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= 2690°

•2467 lute (69.9 x 144 being the pressure per square foot, and

05061 the specific volume of the charge at B).

.2467 At C the pressure is 199 x 144 = the specific volume

·05061 and the value of N after explosion 55.6.

199 x 144 x 2467 The temperature is, therefore,

= 2510°

55.6 x .05061 absolute.

At D the pressure is 199 x 144, and the specific volume •2641

199 x 144 x .2641 ; the temperature is therefore 05061

55.6 x .05061 absolute. At E, the pressure being 44 x 144 pounds per square foot,

•8377 and the specific volume the temperature is

05061' 44 x 144 x .8377

=1885° absolute. 55.6 x .05061

The temperatures at a number of intermediate points have similarly been calculated, and the temperature diagram, fig. 203, plotted, in which the full line corresponds to temperatures calculated from the theoretical diagram, and the dotted line to the actual mean diagram.

Heat added.From these temperatures the units of heat added at each part of the cycle may be found as follows:

During compression from A to B the mechanical equivalent of the heat added will be equal to the increase in the internal energy of the charge during the process, i.e. K(T2-T) x weight of charge=147-5(869 — 550).05061=2,380 foot lbs.

Similarly from B to C the heat added equals 150-2(2,510869).05061=12,470 foot lbs.

From C to D=K,(T.-T,) x weight of charge=205•8(2,690 -2,510).05061=1,845 foot lbs.

During expansion the heat added will be equal to the difference between the work done and the loss of internal energy from D to E. The work done will equal


PV, P,V2 x weight

(199 •2641 – 44 x •8377)144, 2641 and ·8377 being of charger

.31 equal to the total volume occupied by the charge at D and E respectively: that is, the specific volume v, and v, multiplied by .05167, the weight of the charge. This equals 7,290 foot lbs.

The loss of internal energy from Dto E will equal K(T.-T, X weight of charge=150-2(2,690—1,885) * :05061 = 6,105 foot lbs.

During the expansion, therefore, 7,290-6,105=1,185 foot lbs. must have been added.

The total heat accounted for by the diagram is, therefore, 2,380 +12,470 +1,845 +1,185=17,880 foot lbs.

Exhaust.If from this sum we subtract the total amount of work done during the cycle, the remainder will equal the heat rejected in the exhaust. The work done during constant pressure expansion C to D equals P.(V, – v,) = 44 x 199 (-2641 – -2467)=499 foot lbs.

The work done during expansion from D to E is, as above, 7,292 foot lbs. The total work is, therefore, 7,290+499=7,789 foot lbs. Subtracting this from 17,880 foot lbs. above, the remainder (17,880 — 7,789) = 10,096 foot lbs. will represent the heat rejected in the exhaust. This

may be checked by direct calculation from the known temperatures at E and A, for the loss of internal energy in completing the cycle from E to A will evidently be the amount of heat rejected in exhaust. This will equal K, (T.-T,) * weight of charge=150-2(1,885 – 550).05061=10,140 foot lbs., which corresponds with the quantity found indirectly above well within the limits of accuracy possible in measurements of this kind.

Heat rejected in jackets.The jacket water used per explosion being 18 lb. and the rise in temperature 89.2° Fahr., the mechanical equivalent of the heat rejected (taking the mechanical equivalent of one B.T.U. at 778 ft. lbs.) will equal :18 x 89, 2 x 778=12,400 ft. lbs.

Thermal Equivalent of Gas.—The gas analysis gives as the thermal equivalent of the gas used on trial 18:900 B.T.U per pound.

Heat Account.-A heat account can therefore be constructed, and would stand thus:

Heat expended (by analysis) per explosion=:00191 lb. of gas x 18.900 x 778=28,100 ft. lbs.

Heat Accounted for by Diagrams.---(B to C)12,470+ (C to D) 1,845+ (D to E) 1,185+ (heat rejected in jackets)12,400=27,900. Heat Accounted for on trial :

Mechanical equivalent of indicator diagrams 5,140
Heat rejected in jacket water

12,400 Heat rejected in exhaust .

10,140 Total

. 27,680

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Particulars of Engine 1. Diameter of cylinder, in inches 2. Stroke, in inches 3. Diameter of flywheel, in feet 4. Effective circumference of flywheel, in feet 5. Duration of trial, in minutes .

18 5.5 17.458 40


Indicated Horse-power 6. Cylinder constant, per lb., mean pressure, per explosion 7. Number of indicator diagrams taken 8. Average mean pressure during working stroke, in lbs. 9. Average mean pressure during pumping stroke, in lbs. 10. Nett average pressure 11. Total explosions 12. Explosions per minute 13. I.H.P., gross, from lines 6, 8, and 12 14. I.H.P., nett, from lines 6, 10, and 12


Brake Horse-power 15. Wheel, constant, per lb. per revolution 16. Load on brake, gross lbs. 17. Load on spring balance, mean lbs. . 18. Load, nett average, in lbs 19. Total revolutions 20. Revolutions per minute . 21. Brake horse-power lines 15, 18, and 20.



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Gas Consumption 22. Gas total, in cubic feet (without ignition) 23. Gas per explosion . 24. Gas ignition, total, in cubic feet 25. Pressure of gas at meter, in inches of water 26. Pressure, in lbs. per sq. inch . 27. Temperature of gas at meter, Fahr. 28. Specific volume of gas, in cubic feet per lb. 29. Gas used per explosion (without ignition), in lbs. 30.

I.H.P. per hour (without ignition), in cubic feet 31.

(including ignition), in cubic feet 32.


(without ignition) 33.

(including ignition).

201 ·0593 5.8 1.85 14.68 51° 30.95 ·00191 22:8 23.5 26•2

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1. Assumed temperature of charge at end of suction

stroke, in degrees Fahr. 2. Temperature of charge at end of compression B on

theoretical card . 3. Temperature on completion of constant volume

addition of heat at C. 4. Temperature on completion of constant pressure

expansion at D 5. Temperature at exhaust E

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Data for Heat Account


1. Gas consumed per explosion, in lbs. (Table 1, line 29)
2. Calorific value of ditto at 18,900 B.T.U. per lb. (Table

3. Mechanical equivalent of ditto, in ft. lbs.
4. Work done by charge, calculated, gross (page 7)
5. Work done on charge during compression (calculated)
6. Nett work in ideal process, A, B, C, D, E
7. Nett work, mean, of all indicator diagrams

36.1 B.T.U
7,789 ft. lbs.

Heat taken up

8. Heat taken up by charge during compression (A to B) 9.

at constant volume (B to C) 10.

at constant pressure (C to D)

2,380 ft. lbs.

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11. Total heat turned into work during expansion above zero pres

sure (page 6) 12. Loss of internal energy during expansion (page 7) 13. Difference (11–12) = heat added during expansion 14. Total heat added during cycle lines 8, 9, 10, and 13 15. Total work done above zero pressure (line 4). 16. Difference (14—15) = heat rejected in exhaust 17. Heat rejected in exhaust, direct calculation (page 240)

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Jacket Water

18. Jacket water used per explosion, in lbs. .
19. In-going temperature of jacket water, Fahr.
20. Out-going
21. Rise of
22. Mechanical equivalent of heat rejected in jacket water per ex-

plosion, in ft. lbs.





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Heat Account

Ft. lbs.
Work done.
Effective work

(B.H.P.) per

explosion 4,467 15.9
Do.lostin fric-

673 2:4
Shown on dia.

grams, I.H.P. 5,140 18:3
Heat rejected.
Jacket water . 12,400 44:1
Exhaust. . 10,140 36:1
Radiation and

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otherwise un-
accounted for 420 1.5


28.100 100.0

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