# What happens to a high power inductive motor if you decrease the RPM to half the rated in terms of current, torque and heating?

and what is minimum possible RPM ratio for a high power motor

Induction motors are available with power ratings ranging from about 1/500 Hp (1.5 W) to 30,000 Hp (22MW). It is difficult to say just where in that range the motor in question may be. It is likely that the question is about a low to medium power motor rather than a motor that is more than a meter in diameter.

If the motor is controlled by a reasonably good variable frequency drive (VFD) it can be operated down to practically zero speed at 100% rated torque.

The motor current will not exceed the rated full-load current. The VFD input current will be proportional to the mechanical power output of the motor, so it will decline as the motor speed declines. The mechanical power output of the motor is proportional to load torque multiplied by speed. There will be some harmonic current content in the VFD input current. That will increase the current somewhat.

The heat lost in the motor will decline a little bit as speed declines, but the ability of the motor to dissipate losses will decline due to the reduction of the speed of the motor's self-cooling fan. That will limit the time that it is safe to operate the motor at rated torque and any given speed. At half speed, most motors will probably be able to operate at rated torque continuously. If the motor is cooled by a separate blower that is selected and furnished as part of the motor by the manufacturer, it is possible to get a motor that operates continuously at or near zero speed.

In loose terms for all motors $$speed \propto voltage$$ $$torque \propto current.$$

Thus:

• if you half the speed you must half the voltage;

• you can ensure the same current by ensuring the same torque.

The $$\power = voltage \cdot current\$$. Watch out for $$\I^2R\$$ losses though.

The minimum is a difficult question. Testing might be the answer here.

Overspeeding can destroy the squirrel-cage.

• this is true for a DC-Motor but not for an inductive motor. – Findus Mar 31 at 14:13
• @Findus it is approximately true for all motors. – skvery Mar 31 at 14:28

An induction motor, controlled by a variable frequency drive, may be run at 50 % of rated speed applying 50% of the rated voltage at 50% of the rated frequency to obtain 50% of the rated HP and 50% of the rated torque.

Up to the rated speed V/f would be constant. Above rated speed V would be held at rated voltage with only the frequency being increased for higher speeds. The torque would naturally reduce.

Up to the rated speed the V/f ratio is intended to care of motor cooling. However it is not advisable to go below 20% of rated speed without external cooling.

• The torque depends on the load. – skvery Mar 31 at 14:06
• speed depends on frequency, slip and count of poles. – Findus Mar 31 at 14:11
• Hi skvery, Of course! – vu2nan Mar 31 at 14:15
• For sure, Findus! – vu2nan Mar 31 at 14:17
• Sorry, Findus, I couldn't edit my previous comment.You are correct about frequency and slip. However only percentages have been mentioned and not actual values and the number of poles is not relevant. – vu2nan Mar 31 at 14:24

The question is difficult to answer. Holding frequency constant you can only change speed by changing the poles (nearly impossible) or changing the slip due to overload. Using an inverter you can decrease the speed with nearly no effects to current and torque. But reducing motor speed also reduces the fan speed, so machine could overheat. Calculating torque depends on different parameters and is hard to do. Simplyfied you can say it depends on current and current depends on load.

• Since the voltage is also reduced, the problem is only faced at around 20% of rated speed and could be taken care of by external cooling. – vu2nan Mar 31 at 14:33