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Main Bonding Jumper size?

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!

2. Take your conductor size and multiply by 12.5% that will give you the main bonding jumper size. If your numbers are correct:

500 kcmil * 6 sets = 3000 kcmil * 12.5% = 375 kcmil minimum size MBJ, next standard size 400 kcmil.

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Oh thanks,*** so I am actually right? The steps I stated are the same steps you stated, didn't think I actually did it correctly but hey I'll take it! Only question though is that you are also going to chapter 9 table 8 correct? Then going to the Overall Area part and down to .416, and thats how you got 400kcmil. Is that correct? This is the only part I am actually unsure if I did correct.

edited for language
Last edited by augie47; 03-20-17 at 09:14 PM.

4. You took CMil (500) and multiplied it. You don't need to be concerned with the overall area just a conductor that meets or exceeds 375 kcmil.

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Originally Posted by augie47
You took CMil (500) and multiplied it. You don't need to be concerned with the overall area just a conductor that meets or exceeds 375 kcmil.
I'm confused, they are in parallel so if I take 375kcmil and go to chart 250.102. It says I'll only need 2Awg for my Main Bonding Jumper, which is wayyy to small for a 1800amp service.

6. Originally Posted by JasonCo
I'm confused, they are in parallel so if I take 375kcmil and go to chart 250.102. It says I'll only need 2Awg for my Main Bonding Jumper, which is wayyy to small for a 1800amp service.
There is no chart to look at 375 kcmil is the actual minimum MBJ size. Once you go past the last row of 250.102(C) you use the 12.5% formula.

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