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Ideal diesel engine CR 20: efficiency and MEP

Short answer

Given r=20, T_max=2200 K, cp=1005 J/kgK, gamma=1.4. T2=971.6 K, cutoff ratio rc=T3/T2=2.264. Thermal efficiency = 63.5%. MEP = 932 kPa.

An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95kPa and 20°C. If the maximum temperature in the cycle is not to exceed 2200K, determine a) the thermal efficiency and b) the mean effective pressure. [Take Cp = 1005J/kgK, and γ = 1.4]

Given: r = 20, P₁ = 95 kPa, T₁ = 20 + 273.15 = 293.15 K, T₃ = T_max = 2200 K, cp = 1.005 kJ/kg·K, γ = 1.4. So cv = cp/γ = 0.7179 kJ/kg·K and R = cp − cv = 0.2871 kJ/kg·K.

State temperatures

  • T₂ = T₁ r^(γ−1) = 293.15 × 20^0.4 = 293.15 × 3.3145 = 971.6 K
  • T₃ = 2200 K
  • Cut-off ratio: r_c = T₃/T₂ = 2200/971.6 = 2.264
  • T₄: v₄ = v₁, so T₄ = T₃ (r_c/r)^(γ−1) = 2200 × (2.264/20)^0.4 = 920.4 K

(a) Thermal efficiency

η = 1 − (1/r^(γ−1)) × (r_c^γ − 1)/(γ (r_c − 1))

r_c^γ = 2.264^1.4 = 3.1397

η = 1 − (1/3.3145) × (3.1397 − 1)/(1.4 × 1.264) = 63.5 %

Check by heat: q_in = cp (T₃ − T₂) = 1.005 × (2200 − 971.6) = 1234.5 kJ/kg; q_out = cv (T₄ − T₁) = 0.7179 × 627.3 = 450.3 kJ/kg; w_net = 784.2 kJ/kg; η = 784.2/1234.5 = 63.5 %.

(b) Mean effective pressure

v₁ = RT₁/P₁ = 0.2871 × 293.15/95 = 0.8861 m³/kg v₂ = v₁/20 = 0.0443 m³/kg

MEP = w_net/(v₁ − v₂) = 784.2/(0.8861 − 0.0443) = 932 kPa

Answer: η ≈ 63.5 %, MEP ≈ 932 kPa (about 9.3 bar).

✓ Answer checked by Claude Opus 5.5 (high effort)

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