multiwire branch circuts

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ronmath said:
Don't forget in the 2008 NEC 210.4 (B) every multiwire branch circuit must now have either a multipole breaker or handle tie to disconnect all ungrounded conductors simultaneously. The option to this of course is to provide a neutral conductor for each hot. At this point, every wire is considered a CCC so watch 310.15(B)(2)(a).
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Also in the 2008 code is that all conductors that share a neutral leaving a panel where there are multiple conductors in conduit, must be wire tied together.
 
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http://www.roubaixinteractive.com/PlayGround/Binary_Conversion/Binary_To_Text.asp
 
domelectric said:
Also in the 2008 code is that all conductors that share a neutral leaving a panel where there are multiple conductors in conduit, must be wire tied together.
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Good point. I would add one more thing, that there would have to be more than just a MWBC in that conduit for the rule to apply. So a 4 wire MWBC in a conduit by itself would not require any type of grouping. Also grouping, when required, can be achieved by using "similar means" which could be tape.
 
How about if you have a 120/240v 3 phase delta transformer and you are taking two hots and a neutral to linear or nonlinear loads do you count the neutral as a current carrying conductor? How about three hots and a neutral? Would it carry just the imbalaced load per 310.15(4)(a) and not be required to be counted?
 
wawireguy said:
How about if you have a 120/240v 3 phase delta transformer and you are taking two hots and a neutral to linear or nonlinear loads do you count the neutral as a current carrying conductor?
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Not necessary, because you'll be using only the two 120v-to-N lines, and the single-phase neutral isn't succeptable to harmonic issues.
How about three hots and a neutral?
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No again, because you'll use either the three phases without the neutral for 3-phase loads, or the two 120v-to-N lines for 1-ph loads, but never both to the same load.
Would it carry just the imbalaced load per 310.15(4)(a) and not be required to be counted?
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The neutral is only in use for 1-ph loads, so correct.
 
I tossed out the 3 phase one as a what if. Maybe some motor load or something I don't know about ; )

I appreciate the heads up on this Larry, I've never gotten a clear answer from anyone in the field on counting neutrals.
 
domnic said:
ON a 208 v 4 wire if i use two hots and a neutral how many CCC do i have? ON A 240 V SINGLE phase two hots one neutral how many CCC? (no harmonics)
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Wouldn't it depend on the load? If it is a 208v load then the neutral would serve no purpose and not be current carrying. If it is 2 120v loads and they are balanced, then the neutral would still not carry any current. Or am I missing something?

Oops...I guess i should have read the title...Multiwire branch circuits...that eliminates the 208v loads. Still, if the loads are balanced, maybe 2 identical lighting circuits, would the neutral be current carrying?
 
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ray cyr said:
Multiwire branch circuits...that eliminates the 208v loads. Still, if the loads are balanced, maybe 2 identical lighting circuits, would the neutral be current carrying?
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It depends on the system they're supplied from. Two identical loads supplied from MWBC on a 208Y/120 volt system would have a neutral current of about the same current as the phase current. Take a look at the formula and plug in the numbers.

NeutralCurrentFormula.gif
 
ray cyr said:
Wouldn't it depend on the load? If it is a 208v load then the neutral would serve no purpose and not be current carrying. If it is 2 120v loads and they are balanced, then the neutral would still not carry any current. Or am I missing something?
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The first one is correct, the second is not (if we're talking about a 3ph source.)

Oops...I guess i should have read the title...Multiwire branch circuits...that eliminates the 208v loads. Still, if the loads are balanced, maybe 2 identical lighting circuits, would the neutral be current carrying?
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Again, think of a balanced-loaded 3 x 120v system. The neutral carries no current. Now, for each amp you reduce one phase's current by, the neutral current increases by the exact same amount.

When that one phase's current is zero (or if that phase is not even used), the neutral current equals the load of the other two phases; the neutral is a CCC. You can add the third phase without penalty.
 
I think I might be getting it. Since it is a 3 phase system the sine waves are not exactly opposite, as they would be on the neutral in a single transformer system, therefore the currents do not exactly oppose each other so there is actually current flow on the neutral.
Something like that?
 
ray cyr said:
I think I might be getting it. Since it is a 3 phase system the sine waves are not exactly opposite, as they would be on the neutral in a single transformer system, therefore the currents do not exactly oppose each other so there is actually current flow on the neutral.
Something like that?
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You've got it.

Roger
 
ray cyr said:
Something like that?
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Very much like that. With single-phase, the opposing lines ("phases") reach their peaks at the same moment, so any current traveling equally through both lines send no current through the neutral.

With 3-ph, there are no directly-opposing peaks, but when one phase reaches, say, its positive peak, one of the others has just passed its negative peak, and the third is just approaching its negative peak.

That's also why 120 + 120 only = 208. If you could somehow grab two vectors at 120 degrees apart, and forcibly spread the arrows until they were 180 degrees opposed, the total difference would rise to 240v.
 
I think Larry has this nailed down. Big thanks! Larry for explaining stuff to those like me that are foggy on the theory.
 
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