---
title: "Single-stage compressor power numerical"
subject: "Applied Thermodynamics"
chapter: 2
key: "semester_3/applied_thermodynamics:single-stage-compressor-numerical"
source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/single-stage-compressor-numerical
site: BCT Engineering
---

# Single-stage compressor power numerical

Applied Thermodynamics · Chapter 2: Air Compressors

- Asked in: year not given, 8 marks
- Video: [Lecture 5.3: Problem on Single stage reciprocating compressor](https://www.youtube.com/watch?v=tPzWlnpJitw) — Mech-D
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/single-stage-compressor-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-2

## Question

A single-stage reciprocating compressor takes in 4 m³/min of free air at 1 bar and 27 °C and delivers it at 7 bar. Compression follows $pV^{1.3} = C$ and the clearance is negligible. Calculate (a) the indicated power, (b) the isothermal power and the isothermal efficiency, and (c) the delivery temperature.

## Short answer

Using pV^1.3 = C: indicated power = 16.4 kW; isothermal power = p1V1 ln(p2/p1) = 12.97 kW; isothermal efficiency = 79.2%. Delivery temperature = T1 (p2/p1)^((n‑1)/n) = 470 K (197 °C).

## Answer

#### Given
$p_1 = 100$ kPa, $V_1 = 4$ m³/min $= 0.0667$ m³/s, $T_1 = 300$ K, $p_2 = 700$ kPa, $n = 1.3$.

$$
p_1V_1 = 100 \times 0.06667 = 6.667\ \text{kW}
$$

#### (a) Indicated (polytropic) power
Pressure-ratio factor: $(p_2/p_1)^{\frac{n-1}{n}} = 7^{0.2308} = 1.5668$

$$
\begin{aligned}
P &= \frac{n}{n-1}\,p_1V_1\left[\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} - 1\right]\\
&= \frac{1.3}{0.3} \times 6.667 \times (1.5668 - 1) = 16.38\ \text{kW}
\end{aligned}
$$

#### (b) Isothermal power and efficiency
$$
P_{iso} = p_1V_1 \ln\frac{p_2}{p_1} = 6.667 \times \ln 7 = 12.97\ \text{kW}
$$

$$
\eta_{iso} = \frac{P_{iso}}{P} = \frac{12.97}{16.38} = 0.792
$$

#### (c) Delivery temperature
$$
T_2 = T_1\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} = 300 \times 1.5668 = 470\ \text{K} = 197\ \text{°C}
$$

**Answer:** (a) 16.4 kW; (b) 12.97 kW, $\eta_{iso}$ = 79.2 %; (c) 470 K (about 197 °C).

## Check yourself

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

1. Which formula gives the indicated power here?
   A. p1V1 ln(p2/p1)
   B. (p2 - p1) V1
   C. p1V1 (p2/p1)^n
   D. n/(n-1) p1V1 [(p2/p1)^((n-1)/n) - 1]
   Answer: D. Polytropic compression with negligible clearance uses n/(n-1) p1V1[...]. (Key point: Indicated power formula)

2. What is the indicated power?
   A. About 16.4 kW
   B. About 13.0 kW
   C. About 6.67 kW
   D. About 25 kW
   Answer: A. (1.3/0.3) x 6.667 x 0.5668 = 16.38 kW. (Key point: Indicated power value)

3. What is the isothermal efficiency?
   A. About 126 %
   B. About 79.2 %, i.e. 12.97/16.38
   C. About 50 %
   D. About 92 %
   Answer: B. Isothermal power is p1V1 ln 7 = 12.97 kW, divided by 16.38 kW. (Key point: Isothermal efficiency)

4. What is the delivery temperature?
   A. About 300 K
   B. About 700 K
   C. About 470 K (197 °C)
   D. About 390 K
   Answer: C. T2 = 300 x 7^0.2308 = 300 x 1.5668 = 470 K. (Key point: Delivery temperature)

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