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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

 

 

Parents
  • Hmm, well 64 samples second is not very fast, and is just one and a bit samples per cycle of the mains. Rather like the strobe at the disco, or that funny effect on films when the wagon wheels seem to go backwards at certain speeds, that is going to make accurate assessment of the current tricky. I assume on a wave that repeats perfectly at 50Hz, the sampling instant being sightly earlier on each successive cycle means the internal processor does  its sums based on  a ‘slow wave’ version of what is happening. Looking at  the plot below imagine the black is the 50Hz and the red dots are the 64 Hz sample instants, then the best guess reconstructed (assumed) slow waveform the internal processor will calculate with is the red one. When both are steady state sines, all is well, as the RMS is the same for both. I'm less sure how such a thing will handle a vaveform with stepped edges and fine structure between the samples.  

    You are of course correct, at the mains side the voltage has no choice but to be a sine wave (!) , but the current will be almost a square wave whose mark-to-space ratio is modulated depending on the power available. The RMS power will be whatever is available on the DC bus, but what the meter makes of it who knows. Perhaps a call to the makers of the meter.

    And if you are not making toast or frying eggs on it, it is not really losing 2kW, so not that inefficient.

     

    Mike.

    98f7fdbd946a0c80b611f8fb2cb4eaba-original-alias.jpg
Reply
  • Hmm, well 64 samples second is not very fast, and is just one and a bit samples per cycle of the mains. Rather like the strobe at the disco, or that funny effect on films when the wagon wheels seem to go backwards at certain speeds, that is going to make accurate assessment of the current tricky. I assume on a wave that repeats perfectly at 50Hz, the sampling instant being sightly earlier on each successive cycle means the internal processor does  its sums based on  a ‘slow wave’ version of what is happening. Looking at  the plot below imagine the black is the 50Hz and the red dots are the 64 Hz sample instants, then the best guess reconstructed (assumed) slow waveform the internal processor will calculate with is the red one. When both are steady state sines, all is well, as the RMS is the same for both. I'm less sure how such a thing will handle a vaveform with stepped edges and fine structure between the samples.  

    You are of course correct, at the mains side the voltage has no choice but to be a sine wave (!) , but the current will be almost a square wave whose mark-to-space ratio is modulated depending on the power available. The RMS power will be whatever is available on the DC bus, but what the meter makes of it who knows. Perhaps a call to the makers of the meter.

    And if you are not making toast or frying eggs on it, it is not really losing 2kW, so not that inefficient.

     

    Mike.

    98f7fdbd946a0c80b611f8fb2cb4eaba-original-alias.jpg
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