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Brayton cycle pressure ratio 12: mass flow for 90 MW

Short answer

r_p=12 gives T2=300×12^0.2857=610.2 K, T4=1000/12^0.2857=491.7 K. Net work w_net = cp(T3 - T4) - cp(T2 - T1) = 199.2 kJ/kg. Mass flow rate ṁ = 90,000/199.2 = 452 kg/s. Efficiency η = 1 - 1/12^0.2857 = 50.8%.

Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature of 300K, and a turbine inlet temperature of 1000K. Determine the required mass flow rate of air for a net power output of 90MW also calculate thermal efficiency of the cycle.

Assumption: cp = 1.005 kJ/kg·K and γ = 1.4 for air (not stated in the question).

Given: r_p = 12, T₁ = 300 K, T₃ = 1000 K, net power Ẇ_net = 90 MW = 90 000 kW.

Temperatures (ideal Brayton cycle)

r_p^((γ−1)/γ) = 12^0.2857 = 2.0339

T₂ = T₁ r_p^0.2857 = 300 × 2.0339 = 610.2 K T₄ = T₃/r_p^0.2857 = 1000/2.0339 = 491.7 K

Work and heat per kg

  • Compressor work: w_c = cp (T₂ − T₁) = 1.005 × 310.2 = 311.7 kJ/kg
  • Turbine work: w_t = cp (T₃ − T₄) = 1.005 × 508.3 = 510.9 kJ/kg
  • Net work: w_net = 510.9 − 311.7 = 199.2 kJ/kg
  • Heat added: q_in = cp (T₃ − T₂) = 1.005 × 389.8 = 391.8 kJ/kg

Mass flow rate

ṁ = Ẇ_net/w_net = 90 000/199.2 = 452 kg/s

Thermal efficiency

η = w_net/q_in = 199.2/391.8 = 50.8 %

Check: η = 1 − 1/12^0.2857 = 1 − 1/2.0339 = 50.8 % ✓.

Answer: ṁ ≈ 452 kg/s; η ≈ 50.8 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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