VFD Power Conservation

[Jieve]

I’ve been doing some voltage/current measurements on VFDs and have come across a peculiarity that I do not yet fully understand. The specific VFD I’m working with at the moment is a 2hp drive powered by single-phase (240V), with output to a 3-phase 2hp (230V) motor. I’ve looked at both current and voltage waveforms with a scope, input and output, and am trying to reconcile what seems to be a significantly higher apparent output power than apparent input power (kVA).

Looking at the VFD panel for example, at 3000 rpm I read 185.5 Vrms and 2.1 Arms. This gives an apparent power of 1.732 x 185.5 V x 2.1 A = 674.7 VA. At the drive input, using a Fluke 87V meter, I get 241.2 Vrms and 1.15 Arms. This gives 241.2V x 1.15 A = 277.38 VA. And the operator screen and fan use an additional 162mA that I haven’t subtracted out.

In this case the motor is unloaded, so current is basically the magnetizing current.

I was always under the impression that power in = power out, and that this held for apparent power as well as real power (I thought I had seen this with kVA in transformer calculations, as well as real power for power supplies). It seems that because the VFD is a power converter, somehow it’s not quite that simple. But then how does this play out in the VFD case? And is it only real power in = real power out? What about reactive power?

I appreciate any input.

 

[iop95]

As such VFDs are voltage source inverters (VSI), current value and current phase shift are imposed by load (motor in this case).
Even if current is close to sinusoid, voltage is away (square variable width DC bus level), so real output power factor is much lower than what may be calculated/measured as fundamental voltage - current phase shift.
Induction motor it's not a passive load and high value of reactive power may exchange between motor and inverter without any input current (or minimal to compensate higher losses at higher current on output stage).

 

[bacon4life]

I suspect the confusion comes from attempting to simplify a 3 phase circuit into a per phase analysis. In a traditional motor, the magnetic energy sloshes back and forth to the nearest capacitor on the same phase. However, with a VFD we cannot simplify to a per phase equivalent. The amount of magnetic energy stored in a 3 phase motor operating in steady state is constant, even though the magnetic energy stored in an single winding is constantly changing. A VFD forces the magnetic energy to slosh between phases of the motor rather than to an external capacitor.

Most "laws" in electricity are not actually laws, but rather have numerous specific conditions that have to be true in order for the law to be accurate. In this case the text book is referring to impedances whereas the VDF in much more than a simple impedance.

Perhaps imagine that instead of power electronics inside the VFD box, there was a motor-generator set inside the VFD box. The reactive power drawn by the motor-generator would be uncorrelated with the drawn by the load.

 

[Jieve]

Coming back to this as I’ve been taking additional measurements and thinking about the responses here.

VFDs are generally said to be power factor correctors with very high input power factors, almost at unity. Taking data from some scope measurements of VFD input voltage and current with single-phase input power, I calculated power factor based on the equation pf = (V x I)avg / (Vrms x Irms) over a single cycle. In this particular case (details in the original post), the VFD input pf = 0.45.

While in my particular example above, the VFD is powered by single-phase and unloaded, the fact that VFD input power factor is generally not high is confirmed in the general case in this white paper:

https://acim.nidec.com/drives/kbele...-load-reactors-with-variable-frequency-drives

“VFDs typically have a poor power factor (0.6-0.65). The high kVA demand from VFDs, which related to poor factor is due to the harmonic content in the current waveshape.”

I noticed this in an air compressor application I have on single-phase VFD at full load that demonstrates this phenomenon: if all input power to the VFD were real for the 7.5hp motor, then input current to VFD should be ~32A rms at full motor load. Instead I measure 46A rms.

Furthermore, the response in this post describes how it is not possible to have a unity power factor (with common AC input power) without a purely sinusoidal current at the voltage fundamental frequency due to harmonic content:

https://electronics.stackexchange.com/questions/52347/non-linear-load-rectifier-and-power-factor

This becomes clear looking at manufacturer datasheets on line reactors. Often there are tables showing VFD rms input current with and without line reactors for a particular drive. If a VFD driven motor is drawing the same current at same load, and VFD efficiency hasn’t changed, yet the rms input current is significantly reduced with a line reactor, then it becomes clear that there is no way the VFD input power factor is near unity, and a line reactor acts as a power factor corrector by reducing harmonics in the input waveform.

The voltage and current are in-phase at the VFD input (both are positive or negative at the same time), but this does NOT mean that the input power factor is near unity.

In this post, it has commonly been said that the VFD does not take in any apparent power; all power into the VFD is real. But by the definition of power factor, there therefore has to be reactive power being exchanged between the VFD and source (or used in charging the DC bus capacitor?), and this can be seen when breaking the non-linear input current waveform down into its fourier series. So to say that a VFD is only taking in real power is incorrect; in fact there is a significant amount of reactive power being taken in as well.

Anyone care to fight me on this?

 

[bacon4life]

This article describes the importance to specify which power factor (distortion, displacement, or total) is being discussed. For example, adding input reactors will improve distortion power factor but will degrade displacement power factor.

Regarding your current measurements, current measures on VFDs with a standard multimeter can be misleading due to high harmonic content.

 

[Jieve]

This article describes the importance to specify which power factor (distortion, displacement, or total) is being discussed. For example, adding input reactors will improve distortion power factor but will degrade displacement power factor.

Interesting read, will dig deeper into the difference power factor types ... none of the sources I have come across have specified this; nor in the white paper I linked. Although the white paper says this "This means that the ratio of kW/kVA defined as “power factor” will be improved"

Regarding your current measurements, current measures on VFDs with a standard multimeter can be misleading due to high harmonic content.

I used a Fluke i310s clamp meter for the compressor measurements, so not a cheapy; but I thought of this as well. For that reason also used a scope, exported to csv, extracted the waveform data and calculated my own rms current based on a single cycle to confirm. Results were more-or-less the same.

 

[waross]

(V x I)avg / (Vrms x Irms)

Vrms/Vavg is the equation for the form factor. For a sine wave this is 1.11
Vavg/Vrms would be the reciprocal of the Form factor = 1/1.11 = .9
Multiplying the Voltage form factor times the current form factor would give .81 for sine waves.
But you don't have sine waves.
The form factor is not related to the power factor.
Before digital meters, we used mostly multi-meters based on d'Arsonval meter movements.
The measured value was scaled with resistors and then rectified before being applied to the meter movement.
The meter responded to Average Current, not RMS current. The DC reading was scaled by the form factor to indicate RMS values.
But the form factor of 1.11 was only valid for sine waves.
Other wave forms had different form factors.
When a distorted wave form was encountered, the reading was in error by the ratio of the actual form factor of the wave form under test to the sine wave form factor.
Neither rectifier input currents nor PWM Voltages will have the same form factor as a sine wave.
The form factor is related to the form of one quantity, be it voltage or current.
Power factor is related to the interaction between the voltage and the current and to the effects of phase shifts between voltage and current and to the effects of harmonics.
Your formula may yeild interesting results, but it is not related to power factor

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Jieve]

Your formula may yeild interesting results, but it is not related to power factor

The formula was taken from this post on electronics stack exchange describing how modern power meters calculate power factor with non-linear loads:

https://electronics.stackexchange.c...ing-the-true-power-factor-in-non-linear-loads

Do you disagree with the responses?

 

[LionelHutz]

What are you using to do this metering?

pf = W/VA
W is not (V x I)avg
VA = Vrms x Irms

 

[Jieve]

For the case where I calculated displacement power factor using the equation in the link that you responded to, I used a fluke 87V multimeter directly measuring input current, then as a sanity check on the rms current value, I used a Fluke i30s clamp with oscilloscope, exported the data from the scope, extracted out a single cycle and calculated the rms current value from there. The rms values read by the Fluke meter and spreadsheet calculated values were pretty much the same.

For the air compressor motor VFD Irms-in I used a Fluke 87V and a Fluke i310s clamp.

The above link is for displacement power factor; the respondent's logic being that (VxI)avg is zero if the current and voltage are 90 degrees out of phase, and equal to Vrms x Irms if the voltage and current are in phase. I just tested this using a matlab script and it pans out. Average power delivered to a load should be zero if the power is purely reactive, and should be the apparent power if the power is fully real. Again, this is for the displacement factor only, not distortion factor. And works for pure sine waves, and the respondent claims for any non-linear combination of voltage and current as well, although does not site a source.

My calculated displacement power factor does seem low though for the 2hp motor in my original post, I think I need to revisit that.

 

[LionelHutz]

Ah, so your (VxI)avg actually means a power calculation like this?

I think the above is the correct formula, but I didn't put too much effort into finding it. Saying (VxI)avg didn't explain what you were trying to do since that isn't a power calculation formula.

So, you used the scope and got a table of instantaneous voltages and currents with a time step and then used that to calculate the power?