non linear loads and how they affect neutral

[Smart $]

With respect, what you seem to have missed is that isn't a single component. It's 3-p+n. All three phases have non-linear load. It's their combined triple harmonics that flow in the neutral.

Oh, and the CBRD was a 3MW lighting project.

Not missing anything. I'm aware it is 3Ø. What I'm saying is all three line currents add up arithmetically to 25.8A. That's saying all three line currents add up on the neutral and there is 0 cancellation. Is that possible? Yes, it is. But the circuit would have to be set up intentionally to do this. So the question is, how much cancellation will occur. This is taking the opposite approach to how much harmonic current will add in the neutral. If 23% cancellation occurs, the neutral conductor current will be at its limit of ampacity (20A is very conservative for 12AWG 90°C-rated wire)... but functionally acceptable.

If the CBRD neutral bars overheated, someone messed up. :happyyes:

 

[Besoeker]

Not missing anything. I'm aware it is 3Ø. What I'm saying is all three line currents add up arithmetically to 25.8A. That's saying all three line currents add up on the neutral and there is 0 cancellation. Is that possible? Yes, it is. But the circuit would have to be set up intentionally to do this. So the question is, how much cancellation will occur.

None at all in the case of triple harmonics. None. That's the point you seem to be missing.

If the CBRD neutral bars overheated, someone messed up. :happyyes:

Yes. They didn't take account of triple harmonics. That's the long and the short of it.

 

[Ingenieur]

None at all in the case of triple harmonics. None. That's the point you seem to be missing.


Yes. They didn't take account of triple harmonics. That's the long and the short of it.

He assumed 0 cancellation in his calcs, so no, he did not 'miss it'
and proved you would need 77% thd to ol the conductor in the most conservative approach
next to impossible with led lighting which is usually less than 20% and never greater than 30% (if listed)

the long and short is the neut bar was smaller than the ph bars
bad practice
if line sized 100% it would not overheat

 

[Besoeker]

He assumed 0 cancellation in his calcs,

Not so.
"That's saying all three line currents add up on the neutral and there is 0 cancellation. Is that possible? Yes, it is."

and proved you would need 77% thd to ol the conductor (if listed)

Might I gently suggest that you read his post and my response again?

the long and short is the neut bar was smaller than the ph bars
bad practice
if line sized 100% it would not overheat

Your assumption is incorrect.
It was line sized as is common practice. And did overheat.
It's a real life example of one of the projects I got dragged into.

 

[Ingenieur]

Not so.
"That's saying all three line currents add up on the neutral and there is 0 cancellation. Is that possible? Yes, it is."

Might I gently suggest that you read his post and my response again?

Your assumption is incorrect.
It was line sized as is common practice. And did overheat.
It's a real life example of one of the projects I got dragged into.

incorrect
in is calcs he used 0 cancellation
he made an observation that is unlikely, which it is

you should
3 ckts x 0.77 tdh x 8.6 load/ckt ~ 20A common neut

I question the accuracy of your assertion
imo configured to validate your point
that would imply poor drive design or overloaded phase bars
bad design, not of the neut

 

[Besoeker]

incorrect
in is calcs he used 0 cancellation
he made an observation that is unlikely, which it is

But nothing to do with harmonics which is what this thread is about.

I question the accuracy of your assertion
imo configured to validate your point

Fine. Question it. I just happen to have been on that site and observed what I observed.

that would imply poor drive design or overloaded phase bars
bad design, not of the neut

Didn't I mention earlier that it was a 3MW lighting project. No drives involved in our scope of supply. Had there been, that would have fallen into my lap and I would have had to do the calcs up front. It wasn't and I got dragged into it after the event.

Let me reiterate. It was the triple harmonics that can caused neutral overloading. Not the line frequency imbalanced currents.

 

[Smart $]

None at all in the case of triple harmonics. None. That's the point you seem to be missing.

How it appears to you is in error. I understand triplen harmonics... at least the basics. What I'm saying is the additive triplen harmonics must comprise over 77% of the each line's current for it to be an issue.

 

[winnie]

Beo,

Not discounting the reality of the situation you were involved with, but rather focusing on the original post of this thread:

22watts per lamp, 36 lamps per row, 3 rows per 'hot', 277V l-n. Assuming unity displacement pf, we are looking at 8.6A/phase.

Now, assume that these lamps are such wildly non-linear loads that the triplen current drawn is equal to the fundamental.

What is the current on the shared neutral conductor given these assumptions?

In this particular system, is it likely that the neutral is overloaded?

I am pretty sure that Smart$ is talking about the OP, not about harmonics on systems in general.

-Jon

 

[Besoeker]

How it appears to you is in error. I understand triplen harmonics... at least the basics. What I'm saying is the additive triplen harmonics must comprise over 77% of the each line's current for it to be an issue.

No. It is the arithmetic sum from all three lines.

 

[Besoeker]

Beo,

Not discounting the reality of the situation you were involved with, but rather focusing on the original post of this thread:

22watts per lamp, 36 lamps per row, 3 rows per 'hot', 277V l-n. Assuming unity displacement pf, we are looking at 8.6A/phase.

Now, assume that these lamps are such wildly non-linear loads that the triplen current drawn is equal to the fundamental.

What is the current on the shared neutral conductor given these assumptions?

Unlikely scenario but it would be three times the fundamental.

 

[Smart $]

No. It is the arithmetic sum from all three lines.

L1 – 8.6@ x 77% = 6.6A
L2 – 8.6@ x 77% = 6.6A
L3 – 8.6@ x 77% = 6.6A
N – 6.6A + 6.6A + 6.6A = 19.8A
N ampacity 20A.

 

[winnie]

benmin,

Here is a paper that I found which provides some food for thought...but as I said way back at the beginning to actually calculate anything you will need the specs for the particular lamps you are using.

http://www.pe.org.pl/articles/2012/11a/61.pdf

The first point is the paragraph on the lower right of the first page, where it states that IEC 61000-3-2 requires that 3rd harmonic current not exceed 86% of fundamental. (I find this astounding. Calling Electric Light!)

The second point is figures 4 and 5, where 3rd harmonic current appears to be up to 80% of fundamental for some lights.

Now on to calculations:

You cannot have 'all' 3rd harmonic current, because then no power would be delivered to the lamp. But you apparently can have harmonic currents that are equal in magnitude and added to the fundamental current that is powering the lamp.

As far as the neutral is concerned, what you need to do is calculate the magnitude of all 'triplen' harmonics, meaning harmonics which are a multiple of 3, and then simply add them up.

Lets take a concrete but simplified example. Say you have a lamp that draws a fundamental current of 1A, and has a 3rd harmonic of 50%, 9th of 25%, and 15th of 10%. The lamp will produce a 'triplen' content of 0.5 + 0.25 + 0.1 = 0.85A.

Your total line current (in this simplified lamp with just fundamental, 3rd, 9th, and 15th harmonics) would be 1.85A.

Triplen currents _add_ on the neutral.

So if you have 3 of these lamps on a 3 phase wye circuit, the fundamental current would balance out just as normal, but the triplens would add, so you would have 2.55A on the neutral.

-Jon

 

[Besoeker]

L1 – 8.6@ x 77% = 6.6A
L2 – 8.6@ x 77% = 6.6A
L3 – 8.6@ x 77% = 6.6A
N – 6.6A + 6.6A + 6.6A = 19.8A
N ampacity 20A.

Without the harmonic spectrum we can't work out the harmonic contribution. Surely you can understand that?

 

[winnie]

Oh, based on the quoted IEC standard I would not be surprised by LED lamps where the harmonic current exceeds the fundamental.

On the other hand, the only 22W 277V LEDs that google finds are like these with a power factor rating of > 0.9:
http://www.lightingsupply.com/22W-LED-HID-5000K-E26-100-277V.aspx

This implies that the 'extra current' (from harmonics or displacement) does not exceed 11% of the fundamental.

-Jon

 

[winnie]

Without the harmonic spectrum we can't work out the harmonic contribution. Surely you can understand that?

That is perfectly clear.

Smart$ was doing the calculation from the opposite direction, asking what the maximum tolerable harmonic contribution without overloading the neutral could be.

Prior to finding the paper I quoted I assumed that the 'maximum tolerable harmonic current' which he determined would exceed what any lamp would produce. Now I am nor so sure...

-Jon

 

[Smart $]

Without the harmonic spectrum we can't work out the harmonic contribution. Surely you can understand that?

Sure can. Doesn't prevent me from speculating, though.

 

[Ingenieur]

L1 – 8.6@ x 77% = 6.6A
L2 – 8.6@ x 77% = 6.6A
L3 – 8.6@ x 77% = 6.6A
N – 6.6A + 6.6A + 6.6A = 19.8A
N ampacity 20A.

:lol:

 

[Ingenieur]

But nothing to do with harmonics which is what this thread is about.


Fine. Question it. I just happen to have been on that site and observed what I observed.


Didn't I mention earlier that it was a 3MW lighting project. No drives involved in our scope of supply. Had there been, that would have fallen into my lap and I would have had to do the calcs up front. It wasn't and I got dragged into it after the event.

Let me reiterate. It was the triple harmonics that can caused neutral overloading. Not the line frequency imbalanced currents.

Everything to do with harmonics
those comprise the thd, h= harmonics
This thread is about led ltg thd ol'ing the neut
pretty much EVERYTHING about this tread

ok

Lighting producing more than 40% thd?
assume 80 loading, sized for 100 (125%)
3 x 0.40 x 80 = 96 < 100
I do not buy modern listed ltg fixtures with >40% thd
not to mention 0 cancelling

 

[Electric-Light]

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. Doubt computers with uncorrected PSUs are being used in enough quantity these days where neutral overload can become a concern.

Although power factor and harmonics have always been kept under control for lighting ballasts. So that issue seldom affects lighting circuits. What's the spec on the luminaires?

 

[Ingenieur]

That is perfectly clear.

Smart$ was doing the calculation from the opposite direction, asking what the maximum tolerable harmonic contribution without overloading the neutral could be.

Prior to finding the paper I quoted I assumed that the 'maximum tolerable harmonic current' which he determined would exceed what any lamp would produce. Now I am nor so sure...

-Jon

a typical 277 led is <20%
I believe that is a std to avoid this very issue
it will never be more than 60% of a lines i, and assume a full neut is no issue
higher quality/cost ones are rated <15 and <10
I looked at more than a few catalogs
some cheaper unlisted ones were between 25 and 30
still less tham 100
and all this assumes 0 cancelling of harmonics, unlikely