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Ideal Brayton cycle with 30 MW heat input

Short answer

Pressure ratio 10 gives T₂=579.2 K, T₄=621.5 K. Efficiency η = 1 − 1/10^0.2857 = 48.2%. Heat addition q_in = 623.9 kJ/kg, mass flow = 48.1 kg/s; net power output = 14.5 MW.

In an Ideal Brayton cycle, air enters the compressor at 100 kPa and 300K and the turbine at 1000 kPa and 1200K. Heat is transferred to the air at a rate of 30 MW. Determine the efficiency and Power output of the plant. [Take Cp = 1005 J/kg K, and γ = 1.4]

Given: T₁ = 300 K, P₁ = 100 kPa; P₂ = P₃ = 1000 kPa, T₃ = 1200 K; heat input Q̇_in = 30 MW; cp = 1.005 kJ/kg·K, γ = 1.4. Ideal Brayton cycle: 1-2 isentropic compression, 2-3 constant-pressure heat addition, 3-4 isentropic expansion, 4-1 constant-pressure heat rejection.

Temperatures

Pressure ratio r_p = 1000/100 = 10, and (γ−1)/γ = 0.2857.

r_p^0.2857 = 10^0.2857 = 1.9307

T₂ = T₁ r_p^0.2857 = 300 × 1.9307 = 579.2 K T₄ = T₃/r_p^0.2857 = 1200/1.9307 = 621.5 K

Efficiency

η = 1 − 1/r_p^((γ−1)/γ) = 1 − 1/1.9307 = 0.482

η = 48.2 %

Work and mass flow

q_in = cp (T₃ − T₂) = 1.005 × (1200 − 579.2) = 623.9 kJ/kg w_comp = cp (T₂ − T₁) = 1.005 × 279.2 = 280.6 kJ/kg w_turb = cp (T₃ − T₄) = 1.005 × 578.5 = 581.4 kJ/kg w_net = 581.4 − 280.6 = 300.7 kJ/kg (check: η = 300.7/623.9 = 0.482)

ṁ = Q̇_in/q_in = 30 000/623.9 = 48.1 kg/s

Power output

Ẇ_net = ṁ w_net = 48.1 × 300.7 = 14 460 kW (or η Q̇_in = 0.482 × 30)

Answer: efficiency = 48.2 %, net power output ≈ 14.5 MW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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