Derating - Eight #12 THHN in a 3/4" EMT

[Unforgiven]

Drawing calls for eight each #12 THHN conductors to be installed in an existing 3/4" EMT home run. There are five hot conductors (ABCAB) two neutrals and a ground, (120/208 3 phase).

Counting the two neutrals as current carrying conductors, there are then 7 current carrying conductors. (Do they count or don't they? I've seen it argued both ways - what is the forum consensus?).

(2002 NEC) Table 310.15(B)(2)(a) indicates that for between 7 - 9 current carrying conductors the circuits must be derated to 70% of the allowed values in Table 310.16 through 310.19.

Table 310.16 indicates that for #12 THHN the maximum allowed ampacity is 30 Amps.

70% of 30 Amps is 21 Amps.

Footnote to Table 310.16 refers to 240.4(D) which states OCP cannot exceed 20 Amps for a #12 conductor.

The OCP required by 240.4(D) is less than the derated circuit calculation by one amp.

There is no need to take additional measures to derate the circuits in this case as the requirements of 240.4(D) accomplish the same thing.

Is this correct interpretation of the code requirement?

 

[sandsnow]

Looks good.

I only have one comment regarding your question of neutral being current carrying.

ABC+N - Neutral is NOT current carrying except if the major portion of the load is non-linear

AB+N - Neutral is always current carrying when derived from a 3phase 4wire Wye system

 

[Pierre C Belarge]

Only one of the neutrals will be counted as a CCC [take a look at 310.15(B)(4)(b)]. This does not change your calculation though... you are correct the 12 AWG Type THHN will still be permitted to be protected at 20 amps.


Larry snuck in there right before me!! Larry try not to do that again

 

[brownout]

I?ve heard it argued both ways as well. The question seems to center around the term ?unbalanced? 310.15(B)(4)(a). With a 208Y/120V 3 phase system, if you have 16 amps on A, 16 amps on B, and 10 amps on C, then you will have 6 amps on the neutral right? Seems to me that that would make 4 current carriers. My foreman said to read 310.15(B)(4)(a) carefully and I will see that it only makes 3. I asked him what about a 3 wire MWBC on a 120/240V 1 phase system with say 16 amps on A and 8 amps on B. I say 3 current carriers, he says 2. ???

 

[infinity]

Here it is in black and white:

310.15(B)(4) 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)(2)(a).
(b) In a 3-wire circuit consisting of two phase wires and the neutral 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)(2)(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 shall therefore be considered a current-carrying conductor.

 

[LarryFine]

I asked him what about a 3 wire MWBC on a 120/240V 1 phase system with say 16 amps on A and 8 amps on B. I say 3 current carriers, he says 2. ???

If it's 120/240, whether single-phase or part of a high-leg Delta (which are actually the same thing), the neutral carries only the difference current, and need not be counted.

The total amperage shared among the three conductors can never reach above 40 amps (theoretically, of course).

However, when it's two legs of a 120/208 system, the neutral current is not just a difference current, because of the 120 degree (as opposed to 180) phase angle between the hots.

The neutral current can easily equal that of the two hot conductors. Here's why:

Inagine an equally loaded ABCN 4-wire circuit, 20 amps on all phases. Neutral current zero, total current 60 amps. Now, reduce the current on A phase by 5 amps. B and C phases remain at 20 amps.

The neutral current is now 5 amps. A phase's current is what balances the other two phases. Keep reducing A phase to zero amps. Neutral current = 20 amps.

You now (or I should say "still") have three current-carrying current, with a total of 60 amps. This is the equivalent of a 3-wire, 120/208 circuit, with two hots and a neutral.

 

[charlie b]

With a 208Y/120V 3 phase system, if you have 16 amps on A, 16 amps on B, and 10 amps on C, then you will have 6 amps on the neutral right?

Right.

Seems to me that that would make 4 current carriers.

No. It does make the total number of conductors that happen to be carrying current equal to four. But that is not important. The 6 amps on the neutral represents the imbalanced current from the three phase conductors. So the neutral does not count as a "current carrying conductor," for the purposes of derating.

I once did a calculation that proved that, no matter how imbalanced the currents were, presuming that all loads are linear (no harmonics), the heat generated by four conductors (A, B, C, N) can never be higher than the heat that would be generated by the same load, if that load were balanced, so that the neutral current was zero.

I asked him what about a 3 wire MWBC on a 120/240V 1 phase system with say 16 amps on A and 8 amps on B. I say 3 current carriers, he says 2. ???

I say 2, for the same reason. The current on the neutral represents the imbalance from the two phase conductors, so it does not count for derating purposes.

 

[brother]

If it's 120/240, whether single-phase or part of a high-leg Delta (which are actually the same thing), the neutral carries only the difference current, and need not be counted.

The total amperage shared among the three conductors can never reach above 40 amps (THEORETICALLY, of course).

Even though this is code, and my be legal, (im going to have to get my code book and reread that section, i hear it argued too alot) As he stated, it is THEORETICAL!! In real world scenerios current does fluctuate, in theory the neutral should not see no more than 20 amps but in real world it can get higher on certain occassions. Me personally i like to count the neutral as a current carrying conductor on all scenerios, ESPECIALLY if it has the potential of being shared more than what i was intended.

Thats where the REAL world scenerio comes in. Of course we wont go there with the shared neutral issue again LOL thats a whole differeent thread with a record 145 responses

 

[LarryFine]

My point is that, to what ever degree a 240v 1ph load may present to a 2-wire circuit, the same amount of power, spread evenly OR unevenly between two hots and a neutral, will never exceed that value.

The total current will be the same. With a 208/120 3-wire circuit, the same is true: the total current among the three wires (B-C-N) will be the same as it would be on only 208 (B-C) for the same total power.

The same cannot be said for a run comprised of two hots, each with its own neutral, sharing a conduit. There are definitely four CCC's here; a 3-wire 208/120 is better (in my opinion), and more economical.

 

[Unforgiven]

The circuits are serving dimming flouresent light fixtures. So The neutral must then be counted as these are non-linear loads.

 

[brownout]

Thanks Larry and Charlie, I suppose it does make sense when I look at it that way. My foreman was not very good at explaining things.

 

[don_resqcapt19]

Brownout,

With a 208Y/120V 3 phase system, if you have 16 amps on A, 16 amps on B, and 10 amps on C, then you will have 6 amps on the neutral right? Seems to me that that would make 4 current carriers.

While they are all carrying current, they are not all counted as current carrying conductors for the purpose of derating. The reason for the derating is because of the heat produced by the current flow in the conductors. The heat is watts and is found by I^2R (I^2 = amps squared). The R is the resistance of the conductors. Assuming that the conductors are all the same in the MWBC, you can ignore R at this point. If you look at the math, you will find that the most heat is produced when all three of the ungrounded conductors are fully loaded or when 2 of the three are fully loaded. All other cases will produce less heat. This fact is why you don't have to count the neutral of a 3 phase 4 wire circuit (unless the loads are mostly nonlinear) or the neutral of a 3 wire single phase circuit.
If the three phase conductors are loaded to 16 amps, then you will have 768R watts of heat. In your example of 2 phase at 16 amps and the third at 10, you will have 648R watts of heat.
The same for your single phase example. With both hots at 16 amps there would be 512R watts of heat and with the load in your example there is 385R watts of heat.
Don

 

[brother]

Brownout,

While they are all carrying current, they are not all counted as current carrying conductors for the purpose of derating. The reason for the derating is because of the heat produced by the current flow in the conductors.

While this maybe true, I still call a current carrying conductor, 'a conductor that is intended to carry current for the purpose of utilization of electrical equipment.'

Thats my definition. So if it is there, then count. but good formulas and theory Don.

 

[paul]

The circuits are serving dimming flouresent light fixtures. So The neutral must then be counted as these are non-linear loads.

Correct, but you still come out with the same answer to the original question. No need to derate the breaker or oversize the wire...technically. I would, however, look into upsizing the neutral in this scenario for harmonics purposes.

 

[sandsnow]

Only one of the neutrals will be counted as a CCC [take a look at 310.15(B)(4)(b)]. This does not change your calculation though... you are correct the 12 AWG Type THHN will still be permitted to be protected at 20 amps.


Larry snuck in there right before me!! Larry try not to do that again

Oh how terribly rude of me!!!!!!!! I wasn't paying attention. Next time I'll PM you before I post ;-)

 

[don_resqcapt19]

Brother,

While this maybe true, I still call a current carrying conductor, 'a conductor that is intended to carry current for the purpose of utilization of electrical equipment.

However 310.15(B)(4)(a) makes it clear that for the purposes of 310.(B)(2)(a) the neutral conductor carrying only the unbalanced current from other conductors of the same circuit shall not be required to be counted.(as a current carrying conductor)
Don

 

[winnie]

I believe that 310.15(B)(4) gets quite confusing because of the focus on the neutral conductor.

The real issue, as Don and Charlie both point out is that in the 'worst case scenario for self heating' in a MWBC of the sort being considered, one of the conductors is not being utilized. Because of the square law relation between current flow and heat produced, you get the most heat production when current is concentrated in the fewest conductors.

In these particular circuits, with current concentrated in the fewest conductors, one of the conductors will not be utilized, so the 'total effective number of current carrying conductors' is reduced by one. It might be the neutral or one of the 'hots' that is unused. In the event that all conductors are carrying current, you will find that some of these conductors are carrying less than rated current, thus producing less total heat (harmonics excluded).

I believe that the same sort of logic should apply to other situations where current flowing in one conductor reduces the available current flowing in other conductors, eg. multiple switch legs going to different loads all supplied by the same circuit. However this is not supported in code.

-Jon