---
title: "Engine test: 4-cylinder, B.P. 31.4 kW"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:engine-test-bp-31-4kw"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/engine-test-bp-31-4kw
site: BCT Engineering
---

# Engine test: 4-cylinder, B.P. 31.4 kW

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2080 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-bp-31-4kw
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

A 4-cylinder, 4-stroke cycle engine having cylinder diameter 100 mm and stroke 120 mm was tested at 1600 rpm and the following readings were obtained. Fuel consumption = 0.2 kg/min, B.P. = 31.4 kW, Mechanical efficiency = 80%, Calorific value of fuel = 44000 kJ/kg. Determine: i) BSFC (Brake Specific Fuel Consumption) ii) IMEP (Indicated Mean Effective Pressure), and iii) Brake thermal efficiency.

## Short answer

BSFC = (0.2×60)/31.4 = 0.382 kg/kWh. IP = 31.4/0.8 = 39.25 kW, and for four-stroke n=13.333, giving IMEP = 7.81 bar. ηbth = 31.4/146.67 = 21.41%.

## Answer

**Data:** 4 cylinders ($k=4$), four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1600$ rpm, $\dot m_f = 0.2$ kg/min, $BP = 31.4$ kW, $\eta_m = 0.8$, $CV = 44{,}000$ kJ/kg.

#### i) Brake specific fuel consumption

$$
BSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{31.4} = \frac{12}{31.4} = 0.382\ \text{kg/kWh}
$$

#### ii) Indicated mean effective pressure

$$
\begin{aligned}
IP &= \frac{BP}{\eta_m} = \frac{31.4}{0.8} = 39.25\ \text{kW} \\
n &= \frac{N}{2\times60} = \frac{1600}{120} = 13.333\ \text{power strokes/s per cylinder} \\
A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\
p_{mi} &= \frac{IP}{L\,A\,n\,k} = \frac{39.25\times10^3}{0.12\times7.854\times10^{-3}\times13.333\times4} \\
&= 7.809\times10^5\ \text{N/m}^2 = 7.81\ \text{bar}
\end{aligned}
$$

#### iii) Brake thermal efficiency

$$
\begin{aligned}
Q &= \frac{0.2}{60}\times44{,}000 = 146.67\ \text{kW} \\
\eta_{bth} &= \frac{BP}{Q} = \frac{31.4}{146.67} = 0.2141 = 21.41\ \%
\end{aligned}
$$

**Answer:** BSFC = **0.382 kg/kWh**, IMEP = **7.81 bar**, $\eta_{bth}$ = **21.41 %**.

## Check yourself

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

1. What is BSFC for 0.2 kg/min and BP = 31.4 kW?
   A. 0.382 kg/kWh
   B. 0.006 kg/kWh
   C. 12 kg/kWh
   D. 0.2 kg/kWh
   Answer: A. Convert to 12 kg/h and divide by BP: 12/31.4 = 0.382 kg/kWh. (Key point: BSFC)

2. With BP = 31.4 kW and eta_m = 0.8, what is IP?
   A. 25.1 kW
   B. 39.25 kW
   C. 31.4 kW
   D. 0.8 kW
   Answer: B. IP = BP/eta_m = 31.4/0.8 = 39.25 kW. (Key point: Indicated power)

3. At 1600 rpm for a four-stroke cylinder, n equals:
   A. 26.67 per second
   B. 13.33 per second
   C. 1600 per second
   D. 6.67 per second
   Answer: B. n = N/(2 x 60) = 13.33 power strokes per second per cylinder. (Key point: Power strokes per second)

4. How is brake thermal efficiency found?
   A. BP / heat supplied = 31.4/146.67 = 21.41%
   B. IP / heat supplied
   C. Heat supplied / BP
   D. BP x eta_m
   Answer: A. eta_bth = BP/(m_f x CV) with Q = 146.67 kW gives 21.41%. (Key point: Brake thermal efficiency)

## Close variants set in other exams

- [Engine test: fuel measured in litres](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-fuel-litres) — Power Plant Equipment, chapter 5, asked 2×

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