---
title: "Engine test: four-cylinder at 1800 rpm"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:engine-test-1800-rpm"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/engine-test-1800-rpm
site: BCT Engineering
---

# Engine test: four-cylinder at 1800 rpm

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2081 Bhadra, 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-1800-rpm
- 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 100 mm, Stroke = 120 mm, 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 300 cm. if the mechanical efficiency is 90%, determine i) Brake thermal efficiency. ii) Indicated thermal efficiency. iii) Indicated mean effective pressure. iv) Brake specific fuel consumption.

## Short answer

Using net brake load 392.4 N and brake circumference 3 m at 1800 rpm gives BP=35.32 kW; fuel heat=146.67 kW. With ηm=90%, IP=39.24 kW, giving ηbth=24.08%, ηith=26.75%. For four-stroke n=15, IMEP=6.94 bar; BSFC=0.340 kg/kWh.

## Answer

**Data:** 4 cylinders ($k = 4$), four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1800$ rpm, $\dot m_f = 0.2$ kg/min, $CV = 44000$ kJ/kg, net brake load $(W-S) = 40$ kg, brake circumference $\pi D_b = 3$ m, $\eta_m = 0.9$. Take $g = 9.81\ \text{m/s}^2$.

#### Brake power

$$
\begin{aligned}
(W-S) &= 40\times9.81 = 392.4\ \text{N} \\
BP &= \frac{(W-S)\,\pi D_b\,N}{60} = \frac{392.4\times3\times1800}{60} = 35.32\ \text{kW}
\end{aligned}
$$

#### Heat supplied by fuel

$$
\dot m_f = \frac{0.2}{60} = 0.003333\ \text{kg/s},\qquad Q = \dot m_f\times CV = 0.003333\times44000 = 146.67\ \text{kW}
$$

#### i) Brake thermal efficiency

$$
\eta_{bth} = \frac{BP}{Q} = \frac{35.32}{146.67} = 24.08\ \%
$$

#### ii) Indicated thermal efficiency

$$
\begin{aligned}
IP &= \frac{BP}{\eta_m} = \frac{35.32}{0.9} = 39.24\ \text{kW} \\
\eta_{ith} &= \frac{IP}{Q} = \frac{39.24}{146.67} = 26.75\ \%
\end{aligned}
$$

(Check: $\eta_{ith} = \eta_{bth}/\eta_m$.)

#### iii) Indicated mean effective pressure

For a four-stroke engine, power strokes per second per cylinder $n = N/(2\times60) = 15$.

$$
\begin{aligned}
A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\
IP &= p_{mi}\,L\,A\,n\,k \\
p_{mi} &= \frac{39.24\times10^3}{0.12\times7.854\times10^{-3}\times15\times4} = 6.939\times10^5\ \text{N/m}^2 = 6.94\ \text{bar}
\end{aligned}
$$

#### iv) Brake specific fuel consumption

$$
BSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{35.32} = \frac{12}{35.32} = 0.340\ \text{kg/kWh}
$$

**Answer:** $\eta_{bth}$ = **24.08 %**, $\eta_{ith}$ = **26.75 %**, IMEP = **6.94 bar**, BSFC = **0.340 kg/kWh** (BP = 35.32 kW).

## Check yourself

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

1. What is BP at 1800 rpm with net load 40 kg and brake circumference 3 m?
   A. 35.32 kW
   B. 39.24 kW
   C. 24.08 kW
   D. 146.67 kW
   Answer: A. BP = 40 x 9.81 x 3 x 1800/60 = 35.32 kW. (Key point: Brake power)

2. What is the heat supplied by fuel (0.2 kg/min, 44,000 kJ/kg)?
   A. 146.67 kW
   B. 8800 kW
   C. 35.32 kW
   D. 2.2 kW
   Answer: A. Q = (0.2/60) x 44,000 = 146.67 kW. (Key point: Heat input)

3. How are brake and indicated thermal efficiency related?
   A. eta_ith = eta_bth/eta_m, so 24.08% gives 26.75%
   B. eta_ith = eta_bth x eta_m
   C. They are always equal
   D. eta_ith = eta_bth - eta_m
   Answer: A. IP = BP/eta_m so eta_ith = eta_bth/eta_m = 24.08/0.9 = 26.75%. (Key point: Indicated efficiency relation)

4. What does IP = p_mi L A n k use for k and n?
   A. k = 4 cylinders, n = N/120 = 15 power strokes/s
   B. k = 1, n = 1800
   C. k = 4, n = N/60 = 30
   D. k = 2, n = 15
   Answer: A. Four cylinders, one power stroke per two revolutions; this gives p_mi = 6.94 bar. (Key point: IMEP calculation)

## 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, 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×
- [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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