A 10 H.P. 3-phase induction motor with full load efficiency and p.f. of 0.83 and 0.8 respectively has a short circuit current of 3.5 times full load current. Estimate the line current at the instant of starting the motor from a 500 V supply by means of star delta starter.
Question 10.3: A 10 H.P. 3-phase induction motor with full load efficiency ...
Output of the motor = 10 H.P. = 7355 W
Power factor, \cos \phi =0.8
Efficiency, \eta = 0.83
Supply voltage (line value), V_{L} = 500 V
Input to the motor = \frac{output}{\eta } = \frac{7355}{0.83} = 8861.45 W
Full load current, I_{fl} = \frac{Input}{\sqrt{3} V_{L} \cos \phi } = \frac{8861.45}{\sqrt{3} \times 500 \times 0.8} = 12.79 A
Ratio of, \frac{I_{sc}}{I_{fl}} = 3.5
∴ Short circuit current, I_{sc} = 3.5 \times 12.79 = 44.76 A
Staring current, I_{s} = \frac{I_{sc}}{3} = \frac{44.76}{3} = \mathbf{14.92 A}
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