multi wire branch ciruit

[domnic]

In a multi wire branch circuit 120/240 one wire pulls 15 amps the other wire pulls 2 amps and the neutral pulls 13 amps must i count the neutral as a current caring conductor ? what code article 2008. and it change in the 2014 code?

 

[fmtjfw]

In a multi wire branch circuit 120/240 one wire pulls 15 amps the other wire pulls 2 amps and the neutral pulls 13 amps must i count the neutral as a current caring conductor ? what code article 2008. and it change in the 2014 code?

No. You need not count the neutral because it is carrying the difference in the currents in the two hot wires. The Code treats all the wires in a MWBC as a group and looks at the total amperage. So in this case you see 15+2+13=30 which is the equivalent of two wires carrying 15 each and the third wire 0. This is not how you arrive at the number of Current Carrying Conductors, but it is the motivation behind the rule.

 

[infinity]

In the 2008 start with 310.15(B)(2).

 

[iwire]

No. You need not count the neutral because it is carrying the difference in the currents in the two hot wires. The Code treats all the wires in a MWBC as a group and looks at the total amperage. So in this case you see 15+2+13=30 which is the equivalent of two wires carrying 15 each and the third wire 0. This is not how you arrive at the number of Current Carrying Conductors, but it is the motivation behind the rule.

:?

That does not work with all MWBCs, consider a MWBC in a conduit phase A, phase C and neutral. How many CCCs are there?

Or a MWBC supplying non-linear loads?

I think the OP is better off looking at the code sections Rob posted.

 

[don_resqcapt19]

Looking at the total current gives you an idea of what is happening, but looking at the amount of heat gives you an even better idea. The amount of heat is the reason for the derating for more than 3 conductors in a raceway or cable.

The watts of heat will be equal to I?R for each conductor.

If we assume the same length and size of conductors, we can say the heat for a conductor carrying 15 amps is 225R watts. The current in the wire carrying 2 amps is 4R watts and in the one carrying 13 amps is 169R watts of heat, for a total of 398R watts of heat.

The worst case for these three conductors would be with either one ungrounded conductor having a 15 amp load or both ungrounded conductors having a 15 amp load. In both of those cases there will be two conductors carrying 15 amps giving you a total of 450R watts of heat. Any other combination of loads will result in less total heat in the raceway or cable.

 

[fmtjfw]

:?

That does not work with all MWBCs, consider a MWBC in a conduit phase A, phase C and neutral. How many CCCs are there?

Or a MWBC supplying non-linear loads?

I think the OP is better off looking at the code sections Rob posted.

1) The situation that the OP gave us was 120/240 voltages -- not possible for a partial 3-phase MWBC.

2) A MWBC serving non-linear loads would have a higher than expected neutral current. That is why the neutral is counted as a CCC in that case. Once again the OP stated the currents in the three conductors as 15, 2, & 13, which indicated the vector and arithmetic answers were the same hence no triplens.

 

[charlie b]

You need not count the neutral because it is carrying the difference in the currents in the two hot wires.

I agree with that part.

So in this case you see 15+2+13=30 which is the equivalent of two wires carrying 15 each and the third wire 0.

I disagree with that part. In this case, what you see is 15 - 2 = 13 (or if you prefer, 15 - 2 - 13 = 0). You have a 15 amp load on one phase, and a 2 amp load on the other phase, and the net result (net imbalance) is the 13 amps in the neutral. If instead of the proposed loading you really did have a 15 amp load on each phase, you would not see 15 + 15 = 30, but rather 15 - 15 = 0. The 15 amps that leave the source on phase A will return to the source on phase B. They are the same 15 amps of current. They don't add up, and you don't see a "total" of 30 amps.

 

[fmtjfw]

I agree with that part.

I disagree with that part. In this case, what you see is 15 - 2 = 13 (or if you prefer, 15 - 2 - 13 = 0). You have a 15 amp load on one phase, and a 2 amp load on the other phase, and the net result (net imbalance) is the 13 amps in the neutral. If instead of the proposed loading you really did have a 15 amp load on each phase, you would not see 15 + 15 = 30, but rather 15 - 15 = 0. The 15 amps that leave the source on phase A will return to the source on phase B. They are the same 15 amps of current. They don't add up, and you don't see a "total" of 30 amps.

The heating effect of the current is the same no matter the "direction". The rules about Ampacity Adjustments are about heating. In your 15-15 case the effective heating is 30.

The fact that the neutral does carry the difference current (because of the opposite directions of the two hot wire currents) and therefore only produces the difference current heat allows us not to count it as a CCC in the 125/250V linear load instance.

 

[GoldDigger]

1) The situation that the OP gave us was 120/240 voltages -- not possible for a partial 3-phase MWBC.

2) A MWBC serving non-linear loads would have a higher than expected neutral current. That is why the neutral is counted as a CCC in that case. Once again the OP stated the currents in the three conductors as 15, 2, & 13, which indicated the vector and arithmetic answers were the same hence no triplens.

In the case of a partial three phase MWBC, the code states that the neutral is to be counted as a CCC, and wire current and heating analysis comes to exactly the same result. I do not see any conflict in that scenario.

 

[charlie b]

The heating effect of the current is the same no matter the "direction". The rules about Ampacity Adjustments are about heating.

That much is true.

In your 15-15 case the effective heating is 30.

That part is not true. The heating effect (from I^2R) of a pair of conductors, each carrying 15 amps, is (2) times (15^2) times (R), or a total of (450)(R). The heating effect of a single conductor carrying 30 amps is (1) times (30^2) times (R), or a total of (900)(R).

 

[fmtjfw]

You are right, I was wrong.

You are right, I was wrong.

That much is true.
That part is not true. The heating effect (from I^2R) of a pair of conductors, each carrying 15 amps, is (2) times (15^2) times (R), or a total of (450)(R). The heating effect of a single conductor carrying 30 amps is (1) times (30^2) times (R), or a total of (900)(R).

OK, I see what you are saying.

Thanks for pointing out the heating to me. I guess what it does say is that with a neutral load the heating is going to be less than that of the maximum heating from a full capacity balanced load.

 

[GoldDigger]

OK, I see what you are saying.

Thanks for pointing out the heating to me. I guess what it does say is that with a neutral load the heating is going to be less than that of the maximum heating from a full capacity balanced load.

And that in a 2 out of 3 plus neutral three phase MWBC the heating in the neutral will be exactly the same as in each of the two line wires because that MWBC cannot be balanced with the third phase missing.
310.15.(B)(5)(b) [2011]

 

[copper chopper]

you do not have to do this for residential, 4 wire services only.3 phase people.... 310.15 (b)5 page 152 (2011 code)

 

[infinity]

Here's a little chart I made up for neutrals as CCC's:

Neutral Conductors:
Here's some examples of when to count and not count the neutral as a current carrying conductor or CCC:

3?- 208Y/120 or 480Y/277 volt system-different circuit types:
A) 2 wire circuit w/ 1 ungrounded, 1 neutral = 2 CCC's
B) 3 wire circuit w/ 2 ungrounded, 1 neutral = 3 CCC's
C) 4 wire circuit w/ 3 ungrounded, 1 neutral = 3 CCC's*

Notes:
A) A normal 2 wire circuit has equal current flowing in each of the circuit conductors so they both count as CCC's.
B) In this circuit the neutral current will be nearly equal to the current in the ungrounded conductors so the neutral counts as a CCC
C) In this circuit the neutral will only carry the imbalance of the current between the three ungrounded conductors so it is not counted as a CCC, with an exception, *if the current is more than 50% nonlinear (see below for NEC article 100 definition) then the neutral would count as a CCC.


1?- 120/240 volt system-different circuit types:
D) 2 wire circuit w/ 1 ungrounded, 1 neutral = 2 CCC's
E) 3 wire circuit w/ 2 ungrounded, 1 neutral = 2 CCC's

Notes:
D) A normal 2 wire circuit has equal current flowing in each of the circuit conductors so they both count as CCC's.
E) In this circuit the neutral will only carry the imbalance between the two ungrounded conductors so the neutral is not counted as a CCC.


Nonlinear Load. A load where the wave shape of the steady-state current does not follow the wave shape of the applied voltage.
Informational Note: Electronic equipment, electronic/electric-discharge lighting, adjustable-speed drive systems, and similar equipment may be nonlinear loads.

 

[domnic]

heat

heat

That much is true.
That part is not true. The heating effect (from I^2R) of a pair of conductors, each carrying 15 amps, is (2) times (15^2) times (R), or a total of (450)(R). The heating effect of a single conductor carrying 30 amps is (1) times (30^2) times (R), or a total of (900)(R).

So your saying if i parallel conductors i get less heat?

 

[don_resqcapt19]

So your saying if i parallel conductors i get less heat?

You would have to run the numbers. Typically the resistance of the parallel conductors will be greater than that of a single conductor that will carry the same amount of current.

The heat comparisons in this thread have all been based on conductors of the same size so that they would have the same resistance per foot.

If you have a 360 amp load and use parallel 3/0s in place of a single 500 kcmil, you would have ~5.3 watts of heat per foot of the paralleled 3/0s (2.65 watts for each of the two conductors) and ~3.5 watts per foot for the single 500 kcmil.

 

[Sierrasparky]

You would have to run the numbers. Typically the resistance of the parallel conductors will be greater than that of a single conductor that will carry the same amount of current.

The heat comparisons in this thread have all been based on conductors of the same size so that they would have the same resistance per foot.

If you have a 360 amp load and use parallel 3/0s in place of a single 500 kcmil, you would have ~5.3 watts of heat per foot of the paralleled 3/0s (2.65 watts for each of the two conductors) and ~3.5 watts per foot for the single 500 kcmil.

Never looked at it thay way. thx

 

[Sahib]

You would have to run the numbers. Typically the resistance of the parallel conductors will be greater than that of a single conductor that will carry the same amount of current. The heat comparisons in this thread have all been based on conductors of the same size so that they would have the same resistance per foot.If you have a 360 amp load and use parallel 3/0s in place of a single 500 kcmil, you would have ~5.3 watts of heat per foot of the paralleled 3/0s (2.65 watts for each of the two conductors) and ~3.5 watts per foot for the single 500 kcmil.

Practically speaking, replacing single conductor by parallel conductors stipulates equal votage drops (for the same current) and so equal watts loss.

 

[don_resqcapt19]

Practically speaking, replacing single conductor by parallel conductors stipulates equal votage drops (for the same current) and so equal watts loss.

That does not happen under the rules of the NEC.

The main reason to parallel conductors is to save copper.

The amount of copper in the parallel runs is always less than the amount of copper in a single conductor with the same ampacity.

The resistance of the parallel runs will always be higher as will its voltage drop, unless the designer calls for larger paralleled conductors than required by the rules in the NEC.

 

[don_resqcapt19]

Practically speaking, replacing single conductor by parallel conductors stipulates equal votage drops (for the same current) and so equal watts loss.

You have to remember that the NEC ampacity tables are very conservative.

In my example of a 360 amp load using a single 500 kcmil or parallel runs of 3/0, for a 3 phase 480 volt system with a run of 250' the voltage drop on the single 500 would be ~0.8% and on the parallel 3/0s it would be ~1.2%.

Both voltage drops are well within normal design standards, but the parallel 3/0s uses ~67% as much copper as using the single 500 kcmil.