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Air standard diesel: efficiency and mean effective pressure

Short answer

Using air-standard diesel cycle with r=16, rc=2, gamma=1.4. Heat added and rejected give net work. Thermal efficiency = 61.4%. Mean effective pressure MEP = w_net/(v1-v2) = 695 kPa.

At the beginning of a compression stroke of an air standard diesel cycle having a compression ratio of 16, the temperature is 300 K and the pressure is 100 kPa. If the cut off ratio for the cycle is 2, determine (a) the thermal efficiency (b) the mean effective pressure. [Take γ = 1.4, R = 287 J/kg.K].

Given: r = 16, r_c = 2, T₁ = 300 K, P₁ = 100 kPa, γ = 1.4, R = 0.287 kJ/kg·K. So cv = R/(γ−1) = 0.7175 and cp = γ cv = 1.0045 kJ/kg·K.

(a) Thermal efficiency

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

r^(γ−1) = 16^0.4 = 3.0314 r_c^γ = 2^1.4 = 2.6390

η = 1 − (1/3.0314) × (2.6390 − 1)/(1.4 × 1) η = 1 − 0.3299 × 1.1707 = 0.6138

η = 61.4 %

(b) Mean effective pressure

State temperatures:

  • T₂ = T₁ r^0.4 = 300 × 3.0314 = 909.4 K
  • T₃ = T₂ r_c = 1818.9 K
  • T₄ = T₁ r_c^γ = 300 × 2.6390 = 791.7 K

Heat added (2-3): q_in = cp (T₃ − T₂) = 1.0045 × 909.4 = 913.5 kJ/kg Heat rejected (4-1): q_out = cv (T₄ − T₁) = 0.7175 × 491.7 = 352.8 kJ/kg Net work: w_net = 913.5 − 352.8 = 560.7 kJ/kg (check: η q_in = 560.7)

Volumes: v₁ = RT₁/P₁ = 0.287 × 300/100 = 0.861 m³/kg v₂ = v₁/16 = 0.0538 m³/kg

MEP = w_net/(v₁ − v₂) = 560.7/(0.861 − 0.0538) = 695 kPa

Answer: η ≈ 61.4 %, MEP ≈ 695 kPa (about 6.9 bar).

✓ Answer checked by Claude Opus 5.5 (high effort)

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