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

# Engine test: four-cylinder at 2000 rpm

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2072 Chaitra, 8 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/engine-test-2000-rpm
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

In a test for four cylinders, four stroke diesel engine has a diameter of 120 mm, stroke = 140 mm, speed of engine = 2000 rpm, fuel consumption rate 15 kg/hour, calorific value of fuel = 42 MJ/kg. Difference in tension on either side of brake pulley = 40 kgf, brake circumference = 3.0m. If the mechanical efficiency is 80% then determine: i) Indicated horse power ii) Brake thermal efficiency iii) Indicated mean effective pressure iv) Brake specific fuel consumption

## Short answer

Brake power = 39.24 kW (net tension 40 kgf, circumference 3 m, 2000 rpm). Indicated power = 49.05 kW (65.75 hp). Brake thermal efficiency = 22.42%, imep = 4.65 bar, bsfc = 0.382 kg/kWh.

## Answer

**Data:** 4 cylinders, four-stroke, $D = 0.12$ m, $L = 0.14$ m, $N = 2000$ rpm, $\dot m_f = 15$ kg/h, CV = 42 000 kJ/kg, net tension $= 40$ kgf, brake circumference $= 3.0$ m, $\eta_m = 0.80$.

Brake drum radius: $R = \dfrac{3.0}{2\pi} = 0.4775$ m (no rope size given, so this is taken as effective radius).
Net force: $F = 40 \times 9.81 = 392.4$ N.

Brake power:
$$
BP = \frac{2\pi N F R}{60} = F \times \frac{\text{circumference} \times N}{60} = 392.4 \times \frac{3.0 \times 2000}{60} = 39\,240\ \text{W} = 39.24\ \text{kW}
$$

#### i) Indicated horse power
$$
IP = \frac{BP}{\eta_m} = \frac{39.24}{0.80} = 49.05\ \text{kW}
$$
With 1 hp = 0.746 kW: $IHP = 49.05/0.746 = 65.75$ hp (66.69 metric hp if 1 hp = 735.5 W).

#### ii) Brake thermal efficiency
$$
\begin{aligned}
\dot Q &= \frac{15}{3600}\times 42\,000 = 175\ \text{kW} \\
\eta_{bth} &= \frac{39.24}{175} = 0.2242 = 22.42\%
\end{aligned}
$$

#### iii) Indicated mean effective pressure
$$
\begin{aligned}
A &= \frac{\pi}{4}(0.12)^2 = 0.011310\ \text{m}^2,\quad n = \frac{N}{2} = 1000,\ k = 4 \\
p_m &= \frac{60\,IP}{L A n k} = \frac{60 \times 49\,050}{0.14 \times 0.011310 \times 1000 \times 4} \\
&= 4.647\times10^5\ \text{N/m}^2 = 4.65\ \text{bar}
\end{aligned}
$$

#### iv) Brake specific fuel consumption
$$
bsfc = \frac{15}{39.24} = 0.3823\ \text{kg/kWh}
$$

**Answer:** IHP = 49.05 kW (65.75 hp), $\eta_{bth}$ = 22.42 %, imep = 4.65 bar, bsfc = 0.382 kg/kWh.

## Check yourself

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

1. How is brake power computed from net tension and brake circumference?
   A. BP = net force x (circumference x N)/60
   B. BP = net force x circumference
   C. BP = net force x N x 2*pi
   D. BP = net force / (circumference x N)
   Answer: A. Force x distance travelled per second at the brake circumference gives power: 392.4 x 100 = 39.24 kW. (Key point: Brake power from rope brake)

2. How is indicated power found here?
   A. IP = BP x mechanical efficiency
   B. IP = BP / mechanical efficiency
   C. IP = BP + 0.8
   D. IP = BP - friction power of 20%
   Answer: B. Mechanical efficiency = BP/IP, so IP = 39.24/0.80 = 49.05 kW. (Key point: Mechanical efficiency)

3. In imep = 60 IP/(L A n k), what is n for a four-stroke engine?
   A. N
   B. 2N
   C. N/2 (one power stroke every two revolutions)
   D. N/4
   Answer: C. A four-stroke cylinder fires once every two revolutions, so n = N/2 = 1000. (Key point: imep formula)

4. What is heat supplied used in the brake thermal efficiency?
   A. Fuel rate in kg/h x CV directly
   B. IP x CV
   C. BP / mechanical efficiency
   D. Fuel rate (kg/s) x CV = 175 kW
   Answer: D. 15/3600 x 42000 = 175 kW; brake thermal efficiency = 39.24/175 = 22.42%. (Key point: Brake thermal efficiency)

## Close variants set in other exams

- [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, 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×
- [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×

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- [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
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