---
title: "Rope dynamometer trial: brake and indicated power"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:rope-dynamometer-trial"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/rope-dynamometer-trial
site: BCT Engineering
---

# Rope dynamometer trial: brake and indicated power

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2073 Shrawan, 8 marks
- Video: [Rope Brake Dynamometer || Construction and Working of Rope Brake Dynamometer - Theory of Machine](https://www.youtube.com/watch?v=bLC0LY55w3A) — Gear Institute Mechanical Engineering Videos and Physics by Gear Institute
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/rope-dynamometer-trial
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

The following observations were recorded during a trial of a four stroke engine with rope dynamometer. Engine speed = 650 rpm, Dia. of brake drum = 600 mm, Dia. of rope = 50 mm, Dead load on the brake drum = 32 kg, spring balance reading = 4.75 kg, Mechanical efficiency = 80%. Calculate the brake power and indicated power.

## Short answer

Effective radius is 0.325 m, net brake load 267.32 N, torque 86.88 Nm. Brake power = 5.91 kW, indicated power = 7.39 kW.

## Answer

**Data:** $N = 650$ rpm, drum diameter $D = 0.6$ m, rope diameter $d = 0.05$ m, dead load $W = 32$ kg, spring balance $S = 4.75$ kg, $\eta_m = 0.80$.

For a rope brake dynamometer the effective radius is measured to the rope centre:
$$
R_e = \frac{D + d}{2} = \frac{0.6 + 0.05}{2} = 0.325\ \text{m}
$$

Net brake load:
$$
(W - S)g = (32 - 4.75)\times 9.81 = 267.32\ \text{N}
$$

#### Brake power
$$
\begin{aligned}
T &= (W-S)g\,R_e = 267.32 \times 0.325 = 86.88\ \text{N m} \\
BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 650 \times 86.88}{60} = 5914\ \text{W} = 5.914\ \text{kW}
\end{aligned}
$$

#### Indicated power
$$
IP = \frac{BP}{\eta_m} = \frac{5.914}{0.80} = 7.392\ \text{kW}
$$

**Answer:** BP = 5.91 kW, IP = 7.39 kW (friction power = 1.48 kW).

## Check yourself

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

1. What is the effective radius of the rope brake?
   A. 0.3 m
   B. (D + d)/2 = 0.325 m
   C. 0.6 m
   D. 0.275 m
   Answer: B. Radius is measured to the rope centre: (0.6 + 0.05)/2. (Key point: Effective radius)

2. How is the net brake load found?
   A. (W + S) g
   B. (W - S) g = 27.25 x 9.81 = 267.3 N
   C. W g only
   D. W/S
   Answer: B. The spring balance reading opposes the dead load, so the difference is the net load. (Key point: Net load)

3. What is the BP?
   A. 5.91 kW
   B. 59.1 kW
   C. 0.59 kW
   D. 7.39 kW
   Answer: A. T = 267.32 x 0.325 = 86.88 N m; BP = 2 pi x 650 x 86.88/60 = 5.914 kW. (Key point: Brake power)

4. How is IP obtained from BP and 80% mechanical efficiency?
   A. IP = BP/0.8 = 7.39 kW
   B. IP = BP x 0.8
   C. IP = BP + 0.8
   D. IP = BP - 0.8
   Answer: A. IP = BP/eta_m = 5.914/0.8; friction power = 1.48 kW. (Key point: Indicated power)

## Related in Power Plant Equipment

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