---
title: "Gas turbine power from isentropic efficiency"
subject: "Turbo Machines"
chapter: 1
key: "semester_7/turbo_machines:gas-turbine-power-isentropic-efficiency"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/gas-turbine-power-isentropic-efficiency
site: BCT Engineering
---

# Gas turbine power from isentropic efficiency

Turbo Machines · Chapter 1: Introduction

- Asked in: year not given, 6 marks
- Video: [Turbines, Compressors, and Pumps - ISENTROPIC EFFICIENCY in 8 Minutes!](https://www.youtube.com/watch?v=K2A5gdROVZE) — 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/gas-turbine-power-isentropic-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-1

## Question

Combustion gases enter an adiabatic turbine at 600 kPa and 1100 K and leave at 100 kPa. The isentropic efficiency of the turbine is 88% and the mass flow rate is 20 kg/s. Taking $c_p = 1.148$ kJ/kg K and $\gamma = 1.333$ for the gas, find (a) the isentropic exit temperature, (b) the actual exit temperature, (c) the specific work and the power developed.

## Short answer

Using isentropic relation: T02s = T01 / r^((γ-1)/γ) = 703.1 K. Actual exit temperature T02 = 750.7 K. Specific work w = cp(T01-T02) = 401.0 kJ/kg, power = 20 × 401.0 = 8.02 MW.

## Answer

**Data:** $T_{01} = 1100$ K, pressure ratio $r = 600/100 = 6$, $\eta_t = 0.88$, $\dot m = 20$ kg/s, $c_p = 1.148$ kJ/kg K, $\gamma = 1.333$. Inlet and exit velocities are taken as negligible (static = stagnation values).

#### (a) Isentropic exit temperature
$$
\frac{\gamma - 1}{\gamma} = \frac{0.333}{1.333} = 0.250
$$
$$
T_{02s} = T_{01}\left(\frac{1}{r}\right)^{(\gamma-1)/\gamma} = 1100 \times (1/6)^{0.250} = 703.1\ \text{K}
$$

Isentropic temperature drop: $T_{01} - T_{02s} = 396.9$ K.

#### (b) Actual exit temperature
$$
T_{01} - T_{02} = \eta_t (T_{01} - T_{02s}) = 0.88 \times 396.9 = 349.3\ \text{K}
$$
$$
T_{02} = 1100 - 349.3 = 750.7\ \text{K}
$$

#### (c) Work and power
$$
w = c_p (T_{01} - T_{02}) = 1.148 \times 349.3 = 401.0\ \text{kJ/kg}
$$
$$
P = \dot m w = 20 \times 401.0 = 8.02\ \text{MW}
$$

**Answer:** $T_{02s} \approx 703.1$ K, $T_{02} \approx 750.7$ K, $w \approx 401.0$ kJ/kg, power $\approx 8.02$ MW.

## Check yourself

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

1. Which relation gives T02s for the expansion with pressure ratio r?
   A. T01 x r^((gamma-1)/gamma)
   B. T01 / r
   C. T01 x (1/r)^((gamma-1)/gamma)
   D. T01 x (1/r)^(gamma)
   Answer: C. For isentropic expansion T02s = T01 (p2/p1)^((gamma-1)/gamma) = T01 (1/r)^0.25. (Key point: Isentropic exit temperature)

2. How is the actual temperature drop found from the isentropic drop?
   A. Divide the isentropic drop by eta_t
   B. Multiply the isentropic drop by eta_t
   C. Add eta_t to it
   D. Subtract eta_t from it
   Answer: B. For a turbine, T01 - T02 = eta_t (T01 - T02s), since actual work is less than isentropic. (Key point: Using turbine efficiency)

3. Approximately what is the isentropic exit temperature in this problem?
   A. 750.7 K
   B. 1100 K
   C. 549 K
   D. 703.1 K
   Answer: D. 1100 x (1/6)^0.25 = 703.1 K. (Key point: Key result)

4. What power is developed, given w = 401 kJ/kg and 20 kg/s?
   A. 8.02 MW
   B. 401 kW
   C. 20 MW
   D. 0.8 MW
   Answer: A. P = m x w = 20 x 401.0 = 8020 kW = 8.02 MW. (Key point: Power)

## Related in Turbo Machines

- [Actual gas turbine cycle with component efficiencies](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aactual-brayton-cycle-numerical) — Turbo Machines, chapter 3
- [Axial gas turbine stage design calculation](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aaxial-turbine-stage-gas-numerical) — Turbo Machines, chapter 2
- [Gas turbine combustion chamber: requirements and types](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acombustion-chamber-requirements-types) — Turbo Machines, chapter 3
- [Fuel-air ratio in a gas turbine combustor](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acombustion-chamber-fuel-air-ratio-numerical) — Turbo Machines, chapter 3
- [Convergent nozzle: choking and mass flow](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aconvergent-nozzle-choking-mass-flow) — Turbo Machines, chapter 4
- [Convergent-divergent nozzle with efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acd-nozzle-efficiency-exit-area-numerical) — Turbo Machines, chapter 4

## Related in other subjects

- [Adiabatic steam turbine 50 MW: mass flow, isentropic efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Asteam-turbine-5mpa-500c-50mw) — Fundamental of Thermodynamics & Heat Transfer, chapter 5
- [Adiabatic steam turbine 70% isentropic efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aadiabatic-steam-turbine-70-percent-moisture) — Fundamental of Thermodynamics & Heat Transfer, chapter 5
- [Adiabatic air turbine: mass flow and areas](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aturbine-1000k-to-400k-areas) — Fundamental of Thermodynamics & Heat Transfer, chapter 4, asked 2×
- [Gas turbine 6:1: power output and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agt-6-1-power-efficiency) — Power Plant Equipment, chapter 6, asked 2×