--- title: "Brayton cycle 80 MW: mass flow, efficiency, compressor fraction" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 6 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brayton-80mw-300k-1500k-fraction" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-80mw-300k-1500k-fraction site: BCT Engineering --- # Brayton cycle 80 MW: mass flow, efficiency, compressor fraction Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles - Asked in: 2068 Magh, 8 marks - Video: [Example 9.1: Calculating the power output from a gas turbine unit working on Open Brayton's cycle](https://www.youtube.com/watch?v=w_smCBy-iYc) — Dr. Arshad Ali Khan Official - 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-80mw-300k-1500k-fraction - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6 ## Question A power plant operating on an ideal Brayton cycle delivers a power output of 80 MW. The minimum and maximum temperatures during the cycle are 300 K and 1500 K respectively. The pressure at the compressor inlet and outlet are 100 kPa and 1400 kPa respectively. (a) Determine the mass flow rate of air. (b) Determine the thermal efficiency of the cycle. (c) Determine the power output from the turbine. (d) What fraction of the turbine power output is required to drive the compressor? [R = 287 J/kg K, Cp = 1005 J/kg K] ## Short answer Pressure ratio 14, r_p^((γ−1)/γ)=2.1247; T2=637.4 K, T4=706.0 K. Net work per kg = 458.9 kJ/kg. Mass flow = 80 000 kW / 458.9 = 174.3 kg/s. Efficiency = 52.9%. Turbine power = 139.1 MW. Fraction of turbine power to compressor = wc/wt = 0.425 (42.5%). ## Answer **Given:** Ẇ_net = 80 MW, T₁ = 300 K, T₃ = 1500 K, P₁ = 100 kPa, P₂ = 1400 kPa, R = 0.287, cp = 1.005 kJ/kg·K. Here γ = cp/(cp − R) = 1.4, so (γ−1)/γ = R/cp = 0.2856. r_p = 1400/100 = 14; r_p^0.2856 = 2.1247 #### Temperatures T₂ = T₁ r_p^0.2856 = 300 × 2.1247 = 637.4 K T₄ = T₃/r_p^0.2856 = 1500/2.1247 = 706.0 K #### Specific work and heat (per kg) - Compressor: w_c = cp (T₂ − T₁) = 1.005 × 337.4 = 339.1 kJ/kg - Turbine: w_t = cp (T₃ − T₄) = 1.005 × 794.0 = 798.0 kJ/kg - Net: w_net = 798.0 − 339.1 = 458.9 kJ/kg - Heat input: q_in = cp (T₃ − T₂) = 1.005 × 862.6 = 866.9 kJ/kg #### (a) Mass flow rate ṁ = Ẇ_net/w_net = 80 000/458.9 = **174.3 kg/s** #### (b) Thermal efficiency η = w_net/q_in = 458.9/866.9 = **52.9 %** (= 1 − 1/2.1247) #### (c) Turbine power Ẇ_t = ṁ w_t = 174.3 × 798.0 = 139117 kW = **139.1 MW** #### (d) Fraction of turbine work used by the compressor Back-work ratio = w_c/w_t = 339.1/798.0 = **0.425 (about 42.5 %)** The compressor takes about 42.5 % of the turbine output; the remaining 57.5 % is the net output. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With cp = 1005 and R = 287 J/kg K, (γ-1)/γ equals A. R x cp B. cp/R = 3.5 C. 1/γ = 0.714 D. R/cp = 0.2856 Answer: D. (γ-1)/γ = R/cp = 0.2856, giving r_p^0.2856 = 2.1247 for r_p = 14. (Key point: Pressure ratio exponent) 2. The mass flow rate for 80 MW net power is A. 80000/866.9 = 92 kg/s B. 80000/798 = 100 kg/s C. 80000/339.1 = 236 kg/s D. 80000/458.9 = 174.3 kg/s Answer: D. m = W_net / w_net with w_net = 458.9 kJ/kg. (Key point: Mass flow rate) 3. Turbine power output is A. m x w_t = 174.3 x 798 = 139.1 MW B. Equal to 80 MW C. m x w_net D. m x q_in Answer: A. The turbine produces the gross work, larger than the net output. (Key point: Turbine power) 4. What fraction of turbine work drives the compressor? A. w_net/w_t = 0.575 B. w_c/w_t = 0.425 C. q_in/w_t = 1.09 D. w_c/w_net = 0.74 Answer: B. 339.1/798.0 = 0.425, so about 42.5 % of turbine work goes to the compressor. 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