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estimating %THD knowing line and neutral currents for a wye load? 1

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Senselessticker

Electrical
May 28, 2004
395
So I have a delta-wye K13 factored 480/208-120 transformer/PDU feeding a server room. I know the line currents and pf for each seconday phase from the display on the PDU. The display also gives the neutral current of the secondary. As expected, the nuetral current is high. The line currents indicate the secondary loads are greatly imbalanced and the pf's also vary greatly for each phase of the secondary.

I'm wanting to estimate out how much neutral current is result of imbalanced load and/or the varying phase pf vs. how much of the neutral current is result of the harmonics being fed from all the switching mode power supplies in the server room.

Knowing the pf and RMS current on each phase of the seconday (wye), I *should* be able to estimate the current being driven into the neutral from imbalance using phasor/vector addition... right?

Ia(@ ph. "A" current angle) + Ib(@ Ph. "B" current angle - 120 degrees) + Ic(@ Ph. "C" current angle + 120 degrees) = neutral current (due mostly from imbalance only)... right?

Will most of the remaining measured neutral current simply be the summation of the harmonic content from the 3 phases? Or at least be in the ballpark?

My hunch is the THD in this systems is >100%, and the K13 transformer maybe should have been a K20...

Thanks for any help!

 
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I see what you are trying to do and it sounds reasonable. You need to be sure that the PDU instrumentation can measure the voltage and the current accurately with a distorted waveform.

I will suggest buying something similar to an Amprobe 41PQ which measures voltage, current, power factor, and THD% on each phase (individually) up to the 51st fundamental frequency and only costs about $260.

You can find them at and many other places on the web.
 
It depends on how the meter measures power factor. If it measures power factor as the ratio of real to apparent power, then you won't be able to determine the phase angle of the currents. The apparent power will include harmonic currents which will make the power factor lower than cosine(theta).
 
Thanks jghrist! I now remember learning about that somewhere way back in a power electronic class... Thanks for spuring my memory!
 
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