A machine is coupled to a two-stroke engine which produces a torque of (T) = [800 + 180 sin 3 θ] (N.m)

Where θ is the crank angle. The mean engine speed is 600 rpm. What will be the power of engine? [π = 3.14]

Question

Question

This question was previously asked in

BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper

Answer 1

Concept:

Power developed by a rotating shaft is given by: \( P = T_{\text{mean}} \cdot \omega \)

Where, \( T_{\text{mean}} \) is the mean torque and \( \omega \) is the angular speed in rad/s.

Given:

Torque, \(T(\theta) = 800 + 180 \sin(3\theta) \, \text{Nm}\)

Speed, N = \(600 \, \text{rpm} , \pi = 3.14\)

Calculation:

Since the average of a sine function over a complete cycle is 0, the mean torque:

\( T_{\text{mean}} = 800 \, \text{Nm} \)

Angular velocity, \( \omega = \frac{2\pi N}{60} = \frac{2 \times 3.14 \times 600}{60} = 62.8 \, \text{rad/s} \)

Power, \( P = 800 \times 62.8 = 50240 \, \text{W} = 50.24 \, \text{kW} \)