Running older induction motors on VFDs

[Guineafowl]

The mill motor in delta.

 

[Guineafowl]

I borrowed a HV differential probe, but it’s telling the same story, ie that the filter is somehow making the waveform messier.

1. No filter, output overview:

2. Pulses (occurring at 8 kHz) have a rise time of 6.56 us:

3. + filter, output overview:

4. Pulses had an irregular appearance:

5. A better one. Slower rise time, but a lot more ‘features’:

From research, it seems that only very long VFD-motor cables need a proper sine filter. For short cables, as here, a simple output reactor would suffice, but these cost more than a replacement motor, which being a modern type would not need the reactor.

 

[Baluncore]

2. Pulses (occurring at 8 kHz) have a rise time of 6.56 us:

I need to see the data points to know if it shows the zoomed in sampling rate, or the real rise time.

3. + filter, output overview:

I can see the motor drive sinewave in that plot.
What is the cutoff frequency of the low-pass filter.
What is the frequency of the PWM output.

4. Pulses had an irregular appearance:

It looks like you zoomed too far into a waveform, until we see the data acquisition rate. That does not show the slope dv/dt, of the motor drive waveform, which should have been reduced by the LPF.

What is the Y filter centre connected to, or is it floating?

 

[Guineafowl]

As before, I can’t get triggering when zoomed in (or out). Is it better to zoom in, and keep taking single shots until a decent picture appears, or zoom out and force full memory depth of 50Mpts?

 

[Guineafowl]

What is the frequency of the PWM output.

8 kHz

What is the Y filter centre connected to, or is it floating?

What is the cutoff frequency of the low-pass filter.

Configured as per post #35, the caps are in delta and had to be replaced with 2.2 nF, as the 0.47 uF ones were tripping the VFD. Do you mean the LPF in the VFD? I haven’t been able to find that out.

 

[Baluncore]

Configured as per post #35, the caps are in delta and had to be replaced with 2.2 nF, as the 0.47 uF ones were tripping the VFD.

The VFD was expecting to drive an inductive motor. You raised the frequency of the LPF by 470 / 2.2 = 200 times. That explains why the dv/dt is higher than I expected. There is little point using that filter in that situation. You would need to increase the inductance to compensate.

 

[Guineafowl]

Looking at this another way, this link: https://vfds.com/blog/how-do-you-size-a-line-reactor/ suggests the standard for VFD load reactors is 3% of the motor’s impedance.

This motor was originally plated for 400/440 V, 50 Hz, 1.8 A. Taking 415V as a common figure in that range, when I extracted the star point to convert it to delta this changed to 240 V, 50 Hz, 3.1 A.

Would this represent a motor impedance per phase of 240/3.1 =77.4 $\Omega$?

3% of this would be 2.3 $\Omega$.

At 50 Hz, this would be 2.3/$\omega$ = 7.4 mH. The only trouble being, the mill is used between 10-100 Hz depending on the operation. However, it’s set up so that 50 Hz is the most commonly used speed.

 

[Baluncore]

Would this represent a motor impedance per phase of 240/3.1 =77.4 Ω ?

No. Impedance is a vector sum of resistance and reactance, ( R + j X ).

An induction motor has a small reactive current in quadrature with the voltage, that current magnetises the field inductance.

There is also an inphase current that provides the real power to the motor. That is shaft power plus windage, friction and resistive I²R loss. The real current is dependent on mechanical load, and is significantly more than the load independent magnetising current.

You need to look at the power factor under no load and full load, to separate those numbers.

 

[Guineafowl]

No. Impedance is a vector sum of resistance and reactance, ( R + j X ).

An induction motor has a small reactive current in quadrature with the voltage, that current magnetises the field inductance.

There is also an inphase current that provides the real power to the motor. That is shaft power plus windage, friction and resistive I²R loss. The real current is dependent on mechanical load, and is significantly more than the load independent magnetising current.

You need to look at the power factor under no load and full load, to separate those numbers.

Isn’t the full-load rms current of 3.12 A a consequence of the effective, or apparent, impedance under these conditions? As far as I can find out from motor control technical articles, this is the figure used to derive the 3% or 5% reactor impedance target.

To add to that, the commercially-available VFD load reactors are sold based on this FLA figure, too.