Supply side bonding jumper size

[hhsting]

I have incoming into trough service conductor 2 sets of 500 kcmil Al. I looked table 250.102(c)(1), 310. However, I am puzzled as to what correct size supply side bonding jumper copper would be for trough?

 

[oldsparky52]

2*500 = 1,000 kcmil Al equivalent area so I see it being 2/0 copper or 4/0 aluminum.

 

[hhsting]

Ok let’s says it’s above 1750kcmil

So incoming is 3 sets of 600 kcmil Al and wanted copper supply side bonding jumper. What would the correct size be?

 

[charlie b]

Look at Note 1 under that table.

 

[hhsting]

I understand note #1 I think. I like to be sure below is correct:

600x3 = 1800 kcmil Al

1800x0.125 = 225kcmil Al. Next size is 250kcmil Al.


Now I want copper so what size would it be it would be 4/0AWG copper?

 

[infinity]

I understand note #1 I think. I like to be sure below is correct:

600x3 = 1800 kcmil Al

1800x0.125 = 225kcmil Al. Next size is 250kcmil Al.


Now I want copper so what size would it be it would be 4/0AWG copper?

Keep reading, see note #2.

 

[hhsting]

Keep reading, see note #2.

Now that note confuses me. Especially the wording. I am lost. How exactly one would do it?

 

[infinity]

I understand note #1 I think. I like to be sure below is correct:
600x3 = 1800 kcmil Al
1800x0.125 = 225kcmil Al. Next size is 250kcmil Al.
Now I want copper so what size would it be it would be?

You would substitute copper with an equal ampacity for your 3 sets of 600 kcmil Al conductors and then do your 12.5% calculation.

Notes:
#2. If the ungrounded supply conductors are larger than 1100 kcmil copper or 1750 kcmil aluminum and if the ungrounded supply
conductors and the bonding jumper are of different materials (copper, aluminum, or copper-clad aluminum), the minimum size of the
grounded conductor or bonding jumper shall be based on the assumed use of ungrounded supply conductors of the same material as the
grounded conductor or bonding jumper and will have an ampacity equivalent to that of the installed ungrounded supply conductors.

 

[oldsparky52]

Now that note confuses me. Especially the wording. I am lost. How exactly one would do it?

I've never worked stuff this large, and I never really read note 2. I agree, it's confusing. After reading it 3 times, I think I've got it (thanks to infinity).

 

[infinity]

Here's an example from HH's post #5:
3 sets of 600 kcmil Al = 340*3 = 1020 amps
3 sets of 400 kcmil Cu = 335*3= 1005* amps (*too small)
3 sets of 500 kcmil Cu =380*3 = 1140 amps
3*500*12.5% = 187.5 kcmil = #4/0 Cu {Chapter 9, Table 8-#4/0=211600 cir mils} (Note: #3/0=167800 cir mils too small)