---
title: "Single cylinder engine: 60-minute trial"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:single-cylinder-60-min-trial"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/single-cylinder-60-min-trial
site: BCT Engineering
---

# Single cylinder engine: 60-minute trial

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2073 Chaitra, 10 marks
- Video: [IC Engine Performance | Numerical | 2021 | GTU Question Paper | Applied Thermodynamic | 3161910](https://www.youtube.com/watch?v=Kt1YJ7bYijk) — krunal khiraiya
- 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-60-min-trial
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

During a 60 minutes trial of a single cylinder four stroke engine the following observations were recorded: Bore = 0.3 m, Stroke = 0.45 m, Fuel consumption = 11.4 kg, Calorific value of fuel = 42000 kJ/kg, Indicated mean effective pressure = 6.0 bar, Net load on brakes = 1500 N, r.p.m. = 300, brake drum diameter = 1.8 m, Brake rope diameter = 20 mm. Calculate: (a) Indicated power; (b) Brake power; (c) Mechanical efficiency; (d) Indicated thermal efficiency.

## Short answer

Indicated power: 47.71 kW (imep 6 bar, bore 0.3 m, stroke 0.45 m, 150 power strokes/min). Brake power: 42.88 kW (effective radius 0.91 m, net load 1500 N, 300 rpm). Mechanical efficiency: 89.88%, indicated thermal efficiency: 35.87%.

## Answer

**Data:** single cylinder, four-stroke, $D = 0.3$ m, $L = 0.45$ m, fuel 11.4 kg in 60 min, CV = 42 000 kJ/kg, $p_m = 6.0$ bar, net brake load $W = 1500$ N, $N = 300$ rpm, drum diameter 1.8 m, rope diameter 20 mm.

Power strokes per minute: $n = N/2 = 150$.

#### (a) Indicated power
$$
\begin{aligned}
A &= \frac{\pi}{4}(0.3)^2 = 0.070686\ \text{m}^2 \\
IP &= \frac{p_m L A n}{60} = \frac{6\times10^5 \times 0.45 \times 0.070686 \times 150}{60} \\
&= 47\,713\ \text{W} = 47.71\ \text{kW}
\end{aligned}
$$

#### (b) Brake power
Effective radius $R_e = \dfrac{1.8 + 0.02}{2} = 0.91$ m.
$$
\begin{aligned}
T &= W R_e = 1500 \times 0.91 = 1365\ \text{N m} \\
BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 300 \times 1365}{60} = 42\,883\ \text{W} = 42.88\ \text{kW}
\end{aligned}
$$

#### (c) Mechanical efficiency
$$
\eta_m = \frac{BP}{IP} = \frac{42.88}{47.71} = 0.8988 = 89.88\%
$$

#### (d) Indicated thermal efficiency
$$
\begin{aligned}
\dot Q &= \frac{11.4}{3600}\times 42\,000 = 133.0\ \text{kW} \\
\eta_{ith} &= \frac{IP}{\dot Q} = \frac{47.71}{133.0} = 0.3587 = 35.87\%
\end{aligned}
$$

**Answer:** IP = 47.71 kW, BP = 42.88 kW, $\eta_m$ = 89.88 %, $\eta_{ith}$ = 35.87 %.

## Check yourself

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

1. What is IP, with n = N/2 = 150?
   A. 47.71 kW
   B. 95.4 kW
   C. 23.9 kW
   D. 477 kW
   Answer: A. IP = 6x10^5 x 0.45 x 0.070686 x 150/60 = 47,713 W. (Key point: Indicated power)

2. Why is R_e = 0.91 m rather than 0.9 m?
   A. Rope diameter 20 mm is added: (1.8 + 0.02)/2
   B. Rope is subtracted
   C. Drum radius is doubled
   D. It is a rounding
   Answer: A. Torque arm is measured to the rope centre. (Key point: Effective radius)

3. What is BP?
   A. 42.88 kW
   B. 47.71 kW
   C. 85.8 kW
   D. 4.29 kW
   Answer: A. T = 1500 x 0.91 = 1365 N m; BP = 2 pi x 300 x 1365/60. (Key point: Brake power)

4. What is eta_m?
   A. BP/IP = 89.88%
   B. IP/BP
   C. BP x IP
   D. 1 - 0.8988 = 10.12%
   Answer: A. 42.88/47.71. (Key point: Mechanical efficiency)

5. What is the fuel heat rate?
   A. 11.4 x 42000 kW
   B. (11.4/3600) x 42,000 = 133 kW, so eta_ith = 35.87%
   C. 11.4/42000
   D. 133 W
   Answer: B. Fuel per second times CV gives kW; IP/133 = 0.3587. (Key point: Heat supplied)

## Close variants set in other exams

- [Single cylinder engine: 20-minute trial](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Asingle-cylinder-20-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