---
title: "Ideal Brayton cycle: net work per kg and efficiency"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brayton-work-per-kg-efficiency"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-work-per-kg-efficiency
site: BCT Engineering
---

# Ideal Brayton cycle: net work per kg and efficiency

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

- Asked in: 2075 Baisakh, 8 marks
- Video: [Brayton Cycle Problem 9-88 Solved | Net Work & Thermal Efficiency | Cengel Thermodynamics 9th Ed](https://www.youtube.com/watch?v=aw6rrmjCIU0) — 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-work-per-kg-efficiency
- 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 300 K and the turbine at 1000 kPa and 1200 K. Determine the net work per kg of air and the cycle efficiency. [Take γ = 1.4 and cp = 1.005 kJ/kg.K]

## Short answer

Isentropic relations yield T₂ = 579.2 K, T₄ = 621.5 K for r_p=10. Net work per kg = 1.005×[(1200−621.5)−(579.2−300)] = 300.7 kJ/kg. Cycle efficiency = 1 − 1/10^0.2857 = 48.2%.

## Answer

**Given:** T₁ = 300 K, P₁ = 100 kPa; P₂ = P₃ = 1000 kPa; T₃ = 1200 K; γ = 1.4, cp = 1.005 kJ/kg·K.

#### Temperatures
r_p = 1000/100 = 10; (γ−1)/γ = 0.2857; r_p^0.2857 = 1.9307

T₂ = 300 × 1.9307 = 579.2 K
T₄ = 1200/1.9307 = 621.5 K

#### Work and heat per kg
- Compressor work: w_c = cp (T₂ − T₁) = 1.005 × (579.2 − 300) = 280.6 kJ/kg
- Turbine work: w_t = cp (T₃ − T₄) = 1.005 × (1200 − 621.5) = 581.4 kJ/kg
- Net work: w_net = 581.4 − 280.6 = **300.7 kJ/kg**
- Heat added: q_in = cp (T₃ − T₂) = 1.005 × (1200 − 579.2) = 623.9 kJ/kg

#### Efficiency
η = w_net/q_in = 300.7/623.9 = 0.482

Check: η = 1 − 1/r_p^0.2857 = 1 − 1/1.9307 = 0.482 ✓.

**Answer: net work = 300.7 kJ/kg of air; cycle efficiency = 48.2 %.**

## Check yourself

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

1. How is net work per kg found?
   A. cp(T3 - T2) only
   B. Turbine work plus compressor work
   C. cp(T4 - T1)
   D. Turbine work minus compressor work
   Answer: D. w_net = cp(T3 - T4) - cp(T2 - T1) = 581.4 - 280.6 = 300.7 kJ/kg. (Key point: Net work formula)

2. T4 is found from:
   A. T1 x r_p
   B. T3 x r_p^0.2857
   C. T2 / r_p
   D. T3 / r_p^0.2857
   Answer: D. Isentropic expansion: T4 = 1200/1.9307 = 621.5 K. (Key point: Turbine exit temperature)

3. Efficiency is:
   A. w_net/q_in = 48.2%
   B. q_out/q_in
   C. w_net/q_out
   D. w_net/w_turbine = 48.2%
   Answer: A. eta = 300.7/623.9 = 0.482, matching 1 - 1/r_p^0.2857. (Key point: Efficiency)

4. Heat added per kg is:
   A. cp(T3 - T2)
   B. cp(T3 - T4)
   C. cp(T3 - T1)
   D. cp(T4 - T1)
   Answer: A. Heat is added in 2-3 at constant pressure: 623.9 kJ/kg. (Key point: Heat added)

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