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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

 

 

  • Former Community Member
    0 Former Community Member

    Hi Mike,

    Thanks for the useful analogy, that's very useful. 

    I am actually a mechanical engineer so I look at power in from the generator and the power out on the meter and consider everything else a loss, I appreciate this isn't technically correct since the energy still exists but it is not useful (at least in this application).

    I think to get to the real answer I need to determine whether there is reactive power at the meter. If there is it seems I can't so anything to improve the output with the current equipment installed  

  • Ah sorry, should have used the mechanical one - the reactive energy is like the energy stored in the flywheel, built up over many cycles and topping up on each explosion and running down a bit in between. In a well running system you do not intend to get all of the energy out of the flywheel at once- if you did it  would come to a dead stop, indeed once running smoothly the level of energy stored in the flywheel is cyclostationary - If you prefer not to use the mathematical shorthand, that means at the same point in the next cycle, it will be back to exactly the same state- so there is no overall gain or loss to the flywheel over a whole cycle, just short term borrowing and paying back at different phases in the cycle. This is not energy you can use without stopping it working.

    Mike.

  • I am actually a mechanical engineer so I look at power in from the generator and the power out on the meter and consider everything else a loss, I appreciate this isn't technically correct since the energy still exists but it is not useful (at least in this application).
     

    It's probably easier to think of ‘reactive power’ as not being real power at all - it exists as current that's repeatedly exchanged between source and load (and back again), but the actual power (in Watts) is always zero (excepting minor cable losses etc).  The old school approach of calling the reactive current ‘wattless current’ was perhaps clearer.

       - Andy.