non linear loads and how they affect neutral

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Electric-Light said:
I don't know if you can calculate it so easily.

The worst loads are relatively large uncorrected DC power supplies like older computers. If you have three 100W, 200VA computers each connected to its own phase, you'd think it balances out in neural but the neutral doesn't cancel out the current and you can end up with more current than any of the individual phases.

Although power factor and harmonics have always been kept in control for lighting ballasts, so that issue seldom affects lighting circuits. What's the spec on the luminaires?
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it is reasonable for order of magnitude and our purposes
http://www.geappliances.com/email/lighting/specifier/downloads/Total_Harmonic_Distortion.pdf

In four wire WYE systems such as 120/208 and 277/480-volt systems; harmonicsmay cause a problem with overheating of the neutral wire. The phase wires shouldalso be designed for the increased harmonic current, but since the “triplens” areadditive, the problem is especially critical on the neutral. The third harmonic andother “triplens” (9th, 15th, etc.) are additive. Total Harmonic Distortion is thepercentage of all of these additive values in relation to the total load. The sum oftriplen harmonics greater than 33 percent will result in neutral current greater than individual line currents. The resultant current exceeds the neutral conductor’s ratingand causes overheating of the neutral and/or transformer.
 
benmin said:
on a job where we are installing approximatly 18 rows of new LED lights. there are 36 lights in each row at 22watts per light and the supply voltage will be 277V. We are attempting to reuse the old lighting circuits which are multi circuit, L1,L2,L3 and neutral.

Although I am aware of non linear loads, I am not so familiar with how to calculate and divvy up the the circuits. Is there a good source of information?

Am planning on having row 3 rows per hot for 9 rows per multiwire circuit sharing a neutral. Sound right?
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I am in the camp of pretty much not worrying about it. I have never had, seen, or heard (other than some anecdotes on this forum) of an overloaded neutral due to harmonics from lighting. Every time I have clamped a neutral on a MWBC, it has never been more than a few amps above what it would be with simple loads. Even if you were to get some real "sloppy" ballasts or drivers, considering the continuous load derating, the fact that the loading will typically be less than max just because how the loads add up, and the 90 degree rating of the conductor, it is very unlikely to have a problem. If I was wiring a skyscraper would I look more carefully at it? Sure.
 
Ingenieur said:
and all this assumes 0 cancelling of harmonics, unlikely
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And that's the point that keeps getting missed. The triple(n) harmonics DO NOT cancel. They add arithmetically in the neutral.
Think about it. The fundamentals are displaced by 120 degrees. Three times that for the lowest order harmonic puts them in phase. It's as simple as that.
 
electrofelon said:
I am in the camp of pretty much not worrying about it. I have never had, seen, or heard (other than some anecdotes on this forum) of an overloaded neutral due to harmonics from lighting. Every time I have clamped a neutral on a MWBC,
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It's not a MWBC. It's 3p+n.
 
Besoeker said:
And that's the point that keeps getting missed. The triple(n) harmonics DO NOT cancel. They add arithmetically in the neutral.
Think about it. The fundamentals are displaced by 120 degrees. Three times that for the lowest order harmonic puts them in phase. It's as simple as that.
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Now that's a simpler way of explaining it!:thumbsup:
 
Besoeker said:
And that's the point that keeps getting missed. The triple(n) harmonics DO NOT cancel. They add arithmetically in the neutral.
Think about it. The fundamentals are displaced by 120 degrees. Three times that for the lowest order harmonic puts them in phase. It's as simple as that.
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no one missed that
that is why he did not deduct anything for cancelling

and if a balanced system

3 x L = L + L + L
 
Ingenieur said:
no one missed that
that is why he did not deduct anything for cancelling
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Unlikely is the word you used:
"and all this assumes 0 cancelling of harmonics, unlikely" Your words.
Triple (n) harmonic currents add in the neutral, not cancel. That's not unlikely - it's a certainty. You need to get your head round that.
 
Besoeker said:
Unlikely is the word you used:
"and all this assumes 0 cancelling of harmonics, unlikely" Your words.
Triple (n) harmonic currents add in the neutral, not cancel. That's not unlikely - it's a certainty. You need to get your head round that.
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he ADDED them, or x 3, same thing
no cancelling involved
his numbers are correct


you need to realize that there are not only 3, 9, 15 harmonics (0 seq)
it depends on what harmonics occur due to interference
some of those may cancel

so the summed total tdh will likely be less than 3 x individual thd

eg, 5th and 7th
 
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Ingenieur said:
he ADDED them, or x 3, same thing
no cancelling involved
his numbers are correct


you need to realize that there are not only 3, 9, 15 harmonics (0 seq)
it depends on what harmonics occur due to interference
some of those may cancel

so the summed total tdh will likely be less than 3 x individual thd

eg, 5th and 7th
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I specifically noted only triple(n). They add in the neutral.
That's all there is to it.
 
Ingenieur said:
That was the flaw in your analysis
he did add them
end of story
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Since we don't actually have the harmonic spectrum, he couldn't have.
Now, do you have anything constructive to add about how or why non-linear loads affect the neutral?
 
Besoeker said:
Since we don't actually have the harmonic spectrum, he couldn't have.
Now, do you have anything constructive to add about how or why non-linear loads affect the neutral?
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still don't get it do you?

everybody has already explained it to you, but you fail to grasp it
neut I <= sum(thd x line i) or 3 x tdh x bal line I (assuming a line sized neutral)
so basically if thd <33% non issue (if the load is 80% of the cond amapcity thd can be 41%, 3 x 0.41 x 0.8 = 1)
in this case it's led lighting and likely less than 20%
non-issue
 
Ingenieur said:
still don't get it do you?

everybody has already explained it to you, but you fail to grasp it
neut I <= sum(thd x line i) or 3 x tdh x bal line I (assuming a line sized neutral)
so basically if thd <33% non issue (if the load is 80% of the cond amapcity thd can be 41%, 3 x 0.41 x 0.8 = 1)
in this case it's led lighting and likely less than 20%
non-issue
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There are none so blind as those who will not see.
 
Dennis Alwon said:
The light jabs at each other needs to stop. If you can't help yourselves than carry on with pm's not on the forum.
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I'm done. I don't care for wilful ignorance and that's what we are seeing here. I do my very best to provide helpful information for the forum on a topic I have been involved with for decades only to be told I don't understand it. Despite that, I still tried. But thank you for the rein check.
 
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