---
title: "Engine test: four-cylinder at 1820 rpm"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:engine-test-1820-rpm"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/engine-test-1820-rpm
site: BCT Engineering
---

# Engine test: four-cylinder at 1820 rpm

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2082 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/engine-test-1820-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 = 1820 rpm, fuel consumption of 0.2 kg/min, calorific value of fuel is 44,100 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

Use net brake load 392.4 N and brake circumference 3 m to get BP=35.71 kW; fuel heat=147.0 kW. With ηm=90%, IP=39.68 kW, so ηbth=24.29%, ηith=26.99%. For four-stroke n=N/120=15.167, IMEP=6.94 bar; BSFC=0.336 kg/kWh.

## Answer

**Data:** 4 cylinders ($k = 4$), four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1820$ rpm, $\dot m_f = 0.2$ kg/min, $CV = 44100$ 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

$$
\begin{aligned}
(W-S) &= 40\times9.81 = 392.4\ \text{N} \\
BP &= \frac{(W-S)\,\pi D_b\,N}{60} = \frac{392.4\times3\times1820}{60} = 35.71\ \text{kW}
\end{aligned}
$$

#### Heat supplied by fuel

$$
\dot m_f = \frac{0.2}{60} = 0.003333\ \text{kg/s},\qquad Q = \dot m_f\times CV = 0.003333\times44100 = 147.0\ \text{kW}
$$

#### i) Brake thermal efficiency

$$
\eta_{bth} = \frac{BP}{Q} = \frac{35.71}{147.0} = 24.29\ \%
$$

#### ii) Indicated thermal efficiency

$$
\begin{aligned}
IP &= \frac{BP}{\eta_m} = \frac{35.71}{0.9} = 39.68\ \text{kW} \\
\eta_{ith} &= \frac{IP}{Q} = \frac{39.68}{147.0} = 26.99\ \%
\end{aligned}
$$

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

#### iii) Indicated mean effective pressure

For a four-stroke engine, power strokes per second per cylinder $n = N/(2\times60) = 15.167$.

$$
\begin{aligned}
A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\
IP &= p_{mi}\,L\,A\,n\,k \\
p_{mi} &= \frac{39.68\times10^3}{0.12\times7.854\times10^{-3}\times15.167\times4} = 6.939\times10^5\ \text{N/m}^2 = 6.94\ \text{bar}
\end{aligned}
$$

#### iv) Brake specific fuel consumption

$$
BSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{35.71} = \frac{12}{35.71} = 0.336\ \text{kg/kWh}
$$

**Answer:** $\eta_{bth}$ = **24.29 %**, $\eta_{ith}$ = **26.99 %**, IMEP = **6.94 bar**, BSFC = **0.336 kg/kWh** (BP = 35.71 kW).

## Check yourself

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

1. How is BP found from the net brake load of 40 kg and brake circumference 3 m?
   A. BP = (W-S) x 9.81 x 3 x N/60 = 35.71 kW
   B. BP = 40 x 3 x 1820 = 218 kW
   C. BP = 40 x 9.81/3 = 130 W
   D. BP = 2 pi N x 40
   Answer: A. Force = 40 x 9.81 = 392.4 N, and BP = force x circumference x rev/s = 392.4 x 3 x 1820/60 = 35.71 kW. (Key point: Brake power from brake load)

2. Given Q = 147 kW and eta_m = 0.9, which pair is correct?
   A. eta_bth = 24.29%, eta_ith = 26.99%
   B. eta_bth = 26.99%, eta_ith = 24.29%
   C. eta_bth = 35.7%, eta_ith = 39.7%
   D. eta_bth = 90%, eta_ith = 24%
   Answer: A. eta_bth = BP/Q = 35.71/147; eta_ith = eta_bth/eta_m which is greater because IP exceeds BP. (Key point: Thermal efficiencies)

3. For a four-stroke engine, what is the number of power strokes per second per cylinder used in IP = p L A n k?
   A. N/60
   B. N/(2 x 60)
   C. 2N/60
   D. N x 60
   Answer: B. A four-stroke cylinder fires once per two revolutions, so n = N/120 = 15.17 per second. (Key point: IMEP formula for four-stroke)

4. How is brake specific fuel consumption calculated here?
   A. Fuel in kg/h divided by BP in kW: 12/35.71 = 0.336 kg/kWh
   B. BP divided by fuel flow
   C. Fuel in kg/s times BP
   D. Fuel in kg/min divided by IP
   Answer: A. BSFC = (0.2 x 60) kg/h / BP = 0.336 kg/kWh. (Key point: BSFC)

## Close variants set in other exams

- [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×
- [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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