Residential Load Center Feeders

[msnyderku]

My question is dealing with single phase, 3 wire, load center feeders (208/120), and in the case I am interested in, it is for a high rise residential project serving condos.

I am confused as to how to size the neutral for a typical feeder. I understand there is a code section that allows you to reduce the neutral wire size (below the size of the phase conductors), but am not sure which code section that is.

But how can you be allowed to reduce the neutral wire size? Isn't the neutral load going to actually be bigger than the load on the phase conductors, due to the fact that there is not a third phase to cancel out the load on the neutral? If, for example, the load on phases A-N is 50amps @ 0 degrees, and the load on phases B-N is 50amps @120 degrees, the addition of these two vectors results in a load on the neutral greater than 50 amps ... I understand that this only applies for the 120V loads, and that a lot of load will be on the phase conductors only (ranges, etc.), but I still would think the above effect would occur on the neutral.

Any help on this topic would be much appreciated.

Thanks,

 

[George Stolz]

Re: Residential Load Center Feeders

To calculate the current borne on a neutral in a 3?, 120/208 system, the formula is as follows:

?((AA)+(BB)+(CC))-((AB)+(BC)+(CA))

So, in your case, where A = 50, B = 50, C = 0

In = ?((2500)+(2500)+(0) - ((2500)+(0)+(0))
In = ?(5000)-(2500)
In = ?(2500)
In = 50 amps.

I must have done something wrong. I always do. Which is secretly why I answered.

 

[luke warmwater]

Re: Residential Load Center Feeders

Originally posted by msnyderku:

I understand there is a code section that allows you to reduce the neutral wire size (below the size of the phase conductors), but am not sure which code section that is.

Start with 310.52(B)(6). At the bottom of the pararaph...
..."The grounded conductor shall be permitted to be smaller than the ungrounded conductors, provided the requirements of 215.2, 220.22, and 230.42 are met."


But how can you be allowed to reduce the neutral wire size?

Read above.

Isn't the neutral load going to actually be bigger than the load on the phase conductors, due to the fact that there is not a third phase to cancel out the load on the neutral?

You are confusing single phase and three phase.

If, for example, the load on phases A-N is 50amps @ 0 degrees, and the load on phases B-N is 50amps @120 degrees, the addition of these two vectors results in a load on the neutral greater than 50 amps ... I understand that this only applies for the 120V loads, and that a lot of load will be on the phase conductors only (ranges, etc.), but I still would think the above effect would occur on the neutral.

Again, it is not 3-phase.

QUOTE]

 

[luke warmwater]

Re: Residential Load Center Feeders

George,
did you mean:

In= Square Root of [Asq.+Bsq.+Csq.-((AxB)+(BxC)+(CxA))] ??

 

[paul]

Re: Residential Load Center Feeders

Your neutral load would be 50 amps if both (A and B) or (A and C) or (B and C) were 50 amps and the third phase was zero amps. A good site for the explanation of vectors and the math, go to http://www.electrician.com/electa1/electa5htm.htm . So no, do not reduce your neutral when running a three wire circuit using only two of the phase conductors from a three phase system.

[ August 26, 2005, 10:38 PM: Message edited by: paul ]

 

[George Stolz]

Re: Residential Load Center Feeders

Originally posted by paul:
Your neutral load would be 50 amps...

I was right?

You must be wrong.

 

[paul]

Re: Residential Load Center Feeders

Originally posted by georgestolz:
I was right?

According to my math, but I used the following formula to derive my answer.

 

[luke warmwater]

Re: Residential Load Center Feeders

Paul,
your formula is so much harder than mine with all of that sin and cosin stuff thrown in. The answer is still the same though.

The problem is,
he is trying to size the neutral for single phase and using 3-phase vectors.

msnyderku,
Single phase is not (2) phases of a 3-phase system.

 

[msnyderku]

Re: Residential Load Center Feeders

I am not confusing single phase with three phase power, I understand the difference. But in this case, when the power to your load center is 2 phases (A and B for example) coming off a three phase system, the line voltages, and therefore the currents (assuming relatively equal impedance on each phase), are all 120 degrees out of phase. It is for this reason that in a three-phase system, the phase currents will, for the most part, cancel each other out and the resultant current on the neutral will be close to zero (again, assuming a balanced system).

In the case we are talking about, there is not a third phase, so the current on the neutral will be the resultant vector sum of the currents on the phase to neutral loads...

I spoke too soon by assuming that this vector addition would result in a magnitude greater than what was on each phase, but after doing the math myself, (which is actually the same math done by "paul"), I see the resultant is not larger than the phases.

In addition to this, nearly half of the loads on a residential load center will be 208V loads (line to line),i.e. ranges, which have No impact on the current in the neutral, therefore lowering our overall neutral load to that below the load on the phase conductors.

And this may be the answer to my own question.

thank you all for your input!

 

[rattus]

Re: Residential Load Center Feeders

Assuming that the 120V loads are equal in magnitude and phase, the neutral will supply the current normally supplied by the missing third phase in a balanced system. That is, the neutral current will be equal to the 120V phase current of either leg but at a different phase angle. No trig, no square root, just the facts, ma'am or sir as the case may be.

 

[charlie b]

Re: Residential Load Center Feeders

Originally posted by msnyderku:. . . nearly half of the loads on a residential load center will be 208V loads (line to line),i.e. ranges, which have No impact on the current in the neutral. . . And this may be the answer to my own question.

It is.

 

[don_resqcapt19]

Re: Residential Load Center Feeders

Luke,

The problem is, he is trying to size the neutral for single phase and using 3-phase vectors.

He is using two hots and the grounded conductor from a 208/120 wye system to feed the panels. You must use the 3 phase vectors to calculate the grounded conductor current in this case.
Don

 

[steve66]

Re: Residential Load Center Feeders

You have answered your own question in regard to the math. But you must also check the electric code to find out what size neutral is actually required. 220.22 will tell you exactly how to calculate the neutral load and size.

One caveat, you can not size the neutral smaller than allowed by 250.66.

Steve

 

[augie47]

Re: Residential Load Center Feeders

One caveat, you can not size the neutral smaller than allowed by 250.66

huh?
How would that apply if the neutral is not part of a service and being sized as a grounding electrode conductor ?

 

[luke warmwater]

Re: Residential Load Center Feeders

Originally posted by don_resqcapt19:
Luke,

The problem is, he is trying to size the neutral for single phase and using 3-phase vectors.

He is using two hots and the grounded conductor from a 208/120 wye system to feed the panels. You must use the 3 phase vectors to calculate the grounded conductor current in this case.
Don

Yeah, I read right over his 208 part. :roll:

 

[steve66]

Re: Residential Load Center Feeders

posted by augie:

huh?
How would that apply if the neutral is not part of a service and being sized as a grounding electrode conductor ?

I didn't phrase that very well. That only applies to the service conductors. I thought the origonal post mentioned services and feeders, but now that I look back, I see that it refers to feeders only.

As I read 220.22 (of the 2002 code) again, it is clear that in the case the posted described, the neutral conductor cannot be reduced smaller than the phase conductors. See the very last sentance.

Steve

 

[rattus]

Re: Residential Load Center Feeders

Whatever the Code says, it is theoretically possible for the neutral current to be the sum of the magnitudes of the single phase currents.

Imagine one current leading by 60 degrees and the other current lagging by 60 degrees. In that unlikely case, the phase angles of Ia and Ib would be equal, then,

|In| = |Ia| + |Ib|

That is, In could be as much as 2X that of either phase current.