Question

A petrol engine develops a torque of 10 N-m, at a speed of 1500 rpm. The indicated power of the engine is 1.85 kW. Find the friction power.

Answer 1

A petrol engine develops a torque of 10 N-m, at a speed of 1500 rpm. The indicated power of the engine is 1.85 kW. The friction power. is 0.28 kW

Explanation:

This is a question of Mechanical Engineering

Given

Torque developed due to brake T = 10 N-m

Petrol Engine Speed N = 1500 rpm

Brake Power is given by

$\boxed{\text{BP}=\frac{2\pi NT}{1000\times 60}}$ kW

$\implies \text{BP}=\frac{2\times 3.1415\times 1500\times 10}{1000\times 60}$ kW

$\implies \text{BP}=1.57$ kW

Given,

Indicated Power of the engine

IP = 1.85 kW

We know that Friction Power (FP) is given by the following formula

FP = IP - BP

Therefore, Friction Power

FP = 1.85 - 1.57 = 0.28 kW

Hope this answer is helpful.

Know More:

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