# How to determine full-load RPM of an induction motor at any arbitrary frequency?

Lets say an induction motor's spec sheet states the following:

1/ 50 Hz, 2-pole, full-load rotations-per-minute (RPM) = 2850.

2/ 60 Hz, 2-pole, full-load RPM = 3450.

Can we extrapolate this information to find out the full-load RPM for other frequencies (i.e.: 20 Hz, 30 Hz, 40 Hz, etc.)?

(SPECIFIC EXAMPLE)

Like for instance, as per here, the synchronous speed of the motor under 50 Hz is 3000 RPM. The full-load RPM then, is 95% (2850 / 3000) of the synchronous speed.

Doing the same thing for 60 Hz @ 3600 RPM, the full-load RPM is ~95.8% (3450/3600) of its synchronous speed.

Would it be reasonable to make an assumption that under different frequencies, the full-load RPM is ~95% of its corresponding synchronous speed?

EDIT From a suggestion by @Transistor, I am using a "VFD-B" variable frequency drive on my induction motor. (manual, website)

• Note that both examples have the same slip frequency. This will depend on supply voltage and load but it makes some sense that it shouldn't be very dependent on frequency. So rather than a percentage, why not compute speed as synchronous speed - slip?
– user16324
Commented Sep 27, 2017 at 19:02
• Does that mean I should assume the same slip frequency for all other operating frequencies? I've got an induction motor hooked up to a VFD and just wanted to be sure of what the full-load limits are for any given frequency (from about 10 Hz to 60 Hz).
– plu
Commented Sep 27, 2017 at 19:14
• It's more likely to be correct than a fixed percentage. If you have the setup you can measure at some fixed (not necessarily full) load.
– user16324
Commented Sep 27, 2017 at 19:29

You can, but it makes little sense.

An induction motor behaves the same as a transformer, if you reduce the frequency, you had to reduce the voltage applied, too, otherwise the core —both the outer shell and the rotor— get overexcited and heats up.

If you reduce the voltage, the torque/speed characteristic shrinks proportionally in the torque direction. The actual full-load speed depends on the working point made from crossing the load characteristic with the motor characteristic.

So if an induction motor is built for 240V 50/60Hz, in reality it's a 240V 50Hz motor which would also work at 60Hz.

• May I ask what you meant by "crossing the load characteristic with the motor characteristic"?
– plu
Commented Mar 26, 2018 at 22:25
• Please see the picture I added. Commented Mar 26, 2018 at 22:28
• Thanks for the clarification, from the origin of the picture, I'll look more into the concept of "Operating points" (en.wikipedia.org/wiki/Operating_point).
– plu
Commented Mar 26, 2018 at 23:40
• Uh, that's a really crude translation of the German article I wrote years ago. Commented Mar 30, 2018 at 20:43
• nptel.ac.in/courses/108106072/7 This is just FYI for anyone else, but a less crude explanation of "operating points" is in that web-course link, showing how that torque vs. speed curve for the induction motor is based on how its modeled, the fact that if the modeled torque is higher than the torque required by the load, the speed will increase until a stable operating point is reached.
– plu
Commented Jul 11, 2018 at 22:32

It may be irrelevant. Many VFDs use slip compensation.

Slip Compensation Slip compensation is actually a sophisticated version of the open loop concept. The slip compensation method of speed control does not monitor the actual shaft RPM. Rather, it utilizes drive output current transducers to monitor the current drawn by the connected motor. As discussed earlier, when a load is placed on a NEMA B design motor during a situation where the output frequency is held constant, the slip increases, the shaft RPM slows and the motor current increases. The difference here is that the “slip” function “compensates” for the reduction in shaft RPM by increasing the voltage and frequency applied to the motor. Figure 2 illustrates an application that requires the motor to supply full torque at 1500 RPM.

The top portion shows what occurs without slip compensation. The applied frequency is 50Hz, but the motor actual shaft RPM, due to slip, has a value of 1455. The bottom portion shows how slip compensation automatically "compensates" this situation by applying 1.5Hz additional output frequency to the existing output frequency of 50Hz, resulting in a new output frequency of 51.5Hz. The motor shaft still "slips" back, but now the actual shaft speed is the desired 1500 RPM. The amount of slip does not actually decrease. It is simply shifted so that the actual RPM now is the desired RPM. Remember that the drive monitors current drawn by the motor, not the actual shaft RPM.