lute (69.9 × 144 being the pressure per square foot, and the specific volume of the charge at B). At C the pressure is 199 x 144 the specific volume and the value of N after explosion 55.6. The temperature is, therefore, absolute. •2641 199 × 144 × ⚫2467 At D the pressure is 199 x 144, ; the temperature is therefore ·05061 absolute. •2467 ·05061 •2467 ·05061 = 2510° At E, the pressure being 44 × 144 pounds per square foot, 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: 2 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—T1) × 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,—T1) × 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 P1 v1 = P 1 V 2 from D to E. The work done will equal n-1 x weight (199 × 2641-44 × 8377)144 .2641 and 8377 being •31 of charge= equal to the total volume occupied by the charge at D and E respectively that is, the specific volume v, and v2 multiplied by 05167, the weight of the charge. This equals 7,290 foot lbs. The loss of internal energy from D to E will equal K1(T1—T2) × weight of charge=150·2(2,690—1,885) ×·05061 =6,105 foot lbs. 1 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 p1(v2 — v1)=44 × 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. V 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、 (T1— T2) × 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 × 89, 2 × 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 × 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 Total TABLE I 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. 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. ⚫00191 29. Gas used per explosion (without ignition), in lbs.. I.H.P. per hour (without ignition), in cubic feet (including ignition) TABLE II 1. Assumed temperature of charge at end of suction stroke, in degrees Fahr. 550° 89° 2. Temperature of charge at end of compression B on theoretical card . 3. Temperature on completion of constant volume addition of heat at C. 2,510° 2,049° 4. Temperature on completion of constant pressure expansion at D. 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 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 Heat taken up 8. Heat taken up by charge during compression (A to B) 11. Total heat turned into work during expansion above zero pres 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) 22. Mechanical equivalent of heat rejected in jacket water per explosion, in ft. lbs. TABLE IV Efficiencies 12.400 |