<^SlAT10MARr 
 
 
 ENGINEERS 
 
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 National Association of Stationary Engineers. parr\^ 
 
 Questions and Answers 
 
 1897-1898 
 
 Educational Competition. 
 
 Question 1. How much heat is pro- 
 duced by the complete combustion ol’ 
 — (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. 
 
 Ans. 1 
 
 KIND OF COAL. 
 
 PENNSYI.VANIA anthracite. 
 
 Beaver Meadow 
 
 Peach M. Connellsville. . . 
 
 Lehigh 
 
 Pittsburgh, Average 
 
 Lackawanna 
 
 Pennsylvania Buckwheat. 
 Honeybrook Lehigh ...... 
 
 Massachusetts Anthracite , 
 
 BITUMINOUS, SEMI-BITUMINOUS, ETC. 
 
 New River 
 
 Indiana Block 
 
 Kentucky Coking 
 
 Pittsburg Coking 
 
 Youghiogheny 
 
 Pocahontas, run of mine 
 
 Cumberland . . . 
 
 Pennsylvania, semi -bituminous 
 
 Rock Spring, Wyoming 
 
 Bureau Co., Til 
 
 Hocking, slack 
 
 Jackson (nut and slack) 
 
 Bellmore Pea 
 
 Big Muddy, Jackson Co., Ill 
 
 Bituminous— 1896 tests— heat value for 
 ten mines scattered over Eastern and 
 Southern Ohio, (data furnished by 
 
 Akron, Ohio, No. 28) 
 
 Wood (assumed as .4 of coal used) 
 
 Lignite 
 
 Crude Petroleum 
 
 Heat 
 
 Value 
 
 12,000 
 
 to 
 
 14,900 
 
 Average 
 
 13,916 
 
 14,098 
 
 13,551 
 
 13,800 
 
 13,104 
 
 14,500 
 
 12,200 
 
 15,200 
 
 14,000 
 
 14.000 
 14,400 
 14,400 
 14,265 
 14,289^ 
 13,614 
 13,368 
 
 13.000 
 13,025 
 11,083* 
 12,139* 
 12,240* 
 12,600 
 
 13,100 
 
 10,300 
 
 19,200 
 
 *1897 
 
 (See Explanation, page 21.) 
 
 
 Theoretical Evap- 
 
 O (It 0 
 
 T3 
 
 u- ^ 
 
 0 f-i 1/5 
 
 oration 
 
 
 ^ p 
 
 from and at 212° 
 
 2 0^ 
 0 
 
 C J- 
 
 
 
 
 o 
 
 m 
 
 S0% 
 
 > >.h 
 
 W.O' 
 
 ■V tJ u 
 A 
 
 7.45 
 
 9.93 
 
 
 
 9.245 
 
 12 32 
 
 10.07 
 
 70 
 
 8.63 
 
 11.51 
 
 
 
 8.74 
 
 11.52 
 
 
 
 8.4 
 
 11.21 
 
 
 
 8.56 
 
 11.42 
 
 11.17 
 
 78 
 
 8.13 
 
 10.84 
 
 
 
 9. 
 
 12. 
 
 11.13 
 
 74 
 
 7.57 
 
 10.09 
 
 
 
 
 
 10.75 
 
 
 9.44 
 
 12.59 
 
 
 
 
 
 
 
 8.7 
 
 11.6 
 
 
 
 8.7 
 
 11.6 
 
 
 
 8.94 
 
 11 93 
 
 
 
 8.94 
 
 11.93 
 
 
 
 8.87 
 
 11.82 
 
 
 
 8.87 
 
 11.82 
 
 11.53 
 
 78 
 
 8.46 
 
 11.27 
 
 8.74 
 
 62 
 
 8.30 
 
 11.07 
 
 
 
 8.07 
 
 10.76 
 
 
 
 8.07 
 
 10.76 
 
 
 
 6.88 
 
 9.18 
 
 7.78 
 
 68 
 
 7.54 
 
 10.05 
 
 8.25 
 
 66 
 
 7.60 
 
 10.13 
 
 8.98 
 
 71 
 
 7.83 
 
 10.43 
 
 1 
 
 
 8.14 
 
 10.85 
 
 
 
 3.48 
 
 4.68 
 
 
 
 6.40 
 
 8.50 
 
 
 
 11.93 
 
 15.90 
 
 
 
Q. 2. How much heat will be pro- 
 duced by the incomplete combustion of 
 one pound of coal of the kind assumed 
 in 9. V (a)? 
 
 Ans. 2. Assuming 90 % carbon, the 
 heat produced will be 4,400X.90 = 3,- 
 960. 
 
 Q. 3. What are the products of 
 combusition under the conditions of Q. 
 1 (a) and what are their approximate 
 proportions? 
 
 Ans. 3. Carbonic acid gas (CO.), al- 
 so called carbon dioxide, a colorless 
 gas consisting of 3/11 carbon and 8/11 
 oxygen by weight; formed by the un- 
 ion 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 (HoO) 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. 
 
 Q. 4. What are the products of 
 combustion under the conditions 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. What gas in large proportions 
 in the chimney gases indicates that 
 an insufficient quantity of air is be- 
 ing supplied to thie whole or a part of 
 the fuel? 
 
 Ans. 5. Carbon monoxide (CO). 
 
 Q. 6. 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 nat- 
 ural draft? 
 
 Ans. 6. Oxygen (O,). About 10 % 
 by volume. 
 
 Q. 7. What would be the weight of 
 chimney gases per pound of coal 
 burnt, in good practice, with natural 
 draft? 
 
 Ans. 7. About 25 lbs. 
 
 Q. 8. What is the approximate spe- 
 cific heat of chimney gases? What do 
 you mean by this answer? 
 
 Ans. 8. Twenty- four one h u n- 
 dredths (.24). Each pound of the chim- 
 ney gases raised 1° F. in temperature, 
 absorbs twenty-four one hundredths 
 (.24) of a unit of heat. 
 
 Q. 9. 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, prop- 
 
 erly 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 answers to 
 last year’s questions). 
 
 (3) By the melting point of alloys 
 or metals exposed to the gases. 
 
 Q. 10. What proportion of the heat 
 produced by the burning of the coal 
 should go to the formation of steam, 
 in good practice and with a good boil- 
 er plant? 
 
 Ans. 10. From 60 % to 80 % (.60 to 
 .80). 
 
 Q. 11. How can the amount of heat 
 produced by the furnace be approxi- 
 mately estimated and how can the pro- 
 portion of said heat that goes to the 
 production of steam be approximately 
 estimated? 
 
 Ans. 11. By multiplying the heat 
 value of the coal used by the number 
 of pounds of the same burned. 
 
 By multiplying the number of pounds 
 of water evaporated by the heat re- 
 quired to evaporate one pound. The 
 evaporation is usually reduced to an 
 equivalent evaporation from and at 
 212°, and then multiplied by 966, the 
 number of heat units required to 
 change one pound of water at 212° to 
 one pound of steam at the same tem- 
 perature. — See answers to last year’s 
 questions. 
 
 Q. 12. Wihat would be the weight of 
 the chimney gases per pound of coal 
 burnt, in good practice, with a forced 
 draft? 
 
 Ans. 12. About 19 lbs. 
 
 Q. 13. What substance is used — 
 
 (a) For the absorption of carbonic 
 
 2 
 
acid gas (COJ from a mixture of 
 gases; 
 
 (b) For the absorption of free oxy- 
 gen (O) froim a mixture of gases? 
 
 Ans. 13. (a) A solution of caustic 
 
 potash (commercial) in water from the^ 
 hydrant. Proportions about 1 lb. of 
 potash to 2 lbs. of water. Caustic soda 
 may also be used. 
 
 (b) A solution of pyrogallic acid 
 (pyrogallol) in water mixed with the 
 above solution of caustic potash. Prof. 
 Thurston used 5 % pyrogallic acid. 
 Phosphorus may also be used. 
 
 Q. 14. Describe a cheap and simple 
 apparatus by which the proportions, 
 by volume, of carbonic acid gas (CO 2 ) 
 and of free oxygen (0) in the chimney 
 gases may be determined. Describe 
 method of use. 
 
 Ans. 14. Ga. No. 1 sends the follow- 
 ing description: 
 
 “Fill a graduated test-tube with 
 chimney gases. Close the mouth, in- 
 vert the tube and put the mouth un- 
 der water. Arrange the tube so that 
 the level of the water inside and out 
 is the same. Now introduce a piece ot 
 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 CO 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 described 
 and illustrated in a separate chapter. 
 
 Q. 15. Assuming that a sufficient 
 amount of air is supplied to the fuel 
 and that the percent by volume of 
 carbonic acid (CO 2 ) and of free oxygen 
 (0) in the chimney gases is known, 
 how can the weight of the chimney 
 gases per pound of coal be approxi- 
 mately determined? 
 
 Ans. 15. Add the percent by volume 
 of carbonic acid (CO 2 ) to the percent 
 by volume of oxygen (O 2 ) ; divide this 
 result by the percent, by volume, of 
 carbonic acid and multiply the quo- 
 tient by the constant number ten and 
 one-half (10.5). 
 
 (2d) Divide the constant number two 
 
 hundred and ten (210) by the number 
 indicating the percent by volume of 
 the carbonic acid (CO 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. Knowing the percent by vol- 
 ume of carbonic acid gas (CO 2 ) and of 
 free oxygen (O) in the chimney gases, 
 how can the percent by volume of 
 carbon monoxide (CO) be calculated, 
 neglecting water vapor and hydro- 
 carbons? 
 
 Ans. 16. Add the percent by vol- 
 ume, of carbonic acid (CO 2 ) and of oxy- 
 gen (O 2 ), multiply the result by five, 
 subtract the product from 100 and di- 
 vide the remainder by 3. 
 
 Q. 17. If coal, having a heat value 
 of 14,000 units per lb. is used, the feed- 
 water is at 100 degrees, the pressure is 
 70 lbs. gage, it is found that 30 
 lbs. of water is evaporated by 4 
 lbs. of coal burned: — What percent 
 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 14,000X4=56,000. 
 The factor of evaporation for feed- 
 water at 100° and pressure 70 lbs. is 
 1.149. Therefore the equivalent evap- 
 oration from and at 212° is 30X1.149 
 which is 34.47, 34.47X966=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. 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 lbs. of air was being supplied 
 per pound of coal burned, what percent 
 of the heat produced by the combustion 
 of the coal would be going out of the 
 chimney with the gases? 
 
 Ans. 18. If 40 lbs. of air was sup- 
 plied per pound of coal the weight of 
 the chimney gases would be about 41 
 lbs. per pound of coal; the heat car- 
 ried up the chimney by this would be 
 
 3 
 
about 41 X 450 X. 24=4, 428, and the per- 
 cent is 4,428-^14,000=31.6, about. 
 
 Q. 19. What is the relative approx- 
 imate permeability to heat of clean 
 boiler plate, %" thick, and the same 
 plate covered with Ys" of hard sul- 
 phate scale? 
 
 Ans. 19. From two to one (2 to 1), 
 to two to one and a half (2 to 1.5). 
 
 Q. 20. How is the condition of tue 
 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 gases. 
 
 The Engine — Its Consumption of Steam. 
 
 Q. 21. A horizontal boiler is 18 ft. 
 long, 72" diameter, pressure 70 lbs. 
 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" below the upper end 
 of the glass, the feed is then stopped. 
 At 9:45 a. m. the water is 5" from the 
 lower end of the glass. How many 
 pounds of steam is the boiler supply- 
 ing per hour? 
 
 Ans. 21. The water has fallen 3" or 
 one-quarter of a foot. The average 
 width of the surface while falling, mul- 
 tiplied by the length of the boiler, 18 
 ft., and by the distance through which 
 the level has fallen gives the volum’e 
 of the water used. This expressed in 
 cubic feet and multiplied by 56.8 the 
 weight of water at the temperature 
 corresponding to 70 lbs. gage (316°) 
 will give the weight of water evapor- 
 ated 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 sur- 
 face one might take the width at the 
 higher and at the lower levels, add the 
 two together and divide by 2. 
 
 The width of the surface (a d Fig. 
 1) at the higher level may be found 
 by subtracting the square of its hight 
 (b c) above the center of the boiler 
 from the square of the rauius (36") 
 
 extracting the square root of the re- 
 sult, and multiplying by 2. 
 
 Thus: — Radius equals 36", square of 
 radius equals 1296. Hight from center 
 of boiler 20", square of hight 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-f-63.46^2= 
 
 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 between the lines 
 a d and h e, which indicate the higher 
 and lower levels. Then carefully meas- 
 ure the line g f. With the scale sug- 
 gested, the line will be found to be 
 about 30 13/16 long, which multiplied 
 by two, 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 evaporaced 
 would be 5. 14 X 18 X. 25=23. 13 cubic feet, 
 and its weight is 23.13X56.8= about 
 1313 lbs. This multiplied by 4 and di- 
 vided by 3 gives about 1751 lbs. of 
 water per hour. 
 
 Q. 22. What is a calorimeter, how 
 constructed, and for what purpose 
 used? 
 
 4 
 
Ans. 22. The 'calorimeter is, as its 
 name implies, an apparatus for meas- 
 uring heat, in stauonaiy engineering 
 there are two kinds of calorimeters; 
 one for the purpose of ascertaining tJie 
 amount of moiature in steam, the other 
 for ascertaining the heat value of coal. 
 
 A type of the hrst kind is tiie barrel 
 calorimeter. This consists essentially 
 of a barrel of. water with a steam pipe 
 leading into it. The sieam to be tested 
 is allowed to run into and be con- 
 densed in the water. A comparison of 
 the increase in weight and in temper- 
 ature of the water will show the per- 
 centage of water carried in with the 
 steam. 
 
 The second form consists essentially 
 of one vessel placed within another. 
 The outer vessel contains water. A 
 small portion of tne coal to be tested 
 is placed in the inner vessel whicff is 
 supplied with oxygen gas. .The coal 
 is rapidly and completely burned in 
 this gas and the amount of heat gen- 
 erated by its combustion is measured 
 by the rise in temperature of the water. 
 
 Q. 23. What is the average weight 
 of steam per horse power per hour 
 consumed by^ 
 
 (a) Automatic cut-off single cylinder 
 non-condensing engines of about 100 
 HP. 
 
 (b) Double expansion condensing en- 
 gines of about 500 HP, in good prac- 
 tice? 
 
 Ans. 23. (a) About 25 to 30. 
 
 (b) About 12 to 15. 
 
 Q. 24. A single cylinder condensing 
 engine has a terminal pressure of one 
 atmosphere absolute; the work re- 
 quired of it is gradually increased; 
 after a time it is found that the ter- 
 minal pressure is three atmospheres 
 absolute — what alteration of the engine 
 would you recommend to better adapt 
 it to its work? 
 
 Ans. 24. Compound the engine by 
 adding a low pressure cylinder. 
 
 Q. 25. The gage pressure being 120 
 pounds, the feed water 110° and the en- 
 gine using 15 lbs. of steam per horse- 
 power-ihour, what part of the heat that 
 went to the making of steam is trans- 
 formed into useful work by the en- 
 gine? What percent is this of tne heat 
 generated in the furnace, the efl3.ciency 
 
 of the furnace and boiler being .70? 
 
 Ans. 25. (a) The total heat in steam 
 
 above 32° due to 120 lbs. 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 lb. water at 
 110° into steam at 120 lbs, gage pres- 
 sure is 1188—78=1110 H.U. The en- 
 gine in question uses 15 lbs. of steam 
 per horsepower which is equal to 15 X 
 1110=16,650 H.U. As 1,980,000 is equal 
 to one horsepower per hour, and 1 
 H.U. equals 778 ft. lbs. of work, 
 1,980,000 
 
 =2,545 
 
 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 engine is 
 
 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 lbs. of steam, the 
 heat of the furnace would have to pro- 
 duce for each 15 lbs. of steam 
 
 16,650 
 
 =23,785.7 H.U. 
 
 .70 
 
 and the per 'cent of this heat that went 
 to do useful work is 
 2545 
 
 X 100=10.7. 
 
 23,785.7 
 
 Q. 26. What is meant by the “vis- 
 cosity” of oil? How can the relative 
 viscosities of two specimens be con- 
 veniently determined and to what ex- 
 tent is it a criterion of the value of 
 the oil? 
 
 Ans. 26. Viscosity is the friction of 
 the particles upon each other. The 
 relative viscosity of two specimens of 
 oil may be approximately determined 
 by filling a small vessel with the oils, 
 one after the other, allowing them to 
 run out through small orifices and not- 
 ing the time required to discharge 
 equal quantities. The specimen taking 
 the longer time to run out is propor- 
 tionately more viscous. 
 
 The viscosity is often taken as an 
 index of the value of the oil for heavy 
 work. 
 
 Q. 27. What is the rule for finding 
 the number of foot-pounds of work 
 necessary to change the velocity of a 
 
 5 
 
body of known weight from one value 
 to another? 
 
 Ans. 27. Subtract the square of 
 the lesser velocity from the square of 
 the greater velocity, and multiply the 
 remainder by the weight. Divide that 
 product by 64.4 and the result will be 
 the required number of foot-pounds. 
 Velocities are taken in feet per sec- 
 ond, and weight in pounds. 
 
 This is expressed algebraically as 
 follows: 
 
 W(V^— 
 
 E= 
 
 64.4 
 
 in which W=weight in pounds, Vi the 
 less and V the greater velocity in feet 
 per second, E=:energy in foot pounds 
 (work). 
 
 Q. 28. An engine has a twelve (12") 
 inch stroke. Its connecting rod is three 
 (3) feet long. It is running at six hun- 
 dred (600) revolutions per minute. The 
 weight of its piston, piston-rod, and 
 cross-head is 250 lbs. 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 
 
 pin, and the vertical lines at other 
 points of the stroke velocities propor- 
 tional to their lengths. 
 
 Q. 29. 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, tne scale is 
 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 veloc- 
 ity, therefore, at this point is found by 
 the proportion 1.5 : .992 : : 31.4 : an- 
 swer. 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 com- 
 mencement 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. With the engine of Q. 28 
 what is the approximate average force 
 
 stroke of the piston. Draw upon this 
 line a semi-circle A, B, C. Then will 
 points upon the horizontal line rep- 
 resent piston positions and the verti- 
 cal distances above said points to the 
 semi-circle will represent piston veloci- 
 ties at the corresponding positions of 
 the piston. 
 
 The radius B, D, of the semi-circle 
 represents the velocity of the crank- 
 
 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 calcu- 
 lation. 
 
 Ans. 30. The weight of the parts is 
 
250 lbs. The velocity at 1.5" is 20.77 
 ft. per second. The velocity at 2" is 
 23.4 ft. per second. Therefore, accord- 
 ing to the rule of Ans. 27, the work 
 done upon the parts to change their ve- 
 locity from one value to the other is 
 250X[(23.4)-— (20.77)=^]-^ by 64.4, tkat 
 is 250X (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 i/^" 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 lbs. as the average force due to 
 inertia. 
 
 The change of angularity of the 
 connecting rod may be taken into ac- 
 count by multiplying the result by one 
 plus the ratio between the crank and 
 connecting rod, that is, in this in- 
 stance, by one and one-sixth. If The 
 calculation had been made on the lat- 
 ter 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. 31. What is the rule for obtain- 
 ing the centrifugal force of a given 
 weight, moving in a circle of a given 
 radius, with a given velocity? 
 
 Ans. 31. The formula for centrifu- 
 gal force is 
 
 MV^ , 
 
 32.2 R 
 
 in which R is expressed in feet and 
 fractions thereof. 
 
 Multiply the weight by the square of 
 the velocity and divide the product by 
 32.2 times the radius. The radius is 
 to be taken in feet, velocities in feet- 
 per-second, and weight in pounds. 
 
 Q. 32. For the engine of Q. 28 draw 
 a diagram to scale, by help of the rule 
 of Ans. 31, such that horizontal dis- 
 tances shall represent piston-positions, 
 and vertical distances the force exerted 
 by the inertia of the piston, piston-rod, 
 and cross-head, at each piston-position. 
 
 Ans. 32. Lay off as before a hori- 
 zontal line a, b. Fig. 3, to represent the 
 stroke. Imagine that the total weight 
 of the parts is upon the crank pin and 
 calculate what the centrifugal force 
 would be. Lay off a line a, c, propor- 
 
 tional to this force, vertically down- 
 ward from the end of the line a, b, 
 which represents the commencement 
 of the stroke. Lay off a line b, d, ver- 
 tically upward from the other end of 
 the line a, b, and connect the points c, 
 and d, by a straight line. Then will 
 points in the line a, b, represent piston 
 positions and the vertical distances 
 from said points to the line c, d, will 
 represent the force of inertia at the 
 corresponding points in the stroke. 
 
 Distances downward from the line a, b, 
 represent pulls on the connecting rod; 
 distances upward, pressures upon the 
 same. The above diagram neglects the 
 effect of the change in angularity of 
 the connecting rod. This may easily be 
 taken account of, however. If to the 
 line a, c, you add the proportion of its 
 length equal to the ratio of the crank, 
 to the connecting rod, in this case 1/6, 
 and subtract the same amount from 
 the line a, d, and then draw the regu- 
 lar curve e, f, g, through the points 
 thus found and through the center of 
 
 7 
 
the line a, b. Then vertical distances 
 to the line e, f, g, will represent, to the 
 scale chosen, the force of inertia at 
 every point of the stroke. 
 
 The application of the forces of in- 
 ertia to cause the proper action of an 
 engine was first worked by a practical 
 man having little or no book learning. 
 The theory and method of measuring 
 and estimating such forces has since 
 been so simplified that a child may un- 
 derstand it, if he will try. 
 
 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 lbs. Now draw 
 a line a, b. Fig. 4, to represent tiie po- 
 sition of the connecting rod. Let the 
 line a, b, represent, by its length, 14,000 
 lbs., which is the tension upon the rod. 
 Complete the rectangular parallelo- 
 gram a, b, c, d, then will the line a, c. 
 
 Q. 33. What is the principle of the 
 parallelogram of forces? 
 
 Ans. 33. If two forces are repre- 
 sented in intensity and direction by 
 two lines, the resultant of said forces 
 will be represented in intensity and di- 
 rection by the diagonal of a parallelo- 
 gram of which the lines representing 
 the first two forces are the sides. If 
 the resultant of two forces is repre- 
 sented by a line, the forces themselves 
 will be represented by the adjacent 
 sides of a parallelogram formed with 
 the first mentioned line as a diagonal. 
 
 The formula for the stored work in 
 a moving body, which is called the 
 fundamentaj formula of mechanics, 
 and the formula for obtaining the cen- 
 trifugal force of a body should be re- 
 membered. The ’ similarity between 
 these two formulae will help to fix 
 them in the memory. 
 
 The formula for stored force is 
 WV^ 
 
 =foot lbs. 
 
 64.4 
 
 The formula for centrifugal force is: 
 
 32.2 R 
 
 =lbs. 
 
 WV'’ 
 
 Q. 34. 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? Ex- 
 p’ain method of calculation. 
 
 represent by its length the up- 
 ward 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 lbs. 
 to keep the cross-head from pressjng 
 upon the upper guide at this point. 
 
 Q. 85: If the engine of Q. 28 has a 
 
 cylinder 10" internal diameter and is 
 running with a gage pressure of 70 lbs., 
 what is the greatest strain brought 
 upon the connecting rod? The weight 
 and angularity of the connecting-rod, 
 lead, back-pressure and compression 
 being neglected. 
 
 Ans. 35. We have the force of In- 
 ertia and the steam pressure to take 
 into account. By referring to Fig. 3 
 we find that there is at the commence- 
 ment of the stroke a tension of about 
 18,000 lbs, on the connecting rod due 
 to the inertia of the parts; but the ac- 
 tion of the steam is to produce a com- 
 pression upon the connecting rod. That* 
 is, at the commencement of the stroke 
 the steam pressure acts in the oppo- 
 site direction to the force of inertia. 
 Therefore the strain upon the connect- 
 ing rod at the commencement of the 
 stroke is 18,000 — 70X78.54 (area of the 
 10" piston) —18,000—5498=12,502 lbs. 
 tension upon the connecting rod. At 
 the end of the stroke the force of in- 
 ertia is 15,310 — 2552=12,758 — a force of 
 compression; to this should be added 
 the final pressure of the steam upon 
 
 8 
 
the piston, If any. Therefore, the great- 
 est strain brought upon the connecting 
 rod is equal to about 12,758 lbs. and 
 steam pressure, if any, and is a force 
 of compression at the end of the stroke. 
 
 Q. 36. 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. 
 
 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 pis- 
 ton travel will be shown for each an- 
 gular position of the crank, by the 
 same diagram. 
 
 Q. 37. What per cent of the steam 
 is condensed by the cylinder walls in 
 
 C 
 
 c, d, i, 1, Fig. 5. Upon the same disftn- 
 eter 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 cor- 
 responding 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 
 
 simple non-'condensing, fast running 
 engines, without steam jackets, of from 
 50 to 100 HP. 
 
 Ans. 37. From 25 to 50 percent. 
 
 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 cylinder, and 
 only to be accounted for upon the sup- 
 position that much of the heat goes 
 to re-evaporating the condensed steam. 
 
 9 
 
Q. 38. What is the greatest hight to 
 which a pump may draw water that is 
 at a temperature of 191° F.? 
 
 Ans. 38. The pressure of the atmos- 
 phere is about 14.7 lbs. per square inch. 
 The pressure of steam at 190° is about 
 9.5 lbs. per square inch. There could 
 never be a vacuum less than the pres- 
 sure of the steam. So that the pres- 
 sure possibly available to raise the 
 water is 14.7 — 9. 5=5. 2 lbs. per square 
 inch. This corresponds to 
 5.2 
 
 =12 ft. 
 
 .434 
 
 nearly. (See No. 28 of last year’s 
 questions.) 
 
 Q. 39. What effect will the required 
 velocity of the water in the inlet pipe 
 to the pump have on tlie hight to 
 which the water may be raised, neg- 
 lecting the friction of the water on the 
 pipe? 
 
 Ans. 39. It will lessen the hight by 
 an amount equal to the “head” re- 
 quired to give it the velocity. This 
 head would be about equal to the 
 square of the velocity (in feet per sec- 
 ond), divided by 64. 
 
 y2 
 
 h=— f- 
 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 hight 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 vac- 
 uum, as well as slip of valves, all tend 
 to prevent the cylinder from filling. — 
 Ed.] 
 
 Q. 40. What is meant by the flash- 
 ing point of an oil? How may the flash- 
 ing point of a sample be determined? 
 
 Ans. 40. The flashing point is that 
 temperature of the oil at which , it 
 gives off vapor at a sufficiently high 
 rate to form an inflammable mixture 
 with the air above it. 
 
 The point may be determined by 
 gradually raising the temperature of 
 the oil and occasionally passing a 
 flame above it. The temperature at 
 
 which the vapor takes Are is the 
 flashing-point. 
 
 Q. 41. Does the fly-wheel increase 
 the power of the engine? 
 
 Ans. 41. If regulates but does not 
 increase the power. 
 
 Strength of Materials. 
 
 Q. 42. Give three ways of calculat- 
 ing 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 hight of 
 the base of the segment above the cen- 
 ter 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 re- 
 sult is a little too small. 
 
 SECOND method. 
 
 Subtract twice the distance from the 
 base of the segment to the center 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 hight of the 
 base of the segment above the center. 
 Call this result No. 2. Subtract re- 
 sult 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 sometimes 
 useful as a check upon each other, and 
 as some will prefer to use one way 
 and some another, the graphical meth- 
 ods of solving the segment 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 of the radius 
 of the circle, in calculating the trian- 
 gular area. 
 
 10 
 
AN EXAMPI.R. 
 
 The circle is 53" in diameter. The 
 hight of the base of the segment above 
 the center of the circle is 7.5". (See 
 Fig. 6.) 
 
 First Method: — Tlie area of the cir- 
 cle, 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 hight, 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 
 
 Fig. 6. 
 
 area g a b c h, you have left the area 
 of the segment, a, b c e. Therefore 
 1103.09 — 397.5 equals 705.59 sq. ins., the 
 approximate area of the segment. 
 
 Second Method: — If you subtract 
 twice the hight e d, from the semi-cir- 
 cumference g a b c h, you have ap- 
 proximately 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 ra- 
 dius by the hight e d, you have ap- 
 proximately (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. 
 
 Referring again to Fig. 6: — The cir- 
 cumference of a circle having a di- 
 ameter of 53" is 166.5". One-half this 
 (the line g b h,) is 83.25". If from this 
 you subtract twice the hight d e, that 
 is twice 7.5" or 15", you have 83.25 — 
 15=68.25", which is approximately the 
 length of the line a b c. If you multi- 
 
 ply this by one-half the radius, you 
 have 68.25X13.25=904.31 sq. ins., or the 
 area d a b c. This is result No. 1. 
 
 If you multiply the hight d e, by the 
 radius of the circle, you have 7.5X26.5 
 =198.75 sq. ins., which is nearly the 
 area of the triangle a c d. A little 
 large. This is result No. 2. 
 
 Subtracting result No. 2 from result 
 No. 1 you have 904.31—198.75=705.56 
 sq. ins., for the area of the segment, 
 which is a little small because the 
 area of the triangle was a little large. 
 
 By the formula of the answer to 
 last year’s Q. No. 15, and by accurately 
 figuring by the second method, the 
 area will be found to be about 710 sq. 
 ins. 
 
 Q. 43. 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 mentioned brace 
 is to the distance of its point of at- 
 tachment to the shell from the end of 
 the boiler. In this instance 1.118. 
 
 Q. 44. How is the safe torsional 
 strength of a round shaft of machine 
 steel calculated? 
 
 Ans. 44. The force multiplied by the 
 lever arm at which it acts (the mo- 
 ment) is equal to 12,000 times the cube 
 of the diameter of the shaft. The 
 units are in inches and pounds. Let 
 p = the force in pounds. 
 
 A = the length of the lever arm in 
 inches. 
 
 d = diameter of the shaft in inches 
 and fractions thereof. 
 
 This would be expressed arithmeti- 
 cally as: 
 
 P = A 12000XdXdXd. 
 Algebraically: 
 
 PA = 12000d^ 
 
 For ordinary working, about one- 
 eighth of this value would be taken: 
 PA = 1500d‘\ 
 
 One will find a variety of answers 
 to question 44, which may weaken his 
 faith in this kind of calculation. Ex- 
 perience in the use of the formula will 
 certainly restore his confidence. The 
 varieties of values given are probably 
 
 11 
 
owing to a difference of opinion as to 
 what is safe. Sometimes one only 
 wants a part to last a comparatively 
 short time, and sometimes it does not 
 matter much whether a part breaks or 
 not. In these cases one might wish 
 to allow a greater strain than in other 
 cases. We have used machine parts 
 subject to a strain 75% greater than 
 the largest given. These have 
 stretched, bent, or broken, more or 
 less, sooner or later, however. 
 
 The modern understanding of the 
 nature of metals is that they will 
 always give way sooner or later un- 
 der repeated strains. When a metal 
 part will break is merely a question 
 of time. The further the strains are 
 below the elastic limit the longer the 
 part will last. It is for this reason 
 that a considerable factor of safety 
 should be taken in most cases. We 
 believe the best statement of the the- 
 ory, and digest of experience, relating 
 to the so-called “Fatigue of Metals” 
 is to be found in Prof. Weyrauch’s 
 book translated by Prof. A. J. DuBois. 
 We think it is published by Wiley. 
 
 Q. 45. If the rule is taken that a 
 shaft must not twist more than one 
 degree in 10 ft., what is the allowable 
 moment, or torque, for a machine steel 
 shaft in diameter? 
 
 Ans. 45. For soft steel and wrought 
 iron, the following rule may be taken: 
 
 The allowable moment is equal to 
 160 times the fourth power of the di- 
 ameter of the shaft. Inches and pounds 
 are used. 
 
 This is expressed arithmetically as; 
 IGOXdXd^dXd. 
 
 Algebraically as: 
 leOd^M. 
 
 With a ly^!' shaft, the allowable mo- 
 ment under the given conditions is: 
 160X1.5X1.5X1.5X1.5 = 160 X (1.5)^ = 
 160X (5.0625)=810 statical inch-pounds. 
 
 The formula given will be found to 
 correspond very closely with practice. 
 If tempered spring-steel is used the 
 constant 185 should be used instead of 
 160. 
 
 If one wishes to know how many 
 degrees a given force upon a lever arm 
 of a given length will spring a shaft 
 10 ft. long of a given diameter he can 
 calculate it by the following formula: 
 M 
 
 160d*. 
 
 in which D is the angle in degrees, M 
 is the moment (i. e., the force multi- 
 plied by the lever arm), 160 a constant 
 for wrought iron and soft steel, and 
 d, the diameter of the shaft in inches. 
 For other lengths of shaft the angle 
 will be proportional to the length. 
 
 The utility of the coefficient of elas- 
 ticity referred to in Q. 46 is believed 
 to be that it obviates the necessity of 
 taking into account the length of the 
 particular rod. This is illustrated in 
 Ans. 47. This coefficient is quite con- 
 stant for different specimens of steel, 
 but varies somewhat, especially be- 
 tween hard and soft metal. The First 
 German association. No. 15, of Ohio, 
 find it to be 29,200,000; your commit- 
 tee has found it greater than 30,000,-’* 
 000 in tempered steel. 
 
 Q. 46. If it is assumed that the ex- 
 tension of a rod is always proportional 
 to the load, what load would extend a 
 machine steel rod of one square inch 
 cross-section its own length? What 
 is this number called? 
 
 Ans. 46. About 30,000,000 lbs. This 
 number is called the coefficient or mod- 
 ulus of elasticity. 
 
 Q. 47. If a machine steel rod is im- 
 movably attached at its ends when it 
 is at a temperature of 100° what will 
 be the strain upon it when it has 
 cooled down to 50° ? 
 
 Ans. 47. If 1° fall in temperature 
 contracts a rod .0000065 of its length, 
 a fall of 50° would contract it .000325 
 of its length. If the rod was pre- 
 vented from contracting it would exert 
 a force equal to that which would ex- 
 tend the rod that distance. That is 
 30,000, 000 X. 000325=9750 lbs. per square 
 inch. 
 
 Q. 48. What are the different ways 
 in which a riveted joint may give way? 
 
 Ans. 48. The plate may tear across 
 along the 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. What is the general rule for 
 selecting the diameter of rivets for a 
 given thickness of boiler plate? What 
 
 12 
 
should be 'considered 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 ol 
 the plate, otherwise it is liable to be 
 broken. Drilled holes are not made 
 less in diameter than the thickness oi 
 the plate. 
 
 a 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 1 
 
 1 
 
 (a) (b) (c) 
 
 Fig. 7. 
 
 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 in- 
 creases 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=1.2Vt 
 
 Q. 50. 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, be- 
 cause a short tube will retain its cir- 
 cular form better than a long one 
 owing to the tendency of the long 
 tube to sag in the middle and 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 boil- 
 ers 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 be- 
 yond ten or twelve diameters may be 
 neglected. 
 
 Q. 51. 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 en- 
 gines should be rectangular and great- 
 er in the plane of motion than per- 
 pendicular 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. Assuming a factor of safety 
 of 20 how would you determine the di- 
 ameter 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 pres- 
 sure 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 lengtli 
 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: — 
 
 ^ "VpfVl 
 
 d = 
 
 61.75 
 
 Suppose we take the following data 
 for example— Area of piston 100 sq. 
 
 13 
 
ins. Weight of reciprocating parts 
 250 lbs. Speed 250 turns per minute. 
 Maximum steam pressure 50 lbs. 
 Stroke 1 ft. Connecting-rod 36" long; 
 then the greatest pressure of the 
 steam on the piston would be 100X50 
 =5000 lbs. The pressure due to in- 
 ertia would be about 250X13Xl3-^16 
 =2641 lbs. Therefore the maximum 
 pressure to adapt the rod to is 50004- 
 2641=7641 lbs. 
 
 Using a factor of safety of 20, we 
 would have, pressure X factor of safety 
 =7641X20=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" cir- 
 cular connecting-rod. 
 
 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 mater ial, 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 strengtii 
 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. 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 hight (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 hight 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 pres- 
 
 sure, in pounds, 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 (ex- 
 pressed in inches). Call this result 
 No. 2. 
 
 Third. Multiply result No. 1 by re- 
 sult No. 2 and divide this product by 
 127.8. 
 
 This will give the breadth of the rod 
 in inches. The hight is 2.7 times the 
 breadth. 
 
 This is expressed algebraically as 
 follows: — ' 
 
 _ y]^vu 
 
 ® “ 127.8 
 
 In which, as before, F is the factor 
 of safety, P the maximum pressure that 
 may be brought upon the rod, L the 
 length of the rod in inches, 127,8 is a 
 constant. 
 
 Q. 54. What would be the dimen- 
 sions of a proper connecting rod for 
 the engine of Q. 28 assuming a gage 
 pressure of 70 lbs.? 
 
 Ans. 54. The pressure due to the 
 inertia of the parts is 12758, the maxi- 
 mum pressure due to the steam would 
 be 78.54X70=5498. Therefore the 
 greatest pressure might be 12758-^5498 
 =18256. Assuming a factor of safety 
 of 20 the calculation would be as fol- 
 lows: 20X18256=365,120. The square 
 
 root of 365,120 is about 604 and the 
 square root of 604 is about 24.58. Re- 
 sult No. 1. The length of the con- 
 necting-rod is 36"; the square root 
 of this is 6: 6X24.58=147.48, and this 
 divided by the constant, 127.8, gives 
 1.154 as the breadth of the connecting- 
 rod; 2.7X1.154=3.1158 for the hight 
 of the cross-section. 
 
 Q, 55. (a) A round wrought iron 
 
 rod 3 ft. long and %" diam. is rigidly 
 fastened at one end and extends hori- 
 zontally, what weight will it support 
 at the outer end? 
 
 (b) What weight can be put on its 
 outer end for safe working under or- 
 dinary conditions of practice? 
 
 Ans. 55. (a) The greatest weight the 
 rod could be expected to hold may be 
 found by the folowing: — 
 
 Rule: — Multiply the cube of the di- 
 
 14 
 
ameter of the rod (in inches and frac- 
 tions thereof) by the constant num- 
 ber 4900 and divide by the length of 
 the rod in inches. 
 
 The cube of .75 is .422; .422X4900= 
 2068, and this divided by 36, the length 
 of the rod, is 57.8, which is the largest 
 weight in pounds that the rod could 
 be expected to hold at its end. 
 
 (b) The largest weight the rod couid 
 be expected to hold without being per- 
 manently bent, may be found by the 
 following: — 
 
 Rule: — Multiply the cube of the di- 
 ameter of the rod by 1960 and divide 
 by the length of the rod. 
 
 This would give about 23 lbs. as the 
 weight that would bend the rod to its 
 limit of elasticity. About half this, 
 or say 12 lbs., would be taken in prac- 
 tice for ordinary constructions. 
 
 Q. 56. (a)” A wrought iron rod of 
 
 rectangular cross-sect on 1.5" vertically 
 and %" broad, is rigidly fixed at one 
 end so as to extend horizontally, what 
 weight will it support 4.5 ft. from the 
 fixed end? (b) What weight will it 
 safely carry under ordinary working 
 conditions? 
 
 Ans. 56. (a) The rule for obtaining 
 the greatest weight the rod could be 
 expected to hold may be taken as : — 
 Multiply the breadth by the square of 
 the hight and this product by the con- 
 stant number 14000, and then divide 
 this last product by the length of the 
 rod. 
 
 Thus .75X1.5Xl.5X8311-f-54 = 260 
 lbs., as the greatest weight the rod 
 could be expected to sustain. 
 
 (b) The weight that will strain the 
 rod to its limit of elasticity may be 
 found by the following: — 
 
 Rule: — Multiply the breadth of the 
 rod by the square of its hight and this 
 product by 3350 and divide by the 
 length of the rod in inches. In this 
 case the result would be 104 lbs. as the 
 weight that would stretch the rod to 
 the limit of elasticity. About half this 
 last result is taken in practice. This 
 would give 52 lbs. 
 
 In rods subjected to a bending force 
 (Ans. 55-56) the upper and lower fibers 
 may be overstrained while the rest of 
 the rod is within the allowable stress. 
 In special experiments for the purpose 
 of this article the round rods bent and 
 staid bent when subjected to a force a 
 little greater than given by the first 
 
 rule. If the force had been continued 
 they would have broken. Probably the 
 rods would have gradually lient in use 
 with a strain much greater than given 
 by the second rule. 
 
 Q. 57. 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 specifica- 
 tions. 
 
 Ans. 57. Diameter of wheel, di- 
 ameter of shaft, d ametrical pitch, 
 width of face, width and depth of key- 
 way. 
 
 Q. 58. If a boiler, we ghing, 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 44 of the entire weight (10 
 tons) or 5000 lbs. Allowing 10,0^J0 
 lbs. 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 tempera- 
 ture and corrosive action to which the 
 rods would be subjected, the rods 
 would probably best be made 1" or 
 1.125" in diameter. 
 
 Q. 59. What is the relative strength 
 of wrought and cast iron — 
 
 (a) When subjected to a crushing 
 force? 
 
 (b) When subjected to a tensile 
 force ? 
 
 Ans. 59. (a) Wrought iron is from 
 
 ^2 to 1/3 as strong as cast iron when 
 subjected to a compressive strain. 
 
 (b) Wrought iron is from 2 to 3 
 times as strong as cast iron when sub- 
 jected to a strain in tension. 
 
 Q. 60. There is a wrought iron bolt 
 %" diam. having ten U. S. standard 
 threads to the inch cut in it. What is 
 its breaking strength? 
 
 Ans, 60. About 12,000 lbs. 
 
 Electrical. 
 
 Q. 61. (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 
 
 16 
 
pressure (head) or difference of po- 
 tential. It is equal to 10* absolute, or 
 C.G.S. units. An electro-motive force 
 of one volt will produce a current of 
 one ampere when applied to a con- 
 ductor having a resistance of one ohm. 
 
 (b) Practically one cell of ordinary 
 battery gives an electro-motive force 
 of one volt. T^atimer Clark’s stand- 
 ard cell gives 1.436 volts under favor- 
 able 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 pres- 
 sures, 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 conductor 
 [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 concep- 
 tion in dealing with dynamos, but 
 would seem more real to the writer if 
 “a field capable of an aggregate pull 
 of 225 lbs.” 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. (a) 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 prac- 
 
 tical 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 
 82° P. 
 
 We have thought that the ohm was 
 best conceived of, as the resistance 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. 
 
 An idea of the resistance of one ohm 
 may be had from the following table, 
 which gives the number of feet and 
 
 number (B. &. S.) of copper wire hav- 
 ing a resistance of one ohm: 
 
 Fret Per Ohm oe Wire (B. & S. ). 
 
 94 feet 
 
 of 
 
 No. 20. 
 
 150 
 
 “ 
 
 18. 
 
 239 
 
 ii 
 
 16. 
 
 380 
 
 <i 
 
 14. 
 
 605 
 
 
 12. 
 
 961 
 
 
 10. 
 
 1529 
 
 66 
 
 8. 
 
 2432 
 
 66 
 
 6. 
 
 3867 
 
 66 
 
 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. (a) What is an ampere? 
 
 (b) How do you fix its value 
 in your mind for practical purposes? 
 
 Ans. 63. (a) The ampei;e is a cer- 
 
 tain 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, fiowing 
 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 candle- 
 power lamp requires a current of one 
 ampere; a 110-volt lamp, .5 ampere; 
 the ordinary 2000 C.P. arc circuit re- 
 quires 9.6 amperes, or practically one 
 C.G.S. unit. 
 
 Q. 64. (a) 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 prac- 
 
 tical unit of power. It is equal to 10,- 
 000,000 units of power in the C.G.S. sys- 
 tem. To measure the power carried 
 by an electrical circuit we must meas- 
 ure both the volts and the amperes. 
 The product of the volts and amperes 
 
 10 
 
is the watts, or electrical power; 74G 
 watts=one horse-power. 
 
 (b) The value of a watt in foot- 
 pounds per sec. is 
 33,000 
 
 =.7373 ft. lbs. 
 
 60X746 
 
 The value of a watt in B.T.U. per 
 sec. is .7373-J-778=.0009477 H.U. per 
 sec. 
 
 We have thought that the watt was 
 best thought of in reference to its val- 
 ue 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 P., that its temperature is above 
 that of the surrounding atmosphere. 
 Knowing the square feet cf 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. 
 
 Q. 65. What is Ohm’s law? 
 
 Ans. 65. Ohm’s law: — 
 
 V 
 
 (1) Amperes=volts-^ohms=A=: — 
 
 O 
 
 (2) Volts=amperesXohms=V=AO 
 
 V 
 
 (3) Ohms=volts-l-amperes=0= — 
 
 A 
 
 Q. 66. What is meant by the differ- 
 ence of potential of a current? 
 
 Ans. 66. The potential of the cur- 
 rent refers to the voltage between the 
 terminals of the circuit, generally 
 speaking. If we take, for example, an 
 arc light circuit supplying 20 lamps 
 requiring 50 volts per lamp, we will 
 have a 1000 volt circuit, or the differ- 
 ence of potential is said to be 1000 
 volts. The potential decreases along 
 the circuit in proportion to the resist- 
 ance. We would say, for instance, the 
 positive terminal of the dynamo has 
 1000 volts potential, the negative ter- 
 minal has zero potential. After the cur- 
 rent has passed ten lamps the differ- 
 ence 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 “ground- 
 ed.” 
 
 Q. 67. What are the advantages and 
 disadvantages of high tension cur- 
 rents? 
 
 Ans. 67. The advantages of high 
 potential currents consist in the abil- 
 ity 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. What is a transformer and 
 how constructed? 
 
 Ans. 68. A transformer is an appa- 
 ratus for transforming currents of high 
 potential and small amperage into cur- 
 rents of low potential and large am- 
 perage, and vice versa. 
 
 The kind commonly used for trans- 
 forming alternating currents consists 
 of a core cf laminated iron plates with 
 two sets of windings. The primary 
 current is sent by the dynamo through 
 the primary coil and an induced cur- 
 rent is generated thereby in the sec- 
 ondary coil. The change in potential 
 and amperage is regulated by the size 
 of wire and number of turns in the 
 different windings. 
 
 Q. 69. Can a direct current be trans- 
 formed? 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 winding and the 
 armature turns, generating a second- 
 ary current in the other winding. By 
 properly proportioning the windings a 
 current of the desired voltage can be 
 obtained from the secondary. These 
 machines are in common use in large 
 central stations, where they are com- 
 monly termed “boosters.” Stationary 
 or static transformers are used for al- 
 ternating curents. 
 
 Q. 70. 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 proportional 
 to the resistance of the circuits. 
 
 In case the shunted circuit has ten 
 
 17 
 
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 connection between 
 the terminals of the shunt. We may 
 say that the current is therefore di- 
 vided into eleven parts, one-eleventh 
 going through the shunt and ten-elev- 
 enths going through the main circuit. 
 
 Where the main current flowing in 
 the main circuit is too large to be con- 
 veniently 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. What is meant by the carry- 
 ing 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 burn- 
 ing the insulation or wood work. 
 
 Q. 72. What is the allowable carry- 
 ing 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: 
 Rubber W’ther- 
 
 No. of Wire, Circular Covered proof 
 
 B. & S.Gage. Mils. Wire. Wire. 
 
 Amperes. Amperes. 
 
 18 
 
 1624 
 
 3 
 
 5 
 
 16 
 
 2583 
 
 6 
 
 8 
 
 14 
 
 4107 
 
 12 
 
 16 
 
 12 
 
 6530 
 
 17 
 
 23 
 
 10 
 
 10382 
 
 24 
 
 32 
 
 8 
 
 16510 
 
 33 
 
 40 
 
 6 
 
 26250 
 
 46 
 
 64 
 
 5 
 
 33102 
 
 54 
 
 77 
 
 4 
 
 41743 
 
 65 
 
 92 
 
 3 
 
 52634 
 
 76 
 
 no 
 
 2 
 
 66373 
 
 90 
 
 131 
 
 1 
 
 83694 
 
 107 
 
 156 
 
 0 
 
 105593 
 
 129 
 
 185 
 
 00 
 
 133079 
 
 150 
 
 220 
 
 000 
 
 167805 
 
 177 
 
 262 
 
 0000 
 
 211600 
 
 210 
 
 312 
 
 Q. '73. How much current is 
 a 16 c-p lamp on an average? 
 
 used by 
 
 Ans. 73. 
 quires one 
 
 A 55-volt 
 ampere. 
 
 16 C.P. lamp re- 
 
 A 110-volt 16 C.P. lamp requires .5 
 ampere, on an average. 
 
 Q. 74. 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 cor- 
 responds to the drop in pressure along 
 a steam pipe or water main; in the 
 latter case sometimes called “loss of 
 head.” 
 
 Q. 75. 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 per- 
 centages 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. How can the size of wire to 
 be used be determined for conveying 
 a given number of amperes a given dis- 
 tance 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, mul- 
 tiplied 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. length. By taking 
 the resistance of two feet length in 
 this calculation it is then necessary to 
 take only the distance in feet one way 
 in the calculation. 
 
 Area in 
 
 21 X Amperes X Dist. in ft. 
 
 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 pres- 
 sure. By the rule: 21X150X150=472,- 
 500. The number of volts drop is 120 
 X. 03=3.6; then 472,500^3,6=131,23:^ 
 
 18 
 
mils. Turning to answer No. 72, we 
 find that this corresponds to 00 wire 
 very nearly. 
 
 Q. 77. What is a dynamo? How 
 does it generate electricity? 
 
 Ans. 77. A dynamo is a machine 
 for converting energy, in the form of 
 mechanical power, into energy in the 
 form of an electric curent, or vice 
 versa, by the operation of setting con- 
 ductors, usually coils of copper wire, 
 to rotate in a magnetic field, or by 
 varying a magnetic field in the pres- 
 ence of conductors. A dynamo con- 
 sists 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 is constantly increasing or 
 decreasing. This action causes cur- 
 rents 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 conduct- 
 ors. 
 
 Q. 78. What is meant by the mag- 
 netic circuit? 
 
 Ans. 78. The magnetic circuit is the 
 path or space through which the mag- 
 netic lines or force travel. 
 
 Take for example the Edison dyna- 
 mo. The field-magnets of this ma- 
 chine 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 be- 
 tween the poles, D and E, but the mag- 
 netic circuit follows around through 
 all the parts, D, B, A, C, E, and across 
 to D. If an armature with an iron 
 core be placed in the space between 
 the pole-pieces the magnetic circuit be- 
 tween 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 impel- 
 ling force [ampere-turns] more mag- 
 netism would pass through the circuit 
 because the resistance was less. — Ed. 
 Com.) 
 
 Q. 79, What is the relative permea- 
 bility to magnetism of iron and air? 
 
 Ans. 79. S. P. Thompson’s third 
 edition, pages 55 and 56, says: — 
 
 “If we take the magnetic permea- 
 bility of air as one then the permea- 
 bility of iron will be represented by 
 values lying between 5 and 20,000, ac- 
 cording to the quality of the iron” 
 [and the density of the lines of force]. 
 
 The conception of the magnetic cir- 
 cuit and its close analogy to the elec- 
 tric 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 permeability 
 of annealed wrought iron is from 54 
 to 25,000 times that of air. 
 
 Fig. 8. 
 
 Q. 80. Give a table of diameters for 
 fuses for different currents. 
 
 Ans. 80. The diameter of a fuse 
 may be calculated as follows: — 
 
 in which d is the diameter in inches 
 of the 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 con- 
 stant, a, for the metals named: Cop- 
 
 10 
 
per=10,224; tin=1642; lead=1379; 
 
 aliiminum=7585; iron=3150. 
 
 From the above formula may be de- 
 duced the following: 
 
 Rule: — Divide the constant, for the 
 given metal, by the number of am- 
 peres required; multiply this quotient 
 by itself (square it) and 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 de- 
 nominator of a fraction. 
 
 This fraction expresses in inches the 
 diameter required. 
 
 which is the diameter required, ap- 
 proximately. 
 
 If one wishes to obtain the constant 
 for a given metal or alloy, he may do 
 so by the following formula: 
 c 
 
 a= 
 
 Vd» 
 
 Example: A tin wire .021" diam. 
 
 melts with a current of 5 amperes, 
 what is the constant for tin? 
 
 c— 5 
 
 d— .021 . • . d*=.000009261 
 
 Vd-’=V.000009261=.003044 
 
 Rupturing Current of Fuses. 
 
 Current 
 
 Tin Wire 
 
 Lead Wire. 
 
 Copper Wire. 
 
 Iron Wire. 
 
 Aluminum 
 
 in 
 
 Dia. 
 
 Approx. 
 
 Dia. 
 
 Approx. 
 
 Dia. 
 
 Approx. 
 
 Dia Approx. 
 
 Wire 
 
 Amperes. In. 
 
 B.W.G. 
 
 In. 
 
 R.W.G. 
 
 In. 
 
 R.W.G. 
 
 In. B 
 
 W,G. 
 
 Dia. In. 
 
 1 
 
 .0072 
 
 36 
 
 .0081 
 
 35 
 
 .0021 
 
 47 
 
 .0047 
 
 40 
 
 .0026 
 
 2 
 
 .0113 
 
 31 
 
 .0128 
 
 30 
 
 .0034 
 
 43 
 
 .0074 
 
 36 
 
 .0041 
 
 3 
 
 .0149 
 
 28 
 
 .0168 
 
 27 
 
 .0044 
 
 41 
 
 .0097 
 
 33 
 
 .0054 
 
 4 
 
 .0181 
 
 26 
 
 .0203 
 
 25 
 
 .0053 
 
 39 
 
 .0117 
 
 31 
 
 .0065 
 
 5 
 
 .0210 
 
 25 
 
 .0236 
 
 23 
 
 .0062 
 
 38 
 
 .0136 
 
 29 
 
 .0076 
 
 10 
 
 .0334 
 
 21 
 
 .0375 
 
 20 
 
 .0098 
 
 33 
 
 .0216 
 
 24 
 
 .0120 
 
 15 
 
 .0437 
 
 19 
 
 .0491 
 
 18 
 
 .0129 
 
 30 
 
 .0283 
 
 22 
 
 .0158 
 
 20 
 
 .0529 
 
 17 
 
 .0595 
 
 17 
 
 .0156 
 
 28 
 
 .0343 
 
 20.5 
 
 .0191 
 
 25 
 
 .0614 
 
 16 
 
 .0690 
 
 15 
 
 .0181 
 
 26 
 
 .0398 
 
 19 
 
 .0222 
 
 30 
 
 .0694 
 
 15 
 
 .0779 
 
 14 
 
 .0205 
 
 25 
 
 .0450 
 
 18.5 
 
 .0250 
 
 35 
 
 .0769 
 
 14.5 
 
 .0864 
 
 13.5 
 
 .0227 
 
 24 
 
 .0498 
 
 18 
 
 .0277 
 
 40 
 
 .0840 
 
 13.5 
 
 .0944 
 
 13 
 
 .0248 
 
 23 
 
 .0545 
 
 17 
 
 .0308 
 
 45 
 
 .0909 
 
 13 
 
 .1021 
 
 12 
 
 .0268 
 
 22 
 
 .0598 
 
 16.5 
 
 .0328 
 
 50 
 
 .0975 
 
 12.5 
 
 .1095 
 
 11.5 
 
 .0288 
 
 22 
 
 .0632 
 
 16 
 
 .0352 
 
 60 
 
 .1101 
 
 11 
 
 .1237 
 
 10 
 
 .0325 
 
 21 
 
 .0714 
 
 15 
 
 .0397 
 
 70 
 
 .1220 
 
 10 
 
 .1371 
 
 19.5 
 
 .0360 
 
 20 
 
 .0791 
 
 14 
 
 .0440 
 
 80 
 
 .1334 
 
 9.5 
 
 .1499 
 
 9.5 
 
 .0394 
 
 19 
 
 .0864 
 
 13.5 
 
 .0481 
 
 90 
 
 .1443 
 
 9 
 
 .1621 
 
 8 
 
 .0426 
 
 18.5 
 
 .0935 
 
 13 
 
 .0520 
 
 100 
 
 .1548 
 
 8.5 
 
 .1739 
 
 7 
 
 .0457 
 
 18 
 
 .1003 
 
 12 
 
 .0558 
 
 120 
 
 .1748 
 
 7 
 
 .1964 
 
 6 
 
 .0516 
 
 17.5 
 
 .1133 
 
 11 
 
 .0630 
 
 140 
 
 .1937 
 
 6 
 
 .2176 
 
 5 
 
 .0.572 
 
 17 
 
 .1255 
 
 10 
 
 .0698 
 
 160 
 
 .2118 
 
 5. 
 
 .2379 
 
 4 
 
 .0625 
 
 16 
 
 .1372 
 
 9.5 
 
 .0763 
 
 180 
 
 .2291 
 
 4 
 
 .2573 
 
 3 
 
 .0676 
 
 16 
 
 .1484 
 
 9 
 
 .0826 
 
 200 
 
 .2457 
 
 3.5 
 
 .2760 
 
 2 
 
 .0725 
 
 15 
 
 .1592 
 
 8 
 
 .0886 
 
 250 
 
 .2851 
 
 1.5 
 
 .3203 
 
 0 
 
 .0841 
 
 13.5 
 
 .1848 
 
 6.5 
 
 .1028 
 
 300 
 
 .3220 
 
 0 
 
 .3617 
 
 0.5 
 
 .0950 
 
 12.5 
 
 .2086 
 
 5 
 
 .1161 
 
 
 Con, 
 
 1642. 
 
 Con. 1379. 
 
 Con. 
 
 10244. 
 
 Con. 3150. 
 
 Con, 7585. 
 
 ^ , , Omaha No. 1, Nebraska. 
 
 Table from Kent s M. E. Handbook, 2d Ed. j Bramhall, instructor. 
 
 (Ex. Q. What is the diameter re- 
 quired for a lead fuse to carry thirty 
 (30) amperes of current? 
 
 The constant for lead is 1379; this 
 divided by 30 is a little less than 46; 
 46 multiplied by itself (square) is 
 2116; the cube root of 2116 is a little 
 greater than 12,8. 
 
 Therefore the fraction is: 
 
 1 
 
 =.078, 
 
 12.8 
 
 Therefore constant = a = — - 
 
 5 5000000 
 
 = =1642 
 
 .003044 3044 
 
 In reference to Q. 80 some of the as- 
 sociations have thought that no useful 
 table could be constructed because the 
 material of the wire varied, and the 
 result was influenced by the length of 
 the fuse. It seems to us that the only 
 
 20 
 
care in reference to the length of the 
 fuse should be, that the fuse should 
 not be so short that its points of at- 
 tachment will have an appreciable 
 
 effect in conducting away and radiat- 
 ing the heat generated in the fuse. 
 We think the table and formula given 
 should prove useful. 
 
 BOILER AND EliRNACE. 
 
 The efficiency of the furnace and 
 boiler is always directly dependent 
 upon the care, skill, and knowledge, of 
 the engineer. If he properly handies 
 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 chim- 
 ney with the gases and is radiated, or 
 both these sources of loss occur at the 
 same time. 
 
 A good quality of coal should pro- 
 duce about 14,000 heat units per pound, 
 -when properly burned, and may gen- 
 erate as little as 3,500, when improp- 
 erly 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 fifty pounds of gases 
 passing up the chimney per pound of 
 coal burned, and these gases were at 
 a temperature of 500° above the tem- 
 perature of the engine room. Then, as 
 raising one pound of the gases one de- 
 gree takes .24 of a heat unit, raising it 
 500° would take .24X500 or 120 heat 
 units. Raising 50 lbs. to that tempera- 
 ture would take 50X120 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 per cent would be radiated and 
 lost in other minor ways. 
 
 Twenty-five pounds of chimney gases 
 per pound of coal is sufficient (0.7) 
 (practical trials with specially skillful 
 firemen generally show less) and the 
 average temperature of the chimney 
 gases above the outside air in the 137 
 tests recorded by Mr. Geo. H. Barrus 
 is not over 400°. Taking these values 
 only about 17% of the heat would be 
 carried up the chimney in this way. 
 
 An understanding of the operation 
 
 of the furnace and boiler requires that 
 we should know the following facts: 
 
 (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 pro- 
 duced 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 an- 
 swer the first column contains the 
 name of the coal, the second the num- 
 ber 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 
 per cent was used for this purpose: 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 
 has been evaporated by one pound of 
 some of the coals mentioned, in actual 
 practical tests; the sixth column gives 
 the percent efficiency, corresponding 
 to the actual evaporations named in 
 column five. 
 
 Second: — The amount of water evap- 
 orated 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 num- 
 ber 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 
 
 
 21 
 
steam. Or the evaporation may be re- 
 duced to an equivalent evaporation 
 from and at 212° by multiplying it by 
 the proper number (factor of evapor- 
 ation) 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. 
 
 CoAiv Used, Pocahontas (Run of Mine) 
 Heat Value, 14,289. 
 
 
 Heat 
 
 Units 
 
 Per Cent 
 of 
 
 Total 
 
 Heat 
 
 Useful Evaporation.. 
 
 11374.044 
 
 79.6 
 
 Loss by Chimney 
 Gases 
 
 1900.437 
 
 13.3 
 
 Unconsumed Car- 
 bonic Oxide 
 
 271.491 
 
 1.9 
 
 Loss by radiation, 
 unconsumed hy- 
 drocarbons, evap- 
 oration of mois- 
 ture in coal (2.3) 
 per cent.) and 
 unaccounted for. . . 
 
 743.028 
 
 5.2 
 
 Total 
 
 14289 
 
 100 
 
 The above gives a tabulated result 
 of an 1897 practical test. 
 
 ney gases per pound of coal in our own 
 plants, and to ascertain if we are prop- 
 erly burning the coal. 
 
 Test of Boiler and Furnace. 
 
 A complete test of a boiler and fur- 
 nace would involve ascertaining the 
 following facts: — 
 
 First. The quantity of heat generat- 
 ed by the combustion of the coal. 
 
 Second. The proportion of this heat 
 used in making steam. 
 
 Third. The proportion of the heat 
 lost by — 
 
 (a) Incomplete combustion. 
 
 (b) Hot gases passing up the chim- 
 ney. 
 
 (c) Radiation, and dropping of fuel 
 through the grate. 
 
 The second divided by the first is the 
 efficiency of the furnace and boiler. 
 
 The most considerable loss of heat 
 is (usually) through the hot gases 
 passing up the 'chimney. This may be 
 easily estimated. It is also easy to esti- 
 mate whether or not the loss from in- 
 complete combustion is considerable. 
 
 If the fuel is not evenly distributed 
 on the grate, one may have too much 
 air supplied to a part of the furnace 
 and not enough to the remainder, so 
 that excessive quantities of free oxy- 
 gen (O 2 ) 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 
 
 Temperature 
 
 OF 
 
 Pressure of steam by Gauge, in Lbs. per Sq. Inch. 
 Factors of Evaporation. 
 
 Feed Water 
 
 0. P.‘ 
 
 40 P. 
 
 50 P. 
 
 60 P. 
 
 70 P. 
 
 80 P. 
 
 90 P. 
 
 100 B. 
 
 HOP. 
 
 120 P. 
 
 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 
 
 The eight questions on this subject 
 still unanswered will bring out an- 
 will bring out answers that shall teach 
 us how to estimate the weight of chim- 
 
 very considerable it is owing to inex- 
 cusable mismanagement, or imperfect 
 apparatus. One author remarks that 
 the remedy is to get a new fireman, 
 
 22 
 
first. 
 
 The Quantity of Heat Generated. 
 
 Mr. R. S. Hale, in his paper on “Fuel 
 Gas Analysis in Boiler Tests” read be- 
 fore the American Society of Mechan- 
 ical Engineers in 1897 remarks that 
 it is not necessary to have an analysis 
 of the coal, inasmuch as the data of 
 the text books can be relied upon with- 
 in 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 fur- 
 nace, 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 generated 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 pro- 
 duced during that time. The coal is 
 usually weighed in the barrow upon 
 platform scales. 
 
 SECOND. 
 
 The Proportion of Heat Used in Mak- 
 ing 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 as- 
 certained 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 which is 
 emptied into a tank from which the 
 boiler draws its supply. 
 
 The amount may be approximately 
 estimated as indicated in Ans. 21. 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 on the water glass, 
 when the boiler was not in use. It 
 might also be measured by a water- 
 meter. 
 
 Having got the water evaporated fliis 
 amount would be divided by the num- 
 ber indicating the weight of coal used 
 and the result reduced to an equivalent 
 evaporation from and at 212° by multi- 
 plying by the factor of evaporation 
 taken from one of the numerous tables 
 published. This committee regrets that 
 it did not furnish a better table of fac- 
 tors of evaporation, or rather a more 
 complete one. 
 
 The above last result would then be 
 multiplied by 966, and the product di- 
 vided 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 per- 
 formance 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. 
 
 Second. Whether or not a consider- 
 able amount of combustible gas is go- 
 ing up the chimney. 
 
 The first may be determined approx- 
 imately by the ruies 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 un- 
 changed, that as much of the oxygen 
 as is properly used is changed to an 
 
 equal volume of CO 2 (carbonic acid). 
 Therefore if all the oxygen of the air 
 is either unchanged or is changed to 
 carbonic acid, the volume of the car- 
 bonic acid and oxygen taken together 
 must still be 20%, so that if we meas- 
 ure the volume of carbonic acid and 
 oxygen, find what percent they are of 
 the whole taken together, subtract this 
 result from 20 we shall have as a re- 
 mainder a percent which is roughly 
 proportional 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. 
 
 For the volumetric analysis of the 
 chimney gases the Orsat apparatus is 
 generally recommended as accurate 
 
 23 
 
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. say: — 
 
 “For the past year the writer has 
 made extensive use of the Orsat ap- 
 paratus in his boiler testing and has 
 
 .found the work not only interesting, 
 but exceedingly instructive and valu- 
 able. 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 instrument 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 invari- 
 ably attended with an increase in the 
 percentage of carbonic oxide and a 
 reduction in the percentage of oxy- 
 gen, 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 engineers’ outfit.” 
 
 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 illus- 
 trated in Figs. 1 and 2. Anyone can 
 
 24 
 
make it and the apparatus itself ought 
 to cost about fifty cents. 
 
 Fig. 1, A, is a stick held in a verti- 
 cal position by a vice. 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, five-eighths 
 of an inch in diameter and about 15 ' 
 long, drawn down in a gas flame at the 
 ends, so that small rubber (black pre- 
 ferred) 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 posi- 
 tions on the tube to serve as marks. 
 
 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 measure- 
 ment of this water determines the vol- 
 umetric contents of the tube once for 
 all. Suppose for convenience it is 100 
 cubic centimeters (though it may be 
 anything; the larger, the more accu- 
 rate will be the measurements). 
 
 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 con- 
 venient device until all the air is re- 
 moved and the tube is full of chimney 
 gases. 
 
 The pinch-cock G, is now closed, the 
 other end of the tube E, 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 funnel H, need not be on. The 
 pinch-cock G, is now opened for an in- 
 stant until some of the fluid in the cup 
 has run into the tube, and a rubber 
 band J, moved down to mark the sur- 
 face 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 glass, 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 min- 
 ute 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 percent 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 pyro- 
 gallic acid that it will absorb. Fill the 
 funnel with this pyrogallic acid solu- 
 tion 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 over-do this, enough 
 to fill the tube to Vz" is sufficient;. 
 On repeating the operation indicated 
 by, and described in reference 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 indi- 
 cates 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 meas- 
 ures 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 percent 
 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 wa- 
 ter 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 
 
 25 
 
drain long enough before measuring, 
 and also consider the pressures of the 
 water vapor. The temperature of the 
 tube and contents must not vary great- 
 ly 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. 
 
 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 pur- 
 pose — something specially made to re- 
 sist the action of corrosive fluids. 
 [Rubber finger-stalls can be purchased 
 from dealers in rubber goods. — Ed.] 
 
 Unless one has particularly thin 
 skin, or unless the operation is un- 
 necessarily protracted, the caustic will 
 do no harm. Running water, olive oil, 
 and muriatic acid will neutralize or re- 
 move the potash. 
 
 The writer has protected his finger 
 by placing over it a thin rubber bag 
 
 Cost of Apparatus. 
 
 Glass tube $0.10 
 
 1 ft. black rubber hose 0.10 
 
 Glass graduate 0.25 
 
 Pinch-co'ck 0.10 
 
 1 lb. caustic potash 0.10 
 
 1 can pyrogallic acid 0.35 
 
 Total ..$1.00 
 
 26 
 
An Illustrative Test. 
 
 Duration of test, 10 hours. 
 
 Kind of coal used, Pocahontas. Heat 
 value, 14,290. 
 
 Weight of coal consumed, 6,507 Ihs. 
 
 Water fed to boiler, 64,877 lbs. 
 
 Water per pound of coal, 64,877 -j- 
 * 6,507=9.97 lbs. 
 
 Equivalent evaporation from and at 
 212° 9.97X1.19=11.80 lbs. 
 
 Heat used in making steam per 
 pound of coal used 11.8X966=11,398 
 
 B. T. U. 
 
 Efficiency of boiler and furnace 
 11,398^14,290—80. 
 
 Gas analysis percent 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 IIX (15.1+4) -M5.1=13.92. 
 
 Temperature of flue gases above out- 
 side air, 529°. 
 
 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 
 the 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 chim- 
 ney gases, were not very available, and 
 practical for our purposes. Below will 
 be found described a device for that 
 purpose, which we believe accurate 
 simple and convenient. We also ven- 
 ture to speak of the methods of getting 
 the chimney gases into the analyzing 
 tube, etc. 
 
 The drawings are the work of one of 
 the committee and are intended simply 
 to be illustrative, and no great ac- 
 curacy nor proportion in sizes, have 
 been striven for. 
 
 Sensible heat in flue gases 529X13.92 
 X. 24=1, 767 h. u. 
 
 Per cent of heat in flue gases 1,767-i- 
 14,290=12.3. 
 
 Steam making 
 Hot gases 
 
 Heat Units. Pr.Ct. 
 .11,398 80 
 
 . 1,767 12.3 
 
 Balance; radiation, 
 unconsumed gases, 
 dropping coal 
 through grate, etc. . 
 
 Total 
 
 1,125 
 
 ,14,290" 
 
 7.7 
 
 100 
 
 Getting the Gases into Tubes. 
 
 It is generally thought best to lead 
 a 1/4" or %" lead pipe from the “breech- 
 ing” or a part of the flue near the up- 
 take to a convenient place for filling 
 the tube. We have used a small rub- 
 ber hose. It is said that if rubber gets 
 too hot it will generate gases that will 
 effect the results. 
 
 37 
 
To get the gases into the tube (see 
 Fig. 3) one simply connects the tube C, 
 in line with the pipe A, to the breech- 
 ing X, and also connect said tube to the 
 inlet port of a hand air-pump B, and 
 draws a dozen charges of the pump 
 
 to let the water get low enough so 
 that air will be drawn back by the 
 draft. 
 
 We have experimented, successfully, 
 we believe, with the following de- 
 scribed apparatus: 
 
 through the tube. This will clear out 
 the air that was in the tube and re- 
 place it with flue gases. Perhaps it 
 might be better to take a larger num- 
 ber of strokes of the pump. 
 
 A sort of ejector, like that shown in 
 Fig. 4, may be used instead of a pump. 
 
 Fig. 4. 
 
 By the aid of a small jet of steam a 
 steady stream of flue gases may be 
 drawn through the tube. 
 
 A barrel, keg, or large can, fllled 
 with water may be used for this pur- 
 pose, 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 EOt 
 
 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 secured, drawn 
 
 % 
 
 together at one end, b, in a gas flame; 
 the other end, which is open, or slight- 
 ly 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. 
 
 28 
 
It is then taken out and placed with 
 its open end under the surface of some 
 high boiling point (flashing point) oil 
 (we used 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 column 
 of air in the tube has contracted in 
 cooling, and therefore proportional to 
 the change in the temperature of the 
 tube. The column of air, b, e. Pig. 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 tem- 
 perature 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 tem- 
 perature (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-f 90=550° absolute. The oil 
 was drawn up 7", leaving 15.5 — 7=8.5" 
 as the length of the column, b, e, above 
 the oil. Then 8.5 : 15.5 : : 550 : the ab- 
 solute temperature of the flue gases. 
 Therefore the absolute temperature of 
 the flue gases was 550X15.5-^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 
 hight of the column for this effect, one 
 multiplies the hight 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 hight of the oil column. 
 
 In the above example the correction 
 would be as follows: 7X8.5X.004=.238. 
 Therefore to the measured hight of the 
 oil column should be added about .24 
 of an inch. 
 
 One might use a metal tube, for in- 
 stance a piece of gas pipe, with one 
 end closed air tight and the other near- 
 ly 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 originally, and the volume 
 after contraction. For example: Sup- 
 pose we had a tube of 2" inside diam- 
 
 eter, and 5" long. We 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 as the gases. Then 
 take it out and drop it into water, 
 open end down. As it cools it will 
 draw in some water. We then pour 
 out the water it has drawn in and 
 measure that, and find that it is 7 cu. 
 ins. We would then have the same cal- 
 culations as for the glass tube and the 
 same figures, except for the correction 
 for the weight of the column. 
 
 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 perfectly 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. 
 
 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. Ordi- 
 narily the weight of the gases per 
 pound of coal would run about as fol- 
 lows for the corresponding per cents 
 of COo. 
 
 29 
 
Percent 
 
 Weight of 
 Flue Gases 
 
 
 chimney gases. (See Ans. 16 and re- 
 marks under the heading “Analysis of 
 
 carbon dioxide. 
 
 per lb. coal. 
 
 
 flue gases.” That would have to be 
 
 3. 
 
 61 
 
 
 estimated as previously described. 
 
 4. 
 
 50 
 
 
 For the purpose of finding the vol- 
 
 5. 
 
 40 
 
 
 ume of CO 2 only, the tube would 
 
 6. 
 
 33.3 
 
 
 not have to be manipulated, nor 
 
 7. 
 
 28.6 
 
 
 would it be necessary to put 
 
 8. 
 
 25 
 
 
 the hand nor any part of it into the 
 
 9. 
 
 22.2 
 
 
 alkali. A little of the potash solution 
 
 10. 
 
 20 
 
 
 would be put in the cup, the fun- 
 
 11. 
 
 18.2 
 
 
 nel would be filled with it, and a little 
 
 12. 
 
 16.7 
 
 
 let down occasionally by opening the 
 
 13. 
 
 15.4 
 
 
 pinch-cock so as to keep the walls 
 
 14. 
 
 14.3 
 
 
 of the tube covered with the solution. 
 
 15. 
 
 13.3 
 
 
 The correction for the weight of the 
 
 16. 
 
 12.5 
 
 
 column in the analyzing tube is made 
 
 Under the conditions 
 
 at each end 
 
 of 
 
 as above described in referring to the 
 
 the column there would probably 
 
 be 
 
 oil column except that tne constant 
 
 considerable inflammable gases in the 
 
 would be .0063. 
 
 COMPRESSED AIR FORMILAS. 
 
 We had intended to -'take up the 
 subject of compressed air, in regular 
 order, but time and our hearts failed 
 us. However, upon the assurance of 
 the editor that practical formulae in 
 which the use of logarithms could be 
 dispensed with would be of interest 
 and use to our brothers, we send you 
 herewith the results of our work in 
 this line. 
 
 The usual fundamental formula in 
 the book is 
 
 PV^*" = C 
 
 in which P is the pressure, V the vol- 
 ume, 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 an exponent. Moreover it assumes 
 that no heat is lost by the air which, 
 owing to moisture in the air and the 
 cold cylinder walls, is nevec the case. 
 
 If, therefore, we assume an expon- 
 ent of 1 1/3 or 4/3 we shall have a 
 fundamental formula, that may be 
 solved by the more familiar operations 
 of squaring, cubing, and extracting the 
 cube and square root. We shall also 
 have a formula that will give results 
 approximating closer to practice, be- 
 cause it allows for a small loss of heat. 
 Absolute pressures and temperatures 
 are always used. 
 
 The fundamental formula then is 
 PV^'^ == C 
 
 this may be put in the form 
 PV WV = C 
 
 which is more convenient to use. In 
 words this formula says: — The 
 pressure multiplied by the volume, and 
 this product multiplied by the cube 
 root of the volume, is always the same 
 (i. e. equal to a constant). 
 
 From this may be derived the fol- 
 lowing formula: 
 
 C 
 
 P = (1) 
 
 V wv 
 
 Rule No. 1. To find the pressure, 
 the volume being known: 
 
 Divide the constant by the volume 
 multiplied by its own cube root. 
 
 Rule No. 2. To find the volume, rhe 
 pressure being known: 
 
 (1st) — Divide the constant by the 
 pressure and multiply this quotient 
 by itself three times (cube it). Call 
 this result No. 1. 
 
 (2d) — Find the square root of the 
 square root of result No. 1. This 
 last result is the volume required. 
 
 W = 3 [P^V— PV] (3) 
 
 To allow for atmospheric pressure 
 we must subtract 1 from the above, 
 14.7XPiston areaX(Vi — V), 
 
 80 
 
Rule No. 3. — To find the work done 
 by the piston in compressing the air 
 from one volume to another. 
 
 (1st) Multiply the total final pres- 
 sure upon the piston by the final vol- 
 ume. Call this result No. 1. 
 
 (2d) Multiply the total initial pres- 
 sure upon the piston, by the initial 
 volume. Call this result No. 2. 
 
 (3d) Subtract result No. 2 from re- 
 sult No. 1 and multiply the remainder 
 by 3. 
 
 This gives the total work done by 
 the piston. To allow for atmospheric 
 pressure we must subtract the product 
 of the total pressure of the atmosphere 
 
 Multiply the initial temperature by 
 the square root of the square root, of 
 the quotient obtained by dividing the 
 final pressure by the initial pressure. 
 
 Rule No. G. Under constant pres- 
 sure the volume is proportional to the 
 absolute temperature. 
 
 Rule No. 7. Under constant volume 
 the pressure is proportional to the ab- 
 solute temperature. 
 
 Rule No. 8. The specific heat of air 
 is: Under constant volume .1688; un- 
 der constant pressure .2377. 
 
 The above rules contain about all 
 the law and the “profits” of com- 
 pressed air. 
 
 eo 
 
 upon the piston multiplied by the trav- 
 el of the piston. 
 
 T. = TV^ (4) 
 
 Rule No. 4. To find the final tem- 
 perature, the initial temperature, and 
 the initial and final volume being 
 given. 
 
 Multiply the initial temperature by 
 the cube root of the quotient obtained 
 by dividing the initial, by the final 
 volume. 
 
 P. 
 
 T,=TW— (5) 
 
 P 
 
 Rule No. 5. To obtain the final tem- 
 perature, the initial temperature and 
 the final and initial pressures being 
 given; 
 
 Illustrative Examples. 
 
 Suppose we have a compressor that 
 has a 12" stroke and 10 square inches 
 area of piston, compressing to 74.7 lbs. 
 (60 lbs. gage). The figure shows the 
 indicator diagram. For convenience 
 we measure volumes in inch of cylin- 
 der length. 
 
 We first find the constant, once for 
 all, from the fundamental formula 
 C 
 
 P = 
 
 vwv 
 
 At the commencement of the stroke 
 the air in the cylinder is at atmos- 
 pheric pressure (14.7 lbs.) and the vol- 
 ume is 12" cylinder length. Therefore 
 the constant is 
 
 14.7X12 X ®V12 = 14.7 X 12 X 2.289 = 
 403.7796. Call it 404. 
 
 31 
 
We will first find the volume when 
 the air is compressed to 74.7 lbs. per 
 sq. in. by Rule No. 2. 
 
 The constant divided by the pressure 
 C 404 
 
 = = =5.4083. Call it 5.41 
 
 P 74.7 
 
 The cube of 5.41 is 158. Result No. 1. 
 
 The square root of 158 is 12.57 — and 
 the square root of 12.57 is 3.54+; ans. 
 Therefore the piston will be 3.54'' 
 from the head of the cylinder at the 
 end of compression. 
 
 To prove this we substitute the cal- 
 culated volume in formula No. 1 thus: 
 C 404 404 
 
 P= -= 
 
 vwv 3.54XW3.54 3.54X1.525 
 
 404 
 
 =74.7 which is the final pres- 
 
 5.399 
 
 sure, and therefore proves the correct- 
 ness of our calculation by Rule No. 2. 
 
 We will now find the work done in 
 compressing, by Rule No. 3. 
 
 The initial volume is 12" of cylinder 
 length, the final volume 3.54". 
 
 The total initial pressure is 10X14.7 
 =147 lbs., and the total final pressure 
 is 74.7X10=747 lbs. 
 
 1st. — The total final pressure multi- 
 plied by the final volume is 747X3.54 
 =2644: Result No. 1. 
 
 2d. — The total initial pressure mul- 
 tiplied by the initial volume is 147 X 
 12=1764: Result No. 2. 
 
 3d. — Three times the difference be- 
 tween result No. 1 and result No. 2 is 
 3 X (2644—1764) = 3 X 880 = 2640 inch- 
 pounds for the total work of compres- 
 sion. The result is in inch-pounds 
 instead of footrpounds because we 
 used inches of cylinder length in- 
 
 stead of feet of cylinder length. To re- 
 duce to feet-pounds divide by 12. 
 
 To allow for the pressure of the at- 
 mosphere we must subtract from the 
 above result 147x(12 — 3.54)=147X8.46 
 =1164 inch-pounds, making the net 
 work of compression 2640 — 1164=1476 
 inch-pounds. 
 
 We will now find the final tempera- 
 ture by Rule No. 4 assuming that the 
 initial temperature is 60°, that is, 460 
 +60=520° absolute. 
 
 The quotient obtained by dividing 
 the initial by the final volume is 
 12 
 
 =3.39 nearly. 
 
 3.54 
 
 The cube root of 3.39 is about 1.502. 
 The initial temperature multiplied by 
 1.502 is 520X1.502=781° absolute or 
 about 781 — 460=321° by the thermom- 
 eter. 
 
 To check the above result we will 
 calculate the final temperature by for- 
 mula (5). 
 
 P 
 
 Ti=t w— = 
 
 Pi 
 
 74.7 
 
 Ti=520 W =520 W5.0817 
 
 14.7 
 
 The square root of 5.0817 is 2.254+ 
 and the square root of 2.254+ is about 
 1.502. The initial temperature multi- 
 plied by this result is 520X1.502=781° 
 as before. 
 
 Rules 6, 7 and 8 will be found in 
 the books. 
 
 Elliott J. Stoddard, 
 Edward G. Jacques, 
 Edward P. Gilroy, 
 
 Committee. 
 
 82 
 
BOOKS REFERRED TO BY THE COMVIITTEE. 
 
 Gas and Fuel Analysis, for Engineers 
 — Gill — published by John Wiley & 
 Son. 
 
 Hempel’s Gas Analysis — translated 
 by Dennis — published by Macmillan & 
 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 graphi- 
 cal representation of the forces of in- 
 ertia in steam engines, to Holmes on 
 “The Steam Engine,” published in Ap- 
 pleton’s “Text Books of Science” se- 
 ries. 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 in 
 Machinery,” etc., published by Wiley, 
 or G. H. Hurst on “Lubricating Oils.” 
 
 Gill on Oil Testing. 
 
 The following are standard and re- 
 liable works on the strength of mate- 
 rials: 
 
 Strains in Framed Structures, by 
 Bindon B. Stoney. 
 
 Unwin’s Machine Design. 
 
 Rauleaux’s Constructor, translated 
 by Supplee. 
 
 Weisbach’s Mechanics Vol. 1, trans- 
 lated by Cox. 
 
 We think the works of Silvanus P. 
 Thompson, “Elements of Electricity 
 and Magnetism,” “Electromagnets and 
 Electromagnetic Machinery,” and “Dy- 
 namo-Electric Machinery,” are about 
 as good as any on this subject. 
 
 A book entitled “Mechanical Draft,” 
 published as an advertisement, your 
 committee regard the first hundred or 
 hundred and fifty pages as the best 
 work on the subject of combustion 
 they have met. 
 
 The Educational Committee. 
 
INDKX 
 
 PAGE 
 
 Air in Combustion 2 
 
 Air, Compressed, Formulas 30 
 
 Air, Permeability of 19 
 
 Absorption of Carbonic Acid 2 
 
 Absorption of Oxygen 2 
 
 Apparatus for Testing Chimney Gases. 3 
 
 Area of Segment, Calculating 10 
 
 Analysis of Flue Gases 23 
 
 Apparatus for Analyzing Flue Gases. . . 25 
 
 Acid, Pyrogallic 3 
 
 Ampere 16 
 
 Boiler, Conditions, Indicated by Tem- 
 perature of Chimney Gases 4 
 
 Boiler, Area to be Braced 10 
 
 Boiler Brace, Strain on 11 
 
 Boiler and Furnace Efficiency 21 
 
 Brace, Strain on Boiler 11 
 
 Books Referred to by Committee 33 
 
 Bending of Rod 14 
 
 Combustion of Coal, Heat of 1 
 
 Combustion of Coal, Products of 2 
 
 Coals, Heat Value of 1 
 
 Coals, Evaporative Power of 1 
 
 Carbonic Acid 2 
 
 Carbonic Oxide 2 
 
 Carbonic Acid, Absorption of 2 
 
 Carbon Dioxide 2 
 
 Carbon Monoxide 2 
 
 Chimney Gases 2 
 
 Chimney Gases, Weight of 2 
 
 Chimney Gases, Specific Heat of 2 
 
 Chimney Gases, Temperature of 2 
 
 Chimney Gases, Analysis of 3 
 
 Chimney Gases, Weight of 3 
 
 Chimney Gases, Heat Lost in . 3 
 
 Caustic Soda 3 
 
 Caustic Potash 3 
 
 Calorimeters 4 
 
 Centrifugal Force 7-8 
 
 Connecting Rod, Strain 8 
 
 Connecting Rod, Shape of 13 
 
 Connecting Rod, Diameter of 13 
 
 Connecting Rod, Rule for 14 
 
 Crank and Piston, Relative Positions of, 9 
 
 Condensation, Cylinder 9 
 
 Carrying Capacity of Conductors 18 
 
 Current Used by Lamps 18 
 
 Compressed Air Formula 30 
 
 PAGE 
 
 Co-efficient of Elasticity 12 
 
 Currents, High and Low Tension 17 
 
 Conductors, Carrying Capacity of 18 
 
 Draft, Forced, Weight of Gases by 2 
 
 Difference of Potential 17 
 
 Dynamo, The ... 19 
 
 Evaporation, Determining Rate of. . . .4 
 Engine, Change Required to Improve. . 5 
 Elasticity, Co-efficient or Modulus of. .12 
 Electrical 15 
 
 Forces, Parallelogram of 8 
 
 Furnace, Heat of. Estimating 2 
 
 Fuses, Table and Data 19 
 
 Fuses, Rule for 19 
 
 Forced Draft, Weight of Chimney 
 
 Gases by 2 
 
 Formulas 8 
 
 Flashing Point of Oil 10 
 
 Fly-Wheel, Office of 10 
 
 Flue Gases, Analysis of 23 
 
 Gases, Chimney ’ 2 
 
 Gases, Percent by Volume 3 
 
 Gases, Analysis of 3 
 
 Gases, Heat Lost in Chimney 3 
 
 Gases, Analysis of Flue 23 
 
 Gases, Apparatus for Analyzing Flue . . 25 
 
 Heat Value of Several Coals 1 
 
 Heat of Incomplete Combustion of 
 
 Coal 2 
 
 Heat of Combustion, Proportion of, to 
 
 Form Steam 2 
 
 Heat of Furnace, Estimation of 2 
 
 Heat of Fuel, Percent Used in Making 
 
 Steam 3 
 
 Heat Lost in Chimney Gases 3 
 
 Heat Permeability of Scale and Iron. . . 4 
 Heat, Percentage of, in Useful Work. . 5 
 Heat Waste in Chimney 21 
 
 Inertia of Reciprocating Parts 6 
 
 Joints, Riveted 12 
 
 Lamps, Current Used by 18 
 
 Moment 12 
 
 Modulus of Elasticity 12 
 
 Magnetic Circuit 19 
 
 34 
 
PAGE 
 
 Oxygen 2 
 
 Oils, Viscosity of 5 
 
 Oils, Flashing Point of 10 
 
 Ohm, The 10 
 
 Ohm’s Law 17 
 
 Potash, Caustic 3 
 
 » Pyrogallic Acid 3 
 
 Phosphorus * 3 
 
 Piston Velocity ^ 6 
 
 Piston Inertia * 6-8 
 
 Parallelogram of Forces 8 
 
 Pump Drawing Water 10 
 
 Pump, Velocity of Feed Water in 10 
 
 Pinion, Data Required for 15 
 
 Potential, Allowable Fall of 18 
 
 Potential, Difference of 17 
 
 Permeability of Iron and Air 19 
 
 Riveted Joints 12 
 
 Rule for Size of Wire 18 
 
 Specific Heat of Chimney Gases 2 
 
 Soda, Caustic 3 
 
 Stored Work 8 
 
 Segment of Circle, Calculating Area of, 10 
 
 Shaft, Twisting of 12 
 
 Shaft, Spring of 12 
 
 Strain Produced by Change of Temper- 
 ature 12 
 
 PAGE 
 
 Strength, Relative, of Cast and Wrought 
 
 Iron 15 
 
 Shunts 17 
 
 Steam, Weight of Per H.P 5 
 
 Scale, Heat Permeability of Iron and. . 4 
 
 Temperature of Chimney Gases 2 
 
 Torsional Strength, Calculating 11 
 
 Twisting of Shaft 12 
 
 Tube, Collapsing Strength of 13 
 
 Tensile Strength of Rod Supports 15 
 
 Transformer 17 
 
 Transformation of Currents 17 
 
 Tests of Boiler and Furnace 22 
 
 Useful Work, Percentage of Heat in. . . 5 
 
 Viscosity of Oil 5 
 
 Velocity, Change of. Work Required. . 5 
 
 Velocities of Piston 6 
 
 Volt 16 
 
 Work, Percentage of Heat in Useful ... 5 
 Work Required to Produce Change of 
 
 Velocity 5 
 
 Work, Stored 8 
 
 Water, Hight Pump Will Uift 10 
 
 Watt 16 
 
 Wire, Size of to Carry Given Current. .18 
 
 35