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

# Engine test: four-cylinder with rope radius

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2078 Bhadra, 10 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-test-rope-radius
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

In a test for four-cylinders, four stroke engine has a diameter of 100mm, stroke = 120mm, speed of engine = 1800 rpm, fuel consumption of 0.2 kg/min, calorific value of fuel is 44000 kJ/Kg. Difference in tension on either side of brake pulley = 40 kg, Brake circumference is 300cm and the radius of the pulley rope is 25mm. If the mechanical efficiency is 90%. Determine: (i) Brake-thermal efficiency, (ii) Indicated thermal efficiency, (iii) Indicated mean effective pressure, and (iv) Brake specific fuel consumption.

## Short answer

Drum radius is 0.4775 m, effective radius 0.5025 m, giving torque 197.17 N m; BP=37.165 kW and IP=41.295 kW. With heat input 146.67 kW, brake thermal efficiency is 25.34% and indicated thermal efficiency is 28.16%. Imep is 7.30 bar and bsfc is 0.323 kg/kWh.

## Answer

**Data:** 4 cylinders, four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1800$ rpm, $\dot m_f = 0.2$ kg/min, CV = 44 000 kJ/kg, net rope tension $(W - S) = 40$ kgf, brake circumference = 3.0 m, rope radius = 25 mm, $\eta_m = 0.90$.

#### Brake power
Drum radius from circumference: $R_d = \dfrac{3.0}{2\pi} = 0.4775$ m.
Effective radius (to rope centre): $R_e = 0.4775 + 0.025 = 0.5025$ m.
Net force: $F = 40 \times 9.81 = 392.4$ N.
$$
\begin{aligned}
T &= F R_e = 392.4 \times 0.5025 = 197.17\ \text{N m} \\
BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 1800 \times 197.17}{60} = 37.165\ \text{kW} \\
IP &= \frac{BP}{\eta_m} = \frac{37.165}{0.9} = 41.295\ \text{kW}
\end{aligned}
$$

Heat supplied: $\dot m_f\, CV = \dfrac{0.2}{60} \times 44\,000 = 146.67$ kW.

#### (i) Brake thermal efficiency
$$
\eta_{bth} = \frac{37.165}{146.67} = 0.2534 = 25.34\%
$$

#### (ii) Indicated thermal efficiency
$$
\eta_{ith} = \frac{41.295}{146.67} = 0.2816 = 28.16\%
$$

#### (iii) Indicated mean effective pressure
For a four-stroke, $k$-cylinder engine: $IP = \dfrac{p_m L A (N/2) k}{60}$.
$$
\begin{aligned}
A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\
p_m &= \frac{60\, IP}{L A (N/2) k} = \frac{60 \times 41\,295}{0.12 \times 7.854\times10^{-3} \times 900 \times 4} \\
&= 7.302\times10^5\ \text{N/m}^2 = 7.30\ \text{bar}
\end{aligned}
$$

#### (iv) Brake specific fuel consumption
$$
bsfc = \frac{0.2 \times 60}{37.165} = \frac{12}{37.165} = 0.3229\ \text{kg/kWh}
$$

**Answer:** $\eta_{bth}$ = 25.34 %, $\eta_{ith}$ = 28.16 %, imep = 7.30 bar, bsfc = 0.323 kg/kWh (BP = 37.17 kW, IP = 41.30 kW).

## Check yourself

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

1. How is the effective radius for torque found for a rope brake?
   A. Drum radius from circumference only
   B. Drum radius plus rope radius (to rope centre)
   C. Drum diameter
   D. Drum radius minus rope radius
   Answer: B. R_d = 3.0/(2 pi) = 0.4775 m, plus rope radius 0.025 m gives 0.5025 m. (Key point: Effective radius)

2. What must be done with the net tension of 40 kg?
   A. Use as 40 N
   B. Multiply by 9.81 to get 392.4 N
   C. Divide by 9.81
   D. Square it
   Answer: B. Mass-based load is converted to force: F = 40 x 9.81 = 392.4 N. (Key point: Force from kgf)

3. How is IP obtained here?
   A. IP = BP x 0.9
   B. IP = BP / 0.9 = 41.295 kW
   C. IP = BP + 0.9
   D. IP = BP - 0.9
   Answer: B. Mechanical efficiency = BP/IP, so IP = 37.165/0.9. (Key point: Indicated power)

4. Which factors appear in IP = p_m L A (N/2) k / 60?
   A. N/2 for four-stroke and k = 4 cylinders
   B. N for four-stroke and k = 1
   C. N/2 and k = 1
   D. 2N and k = 4
   Answer: A. Four-stroke fires every two revolutions (N/2) and all 4 cylinders contribute. (Key point: imep for multi-cylinder four-stroke)

5. How is bsfc computed?
   A. 0.2 kg/min divided by BP directly
   B. Fuel in kg/h (0.2 x 60 = 12) divided by BP in kW
   C. BP divided by fuel
   D. Fuel x CV / BP
   Answer: B. bsfc = 12/37.165 = 0.3229 kg/kWh. (Key point: bsfc)

## 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 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×

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