200% Neutrals

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jrohe

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Professional Engineer
I have a general question I am hoping to get some input on. When using 200% neutrals because of high harmonic loads, is the best way to achieve the 200% neutral:

1. Parallel neutrals the same size as the phase conductors, or

2. A single neutral with two times the cmil area of the phase conductors?

I appreciate everyone's thoughts.
 
jrohe said:
I have a general question I am hoping to get some input on. When using 200% neutrals because of high harmonic loads, is the best way to achieve the 200% neutral:

1. Parallel neutrals the same size as the phase conductors, or

2. A single neutral with two times the cmil area of the phase conductors?

I appreciate everyone's thoughts.
Click to expand...
I'm an old guy and would rather wrestle two smaller wires vs one big.
 
Remember that the Code has a minimum size of conductor that is allowed to be paralleled. Tell us more about the circuit whose neutral you are working with.
 
ptonsparky said:
360 amps is a lot of harmonics.
Two 1/0 of 180 amp each
Click to expand...
I agree. One pumping station we produced had four 350kW (470 hp) variable variable speed drives. We added harmonic filters mitigate the neutral loading problem.
 
I agree that a 200% neutral is a very rare animal, indeed, and that there are better ways to address harmonics rather than just oversizing equipment to accommodate the harmonics, letting them live on the system. This goes for K-rated transformers, too - a much better solution is a harmonic mitigating transformer. I just knew that 200% neutrals was "a thing" and I was just curious as to how it is/was accomplished.
 
I believe the largest possible current in the neutral with balanced L-N nonlinear loads would be √3 ≈ 1.73 or 173% of the line current. That would happen if there is no overlap of the three L-N current waveforms at any point during a cycle. As mentioned, this would be an unusual situation. I think they round up the 173% to 200% just for convenience.

jrohe said:
When using 200% neutrals because of high harmonic loads, is the best way to achieve the 200% neutral:

1. Parallel neutrals the same size as the phase conductors, or

2. A single neutral with two times the cmil area of the phase conductors?
Click to expand...

I believe this would depend on the specific situation.
In general, the ampacity does not double when the cmil is doubled. For example, 4/0 has twice the circular mils as 1/0. But the ampacity in table 310.16 for THHN is 205A for 4/0 and 135A for 1/0, or a factor of 205/135 = 1.52 more, not the 2X more with parallel conductors.
On the other hand, having two parallel neutrals instead of one larger neutral will add one more current carrying conductor. If that pushes you into a lower derating factor for all of the conductors in a raceway then that would have to be considered.
 
I think pton misread something. I don't see 360 anywhere in the OP.

If the ungrounded conductors are #1/0 or larger we just parallel the neutrals. For a 100 amp panel we would use #3 ungrounded and #3/0 neutral to achieve the 200%.
 
1/0 cu is rated for 180 amp. Double is 360. Far end of scale for sure.

What table are you getting that ampacity from? I don’t follow.
 
synchro said:
I believe the largest possible current in the neutral with balanced L-N nonlinear loads would be √3 ≈ 1.73 or 173% of the line current. That would happen if there is no overlap of the three L-N current waveforms at any point during a cycle. As mentioned, this would be an unusual situation. I think they round up the 173% to 200% just for convenience.



I believe this would depend on the specific situation.
In general, the ampacity does not double when the cmil is doubled. For example, 4/0 has twice the circular mils as 1/0. But the ampacity in table 310.16 for THHN is 205A for 4/0 and 135A for 1/0, or a factor of 205/135 = 1.52 more, not the 2X more with parallel conductors.
On the other hand, having two parallel neutrals instead of one larger neutral will add one more current carrying conductor. If that pushes you into a lower derating factor for all of the conductors in a raceway then that would have to be considered.
Click to expand...
If the non-linearity is purely third harmonic, the root(3) factor is not appropriate.
The absolute worst case is when the load is entirely third harmonic, with no fundamental.
In that case the line currents simply add as scalars on the neutral, since they are in phase.
That would give 300%.
200% is a reasonable reduction to a more conservative value.
100% third harmonic load is not realistic, of course.
 
GoldDigger said:
If the non-linearity is purely third harmonic, the root(3) factor is not appropriate.
The absolute worst case is when the load is entirely third harmonic, with no fundamental.
In that case the line currents simply add as scalars on the neutral, since they are in phase.
That would give 300%.
200% is a reasonable reduction to a more conservative value.
100% third harmonic load is not realistic, of course.
Click to expand...

I agree that 100% third harmonic would be a theoretical worst case. But I think it would be difficult (and probably impractical) to accomplish this if you wanted to, even aside from having any useful application.

On the other hand, the load current on a rectifier with a capacitive load (which is a typical nonlinear load) will usually have peaks in current near the maximum voltage. If such peaks in L-N currents are not overlapping between the three phases (which is possible if there is minimal inductance in the rectifier circuit) then the mean square of the neutral current will be the sum of the individual mean squares of each L-N current. And therefore the RMS neutral current will be √3 ≈ 1.73 times the RMS L-N current for balanced load currents.
If the L-N currents are overlapping, then there is opportunity for some current cancellation in the neutral and therefore the RMS neutral current would be less than the 1.73 factor.
 
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