An 1800-A, 120/208-volt, 3Ø, 4-wire service is to be installed underground using 6 paralleled Schedule 40 PVC conduits. The service equipment will be a fused switchboard. The minimum size of the aluminum main bonding jumper for this service is ___ AWG or kcmil. All conductors are THHN/THWN aluminum.
This is a question I have in school but I am very confused right now... If I go to table 310.15, there is nothing close to 1800 amps.... Do I just divide 1800/6 = 300amp. Then go to 310.15 and get 500kcmil. Then 500 x 6 = 3000kcmil.. Then if I go to Table 250.102(c)(1), it is over 1750... So then do I do 3000 x .125 = 375kcmil. Then go to Chapter 9 table 8, and look at Overall Area and go down to .416 and then that would be 400AWG wire size.
I'm almost positive I am not right, this is just a good guess if I were to really think it over, which I have been trying to for the last 1 hour for 1 single question. If anyone could list the steps that would be incredible! My teacher likes to give us a 70 question quiz during our class so he can sit there on the phone, and the quiz takes about an hour and a half out of a 3.5 hour class. Time that could have been used to teach us this stuff. This homework is so frustrating, wish he could have taught us. Anyways just ranting.. Thanks for your help!