System bonding jumper sizing

[hhsting]

I have quick question on sizing system bonding jumper.

I have 3 sets of 600kcmil AL transformer secondary conductor

By calculation I would get 3x600 = 1800 kcmil. 1800x0.125 = 225 kcmil AL.

Their is no 225 kcmil AL shown on NEC 2017 Table 310.16. Next standard size is 250kcmil AL.

Changing to copper I would get equivalent ampacity for 250kcmil AL to be 4/0 copper.

However engineer has provided 3/0 copper. I can see from where since NEC 2017 Table 250.102(C)(1) goes up to 1750 kcmil AL which is 20/ awg. Assuming the table where to continue it would be over 1750 kcmil Al to over xxx kcmil Al would equal 3/0 awg copper.

I am not sure would 3/0 awg copper be ok or 4/0 awg copper? There is no ampacity table for NEC 2017 Table 310.16

 

[infinity]

I am not sure would 3/0 awg copper be ok or 4/0 awg copper? There is no ampacity table for NEC 2017 Table 310.16

The SBJ is required to be 250 kcmil aluminum or the equivalent ampacity of that size if it's copper? I don't have the book but I believe that one of the notes to T250.102(C) says something like that. Is that what is says?

 

[hhsting]

The SBJ is required to be 250 kcmil aluminum or the equivalent ampacity of that size if it's copper? I don't have the book but I believe that one of the notes to T250.102(C) says something like that. Is that what is says?

I pasted it below. My question is 225 kcmil AL their is no ampacity in NEC 2017 Table 310.16. So nkw what?




Sent from my iPhone using Tapatalk

 

[don_resqcapt19]

I would be happy with the 3/0 copper. If you interpolate the ampacity of 225 kcmil with 4/0 aluminum it would be less than the ampacity of 3/0 copper

 

[hhsting]

I would be happy with the 3/0 copper. If you interpolate the ampacity of 225 kcmil with 4/0 aluminum it would be less than the ampacity of 3/0 copper

4/0 awg AL is 212 kcmil not 225 kcmil???

 

[infinity]

4/0 awg AL is 212 kcmil not 225 kcmil???

The minimum size aluminum SBJ is 250 kcmil because it's a standard size. You cannot use #4/0 because it's not large enough. Use a 250 kcmil conductor size in your comparison calculation.

 

[hhsting]

The minimum size aluminum SBJ is 250 kcmil because it's a standard size. You cannot use #4/0 because it's not large enough. Use a 250 kcmil conductor size in your comparison calculation.

In that case the equalivelnt size sbj for 250kcmil AL would be 4/0 awg copper?

 

[hhsting]

The minimum size aluminum SBJ is 250 kcmil because it's a standard size. You cannot use #4/0 because it's not large enough. Use a 250 kcmil conductor size in your comparison calculation.

In that case the equivalent size sbj for 250kcmil AL would be 4/0 awg copper?

 

[Joe Villani]

The way I read it is you have to come up with an equivalent ampacity using copper conductors. Then size the copper system bonding jumper

As configured the ampacity of the 3 sets of 600 AL is 1020 amps.

4 sets of 250 copper will get you an equivalent ampacity of 1020 amps.

Based on table 250.102(C)(1) 4 sets of 250kcmil a 2/0 copper system bonding jumper would be sufficient.

IMO a 3/0 copper conductor would be more than adequate.

 

[infinity]

The way I read it is you have to come up with an equivalent ampacity using copper conductors. Then size the copper system bonding jumper

As configured the ampacity of the 3 sets of 600 AL is 1020 amps.

4 sets of 250 copper will get you an equivalent ampacity of 1020 amps.

Based on table 250.102(C)(1) 4 sets of 250kcmil a 2/0 copper system bonding jumper would be sufficient.

IMO a 3/0 copper conductor would be more than adequate.

I think you got it. It would have helped me if I actually perused Note 2 in post #3.

 

[hhsting]

The way I read it is you have to come up with an equivalent ampacity using copper conductors. Then size the copper system bonding jumper

As configured the ampacity of the 3 sets of 600 AL is 1020 amps.

4 sets of 250 copper will get you an equivalent ampacity of 1020 amps.

Based on table 250.102(C)(1) 4 sets of 250kcmil a 2/0 copper system bonding jumper would be sufficient.

IMO a 3/0 copper conductor would be more than adequate.

Really? Learn something new everyday. There must be more than one way to get equal ampacity of the same material besides combination of 4 sets of 250kcmil copper? In that case what should one do?

 

[hhsting]

The way I read it is you have to come up with an equivalent ampacity using copper conductors. Then size the copper system bonding jumper

As configured the ampacity of the 3 sets of 600 AL is 1020 amps.

4 sets of 250 copper will get you an equivalent ampacity of 1020 amps.

Based on table 250.102(C)(1) 4 sets of 250kcmil a 2/0 copper system bonding jumper would be sufficient.

IMO a 3/0 copper conductor would be more than adequate.

One can also do the following:

600kcmil AL amapcity is 340A. The next equivalent standard sized copper conductor is 500 kcmil which is 380A.

So 380x3*.125 = 142.5 kcmil which is 4/0 awg copper.

I dont understand why would above be not valid?

 

[infinity]

Really? Learn something new everyday. There must be more than one way to get equal ampacity of the same material besides combination of 4 sets of 250kcmil copper? In that case what should one do?

Find the smallest permitted combination and use that as the minimum size. The obvious easiest method is to just use the same material as the ungrounded conductors and avoid the calculation altogether. I believe that Joe has found the smallest size copper equivalent.

600kcmil AL amapcity is 340A. The next equivalent standard sized copper conductor is 500 kcmil which is 380A.

So 380x3*.125 = 142.5 kcmil which is 4/0 awg copper.

This is incorrect, #3/0 is 167,800 CM.

If you had 3 sets of 500 kcmil the calculation for the SBJ would be:
500*3=1500*12.5%=187,500 CM equals a minimum #4/0.

 

[don_resqcapt19]

Here is what I did in post #4. The calculation shows a required conductor size of 225 kcmil, a size that is not standard.
The 75°'c ampacity of 250 kcmil aluminum is 205 amps, and the ampacity of 4/0 aluminum is 180 amps for a difference of 25 amps. The area of 4/0 is 211,600 making the 250 kcmil, larger than the 4/0 by 38,400 cm. If you divide 38,400 by 25 amps, you get 1536 cm per amp. 225 kcmil would be 13,400 cm larger than 4/0 and 13,400/1536 is 8.75. The interpolated ampacity of 225 kcmil aluminum would be 180 + 8.75 or 189 amps The ampacity of 3/0 copper is 200 amps, greater than the minimum required ampacity for the system bonding jumper for this circuit,.
However this is not exactly what the note to the table says to do, but the result is the same.

 

[infinity]

However this is not exactly what the note to the table says to do, but the result is the same

Yes the note says to use the ampacity of the ungrounded conductors which I missed in my earlier posts.

 

[wwhitney]

As the language of Note 2 is silent as to whether this "assumed use of ungrounded supply conductors of the same material" should assume the same number of sets as actually installed or not, there is some ambiguity in the outcome here.

As per post 13, if we stick with 3 sets, we would need 500 kcmil Cu to meet or exceed 1020A ampacity, so we would need a 4/0 Cu SBJ.

If we are allowed to assume more sets, the required size will almost always be minimized by just using sets of 1/0. For 1020A, this would require 7 sets of 1/0 Cu, which is 7*105.6 = 739 kcmil, so a 2/0 Cu SBJ would suffice.

Since the point of increasing the size of the SBJ as the ungrounded supply conductors gets larger is presumably to handle the resulting larger fault currents, and as fault currents depend on conductor impedance, not ampacity, it seems to me appropriate to require the same number of sets as actually installed. Just like if there was no change in material, 9 sets of 1/0 Al (950 kcmil, 1080A ampacity) would require a smaller SBJ at 4/0 Al than 3 sets of 600 kcmil Al (1800 kcmil, 1020A ampacity) does at 250 kcmil Al.

Cheers, Wayne

 

[infinity]

Sonc

As the language of Note 2 is silent as to whether this "assumed use of ungrounded supply conductors of the same material" should assume the same number of sets as actually installed or not, there is some ambiguity in the outcome here.

Since it's not specific you should be able to calculate finding the smallest possible set of ungrounded conductors/SBJ and that would tell you the minimum size. The minimum size would still serve the purpose of the SBJ.

 

[wwhitney]

Since it's not specific you should be able to calculate finding the smallest possible set of ungrounded conductors/SBJ and that would tell you the minimum size. The minimum size would still serve the purpose of the SBJ.

Yet if you weren't changing materials between the ungrounded conductors and the SBJ, you wouldn't get the latitude to recalculate your ungrounded conductors as a different number of sets of a smaller size to get a smaller cross sectional area for the same ampacity, and therefore a smaller SBJ. So I don't see why you should get that extra latitude just because you must do some recalculating because you are switching conductor materials.

While I agree that Note 2 doesn't specify that you must use the same number of sets, I think an AHJ making that interpretation with respect to Note 2 is perfectly reasonable, and for the reason described, that interpretation is more logical than the alternative.

Cheers, Wayne

 

[infinity]

So I don't see why you should get that extra latitude just because you must do some recalculating because you are switching conductor materials

The point is if the SBJ is sized to the smallest possible conductor size and is still code compliant then how could anyone require you to use one that is larger than the minimum size? There is no wording in that section to suggest otherwise, it just wants the calculation to ensure that the SBJ is large enough to do the job.

 

[wwhitney]

The point is if the SBJ is sized to the smallest possible conductor size and is still code compliant then how could anyone require you to use one that is larger than the minimum size?

Because you didn't install 7 sets or 9 sets, you installed 3. Look at these 4 possible installs where we don't mix materials, each one is a minimum 1000A ampacity for a 1000A OCPD:

3 sets Cu 400 kcm, area = 1200 kcm, SSBJ size = 1200/8 = 150 kcm, next size up is 3/0 Cu
7 sets Cu 1/0, area= 7*105.6 = 739 kcmil, SSBJ size = 2/0 per table
3 sets Al 600 kcm, area = 1800 kcm, SSBJ size = 1800/8 = 225 kcm, next size up is 250 kcm
9 sets Al 1/0, area = 9*105.6 = 950 kcm, SSBJ size = 4/0 per table

Notice that for Cu, using 3 sets gives you over 1.5x the area versus sticking with 1/0, and for Al it's almost 2x the area. More conductor area = less impedance = more fault current. So when we have just 3 sets, we have to use a larger size SBJ than when we have 7 or 9 sets.

Now, why should mixing materials let you get the benefit of a smaller SBJ that using more sets would give you, when you only used 3 sets? It shouldn't, you used 3 sets and have a bigger total area and a lower impedance than if you used 7 or 9 sets, and your SBJ should be larger accordingly.

The only remaining question I have when applying Note 2 is whether it's fair game to say "the OCPD is 1000A, if I had used 3 sets of Cu they would have been 400 kcm, so that's the basis for comparison." Or whether you have to say "the ampacity installed is 1020A, to match that with 3 sets of Cu we'd need to use 500 kcm, so that's the basis for comparison." The former gives you a 3/0 Cu SBJ, the latter a 4/0 Cu SBJ.

Cheers, Wayne