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

# Ideal Brayton cycle: diagrams and efficiency

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

- Asked in: 2072 Asoj, 6 marks; 2071 Bhadra, 6 marks (asked 2 times)
- 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_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-cycle-efficiency-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

Sketch an ideal Brayton cycle on P-v and T-s diagrams. Also derive an expression for its efficiency in terms of pressure ratio.

## Short answer

The ideal Brayton cycle has two isentropic and two constant‑pressure processes. Efficiency derivation: η = 1 − q_out/q_in = 1 − (T₄−T₁)/(T₃−T₂). Using isentropic ratios T₂/T₁ = T₃/T₄ = r_p^((γ−1)/γ) yields η = 1 − 1/r_p^((γ−1)/γ).

## Answer

The ideal Brayton cycle is the air-standard cycle of a gas turbine. It has two isentropic and two constant-pressure processes.

#### Diagrams
```text
 P-v diagram                      T-s diagram
 P                                T
 |  2 ____________ 3              |               3
 |   \             \              |              /|
 |    \             \             |             / |
 |     '.            '.           |           /   |
 |       '.            '.         |    2 _.-'     |
 |         '-._          '-._     |    |          4
 |             '- 1 _________ 4   |    |       .-'
 |                                |    |   _.-'
 |                                |    1.-'
 +----------------------> v       +--------------------> s
 1-2 isentropic compression       2-3 constant-P heat addition
 3-4 isentropic expansion         4-1 constant-P heat rejection
```

#### Derivation of efficiency
Take air as an ideal gas with constant cp. Per kg of air:

- Heat added (2-3, constant P): q_in = cp (T₃ − T₂)
- Heat rejected (4-1, constant P): q_out = cp (T₄ − T₁)

Thermal efficiency:

η = w_net/q_in = 1 − q_out/q_in = 1 − (T₄ − T₁)/(T₃ − T₂)

For the isentropic processes 1-2 and 3-4, with P₂ = P₃ and P₁ = P₄, the pressure ratio is r_p = P₂/P₁ = P₃/P₄:

T₂/T₁ = (P₂/P₁)^((γ−1)/γ) = r_p^((γ−1)/γ)
T₃/T₄ = (P₃/P₄)^((γ−1)/γ) = r_p^((γ−1)/γ)

Hence T₂/T₁ = T₃/T₄, which gives T₃/T₂ = T₄/T₁. Then

(T₄ − T₁)/(T₃ − T₂) = (T₁/T₂) × (T₄/T₁ − 1)/(T₃/T₂ − 1) = T₁/T₂

So

η = 1 − T₁/T₂

**η_Brayton = 1 − 1/r_p^((γ−1)/γ)**

The efficiency depends only on the pressure ratio and γ: it rises as r_p rises. For example, r_p = 10 and γ = 1.4 gives η = 1 − 1/1.931 = 48.2 %.

## Check yourself

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

1. Which processes make up the ideal Brayton cycle?
   A. Two constant-pressure and two constant-volume
   B. Two isothermal and two isentropic
   C. Two isentropic and two constant-pressure
   D. Two isentropic and two constant-volume
   Answer: C. Compression and expansion are isentropic; heat addition and rejection are at constant pressure. (Key point: Brayton processes)

2. Which gives heat rejected per kg?
   A. cp(T4 - T1)
   B. cp(T2 - T1)
   C. cp(T3 - T2)
   D. cv(T4 - T1)
   Answer: A. Heat rejection 4-1 is at constant pressure, so q_out = cp(T4 - T1). (Key point: Heat expressions)

3. Which relation lets the temperatures cancel in the derivation?
   A. T2 = T4
   B. T2/T1 = T4/T3 only for reversible heat
   C. T2/T1 = T3/T4, since both isentropic processes have the same pressure ratio
   D. T1 = T3
   Answer: C. Equal pressure ratios give T2/T1 = T3/T4, hence (T4 - T1)/(T3 - T2) = T1/T2. (Key point: Key step)

4. Final efficiency expression?
   A. 1 - 1/r_p^((g-1)/g)
   B. 1 - T3/T1
   C. 1 - 1/r_p^g
   D. 1 - r_p^((g-1)/g)
   Answer: A. eta = 1 - T1/T2 = 1 - 1/r_p^((g-1)/g). (Key point: Final result)

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