In this topic, you study Efficiency of Induction Motor.
The ratio of net power output at shaft to the power input to motor is known as efficiency of an induction motor.
Various stages of this power conversion and the accompanying losses are diagrammatically represented in Fig. 1. Thus,
Efficiency,
or,
At light loads, for the given input, the constant losses being large as compared with the output, motor efficiency is low. As the load increases, the efficiency increases and ultimately attains its maximum value. The constant and variable losses are equal under this condition. After this, the variable losses become relatively large and hence the efficiency Starts decreasing. Thus
the shape of the efficiency curve for the motor is as shown in Fig. 2.
Solved problems on EFFICIENCY OF AN INDUCTION MOTOR
Example 1
A 3-phase, 4 pole, 50 Hz induction motor runs at 1440 r.p.m. while delivering the power output of 40 kW. The Motor losses at this load are equal to rotor losses and mechanical losses amount to 3500 W. Calculate : (i) Slip, (ii) Total mechanical power developed by the rotor, (iii) Rotor copper loss, (iv) Rotor input, (v) Stator input, (vi) Efficiency.
Solution : (i) With f = 50 Hz, P = 4
Synchronous speed,
Thus Slip,
(ii) Total mechanical power developed by the rotor,
(iii) Now,
Thus,
(iv) Rotor input,
(v) Assuming the rotor core loss to be negligibly small, we have
Stator losses = Rotor losses = Rotor copper loss
= 1812.5 w
Thus, Stator input = Rotor input + Stator losses
(vi) Efficiency,
