---
title: "Two-stroke motoring test: 4.5 kW motoring"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:motoring-test-4-5kw"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/motoring-test-4-5kw
site: BCT Engineering
---

# Two-stroke motoring test: 4.5 kW motoring

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2076 Chaitra, 8 marks
- Video: [MOTORING TEST & WILLIAM'S LINE METHOD FOR MEASURING FRICTION POWER](https://www.youtube.com/watch?v=AaGAOH6N7UU) — Professor kartik
- 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-4-5kw
- 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 4.5 KW. Then the test on the engine was carried out for one hour run and the following observations were recorded: (i) Brake torque = 250Nm; (ii) Speed=1500rpm; (iii) Fuel consumed = 5 kg and (iv) Calorific value of fuel = 40MJ/kg. Determine: a) Mechanical efficiency b) Indicated thermal efficiency c) Brake thermal efficiency

## Short answer

Use motoring power as friction power, 4.5 kW. BP=2π(1500)(250)/60=39.27 kW, IP=43.77 kW, fuel heat=55.56 kW. Mechanical efficiency=89.72%, indicated thermal efficiency=78.79%, brake thermal efficiency=70.69%.

## Answer

In a **motoring test**, the engine (without fuel) is driven by the dynamometer acting as a motor. The power needed to turn it at the test speed equals the **friction power** of the engine. So $FP = 4.5$ kW.

**Data:** $T = 250$ N m, $N = 1500$ rpm, fuel = 5 kg in 1 h, CV = 40 MJ/kg.

Brake power:
$$
BP = \frac{2\pi N T}{60} = \frac{2\pi \times 1500 \times 250}{60} = 39\,270\ \text{W} = 39.27\ \text{kW}
$$
Indicated power:
$$
IP = BP + FP = 39.27 + 4.5 = 43.77\ \text{kW}
$$
Heat supplied by fuel:
$$
\dot Q = \frac{5}{3600} \times 40\,000 = 55.56\ \text{kW}
$$

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

#### b) Indicated thermal efficiency
$$
\eta_{ith} = \frac{IP}{\dot Q} = \frac{43.77}{55.56} = 0.7879 = 78.79\%
$$

#### c) Brake thermal efficiency
$$
\eta_{bth} = \frac{BP}{\dot Q} = \frac{39.27}{55.56} = 0.7069 = 70.69\%
$$

**Answer:** $\eta_m$ = 89.72 %, $\eta_{ith}$ = 78.79 %, $\eta_{bth}$ = 70.69 %.

(These thermal efficiencies are far higher than any real diesel engine, about 30–40 %, because the given fuel consumption is low; the method is what matters.)

## Check yourself

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

1. What does the 4.5 kW motoring reading represent?
   A. Brake power
   B. Indicated power
   C. Friction power
   D. Heat supplied
   Answer: C. The dynamometer drives the unfuelled engine; the power needed equals its friction power. (Key point: Motoring test meaning)

2. How is BP found from torque 250 N m at 1500 rpm?
   A. BP = 2 pi N T / 60 = 39.27 kW
   B. BP = N x T
   C. BP = T / N
   D. BP = 250 x 1500 W
   Answer: A. BP = 2 pi x 1500 x 250 / 60 = 39,270 W. (Key point: Brake power)

3. How is IP calculated?
   A. IP = BP - FP
   B. IP = BP + FP = 43.77 kW
   C. IP = BP x FP
   D. IP = FP / BP
   Answer: B. Indicated power is brake power plus friction power. (Key point: Indicated power)

4. How is heat supplied found for 5 kg in 1 h with CV 40 MJ/kg?
   A. 5 x 40 = 200 kW
   B. (5/3600) x 40,000 = 55.56 kW
   C. 5 x 3600
   D. 40,000/5
   Answer: B. Fuel rate in kg/s times CV in kJ/kg gives kW. (Key point: Heat supplied)

5. Which pair is correct?
   A. eta_m = IP/BP
   B. eta_m = BP/IP = 89.72%; eta_ith = IP/Q
   C. eta_bth = IP/Q
   D. eta_ith = BP/Q
   Answer: B. eta_m = 39.27/43.77 = 89.72%, eta_ith = 43.77/55.56 = 78.79%. (Key point: Efficiencies)

## Close variants set in other exams

- [Two-stroke motoring test with specific fuel consumption](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Amotoring-test-sfc) — 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×

## Related in Power Plant Equipment

- [Diesel engine 75 kW: fuel consumption and efficiencies](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adiesel-75kw-fuel-consumption) — Power Plant Equipment, chapter 5
- [Engine test: 4-cylinder, B.P. 31.4 kW](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-bp-31-4kw) — Power Plant Equipment, chapter 5, asked 2×
- [Rope dynamometer trial: brake and indicated power](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Arope-dynamometer-trial) — Power Plant Equipment, chapter 5