---
title: "Engine dimensions for 5 kW at 2000 rpm"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:engine-dimensions-5kw"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/engine-dimensions-5kw
site: BCT Engineering
---

# Engine dimensions for 5 kW at 2000 rpm

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2076 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/engine-dimensions-5kw
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

A 4-stroke diesel engine develops 5 kW at 2000 RPM when its mean effective pressure is 7.5 bar. If for engine, L = 1.25 D, find their dimensions.

## Short answer

For a four-stroke engine, power=pm L A (N/2)/60 with L=1.25D. Substituting 5 kW, 7.5 bar and 2000 rpm gives D≈74.1 mm and L≈92.7 mm.

## Answer

**Assumptions:** single-cylinder engine; the 5 kW is the power developed with the given mean effective pressure (brake mep basis), so $P = \dfrac{p_m L A n}{60}$ with $n = N/2$ for a four-stroke engine.

**Data:** $P = 5$ kW $= 5000$ W, $N = 2000$ rpm, $n = 1000$ per min, $p_m = 7.5$ bar $= 7.5\times10^5$ N/m², $L = 1.25D$.

$$
\begin{aligned}
P &= \frac{p_m \, L \, \frac{\pi}{4}D^2 \, n}{60} = \frac{p_m (1.25D)\frac{\pi}{4}D^2 n}{60} \\
5000 &= \frac{7.5\times10^5 \times 1.25 \times 0.7854 \times D^3 \times 1000}{60} \\
D^3 &= \frac{5000 \times 60}{7.5\times10^5 \times 1.25 \times 0.7854 \times 1000} = 4.074\times10^{-4}\ \text{m}^3 \\
D &= 0.07413\ \text{m} = 74.1\ \text{mm} \\
L &= 1.25 \times 74.13 = 92.7\ \text{mm}
\end{aligned}
$$

Check: $\dfrac{7.5\times10^5 \times 0.09267 \times \frac{\pi}{4}(0.07413)^2 \times 1000}{60} = 5000$ W.

**Answer:** Bore $D \approx 74.1$ mm, stroke $L \approx 92.7$ mm.

## Check yourself

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

1. Which power expression is used for the four-stroke engine?
   A. P = p_m L A N /60
   B. P = p_m L A (N/2) /60
   C. P = 2 p_m L A N /60
   D. P = p_m L A /N
   Answer: B. Four-stroke has n = N/2 = 1000 power strokes per minute. (Key point: Power formula)

2. What are the correct units for p_m = 7.5 bar?
   A. 7.5 N/m^2
   B. 7.5x10^5 N/m^2
   C. 7.5x10^3 N/m^2
   D. 7.5x10^6 N/m^2
   Answer: B. 1 bar = 10^5 N/m^2. (Key point: Unit conversion)

3. What is the effect of substituting L = 1.25D?
   A. It gives D^2 in the equation
   B. It gives D^3: P = p_m(1.25D)(pi/4)D^2 n/60
   C. It removes D
   D. It gives D^4
   Answer: B. Length times area is proportional to D x D^2 = D^3, which is solved for D. (Key point: Using L = 1.25 D)

4. What are the resulting dimensions?
   A. D = 92.7 mm, L = 74.1 mm
   B. D = 74.1 mm, L = 92.7 mm
   C. D = 100 mm, L = 125 mm
   D. D = 74.1 mm, L = 74.1 mm
   Answer: B. D^3 = 4.074x10^-4 gives D = 74.1 mm and L = 1.25 x 74.1 = 92.7 mm. (Key point: Result)

## Related in Power Plant Equipment

- [Engine test: four-cylinder at 2000 rpm](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-2000-rpm) — Power Plant Equipment, chapter 5, asked 5×
- [Diesel engine 75 kW: fuel consumption and efficiencies](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adiesel-75kw-fuel-consumption) — Power Plant Equipment, chapter 5
- [Two-stroke motoring test: 5 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-5kw) — Power Plant Equipment, chapter 5, asked 6×
- [Two-stroke motoring test: 2.2 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-2-2kw) — Power Plant Equipment, chapter 5, asked 6×
- [Two-stroke motoring test: 2 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-2kw) — Power Plant Equipment, chapter 5, asked 6×
- [Two-stroke motoring test: 1.5 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-1-5kw) — Power Plant Equipment, chapter 5, asked 6×

## Related in other subjects

- [Air standard diesel: efficiency and mean effective pressure](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Adiesel-cycle-cutoff-2-efficiency-mep) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Engine crank 2000 rpm, ICR, piston acceleration](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aengine-crank-2000rpm-acceleration-icr) — Applied Mechanics (Dynamics), chapter 4
- [Main dimensions of 15 kW 2810 rpm cage motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Amain-dimensions-15kw-2810rpm) — Electric Machine Design, chapter 4, asked 2×
- [Coriolis meaning; engine crank 2000 rpm](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aengine-crank-2000rpm-coriolis) — Applied Mechanics (Dynamics), chapter 4