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Ideal Brayton cycle: net work per kg and efficiency

Short answer

Isentropic relations yield T₂ = 579.2 K, T₄ = 621.5 K for r_p=10. Net work per kg = 1.005×[(1200−621.5)−(579.2−300)] = 300.7 kJ/kg. Cycle efficiency = 1 − 1/10^0.2857 = 48.2%.

In an ideal Brayton cycle, air enters the compressor at 100 kPa and 300 K and the turbine at 1000 kPa and 1200 K. Determine the net work per kg of air and the cycle efficiency. [Take γ = 1.4 and cp = 1.005 kJ/kg.K]

Given: T₁ = 300 K, P₁ = 100 kPa; P₂ = P₃ = 1000 kPa; T₃ = 1200 K; γ = 1.4, cp = 1.005 kJ/kg·K.

Temperatures

r_p = 1000/100 = 10; (γ−1)/γ = 0.2857; r_p^0.2857 = 1.9307

T₂ = 300 × 1.9307 = 579.2 K T₄ = 1200/1.9307 = 621.5 K

Work and heat per kg

  • Compressor work: w_c = cp (T₂ − T₁) = 1.005 × (579.2 − 300) = 280.6 kJ/kg
  • Turbine work: w_t = cp (T₃ − T₄) = 1.005 × (1200 − 621.5) = 581.4 kJ/kg
  • Net work: w_net = 581.4 − 280.6 = 300.7 kJ/kg
  • Heat added: q_in = cp (T₃ − T₂) = 1.005 × (1200 − 579.2) = 623.9 kJ/kg

Efficiency

η = w_net/q_in = 300.7/623.9 = 0.482

Check: η = 1 − 1/r_p^0.2857 = 1 − 1/1.9307 = 0.482 ✓.

Answer: net work = 300.7 kJ/kg of air; cycle efficiency = 48.2 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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