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Ideal Brayton: efficiency and net power, pressure ratio 10

Short answer

Pressure ratio 10, r_p^((γ−1)/γ)=1.9307; T2=579.2 K, T4=725.1 K. η=1−1/1.9307=48.2%. Mass flow = P1V1/(RT1) = 5.807 kg/s. Net work = cp(T3−T4)−cp(T2−T1)=397.6 kJ/kg. Net power = 5.807×397.6 = 2309 kW (2.31 MW).

Air enters the compressor of an ideal air standard Brayton cycle at 100kpa, 300k, with a volumetric flow rate of 5m³/s. The compressor pressure ratio is 10. The turbine inlet temperature is 1400k. Determine: a) The thermal efficiency of the cycle b) The net power developed, in kW. [Take R = 287J/kg K, cp = 1005J/kg K, γ = 1.4]

Given: P₁ = 100 kPa, T₁ = 300 K, V̇₁ = 5 m³/s, r_p = 10, T₃ = 1400 K, R = 0.287, cp = 1.005 kJ/kg·K, γ = 1.4.

Temperatures

r_p^((γ−1)/γ) = 10^0.2857 = 1.9307

T₂ = 300 × 1.9307 = 579.2 K T₄ = 1400/1.9307 = 725.1 K

(a) Thermal efficiency

η = 1 − 1/r_p^0.2857 = 1 − 1/1.9307 = 48.2 %

(b) Net power

Mass flow rate from the gas law: ṁ = P₁V̇₁/(RT₁) = 100 × 5/(0.287 × 300) = 5.807 kg/s

  • Compressor work: w_c = cp (T₂ − T₁) = 1.005 × 279.2 = 280.6 kJ/kg
  • Turbine work: w_t = cp (T₃ − T₄) = 1.005 × 674.9 = 678.2 kJ/kg
  • Net work: w_net = 678.2 − 280.6 = 397.6 kJ/kg

Ẇ_net = ṁ w_net = 5.807 × 397.6 = 2309 kW (about 2.31 MW)

Check: Q̇_in = ṁ cp (T₃ − T₂) = 5.807 × 824.9 = 4790 kW; η = 2309/4790 = 0.482 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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