SSBJ vs EGC

[NNN]

I understand the difference between SSBJ and EGC but I have following questions:
1. Why is SSBJ sized differently than EGC? Is it because the magnitude of fault current on both the sides is different?
2. What is the fault current path if the SLG happens line side of service disconnect?

 

[Elect117]

1. Why is SSBJ sized differently than EGC? Is it because the magnitude of fault current on both the sides is different?

EGC takes into account the trip time of the OCPD. There is no OCPD on the supply side bonding jumper. So the ensure it can sustain the fault current for the duration of the fault it is sized to the conductors.

2. What is the fault current path if the SLG happens line side of service disconnect?

That is kind of a loaded question since there are a lot of scenarios for a SLG and there are just as many paths. But it will take all paths it can take to complete the circuit. So any path that brings it back to the neutral of the transformer. A majority of current will flow on the least resistant path but, nonetheless, all the paths it can.

 

[jaggedben]

...
1. Why is SSBJ sized differently than EGC?

Well, one way to look at it is that they’re not sized very differently, once you get past the 8awg minimum for table 205.102. Lots of times they would be the same size (e.g. 1/0 cu on a 150A breaker would get a 6awg green wire by either table.) Most of the time the sizes wouldn't be off by more than one. It's just that table 250.102 is used when you don’t have an OCPD, and the charts don't always line up perfectly.

 

[NNN]

EGC takes into account the trip time of the OCPD. There is no OCPD on the supply side bonding jumper. So the ensure it can sustain the fault current for the duration of the fault it is sized to the conductors.



That is kind of a loaded question since there are a lot of scenarios for a SLG and there are just as many paths. But it will take all paths it can take to complete the circuit. So any path that brings it back to the neutral of the transformer. A majority of current will flow on the least resistant path but, nonetheless, all the paths it can.

I have gone through the rabbit hole of and done a lot of reading but you are the first one to bring in trip time of the OCPD which makes so much sense.

 

[NNN]

Well, one way to look at it is that they’re not sized very differently, once you get past the 8awg minimum for table 205.102. Lots of times they would be the same size (e.g. 1/0 cu on a 150A breaker would get a 6awg green wire by either table.) Most of the time the sizes wouldn't be off by more than one. It's just that table 250.102 is used when you don’t have an OCPD, and the charts don't always line up perfectly.

True. I have done this comparison before and they dont line up perfectly.

 

[NNN]

Where I am getting at is, I have a MV transformer. My SSBJ is running to the OCPD. I have parallel sets of conductors in separate conduits. Ungrounded conductor size is 750kCMIL. SSBJ is 2/0AWG in each conduit. My OCPD is 3500A. If I have parallel sets of conductors even on the load size, my EGC needs to be 500kcmil in each conduit where my SSBJ on the line side only needs to be 2/0AWG in each conduit. This is something hard to grasp.

 

[Tyler C]

Where I am getting at is, I have a MV transformer. My SSBJ is running to the OCPD. I have parallel sets of conductors in separate conduits. Ungrounded conductor size is 750kCMIL. SSBJ is 2/0AWG in each conduit. My OCPD is 3500A. If I have parallel sets of conductors even on the load size, my EGC needs to be 500kcmil in each conduit where my SSBJ on the line side only needs to be 2/0AWG in each conduit. This is something hard to grasp.

I was going to bring this up when I read your initial two questions. The question about this really is:
Why must each EGCs be sized to the 250.122 table value in parallel conduits (that is, every EGC must be sized to the table, regardless of the number of parallel conduits), but for SSBJs 250.102(C)(2) allows us to have parallel SSBJs that are each sized based on the size of the ungrounded conductors in each conduit individually, or have a single SSBJ sized based on the sum of the ungrounded conductors in all of the conduits?


I think this is due to their different role in carrying faults.

In the case of the SSBJ, let's say we have a typical SDS with a system bonding jumper in the transformer, and transformer secondary conductors which terminate in a MCB panelboard. When there is a ground fault in the panelboard or on any of its branch circuits, the fault current is carried back to the transformer along the SSBJ, through the system bonding jumper, and then to the X0 terminal of the transformer. Having a single large SSBJ or multiple smaller SSBJs are both sufficient to reduce the impedance of the ground path enough to clear the fault.

Now take the case of a EGC. If there is a ground fault inside one of the conduits, a single EGC is carrying all of the fault current. A smaller sized EGC would be less likely to clear the fault in this case.

You could make the argument about what if there is a fault in one of the transformer secondary conduits. However I think this is far less common, given the limitations of 240.21, and the locations they are typically installed in compared to feeders and branch circuits in general. Also, in that sort of fault there is no OCPD on the line side, you would be hoping that the fault manages to open the OCPD protecting the primary.

 

[wwhitney]

Now take the case of a EGC. If there is a ground fault inside one of the conduits, a single EGC is carrying all of the fault current. A smaller sized EGC would be less likely to clear the fault in this case.

While there will only be one EGC on the shortest path back to the MBJ/SBJ, there will be other paths through the EGCs in the other conduits, which are tied together at both ends of the parallel runs.

Looking only at the parallel run of n conduits, if we call the impedance of the full length of one EGC in one conduit 1, then suppose we have a fault to one of the EGCs in one conduit at a fraction x of the length from the source end of the conduit, where x is in [0,1]. Then the impedance of the short path is x, and the impedance of the long path is (1-x) + 1/(n-1).

If I did the math right, the parallel impedance becomes x - (1-1/n) * x², and it takes on a maximum when x = 1/(2-2/n), where the impedance is 1/(4 - 4/n).

For n=2, the worst case is when the fault is at the supply end, and the impedance is 1/2, the impedance of the parallel EGCs. For n =3, the worst case is with the fault at 3/4 of the way to the load end, and the impedance is 3/8, greater than 1/3, the impedance of the parallel EGCs. For n=4, the worst case is with the fault at 2/3 of the way to the load end, and the impedance is 1/3, greater than 1/4, the impedance of the parallel EGCs. Etc.

So there is an effect, but it is no so large as to require an n-fold increase in total EGC cross section. For 2 conduits, no increase is required for the worst case impedance to be the same as the impedance for a fault at the load end of the conduits; for 3 conduits, a 9/8 upsizing factor would be required; for 4 conduits, a 4/3 upsizing factor would required, etc.

Cheers, Wayne

 

[Tyler C]

While there will only be one EGC on the shortest path back to the MBJ/SBJ, there will be other paths through the EGCs in the other conduits, which are tied together at both ends of the parallel runs.

Yes, unless you are extremely unlucky with the fault where the EGC is completely severed and the fault only occurs on one side of the break. Which side is the worse case depends on the location, because in this event the long path has the higher impedance if the fault is close to the source, and the short path has the higher impedance if the fault is close to the load end. Since the short path and long path have the same impedance at x = (1-x) + 1/(n-1), as n increases, the equal point moves closer to the source. However, within the fault conduit the only fault path is the single EGC, so as n approaches infinity the equal point converges to x = 1/2, it cannot be any lower.

If I did the math right, the parallel impedance becomes x - (1-1/n) * x², and it takes on a maximum when x = 1/(2-2/n), where the impedance is 1/(4 - 4/n).

I think you got it right, I calculated the parallel impedance, the d/dx[ ] = 0 to find the max, and plugged the max back into the impedance and got all the same results as you.

For n=2, the worst case is when the fault is at the supply end, and the impedance is 1/2, the impedance of the parallel EGCs.

I think you mean at the load end. Besides that, I agree for all the n = scenarios.

Another thing to consider is that this all changes if using metallic conduit, since the EGCs are bonded to the enclosure at both ends, the long path becomes an even lower impedance since it is all the wire-type EGCs in the non faulted conduits paralleled with all of the conduits themselves (plus x-1).

So there is an effect, but it is no so large as to require an n-fold increase in total EGC cross section.

Agreed. I suppose that when writing 250.122(F)(1)(b), CMP-5 wanted to leave the calculus out of it.

 

[Carultch]

Agreed. I suppose that when writing 250.122(F)(1)(b), CMP-5 wanted to leave the calculus out of it.

If we were to match the physics behind what an equipment grounding conductor should be, I would expect there'd be a minimum size as a function of the OCPD (as currently is the rule), along with a maximum number of ohms of resistance for the EGC. This would mean that if you have an excessive length circuit, you'd upsize your EGC, until its resistance is below the maximum. There needs to be at least enough conductance to have an effective ground-fault current path back to the source.

However, from an inspector's point of view, how would they rightly know the resistance of the EGC, or even the length? The rule was likely written so they could just do a visual inspection, see that the phase conductors are upsized by 4 gauge sizes from the minimum, and look for a corresponding increment of 4 gauge sizes of the EGC. The AWG scale is a logarithmic scale of cross-sectional area, therefore a multiplicative change in area, is the same as incremental changes in the gauge size. If they wrote a rule that only specified incremental changes in gauge sizes, it wouldn't be as easy to apply to sizes 250 kcmil and larger, where we no longer use the AWG scale. The rule was written to be able to apply to the full scope of the size table.

 

[Tyler C]

If we were to match the physics behind what an equipment grounding conductor should be, I would expect there'd be a minimum size as a function of the OCPD (as currently is the rule), along with a maximum number of ohms of resistance for the EGC.

The impedance of the source would also be important to consider. A system with a higher available fault current would be able a higher impedance of the faulted conductors than one with fewer amps available. Although you hit a limit at some point (you only have so many volts) when the available current is sufficiently high.

However, from an inspector's point of view, how would they rightly know the resistance of the EGC, or even the length?

True, although there are many areas where the inspector cannot verify everything, because you can't expect an inspector to re-complete every calculation in a design. Some things are really just an engineering responsibility.

If they wrote a rule that only specified incremental changes in gauge sizes, it wouldn't be as easy to apply to sizes 250 kcmil and larger, where we no longer use the AWG scale. The rule was written to be able to apply to the full scope of the size table.

Yes, you have to do the proportion for upsizing for the point you mention. It already is a hassle for 250+ since you need to convert the AWG into MCM, and then back to AWG at the end (unless your EGC ends up larger than 4/0).

Ultimately, 122 is conservative and can result in larger EGCs than appear to be necessary. However, this further reduces clearing time, which in something like an arc event everyone can be thankful for.