Question

An electric motor consumes 12.0kJ of electrical energy in 1.00min . If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500rpm ?

Answer 1

Concepts and reason

The concepts required to solve the given question are power and torque.

First, calculate the portion of the power that appears as the output of the motor. Convert the angular velocity from rotations per minute to radians per minute. Finally, calculate the torque developed by the engine by using the relation between the power, torque and angular velocity.

Fundamentals

Power is defined as the “rate of doing work”.

The expression for power is given by,

Here, is the power, is the work done, and is the time taken.

Torque is the “twisting force that tends to cause the rotation of an object”. Angular velocity may be defined as the rate of change of angular position of an object.

The power, torque and angular velocity of the object can be related as,

Here, is the power, is the torque, and is the angular velocity.

Here, one – third of the total energy gets transformed into heat and other forms of the internal energy of the motor. Power appears as the output of the motor is only the two – third of the total energy is going to the output of the motor.

Total power generated is,

Substitute for and for .

The power delivered as the motor output is two – third of the total power. It is given by,

Here, is the power delivered as the motor output.

Substitute for .

The torque developed by the engine is,

Substitute for and for to find .

Ans:

The torque developed by the engine is .