donf
Sep 9, 2009, 06:22 PM
Okay, I'm working with a 2008 NEC study guide and I want to make sure I understand the high temperature correction process with respect to Ambient Temperatures.
I 'm given a 200 amp sunlight exposed, service drop. The area of the installation calls for an average 113(F) daily summer temp.
Using Table 310.15 (B) (6) - 200 Amp a 2/0 AWG USE cable.
To conductor sizing requirement, you divide 200 by 0.82 (for USE cable) <table 310-16
75 (C) column>. The factor is based on 45 (C) 0.82 from the table below table 310.16
So - amps 200 /0.82 = 243.9 or 244 amps. According to table 310.15 (B) (6) I would need to use a 250 kcmil cable rated for 255 Amps to meet the requirements for this installation.
255 amps X 0.82 correction factor = 209.1 amps or 209 amps.
Did I even come close to getting the process correct?
I 'm given a 200 amp sunlight exposed, service drop. The area of the installation calls for an average 113(F) daily summer temp.
Using Table 310.15 (B) (6) - 200 Amp a 2/0 AWG USE cable.
To conductor sizing requirement, you divide 200 by 0.82 (for USE cable) <table 310-16
75 (C) column>. The factor is based on 45 (C) 0.82 from the table below table 310.16
So - amps 200 /0.82 = 243.9 or 244 amps. According to table 310.15 (B) (6) I would need to use a 250 kcmil cable rated for 255 Amps to meet the requirements for this installation.
255 amps X 0.82 correction factor = 209.1 amps or 209 amps.
Did I even come close to getting the process correct?