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Regenerative Drives - Effect of Power Factor

Former Community Member
Former Community Member

I am working on a small 16kW hydro system which is experiencing about 15% loss in the Regenerative VSD. To maximise efficiency the turbine operates at variable speed. The VSD controls a synchronous generator and supplies the grid.

The VSD is a Siemens G120. The datasheet states that the efficiency should be around 96%, whilst also stating that the Power Factor is 0.9. I am looking to replace this drive for an ABB ACS 880-11 which has similar efficiency but a unity power factor.

Firstly can I trust these datasheets since, I assume, they relate to the VSD delivering electrical energy to a motor rather receiving it from a generator? Is there an efficiency penalty for regenerative generation?

Secondly, with all else being equal, will the drive with a unity power factor equate to more electrical energy on the meter than the drive with a 0.9 power factor? 

Thanks

 

 

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    I think it is likely that you are misleading yourself with the oscilloscope too. The waveforms are not “true RMS” they are sampled at a large number of points and then a calculation is carried out on the numbers. I suggest you refer to a textbook here and make sure you understand this properly. RMS power, the heating effect, is more complex because multiplying an RMS voltage by an RMS current does not give the correct heating effect, depending on the exact phase relationship between the two. It is even more complex where the two waveforms are not similar, and your current waveform is certainly not a sine wave. A current of 1 amp at a point and a voltage at the same point of 1 volt gives a power of 1 Watt. However, if the voltage is zero, there is a current but no real power. We call this reactive power, and it gives the definition of power factor. I suspect that the power factor in “brake” mode is not very good, and thus your apparent inefficiency is simply caused by this error. I am sure that it is not real, as Mike says, something would be getting very hot!

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  • 0

    I think it is likely that you are misleading yourself with the oscilloscope too. The waveforms are not “true RMS” they are sampled at a large number of points and then a calculation is carried out on the numbers. I suggest you refer to a textbook here and make sure you understand this properly. RMS power, the heating effect, is more complex because multiplying an RMS voltage by an RMS current does not give the correct heating effect, depending on the exact phase relationship between the two. It is even more complex where the two waveforms are not similar, and your current waveform is certainly not a sine wave. A current of 1 amp at a point and a voltage at the same point of 1 volt gives a power of 1 Watt. However, if the voltage is zero, there is a current but no real power. We call this reactive power, and it gives the definition of power factor. I suspect that the power factor in “brake” mode is not very good, and thus your apparent inefficiency is simply caused by this error. I am sure that it is not real, as Mike says, something would be getting very hot!

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