Derating bundled wires on same circuit

[rerickson]

When over three wires from the same branch circuit are run in a conduit, NEC appears to require them to be derated just as if they were on separate circuits, all capable of carrying full current at the same time (310.15B2a). Is there a Code paragraph I’m missing?

Because all these #12’s are on one 20A breaker, the maximum I2R heat comes when a single wire carries the full current, in which case the other wires are just passive conduit fill. Spreading the 20A among the wires actually cuts down on the total heat production. I would expect the derating computation to be based on the maximum possible heating, but haven’t found provision for this, and the exceptions of 310.15B2 don’t seem to apply. This can't be such an unusual situation!

The formula of 310.15C is plainly not going to be engineered for a residential application, unless there’s a much simpler calculation that an inspector’s likely to accept. Perhaps Annex B.310.11 can treat this as a diversity computation, but I'm not clear how that would apply for <10 cond, and it's all rather terse... Is there a simple solution?

My actual situation is slightly more complex than the simple case above. There’ll be a multiwire circuit pair with shared neutral, so max total current through the conduit is 40A, with max heating in two wires (either H-N or opposite-phase hots). Possibly another 20A circuit shares the conduit. Nine total conductors (including neutrals) will be split up as needed. The environment is a warm 50C, which is why derating becomes such an issue; .82 times .8 puts #12 THHN below 20A.

P.S. I hope I’m not wearing out my welcome with all these questions, but you all are a very helpful resource! Thank you so much.

[rerickson]

I think I finally tumbled to what Annex B is saying about how load diversity (the percentage of active conductors) applies to the derating of multi-conductor bundles, and this explains how to handle bundles with wires sharing a branch circuit.
Do you agree? Is the inspector likely to agree?

It turns out that Table 310.15(b)(2)(a) is placing a limit on the total I*I*R heat production, based on the geometry of ‘n’ wires. So if A1 is the ampacity coming out of this table, and it expects ‘m’ conductors to be active, then m*A1*A1 is the number this table is limiting.

FPN 1 points us to Annex B Table B.310.11 for raceways with load diversity; the FPN on that annex table explains that it assumes 50% diversity for n of 10 or more (so m=0.5*n), and for those it gives us the formula A2 = A1 * sqrt(m/e), where ‘e’ is the actual number of active conductors, and A2 is your resulting actual ampacity. In other words, if we look at current squared, A2*A2 = A1*A1*m/e, or e*A2*A2 = m*A1*A1! Your actual I-squared, times actual number of active conductors, equals the table’s I-squared, times its assumed number of active conductors.

For n less than 10, the table assumes 100% of conductors to be active, hence m=n, but the same underlying formula should apply: e*A2*A2 = m*A1*A1. Solving for your actual max current A2, we still get A2 = A1 * sqrt(m/e). For example, if you have 9 conductors and (for some reason) we know only 6 can be active at once, then A1 comes from the table for n=9:
A1 = 70% * A (the ampacity for an unbundled conductor from all the other tables), and we adjust for e=6, so A2 = A1 * sqrt(9/6). This gives us A2 = 85% * A. To be cautious, A2 should not exceed the table’s figure for a simple bundle of 'e' wires, so we limit this to 80%, the figure for n=6.

Whew!
Now let’s apply this to shared breakers. With ‘n’ conductors, fed by ‘b’ breakers of uniform size, any or all of those wires might be active. The total heat production is greatest when the current is concentrated into the smallest number of wires (because it’s I-squared), so that’s what ‘e’ will be based on: e = 2*b, allowing for the returning neutral current. In the case of multwire circuits, I believe 310.15(b)(4)(a) lets us exclude the neutral if we’re counting its opposite-polarity hots. So let ‘r’ be the number of neutral conductors that carry return current in the full-load case: any neutrals that are not in multiwire circuits, or for which both hot poles are not included here. This gives us e = b + r.

Example: we have a NEC 2008 multiwire circuit (a pair of opposite-pole breakers), plus one more breaker, all the same size. Suppose our conduit carries switched and unswitched for each, so two hot conductors for each breaker (six total) plus two neutrals. For n=8, the table gives us A1 = 70% * A, and it uses m=n (no diversity). For our adjustment, b = 3 and r = 1, so e = 4 active conductors at max load, and A2 = A1 * sqrt(8/4) = .99 * A. We therefore use A2 = .8 * A, the limit for 4 conductors.

Looking at the actual numbers, it appears that the table doesn't actually care much about geometry for n<10, and is mostly just limiting the total heat production, perhaps even allowing a little more heat for bigger bundles. When n reaches 10, the adjusted A2 starts to be cut back, so the table is taking into account resistance of heat flow to the outside. That's just my interpretation of how the formulas are turning out.

Summing all that up:
n = number of conductors in conduit or bundle
m = number of those which B.310.11 assumes to be active at one time: m = n for n < 10, and m = 0.5 * n for larger n.
b = number of equal breakers
r = number of active neutrals
e = actual number of active conductors for maximum heat = b + r
A = ampacity coming from the other tables
A1 = ampacity coming from B.310.11, based on n; A1 = A * a stated percentage. (This equals Table 310.15(B)(2)(a) when n<10)
A2 = actual ampacity limit = A1 * sqrt(m/e), not to be less than the A1 for e simple conductors

I’ve been chewing on this question for a while, but posting about Annex B is what got me thinking overnight.

Should we even be less conservative, and drop the part about figure A2 not being larger than the e-conductor figure A1? (That is, do we permit the ampacity for a load-diversity medium-sized bundle to actually be larger than for a small bundle?)

[jadnashua]

At least two of those wires will have current on them, the hot and the neutral, not just the hot. On a shared circuit, the neutral current may cancel, but then you have current on the two hots - still at least two wires.

[rerickson]

At least two of those wires will have current on them, the hot and the neutral, not just the hot. On a shared circuit, the neutral current may cancel, but then you have current on the two hots - still at least two wires.

I'm guessing you saw the first posting in this thread but not the follow-up? The first discusses the trivial case of one wire; the second goes into neutrals (which are counted unless part of a multiwire circuit with both poles included).

[rerickson]

That was awfully lengthy, going into all the Code rationale.
Here's the result; wondering if experts agree with me.

A bundle has several hot wires per breaker; we wish to modify the derating accordingly. Load diversity lets us do this, even if all wires could be active at the same time: shared breakers limit the total current, and ampacity is based on maximum heating (and heat flow out of the bundle).

We assume all conductors and breakers are the same size.

1. Compute ‘e’, the maximum number of conductors that could be carrying full breaker current at any given time.
For single-pole (unbalanced) circuits, this is twice the number of breakers (we count hot and neutral).
For multipole (shared-neutral) 110/220V circuits, this is two per multipole pair (we count both hots, or one hot and the neutral).

2. Consider the total number of conductors, ‘n’.

If n is 9 or less, look up a preliminary bundle derating factor for n in the usual Table 310.15(B)(2)(a).
4-6: .8
7-9: .7
Let ‘m’ = n.

If n is 10 or more, look up the preliminary bundle derating factor for n in Table B.310.11.
10-24: .7
25-42: .6
43-85: .5
Let m = n/2. (That table is based on 50% diversity.)

3. Multiply the derating factor by sqrt(m/e), to adjust for your actual diversity.

4. Now look up Table 310.15(B)(2)(a) for ‘e’ wires. (If e<4, act as if the table says 1.0)
Take the smaller of this or the number we just computed in step 3.