jsinclair
Member
- Location
- Pennsylvania
Gentlemen,
I have just gotten out of a debate with a local contractor about conduit fill for a piece of 1 inch EMT. I have a subpanel that needs to be brought up from a 60A, single phase feed to a 100A single phase feed. Existing there is a 1 inch EMT conduit from the MDP to the sub-panel. Distance is around 140 feet. I asked the contractor to use the existing conduit and pull in the required new wire for the job. He said he would have to run new conduit because the conduit would be too full and its against code. I disagree. Here's my calculations:
Vd = 11.1 x 280 x 100 / (52620) = 5.91 v
5% of 120 v = 6 volts (Can use a #3 for conductors)
From Chapter 9, Table 4 and Table 5 (2002 NEC)
1" EMT has an area of 222 mm^2 (at 40% fill)
#3 THHN/THWN has an area of 62.7 mm^2
#8 THHN/THWN has an area of 23.61 mm^2
so...
3 x 62.7 mm^2 = 188.1 mm^2
1 x 23.61 mm^2 = 23.61 mm^2
Grand Total of 211.71 mm^2 in wire
Since 211.71 mm^2 is less than 222 mm^2, the combination of (3) #3's and (1) #8 can be pulled an remain under the 40% mark required by Chapter 9, Table 1.
Anyone agree or disagree?
PS - I know its not the ideal circumstance and an 1.25" conduit would be nice, but under the circumstances that is not feasible.
I have just gotten out of a debate with a local contractor about conduit fill for a piece of 1 inch EMT. I have a subpanel that needs to be brought up from a 60A, single phase feed to a 100A single phase feed. Existing there is a 1 inch EMT conduit from the MDP to the sub-panel. Distance is around 140 feet. I asked the contractor to use the existing conduit and pull in the required new wire for the job. He said he would have to run new conduit because the conduit would be too full and its against code. I disagree. Here's my calculations:
Vd = 11.1 x 280 x 100 / (52620) = 5.91 v
5% of 120 v = 6 volts (Can use a #3 for conductors)
From Chapter 9, Table 4 and Table 5 (2002 NEC)
1" EMT has an area of 222 mm^2 (at 40% fill)
#3 THHN/THWN has an area of 62.7 mm^2
#8 THHN/THWN has an area of 23.61 mm^2
so...
3 x 62.7 mm^2 = 188.1 mm^2
1 x 23.61 mm^2 = 23.61 mm^2
Grand Total of 211.71 mm^2 in wire
Since 211.71 mm^2 is less than 222 mm^2, the combination of (3) #3's and (1) #8 can be pulled an remain under the 40% mark required by Chapter 9, Table 1.
Anyone agree or disagree?
PS - I know its not the ideal circumstance and an 1.25" conduit would be nice, but under the circumstances that is not feasible.
