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?)

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