- 2070 Magh · 8 marks
Otto cycle CR 8: state points and efficiency
Short answer
Isentropic compression with r=8 gives T2=673.5 K, P2=1838 kPa. Constant-volume heat addition of 2000 kJ/kg yields T3=3459 K, P3=9440 kPa. Expansion to T4=1506 K, P4=514 kPa. Efficiency η = 1 - 1/8^0.4 = 56.5%.
A compression ratio of an air standard otto cycle is 8. At the beginning of the compression process, the pressure and temperature of air are 100 kPa and 20°C respectively. The heat added per kg of air during the cycle is 2000 kJ/kg. Determine: a) The pressure and temperature at the end of each process of the cycle. b) The thermal efficiency. [Take Cv = 718 J/kg.K, γ = 1.4]
Given: r = 8, P₁ = 100 kPa, T₁ = 20 + 273.15 = 293.15 K, heat added q_in = 2000 kJ/kg (in process 2-3), cv = 0.718 kJ/kg·K, γ = 1.4.
(a) Pressure and temperature at each state
r^(γ−1) = 8^0.4 = 2.2974; r^γ = 8^1.4 = 18.3792
State 1: P₁ = 100 kPa, T₁ = 293.15 K
State 2 (after isentropic compression): T₂ = T₁ r^0.4 = 293.15 × 2.2974 = 673.5 K P₂ = P₁ r^1.4 = 100 × 18.3792 = 1838 kPa
State 3 (after constant-volume heat addition): q_in = cv (T₃ − T₂), so T₃ = T₂ + 2000/0.718 = 673.5 + 2785.5 = 3459.0 K P₃ = P₂ T₃/T₂ = 1838 × 3459.0/673.5 = 9440 kPa
State 4 (after isentropic expansion): T₄ = T₃/r^0.4 = 3459.0/2.2974 = 1505.6 K P₄ = P₃/r^1.4 = 9440/18.3792 = 513.6 kPa
| State | P (kPa) | T (K) |
|---|---|---|
| 1 | 100 | 293.1 |
| 2 | 1838 | 673.5 |
| 3 | 9440 | 3459.0 |
| 4 | 514 | 1505.6 |
(b) Thermal efficiency
q_out = cv (T₄ − T₁) = 0.718 × (1505.6 − 293.1) = 870.6 kJ/kg w_net = 2000 − 870.6 = 1129.4 kJ/kg
η = w_net/q_in = 1129.4/2000 = 0.5647
Check: η = 1 − 1/r^(γ−1) = 1 − 1/2.2974 = 0.5647.
Answer: η = 56.5 %.
✓ Answer checked by Claude Opus 5.5 (high effort)
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