A gasoline engine takes air in at 290K, 90kPa and then compresses it. The combus

Question

A gasoline engine takes air in at 290K, 90kPa and then compresses it. The combustion adds 1000 kJ/kg of heat to the air after which the temperature is 2050K. Use constant specific heat at 300K and find: the compression ratio the specific work of compression the highest pressure in the cycle

Explanation / Answer

Air at 300K k=1.4 Cp=1.004 kJ/kgK Cv=0.707 kJ/kgK State 1: T1=290K P1=90kPa 1->2 adiabatic compression State2 : T2=? P2=? 2->3 isochoric combustion State3: T3=2050K P3=? 3->4 adiabatic expansion From 2->3 No work (V=constant) -> q = u3-u2 = Cv(T3-T2) q = 1000 kJ/kg T3=2050 K use Cv=0.707 --> T2 = 635.6 K For adiabatic compression T2/T1= (P2/P1)(k-1)/k P2/P1= (T2/T1)k/(k-1) --> P2/P1 = 15.6 --> compression ratio P2=15.6*90kPa = 1404 kPa From 1->2 No heat (adiabatic) -> 0 = u2-u1 +w --> w = Cv(T1-T2) = 0.707(290-635.6) w = -244.34 kJ/kg (done on the system) 2->3 P2V2/T2 = P3V3/T3 (V2=V3) P3 = P2(T3/T2) = 1404(2050/635.6) = 4528 kPa

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