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A gasoline engine operates on the air standard Otto cycle. The air intake to the engine is at 300K and 95kPa (State 1). The a
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Answer #1

T1 = 300 K

P1 = 95 kPa

Q2-3 = 1100 kJ/kg

T3 = 2200 K

Q2-3 = Cv (T3 - T2)

1100 = 0.717 (2200 - T2)

T2 = 665.8298 K

Isentropic compression 1 - 2

P2/P1 = (T2/T1)k/k-1

P2/95 = (665.8298 / 300)1.4/1.4-1

P2 = 1547.2857 kPa

V1/V2 = (T2/T1)1/k-1

V1/V2 = (665.8298 / 300)1/1.4-1

V1/V2 = 7.33846

W1-2 = Cp (T2 - T1)

W1-2 = 1.004 (665.8298 - 300)

W1-2 = 367.2931 kJ/kg

Isentropic Expansion 3-4

V4/V3 = (T3/T4)1/k-1

7.33846 = (2200 / T4)1/1.4-1

7.338460.4 = 2200 / T4

T4 = 991.2443 K

Wnet = W3-4 - W1-2

Wnet = Cp (T3 - T4) - 367.2931

Wnet = 1.004 (2200 - 991.2443) - 367.2931

Wnet = 846.2976 kJ/kg

Eff.th = Wnet / Q2-3

Eff.th = 846.2976 / 1100

Eff.th = 0.76936 or 76.936%

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