A 12 pole, 3-phase alternator drives at a speed of 500 r.p.m supplies power to an 8 pole, three-phase induction motor. If the slip of the motor at full load speed is 4 %, calculate the full load of the motor.
We can determine this as follows:
Given that, speed of alternator (NS) = 500
Number of poles (P) = 12
\( \Rightarrow 500 = \frac{{120f}}{12} \Rightarrow f = 50\;Hz\)
The number of poles in induction motor = 8
The synchronous speed of the induction motor is,
\({N_s} = \frac{{120 \times 50}}{8} = 750\;rpm\)
Slip (s) = 4 % = 0.04
Full load speed, Nr = 750(1 – 0.04) = 727.5 rpm
Rounded off, we get 728 rpm
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