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The situation is such:

There is a multi-pole, three-phase, permanent magnet generator (rated for 100 KVA, let's say) that feeds into a simple diode rectifier with a capacitive/DC Voltage load (something like a battery).

Simplification of Circuit

Without any active correction, the power factor of this circuitry is going to be pretty poor (maybe down around 0.5).

Now I understand that poor power factor is going to give us much higher \$ I^{2}R \$ losses in our wiring, diodes, etc. What I don't understand is the impact on the generator when running high power for short times with poor power factor.

I have two questions:

  1. Can a multi-pole, 3 phase, permanent magnet generator even supply the power equivalent of its rated KVA at all with such power factor? (i.e. can I ever get 100kW out of this generator this way?)

  2. Supposing I can, how long can I do that for? My limited understanding is that this will produce a lot of heat in the generator and cause it to fail if used this way continuously, but could still, in theory, provide this power. If I am to run it at 100kW at poor power factor for only a couple of minutes or less, am I not damaging the generator?

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It is not not just the power factor that is the issue, it is also the distorted waveform causing poor power factor. It the power factor was low without the waveform being distorted, the power would be reduced, but only because the generator would be producing less power with the same current (W = V X A X Pf X SqRt3). If you know the power factor, you can easily calculate how much power the generator can produce without exceeding the maximum current rating. Check to see if the generator has a kVA, current or minimum power factor rating in addition to the power rating. With waveform distortion, some additional heating may be produced with the same current. However that would be no different than loading a transformer with a rectifier. I believe that some derating may be required for that.

The time that you can safely "overload" the generator depends on the time and amount of additional heating. Only the generator manufacturer can tell you.

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  • \$\begingroup\$ Thanks for your answer. This is all hypothetical, so no generator spec to check. I am still curious as to whether or not "overloading" is even possible. If the engine is spinning at max RPM, and the generator produces a certain KVA, there is no theoretical way to extract all of that power with poor power factor, right? \$\endgroup\$ – Nino Mar 22 '18 at 23:05
  • \$\begingroup\$ To be more clear, essentially what I am asking is if the engine provides a constant 100kW of power, can the generator supply 100kW of average power that involves peaks above 100kW by some fashion of stored energy (i.e. does a generator have implicit energy storage)? \$\endgroup\$ – Nino Mar 22 '18 at 23:20
  • \$\begingroup\$ The engine will only respond to the power demand. It will provide the torque required to deliver the power and not be much affected by excess current that is due to low power factor rather than a power overload. Excess current will cause increased losses in the generator that will increase the generator temperature. However the normal generator losses are only about 5%. If the losses increase by 40%, to 7%, that overheat the generator pretty quickly, but only cause a 2% overload on the engine. \$\endgroup\$ – Charles Cowie Mar 22 '18 at 23:25

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