---
title: "Friction power as percentage of indicated power"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:friction-power-percentage"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/friction-power-percentage
site: BCT Engineering
---

# Friction power as percentage of indicated power

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2074 Chaitra, 4 marks
- Video: [Engine performance parameters _ PART 1/Indicated Power/Brake Power/Frictional power.](https://www.youtube.com/watch?v=jwEWCyPa2xE) — Technical Skills by Sumit kashyap
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/friction-power-percentage
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

A single cylinder engine running at 1800 rpm develops a torque of 8 Nm. The indicated power of the engine 1.8 kW. Find the loss due to friction power as the percentage of indicated power.

## Short answer

BP=2π(1800)(8)/60=1.508 kW; FP=IP-BP=1.8-1.508=0.292 kW. Friction loss is 0.292/1.8×100=16.22% of indicated power.

## Answer

**Data:** $N = 1800$ rpm, $T = 8$ N m, $IP = 1.8$ kW.

Brake power:
$$
BP = \frac{2\pi N T}{60} = \frac{2\pi \times 1800 \times 8}{60} = 1508\ \text{W} = 1.508\ \text{kW}
$$

Friction power:
$$
FP = IP - BP = 1.8 - 1.508 = 0.292\ \text{kW}
$$

Friction loss as a percentage of IP:
$$
\frac{FP}{IP}\times 100 = \frac{0.292}{1.8}\times100 = 16.22\%
$$

**Answer:** Friction power = 0.292 kW, which is **16.22 %** of the indicated power (mechanical efficiency = 83.78 %).

## Check yourself

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

1. How is BP found from T = 8 N m at 1800 rpm?
   A. BP = 2 pi N T / 60 = 1.508 kW
   B. BP = N x T = 14.4 kW
   C. BP = T/N
   D. BP = 8 x 1.8 = 14.4 W
   Answer: A. 2 pi x 1800 x 8/60 = 1508 W. (Key point: Brake power)

2. How is friction power found?
   A. FP = IP + BP
   B. FP = IP - BP = 0.292 kW
   C. FP = BP / IP
   D. FP = BP - IP
   Answer: B. Friction power is the difference between indicated and brake power. (Key point: Friction power)

3. What does the friction loss as % of IP equal?
   A. 16.22%
   B. 83.78%
   C. 8.1%
   D. 1.508%
   Answer: A. 0.292/1.8 x 100 = 16.22%. (Key point: Percentage)

4. What is the mechanical efficiency?
   A. 16.22%
   B. 83.78%
   C. 100%
   D. 92%
   Answer: B. eta_m = BP/IP = 1 - 0.1622 = 83.78%. (Key point: Link to mechanical efficiency)

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