---
title: "IP, BP and efficiencies of a petrol engine"
subject: "Internal Combustion Engines"
chapter: 8
key: "semester_6/internal_combustion_engines:indicated-brake-power-efficiency-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/indicated-brake-power-efficiency-numerical
site: BCT Engineering
---

# IP, BP and efficiencies of a petrol engine

Internal Combustion Engines · Chapter 8: Engine performance and testing of engines

- Asked in: year not given, 8 marks
- Video: [ENGINE PERFORMANCE|| IP, BP, FP, efficiencies, Fuel consumption, BSFC, ISFC etc.](https://www.youtube.com/watch?v=99rE11Yg5NM) — AutoGent
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/indicated-brake-power-efficiency-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-8

## Question

A four-cylinder, four-stroke petrol engine has bore 80 mm and stroke 100 mm and runs at 3000 rpm. The indicated mean effective pressure is 9 bar. A brake with net load of 24 kg acting at an effective radius of 0.5 m is used, and the fuel consumption is 11.2 kg/h of petrol with a calorific value of 44 000 kJ/kg. Calculate the indicated power, brake power, friction power, mechanical efficiency, indicated and brake thermal efficiencies, and the brake and indicated specific fuel consumption.

## Short answer

IP from pmi*L*A*(N/2)*n/1000 = 45.2 kW; BP from torque W*R and 2πNT/60000 = 37.0 kW, FP = 8.26 kW, ηm =81.8%. Fuel energy 136.9 kW gives ηith =33.0%, ηbth =27.0%; bsfc =0.303 kg/kWh, isfc =0.248 kg/kWh.

## Answer

Given: $D = 0.08$ m, $L = 0.1$ m, $n = 4$, $N = 3000$ rpm, $p_{mi} = 9 \times 10^5$ Pa, $W = 24 \times 9.81 = 235.4$ N, $R = 0.5$ m, $\dot{m}_f = 11.2$ kg/h, $CV = 44\,000$ kJ/kg.

#### Indicated power
For a four-stroke engine the number of power strokes per cylinder per second is $N/2 = 25$.

$$
\begin{aligned}
A &= \frac{\pi}{4}(0.08)^2 = 5.027 \times 10^{-3}\ \text{m}^2 \\
IP &= \frac{p_{mi} L A \cdot (N/2) \cdot n}{1000} = \frac{9 \times 10^5 \times 0.1 \times 5.027 \times 10^{-3} \times 25 \times 4}{1000} = 45.24\ \text{kW}
\end{aligned}
$$

#### Brake power
$$
\begin{aligned}
T &= W R = 235.44 \times 0.5 = 117.7\ \text{N m} \\
BP &= \frac{2\pi N T}{60\,000} = \frac{2\pi \times 3000 \times 117.7}{60\,000} = 36.98\ \text{kW}
\end{aligned}
$$

#### Friction power and mechanical efficiency
$$
\begin{aligned}
FP &= IP - BP = 45.24 - 36.98 = 8.26\ \text{kW} \\
\eta_m &= \frac{BP}{IP} = \frac{36.98}{45.24} = 0.8175
\end{aligned}
$$

#### Thermal efficiencies
$$
\begin{aligned}
\text{Fuel energy} &= \frac{11.2}{3600} \times 44\,000 = 136.9\ \text{kW} \\
\eta_{bth} &= \frac{36.98}{136.9} = 0.270 \\
\eta_{ith} &= \frac{45.24}{136.9} = 0.330
\end{aligned}
$$

Check: $\eta_m \eta_{ith} = 0.8175 \times 0.330 = 0.270$.

#### Specific fuel consumption
$$
bsfc = \frac{11.2}{36.98} = 0.303\ \text{kg/kWh}, \qquad isfc = \frac{11.2}{45.24} = 0.248\ \text{kg/kWh}
$$

**Answer:** IP = 45.2 kW, BP = 37.0 kW, FP = 8.26 kW, $\eta_m$ = 81.8%, $\eta_{ith}$ = 33.0%, $\eta_{bth}$ = 27.0%, bsfc = 0.303 kg/kWh, isfc = 0.248 kg/kWh.

## Check yourself

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

1. For a four-stroke engine, how many power strokes per cylinder per second are used for IP?
   A. N
   B. N/2
   C. 2N
   D. N/4
   Answer: B. A four-stroke fires once every two revolutions, so power strokes per second = N/2 = 25 at 3000 rpm. (Key point: Indicated power for four-stroke)

2. How is the brake power found from the brake load?
   A. T = W R = 117.7 N m, BP = 2 pi N T/60000 = 36.98 kW
   B. T = W/R, BP = 2 pi N T
   C. T = W x R, BP = 45.2 kW
   D. T = W R, BP = N T/1000
   Answer: A. W = 24 x 9.81 = 235.4 N, T = 235.4 x 0.5 = 117.7 N m, BP = 2 pi x 3000 x 117.7/60000 = 36.98 kW. (Key point: Brake power calculation)

3. With IP = 45.24 kW and BP = 36.98 kW, what are FP and mechanical efficiency?
   A. 8.26 kW and 81.8%
   B. 82.2 kW and 18.2%
   C. 8.26 kW and 122%
   D. 36.98 kW and 45%
   Answer: A. FP = 45.24 - 36.98 = 8.26 kW and eta_m = 36.98/45.24 = 0.8175. (Key point: Mechanical efficiency and friction)

4. What are the fuel energy rate and brake thermal efficiency?
   A. 136.9 kW and 27.0%
   B. 136.9 kW and 33.0%
   C. 11.2 kW and 27%
   D. 493 kW and 7.5%
   Answer: A. Fuel energy = 11.2/3600 x 44000 = 136.9 kW; eta_bth = 36.98/136.9 = 0.270 (indicated is 0.330). (Key point: Thermal efficiencies)

## Related in Internal Combustion Engines

- [Two-stroke and four-stroke engine comparison](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Atwo-stroke-versus-four-stroke-engine) — Internal Combustion Engines, chapter 2
- [Valve timing diagram of four-stroke engine](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Avalve-timing-diagram-four-stroke) — Internal Combustion Engines, chapter 4
- [Engine displacement and mean piston speed](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Aengine-displacement-piston-speed-numerical) — Internal Combustion Engines, chapter 2
- [Four-stroke SI engine working and components](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Afour-stroke-si-engine-working-components) — Internal Combustion Engines, chapter 2
- [Engine performance parameters and efficiencies](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Aengine-performance-parameters-definitions) — Internal Combustion Engines, chapter 8
- [Otto cycle efficiency and mean effective pressure](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Aotto-cycle-efficiency-numerical) — Internal Combustion Engines, chapter 1

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