# Testing an induction motor with a VFD, difference in voltage and mechanical power vs electric power

I have question about a electro motor powered by a frequency regulator. It is the first time I did some test on a frequency regulator on school.

This are the parameters of the motor

The motor is connected in triangle. The frequency regulator is from Schneider. I forgot to write down the specs. But it's a special module for students. It has the schneider frequency regulator, leroy Somer asynchrone motor and a DC motor the can be used as generator to put some load on the asynchrone motor. There is also some special software that came with test bench . With this software we can do test.

So the test I did: I connected a constant load of 5Nm on the motor. After this I regulated the frequency from 10Hz to 50Hz. The following table gives the results of all signals measured.

Where f is the frequency, n the rotation speed, Tu the connected torque, Pm the mechanical power, U voltage output(internal voltage meter of the system, not so accurate), I current, Pe electric power, los. power loss = Pe-Pm , Um.rms is the same as U but with a more accurate voltage meter, Um.1h is the voltage of the first harmonic of the PWM signal, Im.rms current.

Now there are 2 thinks that I don't understand:

the first question: So when I compare Um.rms and Um.1h, with other words the RMS value of the output and the first harmonic of the output there is a difference difference between them. I think this is because, the other harmonics also have a big influence on the RMS value. I read in an technical article (I can't post the article url because reputation isn't high enough yet.) that the 5th and 7th harmonic could have an amplitude of 10% - 40% of the fundamental harmonic. So I think that this is the reason that there is such a big difference between the RMS value of the output and the first harmonic of the output. Is this true?

The second question: When I compare Pe and Pm there is a big difference between them. The motor efficiency is around 50%. Is this because it is powered by a frequency regulator, because the harmonics in the current and voltage? Or those it have another reason. I have no clue why this is. (I can't post the plot because reputation isn't high enough.)

• – TMJ
Jan 9, 2017 at 19:59
• The terminology used on this site is Variable Frequency Drive (VFD) rather than inverter or frequency regulator. Induction Motor is used here rather than asynchronous motor. Title and tags revised accordingly
– user80875
Jan 9, 2017 at 21:38

1. So when I compare Um.rms and Um.1h, with other words the RMS value of the output and the first harmonic of the output there is a difference difference between them. I think this is because, the other harmonics also have a big influence on the RMS value. I read in an technical article (I can't post the article url because reputation isn't high enough yet.) that the 5th and 7th harmonic could have an amplitude of 10% - 40% of the fundamental harmonic. So I think that this is the reason that there is such a big difference between the RMS value of the output and the first harmonic of the output. Is this true?

The 10-50% value in the WEG document refers to the harmonic content of the current to the inverter. The discussion of motor voltage harmonics starts on page 10, section 6. For modern design inverters, the pulse width modulation strategy makes the lower order harmonics lower. Because of the inductive nature of the motor, the higher order harmonic voltages have less effect on the motor current. It is true that the voltage harmonic content is causing the effect that you observed.

1. When I compare Pe and Pm there is a big difference between them. The motor efficiency is around 50%. Is this because it is powered by a frequency regulator, because the harmonics in the current and voltage? Or those it have another reason. I have no clue why this is. (I can't post the plot because reputation isn't high enough.)

I believe that you are measuring the input power to the inverter, so the losses include both losses in the inverter and in the motor. If there are any fans cooling the inverter and motor, the power to drive them should also be included as losses. The motor is only operating at about half of the 1.5 kW output power. Since both the inverter and motor has some constant losses, not proportional to motor load, that will reduce the efficiency. Read section 6.2 in the WEG document. WEG shows the peak efficiency at 50 to 75 percent of rated load, but that is for larger motors. Also both the motor and inverter are presumably the WEG brand. Your equipment will likely have different results. There are additional losses due to harmonics, but the losses in the inverter and the normal motor losses are probably responsible for most of the losses.

Efficiency is difficult to measure accurately. You may have measurement error also.

• Thanks for the information. I found an interesting article as well. Maybe it helps somebody in the future. digitalcommons.calpoly.edu/cgi/…
– TMJ
Jan 11, 2017 at 19:03