Does an induction motor draw more power at higher RPM?

Increasing the frequency of the AC power supply will increase the RPM of the induction motor. However, that should also increase its impedance (jwL.) If that is the case, then less current would be drawn (due to higher impedance) and consequently less power is consumed at higher RPM.

I must have missed something, what is it?

  • \$\begingroup\$ V/f control.... \$\endgroup\$ – DKNguyen Mar 4 at 5:51
  • \$\begingroup\$ The impedance of the motor depends on the mechanical load that is connected to the motor. That is probably the main thing you are missing. If you drive an induction motor with a fixed voltage and frequency, then the current will go up as the mechanical load goes up. \$\endgroup\$ – mkeith Mar 4 at 6:06
  • \$\begingroup\$ Perhaps you missed the whole chapter on IM, but that's called field weakening. Also less torque is produced, more iron loss is produced, ...consequently more power is consumed for loss covering, the efficiency is decreased. \$\endgroup\$ – Marko Buršič Mar 4 at 7:49

The relationship between RPM and power is Power = Torque X RPM.

Since the impedance of the magnetizing branch of the equivalent circuit increases as the frequency increases, it is necessary to maintain the ratio of voltage to frequency relatively constant over the operating range in order to have the optimum magnetic field in the stator.

In some situations, the frequency is increased without increasing voltage at the upper end of the speed range. The resulting reduction in torque capability results in a constant power capability up to some speed at which the power capability decreases.

You have missed quite a bit. You need to study the induction motor equivalent circuit and the related performance equations and curves. Answers to the following questions may be helpful:

When load increases in rotor of induction motor how does stator draws more current?

Torque-Speed Equation for Induction Motor

Why does the V/Hz ratio of a motor affect the magnetic field strength?

Inductor Machines: why is constant flux achieved before reaching nominal speed?

  • \$\begingroup\$ So more power is consumed just because we increase the voltage at higher RPMs? If that is the case, then we have: 1- More current is drawn because we increase the voltage as we increase frequency. 2- Less current is drawn because the reactance of the equivalent impedance is increased as we increase the frequency. Are you saying that the effect of (1) on current is more than (2)? \$\endgroup\$ – mhmhsh Mar 6 at 18:46
  • \$\begingroup\$ The power consumed is determined by the intersection of the load's torque vs. speed demand curve with the motor's torque vs. speed capability curve. That determines motor slip. Motor slip determines Rr(1-s)/s in the equivalent circuit. See the first link in my answer. \$\endgroup\$ – Charles Cowie Mar 6 at 19:37
  • \$\begingroup\$ Thanks! After examining the torque/current vs speed/slip curve, it makes sense now. I was more focused on the reactance but not the mechanical load resistance Rr(1-s)/s which is a non-linear function of RPM. \$\endgroup\$ – mhmhsh Mar 7 at 1:43

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