--- title: "Ideal Brayton: efficiency and net power, pressure ratio 10" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 6 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brayton-5m3s-pr10-1400k" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-5m3s-pr10-1400k site: BCT Engineering --- # Ideal Brayton: efficiency and net power, pressure ratio 10 Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles - Asked in: 2067 Mangsir, 8 marks - Video: [MECH351: Brayton cycle efficiency](https://www.youtube.com/watch?v=jeLkVhKrovw) — Lyes Kadem - 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-5m3s-pr10-1400k - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6 ## Question 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] ## 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). ## Answer **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 ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the mass flow rate found from the volumetric flow? A. m = V1 x T1 B. m = P1 / (R T1) x 100 C. m = P1 V1 / (R T1) = 5.807 kg/s D. m = V / cp Answer: C. Using the ideal gas law at the compressor inlet: 100 x 5 / (0.287 x 300) = 5.807 kg/s. (Key point: Mass flow from gas law) 2. Thermal efficiency for r_p = 10 is A. 10^0.2857 - 1 B. 1 - 1/10^0.2857 = 48.2 % C. 1 - 300/1400 D. 1 - 10^0.2857 Answer: B. η = 1 - 1/1.9307 = 48.2 %, independent of T3. (Key point: Thermal efficiency) 3. The net work per kg is A. Turbine work 678.2 - compressor work 280.6 = 397.6 kJ/kg B. cp (T3 - T2) = 828 kJ/kg C. 280.6 - 678.2 kJ/kg D. 678.2 + 280.6 kJ/kg Answer: A. w_net = w_t - w_c = 397.6 kJ/kg. (Key point: Net work per kg) 4. The net power developed is A. 5 x 397.6 = 1988 kW B. 397.6 kW C. 5.807 x 397.6 = 2309 kW D. 5.807 x 678.2 = 3938 kW Answer: C. W_net = m w_net = 2309 kW. (Key point: Net power) ## Related in Fundamental of Thermodynamics & Heat Transfer - [Brayton cycle pressure ratio 12: mass flow for 90 MW](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-pressure-ratio-12-90mw) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Brayton cycle 80 MW: mass flow, efficiency, compressor fraction](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-80mw-300k-1500k-fraction) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Ideal Brayton cycle: net work per kg and efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-work-per-kg-efficiency) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Ideal gas turbine 100 kW: pressure ratio, work, efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aideal-gas-turbine-100kw-pressure-ratio) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Air standard Brayton: state points, work, efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-0-18mpa-34c-max-2350) — Fundamental of Thermodynamics & Heat Transfer, chapter 6 - [Ideal Brayton cycle with 30 MW heat input](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-30mw-efficiency-power) — Fundamental of Thermodynamics & Heat Transfer, chapter 6, asked 2× ## Related in other subjects - [Ideal Brayton cycle work and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aideal-brayton-numerical) — Turbo Machines, chapter 3 - [Brayton cycle efficiency and optimum pressure ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Abrayton-cycle-efficiency-optimum-pressure-ratio) — Turbo Machines, chapter 3 - [Ideal gas turbine with regenerator at 1400 K](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-ideal-regenerator-1400k) — Power Plant Equipment, chapter 6 - [Gas turbine 5:1: power, efficiency, work ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-5-1-work-ratio) — Power Plant Equipment, chapter 6