# Thread: Power Loss Due to Harmonics

1. Senior Member
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## Power Loss Due to Harmonics

How one should calculate the power loss due to harmonics? A typical 6 pulse VFD with diodes in rectifier module produces about %30 THD (Total Harmonics Distortion). An "Active Front End" VFD reduces the harmonics to %4 or lower. Is it safe to say that, if you are running a 1000HP motor, using an "Active Front End" VFD saves you about %24 of 1000HP in electric bills?
I am just looking or a ballpark estimate.

2. That's an interesting question.

Where are you trying to estimate the savings, I'm guessing the utility meter. Are there any transformers between the VFD and the utility meter? Sometimes transformers losses can be 3-4% higher with significant load harmonics (most transformer no-load losses are not significantly effected by harmonics). I'd have to put some further thought into kWH savings if there are no transformers in between the meter and the VFD.

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Originally Posted by ron
That's an interesting question.

Where are you trying to estimate the savings, I'm guessing the utility meter. Are there any transformers between the VFD and the utility meter? Sometimes transformers losses can be 3-4% higher with significant load harmonics (most transformer no-load losses are not significantly effected by harmonics). I'd have to put some further thought into kWH savings if there are no transformers in between the meter and the VFD.
there is a transformer between meter and VFD feeding the motor. Transformer is 12.4k:600V.
in addition to transformer, there is a line-reactor upstream to VFD (between VFD and transformer).
I kind of see how my original thought (%30 power loss due to %30 THD) is nonsensical. I think I need to find the voltage drop due to higher frequency currents and find the power loss.

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If I remember correctly from my college days, harmonics don't actually cause a power loss except for the additional voltage drop along the wiring.

The reason is that the current caused by the harmonic is twice the frequency as the voltage source. So half of the harmonic power is positive, and half is negative. It flows to the load on the first half of the cycle, and back to the source on the second half. (I believe the reasoning was similar for higher harmonics).

So the losses cancel out, and that's why your first guess was way too high. Your second guess looks much better. I think you need to find the voltage drop due to the harmonics along the wiring, and in the transformer windings. THis voltage drop will have the same frequency as the harmonic. Then you can multiply voltage x current, and find the power lost.

5. Originally Posted by LMAO
...
I kind of see how my original thought (0 power loss due to 0 THD) is nonsensical.
...
After reading OP I was thinking [input] THD is not a direct indication of power loss. Some of the THD current is likely output and transmitted through the motor.
Last edited by Smart \$; 05-14-12 at 10:19 AM.

6. This link / PDf is a bit technical but discusses the transformer and cable losses.

7. Originally Posted by ron
This link / PDf is a bit technical but discusses the transformer and cable losses.
Clara Peller wants to know where's the beef but we want to know where's the pdf? Not that I will read it but....

8. Originally Posted by LMAO
How one should calculate the power loss due to harmonics? A typical 6 pulse VFD with diodes in rectifier module produces about %30 THD (Total Harmonics Distortion). An "Active Front End" VFD reduces the harmonics to %4 or lower. Is it safe to say that, if you are running a 1000HP motor, using an "Active Front End" VFD saves you about %24 of 1000HP in electric bills?
I am just looking or a ballpark estimate.
The 30% THD is the current waveform. The lowest order harmonic for this application is the 5th then 7th (6n±1 series).
Generally the magnitude of the harmonic varies roughly as the reciprocal of the harmonic number.
The non-linear current will increase the total RMS in the supply conductors a little above what they would be on a pure sine wave but in any case you would rate them for the VSD RMS input current.

To answer your original question no, the active front end won't save anything like 24% and it may actually consume more. The AFE will have more losses than a six-pulse plain rectifier input and my initial reaction is that this would likely exceed any losses saved in the supply.

One other thing to think about on losses. If there was 24% of the energy getting lost, something would be getting almighty hot!

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Originally Posted by Besoeker
The 30% THD is the current waveform. The lowest order harmonic for this application is the 5th then 7th (6n±1 series).
Generally the magnitude of the harmonic varies roughly as the reciprocal of the harmonic number.
The non-linear current will increase the total RMS in the supply conductors a little above what they would be on a pure sine wave but in any case you would rate them for the VSD RMS input current.

To answer your original question no, the active front end won't save anything like 24% and it may actually consume more. The AFE will have more losses than a six-pulse plain rectifier input and my initial reaction is that this would likely exceed any losses saved in the supply.

One other thing to think about on losses. If there was 24% of the energy getting lost, something would be getting almighty hot!
read my next comments; I immediately found out how dumb my first assumption was! imagine %30 of 1000HP being wasted as heat; that's about 225kW heat!

10. Originally Posted by ron
This link / PDf is a bit technical but discusses the transformer and cable losses.
http://activeharmonicfilters.com/Con...treduction.pdf

It would have been nice if I posted the link ......

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