oldsparky52
Senior Member
- Location
- Wilmington, NC USA
The job is to replace 2 disconnects. The disconnect on the left is going from a 1,000-amp rating to a 1,200-amp rating. We are limited to using no greater than 500 kcmil conductors. The EE has it drawn up for 4 sets of 350kcmil.
So, it looks like it's going to be very difficult to get a 4th conduit installed between the disconnect on the left at the CT cabinet, and if we did, the conductors would most likely be different lengths and in this short of a run, that might be a high %.
I'm thinking about recommending a wireway be installed between the CT cabinet and new disconnect. Connect the wireway with either a 5" nipple or (2) 3.5" nipples (at each end). The reasoning is if we put more than 3 CCC's in a conduit we have to derate which kills our ampacity rating, but since there are less than 30 CCC's in that wireway, no derating would be required, and the connections at each end of the trough are short enough that derating would not be required (less than 24").
In order to determine the minimum dimensions of the wireway, I did the following:
350kcmil @ 0.5242 x 12 = 6.5583 sq in
3/0 @ 0.2679 x 1 = 0.2679 sq in
6.5583 + 0.2679 = 6.5583 sq in
6.5583 / 20% = 32.79 sq in
So a 6" x 6" wireway would suffice.
Did I miss something?
So, it looks like it's going to be very difficult to get a 4th conduit installed between the disconnect on the left at the CT cabinet, and if we did, the conductors would most likely be different lengths and in this short of a run, that might be a high %.
I'm thinking about recommending a wireway be installed between the CT cabinet and new disconnect. Connect the wireway with either a 5" nipple or (2) 3.5" nipples (at each end). The reasoning is if we put more than 3 CCC's in a conduit we have to derate which kills our ampacity rating, but since there are less than 30 CCC's in that wireway, no derating would be required, and the connections at each end of the trough are short enough that derating would not be required (less than 24").
In order to determine the minimum dimensions of the wireway, I did the following:
350kcmil @ 0.5242 x 12 = 6.5583 sq in
3/0 @ 0.2679 x 1 = 0.2679 sq in
6.5583 + 0.2679 = 6.5583 sq in
6.5583 / 20% = 32.79 sq in
So a 6" x 6" wireway would suffice.
Did I miss something?