---------- answer the above q and down q is exmple for it During the power strok
ID: 2156685 • Letter: #
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answer the above q and down q is exmple for it
During the power stroke in a four-stroke auto-mobile engine, the piston is forced down as the mixture of gas and air undergoes a reversible adiabatic expansion. Find the average power generated during the expansion by assuming the engine is running at = 1500 rpm, the gauge pressure right before the expansion is P gauge = 15 atm, the volumes of the mixture right before and after the expansion are Vi = 57 cm3 and Vf = 410 cm3, respectively, the time involved in the expansion is one-fourth that of the total cycle, and the mixture behaves like an ideal diatomic gas. Answer in units of kW During the power stroke in a four-stroke auto-mobile engine, the piston is forced down as the mixture of gas and air undergoes a reversible adiabatic expansion. Find the average power generated during the expansion by assuming the engine is running at = 1900 rpm, the gauge pressure right before the expansion is P gauge = 15 atm, the volumes of the mixture right before and after the expansion are Vi = 55 cm3 and Vf = 420 cm , respectively, the time involved in the expansion is one-fourth that of the total cycle, and the mixture behaves like an ideal diatomic gas. Correct answer: 15. 7108 kW. Let: P gauge = 15 atm, Vi = 55 cm3, and Vf = 420 cm3. We suppose the air plus gasoline behaves like a diatomic ideal gas. From the equation for the adiabatic process for an ideal gas we find its final pressure PiVi gamma = PfVf gamma , so we have Pf = Pi(Vi / Vf) gamma = (16 atm)(55 cm3 / 420 cm3)1. 4 = 0. 929135 atm, where Pi = P gauge + Patmospheric = (15 atm) + (1 atm) = 16 atm. Since in an adiabatic process we have Q = 0. work done by the gas equals the decrease of its internal energy W = - Delta U = -nCv(Tf - Ti), or exploiting the equation of state for an ideal gas W = 5 / 2[-PfVf + PiVi] = 5 / 2 [-(39. 531 J) + (89. 144 J)] = 124. 033 J. The time for this stroke is Finally, power equals P = W / t = 15. 7108 KW.Explanation / Answer
P1 = 15+1 = 16 atm
P2/P1 = (V1/V2)^
P2/16 = (57/410)^1.4
P2 = 1.01 atm
W = 5/2*[P1V1 - P2V2] = 5/2*[16*57 - 1.01*410]*(101325*10^-6) = 126 J
Time for stroke t = 1/4*(1/1500)*60 = 0.01 s
Power = W/t = 126/0.01 = 12.6 kW
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