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

# Brayton cycle efficiency and optimum pressure ratio

Turbo Machines · Chapter 3: Gas Turbine

- Asked in: year not given, 8 marks
- Video: [Gas turbine derivation - Optimum pressure ratio for specific work output](https://www.youtube.com/watch?v=YHLdTzNfb9I) — CHINMAY ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/brayton-cycle-efficiency-optimum-pressure-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-3

## Question

Draw the p-v and T-s diagrams of the ideal Brayton cycle and derive an expression for its thermal efficiency in terms of the pressure ratio. Also derive the pressure ratio for maximum specific net work output for given maximum and minimum temperatures.

## Short answer

For ideal Brayton cycle, thermal efficiency is η = 1 - 1/r_p^((γ-1)/γ). The pressure ratio for maximum specific net work at given T_max and T_min is r_p,opt = (T_max/T_min)^(γ/[2(γ-1)]).

## Answer

#### Ideal Brayton cycle
Processes: 1-2 isentropic compression; 2-3 constant-pressure heat addition; 3-4 isentropic expansion; 4-1 constant-pressure heat rejection.

```text
 T-s diagram                      p-v diagram
 T                                p
 |             3                  |  2 ___________ 3
 |            /|                  |   \             \
 |           / |                  |    \             \
 |   p2     /  |                  |     \             \
 |         2   |                  |      1 ___________ 4
 |         |   4                  +------------------------ v
 |         |  /   p1
 |         | /
 |         1
 +-------------------- s
```
On the T-s diagram 1-2 and 3-4 are vertical (isentropic); on the p-v diagram 2-3 and 4-1 are horizontal (constant pressure).

For a perfect gas with constant $c_p$, and pressure ratio $r_p = p_2/p_1 = p_3/p_4$:

$$
q_{in} = c_p (T_3 - T_2), \qquad q_{out} = c_p (T_4 - T_1)
$$

$$
\eta = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2}
$$

From the isentropic relations, $\dfrac{T_2}{T_1} = \dfrac{T_3}{T_4} = r_p^{(\gamma-1)/\gamma}$, hence $\dfrac{T_4}{T_1} = \dfrac{T_3}{T_2}$ and $\dfrac{T_4 - T_1}{T_3 - T_2} = \dfrac{T_1}{T_2}$:

$$
\boxed{\eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}}
$$

The efficiency depends only on the pressure ratio and rises with it.

#### Pressure ratio for maximum net work
Let $x = r_p^{(\gamma-1)/\gamma}$, $T_2 = T_1 x$, $T_4 = T_3/x$:

$$
w_{net} = c_p (T_3 - T_4) - c_p (T_2 - T_1) = c_p \left[T_3\left(1 - \frac1x\right) - T_1(x - 1)\right]
$$

$$
\frac{dw_{net}}{dx} = c_p\left[\frac{T_3}{x^2} - T_1\right] = 0 \;\Rightarrow\; x^2 = \frac{T_3}{T_1}
$$

$$
\boxed{r_{p,opt} = \left(\frac{T_3}{T_1}\right)^{\gamma/[2(\gamma-1)]}}
$$

At this ratio $T_2 = T_4 = \sqrt{T_1 T_3}$ and the maximum work is

$$
w_{net,max} = c_p\left(\sqrt{T_3} - \sqrt{T_1}\right)^2
$$

The efficiency keeps rising above this pressure ratio, but the net work falls and becomes zero at $x = T_3/T_1$. Practical gas turbines are therefore designed at a pressure ratio between the maximum-work value and this limit, usually closer to the maximum-work value.

## Check yourself

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

1. The ideal Brayton cycle consists of:
   A. Two isothermal and two isentropic processes
   B. Two isentropic and two constant-pressure processes
   C. Two constant-volume and two isentropic processes
   D. Isothermal heat addition only
   Answer: B. Isentropic compression, constant-pressure heat addition, isentropic expansion, constant-pressure heat rejection. (Key point: Cycle processes)

2. The ideal Brayton efficiency is:
   A. 1 - r_p^((gamma-1)/gamma)
   B. 1 - r_p
   C. 1 - 1/r_p^((gamma-1)/gamma)
   D. 1 - T3/T1
   Answer: C. Since (T4 - T1)/(T3 - T2) = T1/T2, eta = 1 - 1/r_p^((gamma-1)/gamma), depending only on pressure ratio. (Key point: Efficiency formula)

3. For maximum net work, with x = r_p^((gamma-1)/gamma):
   A. x = T3/T1
   B. x = T1/T3
   C. x = (T3/T1)^2
   D. x^2 = T3/T1
   Answer: D. Setting dw/dx = 0 gives T3/x^2 = T1, so x^2 = T3/T1. (Key point: Condition for max work)

4. At the pressure ratio for maximum net work:
   A. T2 = T4 = sqrt(T1 T3)
   B. T2 = T3
   C. T4 = T1
   D. T2 = 2 T1
   Answer: A. x = sqrt(T3/T1) gives T2 = T1 x = sqrt(T1 T3) and T4 = T3/x = sqrt(T1 T3). (Key point: Max-work condition result)

## Related in Turbo Machines

- [Ideal Brayton cycle work and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aideal-brayton-numerical) — Turbo Machines, chapter 3

## Related in other subjects

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