# why no load current of induction motor is high

It depends on the voltage and power factor. If you were using 240 single phase, you could divide 10KW by 240V and get 42A. Then you divide by power factor, le
ts assume a typical value of 0. 85, and you get 49A. Now, 49A on 240 single phase is a bit unrealistic, though doable, so lets look at 480 three phase running delta. You divide 49A by 6 (3 for the 3 phase, and 2 for the 480 step up) getting 8A per winding current, then you multiply by square root of 3 to get the per phase current of 14A - call it 15A - and size the circuit for 20A. Small motors operate with a higher flux in the iron than large motors.

This results in increased iron loss and increased magnetising current, but helps reduce the copper loss - less turns of larger wire. On large motors, it is important to minimise the losses in order to prevent excess temperature rise with a given frame size.

This is not such an issue with small motors. I would try to run the motor open shaft to ensure that you are measuring the magnetising current only. It is possible that there is sufficient shaft torque to increase the current. I have seen many small motors with the iron loss as high as 60%. I have also experienced US made motors designed for 60Hz but stamped 50Hz. These do have a very high magnetising current and iron loss and usually overheat.

I would run the motor, preferably open shaft, and monitor the temperature rise. If the flux is excessive, the motor will heat significantly under open shaft conditions. If the temperature rise is acceptable, there is no problem. Remember that the cooling is proportional to the surface area of the motor, but the power is proportional to the volume. Small motors can tolerate a higher iron loss. Best regards,

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