---
title: "Single cylinder engine: 20-minute trial"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:single-cylinder-20-min-trial"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/single-cylinder-20-min-trial
site: BCT Engineering
---

# Single cylinder engine: 20-minute trial

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2080 Baisakh, 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/single-cylinder-20-min-trial
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

During a 20 minutes trial of a single cylinder four stroke engine the following observations were recorded: Bore = 0.2 m, Stroke = 0.28 m, Fuel consumption = 1.52 kg, Calorific value of fuel = 43900 kJ/kg, Indicated mean effective pressure = 3.1 bar, Net load on brakes = 640 N, r.p.m. = 350, brake drum diameter = 1 m. Calculate: (i) Indicated power; (ii) Brake power; (iii) Mechanical efficiency; (iv) Indicated thermal efficiency.

## Short answer

For a four-stroke engine, use n=N/2=175 power strokes per minute. IP=7.953 kW and BP=11.729 kW, giving mechanical efficiency 147.5%, which is impossible, so the data are inconsistent. Indicated thermal efficiency is 14.30%; if treated as a two-stroke engine, IP=15.906 kW, mechanical efficiency 73.7%, and indicated thermal efficiency 28.6%.

## Answer

**Data:** $D = 0.2$ m, $L = 0.28$ m, $p_m = 3.1$ bar, $N = 350$ rpm, net brake load $W = 640$ N, drum diameter 1 m (no rope diameter given, so effective radius $R = 0.5$ m), fuel 1.52 kg in 20 min, CV = 43 900 kJ/kg.

For a four-stroke engine the number of power strokes per minute is $n = N/2 = 175$.

#### (i) Indicated power
$$
\begin{aligned}
A &= \frac{\pi}{4}D^2 = \frac{\pi}{4}(0.2)^2 = 0.031416\ \text{m}^2 \\
IP &= \frac{p_m L A n}{60} = \frac{3.1\times10^5 \times 0.28 \times 0.031416 \times 175}{60} \\
&= 7953\ \text{W} = 7.953\ \text{kW}
\end{aligned}
$$

#### (ii) Brake power
$$
\begin{aligned}
T &= W R = 640 \times 0.5 = 320\ \text{N m} \\
BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 350 \times 320}{60} = 11\,729\ \text{W} = 11.729\ \text{kW}
\end{aligned}
$$

#### (iii) Mechanical efficiency
$$
\eta_m = \frac{BP}{IP} = \frac{11.729}{7.953} = 1.475 = 147.5\%
$$

#### (iv) Indicated thermal efficiency
$$
\begin{aligned}
\dot m_f &= \frac{1.52}{20\times60} = 0.0012667\ \text{kg/s}\ (4.56\ \text{kg/h}) \\
\text{Heat supplied} &= \dot m_f \times CV = 0.0012667 \times 43\,900 = 55.61\ \text{kW} \\
\eta_{ith} &= \frac{IP}{\dot m_f\,CV} = \frac{7.953}{55.61} = 0.1430 = 14.30\%
\end{aligned}
$$

**Answer:** IP = 7.953 kW, BP = 11.729 kW, $\eta_m$ = 147.5 %, $\eta_{ith}$ = 14.30 %.

**Note on the data:** a mechanical efficiency above 100 % is impossible (BP can never exceed IP), so the figures as printed are inconsistent. If the engine is taken as **two-stroke** ($n = N = 350$), IP doubles to 15.906 kW, giving $\eta_m = 11.729/15.906 = 73.7\%$ and $\eta_{ith} = 15.906/55.61 = 28.6\%$, which are realistic values. In the exam, show the working as above and point out the inconsistency.

## Check yourself

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

1. For the four-stroke engine at 350 rpm, how many power strokes per minute are used in IP?
   A. 350
   B. 700
   C. 87.5
   D. 175
   Answer: D. A four-stroke engine fires once every two revolutions, so n = N/2 = 175. (Key point: Power strokes per minute)

2. Which expression gives the indicated power here?
   A. p_m L A n / 60 with p_m in N/m^2
   B. 2 pi N T / 60
   C. BP x mechanical efficiency
   D. p_m L A N / 2
   Answer: A. IP = p_m L A n /60 = 3.1x10^5 x 0.28 x 0.031416 x 175 / 60 = 7.953 kW. (Key point: Indicated power formula)

3. With a 1 m drum and net load 640 N, what is the torque used for BP?
   A. 640 N m
   B. 160 N m
   C. 320 N m
   D. 1280 N m
   Answer: C. T = W x R with R = 0.5 m, so T = 320 N m, giving BP = 11.729 kW. (Key point: Brake power)

4. Why does the answer point out the data are inconsistent?
   A. Fuel use is too high
   B. Mechanical efficiency came out 147.5%, but BP cannot exceed IP
   C. The bore is too small
   D. Indicated thermal efficiency is above 100%
   Answer: B. BP (11.73 kW) exceeds IP (7.95 kW), which is impossible; treating it as two-stroke gives realistic 73.7%. (Key point: Inconsistent data)

5. How is indicated thermal efficiency found?
   A. BP divided by fuel mass
   B. IP divided by (fuel mass flow x calorific value)
   C. IP divided by BP
   D. Fuel energy divided by IP
   Answer: B. eta_ith = IP / (m_f x CV) = 7.953/55.61 = 14.30%. (Key point: Indicated thermal efficiency)

## Close variants set in other exams

- [Single cylinder engine: 60-minute trial](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Asingle-cylinder-60-min-trial) — Power Plant Equipment, chapter 5, asked 2×

## Related in Power Plant Equipment

- [Two-stroke motoring test with specific fuel consumption](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-sfc) — Power Plant Equipment, chapter 5, asked 6×
- [Engine test: four-cylinder at 1820 rpm](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-1820-rpm) — Power Plant Equipment, chapter 5, asked 5×
- [Engine test: four-cylinder at 1800 rpm](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-1800-rpm) — Power Plant Equipment, chapter 5, asked 5×
- [Engine test: four-cylinder with rope radius](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-rope-radius) — Power Plant Equipment, chapter 5, asked 5×
- [Engine test: four-cylinder, 85% mechanical efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-85-percent) — Power Plant Equipment, chapter 5, asked 5×
- [Rope dynamometer trial: brake and indicated power](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Arope-dynamometer-trial) — Power Plant Equipment, chapter 5

## Related in other subjects

- [Efficiency ratio and specific fuel consumption](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Arelative-efficiency-sfc-numerical) — Internal Combustion Engines, chapter 8
- [IP, BP and efficiencies of a petrol engine](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Aindicated-brake-power-efficiency-numerical) — Internal Combustion Engines, chapter 8
- [Calorific value of fuel by Dulong's formula](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Acalorific-value-dulong-formula-numerical) — Internal Combustion Engines, chapter 3