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Air standard Brayton: state points, work, efficiency

Short answer

Pressure ratio r_p = 2.3/0.18 = 12.78. Temperatures: T1=307.15 K, T2=636.0 K, T3=2623.15 K, T4=1266.8 K; pressures P1=P4=0.18 MPa, P2=P3=2.3 MPa. Compressor work 330.5 kJ/kg, turbine work 1363.2 kJ/kg, thermal efficiency 51.7%.

In an air standard Brayton cycle the air enters the compressor at 0.18 MPa, 34°C. The pressure leaving the compressor is 2.3 MPa, and the maximum temperature in the cycle is 2350°C. Determine: a) The pressure and temperature at each point cycle b) The compressor work, turbine work, and cycle efficiency [Take Cp = 1005 J/kgK, γ = 1.4]

Given: P₁ = 0.18 MPa, T₁ = 34 + 273.15 = 307.15 K, P₂ = 2.3 MPa, T_max = T₃ = 2350 + 273.15 = 2623.15 K, cp = 1.005 kJ/kg·K, γ = 1.4. (The ideal Brayton cycle has no pressure drops: P₃ = P₂, P₄ = P₁.)

(a) Pressure and temperature at each point

Pressure ratio r_p = 2.3/0.18 = 12.78; r_p^((γ−1)/γ) = 12.78^0.2857 = 2.0708

StatePressureTemperature
1 (compressor inlet)0.18 MPa307.15 K (34 °C)
2 (compressor exit)2.3 MPa636.0 K (362.9 °C)
3 (turbine inlet)2.3 MPa2623.15 K (2350 °C)
4 (turbine exit)0.18 MPa1266.8 K (993.6 °C)

with T₂ = T₁ r_p^0.2857 = 307.15 × 2.0708 = 636.0 K and T₄ = T₃/r_p^0.2857 = 2623.15/2.0708 = 1266.8 K.

(b) Work and efficiency (per kg of air)

  • Compressor work: w_c = cp (T₂ − T₁) = 1.005 × (636.0 − 307.15) = 330.5 kJ/kg
  • Turbine work: w_t = cp (T₃ − T₄) = 1.005 × (2623.15 − 1266.8) = 1363.2 kJ/kg
  • Net work: w_net = 1363.2 − 330.5 = 1032.6 kJ/kg
  • Heat added: q_in = cp (T₃ − T₂) = 1.005 × (2623.15 − 636.0) = 1997.1 kJ/kg

η = w_net/q_in = 1032.6/1997.1 = 51.7 %

Check: η = 1 − 1/r_p^0.2857 = 1 − 1/2.0708 = 51.7 % ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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