System bonding jumper sizing

[infinity]

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.

It clearly states "will have an ampacity equivalent to that of the installed ungrounded supply conductors." The conductors installed are 1020 amps so that's the number you would use for the comparable conductors of another material, not the OCPD of 1000 amps.

Note 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.

 

[hhsting]

Well even the minimum required is subject to interpretation.

For example 4/0 awg copper sbj is larger than 2/0 awg copper.

Whose to sy minimum required cannot be 4/0 awg copper since it can handle more current.

Minimum required may not always be smallest you know.

 

[infinity]

Well even the minimum required is subject to interpretation.

For example 4/0 awg copper sbj is larger than 2/0 awg copper.

Whose to sy minimum required cannot be 4/0 awg copper since it can handle more current.

Minimum required may not always be smallest you know.

If the #2/0 is large enough under the scenario of using 4 sets of 250 kcmil as Joe outlined in post #9 why would the minimum size ever need to be larger than that?

For example, if you came up with 5, 10, or 100 different parallel conductor combinations for the minimum of 1020 amps and used the smallest SBJ (based on the the calculation of those combinations) how would that not be the minimum size required? Also when you're trying to find the minimum SBJ size there is nothing in Note 2 that says you need to use the same number of sets as the installed conductors.

 

[wwhitney]

It clearly states "will have an ampacity equivalent to that of the installed ungrounded supply conductors." The conductors installed are 1020 amps so that's the number you would use for the comparable conductors of another material, not the OCPD of 1000 amps.

Well, the point is that it uses the word "equivalent," rather than just saying "not less than," which is certainly a term the NEC uses often. So the meaning intended is different from just "not less than".

As a practical matter, if you have 1000A OCPD, and you choose 3 sets in separate conduits, you're going to choose the smallest conductor size whose ampacity is at least 1000/3. Making the same choice for the other material is, to me, equivalent.

Cheers, Wayne

 

[wwhitney]

there is nothing in Note 2 that says you need to use the same number of sets as the installed conductors.

There isn't anything explicit in the text. But post #20 illustrates why there are good technical/physics reasons to use the same number of sets.

Cheers, Wayne

 

[infinity]

Well, the point is that it uses the word "equivalent," rather than just saying "not less than," which is certainly a term the NEC uses often. So the meaning intended is different from just "not less than".

Without checking every possible scenario what Joe posted in post #9 is probably the only combination that had the equivalent ampacity to the 3 sets in the OP of 1020 amps. The words not less than although not written are likely implied but even if we reject that argument then a strict reading of the note would require the calculation using only an equivalent ampacity of copper conductors.

 

[david]

Interesting discussion, there should not be that many variables and choices.

250,000 × 63% = 157,500= min of 3/0 (167,800)

It should be that simple

 

[david]

The table dosnt say 63%, but when you compare the alum circular mils to the copper circular mils of the required bonding jumpers, the copper circular mils is 63 % that of the alum.

If uou want to be consistent after you determine the alum bonding jumper to be 250,000 than a copper bonding jumper should be a min of 63%

 

[LarryFine]

The table dosnt say 63%, but when you compare the alum circular mils to the copper circular mils of the required bonding jumpers, the copper circular mils is 63 % that of the alum.

Is that a consistent ratio across the board?

 

[infinity]

Interesting discussion, there should not be that many variables and choices.

250,000 × 63% = 157,500= min of 3/0 (167,800)

It should be that simple

It should be that simple but the math proves otherwise as Joe pointed out.

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.

 

[david]

Is that a consistent ratio across the board?

I compared all the ones in the table and I came up with 63% for them

 

[david]

It should be that simple but the math proves otherwise as Joe pointed out.

Yes I agree but the engineer said 3/0 and that seems to be consistent with the table, and it falls between the 2/0 and 4/0 being discussed here

 

[Elect117]

Table 250.102(C)

Note 3: If multiple sets of service-entrance conductors are used as permitted in 230.40, Exception No. 2, or if multiple sets of ungrounded supply conductors are installed for a separately derived system, the equivalent size of the largest ungrounded supply conductor(s) shall be determined by the largest sum of the areas of the corresponding conductors of each set.

I think Wayne is right. You should take into account the number of runs being used into account when making the conversion.

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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).

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Now I don't know what the answer is.

 

[wwhitney]

Is that a consistent ratio across the board?

Yes, if you look at the table, on each row the required Cu SBJ size is 2 sizes smaller in AWG than the Al SBJ size. AWG sizing is geometric, and it's very close to being true that going up 3 sizes doubles the cross sectional area. So going down 2 sizes is a ratio of 2^(-2/3) = 0.630.

Conveniently, the ratio of the resistivities of copper and aluminum is 1.724/2.65 = 0.651 (numbers from engineeringtoolbox.com), which is pretty close to 0.63, the ratio of areas for a difference of two AWG sizes.

Agreed that Note 2 should just be replaced by a requirement to multiple/divide by 0.63 when mixing materials.

Cheers, Wayne

 

[infinity]

Agreed that Note 2 should just be replaced by a requirement to multiple/divide by 0.63 when mixing materials.

I think that you're onto something big.

 

[Elect117]

So then what are we talking, for the size:

600x3=1800kcmil

1800 * .63 = 1134

1134*.125=141.25

So then 3/0 copper?

 

[hhsting]

I compared all the ones in the table and I came up with 63% for them

Which table you talking about?


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[infinity]

So then what are we talking, for the size:

600x3=1800kcmil

1800 * .63 = 1134

1134*.125=141.25

So then 3/0 copper?

The minimum size copper according to Note #2 is #2/0.

 

[hhsting]

Yes, if you look at the table, on each row the required Cu SBJ size is 2 sizes smaller in AWG than the Al SBJ size. AWG sizing is geometric, and it's very close to being true that going up 3 sizes doubles the cross sectional area. So going down 2 sizes is a ratio of 2^(-2/3) = 0.630.

Conveniently, the ratio of the resistivities of copper and aluminum is 1.724/2.65 = 0.651 (numbers from engineeringtoolbox.com), which is pretty close to 0.63, the ratio of areas for a difference of two AWG sizes.

Agreed that Note 2 should just be replaced by a requirement to multiple/divide by 0.63 when mixing materials.

Cheers, Wayne

If only you can figure out how NEC 2017 Tabes 250.66 GEC size from service conductors and Table 310 ampacities are derived

 

[wwhitney]

The minimum size copper according to Note #2 is #2/0.

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