Find out which of the two motors:
Two three-phase induction motors while connected across a 400 V, 50 Hz supply are running at 1440 and 940 RPM respectively. Find out which of the two motors is running at higher slip.
Solution
We know that f = 50 Hz and we may determine the synchronous speed for
P = 2, N_{ s} = (120 × 50 )/2 = 3000 RPM
P = 4, N _{s } = (120 × 50 )/4 = 1500 RPM
P = 6, N_{ s} = (120 × 50 )/6 = 1000 RPM
An induction motor runs at a speed somewhat less than the synchronous speed. Numbers of poles of the motor running at 1440 rpm must be 4 and N_{s} = 1500 rpm.
% Slip of the motor (A),
S_{A} = ((N_{s }- N_{r}) /N_{s}) × 100 = (1500 - 1440/1500) × 1000 = 4%
Numbers of poles of the motor running at 940 rpm must be six and
N_{s} = 1000 rpm.
The slip of the motor (B),
s _{B} = (1000 - 940 )/1000= 6%
Therefore the slip of the motor running at 940 rpm is higher than the slip of the motor running at 1440 rpm.