---
title: "Ideal Brayton cycle work and efficiency"
subject: "Turbo Machines"
chapter: 3
key: "semester_7/turbo_machines:ideal-brayton-numerical"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/ideal-brayton-numerical
site: BCT Engineering
---

# Ideal Brayton cycle work and efficiency

Turbo Machines · Chapter 3: Gas Turbine

- Asked in: year not given, 6 marks
- Video: [Ideal BRAYTON CYCLE Explained in 11 Minutes!](https://www.youtube.com/watch?v=fZ80tkxWa-4) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/ideal-brayton-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-3

## Question

An ideal Brayton cycle takes in air at 100 kPa and 300 K. The pressure ratio is 8 and the maximum cycle temperature is 1200 K. The mass flow rate is 5 kg/s. Taking $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$, find (a) the temperatures at all cycle points, (b) the compressor and turbine work, (c) the net work and power, (d) the thermal efficiency and (e) the back work ratio.

## Short answer

Using isentropic relations: T2=300×8^0.2857=543.4 K, T4=1200/8^0.2857=662.5 K. Compressor work 244.7 kJ/kg, turbine work 540.2 kJ/kg, net work 295.6 kJ/kg, power 1478 kW, thermal efficiency 44.8%, back work ratio 0.453.

## Answer

$(\gamma-1)/\gamma = 0.2857$ and $r_p^{0.2857} = 8^{0.2857} = 1.8114$.

#### (a) Temperatures
$$
T_2 = T_1 r_p^{0.2857} = 300 \times 1.8114 = 543.4\ \text{K}, \qquad T_4 = \frac{T_3}{r_p^{0.2857}} = \frac{1200}{1.8114} = 662.5\ \text{K}
$$

| Point | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| T (K) | 300 | 543.4 | 1200 | 662.5 |

#### (b) Compressor and turbine work
$$
w_c = c_p (T_2 - T_1) = 1.005 \times 243.4 = 244.7\ \text{kJ/kg}
$$
$$
w_t = c_p (T_3 - T_4) = 1.005 \times 537.5 = 540.2\ \text{kJ/kg}
$$

#### (c) Net work and power
$$
w_{net} = w_t - w_c = 540.2 - 244.7 = 295.6\ \text{kJ/kg}, \qquad P = 5 \times 295.6 = 1478\ \text{kW}
$$

#### (d) Thermal efficiency
$$
q_{in} = c_p (T_3 - T_2) = 1.005 \times 656.6 = 659.8\ \text{kJ/kg}, \qquad \eta = \frac{w_{net}}{q_{in}} = 0.448
$$
Check: $1 - 1/8^{0.2857} = 0.448$.

#### (e) Back work ratio
$$
\text{BWR} = \frac{w_c}{w_t} = \frac{244.7}{540.2} = 0.453
$$

**Answer:** $T_2 = 543.4$ K, $T_4 = 662.5$ K; $w_c = 244.7$, $w_t = 540.2$, $w_{net} = 295.6$ kJ/kg; $P = 1478$ kW; $\eta = 44.8\%$; BWR = 0.453.

## Check yourself

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

1. What is the compressor exit temperature T2?
   A. 300 K
   B. 662.5 K
   C. 543.4 K
   D. 1200 K
   Answer: C. T2 = 300 x 8^0.2857 = 300 x 1.8114 = 543.4 K. (Key point: Temperatures)

2. How is T4 found?
   A. T4 = T3 x r_p^0.2857
   B. T4 = T3 / r_p^0.2857 = 662.5 K
   C. T4 = T3 - T2
   D. T4 = T1
   Answer: B. Isentropic expansion over the same pressure ratio gives T4 = 1200/1.8114 = 662.5 K. (Key point: Turbine exit)

3. What is the net specific work?
   A. 540.2 kJ/kg
   B. 244.7 kJ/kg
   C. 659.8 kJ/kg
   D. 295.6 kJ/kg
   Answer: D. w_net = 540.2 - 244.7 = 295.6 kJ/kg. (Key point: Net work)

4. The back work ratio is:
   A. wc/wt = 0.453
   B. wnet/wt
   C. wt/wc
   D. wnet/qin
   Answer: A. BWR = compressor work / turbine work = 244.7/540.2 = 0.453. (Key point: Back work ratio)

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