---
title: "Air standard Brayton: state points, work, efficiency"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brayton-0-18mpa-34c-max-2350"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brayton-0-18mpa-34c-max-2350
site: BCT Engineering
---

# Air standard Brayton: state points, work, efficiency

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

- Asked in: 2070 Bhadra, 8 marks
- 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-0-18mpa-34c-max-2350
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

In an air standard Brayton cycle the air enters the compressor at 0.18 MPa, 34°C. The pressure leaving the compressor is 2.3 MPa, and the maximum temperature in the cycle is 2350°C. Determine:
a) The pressure and temperature at each point cycle
b) The compressor work, turbine work, and cycle efficiency
[Take Cp = 1005 J/kgK, γ = 1.4]

## Short answer

Pressure ratio r_p = 2.3/0.18 = 12.78. Temperatures: T1=307.15 K, T2=636.0 K, T3=2623.15 K, T4=1266.8 K; pressures P1=P4=0.18 MPa, P2=P3=2.3 MPa. Compressor work 330.5 kJ/kg, turbine work 1363.2 kJ/kg, thermal efficiency 51.7%.

## Answer

**Given:** P₁ = 0.18 MPa, T₁ = 34 + 273.15 = 307.15 K, P₂ = 2.3 MPa, T_max = T₃ = 2350 + 273.15 = 2623.15 K, cp = 1.005 kJ/kg·K, γ = 1.4. (The ideal Brayton cycle has no pressure drops: P₃ = P₂, P₄ = P₁.)

#### (a) Pressure and temperature at each point
Pressure ratio r_p = 2.3/0.18 = 12.78; r_p^((γ−1)/γ) = 12.78^0.2857 = 2.0708

| State | Pressure | Temperature |
|---|---|---|
| 1 (compressor inlet) | 0.18 MPa | 307.15 K (34 °C) |
| 2 (compressor exit) | 2.3 MPa | 636.0 K (362.9 °C) |
| 3 (turbine inlet) | 2.3 MPa | 2623.15 K (2350 °C) |
| 4 (turbine exit) | 0.18 MPa | 1266.8 K (993.6 °C) |

with T₂ = T₁ r_p^0.2857 = 307.15 × 2.0708 = 636.0 K and T₄ = T₃/r_p^0.2857 = 2623.15/2.0708 = 1266.8 K.

#### (b) Work and efficiency (per kg of air)
- Compressor work: w_c = cp (T₂ − T₁) = 1.005 × (636.0 − 307.15) = **330.5 kJ/kg**
- Turbine work: w_t = cp (T₃ − T₄) = 1.005 × (2623.15 − 1266.8) = **1363.2 kJ/kg**
- Net work: w_net = 1363.2 − 330.5 = 1032.6 kJ/kg
- Heat added: q_in = cp (T₃ − T₂) = 1.005 × (2623.15 − 636.0) = 1997.1 kJ/kg

η = w_net/q_in = 1032.6/1997.1 = **51.7 %**

Check: η = 1 − 1/r_p^0.2857 = 1 − 1/2.0708 = 51.7 % ✓.

## Check yourself

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

1. Pressure ratio is:
   A. 2350/34
   B. 2.3/0.18 = 12.78
   C. 2.3 - 0.18
   D. 0.18/2.3
   Answer: B. r_p = P2/P1 with P3 = P2 and P4 = P1. (Key point: Pressure ratio)

2. T2 is:
   A. T3/2.0708
   B. T1 x 12.78
   C. T1 x 2.0708 = 636.0 K
   D. T1 + 2.3
   Answer: C. r_p^0.2857 = 2.0708. (Key point: T2)

3. Turbine work per kg:
   A. cp(T4 - T1)
   B. cp(T2 - T1)
   C. cp(T3 - T4) = 1363.2 kJ/kg
   D. cp(T3 - T2)
   Answer: C. T4 = 1266.8 K. (Key point: Turbine work)

4. Efficiency is:
   A. 75%
   B. 30%
   C. 51.7%
   D. 12.78%
   Answer: C. 1032.6/1997.1 = 1 - 1/2.0708. (Key point: Efficiency)

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