Correct : 4500
Given:
Displacement volume Vd = 250 cm³
Clearance volume Vc = 35.7 cm³
P1 = 100 kPa, T1 = 300 K
Qin = 800 kJ/kg, Cv = 0.718 kJ/kg.K, γ = 1.4
Compression Ratio (r):
V1 = Vd + Vc = 250 + 35.7 = 285.7 cm³
V2 = Vc = 35.7 cm³
r = V1 ÷ V2 = 285.7 ÷ 35.7 = 8.0
Temperature after isentropic compression (T2):
T2 = T1 × rγ−1
T2 = 300 × (8)0.4
(8)0.4 = 2.2974
T2 = 300 × 2.2974 = 689.2 K
Pressure after isentropic compression (P2):
P2 = P1 × rγ
P2 = 100 × (8)1.4
(8)1.4 = 18.379
P2 = 100 × 18.379 = 1837.9 kPa
Temperature after constant volume heat addition (T3):
Qin = Cv × (T3 − T2)
800 = 0.718 × (T3 − 689.2)
T3 − 689.2 = 800 ÷ 0.718 = 1114.2
T3 = 689.2 + 1114.2 = 1803.4 K
Maximum Pressure (P3):
Since heat addition is at constant volume, V2 = V3
P3 ÷ P2 = T3 ÷ T2
P3 = P2 × (T3 ÷ T2)
P3 = 1837.9 × (1803.4 ÷ 689.2)
P3 = 1837.9 × 2.616
P3 = ≈ 4500 kPa
∴ The maximum pressure in the cycle = 4500 kPa
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