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I have the following setup:

480V, 200A service fed into PWM Drive, Drive output going to 1000kVA step up transformer (8.4 ratio) feeding a 2380V/53A 3phase induction motor. The motor is UNLOADED.

When running at NP, I measure (with multiple devices) 477V/28A into my drive input. I measure roughly .65 pf

However, I push 283V/158A out of my drive to reach my NP motor voltage and current requirement. I don't have a PF but I imagine it's somewhere in the .13 range.

S = sqrroot(P^2 + Q^2) P = VIcos(theta) Q = VIsin(theta) (might be forgetting the 3)

IN P = 8681W Q = 10079VAR S = 13302 VA

OUT P = 5812W Q = 44278VAR S = 44657VA

By my calcs, I cannot figure out how this is possible. The real power (W) is accurate if you account for losses, but the apparent power (S) proportions are way off. I'm stuck believing it's a function of the drive but is there an explanation or a next step to take to conform this?

ejjman1
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    Just adjust your imagination and the values get better. Btw. where do you need a 2.38kV/53A motor? – PlasmaHH Nov 10 '15 at 13:46
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    What exactly do you expect when the power factor goes from 0.65 to 0.13? If you don't trust your measurements, you should describe your setup more clearly so someone can give a recommendation on how to improve it. – Arsenal Nov 10 '15 at 13:52

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There are two things that will make drive current, power and power factor calculations difficult.

The drive input current will have a high percentage harmonic content, particularly with a light load. That means that means that the input power factor of the fundamental will be about .95 while the total power factor will be much lower.

The DC bus capacitors in the drive carry all of the reactive current of the motor. None of the motor's reactive current will be reflected in the drive input.

  • Additionally, How are you actually measuring the input and output currents? You cannot simply use a clamp on meter to measure the output of the drive. You will not get anything resembling an accurate value. Use ONLY the drive's own instrumentation to determine its actual load. Also, you seem to be ignoring any transformer losses. – R Drast Nov 10 '15 at 17:02