VFD driven motor on Generator power mismatch

..."In an inductive circuit, when frequency increases, the circuit current decreases and vice versa"...

..." in case of AC circuits:

I = V / Z

Where “total resistance of AC circuits = Impedance = Z = √ (R2 + (XL – XC2)”

In case of Inductive circuit:

• Z = √ (R2 + XL2)
• I = V / XL or I = V / Z

It shows that in inductive circuit, Current is inversely proportional to the inductance “L” as well as inductive reactance “XL” as inductance and inductive reactance are directly proportional to each others."...

https://www.electricaltechnology.org/2019/09/inductive-circuit-current-increases-frequency-decreases.html

http://www.davidsonsalesshop.com/catalog/files/Products/Deep%20Sea/DS%20Download%20Instructions.pdf

https://oaktrust.library.tamu.edu/handle/1969.1/192571?show=full

https://www.practicalmachinist.com/vb/transformers-phase-converters-and-vfd/120-hz-three-phase-generator-102096/

There are two factors that may not be taken into account in your calculations.

Firstly, you can't multiply voltage by current to get power in an AC circuit with an inductive load. The current lags the voltage by some angle θ, and power equals voltage x current x cos(θ). The factor "cos(θ)" is the power factor.

Secondly, because the impedance of an inductive load decreases with decreasing frequency, a VFD reduces the output voltage at lower than rated frequency, as shown in the plot below, to prevent the load from drawing excessive current.

So in calculating the power output, the reduced voltage and power factor need to be taken into account.

They can theoretically both be correct under certain load conditions that from rough calculations I would say motor would be considerably overloaded (about max current al rouglly 2/3 full power rated RPM plus considerably worse air cooling) But in this specific case I would just trust the drive readings being correct (considering actual power factor on your working conditions) because between gen and drive chopping cirquitry, you have rectification bridge that makes the VFD input current a full-of-peaks mess of a waveshape that not all instruments can read correctly. If you are puzzled by the VFD current being higher than input, also understand that VFD also outputs lower voltage for lower RPM, not only lower frequency. Actualy programming VFD for pump applications, V/RPM out ratio is more or less linear. You can use an AC RMS clamp meter on VFD input to verify what I said. S.M.

The VFD output voltage is lower- 460.

I know with all the harmonics generated some measurements are not accurate.

But the VFD rectifier should be the only load the generator sees so is that an inductive load? How does the generator see an inductive, ie. motor load?

I was thinking the work done by the drive output I*E*1.73/1000 would be the KW output (roughly) and the generator input would have to be similar (plus conversion losses)

Well, the current the generator sees drawn is a bit more complicated than that. It is not exactly phase leading or lagging. (Still does not have a good power factor). On each phase and for simplicity reasons showing the positive part only, the gen sees something like the pic below. The current is the green wave, not a sine, not half a sine, nothing of the sort. It's width and peak value depends on gen output impedance, VFD input and in-between wiring impedance, and of course the VFD loading, i.e. how much the bulk capacitors (their voltage is the red line here) have discharged since the last cycle. You can only measure that current correctly with an RMS meter.

Now advanced vector control VFDs sense the motor current and load and for total power saving reasons at lower than expected load can lower their output voltage even lower than the initial V/F scheme and increase the output frequency a bit to compensate for the increased slip. This lowers the total consumed power somewhat, but unfortunatly this can increase motor copper losses. This looks like an oxymoron but it really isn't. The output current just increases more than the voltage lowers. Tricky eh? And all this behavior is on most times programmable. If it were that simple everybody would design VFD's, no? (LOL) Have fun. S.M.

The generator seems undersized for a 500hp motor....or the motor is substantially oversized.....is the pump operating properly? What is the gpm? What is the generator voltage?

https://www.eng-tips.com/viewthread.cfm?qid=448686

https://www.generatorsource.com/Power_Calculator.aspx

Increasing the speed of the original poster's pump may well result in the generator tripping on overload, as the power drawn by a centrifugal pump varies by something like the cube of the tip speed of the impeller.

Just a thought.

The 119 KW is based on 460 volts to the motor 460*150*√3/1000=119 KW. At 38 Hz the voltage is about 460/60*38 = 291 V; 291*150*√3/1000 = 76 KW. The ratio of power in verses power out is about 55%. The power supplied by the generator is lower, in part, because the generator AC is converted to DC in the VFD and there are reactors between the VFD and generator to take into account. As the load gets closer to the full power the discrepancy will become less and less until the efficiency reaches the maximum.

VFDs have internal losses. When VFD is operated at a low load, the losses tend to be higher than those at rated VFD capacity in % terms. Generator has to cater to those VFD losses as well (apart from the motor power).

Thanks, that makes much more sense. This is a stacked horizontal centrifugal pump. First one I've messed with, but the vendor came out and verified we were running on the pump curve.

So is that rule of the VFD output being the square of the speed change apply to Variable Torque applications or just a function of the V/Hz ratio?

We have noticed that we have to keep the speed below 40 Hz to avoid high vibration and we load the generator to 130 Kw (Drive showing 39 Hz at 209 A)

Thanks everyone, this was a great learning experience. Even the generator folks couldn't explain this, though they did admit it was something going on inside the drive.

So is that rule of the VFD output being the square of the speed change apply to Variable Torque applications or just a function of the V/Hz ratio?

It is based on the LOAD being Variable Torque. You tell the VFD it is a VT load, so the VFD uses a V/Hz² curve in order to conserve energy.