---
title: "Two-stroke motoring test with specific fuel consumption"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:motoring-test-sfc"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/motoring-test-sfc
site: BCT Engineering
---

# Two-stroke motoring test with specific fuel consumption

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2072 Kartik, 10 marks
- Video: [IC Engine Performance Parameters (Indicated Power, Brake Power, Friction Power & Efficiency)](https://www.youtube.com/watch?v=uLqdBl21Rk0) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/motoring-test-sfc
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

A diesel engine power plant operated by a two stroke diesel engine was motored when the meter reading was 4.5kW. Then the test on the engine was carried out for one hour run and the following observations were recorded: (i) Brake torque = 250 Nm; (ii) Speed = 1500rpm; (iii) Fuel consumed = 5kg/hr and (iv) Calorific value of fuel = 40MJ/kg. Determine: i) Mechanical efficiency ii) Indicated thermal efficiency iii) Brake thermal efficiency iv) Indicated specific fuel consumption and v) Brake specific fuel consumption

## Short answer

Friction power from motoring = 4.5 kW. BP = 39.27 kW, IP = 43.77 kW. Mechanical efficiency = 89.72%, indicated thermal efficiency = 78.79%, brake thermal efficiency = 70.69%, isfc = 0.114 kg/kWh, bsfc = 0.127 kg/kWh.

## Answer

In a **motoring test** the engine is driven by the dynamometer without fuel; the motoring power equals the **friction power**: $FP = 4.5$ kW.

**Data:** $T = 250$ N m, $N = 1500$ rpm, fuel = 5 kg/h, CV = 40 MJ/kg = 40 000 kJ/kg.

Brake power and indicated power:
$$
\begin{aligned}
BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 1500 \times 250}{60} = 39.27\ \text{kW} \\
IP &= BP + FP = 39.27 + 4.5 = 43.77\ \text{kW}
\end{aligned}
$$
Heat supplied:
$$
\dot Q = \frac{5}{3600}\times 40\,000 = 55.56\ \text{kW}
$$

#### i) Mechanical efficiency
$$
\eta_m = \frac{39.27}{43.77} = 0.8972 = 89.72\%
$$

#### ii) Indicated thermal efficiency
$$
\eta_{ith} = \frac{43.77}{55.56} = 0.7879 = 78.79\%
$$

#### iii) Brake thermal efficiency
$$
\eta_{bth} = \frac{39.27}{55.56} = 0.7069 = 70.69\%
$$

#### iv) Indicated specific fuel consumption
$$
isfc = \frac{\dot m_f}{IP} = \frac{5}{43.77} = 0.1142\ \text{kg/kWh}
$$

#### v) Brake specific fuel consumption
$$
bsfc = \frac{\dot m_f}{BP} = \frac{5}{39.27} = 0.1273\ \text{kg/kWh}
$$

**Answer:** $\eta_m$ = 89.72 %, $\eta_{ith}$ = 78.79 %, $\eta_{bth}$ = 70.69 %, isfc = 0.114 kg/kWh, bsfc = 0.127 kg/kWh.

(The efficiencies are unrealistically high for a real engine because the given fuel rate is low; the method is the point of the problem.)

## Check yourself

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

1. In a motoring test, what does the power reading of the dynamometer give?
   A. Friction power of the engine
   B. Indicated power
   C. Brake power
   D. Heat supplied by the fuel
   Answer: A. The engine is driven without fuel, so the power absorbed equals the friction power. (Key point: Motoring test)

2. How is indicated power found in this problem?
   A. IP = BP - FP
   B. IP = BP + FP = 39.27 + 4.5
   C. IP = BP x mechanical efficiency
   D. IP = FP / BP
   Answer: B. Indicated power is brake power plus friction power. (Key point: Indicated power)

3. Which formula gives the brake power from torque and speed?
   A. BP = 2*pi*N*T
   B. BP = N*T/60
   C. BP = 2*pi*N*T/60
   D. BP = T/(2*pi*N)
   Answer: C. With N in rpm and T in N m, 2*pi*N*T/60 gives watts; here 39.27 kW. (Key point: Brake power)

4. How are the brake thermal efficiency and bsfc obtained?
   A. Brake thermal = IP / heat supplied; bsfc = fuel per hour / IP
   B. Brake thermal = BP / IP; bsfc = BP / fuel
   C. Brake thermal = heat supplied / BP; bsfc = BP / fuel per hour
   D. Brake thermal = BP / heat supplied; bsfc = fuel per hour / BP
   Answer: D. Brake quantities use BP; indicated ones use IP. Heat supplied is fuel rate x CV (55.56 kW). (Key point: Efficiencies and sfc)

## Close variants set in other exams

- [Two-stroke motoring test: 4.5 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-4-5kw) — Power Plant Equipment, chapter 5, asked 6×
- [Two-stroke motoring test: 5 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-5kw) — Power Plant Equipment, chapter 5, asked 6×
- [Two-stroke motoring test: 2.2 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-2-2kw) — Power Plant Equipment, chapter 5, asked 6×
- [Two-stroke motoring test: 2 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-2kw) — Power Plant Equipment, chapter 5, asked 6×
- [Two-stroke motoring test: 1.5 kW motoring](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-1-5kw) — Power Plant Equipment, chapter 5, asked 6×

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