--- title: "Ideal Brayton cycle with 30 MW heat input" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 6 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brayton-30mw-efficiency-power" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-30mw-efficiency-power site: BCT Engineering --- # Ideal Brayton cycle with 30 MW heat input Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles - Asked in: 2073 Magh, 8 marks; 2072 Magh, 8 marks (asked 2 times) - Video: [Simple Cycle Gas Turbine Efficiency and Net Power Output](https://www.youtube.com/watch?v=nbQkTBRAX1o) — Engineers Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-30mw-efficiency-power - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6 ## Question 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] ## 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. ## Answer **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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is T2 found? A. T2 = T1 + r_p B. T2 = T3/r_p^((g-1)/g) C. T2 = T1 x r_p^((g-1)/g) D. T2 = T1/r_p Answer: C. Isentropic compression gives T2 = T1 r_p^0.2857 = 579.2 K. (Key point: Temperature relation) 2. What does the ideal Brayton efficiency depend on? A. Heat input rate B. Maximum temperature only C. Mass flow rate D. Only the pressure ratio and gamma Answer: D. eta = 1 - 1/r_p^((g-1)/g) = 48.2% for r_p = 10. (Key point: Efficiency) 3. How is mass flow obtained? A. Q_in rate divided by w_net B. Q_in rate divided by cp(T4 - T1) C. Q_in rate divided by q_in = cp(T3 - T2) D. Q_in rate times cp Answer: C. m = 30 000/623.9 = 48.1 kg/s. (Key point: Mass flow from heat rate) 4. What is the net power output? A. About 29 MW B. About 14.5 MW (eta x 30 MW) C. About 48 MW D. About 30 MW Answer: B. W = eta x Q_in = 0.482 x 30 = about 14.5 MW. 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