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How do I calculate the kW output on a generator when changing from 60Hz to 50Hz?
i.e. if I have a 70kVA generator at 60Hz producing 56kW, how do I calculate what the kVA and kW's would be if I converted it to 50Hz?

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  • \$\begingroup\$ Is your system one generator and some load(s)? Or are you connecting this to the mains with lots of other generators connected to it? \$\endgroup\$ – The Photon Oct 2 '18 at 2:57
  • \$\begingroup\$ What does the manufacturer say? Wouldn't it be best to consult some kind of documentation or contact the manufacturer? \$\endgroup\$ – mkeith Oct 2 '18 at 3:01
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    \$\begingroup\$ What is the prime mover of the generator? And how do you propose to change the frequency output? If it is an engine, reducing the prime mover speed directly will result in a lower power output from the engine, so a lower power output from the generator. \$\endgroup\$ – R Drast Oct 2 '18 at 8:46
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You could estimate that reducing the speed to reduce the frequency from 60 Hz to 50 Hz will reduce the voltage to 5/6 of the original voltage. At the reduced speed and voltage too, the generator could likely be loaded with the same current, so that would reduce the kVA in proportion to the voltage reduction. The torque required would likely not change much, so you could assume that whatever is driving the generator could continue to produce the same torque and the available power would also be reduced in proportion to the speed reduction.

If you know the details of the characteristics of the generator and whatever is driving it, you could make a better estimate.

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  • \$\begingroup\$ Wouldn't such a generator have field windings, and so it could produce the exact same voltage at either 50 or 60 Hz? \$\endgroup\$ – mkeith Oct 2 '18 at 5:38
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    \$\begingroup\$ It will need more field current to produce a given voltage at the lower speed, which depending on the design of the magnetics might push something into saturation, so you may well be forced to run at lower output voltage to avoid excessive iron losses. The datasheets for the machine should reveal all. \$\endgroup\$ – Dan Mills Oct 2 '18 at 10:13
  • \$\begingroup\$ It's been a while since I've reviewed this. But I was under the impression that the math requires no fundamental loss of power availability. That is, a prime mover could produce the same torque at a new lower speed, supposing there wasn't a mechanical reason for that loss, like a piston timing thing or some such. Am I mistaken? Is this a torque coefficient or something I should look up? \$\endgroup\$ – Sean Boddy Oct 3 '18 at 17:55
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    \$\begingroup\$ @SeanBoddy: Power is torque multiplied by speed. To produce the same power at a lower speed, a machine must produce a higher torque at that lower speed. \$\endgroup\$ – Charles Cowie Oct 3 '18 at 21:02
  • \$\begingroup\$ Ah. I had forgotten that bit. The turbine generators on my submarine tours were a touch oversized, relatively, to deal with starting currents of large loads, so it was easy for me to forget that not every prime mover is over-specced. \$\endgroup\$ – Sean Boddy Oct 5 '18 at 19:36

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