Electrical: How do you compute conduit fill?

Question

I can see according to the NEC that I can put 4 #250 in a 2 1/2" conduit. Can I add a #4 ground as well?

Answer 1

Why do people with no clue answer questions.

The prior answer that says to calculate the area of the conduit is wrong. The 2 1/2" is a "trade size" and not an actual dimension, like a 2x4 is not 2" by 4". The actual area inside the conduit depends on the type of conduit (which you do not say). I quickly found 4 250's THHN will fit in 2.5" sch 80 PCV conduit from annex C. If that is not your case, you will have to look up in chapter 9 yourself.

You use tables 4 and 5 in chapter 9. In table 4, PVC sch 80 2.5" is 4.119 sq in, and the 40% is 1.647 sq in. In Table 5, 250 THHN has an area of .3970 sq in, so 4 are 1.588 sq in. That leaves 0.059 sq in spare for another wire. A 4 AWG THHN is too large (0.0824) to fit, so you cannot do it (for THHN and sch 80 PCV). Annex C indicates other conduits can hold 5 or 6 250's so they can work.

Answer 2

It has to do with the area of the wire compared to the area of the conduit. The limit is 40% filled.

The area formula is pi times the radius squared. So for a 2-1/2" conduit, the area is 3.9 square inches. 40% is 1.56 square inches so the total area of the individual wires shouldn't exceed that number.

Can't answer the rest of your question because I don't know the insulation thickness of your wires. The manufacturer's specs should tell you that. But it may be nitpicking.

Answer 3

dude i have no idea