driving a lighting or power generator is con- stantly subjected to sudden, and very often considerable, variations of load, and must under these circumstances maintain a constant or nearly constant speed. It must also be economical in the use of steam, run at a comparatively high rotary speed, and be simple in design. In general, these, briefly, are the conditions which have evolved the high- speed automatic cut-off engine, these engines, while not so economical in the use of steam as the Corliss type, are vastly better than the old throttling engines. They consume from thirty to thirty-five pounds of steam per horse-power hour, when operating under an initial pressure of eighty pounds and cutting off at one-quarter stroke, while where compounded, which is frequently done, their steam consumption is about twenty-five pounds. Q. 59. (1900-01.) Explain what is meant by a "four- valve" engine, and why it is more economical than a single-valve engine? What are the advantages and disadvantages of single-acting engines? Ans. 59. The four-valve engine is a type which has been designed to combine some of the principal advantages of the Corliss type and the high-speed engines. It resembles the high-speed type in that the cut- off is varied to meet the changes in load by means of a shaft governor, and it resembles the Corliss engine in having separate passages for the admission and the exhaust, thus avoiding the losses inherent in single- valve engines and at the same time retaining the advantages to be derived from high rotative speed, which may be had on account of the absence of the releasing mechanism. There being separate parts -for admission and exhaust, the fresh steam does not come in contact with the surfaces which have been previously cooled by the exhaust, hence the condensation of the steam i reduced to the minimum. Some of the advantages of the single-acting engine are high rotative speed, simplicity of design and the economy that has been obtained in steam consumption. The results which are attained in this respect are due to high speed, multiple expansion, quick cut-off and short steam passages. The lim- 129 ited amount of floor space occupied makes its application very desira- ble where the available space is limited. The two principal designs of this class of engines are the "Willans" and the Westinghouse. These engines are well adapted for a variety of uses. They are especially valuable in location where the engine is exposed to dust, since the working parts are almost completely enclosed in the casing. The high speed and close regulation make it useful also for electric lighting and railway service, although it is not nearly so economical in the use of steam as many other types of engines. Q. 60. (1900-01.) What are the most frequent causes of knocking in steam engines? How may knocks be located? What are the remedies for different kinds of knocks? Explain how an engine with a fixed eccentric may be reversed. Ans. 60. The most frequent cause of knocking in steam engines are lost motion in the cross-head, wrist and crank pin boxes, looseness in the pillow-block or main bearing boxes, looseness of the piston rod or follower-plate, the crank pin or crank shaft being out of line with the cylinder or the wrist pin, crank pin or main bearing journal being worn oval; the slide valve having too much lead or not enough; the exhaust opening being too soon or too late; the valve being badly proportioned or the exhaust passage outside of the cylinder being contracted. Other causes are shoulders being worn in each end of the cylinder, in consequence of packing rings not traveling over the counter-bore at each end of the stroke; or shoulders being worn on the guides, result- ing from cross-head shoes not overlapping them when the crank is on the dead center; the piston not having sufficient clearance at either end of the cylinder, in consequence of its being altered by taking up the lost motion in the boxes; there not being sufficient draught, in the keys to take up the lost motion in connecting-rod boxes; the packing being screwed too tight round the piston-rod; excessive cushioning, resulting from the leaky condition of the piston, which allows the steam to occupy the space between the cylinder and piston-head, as the crank approaches the center, thereby subjecting the engine to an enormous strain, as at this part of the stroke the fly-wheel is traveling very fast and the crank moving very slowly; lost motion in the connection by which the slide-valve is attached to the rod. Engines out of line frequently knock sideways at the half-stroke, but most generally at the outward and inward upper or lower dead center, as these are the points that the greatest strain is thrown on the bearings in consequence of the direction of the connecting-rod hav- ing to be reversed. The foregoing causes of knocking in engines con- stitute the principal ones. Knocks arising from lost motion in any of the revolving, recipro- cating or vibrating parts of an engine may be located by placing the fingers on the part, while the cross-head is being moved back and forth on the guides by the starting bar; but knocks induced by the valve opeing or closing too soon, by the contraction of the exhaust, or by the valves being improperly set, are the most difficult to discover, as they are different from those induced by lost motion, the sound being a dull, heavy thud in many instances, causing the engine building and even the foundation to vibrate at each stroke. While an intelligent and careful search will in most cases result in successfully locating the knock, some will for a time baffle the most expert engineer. There are instances where the indicator has been applied in order to determine the precise location of the knock or thud. 130 Remedies for knocking in steam engines: While it is possible in most cases to locate the knocking, it is hardly possible to prescribe a remedy for all, as in many instances it must arise out of and be deter- mined by circumstances of the individual case. The most practical method of remedying the knocking induced by the crank-pin being out of line, is to place the crank-shaft at right an- gles with the center of the cylinder, remove the old crank-pin, rebore the hole so as to bring the center of the new pin perfectly in line with the axis of the cylinder, and replace the old pin with a new one. The knocking induced by the wrist-pin and crank-pin becoming worn oval, may be remedied by filing them perfectly round; but the knocking caused by the crank-shaft journal being worn out of round is very dif- ficult to remedy; in fact, there is no remedy for it except remove the shaft, true it up in a lathe and refit the boxes; this operation is attend- ed with considerable difficulty, more especially when the engine is large. Knocking in boxes on the crank-pin and cross-head or valve rod may be remedied by refitting the boxes, readjusting the keys or by putting a liner behind or in front of the boxes when there is not sufficient draught in the keys and gibs. Knocking in the steam-chest caused by looseness in valve connec- tions may be remedied by readjusting the jam-nuts or the yoke. Knock- ing arising from this cause manifests itself more frequently when the steam is shut off from the cylinder, preparatory to stopping the engine, than when the engine is running; the lost motion is taken up in the valve connections by the pressure of the steam on the back of the valve. Knocking in the piston is generally caused by the rod becoming loose in the head, and if it continues for any length of time it destroys the fit of the rod in the hole. The only practical remedy is to remove the rod, rebore the hole, bush it or thicken the rod at that point by welding, and fit to the head after the hole is rebored perfectly true. Knocking in follower-plate is generally caused by bolts being too long or from dust allowed to accumulate in the holes, which prevents them from entering sufficiently far to take up the lost motion in the plate. Remedied by cleaning out the holes or shortening the bolts. Knocking caused by shoulders becoming worn in cylinders at each end can be remedied by reboring the cylinder, making the counterbore sufficiently deep that a part of one of the rings will overlap it at each end of the stroke. Shoulders worn on guides can be remedied by planing the guides and making the shoes sufficiently long that they will overrun the guides when the crank is at either center. The knocking induced by any of the foregoing causes is generally a source of great annoyance, as any attempt to adjust the boxes on cross- head or crank-pin or the piston-packing in cylinder generally aggravates the cause of the knocking, as any adjustment of the connecting rod boxes alters the position of the piston in cylinder and the cross-head on guides, and causes them to strike harder against the shoulders. Knocking caused by the valve or valves being improperly set may be remedied by removing the bonnet of steam-chest and adjusting the valve, so that it may move uniformly on its seat, thereby giving the valve the same and proper amount of lead at each end of the stroke; then if the valve is well proportioned and the connections thoroughly fitted and skillfully adjusted there is no reason why the engine should knock from this cause. The knocks arising from bad proportion in the valves and steam passages are the most difficult of all to remedy, as they are inherent in the machine. Hov/ to reverse an engine: Place the crank orr the dead center and remove the bonnet from the steam chest; observe the amount of lead or opening that the valve has on the steam end; then loosen the eccentric and turn it around on the main shaft in the direction in which it is intended the engine should run until the valve has the same amount of lead on the other end. The engine should then be turned on the other center for the purpose of equalizing the lead; the crank should also be placed half-stroke, top and bottom, for the purpose of determining whether the part opening Is the same in both positions.When the crank is half-stroke the center of crank-pin is plump with the center of crank-shaft. 132 Pumps, Compressors, Hydraulics, Refrig- eration, Heaters, Condensers, Injectors, Etc. Q. 8. (1896-7.) How to properly set duplex steam pump valves? Ans. 8. Place steam pistons in the center of their travel, which will bring the rocker arms perpendicular to rods. Place slide valve over center of steam ports. The lost motion in pumps of small size that usually have non-adjustable blocks, on valve stems, should meas- ure the same on each side of blocks, between valve lugs. If lost mo- tion is equal the valves are set. Move one valve so as to admit steam before putting on steam chest covers. On the medium and larger sizes, where adjustable locknuts are used on valve stems, there are no fixed rules for the exact amount of lost motion, which governs the length of stroke, and this can only be determined satisfactorily by trial and careful adjustment. In the large pumps with the outside slotted link and sliding block, where the valve movement is coincident in time and amount to valve rod movement the distance from each end of link block to each end of link slot should measure the same when slide valves are central over ports. Q. 9. (1896-7.) HP. required to raise 300 tons water 150 ft. in one and one-half hours, barring friction? Ans. 9. 300 (tons) X 2,000 (Ibs.) X 150 (ft.) -r- 1.5 (hours) -=- 60 (min.) -T- 33,000 (ft. Ibs.) = 30 10/33 HP. Q. 12. (1896-7.) How to find steam cylinder diameter for direct acting boiler feed pump? Ans. 12. Find the maximum amount of water required. Find the size of the plunger that will displace this amount of water at a fair piston speed and make the steam end 2*4 times the area of the water end. [Pump cylinder should allow 2% to 15% for slip.] Q. 13. (1896-7.) How many tons of ice (spec. gr. .9188) to fill a room 15 ft. X 16 ft. X 10 ft.? Ans. 13. 15 (ft.) X 16 (ft.) X 10 (ft.) X 62.5 (Ibs.) X .9188 (spec, grav.) -r- 2,000 (Ibs.), = 68.91 (tons). [Ice should be packed with 1" strips of wood between, hence allowance must be made for the strips.] Q. 16. (1896-7.) What is efficiency of direct acting steam pump from coal in furnace? Ans. 16. A direct acting steam pump requires about 120 Ibs. steam per hour per HP. Assuming that the boiler evaporates 10 Ibs. of watef per pound of coal there would be 12 Ibs:. coal used per HP. per hour. 133 The number of foot pounds used per hour in the furnace would he 13 000 X 778 X 12, assuming the value of each pound of coal to be 13,000 heat units. This divided by 60 would give the heat units used per min- ute and this result multiplied by 100 and divided into 33,000 would give the percentage of efficiency. = 1.66 % efficiency. Q. 30. (1896-7.) What is the cause of water hammer in the dis- charge pipes of pumps, and how is it avoided? Ans. 30. The inertia of the water, which, when in motion, tends to continue in motion, after the impelling force has ceased to act and its energy of motion is expended in giving a blow to the containing p:pe. Any means which will induce an uninterrupted flow, or which will bring the water gradually to rest, will prevent water hammer. The first result can be obtained by the use of a large air chamber, or, in the case of a duplex pump, by adjusting the steam valves so that one piston begins its stroke before the other has stopped. The second result (gradually checking the flow) can be secured by placing an air chamber on the discharge pipe near the end at the pump or by having the pump cushion in such a manner as to bring the piston gradually to rest and not keep up its speed to the extreme end of the stroke and then abruptly stop. The admission of air with the water entering will, in some cases, prevent the trouble and a small snifting valve, opening inward, placed on the suction pipe near the pump, will in case the discharge pressure Is low, sometimes stop the hammer. Q. 48. (1896-7.) What is the most economical for feeding boilers, a direct acting pump, a power pump, or an injector? Has one any ad- vantage over the others under certain conditions? If so, when and how? Ans. 48. In cases where the exhaust from engines can be utilized for heating the feed water the injector cannot be used with economy, because, first, if an open heater is used the injector will not take the water; second, if a closed heater is used the feed will not take up much of the heat in the exhaust steam for the reason that the feed water, after passing through the injector, is nearly as hot as the ex- haust steam itself. A power pump driven by a belt from the shafting is more economical as a boiler feeder than either a direct acting pump or an injector. It has, however, this disadvantage, that the shaft must run in order to get water into the boiler. On account of this it is not generally used. The direct acting steam pump, while it is not as economical as the power pump, is independent of the engine, and can be placed in any part of the boiler or engine room, and is ready to start as soon as there are a few pounds pressure of steam in the boiler. The injector, while it is still less economical than either the power pump or direct acting pump, is, in cases where the feed water is cold and no waste heat is available for heating it, preferable and more eco- nomical than either the steam pump or power pump, first, because it furnishes hot feed water for the boiler, and, second, because the injector, taken as a pump and feed water heater, has a very high effi- ciency. But, all things taken into account, we consider the direct acting steam pump the most economical method of feeding boilers. Q. 54. (1896-7.) How do we find the amount of injection water required per HP per hour? Ans. 54. The amount of injection water required per indicated horse-power will depend upon, first, the temperature of the injections; second, the economy of the engine; third, the final temperature of the hot well. The unit of water required per unit of steam condensed can be found by the following formula: Let I = temperature of injection. Let U temperature of discharge. Let S = total heat of steam discharged at release. S D = lbs. of water per Ib. of steam condensed. D I Example: Temperature of injection, 60; temperature of discharge, 120; total heat above 32 in 1 Ib. of steam of 20 Ibs. pressure equals 1150 units; then 1150 120 1030 = = 17.16 Ibs. water per pound of steam condensed. 120 60 60 If 20 Ibs. of steam are used per horse-power per hour, then 20 X 17.16 = 343.20 Ibs. of water used per HP per hour. This amount increases with temperature of injection and inversely with the temperature of the discharge. Q. 81. (18^6-7.) A pumping engine works against a gage pressure of 43.4 Ibs., and a vacuum of 25" is needed to raise water to the pump: What will be the total height in feet against which the pump works? Ans. 81. No temperature being given, we will assume that 1 ft. of water exerts a pressure of .434 Ibs. per sq. in., also that 1" mercury equals 1.133 ft. of water, then: 43.4 = 100 ft. on discharge side and 1.133 X 25 inches vacuum = 28.325 .434 ft. on suction side, or 43.4 h [1.133 X 25] = 128.325 ft. .434 Q. 91. (1896-7.) An air compressor takes air at atmospheric pres- sure (15 Ibs.) and 60 F., and compresses same adiabatically to 120 Ibs. gage pressure. Find the temperature at the end of compression. The ratio of relative specific heats is 1.4. Ans. 91. T 1 = absol. temp, after compression. T == absol. temp, before compression, or 460+ 60= 520. P 1 = absol. pressure after compression, 120 + 15 = 135 Ibs. P = absol. pressure before compression = 15 Ibs. Tl /P 1 \ 29 /P 1 \ 29 /lSo\ 29 = c,T,=T- =T'=5- =520X9." /P 1 \ 29 /lSo\ 29 ,=T(|)- =T'=5Q- = log 9 = .954243 X .29 = .276731 representing 1.8915: 520 X 1.8915 = 983.32, total tempr., less 460 = 523.3. Q. 94. (1896-7.) Given a piston force pump with a lever of the second order, having a 2" piston and a ram with 25" diameter, pump lever is 5" from fulcrum to center of piston, and 60" from center of piston to end of lever: How much weight needed on end of lever, to equalize a load of 100,000 Ibs. on the ram, ignoring friction or weight of parts? 135 Ans. 94. 25 2 X .7854 = 490.8 sq. in. area of ram. 2 2 X .7854 = 3.1416 sq. in. area of piston. 100,000 Ibs. = 203.72 Ibs. pressure on each inch of ram and piston. 490.8 3.1416 X 203.72 = 640.07 Ibs. total pressure on piston. 640.07 X 5 .-. = 49.23 Ibs., or say 50 Ibs. 65 Or a shorter method: 25 2 = 625, 2 2 = 4. 100,000 640X5 = 160. 160X4 = 640. = 49.23 + 625 65 Q. 100. (1896-7.) In a refrigerating plant it is required to find the thermal units of refrigeration and of condensation, also the efficiency of compression, the specific heat of the liquid being 1.08, when the inferior absolute pressure (suction pressure) of the ammonia is 40 Ibs. to the square inch. The temperature when it enters the compressor is 25 F. and it is compressed to 165 Ibs. per square inch absolute with- out superheating, then condensed into a liquid of 85 F., then admitted to the cooling coils evaporated, and heated to the initial state. Ans. 100: Temperature absolute at 40 Ibs r 2 = 472 Temperature of liquid 472-460 T 2 = 12 Absol. tempr. of saturation 460+25 : . . r = 485 Superheating r-r 2 = 13 Absol. tempr. at end of compression ri = 545 Absol. tempr. of saturation 460 + 25 r = 485 Temperature of liquid at PI T t = 85 Fall of tempr. from Ti to T 2 T,-T 2 = 73 Heat absorbed during this fall 1.08 X 73 = 78.84 Latent heat of evaporation at T 2 h = 548 Condenser heat removed h t = 502 Refrigeration per Ib. of ammonia 548 + (13 X 0.508) 78 84. .h 2 = 475 76 Efficiency h 2 E= 18.13 That is, for every thermal unit of work done by the compressor 18.13 thermal units will be removed from the cold room. The efficiency of a compressor is greatest when the difference between the inferior and superior temperature (^ to r 2 ) is the smallest. Q. 38. (1897-8.) What is the greatest height to which a pump may draw water that is at a temperature of 191 F.? Ans. 38. The pressure of the atmosphere is about 14.7 Ibs. per square inch. The pressure of steam at 190 is about 9.5 Ibs. per square inch. There could never be a vacuum less than the pressure of the KMT' Q f -fS*,^ Pressure possibly available to raise the water is 14.7 9.5 = 5.2 Ibs. per square inch. This corresponds to 5.2 = 12 ft. .434 nearly. (See No. 28 of last year's questions.) 136 Q. 39. (1897-8.) What effect will the required velocity of the water in the inlet pipe to the pump have on the height to which the water may be raised, neglecting the friction of the water on the pipe? Ans. 39. It will lessen the height by an amount equal to the "head" required to give it the velocity. This head would be about equal to the square of the velocity (in feet per second), divided by 64. V 2 h = + 64. A number have answered No. 39 by saying that the velocity of the water in the inlet pipe would have practically no effect upon the height to which the water could be drawn. This is no doubt so if the velocity of the water is small and regular. Nevertheless we believe that very often when a pump fails to take water for its entire stroke the reason is to be found in the inertia of the water entering the inlet pipe. [Air in water and the vapor of water produced by the vacuum, as well as slip of valves, all tend to prevent the cylinder from filling. Ed.] Q. 29. (1898-9.) Define omits of ice making and refrigerating ca- pacity. Anc. 29. In refrigeration, an effect equivalent to the conversion of 2,000 pounds of water at 32 into a ton of ice at 32; that is, the ab- straction of 284,000 B. T. U. is considered a "ton," the latter word in this sense being a unit expressive of the duty just described. Q. 102. (1898-9.) Give the piston speed you consider best adapted for boiler-feed pump, when same is supplying a "continuous" feed. Ans. 102. A desirable piston speed for boiler feed pump when sup- plying a continuous feed approximates 50 ft. per minute. Q. 103. (1898-9.) Name the type of pump and dimensions necessary to supply the boilers referred to in Q. 93. Ang. 103. Dimensions of feed pump for boilers referred to in Ans. No. 93 may be determined as follows: Figuring on 200 HP as the normal demand on boilers but providing 50% for overload = 300 HP. Allowing 30 Ibs. water per HP plus 25% for non-efficiency of pump over its nominal displacement, gives 40 Ibs. 300 X 40 Or = 200 Ibs. of feed water to be "moved" per minute. 60 Assuming 50 ft. of piston travel as per Ans. No. 102 we have 200 -=- 50 = 4 Ibs. or % gallon as! the required displacement per foot of piston travel and which is equivalent to 3%" dia. plunger for single cylinder or 2 l / 2 " dia. for double cylinder pump. Stroke of pump may be any length common to the trade, for the given diameters. If conditions permitted a power pump, one of the "triple" type, would be preferred. The independent steam pump has an "emergency reserve," which is equal to any margin of speed that can be properly attained above the 50 ft. which is usually figured on. When the limit of speed in a power pump is fixed in its driving mechanism, such a reserve is not available and for that reason their capacity should be somewhat greater than figured for the steam pump. Power pumps may be kept in constant motion and arranged to sup- ply either a constant or variable feed by providing a suitable "by-pass" arrangement. '37 Q. 104. (1898-9.) What would probably be the size of the pipe con- nections on any standard injector, capable of supplying boilers of 200 HP? Ans. 104. IW would be the minimum size for the injector-pipe connections for the duty referred to in this question. Q. 105. (1898-9.) Explain the general features claimed for open and closed feed-water heaters. Ans. 105. Peed-water heaters tend to economy and obviate strains caused by forcing cold water into hot boilers. Ordinarily the saving of fuel amounts to about 1% for every 11 of temperature added to the "feed" by the heater. Heaters of the "open" type are preferable for waters that will pre- cipitate solid matter at temperatures near the boiling point. Well con- structed heaters of this style heat the feed water very efficiently; they intercept large quantities of "scale" making matter which is an especial advantage with certain waters. They can also be thoroughly inspected and cleaned when necessary. The exhaust, mingling with the feed water condenses a percentage of the steam but also causes oils from cylinder lubrication to become mixed with the feed. The mixing of the oil and the further disadvantage of pumping hot water do not obtain with "closed" heaters, but as both of these so- called "difficulties" which attend open heaters are readily overcome they are very much in vogue, for the reasons which are noted. Q. 67. (1899-1900.) What is the force against which a pump works, aside from boiler pressure? Ans. 67. Gravity, or the attraction of the earth, which prevents the water from being lifted. This is shown in the fact that water can be led, or trailed, an immense distance, limited only by the friction of the pump. Q. 68. (1899-1900.) In designing or purchasing a pump, what is a safe rule as to capacity? Ans., 68. One should be selected capable of delivering 1 cu. ft. of water per H.P. per hour; or, in other words, about 3 Ibs. of water for each sq. ft. of heating surface. Q. 69. (1899-1900.) What is the most necessary condition for the satisfactory operation of a pump? What is the advantage of a suction chamber? What should long suction pipes be provided with? Ans. 69. A full and steady supply of water. The pipe connections should in no case be smaller than the openings in the pump, and the suction lift and delivery pipe should be as straight and smooth on the inside as possible. A suction chamber (air) eliminates pounding, makes the action of the pump easy and uniform, also facilitates the filling of the barrel of the pump when at high speed. Long suction pipes should be provided with a foot-valve just above the strainer in the well or pit. Q- 70. (1899-1900.) For boiler feed pumps, or pumps doing a sim- ilar duty, approximately what is the proportional area of the steam to the water cylinder? Ans, 70. The steam piston should have about 2% times the area of the water piston. There being no mechanical purchase in favor of nvprttfi a a m P1 S n> ^ mUSt have the greater area of the tw in order to overbalance the pressure on the water piston 138 Q. 71. (1899-1900.) What is the rule for finding the quantity of water, in gallons, pumped in one minute, at 100 piston ft. per minute? Ans. 71. The diameter, in inches, squared, and multiplied by 4, equals the required amount of water in gallons. 4 D 2 = number of gallons. Q. 72. (1899-1900.) How do you find the H.P. necessary to pump water to a given height? How many H.P. are required to pump (10,000,000) ten million gallons of water (150) one hundred and fifty ft. high in (10) ten hours, slippage and friction not taken into account? Ans. 72. Multiply the total weight of the water in pounds by the height in feet, and divide the product by 33000. 10,000,000 gals, pumped in 10 hours. 1,000,000 gals, pumped in 1 hour. 16.666 gals, pumped in 1 minute. 16.666 X 8 1/3 = 138,883 Ibs. in one minute. 138,883 X 150 = 631.3 H.P. 33,000 Q. 73. (1899-1900.) What have you to say in regard to friction and slippage in well designed and constructed pumps? Ans. 73. In well-designed and properly constructed steam pumps the slippage will amount to about 10 % of water pumped, and an allowance of 25 % would be a fair allowance for both the slippage and the friction of the pump. In old or badly constructed pumps, working against a very high pressure, or a very low. lift, the net loss would be increased to twice the percentage given; so that, in calculating the size of a feed pump, allowing for leakage of water, priming of the steam, blqwing off and all other leaks that may occur. For a steam engine the pump should have the capacity of 2 to 2% times the net quantity of feed water required for the work of the engine. For marine engines, liable to use salt water, the size of the pump should be so as to be able to deliver from 3 to 4 times as much. Q. 74. (1899-1900.) From what is power developed in forcing water into a boiler by an injector? Approximately, what is the difference in the velocity of the steam and the water? Ans. 74. From the difference in the velocity of the escaping steam from a boiler under pressure, and the velocity acquired by water from the same boiler, and under the same pressure, and at the same time. Approximately, the steam has a velocity of 16 to 18 times that of the water, varying with the different pressures. Q. 76. (1899-1900.) Give rule for finding the saving which may be expected by heating the feed water a given amount. What percentage of the saving of fuel may be expected by heating feed water from 60 to 200 for a boiler carrying 125 Ibs. gauge pres- sure of steam? Ans. 76. Divide the difference in the total heat of the water above 32 F., before and after heating, by the total heat required to convert it into steam from the given initial temperature; multiply this quotient by 100, and the product will be the percentage of saving to be expected. Water at 200 contains above 32, 168.70 B.T.U. Water at 60 contains above 32, 28.01 B.T.U. 139 16 g 70 28.01 = 140.69 = the number of heat units between the two temperatures of the water (60 and 200). There is in steam, at 125 gauge pressure, 1189.5 B.T.U. above 32 F or 1189.5 28.01 = 1161.49 B.T.U. above 60 F. This is the amount of heat that the boiler would have to supply to make a pound of steam from the given initial temperature of 60 F.; but we have seen that in raising the water to 200 F. 140.69 units are supplied by the heater, thus saving 140.69 * 1161.49 of the heat re- quired. 140.69 X 100 -=- 1161.49 = 12.1 %, the percentage of saving to be ex- The same result may be found from the use of the table showing the increase of temperature for each degree of initial temperature and steam pressure; thus, in the column of 60 and the pressure column of 125 Ibs. we find the increase .0861, the saving for each degree; hence .0861 X 140 = 12.054 or 12.1%. Q. 100. (1899-1900.) With an engine exhausting into a surface condenser, 1,145 Ibs. of steam, requiring 21,526 Ibs. of water for con- densation of the steam, the pressure of the steam entering the con- denser is 4 Ibs. absolute per square inch, the condensing water entering the condenser at temperature of 60 F., discharging at 115 F., what was the temperature of the condensation? How many inches of vacuum was being maintained? Ans. 100. Difference in temperature of the entering and the dis- charged water equals 115 F. 60 F. = 55 3 F. The B. T. U. in water at a temperature of 115 F. = 83.129. The B. T. U. in water at a temperature of 60 P. = 28.009; hence The number of heat units that the water will absorb in changing from a temperature of 60 F. to a temperature of 115 F. equals 83.129 28.009 = 55.12 B. T. U. The total weight of the water used 21526 Ibs. The total weight of the steam condensed 1145 Ibs. Each pound of steam condensed requires as many pounds of water as 1145 is contained in 21526. 21526 -T- 1145 = 18.80, equals the number of pounds of water required to condense one pound of the steam. One pound of water absorbs 55.12 B. T. U. Then 18.80 Ibs. water will absorb 18.80 times 55.12 B. T. U. 18.80 X 55.12 = 1036.256 B. T. U. equals the number of heat units absorbed in condensing 1 pound of the steam. One pound of steam at 4 Ibs. absolute pressure contains a total heat of 1128.6 B. T. U. above 32 F., so that the total heat equals 1128.6 added to 32, 1128.6 + 32 = 1160.6 B. T. U. If the water absorbs 1036.256 B. T. U. in condensing one pound of steam then it must leave in the condenser a temperature equal to the difference between 1160.6 and 1036.256. 1160.6 1036.256 = 124.344 B. T. U. = equiv. temp. 124.04 F., sus- taining a vacuum of about 26.4 inches. In practice this will be some less for the reason that the air gets into the condenser and the pressure will be higher than that due to the temperature. For this reason it is common to see condensers with a discharge temperature of 100 F., which by the tables should give a vacuum of 28 inches and over, while the gage shows but 26 inches and even less. Q. 101. (1899-1900.) Describe a surface condenser. Describe a jet condenser. Which requires the greater amount of water for condensa- tion purposes? In selecting a type of condenser what conditions exist- ing would govern the choice between the two types of condensers? 140 Ans. 101. A surface condenser is a vessel, or a receptacle, con- structed in various shapes, with double heads at each end. The space between the inner heads is filled with small brass tubes varying in sizes in different constructions from y z inch to 1 inch diam- eter. In some constructions the tubes are made fast in one tube sheet and in the other tube sheet made secure by means of stuffing box and gland. In some constructions 'they are made secure by stuffing box and gland at both ends. Some constructions the tubes are made fast in each tube sheet and corrugated their entire length. These different precautions are taken for an allowance of expansion and contraction due to the different temperatures to which they are subjected. The steam is usually passed through the annular space surrounding the tubes and the water forced through the tubes, although a "vice versa" process is allowable and is sometimes practiced. The steam from the exhaust of the engine coming in contact with the cool surfaces of the tubes, is condensed and falls to the bottom of the condenser, causing a partial vacuum, and is removed by the means of an air pump, the same pump also performing the duty of removing any air that may enter the condenser. The circulation of water in the tubes is usually kept up to a normal speed per minute by the means of a circulating pump. Two pumps are generally used with a surface condenser. The jet condenser is also constructed in various shapes, and is generally less bulky than the surface condenser. The steam from the exhaust pipe of engine meeting a spray of cold water, is condensed and with the condensing water drops to the bottom of the condenser and is removed by the aid of an air pump. This form of a condenser requires a much larger air pump than the surface condenser, the air pump for the jet condenser having to perform the double duty of the air and circulating pump for the sur- face condenser. The surface condenser requires a much larger quantity of condens- ing water than the jet condenser. The type of condenser to select for installation in a plant depends upon a great many local circumstances, a few of which are: Space available, condition or quality of water to be used for condensation purposes; also the cost of installation. Q. 102. (1899-1900.) What relative duty does an air-pump perform for a condenser? What relative duty does a circulating-pump perform for a surface condenser? Give rule, or express in formula, for finding the diameter of a sin- gle acting air-pump for a jet condenser (using a stroke of 18 inches). Ans. 102. The relative duty of an air pump to a condenser is to remove the water of condensation and the air that enters the con- denser with the steam, and through minor leaks. For a jet condenser, in addition to the above enumerated duties, it has to remove the condensing water. Rule. Multiply the total number of pounds to be condensed per minute by the number of pounds of cooling water per pound of steam; reduce the pounds of cooling water used per minute to cubic feet. Also the water of condensation; add the two together and their sum equals the volume of water to be handled per minute. Divide this volume of water by the number of strokes the pump makes multiplied by the length of the stroke and the quotient equals the cross-sectional area 141 of the cylinder of the pump; theoretically, for making allowance for the air to be removed, multiply the volume of water by 2.75 and the product is the area of the piston required. When the air and water are removed a partial vacuum is formed in the condenser causing the condensing water to be lifted into it by atmospheric pressure if the vertical distance is not above 20 feet. The relative duty of a circulating pump to a surface condenser is to circulate the condensing water through the tubes of the condenser to absorb the heat in the steam. In addition to the volume of water to be removed there is a large quantity of air; therefore the displacement in the air pump for a jet condenser must be in excess of the volume of water to be displaced. Good authorities place this at 2.75 times the volume of water for single acting pumps. Formula: For the volume of single acting pumps Q + q Volume = 2.75 n In which Q volume of condensing water. q = volume of water for condensation. n = number of strokes of pump per minute. Q. 103. (1899-1900.) Find the diameter of a single-acting air-pump for a condenser to an engine developing 500 I. H. P. using 18 Ibs. of steam per I. H. P.. per hour, exhausting into a condenser (jet) at a temperature of 140 F., using 22 Ibs. of water for condensing purposes per pound of steam used; the temperature of the condensing water 60 F.; air pump with a stroke of 18 inches, making 140 strokes per minute. Ans. 103. 500 X 18 = 9000 Ibs. of steam to be condensed per hour = 9000 -5- 60 = 150 Ibs. per minute; 150 X 22 = 3300 Ibs. of condensing water used per minute. 3300 Ibs. of water at 60 F., volume= 52.91 c. ft. 150 Ibs. of water at 140 F., volume = 2.44 c. ft. Total volume of water to be moved per minute by air pump = 52.91 + 2:44 = 55.35 c. ft. 55.35 H- 70 (strokes of pump) =.79 c. ft. per stroke. .79 c. ft. X 1728 = 1365.12 c. in. = volume of water per stroke of pump. 1365.12 X 2.75 = 208.56 inches, the area of the cross-section of the 18 pump cylinder. /20O6 ^/ = 16.3 inches, the diameter of the cylinder. Or, 16" X 18" pump at 70 strokes per minute. Q. 104. (1899-1900.) Upon what does the amount of condensing water depend? Give the rule, or express the formula, for finding the amount of water required. What quantity of condensing water per pound of steam is required to condense the steam exhausting into a condenser at 140 F the temperature of the condensing water 60 F., and the temperature of t&e hot-well 110 F. Ans. 104. The amount of condensing water depends upon its tem- perature entering the condenser, and the amount of heat to be absorbed 142 from the steam; also upon the type of condenser used; the jet con- denser taking less water than the surface condenser to perform the same work; the quantity depends principally upon the difference of temperature of the water and the steam. Rule. Subtract from the heat of the steam at the terminal pres- sure, the heat in the water of condensation as it leaves the condenser, and divide this remainder by the difference in the temperatures of the water before and after passing through the condenser. Formula HW Q R In which Q quantity of cooling water. H = heat units given up by the steam in condensing. W = weight in pounds of steam condensed. R = difference in temperature of cooling water and that of hot well. There must be sufficient water mixed with the steam to absorb and reduce the heat of the steam at a temperature of 140 F. to 110 F., the temperature of the hot well. The total heat of steam at 140 F. above 32 F. = 1124.64 B. T. U. The total heat of water from the condenser at 110 F. above 32 F. equals 78.11 B. T. U. 1124.64 78.11 = 1046.53 B. T. U. absorbed by the cooling water. The difference in temperature of the cooling water and the dis- charge from the condenser equals 110 60 = 50 F. Then 1046.53 -=- 50 = 20.93 Ibs. water to condense one pound of steam at a temperature of 140 F. to 110 F. 143 Electricity, Dynamos, Motors, Wiring, Etc. Q. 60. (1896-7.) What is the difference between a kilowatt and an electrical horse power? Ans. 60. One electrical horse-power equals 746 watts; one kilowatt equals 1,000 watts. Then 1 kilowatt equals 1,000 divided by 746, or 1.3405 H.P. Or, again, a kilowatt is 1,000 watts. An electrical horse-power is 746 watts; therefore the difference is 254 watts. Q. 61. (1897-8.) (a) What is a "volt"? (b) How do you fix its value in your mind for practical purposes? Ans. 61. (a) The volt is the practical unit of electro-motive force, electrical pressure (head) or difference of potential. It is equal to 10 s absolute, or C.G.S. units. An electro-motive force of one volt will pro- duce a current of one ampere when applied to a conductor having a resistance of one ohm. (b) Practically one cell of ordinary battery gives an electro-motive force of one volt. Latimer Clark's standard cell gives 1.436 volts under favorable conditions. The volt is usually conceived as about the E. M. F. of a Daniels cell, or the ordinary voltaic cell. This last is somewhat inaccurate, however, as the different cells have different pressures, from between .6 and .7 in the Edison Lalande to nearly two volts in the Grove, Bunsen, Grenet, etc. One association defines the volt as the pressure induced in a con- ductor [one foot in length] passing through a magnetic field at such a speed as to cut the lines of force at the rate of 100,000,000 lines per second. This is a useful conception in dealing with dynamos, but would seem more real to the writer if "a field capable of an aggregate pull of 225 Ibs." could be substituted for a field of 100,000,000 lines of force. An idea of the strength of 110 volts may be obtained by putting the end of your finger in a lamp socket having this potential. Q. 62. (1897-8.) What is an "ohm"? (b) What is the resistance of one hundred feet of No. 19 American gage copper wire? Ans. 62. (a) The ohm is the practical unit of electrical resistance. It is equal to 10" absolute, or C.G.S. units of resistance. The legal ohm is the resistance of a column of mercury 1 sq. millimeter in section, and 106 centimeters long, at a temperature of 32 F. We have thought that the ohm was best conceived of, as the resis- tance of a certain length of a certain sized wire. The writer always thinks of it as the resistance of a certain piece of german silver wire that he has strung on a board between binding-posts for the purpose of estimating the resistance (internal) of voltaic cells, etc. 144 An idea of the resistance of one ohm may be had from the follow- ing table, which gives the number of feet and number (B. & S.) of cop- per wire having a resistance of one ohm: FEET PEB OHM OF WIRE (B. & S.) 94 feet of No. 20. 150 239 380 605 961 1529 2432 3867 18. 16. 14. 12. 10. 8. 6. 4. (b) We find in Kent's tables that 124.4 ft. of No. 19 A.W.G. copper wire has a resistance of 1 ohm. Therefore 1 foot has a resistance of .008038 ohm; 100 feet a resistance of .8038 ohm. Q. 63. (1897-8.) What is an ampere? (b) How do you fix its value in your mind for practical purposes? Ans. 63. (a) The ampere is a certain quantity of electricity flowing in a certain time. It is the practical unit of current. It is one-tenth of the C.G.S. unit of current. The C.G.S. unit of current is that current which, flowing through a length of one centimeter of wire, acts with a force of one dyne upon a unit of magnetism distant one centimeter from every point of the wire. (This C.G.S. definition is a good illustration of the un-get-at-ableness of electrical definitions. The unit quantity of magnetism, or unit pole, is an imaginary thing that does not and cannot exist. The centimeter is about .4" and the dyne about 1/445,000 of a pound, however. Is there, then, no iconoclastic balm in Gilead that will mitigate this apparently necessary evil? Ed. Com.) (b) An ordinary 55-volt 16 C.P. lamp requires a current of one am- pere; a 110-volt lamp, .5 ampere; the ordinary 2000 C.P. arc circuit requires 9.6 amperes, or practically one C.G.S. unit. Q. 64. (1897-8.) What is a watt? (b) What is its value in foot-lbs. per second, and in B.T.U. per second? Ans. 64. (a) The watt is the practical unit of electrical power. It is equal to 10,000,000 units of power in the C.G.S. system. To measure the power carried by an electrical circuit we must measure both the volts and the amperes. The product of the volts and amperes is the watts, or electrical power: 746 watts one horse-power. (b) The value of a watt in foot-pounds per sec. is 33,000 =.7373 ft. Ibs. 60 X 746 The value of a watt in B.T.U. per sec. is .7373 -h 778 = .0009477 H.U. per sec. We have thought that the watt was best thought of in reference to its value in foot-pounds and in heat units. It is said that a wire will throw off about .005122 heat units per square foot of surface and per second for each degree F., that its temperature is above that of the surrounding atmosphere. Knowing the square feet of surface of the wire, the number of watts expended in it and the value of a watt in heat units, it is easy to calculate how hot the wire will become, and how much of the power is wasted in heat. ] 45 Q. 5. (1897-8.) Wliat is Ohm's law? Ans. 65. Ohm's law: (1) Amperes = volts -r- ohms = A = ( 2 ) Volts = amperes X ohms =V = AO (3) Ohms = volts :-=- amperes = O = A we will have a 1000-volt circuit, or the difference of potential is said to be 1000 volts. The potential decreases along the circuit in proportion to the resistance. We would say, for instance, the positive terminal of the dynamo has 1000 volts potential, the negative terminal has zero potential. After the current has passed ten lamps the difference of potential is 500 volts, etc. The potential of the earth is zero, and if a connection with the earth and the circuit is made at any point the current runs to earth, or is "grounded." Q. 66. (1897-8.) What is meant by the difference of potential of a current? Ans. 66. The potential of the current refers to the voltage between the terminals of the circuit, generally speaking. If we take, for exam- ple, ^n arc light circuit supplying 20 lamps requiring 50 volts per lamp, Q; 67. (1897-8.) What are the advantages and disadvantages of high tension currents? Ans. 67. The advantages of high potential currents consist in the ability to transmit electricity cheaply, as a smaller wire is required to transmit a given amount of electricity at a higher potential than at a lower one. Small wires cost less than large ones. The disadvantages of high potential currents arise from the difficulty of insulation, requiring more expensive insulators. High potentials are also dangerous to human life. -- Q. 68. (1897-8.) What is a transformer and how constructed? Ans. 68. A transformer is an apparatus for transforming currents of high potential and small amperage into currents of low potential and large amperage, and vice versa. The kind commonly used for transforming alternating currents con- sists of a core of laminated iron plates with two sets of windings. The primary current is sent by the dynamo through the primary coil and an induced current is generated thereby in the secondary coil. The change in potential and amperage is regulated by the size of wire and number of turns in the different windings. Q. 69. (1897-8.) Can a direct current be transformed? If so, how? Ans. 69. A direct current can be transformed by means of a rotary transformer. This consists of a motor having two sets of windings and two commutators. The primary current is sent through one wind- ihg and the armature turns, generating a secondary current in the other windings. By properly proportioning the windings a current of the ired voltage can be obtained from the secondary. These machines are in common use in large central stations, where they are commonly termed "boosters." Stationary or static transformers are used for alternating currents. 146 Q. 70. (1897-8.) In a circuit there is a shunt that has 10 times as much resistance as the part of the main circuit between the terminals of the shunt, what part of the total current will go through the shunt? Ans. 70. In dividing a circuit the current divides inversely propor- tional to the resistance of the circuits. In case the shunted circuit has ten times the resistance of that part of the main circuit between the terminal of the shunt, there will be one- tenth as much current through the shunt as through the main connec- tion between the terminals of the shunt. We may say that the current is therefore divided into eleven parts, one-eleventh going through the shunt and ten-elevenths going through the main circuit. Where the main current flowing in the main circuit is too large to be conveniently measured, the ammeter is frequently placed in a shunt through which a certain definite portion of the current is known to be flowing. Q. 71. (1897-8.) What is meant by the carrying capacity of a wire? Ans. 71. By carrying capacity of a wire is meant the amount of current it will carry without heating to such an extent as to increase the resistance too much or to cause danger of burning the insulation or wood work. Q. 72. (1897-8.) What is the allowable carrying capacity of the usual sizes of wire from No. 0000 to No. 18 B. and S. gage according to the National Board of Underwriters? Ans. 72. The rules of the National Board of Underwriters (1897) gave the following table of carrying capacities: No. of wire, B.&S. gage. 18 16 14 12 10 8 6 5 4 3 . 2 1 00 000 0000 Rubber- Circular covered Mils. Wire. Amperes. 1624 3 2583 6 4107 12 6530 17 10382 24 16510 33 26250 46 33102 54 41743 65 52634 76 66373 90 83694 107 105593 129 133079 150 167805" 177 211600 210 Weather- proof Wire. Amperes. 5 8 16 23 32 40 64 77 92 110 131 156 185 220 262 312 Q. 73. (1897-8.) How much current is used by a 16 c-p lamp on an average? Ans. 73. A 55-volt 16 C.P. lamp requires one ampere. A 110-volt 16 C.P. lamp requires .5 ampere, on an average. Q. 74. (1897-8.) What is meant by the "drop" in a circuit? Ans. 74. The drop in a circuit means the number of volts lost in overcoming the resistance of the circuit. It corresponds to the drop in pressure along a steam pipe or water main; in the latter case some- times called "loss of head." 147 Q. 75. (1897-8.) What is the allowable drop (a) for electric light wiring, (b) for wiring to motors for power? Ans. 75. In electrical wiring for lights various percentages of drop are allowed, from 1% to 10%, depending on circumstances. The larger per cent is used when the distance is long, as it is then cheaper to generate the extra electricity than it is to put in large wires. In wiring for motors various percentages of drop are used, from 1% to 30%, according to circumstances. It often happens that there is as much as 30% drop in street railway circuits at the end of long lines. In ordinary shop practice about 3% drop is allowed for lights and 5% for motors. Q. 76. (1897-8.) How can. the size of wire to be used be determined for conveying a given number of amperes a given distance with a given drop? Ans. 76. A wire which conveys a current of electricity must have an area of as many circular mils in its cross-section as is equal to 21 times the distance transmitted in feet, multiplied by the current in amperes and divided by the drop in volts. The constant 21 equals the resistance, in ohms, of a wire one circular mil in diameter and 2 ft. in length. By taking the resistance of 2 ft. length in this calculation it is then necessary to take only the distance in feet one way in the calculation. 21 X Amperes X Dist. in ft. Area in Cir. Mils = Drop in volts. After finding the circular mils, look up the corresponding size of wire in a table. (See Ans. No. 72.) Example: 150 amperes are carried 150 feet with 3% drop, 120 volts pressure. By the rule: 21 X 150 X 150 = 472,500. The number of volts drop is 120 X .03 = 3.6; then 472,500 H- 3.6 = 131,233 mils. Turning to answer No. 72, we find that this corresponds to 00 wire, very nearly. Q. 77. (1897-8.) What is a dynamo? How does it generate elec- tricity? Ans. 77. A dynamo is a machine for converting energy, in the form of mechanical power, into energy in the form of an electric current, or vice versa, by the operation of setting conductors, usually coils of copper wire, to rotate in a magnetic field, or by varying a magnetic field in the presence of conductors. A dynamo consists of two essential parts, a field-magnet and an armature. The field-magnet must be magnetized either by the dynamo itself or some outside source. The armature consists of coils of insulated copper wire wound upon a core which is mounted on a shaft and caused to rotate between the poles of the field- magnet in such a manner that the lines of magnetic force passing through the coils of the armature are constantly increasing or decreas- ing. This action causes currents of electricity to be generated in the coils of the armature, which are collected from rings or a commutator by suitable brushes and sent out on the circuit through suitable con- ductors. Q. 78. (1897-8.) What is meant by the magnetic circuit? Ans. 78. The magnetic circuit is the path or space through which the magnetic lines or force travel. Take, for example, the Edison dynamo. The field-magnets of this machine consist of five pieces, viz., the pole pieces D, E; cores B, C, and a yoke, A. (Fig. 8.) When the magnet is in action the strongest field is between the poles, D and E, but the magnetic circuit follows around through all the parts, D, B, A, C, E, and across to D. If an armature 148 FIG. 8. with an iron core be placed in the space between the pole-pieces, the magnetic circuit between the pole-pieces is intensified, as magnetism travels through iron easier than through air. (Note: In the last sentence we would say that with the same impelling force [ampere-turns] more magnetism would pass through the circuit because the resistance was less. Ed. Com. 1897-8.) Q. 79. (1897-8.) What is the relative permeability to magnetism of iron and air? Ans. 79. S. P. Thompson's third edition, pages 55 and 56, says: "If we take the magnetic permeability of air as 1, then the permea- bility of iron will be represented by values lying between 5 and 20,000, according to the quality of the iron" [and the density of the lines of force]. The conception of the magnetic circuit and its close analogy to the electric circuit is comparatively modern and exceedingly useful. One can think of the air gaps as affording thousands of times as much resistance as iron of equal length and cross-section. In Kent's handbook we find values of permeability showing that for grey cast iron it is from 37 to 800 times that of air, and that the perme- ability of annealed wrought iron is from 54 to 25,000 times that of air. Q. 80. currents. Ans. 8 (1897-8.) Give a table of diameters for fuses for different The diameter of a fuse may be calculated as follows: a=(c--a)f in which d is the diameter in inches of fuse, c the number of amperes carried, and a, a constant which varies according to the material of which the fuse is composed. The following are values of the constant, a, for the metals named: Copper = 10224; tin = 1642; lead = 1379; aluminum = 7585; iron = 3150. From the above formula may be deduced the following: Rule: Divide the constant, for the given metal, by the number of amperes required; multiply this quotient by itself (square it) and 149 extract the cube root of this product (square). Call this result No. 1. Second. Take unity (one) as the numerator and result No. 1 as the denominator of a fraction. This fraction expresses in inches the diameter required. RUPTURING CURRENT OF FUSES. n M Tin Wire "c ft 11 go u S M tS ' Metals> therefore . are said to be good conductors Experiments have also shown that in the case of the dry glass rod nhT a ^- e( l e v 0f EP** 1 ?** or resin ' that onlv that Part of the substance which has been rubbed will be electrified, the other parts will 156 produce neither attractions nor repulsions. These bodies do not readily conduct electricity; that is, they oppose or resist the passage of elec- tricity through them and are termed non-conductors of electricity. This distinction is not absolute, for all bodies to some extent conduct electricity, while there is no known substance but what offers some resistance to the flow of the electric current. The names of some of the most important in the list of good conduc- tors and non-conductors are given, classified as to their conductivity, etc.: Conductors Silver, copper, aluminum, other metals, charcoal, water, the body. Non-conductors Paper, oils, porcelain, wood, silk, resins, gutta percha, shellac, ebonite, parafflne, glass, dry air, etc. Q. 10. (1899-1900.) Define the word potential. For what is it sub- stituted and to what is it analogous? Ans. 10. Potential is a word substituted for the general and vague phrase, electrical condition. It is analogous with pressure in gases; heads in liquids, and temperature in heat. Q. 11. (1899-1900.) Explain the conditions when an electrified body is connected to earth direction of flow as regards a high or low poten- tial in relation to earth. Ans. 11. If an electrified body, positively charged, is connected to the earth by a conductor, electricity is said to flow from the body to the earth; but if negatively charged and connected to the earth in a similar manner, electricity is said to flow from the earth to that body. This is called the direction of flow of an electric current. That which determines the direction of the flow is the relative electrical potential, or the pressure of the two charges in regard >to the earth. It is impossible to say with certainty in which direction electricity really flows, or, in other words, to declare which of two points has the higher or lower electrical potential, or pressure. All that can be said with certainty is that when there is a difference of electrical potential, or pressure, electricity tends to flow from the higher to that of the lower potential or pressure. For convenience it has been arbitrarily assumed and conventionally adopted that the positive electrical condition is at a higher potential, or pressure, than that called the negative, and that electricity tends to flow from a positively to a negatively electrified body. Q. 12. (1899-1900.) What is meant by zero potential? Ans. 12. The normal level of the water is taken as that of the sur- face of the sea, the normal pressure of air and gases, as that of the atmosphere at the sea level; similarly there is a zero potential, or pressure, of electricity in the earth itself. The earth may be regarded as a reservoir of electricity '^r infinite quantity, and its potential, or pressure, taken as zero. A positive electrical condition may be assumed to be at a higher potential, or pressure, than the earth, and that called negative is as- sumed to be at a lower potential than the earth. Q. 13. (1899-1900.) What is first necessary to do in order to pro- duce an electric current? What is the effect of placing a piece of cop- per and zinc together; when separated slightly and one end of each submerged in saline or acidulated water? Describe the simple voltaic or galvanic cell. Ans. 13. To produce an electric current, it is first necessary to cause a difference of electrical potential between two bodies:, or be- tween two parts of the same body. 157 When two dissimilar substances are placed in contact, one always assumes the positive and the other the negative condition. There is instantly developed a difference of potential between the two bodies. A piece of copper and zinc placed in contact will develop a differ- ence of electrical potential easily detected. Should the plates be slightly separated, and one end of each sub- merged in a vessel containing saline or acidulated water, the same results will follow. The exposed ends of the zinc and copper are now electrified to different degrees; that is, there is a difference of poten- tial between them. The voltaic cell is an apparatus for developing a continuous current of electricity and consists essentially of a vessel containing saline or acidulated water, into which are submerged two plates of dissimilar metals, or one metal and a metalloid. The exposed ends are connected by any conducting material, the potential between the plates tends to equalize and a momentary dis- charge passes between the exposed ends through the conductor and also between the submerged ends through the liquid. During the passage of the current through the liquid, it causes cer- tain chemical changes to take place; these changes in their turn show a fresh difference- of potential between the plates, followed instantly by another equalizing discharge. These changes follow one another so rapidly that they form an almost continuous current. .An electric current becomes continuous when the difference of the potential is constantly maintained. The name given to the liquid in which the metals are submerged is "electrolyte," which, as it transmits the current, is decomposed by it. Of the submerged ends, the zinc assumes the positive condition and the copper the negative; of the exposed ends the copper is the positive and the zinc the negative; hence the flow of the current is from the submerged end of the zinc through the electrolyte to the submerged end of the copper, thence from the exposed end of the copper through a conductor forming any outside circuit to the exposed end of the zinc, flowing continuously when the circuit is closed, or remaining passive when the circuit is open. Q. 14. (1899-1900.) In any voltaic couple which is the positive and which is the negative element? Ans. 14. Two dissimilar metals, when spoken of separately, are called voltaic or galvanic elements; when taken collectively, are known as a voltaic couple. The metal terminals or poles of a cell, to which the conductors are attached, are termed electrodes, the polarity of which are directly opposite to the polarity of the submerged ends of the metal substance. The element which is acted upon by the electrolyte will always be the positive element, and its electrode or terminal the negative elec- trode of the cell in any voltaic couple. Q. 15. (1899-1900.) Electromotive series; name them in the regu- lar order and tell how the positive or negative element may be deter- mined. Give the difference of potential in relation of one element to another of the series. Ans. 15. The following list of voltaic elements composes what is known as the electromotive series; any two of which form a voltaic couple when submerged in an electrolyte, the one standing first on the list being the positive element to the one following it. And the ifference of potential will be the greater in proportion to the distance between the two substances in the list. For example, the difference of potential developed between zinc and 158 platinum is greater than between zinc and nickel, or equal to the sum of the difference of potential between zinc and nickel and between nickel and platinum. The electromotive series: 1, zinc; 2, cadmium; 3, tin; 4, lead; 5, iron; 6, nickel; 7, bismuth; 8, antimony; 9, copper; 10, silver; 11, gold; 12, platinum; 13, graphite. Q. 16. (1899-1900.) Name three ways in which electricity flowing as a current differs from static charges. Ans. 16. 1. The potential is much lower. 2. Its actual quantity is much greater. 3. It is continuous. The potential of a current of electricity is comparatively so small that a voltaic battery composed of a large number of cells is not suf- ficient to produce a spark of more than two-hundredths of an inch in air; whereas, a rapidly moving leather belt will sometimes produce sparks of one to three inches. If, however, the actual quantity of electricity is measured by its effect in decomposing water, then the quantity produced by a small voltaic cell would give greater results than that from a large rapidly moving belt producing static charges several inches in length. Q. 17. (1899-1900.) Can an electric current be developed between two non-conducting substances as in the case of static charges? Will it flow if the conductor is not made entirely of conducting material? Ans. 17. An electric current cannot be developed between two non- conducting substances, as in the case of static charges, and it will never flow unless the conducting path is made entirely of conducting material. Q. 18. (1899-1900.) How many ways are there of grouping the cells of a voltaic battery? Describe each, and the effect upon the potential for each grouping. Ans. 18. There are three different ways or methods of connecting or grouping the cells of a voltaic battery: In series; in parallel or mul- tiple-arc; in multiple series. Cells are connected in series when the positive electrode of the first cell is connected to the negative electrode of the second, and the positive electrode of the second is connected to the negative electrode of the third cell, and so on, as shown in the diagram Fig. 1. This method is used when there is available a large number of low potential cells and a high potential is desired, usually used for bell service and long telegraph lines or other high resistance circuits. The potential is increased on each addition of cell, that is, if the mm* Fig. 1. voltage of each cell is 2 volts; then a grouping of 12 cells in this manner would develop 24 volts, other groupings in the same ratio. Should we wish to produce a strong current at a low voltage or potential, that is, when the external resistance is low, as in electro- plating, the cells may be connected in parallel or multiple-arc; in this method only a part of the total current flowing in the main conductors will pass through each cell. The diagram Fig. 2 will show the method of connection: 159 Fig. 2. In a grouping of this kind, the potential would be equal to the potential of a single cell, no matter how many cells there are in the Should a stronger current and a higher potential be needed, then the grouping can be in what is termed multiple-series, each group being composed of two or more cells as the case may be, connected in series and then connecting all the groups in parallel or multiple-arc. The diagram will readily show the grouping and the effect on the potential, etc., each grouping according to the number of cells in the grouping determining the potential of the current developed, in this case the voltage equals that of two cells or four volts. Fig. 3. Q. 19. (1899-1900.) Describe a circuit broken or open, closed, grounded, external and internal. Ans. 19. A circuit is a path or a conductor, or several conductors joined together, through which an electric current flows from a given point around the conducting path back again to its starting point. A circuit is said to be open, or broken, when its conducting ele- ments are discontinued in such a manner as to prevent the current from flowing. A circuit is closed or completed when its conducting elements are uninterrupted or continuous. A circuit is said to be grounded when the earth forms a part of its conducting path. An external circuit is that part of a circuit which is outside or external to its electric source. An internal circuit is that part of the circuit which is included within the electric source. In a voltaic ce^ the internal circuit consists of the two metallic plates, or elements, and the electrolyte. The external circuit, a conductor connecting the free ends of the electrodes. Q. 20. (1899-1900.) When are conductors said to be in series? What do you understand by a shunt or derived circuit, in parallel or multiple arc? With your explanation, furnish sketch. Ans. 20. Conductors are said to be a series when they are so joined together as to allow the current to pass consecutively through each. In the diagram conductors a, b, c and d are in series. A shunt or derived circuit is one which is divided into two or more branches, each transmitting a part of the main current. The separate branches are said to be in parallel, or multiple-arc: C d The main current flowing through the conductor (a), dividing and passing through the branches (b and c), uniting and completing the circuit through the conductor (d). The two branches (c and b) are said to be in parallel or multiple- arc. Q. 21. (1899-1900.) What are magnets, and to what is' the term magnetism applied? Does magnetism exist in a natural state, and by what name is it known in chemistry? Where and by whom was the ore first found? From what fact was the name lodestone given to the ore? What are artificial magnets? What magnets are called perma- nent magnets? Describe the common form of artificial or permanent magnets? Why is a keeper used? Ans. 21. Magnets are substances which have the property of at- tracting pieces of iron or steel, and the term magnetism is applied to the cause of this attraction. Magnetism exists in a natural state in an ore of iron, which is known in chemistry as magnetic oxide of iron, or magnetite. This magnetic ore was first found by the ancients in Magnesia, a city in Asia Minor; hence substances possessing this property have been called magnets. It was also discovered that when a small bar of the ore is suspended in a horizontal position by a thread it has the property of pointing in a north and south direction. From this fact the name lodestone leading-stone-was given to the ore. When a bar or needle of hardened steel is rubbed with a piece of lodestone, it acquires magnetic properties similar to those of the lode- stone, without the latter losing any of its own force. Such bars are called artificial magnets. Artificial magnets which retain their magnetism for a long time are called permanent magnets. The common form of artificial or permanent magnets is a bar of steel bent in the shape of a horseshoe and then hardened and magnet- ized. A piece of soft iron called an armature, or a keeper, is placed across the two free ends, which helps prevent the steel from losing its mag- netism. Q. 22. (1899-1900.) What is the result of dipping a magnet into iron filings? Explain the neutral line, the poles and the axis of mag- netism? Explain the polarity of a magnet, the attraction and repul- sions in general. Ans. 22. If a bar magnet is dipped into iron filings, the filings are attracted towards the two ends, and adhere to these in tufts; while towards the center of the bar, half way between the two ends, there is no such tendency. That part of the magnet where there is no apparent attraction is called the neutral line; and the parts around the two ends, where the attraction is the greatest, are called poles. An imaginary line drawn through the center of the magnet, from end to end, connecting the two poles together, is the axis of magnetism. 161 Using a compass, which consists of a magnetized steel needle, rest- ing upon a fine point, so as to turn freely in a horizontal plane, the needle, when not in the vicinity of other magnets, or magnetized iron, will always come to rest with one end pointing to the north and the other end towards the south. The end pointing northward is the north pole, and the opposite end is the south pole. This polarity applies as well to all magnets. The attractions and repulsions are governed by the same rule as given in answer to one of the first questions. Electrified bodies with similar charges are mutually repellant, while those with dissimilar charges are mutually attractive; so it is with the polarity of magnets; unlike poles having attraction between them and like poles repelling one another. Q. 23. (1899-1900.) How is the earth considered in relation to magnetism? Is it possible to produce a magnet of one pole? Ans. 23. The earth is a great magnet whose magnetic poles coin- cide nearly, but not quite, with the true geographical north and south poles. It is impossible to produce a magnet with only one pole. If a long bar magnet is cut into a number of pieces each part will still be a magnet and have a north and south pole. Q. 24. (1899-1900.) What have you to say in relation to magnetic substance? What substances are magnetic and what are all other sub- stances called? Ans. 24. Substances which, although not in themselves magnets, not possessing polar and neutral lines, are, nevertheless, capable of being attracted by a magnet. In addition to iron and its alloys, the following elements are magnetic substances: nickel, cobalt, mangan- ese, oxygen, cerium and chromium. These, however, possess magnetic properties in a very inferior degree compared with iron and its alloys. All other known substances are called non-magnetic substances. Q. 25. (1899-1900.) What is the space surrounding a magnet called? The direction of magnetic attractions and repulsions? How may the positions of lines of force be determined? How may their direction be determined? Ans. 25. The space surrounding a magnet, in which any magnetic substance will be attracted or repelled, is called its magnetic field, or simply its field. Magnetic attractions and repulsions are assumed to act in a defin- ite direction, and along imaginary lines, called lines of magnetic force, and every magnetic field is assumed to be traversed by such lines of force in fact, to exist by virtue of them. Their position may be deter- mined (in any plane) by placing a sheet of paper over a magnet and sprinkling fine iron filings over the paper. In the case of a bar magnet lying on its side, the iron filings will arrange themselves in curved lines, extending from the north to the south poles. The nearer to the poles the more dense will be these lines of force, consequently more of the filings will converge around them. The magnetic field looking toward either pole of a bar magnet would exhibit merely radial lines, and as in the case of the bar mag- net, the lines of force will be more dense at the immediate poles. Every line of force is assumed to pass out from the north pole, mak- ing a complete circuit through the surrounding medium and into the south pole thence through the magnet to the north pole again. This is called the direction of the lines of force, and the path which they take is called the magnetic circuit. By the use of a compass the direction of the lines of force can be traced; the north pole of the needle will always point in the direction 162 Q. 26. (1899-1900.) Can lines of force intersect each other? What would be the result of two north poles, or two south poles, being brought near one another; or, in other words, two opposing magnetic fields? What are the resulting poles called? Ans. 26. The lines of force can never intersect each other; when two opposing magnetic fields are brought together the lines of force from each will be crowded and distorted from their original direction until they coincide in direction with those opposing, and form a result- ant field in which the direction of the lines of force will depend upon the relative strength of the two opposing negative fields, the resulting poles thus formed are called consequent poles. This would be the re- sult of two north or two south poles being brought near each other. Q. 27. (1899-1900.) What does every magnetic field possess? When a magnetic substance is brought into a magnetic field, what is; the tendency of the lines of force? If the substance is free to move on an axis, but not bodily towards the magnetic pole, in what direction will it come to rest? What will the body then become? What position will the south pole assume? Ans. 27. In every magnetic field there are certain stresses which produce a tension along the lines of force and a pressure across them; that is, they tend to shorten themselves from end to end, and repel one another as they lie side by side. When a magnetic substance is brought into a magnetic field the lines of force in that vicinity crowd together and all tend to pass through the substance. If the substance is free to move on an axis (but not bodily) towards the magnetic pole, it will always come to rest with its greatest extent or length in direction of the lines of force. The body will then become a magnet, its south pole being situated where the lines of force enter it, and its north pole where they pass out. Q. 28. (1899-1900.) What is understood by magnetic induction? How is the quantity of magnetism expressed? What is meant by the term magnetic density? Ans. 28. The production of magnetism in a magnetic substance in this manner is called magnetic induction. The production of artificial magnetism in a hardened steel needle or bar by contact with lodestone is one case of magnetic induction. The amount or quantity of mag- netism is expressed by the total number of lines of force contained in the magnetic circuit. Magnetic density is the number of lines of force passing through a unit area measured perpendicularly to their direction. Q. 29. (1899-1900.) Explain how the direction of a current can be determined by the use of a compass? If a conductor conveying a cur- rent should be brought up through a piece of card-board and iron filings are sprinkled on the cardboard, how will the filings arrange them- selves? What can every conductor conveying a current be imagined to be surrounded with? How, or in what manner, does the magnetic density increase or decrease? Has the subject of this question any- thing to do with the insurance rules, in regard to the separation of conductors, or in other words, the distance conductors shall be placed from each other? Ans. 29. If a conductor be placed parallel to the magnetic axis of a compass needle, and a current passed through it in either direction, the needle will tend to place itself at right angles to the conductor, or, in general, an electric current and a magnet exert natural force upon each other. Should the conductor be threaded up through a piece of cardboard and iron filings are sprinkled on the cardboard, they will arrange, themselves in concentric circles around the conductor, 163 This effect will be observed throughout the entire length of the conductor, and is caused entirely by the current. In fact every con- ductor conveying a current of electricity can be imagined as completely surrounded by a sort of magnetic whirl, the magnetic density decreas- ing- as the distance from the current increases. The National Board of Fire Underwriters, in their national code for the installation of electric machinery and wiring of buildings have promulgated certain rules for the separation of conductors conveying currents of electricity. This magnetic influence, together with other reasons, have most certainly entered into the problem upon which they have made their decision in the matter. Granting that the proper insulation of the conductor is calculated to eliminate all danger from magnetic influence, contacts are often made through abrasions of the insulation of the conductor, which often happens through accident, carelessness or other means, the wire becoming detached from its fast- Qning, swaying or pushed out of its proper position, etc. In substantiation of this statement we add a quotation from their general suggestions: "In all electric work conductors, however well insulated, should always be treated as bare, to the end that under no conditions, existing or likely to exist, can a grounding or short circuit occur, and so that all leakage from conductor to conductor, or between conductor and ground, may be reduced to the minimum." Q. 30. (1899-1900.) If the current is flowing away from the ob- server, in which direction will the lines of force encircle the conduc- tor? Ans. 30. If the current is flowing in a conductor away from the observer, then the direction of the lines of force will encircle the con- ductor from left to right, or in the same direction as the hands of a clock move. Q. 31. (1899-190X).) Explain in either case how two parallel con- ductors, both transmitting currents of electricity, would be, either mutually attractive or repellant? Suppose the conductor was bent into a loop, what directions would the lines of force take? ', Ans. 31. Two parallel conductors, both transmitting currents of electricity, are, mutually attractive or repellant, depending upon the relative direction of their currents. If the current in both conductors ar^e flowing the same direction the lines of force will tend to surround Tjoth conductors and contract, thus attracting both conductors. If however, the currents are flowing in opposite directions, the lines of orce lying between the conductors will have the same direction and Tore, repel the conductors. the condu ctor be bent into a loop then all the lines of force conductor will thread through the loop in the same direc- s ( 1 8 "- 1900 -) Describe the helix. How would the lines of . coincide with the different loops? How in relation to the helix as a wnoie : r urnish sketch. Ans. 32. A helix is formed by bending a conductor into a series of loops, or coil, the lines of force around each loop will coincide with those around the adjacent loops, forming several long lines of force which thread through the entire helix, entering at one end and passing out at the other end. The same conditions existing in the helix as exist in the bar magnet, the lines of force pass out from one end and enter the other end, in fact having a north and south pole, a neutral line, and all the properties of attraction and repulsion of a magnet. Q. 33. (1899-1900.) If a helix is suspended in a horizontal position, and free to turn, in what direction will it come to rest? Ans. 33. A helix suspended in a horizontal position and free to turn, it will come to rest in a north and south direction if a current be passing through it. Q. 34. (1899-1900.) Describe and furnish sketch of a solenoid. Upon what does the polarity of a solenoid depend? Ans. 34. A helix made in the manner described in Answer 32 and around which a current of electricity is circulating, is called a solen- oid; that is, the solenoid is the magnetizing coil of an electro magnet. LJ ^ L ^ L ^ Wa ^ 1 ~ 1 "" N The polarity of a solenoid, that is, the direction of the lines of force which thread through it, depends upon the direction in which the conductor is coiled, and the direction of the current in the conductor. To determine the polarity of a solenoid, knowing the direction of the current: If in looking at the end of the helix, it is so wound that the current encircles the helix in the direction of the hands of a clock, that end will be the south pole; if in the other direction it will be the north pole. . Q. 35. (1899-1900.) In looking at the end of a helix, if it is so wound that the current circulates around the helix from left to right, which end will be the north pole? If wound so that the current circu- lates around from right to left, which end will be the south pole? In either case, in which direction will the lines of force take through the helix? Ans. 35. In looking at the end of a helix, so wound that the cur- rent encircles: the helix from left to right, or clockwise, the end away from the observer will be the north pole. If wound so that the current encircle the helix from right to left, the end away from the observer will be the south pole. In either case the lines of force will leave the north pole and pass around and in at the south pole, through the helix to the north pole again. Q. 36. (1899-1900.) How can the polarity of a solenoid be changed? Ans. 36. The polarity of a solenoid can be changed by reversing the direction of the current in the conductor. Q. 37. (1899-1900.) Why does a magnetic substance offer a better path for the lines of force than air or other non-magnetic substances? What do you understand by magnetic permeability? 165 Ans 37 A magnetic substance offers a better path for the lines of force for the same reasons explained in relation to magnets in offering a better path than air or other non-magnetic substances. The facility offered by any magnetic substance to the passage through it of the lines of force is called magnetic permeability. The permeability of all non-magnetic substances, such as air, copper, wood, etc., is taken as 1 or unity. The permeability of soft iron may be as high as 2,000 times that of air. If therefore, a piece of soft iron be inserted into the magnetic cir- cuit of a solenoid, the number of lines of force will be greatly increased, and the iron will become highly magnetized. Q. 38. (1899-1900.) What is an electro magnet? What is the sub- stance around which the current circulates called? How is the solen- oid generally termed? Ans. 38. A magnet produced by inserting a magnetic substance into the magnetic circuit of a solenoid is an electro-magnet, and the substance around which the current flows or circulates is called the core. The solenoid is generally termed the magnetizing coil. Q. 39. (1899-1900.) Describe the ordinary form of the electro-mag- net; its winding, insulation and why insulation is necessary. Ans. 39. In the ordinary form of electro-magnets the magnetizing coil consists of a large number of turns of insulated wire, that is, wire covered with a layer or coating of some non-conducting or insulating material, usually cotton or silk; otherwise the current would take a shorter and easier circuit from coil to coil througn the iron core with- out circulating around it; for this reason each coil is thoroughly in- sulated from one another that the current may be forced to make the requisite number of turns around the core for such specific purposes that the coil may be designed for. Q. 40. (1899-1900.) How would the field coil of a motor be affected if at some point in the winding of the coil there should be a defective place in the insulation, and the current jumped through a portion of the winding instead of making the regular number of turns? Would it increase or decrease the magnetism of that particular coil? What effect would it have on the remaining coils and how would it affect the speed of the motor carrying its full rated load? How would you detect this particular trouble? What would be liable to happen if the motor should continue to run until it burned out? Ans. 40. If, in a shunt wound, constant potential motor, one of the field-coils: should be partially, or wholly, short-circuited, the mag- netism in that magnet would be, according to the circumstance, below the normal, the field current would be excessive, causing the remain- ing coils to heat excessively, the defective coil remaining comparatively cool. The counter B. M. P. would drop, the applied B. M. F. would become excessive, and the speed of the motor would be increased to a point where the counter E. M. F. would again balance the load. This excessive applied E. M. F. would cause the armature also to heat, and should the defect not be remedied, and the motor continue to run with a full load, the consequence would be the loss of one or all the good field coils, with, in all probability, the armature. nJV 1 !' (18 "- 1900 -) Wh *t are the three principal units used in practical measurements of a current of electricity, and what do they denote? 1 66 How would you explain them in analogy with the flow of fluids ? How would they correspond with a current flowing through a wire? Ans. 41. The three principal units used in practical measurements of a current of electricity are: The ampere, or the practical unit, denoting the rate of flow of an electric current, or the strength of an electric current. The ohm, or the practical unit of resistance. The volt, or the practical unit of electrical potential or pressure. Explained in analogy with the flow of liquids or water, the force which causes the water to flow through the pipe is due to the head or pressure; that which resists the flow, is friction of the water against the inside of the pipe, and the amount would vary with circumstances. The rate of flow, or the current, may be expressed in gallons per minute, and is the ratio between the head, or pressure, and the resist- ance of the water against the inside of the pipe. For, as head or pres- sure increases, the rate of flow increases in proportion; as the resist- ance increases the current diminishes. In the case of an electric current flowing through a conductor, the electromotive force, or potential, corresponds to the pressure, or the head of water, and the resistance which a conductor offers to the flow of electricity to the friction of the water against the pipe. Q. 42. (1899-1900.) The strength of a current, or the rate of flow of electricity, is a ratio. Explain this ratio? By whom was this ratio first discovered? By whom was it first applied to electricity, and by what name has it since been called? Ans. 42. The strength of an electric current, or the rate of flow of electricity, is also a ratio a ratio between the electromotive force and the resistance of the conductor through which the current is flow- ing. This ratio as applied to electricity was first discovered by Dr. G. S. Ohm, and has since been called "Ohm's Law." Q. 43. (1899-1900.) What is "Ohm's law" and how is it usually expressed algebraically? What is: an ampere? As electricity possesses neither weight nor extension, therefore, it cannot be measured like fluids or gases, how can the strength of an electric current be determined? Ans. 43. Ohm's Law. The strength of an electric current in any circuit is directly proportional to the electromotive force developed in that circuit and inversely proportional to the resistance of the circuit, i. e., it is equal to the electromotive force divided by the resistance. Ohm's Law is usually expressed algebraically, thus: Electromotive force Strength of current = Resistance If C = the current in amperes. E =r the electromotive force in volts. R = the resistance in ohms. The formula will give the strength of the current (C) directly in amperes: E Thus C = R An ampere is the unit of strength of the current. The strength of an electric current can be described as a quantity of electricity flowing continuously every second, or, in other words, it is the rate of flow of electricity, just as the current expressed in gal- lons per minute is the rate of flow of liquids. 167 When one unit quantity of electricity is flowing continuously every second then the rate of flow, or the strength of current, is one ampere; if two units quantities of current are flowing continuously every sec- ond then the rate of flow, or the strength of current, is two amperes, m It S< makes no difference in the number of amperes whether the cur- rent flows for a long time or for only a fraction of a second; if the quantity of electricity that would flow in one second is the same in both cases, then the current strength in amperes is the same. Electricity possesses neither weight nor extension, therefore an electric current can not be measured by the usual method adopted for measuring liquids and gases. In liquids the strength of the current is determined by measuring or weighing the actual quantity of the liquid which has passed between two points in a certain time and dividing the results by that time. Thus, if 100 gallons of water should pass through a pipe in 5 sec., the rate of flow would be equal to 100 H- 5, or 20 gals, per second. The strength of an electric currant, on the contrary, is determined directly by the effect it produces, and the actual quantity of electricity which has passed between two points in a certain time is afterwards calculated by multiplying the strecgth of the current by the time. The principal effects of an electric current were given in answer to the first set of these questions. Of these, the one most generally used for measuring, is the action of the current upon a magnetic needle. Q. 44. (1899-1900.) Give a brief description and the use of a gal- vanometer. Ans. 44. The instrument commonly used in laboratory practice is called a galvanometer. The action of the galvanometer is based upon the principles an- swered in the second series of answers, where a magnetic needle sus- pended (freely) in the center of a looped or coiled conductor, is de- flected by the current of electricity passing around the coil or loop. In ordinary practice, the needle is suspended, either upon a point projecting into an agate cup fixed into the needle, or by a fiber sus- pension. In the simpler form of galvanometers, the magnetic needle itself swings over a dial graduated in degrees; in other forms, a light index needle is rigidly attached to the magnetic needle and swings over a similar dial. In the more sensitive galvanometers a small reflecting mirror is attached to the fiber suspension and reflects a beam of light upon a horizontal scale situated several inches from the galvanometer. In any of these galvanometers, when no current is flowing in the coils, the needle should point in a direction parallel to the length of the coil. The measuring of currents by most galvanometers depends upon the magnetic needle being held in position by the magnetic attraction of the earth's magnetism or the attraction of some adjacent magnet. When a current of electricity passes around the coil, the tendency is to deflect the magnetic needle at right angles to its original posi- tion, while the tendency of the earth's magnetism is to oppose the movement. The couple thereby produced will cause the needle to be deflected a certain number of degrees from its original position, de- pending upon the relative strength of the two magnetic fields. The stronger the current in the coil the greater the deflection. With a galvanometer of standard dimensions and a magnetic field of known strength, such as the earth's magnetism at a convenient place on its surface, a strength of a current can be conveniently adopted as a unit which will produce a certain deflection; all other galvanometers can be calibrated from this standard, and their dials graduated to read the strength of currents directly in the conventional unit adopted. 168 Q. 45. (1899-1900.) What is the ohm? How can electric resistance be denned? What is one of the most important quantities in electrical measure- ments? What is meant by the international ohm? Ans. 45. The ohm is the unit of resistance. It has been previously stated that the resistance varies in different substances; that is, one substance offers a higher resistance to a current of electricity than another. Electrical resistance can, therefore, be denned as a property of matter, varying with different substances, and in virtue of which such matter opposes or resists a pas- sage of electricity. The resistance which all substances offer to the passage of an electric current is one of the most important quantities in electrical measurement. In the first place, it is that which determines the strength of an electric current in any circuit in which a difference of potential is consequently maintained, as shown in Ohm's Law. In the second place, the unit of resistance, the ohm, is the only unit in elec- trical measurements for which a material standard can be adopted, other quantities being measured by the effect they produce. The basis of any system of physical measurements is generally some material standard, conventionally adopted as a unit, physical meas- urements in each system being made in comparison with that unit. The unit of electrical resistance now universally adopted is called the international ohm. One international ohm is the resistance offered by a column of pure mercury 106.3 centimeters in length, 1 square millimeter in sectional area at 32 F., or at a temperature of melting ice. The dimensions of the column expressed in inches are as follows: Length, 41.85 inches; sectional area, .00155 sq. inch. Q. 46. (1899-1900.) What is said of the resistance of conductors at equal temperatures, irrespective of the current flowing through them, or the electro-motive force of the current? In a given conductor, which offers a resistance of 2 ohms to a cur- rent of 1 ampere, what would be the resistance in the same conductor if 12 amperes were flowing through it? Ans. 46. The resistance of a given conductor at equal temperatures is always constant, irrespective of the strength of current flowing through it or the electromotive force of the current. Hence, if a given conductor offers a resistance of 2 ohms to a cur- rent of 1 ampere, it offers the same amount of resistance, no more nor less, to a current of 12 amperes. Q. 47. (1899-1900.) How is the resistance of a given conductor in- fluenced by a change in length? What will be the resistance of two miles of copper wire, if the re- sistance of ten feet of the same wire is .013 ohms? Ans. 47. If the length of a conductor be doubled, its resistance will be doubled; that is, the resistance of a given conductor increases as the length of the conductor increases, the resistance being directly pro- portional to the length of the conductor. When it is required to find the resistance of a conductor of which the length is varied, other conditions remain unchanged, the follow- ing formula may be used: 169 in which r t = the original resistance; r 2 = the required resistance; h = the original length; 1 2 = the changed length. As in all examples of proportion, the two lengths must be reduced to the same unit. By this formula we see that the resistance of a conductor after its length is changed is equal to the original resistance multiplied by the changed length, and the product divided by the original length. In two miles there are 10,560 ft. Then by our formula .013 X 10560 The required resistance = =13.728 ohms. 10 Q. 48. (1899-1900.) How is the resistance of a given conductor in- fluenced by a change of the area of the conductor? If the resistance of a conductor, whose sectional area is .025 sq. in. is .32 ohms, what will be the resistance of a conductor if its sectional area is increased to .25 sq. in.? How does the resistance of a round conductor vary? The resistance of a round copper wire .2 inches diameter is 45 ohms; what will be the resistance of the same kind of wire .4 inches in diam- eter? What is said of the resistance of two or more conductors in series? Ans. 48. If the sectional area of a conductor is doubled, the re- sistance will be halved. We may, then, obtain" the value of the re- sistance of a conductor from any change of sectional area by the fol- lowing formula: r t ai In which ri = the original resistance of the conductor; r 2 = the changed resistance; a! = the original area; az = the changed area. From the relations here expressed it will be seen that the resist- ance varies inversely as the sectional area; that is, the resistance of a given conductor diminishes as its sectional area increases. The resistance of a conductor is independent of the shape of its cross-section. By our formula the conditions as stated Sectional area = .025 sq. in. I Resistance = .32 ohms. changed area .25 sq. m. .32 X .025 The required resistance = = .032 ohms. .25 When comparing resistance of round copper wire the following for- mula is used: l^D 1 r 2 = dz In which r! = the original, or known, resistance; r 2 = the required resistance; D = the original diameter; d = the changed diameter. This formula is based on the rule that, since the sectional area of a round conductor is proportional to the square of its diameter (sectional 170 area diameter squared multiplied by the constant .7854), the re- sistance of a round conductor is inversely proportional to the square of its diameter. By our formula conditions as stated in the question A round copper wire .2 inches diameter; The resistance 45 ohms; The resistance when the diameter is increased to .4 inches, by our formula will be equal to the original resistance multiplied by the diam- eter squared, divided by the changed diameter squared. Thus 45 X .2 2 45 X .4 Required resistance = = = 11.25 ohms. .4' .16 The resistance of two or more conductors connected in series is equal to the sum of their separate resistances. As an example, if four conductors having separate resistances 7, 11, 13 and 19 ohms, respec- tively, are connected in series, their total, or joint, resistance will be equal to the sum of the separate resistances, or 50 ohms. Q. 49. (1899-1900.) What is a microhm? For what was it de- vised? In order to compare resistances of different substances how should the dimensions compare? Why? What metal under like conditions offers the least resistance? What other metal comes next? Does the resistance of a given conductor always remain constant? Why.' Does the resistance increase or decrease as the temperature raises or lowers? Kow is this, in comparison with liquids and carbons? Ans. 49. The microhm is a unit of resistance devised to facilitate calculations and measurements of exceedingly small resistances, and is equal to one-millionth of an ohm (1/1,000,000). In order to compare the resistances of different substances, the di- mensions of the pieces to be measured must be equal; for, by changing its dimensions, a good conductor may be made to offer the same resist- ance as an inferior one. Under like conditions, annealed silver offers the least resistance of all known substances; soft annealed copper comes next on the list, and then follow all other metals and conductors. The resistance of a given conductor, however, is not always con- stant, it changes with the temperature of the conductor. In all metals the resistance increases as the temperature rises. In liquids and carbons the resistance decreases as the temperature rises. Q. 50. (1899-1900.) What is the variation of the resistance of a con- ductor caused by a change of temperature of 1 called? How would you find the resistance of a conductor, after its tempera- ture had risen, knowing its original resistance and the number of de- grees rise in temperature, other conditions remaining the same? The resistance of a piece of copper wire at 32 is 40 ohms; deter- mine its resistance when its temperature is 74 F., the temperature co- efficient for copper is .002155. How would you determine the resistance if the temperature is dropped, other conditions remaining the same? If the original resistance of a German silver wire is 16 ohms, determine its resistance after its temperature has fallen 15 F. Tem- perature coefficient of German silver is .000244. Ans. 50. The amount of variation in the resistance* caused by a change of temperature for one degree is called the temperature co- efficient. 171 These co-efficients, however, hold true for a limited change of tem- nprature and should not be used with extreme changes. P ?o "find the resistance of a conductor after its temperature has risen, knowing its original resistance and the number of degrees rise, other condlSfns remaining unchanged, multiply the original resistance by 1 plus the product of the number of degrees rise and the temperature co-efficient. Formula r 2 = r x ( 1 + t k ) In which ri = the original resistance; r 2 = the resistance after the temperature change; t = rise or fall of temperature in degrees F.; k = temperature co-efficient. Resistance of copper wire at 32 F. is 40 ohms. At a temperature of 74 F. (equals 42 F. rise) the temperature co- efficient is .002155. Then according to our rule, or formula The changed resistance = 40 (1 + 42 X .002155) =43.6204 ohms. To determine the resistance if the temperature drops: Divide the original resistance by 1 plus the product of the number of degrees fall and the temp. coef. Formula r 1 1 + tk The letters having the same value as above. If the original resistance of a German silver wire is 16 ohms; the temp, drops 15 F. the temp. coef. of German silver wire is .000244. By our rule or formula 16 The changed resistance = = 15.941 ohms. 1 + 15 X .000244 Q. 51. (1899-1900.) What is understood by specific resistance of substances? What instrument is usually used in measuring resistances? Give a brief description of this instrument. Ans. 51. Specific resistance is the term given to the resistance of substances of unit length and unit sectional area at some standard temperature. The specific resistance of a substance is the resistance of a piece of that substance one inch in length and one square inch in sectional area at 32 F.; that is, at a temp, of melting ice; this may also oe expressed as the resistance of a cube of that substance taken between two oppos- ing faces. A list of the common metals in the order of their relative resistances beginning with silver as offering the least resistance: Silver, annealed 1.000 Silver, hard drawn 1.086 Copper, annealed 1.063 Copper, hard drawn 1.086 Etc., etc. An instrument known as "Wheatstone Bridge" is usually used in laboratory practice in determining the resistance of different sub- stances, coils, etc. These are different forms of construction, some simple, others elab- orate, with a. galvanometer in circuit. A simple form in laboratory practice, where small currents are used and great accuracy is required, the resistance coils are enclosed in a 172 wooden box and the actual resistance of each coil is carefully deter- mined, the separate coils offering resistances from one ohm upward to 5,000 ohms. The operation of adjusting the resistance is by the means of removal plugs, allowing the current to pass through all or part of the coils, as the case may require. Q. 52. (1899-1900.) What does the term "volt" denote? What three facts are to be carefully noted regarding the application of "Ohm's law" to closed circuits? Ans. 52. The volt is the practical unit of electromotive force. In mechanics, pressures of all kinds are measured by the effects they produce; similarly, in electrotechnics, potential is measured by the effect it produces. By definition the volt is that electromotive force which will main- tain a current of one ampere in a circuit of one ohm's resistance. With a known resistance in ohms and a known strength of current in amperes the electromotive force is determined by "Ohm's" law, in which the current in amperes multiplied by the resistance in ohms equals the electromotive force in volts. Or, E C R. E = electromotive force in volts. C = current in amperes. R = resistance in ohms. The application of Ohm's law to closed circuits, the following facts are to be carefully noted: The current (C) is the same in all parts of the circuit, except in the cases of derived circuits, where the sum of the currents in the separate branches equal to the current in the main, or undivided, branches. The resistance (R) is the resistance of the internal circuit plus the resistance of the external circuit. Q. 53. (1899-1900.) How do you determine the strength of a current in amperes, the E. M. F. and R. being given? If the two electrodes of a simple galvanic cell are connected by a conductor whose resistance is 1.5 ohms; the internal resistance of the cell is 6 ohms, and the total E. M. F. developed is 1.83 volts, what is the strength of the current flowing in the circuit? Ans. 53. The strength of the current (C) in amperes divided by the resistance (R) in ohms equals the electromotive force (E) in volts. Internal resistance 6. ohms. External resistance 1.5 ohms. Total resistance of 7.5 ohms. Then, according to Ohm's law E 1.83 C = = = .244 amperes. R 7.5 The strength of the current flowing in the circuit. Q. 54. (1899-1900.) How would you find the total resistance in ohms of a closed circuit when the E. M. F. and the strength of the cur- rent is known? The total E. M. F. developed in a closed circuit, 1.7 volts, and the current is .7 amperes; find the resistance in ohms. Ans. 54. The total resistance in ohms of a closed circuit equals the electromotive force (E) in volts divided by the strength of the current (C) in amperes. 173 Electromotive force = 1.7 volts. Current = .7 amperes. 1.7 The resistance equals =2.428 ohms. The resistance in the circuit. Q. 55. (1899-1900.) It is desired to transmit a current of 15 am- peres to a receptive device, situated 1,000 feet from the source; the total E M F. generated is 125 volts, and only 1-10 of this potential is to be lost in the conductors, leading to and from the device; find the resistance of the two conductors. How much will be the resistance per foot of the conductors? In a voltaic cell, what is to be. understood by the term, available or external E. M. F.? The internal or generated E. M. F.? Ans. 55. 1/10 of 125 volts = 12.5 volts, which represents the drop, or loss in potential on the two conductors. Let ' = 12.5 volts. C = the current in amps. 15. R 1 = total resistance of the two conductors. Then E 1 12.5 R 1 = = = .8333 + ohms. C 15 The resistance per foot of conductor equals .8333 + r- = .0004166 + ohms. 2000 The difference of potential between the two electrodes of a simple voltaic cell when no current is flowing; that is, when the circuit is open, is always equal to the total E. M. F. developed in the cell, and is called the internal, or generated, E. M. F. When a current is flowing; that is, when the circuit is closed, a certain amount of the potential is expended in forcing the current through the internal resistance of the cell itself. Hence, the difference of potential between the two electrodes when the circuit is closed is always smaller than when the circuit is open. This difference of potential between the two electrodes when the circuit is closed, is called the available, or external, E. M. F., to dis- tinguish it from the internal, or total, generated E. M. F. Q. 56. (1899-1900.) In a voltaic cell, the total generated E. M. F. equals 2.5 volts, and the internal resistance is .7 ohms; if a current of 1.5 amps, flows through the cell when the circuit is closed, what is the available E. M. F. developed by the cell, or what is the difference in potential between the two electrodes? How do you find the E. M. F. (in volts) in a closed circuit when the strength of the current and the total resistance is known? Ans. 56. To find the available E. M. F. of a cell: Let E = the total generated E. M. F.; E^=the available E. M. F. when the circuit is closed; the current in amp. flowing when the circuit is closed; r, = the internal resistance of the cell. Then the drop, or loss, of potential in the cell equals C r,. The available E. M. F. E 1 = E C r^ 174 If the total E. M. F. equals 2.5 volts; the internal resistance equals 7 ohms; with a current flowing of 1.5 amps., the available E. M. F. will equal Ei = E C r, = 2.5 1.5 X .7 = 1.45 volts or the difference in potential between the two electrodes when the cir- cuit is closed. To find the E. M. F. (in volts) in a closed circuit, when the strength of the current and the total resistance is known: The available E. M. F. of a cell is equal to the difference between the total generated E. M. F. and the potential expended in forcing the current through the internal resistance of the cell when the circuit is closed. From Ohm's Law: This drop of potential in the cell itself is equal to the product of the internal resistance in ohms and the strength of the current in amps, flowing through the circuit. Q. 57. (1899-1900.) The internal resistance of a closed circuit is 3 ohms, and the external resistance is 4 ohms, the current flowing is .5 amperes; what is the E. M. F. developed? What is understood by a drop of potential? How is this drop usually expressed? If a circuit in which a current of 4 amperes is flowing having three separate resistances as (in diagram) a to b, b to c ,and c to d, the resistance from a to b 1.6 ohms, b to c is 3.2 ohms and c to d 4.3 ohms, find the difference in potential from a to b, b to c and c to d? And from a to d? Give a general explanation of conductivity in regard to resistance? When the resistance of two branches are unequal, how will the cur- rent divide between them in a derived circuit? Ans. 57 Let R == total resistance = n + ^ = .7 ohms. r l = internal resistance; 3 ohms. r-2 = external resistance; 4 ohms. C = current in amps. .5 amps. The total E. M. F. = C R = .5 X 7 = 3.5 volts. The drop in potential = Cr 1 = .5 X 3 = 1.5 volts. The available E. M. F. = 3.5 1.5 = 2.0 volts. Drop or loss of Potential: By referring again to water flowing through a pipe; though the quantity of water which passes is the same at any cross section of the pipe, the pressure per square inch is not the same. It is this dif- ference of pressure that causes the water to flow between two points against the friction of the pipes. This is precisely similar to a current of electricity flowing through a conductor. Though the quantity of electricity that flows is equal at all cross sections, the E. M. F. is by no means the same at all points along the conductor. It suffers a loss or drop of electrical potential in the direction in which the current is flowing, and it is this difference of electrical po- tential that causes the electricity to flow against the resistance of the conductor. Ohm's law not only gives the strength of a current in a closed circuit, but also the difference of potential in volts along the circuit. The difference of potential (E 1 ) in volts between any two points is equal to the product of the strength of the current (C) in a.m.neres and 175 the resistance (R 1 ) in ohm's of that part of the circuit between those two points, or B 1 = CR 1 E 1 also represents the loss or drop of potential in volts between the two points. If any two of these quantities are known the third can readily be found, for by transposing E 1 E 1 C = and R 1 = . R 1 C See sketch R 1 = separate resistances, E 1 = the drop in potential between a and d, according to Ohm's law. E 1 = C R 1 The difference of potential between a and b = 4 X 1.6 = 6.4 volts. D and c = 4 X 3.2 = 12.8 volts, c and d = 4 X 4.3 = 17.2 volts. aandd = - 36.4 volts. 36.4 volts represents the drop of potential caused by a current of 4 amperes flowing between the points a and d. Conductivity can be denned as the facility with which a body transmits electricity, and is the opposite to resistance. ' For example, copper is high conductivity with low resistance; mer- cury is high resistance and low conductivity. In other words, conductivity is the inverse or reciprocal of resist- ance. The conductivity of any conductor is, therefore, unity divided by the resistance of that conductor; and conversely the resistance of any conductor is unity divided by its conductivity. When the current flows through two branches of a derived circuit and the resistances are equal the current will divide equally between the two branches. When the resistances of the two branches are unequal, the current will divide between them in inverse proportion to their respective re- sistances. i. 58. (1899-1900.) In the accompanying diagram, if the resistance 2 ohms, and the resistance of T2 = 4 ohms, find the separate cur- rent, cf and "ca, in the two branches respectively? When C= (main current), 30 amperes, in the undivided mains, a and b. If the separate resistances of two conductors are equal, what will be their joint resistance when connected in parallel? Give the rule when the separate resistances are unequal? What is the joint resistance when three or more conductors are connected in parallel? fn a derived circuit of any number of branches, what would be the difference of potential between where the branches divide and where they unite? How -can the separate resistances of the branches of a derived circuit be determined? 176 Ans. 58. See sketch: F! = 2 ohms; r 2 = 4 ohms; C = 30 amperes; d = current in one branch of the derived circuit; C 2 = current in the other branch of the derived circuit. The resistances in the two branches are i\ and r 2 , therefore, c t : ca : : T2 : r : . By algebra this proportion gives the two following formulas: Cr 2 30 X 4 120 For the first branch c, = = = = 20 amps n + r 2 2 + 4 6 Cri 30 X 2 60 For the second branch c 2 = = = * = 10 amps r t + r 2 2 + 4 6 In the branch ci 20 amperes of current will flow. In the branch c 2 10 amperes of current will flow. Reducing our formula to rule we have for the first branch. Of two branches in parallel, dividing from a main circuit, the cur- rent in the first branch is equal to the current in the main multiplied by the resistance of the second branch, the product divided by the sum of the resistances of the two branches. For the second branch, The current of the second branch is equal to the current of the main circuit multiplied by the resistance of the first branch, the product divided by the sum of the resistances of the two branches. If the resistances of two conductors arc equal, their joint resistance when connected in parallel is one-half the resistance of either con- ductor. When the separate resistances of two conductors in parallel are unequal, the determination of their joint resistance when connected in parellel involves some calculation. Referring to the diagram, the conductivities of the branches are 1 1 and , respectively. F! Ta Hence their joint conductivity when connected in parallel is 1 1 r 2 + r, Now, since the resistance of any conductor is the reciprocal of its conductivity, then the* joint resistance of the two branches in parallel is the reciprocal of their joint conductivity; r 2 + r 1 FJ r 2 or, 1 -r- - . Hence, joint resistance R 11 = - . Fj + F 2 That is, the joint resistance of two conductors connected in par- allel is equal to the product of their separate resistances divided by the sum of their separate resistances. The joint resistance of three or more conductors, connected in parallel, is equal to the reciprocal of their joint conductivity. If in our diagram we connect in another conductor, making three instead of two branches, then the joint resistance of the three branches F! r 2 Ta jn parallel R 1U = --- r 2 rs + ri r 2 + FJ r 3 177 In a derived circuit of any number of branches, the difference of potential between where they divide and where they unite is equal to the product of the sum of the currents in the separate branches and their joint resistance in parallel. The separate currents in the branches of a derived circuit can be determined by finding the difference of potential between where the branches divide and where they unite again, and dividing the result by the separate resistance of each branch. The separate resistance of the branches of a derived circuit can be determined by finding the difference of potential between where the branches divide and where they unite, dividing the result by the sep- arate currents of each branch. Examples of the three conditions: First condition If the currents in the three branches are 16, 8 and 4 amperes, respectively, and the joint resistance is 2 6-7 ohms, then the difference of potential between a and b 6 20 = (16 + 8 + 4) 2 = 28 X = 80 volts 7 7 Second condition Assume that the separate resistances of the three branches are, re- spectively, 5, 10 and 20 ohms, and the difference of potential from a to b is 80 volts. Then the current in the first branch is equal to 80 ^5 = 16 amps; second branch to 80-^10 = 8 amps; third branch to 80 -=-20 = 4 amps. If the difference of potential between a and b is 80 volts, and the currents in the separate branches are 16, 8 and 4 amperes, respectively, then the resistance of the first branch is 80 -f- 16 = 5 ohms; second branch is 80 ' -=- 8 = 10 ohms; third branch is 80 -f- 4 = 20 ohms. Q. 59. (1899-1900.) How can the strength of an electric current be defined? What is the practical unit of electrical quantity, and what does it represent? What is the rule for calculating the quantity of electricity which has passed in a circuit in a given time when the strength in amperes is known? Explain, briefly, the term electrical work? The principle of the conservation of energy teaches that energy cannot be destroyed; what follows in regard to electrical energy? How would you find the amount of electrical work accomplished, in joules, during a given time in any circuit? . '"- Ans. 59. The strength of an electric current can be defined as a quantity of electricity flowing in one second. The practical unit of electricity is called the coulomb. The coulomb is such a quantity of electricity as would pass in one second through a circuit in which the strength of the current is one ampere. To calculate the quantity of electricity which has passed in a cir- cuit in a given time when the strength of the current in amperes is known: Let Q = the quantity in coulombs; C = the strength of current in amperes; t = the time. Then Q = Ct. If any two of these quantities are known the third can be readily found. Q Q By transposition, C = , or t = . t C 178 Electrical Work: When an electric current flows from a higher to a lower potential, electrical energy is expended, and work is done by the current. The principle of the conservation of energy teaches that energy can never be destroyed; it follows, therefore, that if energy has to be expended in forcing a quantity of electricity against a certain amount of resistance, the equivalent of that energy must be transformed into some other form. This other form is usually heat; that is, when a quantity of electricity flows against the resistance of a conductor, a certain amount of electrical energy is transformed into heat energy. The actual amount of heat developed is an exact equivalent of the work done in overcoming the resistance of the conductor, and varies directly as the resistance. For example, take two wires, the resistance of one being twice that of the other, and send currents of equal strength through each. The amount of heat developed in the wire of higher resistance will be twice that developed in the wire offering the lower resistance. The unit used to express the amount of mechanical work done is known as the foot-pound. The work done, in raising any mass through any height, is found by multiplying the weight of the body lifted by the vertical height through which it is raised; similarly, the practical unit of electrical work is that amount accomplished when a unit quantity of electricity, one coulomb, flows between a potential of one volt. The unit of electrical work is, therefore, the volt-coulomb, and is called the joule. 1 joule .7373 foot-pound. By the means of the following formulas, we may find directly the amount of electrical work accomplished in joules during a given time in any circuit. Let J = electrical work in joules; C = current, in amps; t = time, in seconds; E = potential or E. M. F., of circuit; R = resistance of circuit. When the E. M. F. and current are known, J = C Et. When the current and resistance are known, J = C 2 R t. When the E. M. F. and resistance are known, E 2 t T R To reduce the work as expressed in joules to foot-pounds foot-pounds = Joules X .7373 Q. 60. (1899-1900.) What is the unit of power or rate of doing work called? Give three rules for computing the power in watts. Give four rules for determining the H. P. The H. P. being given, how would the number of watts be determined? Either way, how would the same be expressed in kilowatts? Ans. 60. Power, or rate of doing work, is found by dividing the amount of work done by the time required to do it. In mechanics, the unit of power is called the horse-power. - In electrotechnics, the unit of power is the watt. It is found by dividing the amount of electrical work done by the time required to do it. 179 Let E = E. M. F. in volts; . Q = quantity of electricity in coulombs; C=: current in amps; e,ectr,ca, work, J = C E t. CEt Rule- The power in watts is equal to the strength of the current in amperes multiplied by the E. M. F. in volts. C 2 Rt W = - = C 2 R. Rule: The power in watts is equal to the strength of the current in 'amperes squared multiplied by the resistance in ohms E 2 t E* W = - = . Rt R The power in watts is equal to the quotient arising from dividing the E. M. F. in volts squared by the resistance in ohms. H P = 746 watts or 1 watt = - H P. 746 W HP = - . 746 To express the rate of doing work (electrical) in horse-power units, divide the number of watts by 746. EC C 2 R E 2 H P = - . H P = - . H P = - . 746 746 746 R . To express the electrical power in kilo-watts. 1,000 watts equal 1 kilo-watt. By substituting 1,000 for 746 in the preceding formulas the results obtained will be in kilo-watts. Q. 61. (1900-01.) When is an electric current generated in a con- ductor? Name two experiments from which the foregoing principle is de- ducted. To what are due currents generated in a conductor cutting lines of force and those induced in a coiled conductor by a change in the num- ber of lines of force? Give brief explanation. In calculations how is it convenient to make distinctions between the two cases? In these explanations what caution should be observed? Ans. 61. An electric current is generated in a conductor when that conductor is moved across a magnetic field, so as to cut the lines of force at an angle. If a coiled conductor be straightened out, forming one long conduc- tor, then be moved across the magnetic field at right angles to the lines of force, a current will be generated in the circuit. The current, how- ever, immediately subsides when the motion ceases, no matter whether the conductor is in the magnetic field or not Should the conductor be moved in the magnetic field with its length parallel to the lines of force, no current will be generated in the circuit. In reality currents generated in a conductor cutting lines of force and those induced in a coiled conductor by a change in number of lines of force which pass through the coil, are due to the same movement. For every conductor conveying an electric current forms a closed coil and every line of force is a complete circuit by itself. Consequently, where any part of a closed coil is cutting lines of force the lines of force are passing through the coil in a definite direction and changing at the same rate as the cutting. For example, if a closed coil is represented by a heavy loop and a light loop represents four lines of force. When the two closed loops are brought together, the closed coil is cut at one place by four lines of force and at the same time the number of lines of force passing through the closed coil increases from nothing to four. In calculations, however, it is convenient to make distinctions be- tween the two cases; in the one case, to consider that the current is generated by a conductor of a certain length cutting lines of force at right angles; and, in the other case, to consider that the current in a closed coil is induced by the change in the number of the lines of force passing through the coil. . In these explanations it must not be forgotten that an electric cur- rent is the result of a difference of potential or electromotive force. Consequently it is not actually a current that is generated in the mov- ing wire, but an electromotive force; for, in all of the previous experi- ments in which currents are induced or generated in a conductor by the lines of force, if the circuit is opened at any point, no current will now, but the electromotive force still exists. Q. 62. (1900-01.) How many methods are known of producing an electromotive force by -induction in a coiled conductor? Name them. Explain each method. Ans. 62. There are three methods of producing an electromotive force by induction in a coiled conductor i. e., "a" Electro-magnetic induction. "b" Self induction. "c" Mutual induction. In electro-magnetic induction the change in the number of lines of force which pass through the coil is due to some relative movement between the coil and a magnetic movement. For example, by thrusting a magnet into the coil or withdrawing it; or again, by suddenly thrust- ing the coil into a magnetic field with its plane at right angles to the lines of force. In self induction the change in the number of lines of force is caused by sudden changes in a current which is already flowing through the coil itself, and is supplied from some exterior source. This exterior current produces a magnetic field in the coil itself, and for so long as the strength of the current remained constant there is no change in the number of lines of force which pass through the coil. Should the strength of the current be suddenly increased, a change in the number of lines of force occurs; this change in turn induces an electromotive force in the conductor, which opposes the original current in the coil and tends to keep the current from rising. Its action is similar to that which would take place if some extra resistance were suddenly inserted into the circuit at the instant the strength of the current is increased. The original current eventually reaches its maximum strength in the coil as determined by Ohm's law, but its rise is not instantaneous; it is greatly retarded by this induced electromotive force. If, on the contrary, the strength of the original current is suddenly allowed to decrease, another change is produced in the lines of force which pass through the conductor or coil; this new change induces an electromotive force which acts in the same direction as that of the original current and tends to keep it from falling. 181 As in the previous case, however, the original current will eventual- ly drop to its maximum strength, as determined by Ohm's law, but will :all gradually and a fraction of a second will elapse before it becomes constant. In short, the current flowing through a coiled conductor acts as possessing inertia; any sudden change in the strength of the current produces a corresponding electromotive force which tends to oppose that change and keep the current at a constant strength. Mutual Induction. In mutual induction two separate coiled conductors, one conveying a current of electricity, are placed near each other, so that the mag- netic circuit produced by the one in which the current is flowing is enclosed by the other, as shown by the sketch, where the current cir- culates around the coil P when the circuit is closed at b. The coil P is called the primary or exciting coil, and the' coil S is called the sec- ondary coil. Any sudden change in the strength of the current in the primary coil, as, for instance, breaking the circuit at b, a corresponding change in the number of lines of force in the magnetic circuit which passes through both coils; and hence, an electromotive force is induced in the secondary coil. If the primary circuit is completed at b, and the current tends to rise in the coil, the electromotive force induced in the secondary coil, causes a current to circulate around it in the opposite direction to the current in the primary coil. If, on the contrary, the circuit at b is suddenly broken and the current in the primary decreases, the induced electromotive force in the secondary coil causes a current to circulate around it in the same direction as the current in the primary coil. The directions of an induced current in a coil depends upon the direction of the lines of force in the coil and whether their number is increasing or diminishing. If, then, two facts are known, the direction in which the current circulates around the coil is determined by the following rule: Rule. If the effect of the action is to diminish the number of lines of force that pass through the coil, the current will circulate around the coil from left to right, or clockwise, as viewed by a person looking along the magnetic field in the direction of the lines of force; but t the effect is to increase the number of lines of force that pass through the coil, the current will circulate around in the opposite direction Q. 63. (1900-01.) Give a convenient method for remembering the direction of a current generated in a straight conductor when the con- ductor is moved in a magnetic field at right angles to the lines of force. * ti^ U v mm , ary of electr - m agnetic induction experiments, what law IS 6St3.DllSn.6Q? th/H^t 6 !' 1 * ule -- place th e thumb, forefinger and middle finger of the right hand so that each will be perpendicular to the other two; if 182 the forefinger points in the direction of the lines of force, and the thumb points in the direction in which the conductor is moving, then the middle finger will point in the direction toward which the current generated in the conductor tends to flow. The summary of these electro-magnetic induction experiments can be stated as follows: Electromotive forces are generated in a conductor moving in a magnetic field at right angles to the direction of the lines of force, or are induced in a coiled conductor when a change occurs in the number of lines of force which pass through the coil. Q. 64. (1900-01.) To what is E. M. F. generated in a moving con- ductor cutting lines of force at right angles, directly proportioned? Can you explain this principle by the use of cross-section paper? Ans. 64. The E. M. F. generated in a moving conductor cutting lines of force at right angles is directly proportioned to the rate of cutting. If, for an example, that a magnetic field contains 100,000 lines of force and a conductor is moved across the field at right angles in such a manner as to cut every line of force in one second, then the rate of cutting is 100,000 lines per second; if it occupied two seconds of time, then the rate of cutting would be 50,000 per second, and the electro- motive force in this case would be one-half that of the former case. Q. 65. (1900-01.) Explain the change in direction of the current in a coiled conductor moving in a magnetic field, and give value of this current at different points in one revolution. Ans. 65. When the coil begins to revolve in the magnetic field, a feeble E. M. j?. is generated; this E. M. F. causes a corresponding cur- rent to flow through the circuit in a positive direction; as the E. M. F. becomes larger, the strength of the current in the circuit becomes greater, and vice versa. After the coil is rotated one-half of a revolution and the direction in which the E. M. F. tends to act becomes negative, the direction of tne current in the circuit is also reversed. If there is no self induction to retard the rise and fall of the cur- rent in the circuit the strength of the current in the circuit at any instant is exactly proportional to the E. M. F. that is being generated in the coil at that moment; for, according to "Ohm's" law, the strength of a current in any circuit is equal to the E. M. F. generated in that circuit, divided by its resistance. The rising and falling and also the reversing of the current in all parts of the circuit can be graphically represented on cross-section paper. In which the current at the starting point in each revolution is zero, rising to the maximum strength at one-quarter of the revolution, following again to zero at the one-half point of the revolution, at which point the current is reversed and again rising to its maximum strength at the three-quarter revolution, falling again to zero strength at the completion of the revolution, when, as at the one-half revolution or zero point, the current is reversed, showing that the strength of the current is at its maximum strength and at zero strength twice in each revolution. At intermediate points the strength of the current is pro- portional to the time required for making the full revolution. Q. 66. (1900-01.) Describe an alternating current. Show by the use of cross-section paper the value of the E. M. F. in one complete revolution. What is understood by the term "alternation"? What is understood by the term "frequency"? 183 What is understood by the term "cycle"? A "cycle" represents how many degrees? How in regard to time? x ' ^~^ \ / \ A -B C \ D ^ ^ .. Lx A Time DESCRIPTION OF SKETCH. Ans. 66. The E. M. F. that is generated in a coil at every instant during one complete revolution is graphically shown in the sketch by the use of cross-section paper.' (See sketch.) The sum of the divisions between A and E represents the time occu- pied by the coil in making one complete revolution; the divisions be- tween A and Y represent the E. M. F. which tends to send a current through the coil in one direction during the first half of the revolution, and the divisions between A and ^ represent the E. M. F. which tends to send a current through the coil in an opposite direction as in the last half of the revolution. The divisions between the curved line and the base line AE give the E. M. F. that is being generated in the coil at any instant during the revolution, and the direction in which the E. M. F. tends to act depends upon whether this E. M. F. is above or below the base line AE. For convenience, let the direction of the E. M. F. in the first part of the revolution be called positive and in the last half negative. For example, the E. M. F. that is generated in the coil when it has revolved three-quarters of a revolution is represented by the distance between D and the curved line, which in this case is two divisions; and, since these divisions are below the base line, the direction in which this E. M. F. tends to act is negative. The time of one complete revolution is represented with sub-divi- sions A, B, C, D, E. A B representing one-quarter revolution. A. C. representing one-half revolution. A D representing three-quarter revolution. A E representing complete revolution. The electric current as described in answer 65 is an alternating cur- rent. The term Alternation. In each reversal of current that is, each in- crease of current from zero to maximum and the decrease to zero again is called an alternation, thus two alternations occurring in one revolution. This pair of alternations is called a cycle, and the number of cycles that occur in one second is termed the frequency. A cycle represents 360, regardless of time. Q. 67. (1900-01.) What is meant by the term "commuting"? For what purpose is a commutator used? What is a "pulsating" current? How can the strength of such current be made more uniform and pulsations less noticeable? magnet? ^ ^ * & *' f inserting an iron core between the poles of a 184 Ans. 67. The term commuting means to change, and for the subject which we are treating would mean to change an alternating current into a continuous current, for which purpose a commutator is used. A combination of two segments of copper insulated apart from each other, or, in fact, any number of segments constitute what is called a com- mutator. Two or more copper or carbon brushes press against the segments and are held in the proper position while the coil is rotated. The brushes rub or brush against the segments and make electrical contact only. By this arrangement the current in the external circuit flows continually in the same direction, while the current in the coil flows in two directions during every revolution. But the strength of the current in the external circuit is by no means constant. It rises from zero to maximum and falls again to zero twice in every revolu- tion, but always in the same direction. This effect is produced con- tinually in the external circuit if the coil is rotated at a constant speed. Tnese impulses in the strength of the current give it the name of "pulsating current." The strength of such currents can be made more uniform and the pulsations less noticeable by using more coils connected to segments in the commutator, the planes of the coils being placed at equal angles from each other. In last year's treatment of this subject the term permeability was denned as the facility afforded by any substance to the passage through it of lines 'of force, and that permeability of soft iron may be in the ratio of 2,000 to 1 as compared to air; and that when a magnetic sub- stance is brought into a magnetic field the lines of force in the field crowd together and all try to pass through that substance, etc. Hence, if the coils are wound around a cylindrical drum of iron the number of lines of force passing through the coils is increased. These coils are entirely insulated from the iron core by some non- conducting material, otherwise they would be short circuited on the core that is, the current would pass through the iron instead of pass- ing into external circuit. An iron core inserted between the poles of a magnet not only in- creases the lines of force from the magnet, but attracts nearly all the stray lines of force from the surrounding air. Q. 68. (1900-01.) In a rotated coil, where will the greatest differ- ence in potential be found? What means are used to utilize this difference of potential between each pair of wires? it a comparatively large number of turns and segments are used, what will be the effect of the current? Why? Ans. 68. In a rotated coil the greatest difference in potential will be found between any two turns diametrically opposite one another when they pass through the vertical diameter. To utilize this difference ol potential between each pair of turns as they arrive in a .vertical position. This is accomplished by connecting each turn to a separate seg- ment of a commutator by a small conductor and allowing two brushes to rub against the commutator at two points diametrically opposite each other on the vertical diameter. If a comparatively large number of turns and segments are used the current flowing through the external circuit will practically be con- tinuous that is, non-pulsating; the fluctuations caused by the brushes passing from one segment to another are extremely minute and pro- duce no appreciable change in the strength of the current in the ex- ternal circuit. 185 Q. 69. (1900-01.) Describe a closed coil winding. Describe an open coil winding. What is meant by neutral spaces? What is meant by neutral points? Explain shifting of brushes, etc. Ans. 69. Closed Coil Winding. A conductor wound upon a core in this manner is termed a "closed" coil winding, since all the turns are connected together in one continuous or closed coil and the current is obtained from it by lapping into each turn or set of turns. In the case where the turns or set of turns are separate and distinct from each other and their ends are connected to opposite segments of a commutator the winding is termed an "open" coil winding. The parts of the core where the conductors are not cutting lines of force as the core is rotated are called "neutral spaces." The two opposite parts of the commutator to which the coils are connected are called the neutral point of the commutator. ... - Each individual conductor in a bipolar machine becomes inactive twice in each revolution and passes through two neutral spaces; but this fact does not change the position of the neutral spaces they lie on an imaginary diameter approximately perpendicular to the lines of force. This same effect takes place in the commutator i. e., each segment passes through two neutral spaces during one revolution, but the neu- tral points remain in a fixed position relative to the neutral spaces of the core. The neutral segments of a commutator, at any instant during a revolution, are those segments connected to the conductors passing through two neutral spaces at that instant. The neutral points can be shifted to different points around the commutator by changing the leads from the coil to the segment. In order to collect any current from the commutator the brushes must be at the neutral points. The current flowing through the winding divides at the neutral space and flows through the coil in opposite directions, uniting again at the other neutral space. Q. 70. (1900-01.) Suppose a magnetic field contains 25,000,000 lines of force and a conductor cuts the total number in the same direction in one second. What will be the generated E. M. F.? Suppose there were 25 conductors connected in series, cutting the lines of force at equal rates (as above), what will be the generated E. M. F.? Suppose in the latter case that the 25 conductors were moved across this same magnetic field at the rate of 40 times per second, wnat would be the generated E. M. F.? What application of formula can be made in relation to a closed coil conductor, wound upon a ring or drum core? Ans. 70. In the case of a single conductor moving across a mag- netic field in which the total lines of force is known, the rate of cut- ting is equal to the total number of lines of force cut by the conductor divided by the time required to cut them. This may be expressed in the form of an equation, thus the rate of cutting = , where N = the total lines of force cut and t = the time required to cut them. By definition: One volt is that E. M. F. generated in a conductor when it is cutting lines of force at the rate of one hundred million (100,000,000) per second. 186 Hence E = - when E is the E. M. F. in volts; t = the time in 10t 10 8 In which N = number of lines of force. n = number of times per sec. one conductor cuts the lines of force. t = time in seconds. S = number of conductors. This equation can be applied with some modifications to the closed coil conductor wound upon either the ring or drum core. In case of the ring core, E in the equation represents the maximum E. M. F. in volts that is obtained from the positive and negative brushes when the core is revolved. N is the total number of lines of force passing from the north pole through the core to the south pole (bi-polar dynamos). Each wire, therefore, on the periphery of the core, cuts the total number of lines of force twice in each revolution, or, in other words, outside wire cuts 2N lines of force per revolution. S represents the number of outside wires on the periphery through which the current flows in series, and n is the number of complete revolutions per second of the core. Therefore, the maximum E. M. F. in volts that is obtained from the brushes is found by the formula 2NSn 10 s This formula holds equally true for the drum core. In both cases the number of outside wires through which the current flows in series is equal to one-half of the total number of outside wires. Hence, by using the same magnetic field and rotating the cores at equal speed, the E. M. F. generated in both cases will be equal. The difference of potential between the brushes when the external circuit is closed is somewhat smaller than where no current is flowing; the same as in the case of the voltaic cell, a part of the total E. M. F. developed is required to overcome the internal resistance. Q. 71. (1900-01.) The foregoing questions treat upon the elemen- tary principle and physical theory of a dynamo. What is a dynamo? What are its three most essential features? In all dynamos, how is the magnetic field produced? 187 Ans. 71. A dynamo is a machine for converting mechanical energy into electrical energy by electro-magnetic induction. The three essential features are: First, a magnetic field. Second, a conductor or several conductors, called an armature, in which the electromotive force is generated by some movement relative to the lines of force in the magnetic field. Third, a commutator or collector from which the current is col- lected by two or more conducting brushes. In all dynamos the magnetic field is produced either by a perma- nent magnet or by an electro magnet, and they are classified accord- ingly. For our purpose, however, it is sufficient to consider only the uni- form magnetic field lying between the poles of some large magnet. Q. 72. (1900-01.) Give general rule for a conductor conveying an electric current when placed in a magnetic field. Give a convenient rule for remembering the direction of motion imparted to a conductor conveying an electric current when placed in a magnetic field. If a vertical conductor in which a current is flowing downward, is placed in front of the north pole of a magnet, in which direction will the conductor tend to move? Ans. 72. When a conductor conveying an electric current is placed in a magnetic field, the conductor will tend to move in a definite direc- tion and with a certain force, depending upon the strength and direc- tion of the current and upon the direction and density of the lines of force in that field. Rule. Place thumb, forefinger and middle finger of the left hand each at right angles to the other two; if the forefinger points in the direction of the lines of force and the middle finger points in the direc- tion toward which the current flows, then the thumb will point in the- direction of movement imparted to the conductor. In the case of a vertical conductor in which the current is flowing downward and is placed in front of the north pole of a magnet the conductor will tend to move from left to right. Q. .73. (1900-01.) Compare rule given in question 63 and the rule given in question 72 and explain the opposition seemingly of one to the other in the direction of current. Explain the theory of counter torque of a dynamo. Ans. 73. Comparing the previous rules with the one just given, it will be seen that the two appear to oppose each other; or, in other words, the current which flows in the former case, according to the latter rule, tends to oppose the motion of the conductor and move it in the opposite direction. This is exactly what takes place. When a conductor is moved across the lines of force, an electric current is generated, which tends to send a current in a definite direc- tion. If the circuit is open and no current flows, it requires no force to move the conductor across the field; but if the circuit is closed and a current flows through the conductor, then the action of the lines of force on the current opposes the original motion and tends to stop or retard the conductor. The opposing force is proportional to the strength of the current flowing in the conductor. But, so long as the conductor is moved, the applied force is always larger than the counter force. Hence the stronger the current in the conductor the greater will be the force necessary to keep the conductor moving in the original direc- tion. The counter force would never actually move the conductor in 1 88 its direction, but it exerts a dragging effect upon the conductor which would reduce its speed and almost stop its motion, if the exterior mo- tive force is not increased. The above principles explains the action of converting the mechan- ical energy into electrical energy in a dynamo. If an armature is properly wound and connected to a commutator, an electromotive force is generated in the outside conductors on the core, causing a difference of potential between the brushes. If the brushes are not connected to an external circuit and no current is flow- ing through the armature, it requires no energy to rotate the armature, excepting a small amount to overcome the mechanical friction and the loss in the armature iron by eddy currents. If, however, connected to an external circuit and a current flowing through the armature the conditions are changed. The lines of force react upon the current in the conductors, tending to rotate the core in an opposite direction and to retard its motion; the stronger the current the greater the retarding effect. Hence, to keep the speed constant and to generate a constant B. M. F., more energy must be supplied. This retarding effect of the current is known as the "Counter Torque" of a dynamo. Q. 74. (1900-01.) Can it be mathematically proven that the me- chanical energy delivered to the armature from any exterior source is exactly equal to the electrical energy obtained from the armature plus tne energy lost in mechanical friction, eddy currents in the iron and other small losses? What are the losses that occur in an armature? How can the effect of armature reaction be almost entirely elimi- nated? Ans. 74. It can be mathematically proven that the mechanical en- ergy delivered to an armature from any exterior source is exactly equal to the electrical energy obtained from the armature plus the energy lost in mechanical friction eddy currents in the iron and other small losses. Besides producing a counter torque in the armature, the current tends to distort or crowd the lines of force from their original position in the magnetic field. This effect is termed armature reaction. The greater part of the electrical energy is transmitted to the external cir- cuit, while the rest of the energy, usually the smaller portion, is con- verted directly or indirectly into heat energy in different parts of the dynamo itself. The principal armature loss is that produced by the current flowing against the internal resistance of the armature, that is the resistance of the conductors. The core loss, is the energy converted into heat in the iron discs of the armature core when they are rotated in a magnetic field. A small portion of this loss is due to eddy currents generated in the revolving core discs. A large portion is due to magnetic friction which occurs whenever the direction of the lines of force is rapidly changed in a magnetic substance. This effect is termed "Hysteresis." The energy expended by hysteresis is furnished by the force which causes the change ii the magnetism; and in the case of an electro- magnet when the magnetism is reversed by the magnetizing current, the energy is suppli-i by the magnetizing current. The san under the influence of sunlight, devolves the office of again separating the carbon from the oxygen, and to this source can be traced our present stores of fuel. Fuel can be burned as we will, but each atom thereof represents a part of a universe in which things are muchly shifted about, but nothing is ever completely annihilated. Q. 4. (1898-9.) What is vacuum? Why is it impossible to maintain such over or in connection with water? Ans. 4. Space .devoid of matter. From an engineer's standpoint, a vacuum refers to some chamber partially relieved of atmospheric pressure. A perfect vacuum, though realized, could not be maintained over or in connection with water, because liquids give off vapor, which rapidly refills the empty space; at a temperature of 60 Fahr. the absolute pressure due to the vapor of water is a quarter of a pound per square inch. A mercury gage would give an indication of 29.4 inches, which represents the best permanent "vacuum" obtainable at that tempera- ture. At 100 Fahr. the same gage would read 28 inches, and so up the scale for increased temperature until at 212, the boiling point, the vapor balances the normal pressure of the atmosphere. Q. 5. (1898-9.) Extension or volume expresses dimension and no body can be conceived that does not possess length, breadth and thick- 209 ness; neither can two bodies occupy the same space at the same time, hence matter is impenetrable. Extension and Impenetrability are regarded as the essential prop- erties of matter; in addition to this certain general properties are ac- credited. Name nine other general properties. Ans. 5. No particle or body of matter can be conceived that does not possess length, breadth and thickness; neither can two bodies oc- cupy the same space at the same time. The essential properties of all matter, i. e., extension and impenetra- bility, are defined by the foregoing statement. Nine other general properties accredited to matter are as follows: Weight: Due to the force of gravitation. Mobility: By virtue of which the position of bodies may be changed by the application of suitable force. Inertia: Which defines the persistency of bodies to retain their state,- be it motion or rest. Divisibility: All matter can be divided into distinctly different parts. There are practical limitations to dividing a substance; theo- retically the final limit is assumed to be the atom or molecule, as already explained. , Porosity: The ultimate particles of a substance, though apparently dense, are supposed to be suspended in space, and therefore not in actual contact. All matter is more or less porous. Compressibility: Which means the volume of any body may be diminished; this is a sequence to "porosity." Expansibility: The converse of compressibility, i. e., substances will vary in volume or size at different temperatures. Indestructibility: Matter may be caused to vary in form, but, as already explained, its elementary existence can never be destroyed. Elasticity: Within certain limits, and varying in different sub- stances, bodies tend to recover their original shape when relieved from the strain due to an applied force. Q. 6. (1898-9.) Why are hardness or elasticity considered as spe- cific properties of matter? Ans. 6. Hardness: When used to define the property of a substance, like stone; or elasticity, when used to characterize the peculiar ten- dencies of rubber, are regarded as expressive of specific properties; for not all bodies are hard or elastic in the sense here implied. Q. 7. (1898-9.) Affinity, Cohesion and Gravity are known as natural forces. Explain by familiar examples some of the effects attributed to each. Ans. 7. Affinity, cohesion and gravity are the three great forces of Nature. Affinity is the strongest of the forces, but acts only through in- finitesimal distances that is to say, it is the power that binds to- gether the atoms which constitute the molecule of a compound sub- stance. Water exists because the atoms of hydrogen and oxygen are held together by "affinity." Affinity is considered an atomic force. Cohesion is weaker than affinity acts through greater but still in- sensible distances only. It binds into a mass molecules of similar na- ture and is the power by which a homogeneous body retains its form, tenacity, etc. Cohesion is considered a molecular force. Gravity is the weakest of these three natural forces, but acts through all known distances. It tends to bind "bodies" together, as is mani- fest by the attraction which is known to be mutual between all bodies in the universe. Gravity is known as a molecular force. 219 Q. 8. (1898-9.) Give another and more familiar name for the force known as "adhesion"; give the cause for its action and examples of what would happen if it ceased to exist. Ans. 8. Adhesion is classed as a force causing molecules of dif- ferent kinds of matter to "cling" together. Chalk clings to the black- board; dust of any kind clings to every substance; bricks adhere to mortar, etc. Friction is a form of adhesion, and it is a mistake to assume that the resistances due to this cause are without compensat- ing advantages. The stability of the major portion of everything erected by man depends on friction, and "pell mell" destruction would follow in the wake of a frictionless era. Q. 9. (1898-9.) What are the three states of matter? Name the force which causes a change of state. Ans. 9. Matter exists in three different states, which are distin- guished as the "solid," "liquid" and the "gaseous" state. The state of matter varies with the force of "cohesion," which is strongest in solids, weakest in liquids, and totally lacking in gases. The intensity of cohesion is influenced by temperature. Q. 10. (1898-9.) "Force" define this term in the broad sense in which it is used in science. Ans. 10. "Force," in a broad sense, is "that" which causes any sort of change in matter, or such as may vary the form of substance or alter its condition or position in space. Q. 11. (1898-9.) Mechanics is a branch of science that treats of the effects of force upon matter. Nothing is known of force except through matter; the two may be regarded as inseparable. Motion is one of the effects of force; name another. Ans. 11. Mechanically considered, "force" is the cause of motion, or is "that" which tends to produce motion, either retarding or totally preventing the movement of bodies to which it is imparted. The effect of force, not manifest as motion, is "strain." Q. 12. (1898-9.) . What causes "strain" and how is it measured? Ans. 12. Strain is due to resistance or the reaction of opposed forces. It is measured in units of weight. Q. 13. (1898-9.) What is motion? AnS. 13. "Motion" signifies change of position. A body moving or changing its place, with regard to some point that is fixed, is said to have an "absolute motion." Relative motion refers to the movement of a body when taken with reference to another moving point. Q. 14. (1898-9.) What is "velocity" and how is it usually ex- Ans. 14. Velocity, usually expressed in feet per second, or some convenient multiples of either of these units, is indicative of the "rate of motion," i. e., time and space. INTRODUCTION TO QUESTIONS 15 to 30 (1898-9). The foregoing questions deal primarily with the well established theories concerning the constitution of matter and should have served the purpose for which these questions are intended, i. e., to give the uninformed a partial insight as to how scientists view nature in the 211 abstract. A better comprehension of the Atomic Theory, the salient points of which are referred to, is another of the objects sought, and in no instance can the aims of the committee fall short of the mark, where associations or individual members have taken up the work in the manner and spirit prescribed. A thorough understanding of nature's laws, such as may be ob- tained by an intelligent interpretation of fundamental points, is deemed essential and it is hoped that none seeking enlightenment con- sider themselves so well equipped in a "practical way" as to be exempt in this particular. In thus advocating the cause of theoretical knowledge and insisting that the progressive engineer or mechanic cannot afford to be blind to the teachings of science, we may say further, that the most important and useful of these facts are not at all abstruse and can readily be acquired by energetic believers in the forcible principle of removing an obstacle, in preference to being continually hampered by it. The second paper of the Review is offered, therefore, as additional aid in "smoothing the roadway." The Educational Committee feels confident that its efforts in this direction are not misdirected. In pointing out material which should be mastered, to insure ad- vancement, we cannot overlook the importance of mathematics as a factor concerned in nearly every proposition of a mechanical nature. To beginners in this most needful branch of learning we would say that, in this as in every other branch of science, certain fundamental principles must be absorbed to facilitate progress. Ordinary arithmetic is the stepping stone to such higher "planes' as algebra, etc.; for the present, however, we would advise a sojourn on the lower level, each individual to remain there until such time that he can "reason" him- self to a more exalted position in mathematics. A good practical knowledge of arithmetic means more than an ability to add and subtract, multiply and divide; more especially if these operations are carried out in the mechanical manner, which is not at all uncommon. The impracticability of reaching this subject by any other method has prompted the measure we provide; the article which follows is intended for all who choose to inquire into their mathematical standing; the view being taken from a strictly elemen- tary standpoint. The matter is pertinent at this time and, while di- rected at the rudiments of arithmetic, its broader aim is to inculcate the need of constant reasoning in all that pertains to the attainment of an education. MATHEMATICS. Mathematics is the science of quantity and embraces Arithmetic, Algebra, Geometry, etc. Arithmetic is more particularly the science of numbers and to a partial consideration of the rudiments of this subject the following is directed: The first proposition engaging the attention of a beginner in arith- metic is usually stated thus: "A unit is one," a single thing. An en- gine, for instance, is built up of distinct pieces, but considered collec- tively, we refer to it as a single thing. "One" as a unit, therefore, may represent either a definite quantity or an aggregate of quantities. Quantity is anything which can be increased or diminished. Mag- nitude is quantity, considered in an individual form. Multitude is quantity, when made up of individual or distinct parts. Number is a term covering broadly all answers to the question: How many? When one or more things are "counted," the result is expressed by a number which denotes the sum total of units under consideration. Numbers, applied to any particular thing, as one engine, two boilers, etc., are defined as "concrete numbers"; all numbers not associated with something tangible, are considered "abstract." 212 Notation is the shorthand method of expressing numbers by charac- ters called "figures." Numeration is the art of reading numbers so written. The figures commonly used are 0-1-2-3-4-5-6-7-8-9; combinations of these serve to express any number the mind can conceive. The figure 9, appearing alone, stands for "nine" not because the symbol is especially adapted to represent this number, but because it is T)art of a "system," which assigns a universally recognized value. When appearing alone 9 stands for the greatest possible number of units that can be given expression, by any single figure; counting from one to nine exhausts the significant figures, and at this point the elasticity of the system becomes manifest. Further progress in notation is made by assuming the next higher number as the initial unit of a higher order, i. e., one unit of the new order being the equivalent of ten single things as originally conceived. To express this number we again revert to the figure 1, using in con- nection therewith the auxiliary digit 0, writing 10, to represent ten, and considering same as a single unit of the second order or place. The need of the character 0, referred to as an auxiliary and usually called naught or cipher is self evident; it has no. assigned value, but its posi- tion at the right of any figure increases the numerical value thereof tenfold. At 99 we have nine units of the order of tens and nine simple units, the sum being equivalent to ninety-nine, the largest number capable of expression with two figures. It is plain, however, that the process of building up additional orders can be repeated and the system ex- tended indefinitely. Numbers written with more than three figures, are for convenience divided into groups of three places each, which are known as periods. This grouping of the orders into sections of three is indicated by dots, placed to precede the first figure of each period. The first period consists of Units, Tens, Hundreds, and is called Unit Period; thus the number 583 stands for five hundred and eighty- three units. The second period is also taken to consist of Units, Tens and Hun- dreds, but each unit here has a value one thousand times greater than assigned the figures in the first period; hence 583,000 is read five hun- dred and eighty-three thousand. Additional periods are successively Millions, Billions, Trillions and so on upward into a maze of figures that seldom, if ever, need invade the engineer's mind. The foregoing epitome of the Arabic System of Notation embraces points more dormant than new; it is, however, the "essence" of arith- metic and, as such, the several definitions must not suffer rusty repose in our intellects. Four fundamental operations in arithmetical computations are Addi- tion, Subtraction, Multiplication and Division, represented and indi- cated respectively by the familiar signs +, , X, -f-. The several processes are based upon and carry out the principles set forth in the ground work of Arabic notation and when a correct idea of the system of "unit values" is once firmly rooted in, and thoroughly comprehended, a mathematical stride of no mean importance has been accomplished. * * * * To profit by the experience of others, and to arrive at conclusions for nimself, also to lessen the labor involved in many calculations, the engineer must learn to read abbreviated mathematical propositions and should understand the "short cut" methods which facilitates the opera- tions indicated. Ihis does not necessarily imply a profound knowledge of Algebra, as the operations indicated in some of the most useful engineering formulae are readily solved by arithmetical computation. The fact that a person can add, subtract, multiply and divide is not enough for the purpose; there is much more than this between the 213 Covers of a common school arithmetic which may be absorbed and used with profit and effect. Mathematics is essentially a science of reasoning and the better the cultivation of the reasoning power the less burdensome becomes the mathematical labor attached to the work. To this end, therefore, a plea is made for a recognition of the true value and importance of the principles as set forth in any ordinary arithmetical text book. It may seem tiresome, or may be even frowned upon as "folly" by some, to again take up studies of. the kind here recommended, Many of course are far past this stage, but none have traveled a better or easier route. C. H. F. Note. Anything assumed as One may be taken as a Unit. Some particular thing, or rule, established to fix a permanent value, becomes a "standard"; and as such it serves as the accepted measure for the purpose it is designed for. The absolute need of unchangeable "stand- ards," by which values, etc., may be measured and compared, is self- evident. Every civilized nation defines its units of measurement with accuracy and upholds and enforces them by legal enactments. This paper is largely given to the consideration of the units em- braced on the engineer's "yard stick," and those not already conversant with the subject should cultivate that intimacy which its importance demands. Q. 15. (1898-9.) First Law Every body continues in a state of rest, or of uniform motion in a straight line, unless acted upon by some external force. Second Law Motion, or a change of motion, is proportional to the force impressed and is in the direction of the line in which that force acts. Third Law Action and re-action are always equal and are in oppo- site directions. Who was the first to give expression to the foregoing? Ans. 15. The three "Laws of Motion" are accredited to Sir Isaac Newton. Q. 16. (1898-9.) Weight is due to the force of gravitation, hence every operation of "weighing" is a measurement of that mutual attrac- tion, which is inherent to every particle of matter in the universe. Give the units of weight commonly used in engineering. Ans. 16. Avoirdupois, or Commercial Weight, is the common stand- ard of the United States; the principal unit is the "pound," but great weights are more conveniently expressed in tons, the latter being a multiple of the original unit. The legal definition of the ton is 2,240 pounds; the short ton of 2,000 pounds is customary in many branches of trade. The avoirdupois ounce is 1/16 of a pound; in engineering calcula- tions it is preferable to express such fractions in decimals of a pound. "The standard avoirdupois pound is equivalent to the weight of 27.7015 cubic inches of distilled water, Weighed in air, at 39.83 Fahr., barometer at 30 inches." Haswell. The French, or Metric, system is also a legal standard in this coun- try. Q. 17. (1898-9.) A line is length without breadth or thickness. A straight line is the shortest distance between two points. How is the measurement of length ordinarily expressed and upon what is the unit based? Ans. 17. The foot, and its subdivision, the inch, are the units gen- erally used for expressing "Long Measure." The United States and British standards are the same, and the funda- mental unit is the yard, which is said to have been based originally 214 on the length of a pendulum vibrating seconds at the level of the sea, in the latitude of London, in a vacuum, with Fahrenheit thermometer at 62. The length of such a pendulum is supposed to be divided into 39.1393 equal parts, called inches, and 36 of these inches were adapted as the standard yard. Q. 18. (1898-9.) As a matter of convenience many of the standard units of measurement are either subdivided or used in multiple; thus "Time" is noted in seconds, minutes, hours, etc. It would be awkward to refer to 600 seconds of time, but six seconds is a more ready ex- pression than six-sixtieths of a minute and so on. Frequently certain units are combined and thus form a new meas; ure. Assuming the minute as the unit of time, what important other unit is evolved when expressed in connection with those representing weight and distance? Ans. 18. Using the minute as expressive of the "time"; the pound representing "weight," and the foot indicating the space or distance traversed, gives us the "foot-pound," or the unit of work. Time, weight and space are inseparably linked in the measurement of power, and the elements embraced in this most important unit should be under^ stood in a way inspiring a full confidence as to what they really por- tend. Hazy views on this point will impede progress. Q. 19. (1898-9.) Explain the relation between "foot-pounds" and "horse-power." Ans. 19. 33,000 foot-pounds are the equivalent of one horse-power; both are units for the measurement of work or power the relation between them being similar to that of the pound and the ton; that is. one being a multiple of the other, both expressing the same thing. Q. 20. (1898-9.) Area is a term used to express the superficial con- tents of any figure. Give the units usually used in this connection and define a "plane" surface. Ans. 20. Surface or area is expressed in square inches or square feet, as best befits the case. By a "plane figure" is meant any flat sur- face enclosed by real or imaginary lines which form the outlines or boundary thereof. Such a figure is assumed to have length and breadth, without thickness, and its surface is computed in squares correspond- ing to the units used in measuring its lineal dimensions. Q. 21. (1898-9.) Length, breadth and thickness are essential to volume. How would you express the volume or cubical contents of a solid? Ans. 21. As indicated in the question, three dimensions are essen- tial to volume that is, we measure length, breadth and thickness and ;: express "cubical contents" in cubic inches, cubic feet, and so on, as the case may require. Awkward and ridiculous blunders are a frequent result from the careless confounding of the expressions noted in this and the preceding question. Q. 22. (1898-9.) Give the name of a unit commonly used in measur- ing fluids and state upon what it is based. Ans. 22. The United States standard gallon, containing 231 cubic inches, is a'common measure for liquids. Q. 23. (1898-9.) Sketch diagram representing blocks, measuring in inches, 12" X 12" X 12"; 24" X 12" X6"; 36" X 12 3 ; calculate cu- bical contents of each in feet, and surface in square feet. Demonstrate thoroughly the difference between solid and surface measurement. Ans. 23. See cuts. 215 -fatf j Sfevre />. ANSWER NO. 23. Q. 24. (1898-9.). Geometrically considered the circumference of a circle is a curved line, all points of which are equally distant from a certain point within, called the center. The unit lor measuring angles is derived from the circumference of a circle. Give the recognized name of an angle measuring 89 degrees 59 min- utes and 60 seconds; explain briefly how this measurement is applied and the signs used to abbreviate the units mentioned. Arts. 24. Circular measurement assumes the periphery of the circle divided into 360 equal parts, each of the spaces being known as a "degree." The degree is divided into 60 equal parts, called "minutes," and a further subdivision of the minute into 60 parts gives seconds. The signs or abbreviations used are thus: 89, is read, eighty-nine de- grees; 59', indicates fifty-nine minutes, and 60", refers to sixty seconds. An angle measuring 89, 59', 60" is the equivalent of 90 degrees, and is 360 =4 90 therefore the fourth part of a circle. In more ordinary language, such an angle is a square, and geometrically it is defined as a "right angle." Q. 25. (1898-9.) Temperature is also expressed in "degrees," but it is evident that heat and cold do not enter into a geometrical proposition. Explain the mercurial thermometer and Fahrenheit's scale. Ans. 25. The principle involved in the ordinary mercurial ther- mometer hardly requires special mention; interest centers rather to the scale or graduation by which the expansion and contraction of the mercury is measured and by which the instrument is read. "Fahrenheit" is the common U. S. standard in engineering practice, and this scale, which takes the name of its originator, marks the boiling point of water by 212 and the freezing point by 32. Zero, or 0, was fixed at the temperature acquired by a mixture of ice and salt. Thermometric scales are entirely arbitrary, but once established and recognized, this fact in no way detracts from their general usefulness. Q. 26. (18989.) Explain briefly the principle embodied in a barom- eter; state how this instrument is read and what the reading indicates. Ans. 26. The barometer is an instrument in which the known v.-eight of a column of mercury is placed in opposition to that of the 216 atmosphere, the prime purpose being to ascertain the varying pressure of the latter. The readings are taken in inches of mercury which represents the difference in levels between the surfaces of the mercury contained in the ciste-n and the indication as noied in the tube. It should be understood that the upper end of the tube is sealed; also that the empty space in the upper part thereof is a vacuum, and that the pressure of the atmosphere is acting on the surface of the mer- cury into which the lower end of the tube is immersed. Knowing the weight of a cubic inch of mercury, and also, being conversant with the manner in which pressures are transmitted by liquids and gases we may readily recognize the principle which is involved and also un- derstand how the reading of the barometric inches can be converted into pounds pressure per square inch. Q. 27. (1898-9.) What impression should the reading of a pressure -gauge convey to the engineer? Ans. 27. The reading of the pressure gauge should convey to the engineer a realization that the force indicated in pounds pressure per square inch is acting with that intensity upon all the surfaces or areas which are accessible to the pressure under notice of the gauge. It should not escape us, however, that considered in the sense of causing rupture the ends of a string must be pulled in opposite directions to produce a strain equal to the force applied at one of the ends. Q. 28. (1898-9.) In a concise yet comprehensive manner explain the value of the British Thermal Unit of heat and the connection and importance of this unit in the measurement of heat exchange. Ans. 28. The British Thermal Unit of heat, usually abbreviated to B. T. U., is a measure which defines heat as a quantity irrespective of temperature. The unit is based on and is equivalent to the amount of heat required to raise one pound of water through one degree of Fahren- heit. Similar quantities of different substances require varying quanti- ties of heat to produce the same temperature; the B. T. U. therefore establishes a basis for intelligent comparisons and is used in computing all questions relating to the transmission or absorption of heat. There is believed to exist a definite relation between heat and mechanical energy; this relation expressed in foot pounds has been determined experimentally and is known as the "Mechanical Equivalent of Heat" that is, 778 foot pounds of work done, being regarded as the equivalent of one d. T. U. INTRODUCTION TO QUESTIONS 31 TO 42 (1898-9). ANOTHER STEP IN MATHEMATICS. The value and need of good, sound arithmetical knowledge, ranging upward from the simplest propositions, has been accorded proper recog- nition in these columns. The fact that it is quite possible to "get along" without such further embellishment of the science, as are em- braced in the higher branches of mathematics, has also been conceded. This admission, while pleasing, perhaps, to those preferring to evade the more advanced methods of calculating, must be qualified, however, by the further assertion, that the "kind" of progress which is possible under such limitations, may be likened to the difference in travel, between the stage coach of old, and the "flying" express trains, now so familiar in modern railroading. By either method, of course, the traveler starts with the expectation of reaching his destination, 217 but where the latter mode is available, it is safe to say, that usually the business may be done and time enough remains to forget about the transaction, before the ancient form of conveyance could be "due" at the further end of tfie route. For covering short, easy distances, the simple and more primitive vehicle continues to fit conditions not requiring the excessive elabora- tion of a railway system; so with propositions of a mathematical na- ture easy problems are best solved by simple methods, but when the elements of the case are quite intricate, the question resolves itself into one of two things, i. e., a solution, acquired by a tedious, slow and laborious process, or the same results, reached with ease, rapidity and dispatch. As the railway emanated from the older and slower methods of transportation, so are the higher mathematics evolved from the funda- mental reasoning on which arithmetic is based; both are representative of progress, and as the business people look to the former as the means of facilitating trade, so should progressive engineers view the latter, for algebraic expressions will continue to be used by advanced writers, and to expect anything but this is fully as unreasonable as a demand for a general return to customs and practices which are regarded as ob- solete. It would be a difficult matter to answer satisfactorily "how far" the stationary engineer should pursue mathematics, or to say definitely how little is. necessary to serve his purpose; each individual, rather, should estimate his needs, and referring more particularly to such engineers, not favored with an early training, the deficiency should be made good as fast as it is found that a lack of such knowledge acts as an impediment to progress. Personal experience is one of the best ways to gain knowledge but when dependent on this alone, progress is slow, because life is too short to acquire all things by actual contact; advancement follows with more certainty and greater rapidity to those who aim to couple with their own practice the facts garnered by the multitude of other patient workers in the same field. To distribute accumulated experience is the object of the mechanical press; it is the object of all literature which pertains to the work of the engineer or the artisan, and is the prime object of this, and other, associations organized for the dissemination of technical knowledge. To profit by reading, it is evident that the language used should "strike home,", as published rules and deductions are usually expressed algebraically, it follows that the reader should be conversant with the mathematical signs ordinarily used and should also be familiar with simple equations, for seldom do the most useful formula exceed the scope of this ready method of stating concisely the mathematical opera- tions to be performed on the quantities which represent the elements entering into the problem. As expressed verbally, or in written language, every rule recites the arithmetical operations to be performed with certain known quantities, in order that the value of some other may thereby be determined. Let us assume, now, that we are charged with the duty of investi- gating the strength of a double riveted lap joint and let the mathe- matical reasoning concerned therewith serve to make plain the opera- tions usually involved in any formula, when set forth in the form of a simple algebraic equation. In this particular case, the elements entering into this problem are as follows: The tensile strength of the plate per square inch of section, the thickness of plate in inches and the portion of the plate remaining unimpaired between rivet holes; the product of these three factors represents the ultimate strength of the joint, as far as the plate has to do with the question. On the other hand, we have two rivets, corresponding to the above section of plate; these have a certain shearing resistance per square 218 inch of suction; the combined area of " and multiplied by their shearing strength, gives the ultimate holding power of this second element of the joint... /- (j y T A concise writer would probably express the foregoing. elements as follows, representing each of the several factors by. , means of. -a- single letter, thus: . . T = Tensile strength of pla"te~p r er square inch. . ; p Thickness of plate. r = Length of plate remaining between rivet holes. A = Area of the rivets. s = Shearing strength of rivets per square inch. Then, T X p X r = A X s, is the equation, covering the theory of the riveted joint; interpreted, it reads as indicated in the preceding para- graphs, i. e., the tensile strength of the plate per square inch (T) multiplied by the thickness of the plate (p) and again multiplied^? the length of the plate remaining betw'een rivet holes (r) gives a cer- tain product, which varies of course according to the values which may be given or assigned to each of the several factors. Now if this product is found equal to the value of the area of the two rivets which serve to hold the above section of plate when their combined area (A) is multiplied by (s) their shearing strength per square inch, we may conclude that the j'oint is - at its -best, for It is evident that each side, or element, is affording exactly the -same -resist- ance to a rupturing force. As written, the equation TXpXr = AXs, is general, i. e.,.J leaves the actual value of each factor to be found and substituted fey the person making practical use of the rule, hence if the several dimen- sions and the strength of the plate and ;rivet are known, and after performing the operations indicated, it is found that one is greater or less than the other, the. fault is not with the equation, but with the joint, for the proposition may be such that the products found; do; not maintain the required equality which the theory of the -case demands:;: The investigation of a given joint, according to the method .ovffi- lined, is simply "weighing" the question in the- algebraic "balance," called an equation, and in this connection, it -i necessary 'to firmly x in mind the true meaning of the mathematical sign of equality, thus,-~. The sign of equality means just what its- name 1 indicates, and- -when it appears in a mathematical proposition, it serves as an indication that the values of the huinbers, or -quantities, between which It is placed, are equal, or the equivalent of one another^ just- as the pans of the "old time" balance come to a level when both sides are equally weighted. Equations are solved by performing certain operation otr its terms, the object being to find the value of unknown factors in .the problem, by a method of reasoning directed against others which have a ; known or assigned value. In a general way, the -process may be illustrated by assuming the quantities on each side of the sign =, to be weights, laying on the pans of an ordinary beam balance. The weights being equal, the pans stand at a level and so they will remain, provided equivalent measures only are added, or taken from either side at the same time. y . . ;-",-;'" Let us now again take up the general equation . noted in connection with the theory of riveted joints; where T X p X r A X s. Assume the factors p Xr, to have a balancing value, the exact amount of which we need care nothing about. Simply to. illustrate the proposition, how- ever, we say that p X r represents half the weight contained in ,th,e, left hand pan of the balance. We can remove this half 'of the weight from the right side and not destroy the equilibrium of the balance,, provided the weight in the left hand pan is also divided or diminished by "half." A little reasoning now makes evident'the fact, that p X r, taken from the right hand side of the equation and used as a divisor against 219 *bft terms appearing on the right, will change the reading from A X s TX PX r = AX s to T = (1) p Xr. Further investigation will prove that this operation has in no way disturbed the equality between the two sides; that is, the pans of the balance remain level; we have gained an important point, however, by thus solving "T" and by similar reasoning we deduce. A X s (2) (3) T Xp -also, T X p X r (4) (5) It will be noted the original equation has undergone five variations, that is, each factor has been -arranged to stand alone in some one of these, hence if the value of any four of the five are known, the proper and corresponding value to be assigned to the one which is sought, can be readily determined by performing the operations indicated, substi- tuting for the letters the numerical values which they are representa- tive of; in other words, With the aid of a little mathematical reason- ing, differing but slightly from the kind ordinarily used, the theory of -a riveted joint have been converted into five rules. These five rules in this form, occupy but little space and answer effectually every question that can be raised in connection with the subject to which they pertain. If it be a question of plate thickness, against the other values of a joint, Equation (2) is -used reading thus: thickness of plate (p) equals, (=) product of rivet area and shearing strength of same, divided by the product 'Of tensile strength of plate and length of unimpaired plate, between rivet holes, and so on. It is not the object of this article to instill more than a mere ele- mentary idea of the higher mathematics, neither has it been the aim to deal particularly with the subject of riveted joints, for it should be understood the principle explained has a wide and general applica- tion. Neither of the foregoing subjects can be administered in a single heroic dose; if, however, the importance of this additional step in mathematics is made more apparent by the language here recorded, the object in view will be quite fully accomplished. C. H. F. Q. 31. (1898-9.) Gravity is a force, acting at all distances and tend- ing to attract mutually, all bodies in the universe. Give the law of gravity and state by whom it was first enunciated. Ans. 31. Gravity is an attraction common to all material sub- stances; the law of universal gravitation was discovered by Sir Isaac Newton and is as follows: "The force of attraction between bodies of matter is in exact .proportion to their mass and inversely as the square of their distance aparti" The first of these propositions is quite plain; according to the other, the attractive power diminishes in the same proportion as the "square" of the number which expresses the increase of distance. Thus if the distance is doubled the attraction is lessened fourfold, because 4 = 2 X 2 and is the square of two, etc. Q. 32. (1898-9.) What common instrument serves to show the direc- tion in which gravity acts and what does this indicate to the mechanic? Ans. 32. The plumb line and "bob" is a common instrument which shows the direction in which the force of gravity acts. To the me- chanic it indicates a true perpendicular; it is evident, however, that no two such lines can be called strictly parallel, but for all the ordi- nary purposes of a mechanic, they may be so considered. Q. 33. (1898-9.) Is the force of gravity constant; i. e., does the weight of a body vary according to its position with reference to the surface of the earth? Ans. 33. Gravity is constant in the sense that its. force is always acting. "Weight" is simply a measure of the force of gravity and serves to indicate the intensity of attraction at or neaj the surface of the earth and at which point is realized the maximum effect or great- est weight. At the earth's center weight is nullified and becomes manifest ac- cording to the following laws of weight: Moving outward from the earth's center, weight increases as the distance from the center increases. Above the surface of the earth weight decreases as the square of the distance increases. Therefore, whenever the distance between two bodies varies to a sensible amount, gravity must be considered as a variable force. Q. 34. (1898-9.) The "center of gravity" is a point upon which a body balances in every position. What is the difference between stable and unstable equilibrium? Ans. 34. When a change in position or a slight displacement teod& to elevate the center of gravity the body so moved tends to return to its original position and it is then said to be "stable" or in a state of stable equilibrium. When, however, the center of gravity- is above the point of support and the shape of the body is such, that a jolt is likely to throw the line of direction outside of its base, then we have an un- stable body or one likely to fall because of the constant tendency of the center of gravity to seek a position nearest to the point of support. Q. 35. (1898-9.) Explain the difference between a constant or uni- form rate of motion and motion uniformly "accelerated." Ans. 35. "Uniform motion" means, equal spaces described in equal time; that is, the "rate" being always the same. If the body describes a greater space in each successive movement the motion is "accele- rated"; on the other hand, if the spaces be less, motion is said to be "retarded." FALLING BODIES, First law of falling bodies. The space described by a falling body, in any given second, is equal to the product of twice the number of seconds, minus one, times the space described the first second. Thus, a body will fall during the sixth second, viz: 2 X 6=-l.$i)& 12 minus 1 = 11. Space described the first second is 16.08, Then 16.08 X 11 = 176.88 feet. Second law of falling bodies. The velocity acquired by a falling body at the end of any given second is equal to the product of the number of seconds into twice the space described the first second. Thus, the velocity attained at the end of the sixth second is 32.16 X 6 = 192.96 feet. Third law of falling bodies. . The total space described by a body at the end of any given second is equal to the product of the square of the number of seconds into the space described the first second. Thus, the total fall during six seconds is 16.08 X 36 = 578.88 feet. Q. 36. (1898-9.) By whom were the foregoing laws determined and under what conditions are they true? ,Ans. 36. The "Laws of Falling Bodies" as noted above were dis- covered by Galileo. The laws assume that only the force of 'gravity is acting on the falling bodies; retardation due to resistance of the air or other influ- ences are not considered, as such are always variable and cannot be allowed for 'in a single b.r'oad rule. Falling in a vacuum the lightest substance partakes of the same rate of motion and drops just as promptly as the heaviest metals. An unretarded falling body is an example of accelerated motion as may be noted from the various examples given in Q. 36 and 37. Q. 37. (1898-9.) Knowing the laws of the falling bodies as given, how would you reason therefrom the final velocity acquired by a body falling through a. space of 231 feet? Ans. 37. In the absence of a specific rule, but knowing the laws of falling bodies, an answer to the proposition embodied in Question No. 37 is obtained by reasoning as follows: Wanted "the .final velocity acquired by a body falling through a space of 231 feet." According to the second law of falling bodies the velocity acquired at the end of any given second of time is 32.16 times the number of seconds and V 32.16 X t, is an equation which expresses that fact. The proposition is to find the value of V, but in order to do this it is evident -that a numerical substitute must be found for "t," which in this case stands for "seconds" during which the body is falling: . From the third law we deduce: h = t 2 X 16.08, which means "h," the total height fallen, equals the product of the square of the time and the space traversed the first second. Transposing the equation: f = -^- t = /~ 16.08 -y ns.08 Hence the value of "t" is equal to the square root of the distance "h" divided by 16.08. The expression 16.08 can therefore be used as a substitute for "t" in the first equation, thus: For the case in hand the true value of "h" is 231 and inserting this, we have * 16.08 making the value of "h" = l, we establish a general rule, viz: V = 32 ' 16X -' 16.08 which is equivalent to V = 8.02 X l/h" Using this for the present proposition gives: 8.02xl/^T= 121.88 feet. So also does the more general rule: V = l/2gh~= 121.88 feet. all of which verifies the correctness of the reasoning and indicates that principles, well fathomed, are quite independent of fixed rules. Q. 38. (1898-9.) Which of the values given in the preceding laws do you recognize as the "increment of velocity" due to gravity? What small letter of the alphabet is generally used to represent this value, when such enters into a proposition which is expressed algebraically? Ans. 38. 32.16 is a quantity well known as the "increment" of ve- locity. This value is slightly variable for different parts of the globe, and it is customary therefore to represent it by the small letter "g" in all general formulae in which it occurs; as a factor. Q. 39. (1898-9.) The specific gravity of a substance is expressed by figures which represent the relative or proportionate weight of that substance as compared with an equal bulk of some other which is fixed upon as a standard. Name the standard substances usually assumed and explain how to determine the weight of a cubic foot of metal when its specific gravity is given at 7 7/10. What metal is probably referred to? Ans. 39. Water at 62 Fahr., weighing 62.355 pounds per cubic foot, is the standard usually referred to, in connection with specific gravity. For gases, the density of atmospheric air is made the basis of com- parison. The weight of a cubic foot of any substance may be found by multi- plying its specific gravity, by the weight of a like volume of water; hence if 7.7 represents the S. G. then 62.355 X 7.7 = 480. The product corresponds with the weight of a cubic foot of wrought iron. Q. 40. (1898-9.) What constitutes a "buoyant" substance and what law governs the degree of submersion of a body placed in water? Ans. 40. Any solid, which weighs less, volume for volume, than the fluid in which it is immersed is said to be a "buoyant substance." The law of submersion is that an immersed body loses an amount of weight, equivalent to the weight of the fluid it displaces. Q. 41. (1898-9.) State what proportion of a timber, having a cross section of 12 by 12 inches, will remain above water, when its specific gravity is 0.5. How much additional weight must probably be added for every foot of its length, to cause it to sink until submerged? Ans. 41. The specific gravity of timber being .5, its weight per cubic foot is 62.5 X .5 = 31.25 pounds or just half the weight of like volume of water. In accordance with the law quoted in the preceding case and noting the proportionate weights of the water and the timber, it is evident that one-half of the latter will remain above water and if its section is 12" X 12" an additional load of 31.25 pounds on every foot of its length will be required to balance the "buoyant effort" due to keeping the timber down on the water's surface. Q. 42. (1898-9.) Explain the instrument known as the Hydrometer. Ans. 42. The hydrometer is an instrument devised for determining the density of liquids. The graduations follow an arbitrary scale of degrees in Baume's and other instruments. The hydrometer really indi- cates the specific gravity of fluids and its scale can be arranged to read that way directly, or the arbitrary scale readings in degrees, may be converted into specific gravity by reference to the proper tables. The buoyant power of liquids varies with their weight or density and this is the principle made use of in the hydrometer. The instrument is marked at or zero when immersed in water and the graduations running both ways from this point indicate a greater or lesser density when the instrument is immersed in the liquid which is to be tested. PREFACE TO QUESTIONS OF 1899-1900. FALLING BODIES. To the progressive engineer, a thorough knowledge and understand- ing of this subject is certainly advisable. All computations in relation to bodies in motion, in their different characteristics, are affected by these rules, and without a knowledge of these principles one is in the dark why certain formulae and rules are given. I trust that a few questions on the subject will in no wise appear as a bugbear to our members who are earnestly seeking for a higher standard of improvement, not only in their manual training but in their social life, and I trust that those who are willing to work will be amply repaid for Uie time spent. Refer back to last year's elementary course and refresh your mind on the treatment of the laws of motion, etc., and then grapple with the few questions of this year. Some standard work, for reference, should also be in the hands of all engineers. "Acceleration." When an unrestricted force, acting upon a body, sets it in motion (i. e., gives it velocity), in the direction of the force, this velocity increases as the force continues to act, each equal interval of time (if the force remains constant) bringing its own equal increase of velocity. Thus, if a stone be let fall, the force of gravity gives to it, in the first inconceivably short interval of time, a small velocity down- ward. In the next equal interval of time it adds a second equal velocity, so that at the end of the second interval the velocity of the stone is twice as great as at the end of the first one, and so on. We may divide the time into as small intervals as we please, and each such interval, the constant force of gravity gives to the stone an equal increase of velocity. The rate of acceleration is the acceleration which takes place in a given interval of time, usually one second. The unit rate of acceleration is that which adds unit of velocity in a unit of time, or when English measures are used, one foot per second per second. For a given rate of acceleration the total accelerations are, of course, proportional to the times during which the velocity increases at that rate. Laws of acceleration: First When the forces are equal the rates of acceleration are inversely as the masses. Second When the masses are equal the rates of acceleration are directly as the forces. We thus arrive at the principle that, in any case, the rate of accelera- tion is directly proportional to the force and inversely proportional to the masses. Hence, if we make two forces proportional to two masses, the rates of acceleration will be equal; or for a given rate of accelera.- tion the forces must be directly as the masses. Time and space in our paper would not permit me to extend this study in detail, yet I will call your attention to some headings which should be carefully studied and thoroughly understood. For my own part, I regard the laws of falling bodies one of the most interesting studies that I have ever taken up. I will call our attention to: The constant force of gravity; the acceleration of gravity; the rela- tion between force and mass; the unit of mass impulse. Up to this point we should know definitely what to Consider as the unit of velocity, force, mass and time. From this knowledge we will find that: force X time Vel. = velocity X mass Force = time force X time Mass = velocity mass X velocity Time = force Also force X time mass X velocity. This is simply a problem of four factors, having three of them given to find the fourth, which by transposition we see are easily obtained. Forces in opposite directions; inertia and the densities of masses should be carefully investigated. Force in relation to work and so on until you arrive at the direct laws of falling bodies, which at some future time may be fully explained to you. Q. 61. (1899-1900.) Having the temperature of sensible heat of steam given, give the rule for finding the total heat of the steam. Il- lustrate with an example expressed in formula. Ans. 61. Using Kent for authority, the total heat of saturated steam is found by first subtracting from the temperature of a given pressure, 32, and multiplying the remainder by .305 (the specific heat of saturated steam), and to the product add 1091.7. Example: When H= total heat, and t = the temperature. Then H = . 305 (t 32). + 1091.7. The temperature or sensible heat of steam at 70 Ibs. g.p. is 316 1 The total heat equals H. Then, H = .305 (316.1 32) + 1091 7 = 1178.3 B. T. U. Q. 62. (1899-1900.) What is the total heat of steam at 100 Ibs gauge pressure? Also the latent heat? Ans. 62. The sensible heat of steam 100 Ibs. g.p. is 337.8 F., and the total heat, using the above formula (Kent's) where H = .305 (t 32) +1091.7. H = .305 (337.832) +1091.7. H = .305 (305.8) + 1091.7. H = 93.27 + 1091.7. H = 1185.0, or the total heat of steam at 100 Ibs. g.p. For the latent heat of this given pressure (100 Ibs. g.p.) using "Clark's formula" for finding the latent heat of steam at any given temperature. Where L = latent heat; t = temp, of the steam L = 1092.6 .708 (t 32) L = 1092.6 .708 (337.8 32) L = 1092. 6 (.708 X 305.8) L = 1092.6216.5. L = 876.1 B.T.U., or the latent heat of steam at 100 Ibs. g.p. 225 Q. 63. (1899-1900.) If the steam in the boiler is 270 and the feed water is 110, how many units of heat will be necessary to add to the water to turn one pound of it into steam? Ans. 63. Using "Kent's formula" the number of heat units to be added will be . .305 (t 32) + (1091.7110) = .305 (270 32) + 981.7 (.305 X 238) +981.7 = 72.59 + 981.7 = 1054.29 heat units to be added. Q. 65. (1899-1900.) Which is the better conductor of heat, dry or moist steam? Why? Ans. 65. Moist steam is the better conductor of heat. Dry steam is a poor conductor of heat, as compared with liquid water, or moist steam, for after moist steam has received enough heat to make it dry, or nearly so, it receives additional heat very slowly. Q. 75. (1899-1900.) Does the change from water to steam, by the application of heat, affect the relation of the particles of the fluid? What has this change to do in relation to power? Ans. 75. As water, the particles are strongly cohesive; as steam, the particles are repellant. It is this repellant force existing among the infinitely small atoms of steam which appears to give the energy to the mass of steam and render it serviceable. The fluid, as water, is inexpansive, but the change to steam, by the application of heat, gives it energy or ability to do work, by the reason of its great expansive or elastic tendency. Q. 81. (1899-1900.) What letter of the alphabet denotes accelera- tion? What is the rate of acceleration per second, in feet, of a body falling freely in vacuo, from a state of rest? What is the total acceleration in feet at the end of one, two, three, four, five, six and seven seconds? What will be the distance fallen from a state of rest at the end of each second (as above stated) ? What will be the distance fallen (as above stated) in feet between each consecutive second? Ans. 81. Acceleration is denoted by the letter "g" and equals 32.2 feet. A body, falling freely in vacuo from a state of rest, acquires, by the end of the first second, a velocity of about 32.2 feet per second; and in each succeeding second an addition of velocity, or acceleration, of about 32.2 feet per second. In other words the velocity receives in each second an acceleration of about 32.2 feet per second, or is accelerated at the rate of 32.2 feet per second per second. This rate is generally called simply the acceleration of gravity. 'It increases from about 32.1 feet per second per second at the equator to about 32.5 feet at the poles. These values at the sea level, but at a height of five miles above that level it. diminishes by only one part in 400. For all practical purposes it may be taken at 32.2 feet. The total acceleration in feet of a body falling in vacuo, from a state of rest, may be found by the following rule: Rule: g multiplied by the time the body was falling equals the total acceleration acquired at the end of that second. The acceleration acquired end of 1st second equals 32.2 feet. 2d second equals 64.4 feet. 226 3d second equals 96.6 feet. 4th second equals 128.8 feet. 5th second equals 161.0 feet. 6th second equals 193.2 feet. 7th second equals 225.4 feet. The total distance fallen from a state of rest, at the end of each second, is found by the following rule: Rule: The square of the time of the falling of the body multiplied by V 2 g. A body will fall at the end of the 1st second 16.1 feet. 2d second 64.4 feet. 3d second 144.9 feet. 4th second 257.6 feet. 5th second 402.5 feet. 6th second 579.6 feet. 7th second 788.9 feet. The distance fallen in feet between each consecutive second is found by the following rule: Rule: g multiplied by the number of seconds the body was falling minus % second equals the distance a body will fall from the end of one second to the end of the next succeeding second. A body will fall, from rest in the 1st second 16.1 feet.' 2d second (or from the end of the first to end of second) 48.3 feet. 3d second 80.5 feet. 4th second 112.7 feet. 5th second 144.9 feet. 6th second 177.1 feet. 7th second 209.3 feet. Q. 82. (1899-1900.) In answer to this question give the rule, or state in formula, what will be the acceleration acquired in a given time (from rest) ; in a given fall (from rest) ; in a given fall in a given time (from rest)? What will be the time required for a given acceleration; for a given fall (from rest) ; for a given fall (from -rest or otherwise) ? What will be the fall for a given time (from reat) ; required for a given acceleration (starting from rest); during any given second (counting from rest) ? Ans. 82. The acceleration acquired in a given time = g X time. In a given fall (from rest) = V 2 g X fall. In a given fall from rest in a given time = twice the fall -4- time. The time required for a given acceleration = accel. -4- g. For a given fall (from rest) = V fall -4- y 2 g, or fall -=- % final ve- locity. For a given fall from rest or otherwise = fall -=- mean velocity, or = fall -f- V 2 (initial vel. + final vel.). The fall in a given time (from rest) = time X Vz final vel., or = time 2 xy 2 g. For a given acceleration starting from rest acceleration 2 -r- 2 g. During any given second counting from rest = g (number of sec- onds y 2 second). Q. 83. (1899-1900.) Does the resistance of air affect the strict ac- curacy of these rules? Does the specific gravity of different bodies affect their falling properties? How do these laws apply to bodies thrown upwards vertically in the air with a given velocity? 227 Ans. 83. Owing to the resistance of the air none of the above rules gives perfectly accurate results in practice, especially at great veloci- ties. The greater the specific gravity of the body the better will be the result. The air resists both rising and falling bodies. If a body be thrown vertically upwards with a given velocity, it will rise to the same height from which it must have fallen in order to acquire said velocity; and its velocity will be retarded in each second 32.2 feet per second. Its average ascending velocity will be one-half of tliat with which it started; as in all other cases of uniformly retarded bodies. In falling it will acquire the same velocity that it started up with, and in the same time. Q. 84. (1899-1900.) The time a weight was falling (from rest) was seven seconds, what was the distance in feet, passed through? What was the distance at the end of the fifth second? What was the distance passed through between the sixth and seventh second? Ans. 84. The distance the weight would pass through in 7 seconds would be d = t 2 ^> g = 7 2 X 16.1 = 49 X 16.1 = 788.9 feet ans. Second condition d = t 2 % g = 5 2 X 16.1 = 25 X 16.1 = 402.5 feet ans. Third condition d = g (7 %) = 32.2 X 6V 2 = 209.3 feet ans. In which d = distance fallen in feet. t = time in falling in seconds. g acceleration in ft. per second. Q. 85. (1899-1900.) What will be the impact (striking force) in foot pounds of a weight of 450 pounds with a given velocity of 80 feet per second? Ans. 85. The energy exerted equals the work performed. The fol- lowing rule describes how the energy of a moving body at a given ve- locity can be determined. Rule: The weight of the body in pounds multiplied by the ve- locity squared, and this product divided by twice the rate of accelera- tion (32.2) in feet per second equals the work performed, or energy exerted. Stated in formula Wv 2 450 Ibs. X 80 2 450 X 6400 F = = = - = 44721 ft. Ibs. 2 g 64.4 64.4 Thus F = 44721 foot pounds. Energy exerted, in which F = force, or energy exerted. W = weight of the body = 450 Ibs. v = velocity in ft. per sec. = 80 ft. g=rate of accel. in ft. per sec. = 32.2 ft. By another rule we find that the measure of actual energy is the product of the weight of the moving body multiplied by the height from which it must fall to acquire its actual velocity. Let v = velocity in ft. per sec. and h = height, it must fall then according to the laws of falling bodies V 2 80' h = = = 99.378882 ft, or the height of the fall to acquire a 2g 64.4 vel. of 80 ft. per sec., then 450 Ibs. X 99.378882 ft. = 44721 foot pounds. The same result as obtained by the first rule, or formula. Q. 86. (1899-1900.) In raising a weight of 16,000 pounds by the aid of a screw jack, the screw %-inch pitch, or four thread to the inch 228 (barring friction) how many pounds pull must be applied to a bar 36 inches long from center of screw to center of pull? Approximately, what is the efficiency of a screw? Let W=weight in lbs.=16000.x=power required, then mean circum- ference = 72 inches. W:x ::ir72 : %. Substituting 16000 : x : : 3.1416X72 : %. 16000 : x : : 226.2 : %. Then, as the product of the extremes 16000 X % =4000, equal the product of the means (226.2) x, x = 4000 ^ 226.2 = 17.6 Ibs., the re- quired power. Or, as the weight to be raised is to the power required to raise it in one revolution of the screw, so is the circumference of the circle described by turning of the lever to pitch of the screw. In practice the screw is used as a combination of leverage with an inclined plane; a spiral inclined plane being formed by the threads of the screw. While the power applied to the lever which turns the screw moves around an entire circle, the body moves only the distance between the centers of two threads. The friction of the screw (which under heavy loads becomes very great) has also to be overcome by the power; and this fact makes these calculations of but little use. Q. 93. (1899-1900.) How many pounds of steam at 85 Ibs. pressure per square inch (absolute pressure) will be required to raise the temperature of 760 Ibs. of water from 58 F. to 164 F., the water and steam mingling freely together? Ans. 93. The mixture of steam and water at a temp, of 164 F. con- tains 132.404 B. T. U.; at 58 F. water contains 26.007 B. T. U. 164 F. 58 F. = 106 F., the difference in temperature of the water and the steam and water mixed. 132.404 B. T. U. 26.007 B. T. U. = 106.397 B. T. U., the number of B. T. U. in 106 F. temp. Therefore, the number of B. T. U. that each pound of water will absorb equals 106.397 B. T. U. 760 X 106.397 = 80861.72 B. T. U. equals the number required to raise 760 pounds of water from a temp, of 58 F. to 164 F. The steam mingling with the water will give up its latent heat of evaporation, which, at a pressure of 85 Ibs. absolute per sq. in., is 891.3 B. T. U. per pound. It will also give up a portion of its sensible heat in falling from a temperature due to 85 Ibs., which is 316 F. to a temperature 164 F. Steam at 316 F. contains 287.022 B. T. U. (sensible). Water at 164 F. contains 132.404 B. T. U. (sensible). 287.022 132.404 = 154.618 B. T. U. that the steam gives up in addition to its latent heat (891.3 B. T. U.), thus 891.3 + 154.618 = 1045.918 B. T. U. per pound of steam. The number of pounds of steam required is equal to the number of B. T. U. absorbed by the water, 80861.72 divided by the B. T. U. given up by one pound of steam, 1045.918. 80861.72 -f- 1045.918 = 77.31 pounds of steam required under condi- tions of question. Formula W (ts ti) 760(164 58) S = = 77.22 ans. L+(t 1 2 ) 891.3+ (316 164) In which S = steam required in pounds. W the water to be heated, 760 Ibs. tj = initial temp, of water = 58 F. t 2 = final temp, of water = 164 F. t temp, of the steam at given press. 316 F. L = latent heat of steam at given press. 891.3 B. T. U. 229 Q. 105. (1899-1900.) What is understood by the term "Mechanical Efficiency" of an engine? By the term "Thermal Efficiency" of an engine? What is the mechanical efficiency of an engine developing 500 I. H. P. and 425 brake H. P.? What is the thermal efficiency of an engine using steam at 130 Ibs. absolute pressure, exhausting at 6 Ibs. absolute pressure per sq. in.? Ans. 105. The "mechanical efficiency" of an engine is the ratio of the actual horse-power used in performing the work, to the indicated horse-power, or the mechanical energy, developed in the cylinder real- ized in useful work expressed in percentage. Rule. Multiply the actual horse-power by 100 and divide this prod- uct by the indicated horse-power and the quotient will be the mechani- cal efficiency, expressed in percentage. The "thermal efficiency" of an engine is the ratio of the heat ex- pended in the cylinder, is to the total heat entering the cylinder, or the total heat entering the cylinder that is used in performing work. 425 X 100 = 85 per cent mechanical efficiency. 500 130 Ibs. absolute = 347.1 F. 6 Ibs. absolute = 170.1 F. 347.1 + 460 = 807.1, total temp, of steam entering cylinder. 170.1 + 460 = 630.1, total temp, of steam exhausting from cylinder. 807.1 630.1 = 177.0 F. that the temp, of the steam is reduced to in performing the work. Then 177 X 100 =21.66 per cent thermal efficiency. 807.1 Formula T _ T i X 100 = per cent T. E. T Substituting 807.1 630.1 x 100 = 21.66 per cent of thermal efficiency. 807.1 T = absolute temp, of initial steam. T* = absolute temp, of exhaust steam. Q. 21. (1900-01.) What is the unit of heat and how is it expressed? Ans. 21. Unit of Heat. The British unit of heat, or British ther- mal unit (B. T. U.), is that quantity of heat which is required to raise the temperature of 1 pound of pure water 1 at or near 39.1 F., the temperature of maximum density of water. The French thermal unit, or calorie, is that quantity of heat which is required to raise the temperature of 1 kilogramme of pure water 1 Cent, at about 4 Cent., which is equivalent to 39.1 F. 1 French calorie = 3.968 B. T. U. 1 B. T. U. = .252 calorie. Q. 22. (1900-01.) What is the mechanical equivalent of heat? Why do we say heat has a mechanical equivalent? How many foot pounds are represented by each B. T. U., and what name is given to this equivalent? 230 Ans. 22. The Mechanical Equivalent of Heat. Is the number of foot pounds of mechanical energy, equivalent to 1 British thermal unit? Heat has a mechanical equivalent because they both are mutu- ally convertible. Joule's experiment (1843-50) gave the figure 772, which is known as Joule's equivalent. More recent experiments by Prof. Rowland gives higher figures, and the most probable average is now considered to be 778. That is, 1 B. T. U. = 778 foot pounds, or 778 foot pounds is considered as the mechanical equivalent for 1 B. T. U. Q. 23. (1900-01.) If one heat unit equals 778 foot pounds of energy, what decimal part of a heat unit does one foot pound represent? How many heat units per hour, per minute, per second, are required for one horse-power? One pound of carbon burned to C Oa (carbon dioxide or perfect combustion) equals 14,544 heat units; what is the efficiency of one pound of carbon per horse-power per hour? Ans. 23. If 1 heat unit equals 778 foot pounds of energy, 1 foot pound equals 1/778 = .0012852 heat units. 1 H. P. = 33000 foot pounds per minute. 1 H. P. = 2545 B. T. U. per hour. 1 H. P. = 42.416 + B. T. U. per minute. 1 H. P. = .70694 B. T. U. per second. 1 pound of carbon burned to C 2 = 14544 heat units. 1 pound of carbon per H. P. per hour = 2545 -f- 14544 = 17% per cent efficiency. Q. 25. (1900-01.) What is thermal capacity, and what is the co- efficient of thermal capacity? Ans. 25. Specific Heat. The thermal capacity of a body is the quantity of heat required to raise its temperature 1. The ratio of the heat required to raise the temperature of a given substance 1 to that required to raise the temperature of water 1 from the temperature of maximum density; 39.1 F. is commonly called the specific heat of the substance, or co-efficient of thermal capacity. Q. 26. (1900-01.) How is tne specific heat of a substance obtained or determined according to the method by mixture? Give formula for the same. Ans. 26. Determination of Specific Heat. Method by Mixture: The body whose specific heat is to be determined is raised to a known temperature, and then is immersed in a mass of liquid of which the weight, specific heat and temperature is known. When both the body and the liquid have attained the same temperature, this is carefully ascertained. Now the quantity of heat lost by the body is the same as the quan- tity of heat absorbed by the liquid. Let G = specific heat of the hot body; W = weight of the hot body; t = temperature of the hot body; C* = specific heat of the liquid; W l = weight of the liquid; t 1 = temperature of the liquid. T = temperature the mixture assumes. Then, by the definition of specific heat, CXWX (t T ) = heat units lost by the hot body, and 231 C 1 X W 1 X (T t 1 ) =heat units gained by the cold liquid. If there is no heat lost by radiation or conduction these must be equal, and OW 1 (T t 1 ) C W (t T) =C J W 1 (T t 1 ) or, C = W (t T) Q. 33. (1900-01.) How would you place a globe valve on a steam pipe with the pressure above or below the valve, and why? Would you place the valve so the stem is in a vertical or horizontal position? Ans. 33. A globe valve should be placed on a steam pipe with the pressure below the valve. There are opinions contrary to this, but it is believed that such opinions are entertained principally by cranks and those who like to be on tne off-side. It is advocated that the pressure being on top of the valve that it is not so liable to leak. This is erroneous. If the seat is cut a valve will leak regardless of the pressure. Again, if the valve becomes detached from the stem with the pres- sure on top, operations have to be suspended until the valve is replaced or repaired. For convenience of packing the stem, the pressure should be placed on the under side of the valve. It is proper to have the stems of the valves horizontal. This is also a matter of convenience for opening and closing and also avoids forming a water pocket. Valves so placed that the stems hang down beneath the pipe is not good practice, and should be severely condemned. Q. 35. (1900-01.) Describe the "steam loop," its uses and upon what does its action depend. Furnish sketch. Ans. 35. The "Steam Loop." It is a system of piping by which water of condensation in steam pipes is automatically returned to the boiler. In its simplest form it consists of three pipes, which are called the "riser," the "horizontal," and the "drop-leg." When the steam-loop is used for returning to the boiler the water of condensation and entrainment from the steam pipe through which the steam flows to the engine cylinder, the riser is generally attached to a separator; this riser empties at a suitable height into the hori- zontal, and from thence the water of condensation is led into the drop- leg, which is connected to the boiler, into which the water of conden- sation is fed as soon as the hydrostatic pressure in drop-leg in connec- tion with the steam pressure in the pipes is sufficient to overcome the boiler pressure. The action of the device depends on the following principles: Difference of pressure may be balanced by a water column; vapors or liquids tend to flow to the point of lowest pressure; rate of flow depends on difference of pressure and mass; decrease of static pressure in a steam pipe is proportional to rate of condensation. In a steam current water will be carried or swept along rapidly by friction. The water of condensation runs into a separator. The drip from the separator is below the boiler, and, evidently, were a pipe run direct- ly to the boiler, we would not expect the water to return up hill. Moreover, the pressure in the boiler is say, 100 pounds, while in the separator it is probably about 95 pounds, due to the drop of pressure in the steam pipe, by reason of which difference the steam flows to the engine. Thus the water must not only flow up hill to the boiler, but must overcome the difference in pressure. 232 The device to return it must perform work, and in so doing heat must be lost. The loop, therefore, may be considered as a peculiar motor doing work, the heat expended being radiation from the upper or horizontal portion. From the separator or drain leads the pipe called the "riser," which at a suitable height empties into the "horizontal." This leads to the "drop-leg," connecting to the boiler anywhere under the water line. The "riser," "horizontal" and "drop-leg" form the loop, and usually consist of pipes varying in size from %-inch to 2 inches, and are wholly free from valves, the loop being simply an open connection from sepa- rator to boiler. (For convenience stop and check valves may be used, but they take no part in the loop's action.) Suppose steam is passing, engine running and separator collecting water. The pressure of 95 Ibs. at the separator extends back through the loop, but in the drop-leg meets a column of water which has risen from the boiler when the pressure is 100 Ibs., to a height of about 10 feet; that is, to the hydrostatic head equivalent to the 5 Ibs. difference in pressure. Thus the system is placed in equilibrium. Now the steam in the horizontal condenses slightly, lowering the pressure to 94 Ibs., and the column in the drop-leg rises 6 inches to balance it; but meanwhile the riser contains a column of mixed vapor, spray and water, which also tends to rise to supply the horizontal as its steam condenses, and being lighter than the liquid water of the drop-leg, it rises much faster. If the contents of the riser have a specific gravity of only .1 that of the water in the drop-leg, the rise will be ten times as rapid; and when the drop-leg column rises 1 foot the riser column will lift 10 feet. By this process the riser will empty its contents into the horizontal, whence it is a free run to the drop-leg and thence to the boiler. In brief, the above may be summed into the statement that a de- crease of pressure in the horizontal produces similar effects on con- tents of riser and drop-leg, but in degree inversely proportional to their densities. When the condensation in horizontal is maintained at a constant rate sufficient to give the necessary difference of pressure, the drop-leg column reaches a height corresponding to the constant differ- ence and rises no higher. Thus the loop is in full action, and will main- tain circulation so long as steam is on the system, and the difference of pressure and quantities of water are within the range for which the loop is constructed. No water should accumulate in the separator, as it is the mission ot the loop to remove it before it assembles into a liquid mass. 233 It is here that constant and vigorous action is of great practical utility, enabling the loop to act as a preventive rather than a device for removing water after it has accumulated. The separator evidently must be of such form as to give the sweep toward and through the loop better opportunity to pick up the entrained water than is afforded by the current sweeping toward the engine, pump or steam using device. The loop action is practically independent of the distance the source of supply is above or below the boiler and also independent of the length of return. It is capable of handling such quantities of water as usually exist in steam systems. It is practically limited by excessive differences in pressures, and abnormal quantities of water. 234 The National Association of Stationary Engineers ORGANIZED OCTOBER, 1882. INCORPORATED OCTOBER, 1892 Three hundred and eighty subordinate Associations with sixteen thousand members in forty-eight States and Territories PREAMBLE. This Association shall at no time be used for the furtherance of strikes, or for the purpose of interfering in any way between its mem- bers and their employers in regard to wages; recognizing the identity of interests between employer and employe, and not countenancing any project or enterprise that will interfere with perfect harmony between them. Neither shall it be used for political or religious purposes. Its meetings shall be devoted to the business of the Association, and at all times preference shall be given to the education of engineers, and to securing of the enactment of engineers' license laws in order to prevent the destruction of life and property in the generation and transmission oi steam as a motive power. FORMER PRESIDENTS AND YEARS OF SERVICE. 1882. H. D. Cozens Providence, R. I. 1883. James G. Beckerleg Chicago, 111. 1884. James G. Beckerleg Chicago, 111. 1885. R. J. Kilpatrick St. Louis, Mo. 1886. F. A. Foster Bridgeport, Conn. 1887. G. M. Barker Boston, Mass. 1888. R. O. Smith New York City 1889. John Fehrenbatch Cincinnati, O. 1890. J. J. Illingworth Utica, N. Y. 1891. William Powell Cleveland, O. 1892. C. W. Naylor Chicago, 111. 1893. James D. Lynch Philadelphia, Pa. 1894. M. D. Nagle New York City 1895. Charles H. Garlick Pittsburg, Pa. 1896. J. W. Lane Providence, R. I. 1897. C. A. Collett St. Louis, Mo. 1898. W. T. Wheeler New York City 1899. Herbert E. Stone Cambridge, Mass. 1900. P. E. Leahy New York City 1901. E. G. Jacques Detroit, Mich. 235 INDEX PAGE Acceleration . . . . . . . 226-227 Adhesion . . . . . . . . 211 Adiabatic Expansion . . . . . . .199 Affinity . . . . . . . . 210 Air Compressor Temperature . . . . . 135 Air Pump Capacity ....... 68 Alternating Current . . . . . . .183 Alternation . . " . . . . 183 Ampere . . . . ... . . 145-151-166 Angular Advance . . . . . . 66-128 Angularity of Connecting Rod . . . . . . 67 Armature Energy . . . . . . 189 Losses in ....... 189 Reaction . . . . . . 189 Atom . . . . . . . . . 208 Automatic Cut-off . . '. . . . 120 Pabbitt Bearing . . '.- ' . . . . . 68 Band Wheel Size and Weight . . . . . 89 Barometer . .- . . ; . . . . 216 Bearings Brass or Babbitt . -. . . . . 68 Belts Improper Running ....... 70 Capacity of . ' . .'"".. . . . 98 Size . . . . . . .98 Pull on . ... . . . . 98 Width How to Join - >; -'-'/' . . . . 90 Bo lers and Engine Efficiency . . . . '-'':" 65 Blow-off . . . .':'-. . 4 1-42 Braces, Strains in . :. . . . . 5-35 Bumped Heads Pressure - . . . .54 Butt Straps . . '. ' . . , . 53 Compound, Universal . . . . . .62 Corrosion, External ...... 5 Dimensions of . . . . . .38 Dome ........ 41 Factor of Safety .... 55 Grate and Heating Surface Ratio .... 6 Head of , Sketch . . . . . . 10-11 Heating Surface Efficiency ..... 10 Horse Power . . . . . . .9-12 Horse Power Ratios . ' . . . . . 56 Horizontal, Support for . . . . . .44 Lecture on . . . . . . ... . . 14 Metals Used in . . , : /. 8 Mysterious Gas in . . . . . 6 Number for a Plant . . . . . . 40 Patch on Fire Sheet . . . . . . 10 Pitch in Setting ....... 7 Pitting ....... 6 Plate Formulas ...... 49-52 " Tensile Strength . . . . . 14 " Thickness ....... 13 Riveted Joints Value .... 6 Room Reports . . . . . .48 Safety Valves . . . . . . 7 Scale Preventer . . . . . . -6-7 Segment Area of . . . . . -33-197 Steam per Hour ..... 3 2 Strains in ....... 5-9 236 Boilers Surface Exposed to Heat " Supports, Size of Tests of Test, Report of a Tubes as Stays Tubes, Iron or Steel . Tubular Construction " Setting ' ' Specifications . Water Tube Advantages Where to Feed Working Pressure Books Referred to . B. T. U. . . . Brushes, Shifting of Buckeye Engines . . Palorimeter Purpose Use of " Carbonic Acid Gas Absorbed Carbon Monoxide, Percentage of Center of Gravity Centrifugal Force Chimney Area . . " Capacity Draft Apparatus Specific Heat Temperature Weight of . Waste in ' ' Guys for " Height of " Size of . . " Steel or Iron 1 ' Temperature Circuit Drop in . " Breaker " " versus Fuse Clearance in Engine Coal vs. Oil Fuel . Coal Amount Required " Combustion of " Evaporation from . . " and Combustible Combustible and Coal . Combustion Flame from ' ' Secondary of Coal . . Steam Jet . ' Co-efficient of Elasticity Cohesion . ' . Coil Winding . Compass Use of Compressed Air . . Compression . . Commuting Commutator Condenser Air Pumps Calculation Types of " Water Required PAGE 44 36 15-58 60 52 54 5 57 56 . 6-38-48 9 55 36 202-217-230 . 186 69 . 37-203 30 3i . 221 204 45 45 43-44-46 29 30 29 29-30 . 20-29-30 3i 13 44-47 43 43 10-23 147-148 153 153 . 70-117 10 37 i 7 f 3 64 . 64 8-27-28 13 207 210 186 . 163 24 67 184 . 184 141 . 140 140 . 142 237 ,A PAGE Conductors in Series ....... 160 Connecting Rod Shape of . . . . . . ' 84 Size of ..... 84-85 Strain on ..... 82 Corliss Engine Speed . . . . 75 " Valves ... . . . .67 Corrosion . . . '; . . . - , 61 Crank-Pin Pressure ... . . . . 79 Crank Shaft How to Line-up . . . . 92 Pressure . , . . . . . . 101 Crank and Piston Inertia . ' - '.''" '.-.'.'"'. 81 Cross-Head and Piston Inertia . . . . . . 81 Critical Temperature . . ". ; . \ . 212 Current Direction . '. . . . . .188 " High Tension . . . ... 146 " Potential ". . . . . . . 146 Cut-Off . . . . . . . 120 " Equal or Not . . . . . 71 Cycle Alternating ... . . . . 184 Cylinder Condensation . . . ., . -c 79-84 " Diameter . . . . . ... 71 " Low Pressure Size of . . . .79 F)raft in Chimney . .- . . : . 43-44-46 Dry Pipe . " . . . . . . . . 4I Dynamo . _ 148-187-190-191-192-194-195 Bi-polar -Multi-polar . . - . . . 153 " Brushes . . . . . . . 153 Compound . . . . - . ' . . 193 Capacity . . . . ' . y . .152 Efficiency . . .- . " . ' . . 195 " How Connected . . . . ' ." - . . 153 Shunt or Compound . . . ... 152 " Shunt ; . . ; . . . . 191 ' ' Series . . . - . . . ' ' . 1 92 vSparking . . . . . : ' . . 153 Horse Power of .- . . . ." '. 152 parth a Magnet . . , . . . . . 162 Eccentric -Turning Down ...... 66 Angular Advance . . . . . .66 119 Efficiency of Heating Surface ' . . . . . . i O Elastic Limit . . . . . . . . 44 Elasticity . . . . . .-..,. 210 E. M. F. To What Proportioned . . . . . 183 Electric Connection, Rule . . . . . . , 154 " Conductors, Danger . . ... .184 . . . . .- . . 186 Current, To Produce . . : . . ... . 187 " Developed . . . . . . 159 Strength . . ' . . . . 178 Quantity of . . . . . 179 " Generated . . . . . . 180 " Direction ...... 182-183 Circuit Open or Closed . . . . . .160 " Non- Conductors ...... 156 " Series . . . . . . . 156 Electrical Induction . . . . . . 181-182 Work Units . . . . . . .179 Horse Power 144 Instruments ....... 154 Energy ..... 195 Unit 151 238 Electrical Knowledge Action . Electricity Current of " Static . " Positive and Negative " Definition Electro Static " Dynamic V " Saturation " Magnet Electro-Motive Series . Electrified Body Grounded Elevator Repairs Engine and Boiler Efficiency Engine Clearance " Compound, H. P. of " Cylinder Size . " " Advantage " Connecting Rods Corliss . " Capacity to Increase " Cranks " Cross Compound " Cross Heads " Cylinder Wear 4< " Drips " Ratio . " " Economy of * . * ' Eccentric . . " Four- Valve. Economy . High Speed -When Use " Governor . " " Corliss * ' Horizontal or Vertical . Increased Load ' ' Knocking In ; Multi-Cylinder Most Economical " Mechanical Efficiency- New, How to Start " Receiver Purpose . ' ' Pressure " Rotary " Single-Acting " Size Required " Set Up Cost " Thermal Efficiency ". " Underloaded Engineers, Famous . :"._ . Evaporation , Efficiency of Equivalent ; From Coal Good Temperature of Suspicious Extension . . . ,- pactor of Safety Falling Bodies Laws Feed Water Where Enter Field Winding Defects in Firing Hand 9.V PAGE 155 155 159 i 56- i 59 '55 155 155 158 ' 'I 70-117 106 107 102-108 103 . 118 94-105 118 117 79-94 H9 I2 9 15-129 126 128 93 8o. 130 106 117 230 92 104 105 116 129 87-93 93 230 93 199 10 37 ' 37-6i . 209 55 221-222-224 228 239 PAGE Fly Wheel Centrifugal Force . 96 Functions of . . . . . 84- 1 20 Safe Speed . . . . . . 80 Safe Weights . . . . .120 Construction . . . . . . 120 vs. Band Wheel . . . . . . . 89 Pit . . . . . . . . 91 Flue Area . . . . . . . . . 55 " Gas Analysis . " . . . . . - ; . 17 Foot Pounds and H. P. . . . - ', . . . . 215 Force Definition . . . . . . . 211 Forced Draft . . . . . . . . 43 Foundations For Engine . . . . . . 90-92 Material . . . . . . -91 Template . . . . -,. . . 91 Bolts for . . . . , - - . . 91 Capstone Material . . . * 9 1 Wedges for .... . . -91 Frequency . . . .- ..... 183 Fuels Table of . . . - . . .28 Fuses Diameter . . . . - . . 149-150 " vs. Circuit Breakers . . . . . ; . 153 Fusible Plugs . , . . I .*.':. 42 Furnace Formula . . . . . ,."-'.. . 49 galvanometer . . ... . 168 Gauge Pressure . . . . . -. . 217 Grate Surface Ratio .. . . ' . . . 6-40 " Distance of . . . . 44 Ratio to Chimney . . .. . . . 45-46 " Bar Opening . . . . . . . . 55 Governor Pulley Size of . . . . ..\ . 73-128 Guides Pressure on . ; . . -. . . 82-99-101 Gravity . . . . .' . . 210-220-221 ~LJ eat Utilized in Furnace . . . . .' . 29 " Used in Making Steam . . . . . 29-31 " Definition . . . . . ' . . . 199 " Mechanical Equivalent of . . . '' -. . 230 " Units per H P. . . . . . . 231 Heaters, Open and Closed . . . . . . 138 " Saving by Use . . . . . . . 139 Heating Surface, Efficiency . . . . . 10 Ratio . . . . . . 6-40 Helix . . . . . . . . . 164-165 Horse Power of Boiler . . . . . . 9-12 " " and Cylinder Diameter ... . . 71 " " to Raise Water . . ... 133 " " and Foot Pounds . . .. ... 215 Hysteresis . . ... . . . . 189 Jce Making . . .. .... 137 " Mixture ........ 201 " Amount in a Room ...... 133 Impenetrability ....... 209 Incandescent Lamps, Current Used .... 143. Incrustation . . . . . . . . 61 Indicator Card Sketch . . . . ' ,. . 94 " M. E.P. . . . . . . 112 " Card, Water from . . . . . . 112 " Diagram ....... 109-110 " How Used ...... 108 " Theoretical Curve . . . . . . in 240 PAGE Induced Draft . 43 Inertia of Engine Parts . . . . . . 204 Injector, Economy of . . .- . r . 134'! Size of Pipe . . . .138' " Source of Power . . . '. . . 139 Injection, Water Required . ... 134 Iron, Wrought vs Cast . . ... .'..... 208 Strength .. ...."'. . 208 Isothermal Expansion . . . 200 Tack Shaft . . . . 89 J " " Diameter . '. . . . . 119 Joule . . . .- . . . . 178 Kilo-Watt . 144 patent Heat of Steam . . . . . 200-201 Lecture, Boilers and Furnaces . . . . 14 " Foundation ... . . . .86 Lifting Water ... . . . . 136-137 Lines of Force . .. . . . 162-164-186-190 Liners in Bearings . . . . . .. . 66 Link Motion, Stephenson . . . . . - . 125 Lode Stone . . . . . . . 161 Lubricants, Oils vs. Grease . '. . ' . . . 202 A/Tagnet . .... . . . . 161 Electro . . . . . . .166 Magnetic Circuit . . . . . . . 148 Substance . . . . . . 162 Field . . . . . . 162-163-188 Polarity . . . . . . . 163 Induction . . . . . . 163 Density .... . . , . 163 Magnetism . . . . . . 160-163 Permeability . . ... 149-165 Residual . . . . . . 191 Mathematics ... ; . . . . . 212-217 Matter . . . ' . . . 208 " Indestructibility ...... 209 " Attributes of ...... 210 Three States of . . . . . . 211 Metals, Thermal Conductivity , . . . . . 199 Microhm . . . . . . . . 171 Modulus of Elasticity . . . . . . 207 Molecules ' . . .' . ' . . . . 208 Momentum . . '. . ' . . 198 Motion . . . . .... . 2ii Laws of ... . .. . . . 214 " Ratio of . . . .. . . . 221 Motor, Winding for ... ... 154 " Circuits . . . . . . 154 Starting Box ' ^ J 54 " Cut-out . . . . '. . . . . 154 'NJ A. S. E. Educational Committee . - 3 * ' Officers, Past ... . . . . . 235 Qhm ..... . 144 151-166-169 Ohm's Law . . . . . . ,. 133-145-167 Oil vs. Coal Fuel . . . -. . . . 64 ' ' Viscosity of . ^ . . . ,- 203 " Flashing Point . . . . . ., 205 Oxygen, Absorbed . . ... . '. . 30 241 PAGE parallelogram of Forces . ... 205 Patch on Fire Sheet . . . . . . 10 Permeability, Magnetism ... 149 Pipe, Expansion of . . . . . . 198 Pitting, How Caused ... . 6 Piston, Position and Velocity . . . . . 81 " and Cross Head Inertia . . 81 and Crank Positions ...... 83 Speed, High . .115 Porter Allen Engine . . . . . . . . 73 Potential, Definition . . . . ... 157 Difference of . . .... 146-185 " Zero . . . . . . . . 157 Drop in . . 175 Pulsating Current . ... 184 Pumps, Air, Size of ..... . . . . 142 " Duty of ..... ... . 141 Capacity .... 138-139 Circulating ... . . 141 Cylinder Ratio . . . . 138 Forces Opposing .... 138 Force Pumps, Lever . 135 Friction . . ' . . . . . 139 Head Against . . . . . . 135 H. P. Required . ... 139 Lifting Water . ... 136-137 Dimensions . . . , . . . 137 Most Economical . . . . . . 134 Piston Speed . . . . .-,., . 137 to Set Valves . . ... . . 133 Slippage . . .139 Steam, Cylinder Diameter . . 133 " Efficiency of . . ... . 133 Successful Operation . 138 Suction Pipe ; ;.-=...' . 138 Water Hammer in . . . ; . ; . 134 . Radiator Size required . . . . . . . 203 Refrigeration Calculation . . . . . 136 Capacity ... . . . . 137 Rivet Strength of -55 Diameter of . . . ,. . . 35~53 Pitch of . . . .' . 53 Riveted Joint in Boilers . ..... 6 Pitch of . . . ... 8 How Fail . . . . ... -35 Rod Horizontal Strength of . . . 207 Rope Transmission . ' . . . . . . . , . 79 Cafety Valves . . . . . . . . 7- 1 3-42 Scale Preventer .'... . ' . . 6 " in Boilers . . . . . . 12 " Loss Due to . . ; n . . -31 " When Deposited . . . . . 200 Screw Raising Weight . . . . . . .228 Segment Area of ....... 33~'97 Shaft Torsional Strength . . . r . . 205 Torque . . . ... . .206 Shunt Circuit . . 147-160 Smoke Burning ....... 8 Cause of ........ 63 Solenoid ........ 165 Sparking . . 153 242 Speed of Engine . . , '-... . . 128 Spirit Level How Tested . ... 92 Specific Heat . . , ,-.'". 202-231 " " of Chimney Gas . . . . .29 Specific Gravity . . ^ . . . . 223 Starting Box '. . * . .154 Stays Load on . . . . . . . 49 Tubes as . , . - . . . . . 52 Slide Valve Cut-off . . . . ... 70 Steam Consumption Indicated . . . . . . 71 Distribution . . . .... 1 28 Drums . ... 41 Expansion in Cylinders . . . . . 103-104 Effective Pressure . . . . , . . 88 Heat Used to Make . , . 31 Heat Sensible . 225-226 Jacket . . ... . . . 68-91 Jets Combustion . . , .... 13 Liquified by Pressure . . . , , , 202 Conductor of Heat . , . . 226 Physical Condition . . . . . . 226 To Raise Temperature of Water . . ... 229 Loop . . . . . .... 232 Weight of . . . .' . . ' . . .198 Required per H. P. . . ... . , . 80-88 Jacket Results . . . . . . . -95 Used Expansively . . . . . . . 95 Pipe Specification . .88 " for Engine . . . . . 113 " Drips from . . . . .114 Separator - Purpose . . . . . . 113 Where Put . . . . , . .114 Useful Work from . . . . . ."' 33 Stokers Mechanical . . . . . - . . 44 Steel Not an Element ... . . . 208 Strain . . . . . ... 211 ' ' by Change in Temperature . . . 207 Sublimation . . . . , . . 200 Surface Blow-off . . . . . . . 42 ""Temperature Chimney Gases . . . . . . 10 of Evaporation . . . ... 37 Tensile Strength Boiler Plate . / . , '' . 14 Ultimate . . . . . , v 44 Thermal Conductivity of Metal . . . \ . . . 199 " Capacity ',... . 231 Theoretical Curve . . . . . . " . ... iio-in Transformer Construction . .... , ' 146 Transforming Direct Current . . . . . 146 Terminal Pressure . ... ". . . . . 71 Torque ....... . 188 Tubes Expanding vs. Beading . . . . 52 Tube Area ."''.. . . . -55 |J nits of Area . '. ^ . . . . .. . 215 Circular Measure , . .. . . .216 Length . v . . - . . . 214 Liquid Measure . . . . . . . 215 Heat . . . ;. . . . 230 Temperature . . . . . .216 Time . . . . . . . . 215 Volume ........ 215 Weight . . . . . - . . .214 Uptake Temperature . . . . . . .8 243 PAGE Vacuum What Is It? , . . . . % . . . 209 Valve Diagram Zeuner . . ' . . , . 72-75-120 " Bilgram . . . , . . 76-77 " Sweet . . . . . 77~7& " Setting Corliss . -. .'-''. . . 67 " Buckeye , . . . ... 69 " " Porter Allen . . . , 73 Valves Kind to Use . . . ,. ... 113-232 Valve Gear . . . . . . . 120 " " Functions . . , . . ... , . 120 " Travel . . . . . ' .66 70-75 ' Angle of Advance . . . . , . .121 " Admission . . . . , ... 121 " Balanced . . . . . - . . . 121 " Compression . . . . , . . 121 " Cut-off , . -. . . . 121-125 Exhaust . . . . . ', . . 121 Lap . . 121-124 Lead . ... 121-124 Gridiron . . . . . .121 Piston . . . ' . 124 " Poppet ..... .125 Relief . . .... . . .121 " Rotary . ... .-'-.-. . . . 121 " Slide . . . . . . . .121 " Travel . . . . .-. .121 Velocity . . . . ....... 211 Viscosity of Oil . . . ' . . . . 203 Volt . . .... 144-151-167-173 Voltage Most Suitable , . . . . . .151 Voltaic Battery Grouping . . . . . . . 159 " Cell 174 Couple . 158 Yyater Tube Boilers Advantages , . . ' . , 6-38-48 Columns Connections ' . . . . . 42 ' Velocity of . ... .. . . . ic,7 Evaporated, How . . . ; ... 198 Column Pressure . . . . - . .. 198-199 ' Changed to Steam . . . . , . ... 198 Watt Value of . . . . . . . 145-151 Weight Units of . . ... . 214 Wheatstone Bridge . . .'..... . . . 172 Wire Capacity of . . .- . . . . 147-148 Wires Resistances of . . 144-169-170-171-172-173-174-175-176 Work Definition ....... 199 " Foot Lbs. and Velocity . '. . . . . 203 ^ero Temperature Absolute . . " . 201 244 THE LIBRARY UNIVERSITY OF CALIFORNIA Santa Barbara THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW. 000630889 4