System bonding jumper sizing

[Elect117]

That answer is illogical, interpreting the language in Note #2 with any consideration for the physics involved gives a minimum size of 3/0 Cu.

Cheers, Wayne

I wouldn't say illogical.

----------------------------------------------------------------------------------------------------------------------------------------

I am going to try to calculate the equivalency using their impedances (I am going to use aluminum conduit numbers and hope that is the worst case, highest impedance on the return path).

Aluminum 600kcmil-
wire 1 // wire 2 // wire 3 -> 0.041+j0.039 // ... //... -> 0.01367+j0.013 / 1000 ft (0.01886<43.57°)

12.5% -> 1/.125= a kcmil reduction factor of 8. so the impedance should be acceptable to increase by the same factor.

8*(0.041+j0.039) = 0.1093+0.104i/1000 ft (0.151<43.57°)

closest aluminum conductor with the same or less impedance is 250kcmil Al (0.090+j0.41) (but depending on the rounding made by table 9, maybe 4/0 Al).

closest copper conductor with the same or less impedance is 2/0 (0.10+j0.043).


You could convert the parallel impedance of the ungrounded conductors and multiply by the 12.5% factor (8x) and get 2/0 copper as okay.

I just don't know if it is correct to assume that the SBJ, SSBJ, MBJ's purpose is a low impedance bond or if there are other considerations that I missed.


--------------------------------------------------------------------------------------------------------------------------------------------

But if someone told me I had to choose one for a plan set that was due today, I would go with 4/0 and be done with it. It is just a small piece on a transformer anyway. We aren't talking about running this for hundreds of feet. If someone installed 3/0, I wouldn't loose sleep over it either. I probably wouldn't loose sleep over 2/0 either to be honest.

 

[infinity]

hat answer is illogical, interpreting the language in Note #2 with any consideration for the physics involved gives a minimum size of 3/0 Cu

Are you saying that Joe's calculation in post #9 is incorrect?

 

[Elect117]

I know you are quoting Wayne, but yes. I think Joe incorrectly changed the number of runs.

By that same math, you could end up with 3 copper supply side bonding jumpers that are sized as #2 when they should be 1/0 Cu.

So, if we called the equivalent of 1 run of 600kcmil Al the parallel of 2/0 Cu (340A<175*2) and then the circular mils of 2/0 is 133.1, so multiply that by 2, and that is 266.2kcmil and so that falls under #2 on table 250.102(C).

If we try to replace 600kcmil Al with the copper equivalent (no additional imaginary parallel runs of copper added) then it would be 500kcmil (340<350A) copper and the SSBJ would be 1/0 Cu.


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Some side math I did to try and figure out how the sizes are decided~~~~~~~~~~~~

250kcmil - 4 parallel runs

0.057+j0.041//...//...//... = 0.01425+j0.01025

So then taking that into consideration, (0.0145+j0.01025)*8 = 0.114+j0.82

which 2/0 Cu is still less. 2/0 Cu being 0.10+j0.055.

But the reactance is higher. The magnitude being 0.140<35.73°


------

I realize I made a typo on my math in the earlier post.

The 3 parallel runs of 600kcmil should have been typed as " (0.01367+j0.013) * 8 = 0.1094+j0.104 "

-------



7 parallel - 1/0 - 0.13+j0.044 = 0.01857+j0.00629 --> ans*8= 0.149+j0.0502 which leads to a wire of 1/0 Cu @ (0.13+j.044) because it's impedance is less than the parallel impedance of the circuit.

So by my math, you could incorrectly get 1/0Cu as a SBJ for 7 runs of 1/0Cu. 105600*7=739.2kcmil which should be 2/0Cu per the table.

Maybe the math I did is just dogwater then. Or there is some kind of rounding, adjustment or factor in the table. I don't know.

 

[wwhitney]

Maybe the math I did is just dogwater then. Or there is some kind of rounding, adjustment or factor in the table. I don't know.

The table, certainly the last row but I expect all the rows if you look closely (which I didn't), is going by cross-sectional area. That will not scale as inverse impedance at these conductor sizes, as reactance is close to constant, while inverse resistance does scale by cross sectional area.

That's why your computations based on total impedance are not matching the table. If you use resistance only, it should match the table. [I'm still unclear why the resistance figures in Table 9 change slightly depending on the conduit type.]

Cheers, Wayne

 

[wwhitney]

Are you saying that Joe's calculation in post #9 is incorrect?

Yes, I'm saying that it's more important to match the number of runs (3) than to match the exact ampacity. For these purposes 1000A, 1005A, and 1020A ampacity are all equivalent, you could use any of them with a 1000A OCPD. The closest copper analogue to (3) sets of 600 kcmil Al would be (3) sets of 400 kcmil Cu.

Cheers, Wayne

 

[infinity]

Yes, I'm saying that it's more important to match the number of runs (3) than to match the exact ampacity. For these purposes 1000A, 1005A, and 1020A ampacity are all equivalent, you could use any of them with a 1000A OCPD. The closest copper analogue to (3) sets of 600 kcmil Al would be (3) sets of 400 kcmil Cu.

Cheers, Wayne

I'm not saying that your interpretation doesn't make sense but sorry but none of this matches the wording in Note 2. It doesn't mention the same number of sets or the size of an OCPD.

 

[Joe Villani]

Yes, I'm saying that it's more important to match the number of runs (3) than to match the exact ampacity. For these purposes 1000A, 1005A, and 1020A ampacity are all equivalent, you could use any of them with a 1000A OCPD. The closest copper analogue to (3) sets of 600 kcmil Al would be (3) sets of 400 kcmil Cu.

Cheers, Wayne

Then it should say that in Note 2.

 

[wwhitney]

Then it should say that in Note 2.

Agreed. But as it is silent on how to handle the number of runs, you can still use common sense and size the SBJ based on the same number of runs. You know, using the size of conductor of the other material you would have used to do the job if you had used that material.

Cheers, Wayne