I have 3kw 3phase AC induction motor which will work at 50deg in crane application. As per mfrs. spec. the max. permissible output power should be 93% of the rated power. Is that mean my motor will deliver 2.79kw at 50deg ambient temperature with rated frequency voltage? If so, then what about the rated current, rated torque starting current ratio? All of them will decrease according to the power? Iam totally confused with the word 'de-grading of the motor'. Pls explain
The rated frequency and voltage do not change. The rated speed is determined by the frequency, so rated speed doesn't change either. Rated power is rated torque multiplied by rated speed, so only the rated torque is reduced to 93% of the normal rating. Current is determined by the actual applied load torque, but reducing the torque by 7% will cause the current to decrease by somewhat less than 7% because the motor has a high no-load current. The current may be something like 35% of rated when there nothing connected to the motor shaft.
it means that you should reduce the external load that you put on the motor by 7% so that it doesn't overheat
