Results 1 to 6 of 6

Thread: Main Bonding Jumper size?

  1. #1
    Join Date
    Mar 2016
    Location
    Houston, Texas
    Posts
    54

    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. #2
    Join Date
    Jul 2003
    Location
    New Jersey
    Posts
    22,493
    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.
    Rob

    Moderator

    All responses based on the 2014 NEC unless otherwise noted

  3. #3
    Join Date
    Mar 2016
    Location
    Houston, Texas
    Posts
    54
    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. #4
    Join Date
    Jul 2005
    Location
    Tennessee
    Posts
    15,085
    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.
    At my age, I'm accustomed to restaurants asking me to pay in advance, but now my bank has started sending me their calendar one month at a time.

  5. #5
    Join Date
    Mar 2016
    Location
    Houston, Texas
    Posts
    54
    Quote Originally Posted by augie47 View Post
    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. #6
    Join Date
    Jul 2003
    Location
    New Jersey
    Posts
    22,493
    Quote Originally Posted by JasonCo View Post
    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.
    Rob

    Moderator

    All responses based on the 2014 NEC unless otherwise noted

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •