---
title: "Engine displacement and mean piston speed"
subject: "Internal Combustion Engines"
chapter: 2
key: "semester_6/internal_combustion_engines:engine-displacement-piston-speed-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/engine-displacement-piston-speed-numerical
site: BCT Engineering
---

# Engine displacement and mean piston speed

Internal Combustion Engines · Chapter 2: Engine construction and operation

- Asked in: year not given, 6 marks
- Video: [Piston displacement, Volumetric efficiency and piston Speed](https://www.youtube.com/watch?v=VUQ39oGo4Jc) — Mr Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/engine-displacement-piston-speed-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-2

## Question

A four-cylinder, four-stroke petrol engine has bore 100 mm, stroke 120 mm and compression ratio 9. It runs at 2400 rpm. Calculate (a) the swept volume per cylinder and the engine displacement, (b) the clearance volume of one cylinder, (c) the mean piston speed, and (d) the number of power strokes per second for the engine.

## Short answer

Swept volume per cylinder 942.5 cm³, engine displacement 3.77 L, clearance volume 117.8 cm³, mean piston speed 9.6 m/s, 80 power strokes per second.

## Answer

Given: $D = 0.1$ m, $L = 0.12$ m, $z = 4$, $r = 9$, $N = 2400$ rpm.

#### (a) Swept volume and displacement
$$
\begin{aligned}
V_s &= \frac{\pi}{4} D^2 L = \frac{\pi}{4}(0.1)^2(0.12) = 9.425 \times 10^{-4}\ \text{m}^3 = 942.5\ \text{cm}^3 \\
V_{engine} &= z V_s = 4 \times 942.5 = 3770\ \text{cm}^3 = 3.77\ \text{L}
\end{aligned}
$$

#### (b) Clearance volume
$$
r = \frac{V_c + V_s}{V_c} \Rightarrow V_c = \frac{V_s}{r - 1} = \frac{942.5}{8} = 117.8\ \text{cm}^3
$$

#### (c) Mean piston speed
The piston travels $2L$ in each revolution:

$$
\bar{S}_p = 2LN = 2 \times 0.12 \times \frac{2400}{60} = 9.6\ \text{m/s}
$$

#### (d) Power strokes per second
In a four-stroke engine each cylinder fires once in two revolutions:

$$
\text{Firing rate} = z \times \frac{N}{2 \times 60} = 4 \times \frac{2400}{120} = 80\ \text{per second}
$$

**Answer:** (a) $V_s = 942.5$ cm$^3$, displacement 3.77 L; (b) $V_c = 117.8$ cm$^3$; (c) 9.6 m/s; (d) 80 power strokes per second for the engine (20 per cylinder).

## Check yourself

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

1. How is the swept volume of one cylinder found?
   A. (pi/4) D^2 L
   B. pi D L
   C. (pi/4) D^2 L x r
   D. (pi/4) D L^2
   Answer: A. Vs = piston area x stroke = 942.5 cm^3 here. (Key point: Swept volume)

2. How is the clearance volume obtained from r = (Vc + Vs)/Vc?
   A. Vc = Vs/r
   B. Vc = Vs/(r - 1)
   C. Vc = Vs x (r - 1)
   D. Vc = Vs x r
   Answer: B. Rearranging gives Vc = 942.5/8 = 117.8 cm^3. (Key point: Clearance volume)

3. What is the mean piston speed formula used?
   A. LN
   B. L N / 2
   C. 2LN (N in rev/s)
   D. pi D N
   Answer: C. The piston travels 2L per revolution, giving 2 x 0.12 x 40 = 9.6 m/s. (Key point: Mean piston speed)

4. Why is the firing rate z x N/(2 x 60) for a four-stroke engine?
   A. Each cylinder fires every revolution
   B. Only one cylinder fires at a time
   C. Each cylinder fires twice per revolution
   D. Each cylinder fires once in two revolutions
   Answer: D. With 4 cylinders at 40 rev/s, 4 x 20 = 80 power strokes per second. (Key point: Firing rate)

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