---
title: "Brayton cycle pressure ratio 12: mass flow for 90 MW"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brayton-pressure-ratio-12-90mw"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-pressure-ratio-12-90mw
site: BCT Engineering
---

# Brayton cycle pressure ratio 12: mass flow for 90 MW

Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles

- Asked in: 2069 Bhadra, 8 marks
- Video: [Brayton Cycle Problem 9-86 Solved | Mass Flow Rate for 32 MW Output | Cengel Thermodynamics 9th Ed](https://www.youtube.com/watch?v=hUAYsUK8shg) — Education Shop
- 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-pressure-ratio-12-90mw
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature of 300K, and a turbine inlet temperature of 1000K. Determine the required mass flow rate of air for a net power output of 90MW also calculate thermal efficiency of the cycle.

## Short answer

r_p=12 gives T2=300×12^0.2857=610.2 K, T4=1000/12^0.2857=491.7 K. Net work w_net = cp(T3 - T4) - cp(T2 - T1) = 199.2 kJ/kg. Mass flow rate ṁ = 90,000/199.2 = 452 kg/s. Efficiency η = 1 - 1/12^0.2857 = 50.8%.

## Answer

**Assumption:** cp = 1.005 kJ/kg·K and γ = 1.4 for air (not stated in the question).

**Given:** r_p = 12, T₁ = 300 K, T₃ = 1000 K, net power Ẇ_net = 90 MW = 90 000 kW.

#### Temperatures (ideal Brayton cycle)
r_p^((γ−1)/γ) = 12^0.2857 = 2.0339

T₂ = T₁ r_p^0.2857 = 300 × 2.0339 = 610.2 K
T₄ = T₃/r_p^0.2857 = 1000/2.0339 = 491.7 K

#### Work and heat per kg
- Compressor work: w_c = cp (T₂ − T₁) = 1.005 × 310.2 = 311.7 kJ/kg
- Turbine work: w_t = cp (T₃ − T₄) = 1.005 × 508.3 = 510.9 kJ/kg
- Net work: w_net = 510.9 − 311.7 = 199.2 kJ/kg
- Heat added: q_in = cp (T₃ − T₂) = 1.005 × 389.8 = 391.8 kJ/kg

#### Mass flow rate
ṁ = Ẇ_net/w_net = 90 000/199.2 = **452 kg/s**

#### Thermal efficiency
η = w_net/q_in = 199.2/391.8 = **50.8 %**

Check: η = 1 − 1/12^0.2857 = 1 − 1/2.0339 = 50.8 % ✓.

**Answer: ṁ ≈ 452 kg/s; η ≈ 50.8 %.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is T2 found after the compressor in the ideal Brayton cycle?
   A. T2 = T1 / r_p^((γ-1)/γ)
   B. T2 = T1 x r_p^((γ-1)/γ)
   C. T2 = T3 x r_p^((γ-1)/γ)
   D. T2 = T1 x r_p
   Answer: B. Isentropic compression gives T2/T1 = r_p^((γ-1)/γ) = 2.0339, so T2 = 610.2 K. (Key point: Isentropic temperature)

2. How is net work per kg obtained?
   A. Heat added plus heat rejected
   B. Turbine work plus compressor work
   C. Turbine work minus compressor work
   D. Compressor work minus turbine work
   Answer: C. The turbine produces work and the compressor consumes it, so w_net = 510.9 - 311.7 = 199.2 kJ/kg. (Key point: Net work)

3. What is the required mass flow rate for 90 MW?
   A. 290 kg/s
   B. 452 kg/s
   C. 199 kg/s
   D. 90 kg/s
   Answer: B. m = W_net / w_net = 90000 / 199.2 = 452 kg/s. (Key point: Mass flow rate)

4. Which expression gives the Brayton thermal efficiency directly?
   A. 1 - r_p^((γ-1)/γ)
   B. 1 - T3/T1
   C. 1 - 1/r_p^((γ-1)/γ)
   D. w_c / w_t
   Answer: C. η = 1 - 1/r_p^((γ-1)/γ) = 1 - 1/2.0339 = 50.8 %. (Key point: Efficiency shortcut)

## Related in Fundamental of Thermodynamics & Heat Transfer

- [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: efficiency and net power, pressure ratio 10](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-5m3s-pr10-1400k) — 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×
- [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
- [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
- [Adiabatic air turbine: mass flow and areas](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aturbine-1000k-to-400k-areas) — Fundamental of Thermodynamics & Heat Transfer, chapter 4, 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
- [Two-shaft gas turbine, pressure ratio 7:1](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-two-shaft-7-1) — Power Plant Equipment, chapter 6, asked 2×
- [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