Do I have to Derate

[copper chopper]

whoops

whoops

sorry for the caps,,, this is what we used as far as code articles to make our case... these are right from the 2011 code book and are the same in the 2008.

5) Neutral Conductor.
(a) A neutral conductor that carries only the unbalanced
current from other conductors of the same circuit shall not be
required to be counted when applying the provisions of
310.15(B)(3)(a).This states that I do not have to derate when using the scenario i described. then-

Chapter 9 table c.1 page 746 States that I can put 16 #12awg thhn wires in a 3/4 emt pipe

These are the codes we use to make our case right.....
if its wrong tell us the codes that make it wrong..????hmy:

 

[don_resqcapt19]

sorry for the caps,,, this is what we used as far as code articles to make our case... these are right from the 2011 code book and are the same in the 2008.

5) Neutral Conductor.
(a) A neutral conductor that carries only the unbalanced
current from other conductors of the same circuit shall not be
required to be counted when applying the provisions of
310.15(B)(3)(a).This states that I do not have to derate when using the scenario i described. then-

Chapter 9 table c.1 page 746 States that I can put 16 #12awg thhn wires in a 3/4 emt pipe

These are the codes we use to make our case right.....
if its wrong tell us the codes that make it wrong..????hmy:

The code section you have cited makes it wrong. You do not have a neutral conductor that carries the unbalanced current from other conductors of the same multiwire circuit. You have grounded conductors that are associated with a single ungrounded conductor. These grounded conductors are current carrying conductors and with 7 ungrounded conductors and 7 grounded conductors you have 14 current carrying conductors and a derating factor of 50%. There is no issue with putting the 14 conductors in the raceway...the issue is the overcurrent protection for those conductors. If they are 90?C rated #12s, the maximum permitted OCPD would be 15 amps.

 

[charlie b]

Let me try to explain it this way. Consider a single phase, three wire, 120/240 volt panel, like we have in our houses. Connect a black wire to a breaker on Phase A, and a white wire to the neutral bar, and run them to a 10 amp load. Current leaves the panel on the black wire, and returns on the white wire, and the two currents are the same.

Now connect a red wire to a breaker on Phase B, and a separate white wire to the neutral bar, and run them to a 5 amp load. As before, current leaves on the red wire, and returns on the white wire, and the two currents are the same.

Next, disconnect and remove one of the white wires, and connect the remaining white wire to both loads. You just created a multi-wire branch circuit, and all three wires form part of this one circuit. Here is what you will get for the current flows:

• 10 amps flow away from the panel on the black wire.
• 5 of those amps flow back to the panel on the red wire. That is because the two phases (A and B) are out of phase with each other by 180 degrees. So when current is positive on the A phase, it is negative on the B phase.
• The remaining 5 amps flow back to the panel on the only remaining white wire. It is in this sense that the neutral wire is carrying only the unbalanced load from the other wires in the same circuit.
• If by chance the two loads were the same, rather than one being 10 amps and the other being 5 amps, then there would be no current in the white wire.
• In this set of circumstances, you would not count the white wire as being a current-carrying conductor, for the purposes of derating.

Does this make sense?

 

[copper chopper]

finnally

finnally

thanks all i needed was an example and an explenation in lamens terms.....thanks agian.


but what is an example of (A)

(a) A neutral conductor that carries only the unbalanced
current from other conductors of the same circuit shall not be
required to be counted when applying the provisions of
310.15(B)(3)(a).
(b) In a 3-wire circuit consisting of two phase conductors
and the neutral conductor of a 4-wire, 3-phase, wye-connected
system, a common conductor carries approximately the same
current as the line-to-neutral load currents of the other conductors
and shall be counted when applying the provisions of
310.15(B)(3)(a).
(c) On a 4-wire, 3-phase wye circuit where the major
portion of the load consists of nonlinear loads, harmonic currents
are present in the neutral conductor; the neutral conductor
shall therefore be considered a current-carrying conductor.

 

[david luchini]

thanks all i needed was an example and an explenation in lamens terms.....thanks agian.


but what is an example of (A)

You gave an example of (A) in your original post.

before we would only pull 9 wires,6 hots and 2 neutrals and 1 ground

This multiwire branch circuit would be an example of (A). You would only have 6 current carrying conductors when applying the provisions of 310.15(B)(3)(A).

 

[copper chopper]

isnt multiwire B and C the word multi is not used in A read the code for A and tell me the differnce between A<B and C

 

[infinity]

isnt multiwire B and C the word multi is not used in A read the code for A and tell me the differnce between A<B and C

A MWBC is implied in (A), it says "the unbalanced
current from other conductors of the same circuit". A MWBC is the only thing that would fit that description. But you're right the wording is rather vague. (B) and (C) are self explanatory.

 

[charlie b]

. . . tell me the differnce between A, B, and C

In my post #23, I said that current leaving the panel along the black wire will return to the panel along the red wire. I also said that if the two loads were equal, then all the current would be flowing in the black and red alone, and that the white wire would get no current. That works only because I was talking about a single phase, 120/240 volt, multi-wire branch circuit. For that circuit, the Phase A current will reach its positive peak at exactly the same moment that the Phase B current reaches its negative peak. The two are exactly 180 degrees out of phase with each other.

However, in the subparagraph (B) to which you refer, you are dealing with a three phase system. The three phase currents are not 180 degrees out of phase with each other, but rather are 120 degrees apart. If you build a MWBC using only two of the phases, say Phase A (black wire) and Phase B (red wire) and a common neutral (white wire), what happens is that current leaves the panel along the black wire, and some of it returns via the red wire, and some of it returns on the white wire, even if the loads are equal. That is a consequence of the relative phase angles. The math is too complex for me to explain here. But this fact is exactly what the code is talking about, when it says, ?a common conductor carries approximately the same current as the line-to-neutral load currents of the other conductors.? That is, the black, red, and white wires all carry approximately the same current, even if the loads are balanced between the two phases. That is why that particular neutral wire has to be counted as a current-carrying conductor in that type of installation.

For subparagraph (C), the circumstances are entirely different. This one has nothing to do with balanced loads. It is about harmonic currents and non-linear loads. It is about load currents that do not look like ?pretty curves,? or like smooth sine waves. If you saw the wave form on a digital display, between the beginning of one cycle and the end of that cycle (60 times every second), you would see a lot of jagged peaks and valleys. This is caused by such things as fluorescent lighting ballasts, computer power supplies, and variable speed drive electronics. The essential result is that you get a very high ?extra? current that appears in the neutral wire, regardless of how well the currents are balanced amongst the three phases. That is why that particular neutral wire has to be counted as a current-carrying conductor in that type of installation.

 

[inspector23]

Thank you !

Thank you !

Sorry for screaming, but THANK YOU MR BECK!

Your post is without a doubt the best explanation I have ever read for (B) above. I really do appreciate you taking your time to explain it in such a easily understandable manner.

As an inspector, my first responsibility is to get the call right. My second, in my opinion, is to try to explain why or how in order to help everyone know better the next time. It helps the installer, and it helps me because the better I understand it, the better I am able to communicate.

I see a lot of moderators here spend a great deal of time sharing their knowledge and experience, and while it is often unspoken, the depth of gratitude of the forum readers like myself is always there for you and the others.


Thanks -

Mike