Discussion
A three-phase 50 Hz, 12 pole, 200 kW slip-ring induction motor drives a fan whose torque is proportional to the square of the speed. At full load the motor slip is 0.045. At full load, what is the speed of the motor?
Answer: C
The synchronous speed of the 12 pole motor on a 50 Hz supply is
\[ N_s = \frac{120 \times 50}{12} = 500\ \text{rpm} \]
\[ N = (1 - s)N_s = 0.955 \times 500 = 477.5\ \text{rpm} \]
\[ \begin{aligned} \omega = \frac{2\pi N}{60} \\ &= \frac{2\pi \times 477.5}{60} = 50\ \text{rad/s} \end{aligned} \]
The torque law of the fan (proportional to the speed squared) is not needed to find the speed at the stated slip. The values 45, 55 and 60 rad/s correspond to speeds of 430, 525 and 573 rpm, and the last two exceed the synchronous speed of 52.4 rad/s. The answer is 50 rad/s, option C.
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