---
title: "Compressor work and polytropic efficiency"
subject: "Turbo Machines"
chapter: 1
key: "semester_7/turbo_machines:compressor-exit-temperature-polytropic-efficiency"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/compressor-exit-temperature-polytropic-efficiency
site: BCT Engineering
---

# Compressor work and polytropic efficiency

Turbo Machines · Chapter 1: Introduction

- Asked in: year not given, 8 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/compressor-exit-temperature-polytropic-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-1

## Question

Air at 100 kPa and 300 K is compressed adiabatically in a compressor to a pressure ratio of 8. The isentropic efficiency is 82% and the air flow is 10 kg/s. Taking $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$, find (a) the exit temperature, (b) the specific work and power input, and (c) the polytropic efficiency of the compressor.

## Short answer

Isentropic T02s = T01 r^((γ-1)/γ) = 543.4 K; actual T02 = T01 + (T02s-T01)/η_c = 596.9 K. Specific work 298.4 kJ/kg, power 2984 kW. Polytropic efficiency η_p = (0.2857 ln8)/ln(596.9/300) = 86.4%.

## Answer

**Data:** $T_{01} = 300$ K, $r = 8$, $\eta_c = 0.82$, $\dot m = 10$ kg/s, $c_p = 1.005$, $\gamma = 1.4$, so $(\gamma-1)/\gamma = 0.2857$.

#### (a) Exit temperature
Isentropic exit temperature:
$$
T_{02s} = T_{01}\, r^{(\gamma-1)/\gamma} = 300 \times 8^{0.2857} = 543.4\ \text{K}
$$
Actual temperature rise:
$$
T_{02} - T_{01} = \frac{T_{02s} - T_{01}}{\eta_c} = \frac{243.4}{0.82} = 296.9\ \text{K}
$$
$$
T_{02} = 300 + 296.9 = 596.9\ \text{K}
$$

#### (b) Work and power
$$
w = c_p (T_{02} - T_{01}) = 1.005 \times 296.9 = 298.4\ \text{kJ/kg}
$$
$$
P = \dot m w = 10 \times 298.4 = 2984\ \text{kW}
$$

#### (c) Polytropic efficiency
$$
\eta_p = \frac{\frac{\gamma-1}{\gamma}\ln r}{\ln (T_{02}/T_{01})}
= \frac{0.2857 \times \ln 8}{\ln (596.9/300)} = \frac{0.5941}{0.6879} = 0.864
$$

The polytropic efficiency (86.4%) is higher than the isentropic efficiency (82%) because of the preheat effect: the losses in each small stage heat the air entering the next stage, which then needs more work to compress.

**Answer:** $T_{02} \approx 596.9$ K; $w \approx 298.4$ kJ/kg; $P \approx 2984$ kW; $\eta_p \approx 86.4\%$.

## Check yourself

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

1. For a compressor, how is the actual temperature rise obtained from the isentropic rise?
   A. Multiply by eta_c
   B. Divide by eta_c
   C. Subtract eta_c
   D. Take the square root with eta_c
   Answer: B. Compressor: T02 - T01 = (T02s - T01)/eta_c = 243.4/0.82 = 296.9 K, because real compression needs more work. (Key point: Actual temperature rise)

2. What is the compressor exit temperature?
   A. 543.4 K
   B. 300 K
   C. 443.4 K
   D. 596.9 K
   Answer: D. T02 = 300 + 296.9 = 596.9 K, higher than the isentropic 543.4 K. (Key point: Exit temperature)

3. Which expression gives the polytropic efficiency of the compressor?
   A. ((gamma-1)/gamma) ln r / ln(T02/T01)
   B. ln(T02/T01) / ln r
   C. (T02s - T01)/(T02 - T01)
   D. r^((gamma-1)/gamma) - 1
   Answer: A. eta_p = [(gamma-1)/gamma] ln r / ln(T02/T01) = 0.5941/0.6879 = 0.864. (Key point: Polytropic efficiency formula)

4. Why is the polytropic efficiency (86.4%) higher than the isentropic efficiency (82%)?
   A. Because of friction in the bearings
   B. Because of the reheat effect, where work of one stage heats the next
   C. Because the pressure ratio is low
   D. Because heat is removed during compression
   Answer: B. In multistage compression, losses in early stages raise temperature and increase later stage work (reheat effect), so eta_c < eta_p. (Key point: Why eta_p exceeds eta_c)

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