Question 4.7: A 400 V, 50 Hz, three-phase induction motor is rotating at 9...
A 400 V, 50 Hz, three-phase induction motor is rotating at 960 rpm on full load. Calculate the following for the motor:
Number of poles; full-load slip; frequency of rotor-induced EMF; speed of the rotor magnetic field with respect to the rotor.
Also show that both the stator field and the rotor field are stationary with respect to each other.
f = 50 Hz
P = 2, 4, 6, etc.
N_{s} =\frac{120 f}{p}=\frac{120× 50}{2} =3000 rpm ( for P=2)
=\frac{120× 50}{4}=1500 rpm ( for P=4)
Rotor speed is somewhat less than the synchronous speed N_{s} . Logically, here N_{s} can only be 1500 rpm, when N_{s} = 1500 rpm, P = 4.
Full load slip, S=\frac{ N_{s}- N_{r} }{ N_{s} }=\frac{1500-1400}{1500}=0.04
Frequency of rotor induced EMF f_{r} = S × f = 0.04 × 50
= 2 Hz
Speed of rotor field with respect to rotor, N is
N=\frac{120×f_{r}}{p}=\frac{120× 2}{4} =60 rpm
The rotor rotates at a speed of 1440 rpm. This means the speed of the rotor with respect to stator, which is stationary, is 1440 rpm.
The speed of the rotor field with respect to the rotor is 60 rpm.
Therefore, the speed of the rotor field with respect to the stator is 1440 + 60 = 1500 rpm. And, the speed of the rotating magnetic field produced by the stator rotates at synchronous speed, N_{s} with respect to the stator. In this case the speed of rotating field produced by the stator is 1500 rpm. Thus, we see that both the magnetic fields of the stator and rotor are stationary with respect to each other, which is, of course, the essential condition for production of torque.