Conduit Fill Calculator

Conduit fill is a mechanical limit, not a thermal one. The question it answers is whether a bundle of conductors can be pulled through a raceway without jamming in a bend or having the insulation scraped off against the wall on the way. Heat is dealt with separately, by the ampacity rules, and a pull can comfortably pass one check while failing the other.

The calculation has three inputs, each from its own table in Chapter 9. First,Table 1 sets the percentage of the raceway that may be occupied, based only on how many conductors are in it: 53% for a single conductor, 31% for exactly two, and 40% for three or more. The dip at two is not a misprint — two round conductors lying side by side are the arrangement most likely to wedge against each other in a bend, so the code is stricter there than for a bundle that can self-organise.

Second, Table 5 gives the cross-sectional area of each conductor, keyed by size and insulation type. This is where most errors creep in. Insulation is not a footnote to the conductor: 14 AWG THHN occupies 0.0097 in² while the same copper in XHHW occupies 0.0139 in², because XHHW carries a thicker cross-linked wall where THHN uses a thin nylon jacket over PVC. Mixed pulls are normal, so the areas are summed conductor by conductor rather than assumed uniform.

Third, Table 4 gives the raceway's internal area — and there is a separate Table 4 for every raceway article. EMT, PVC Schedule 40, PVC Schedule 80, RMC and FMC all differ at the same trade size, sometimes substantially. Multiply the internal area by the Table 1 percentage to get the allowed area, compare it against the summed conductor area, and that is the answer.

Take a common branch-circuit pull: three 10 AWG THHN conductors and two 12 AWG THHN conductors, going into 3/4 inch EMT. Two circuits sharing a raceway, near enough.

Step one, the conductor count. Three plus two isfive conductors. Five is more than two, so Table 1 gives the40% row. Nothing about the sizes matters at this stage — the fill percentage depends only on how many conductors are present.

Step two, the conductor areas. From Table 5, a 10 AWG THHN is 0.0211 in² and a 12 AWG THHN is 0.0133 in². So the ten-gauge conductors come to 3 × 0.0211 = 0.0633 in², the twelve-gauge to 2 × 0.0133 =0.0266 in², and the total is 0.0899 in².

Step three, the allowed area. Table 4 gives 3/4 inch EMT a total internal area of 0.533 in². At 40%, the allowable fill is 0.533 × 0.40 = 0.2132 in².

Step four, the comparison. 0.0899 in² of conductor against 0.2132 in² allowed. It fits, and comfortably — the pull uses 42% of what is permitted, and the actual fill is 0.0899 ÷ 0.533 = 16.9% of the pipe against a 40% ceiling.

That headroom is worth acting on. Half-inch EMT has 0.304 in² internal, allowing 0.1216 in² at 40% — still above our 0.0899 in². So this pull is legal in1/2 inch EMT, one trade size down, which is the kind of saving that matters when it repeats across a job. Set the trade size selector to Auto and the calculator reports exactly that.

Two tables here rather than one big one. The first is Chapter 9 Table 1 in full — it is only three rows, and it is the piece worth committing to memory, because the percentage depends on nothing but the conductor count.

The second is deliberately not the whole of Tables 4 and 5. Rendering every conductor size against every insulation type against every raceway type would be several hundred rows and unreadable on a phone, which is exactly the job the calculator above exists to do. Instead it shows a handful of rows that make the point about insulation: at small sizes XHHW is dramatically bulkier than THHN, and above roughly 4 AWG the ordering quietly reverses. If you remember one thing from this page, make it that the insulation printed on the jacket changes the arithmetic.