Equipment grounding conductor sizing for increased phase conductors

[rbalex]

...You will notice that I omitted Step Six, which would be when you explain the reason for Step Five to your apprentice.
...

You got a genuine LOL from me on that one Charlie - your eloquence breaks through once more.

 

[K8MHZ]

To say that a longer wire will decrease the amount of fault current is to speak an obvious truth. To infer that it will not be enough to trip a breaker is to go too far. A calculation could prove it one way or the other, but we are not allowed to use a calculation. Also, if there is no voltage drop concern, and no ambient temperature concern, and fewer than 3 CCCs in a conduit, and if the only reason you upsized the phase conductors is that you had a spare roll of it in your truck, I don't think there is a need to upsize the EGC. It would be hard for an inspector to make a call on such a basis. I say again, this is about making it easy for the inspector, not about safety.
That is backwards. You have already increased the available fault current (by making the phase conductors bigger). Why should you think you need to increase the available fault current further, by also making the EGC bigger?

The way I see it is that if the EGC is not re-sized it is the same as putting a resistor in series with an re-sized one. I don't see the increase in size as increasing the available fault current, but providing a low impedance path for it back to the panel.

You do pose a question that is harder to rationalize, about the increase in size not related to length. I believe the intent of the rule would still be met if the run were short and a GEC sized to the OCPD were used, but I don't think there are any provisions for this.

Again, this is just my 'common sense' explanation. In my feeble little mind it just makes sense to up size all the conductors, not just the CCC's. I admit that I would not be able to easily mathematically prove that I am correct, but I don't need to so long as the NEC requires it.

 

[bob]

The way I see it is that if the EGC is not re-sized it is the same as putting a resistor in series with an re-sized one. I don't see the increase in size as increasing the available fault current, but providing a low impedance path for it back to the panel.

The origional conductor was to be 3/0 cu. It was increased to 300 kcm. I would hope that you would agree that the fault current would be higher with the 300 kcm. If you look at table 250.122 , the egc for a 200 amp breaker is #6 cu. If #6 is ok for a 3/0 ckt it certainly good enough for a 300 kcm ckt because the fault current is higher and will trip the breaker quicker.

You do pose a question that is harder to rationalize, about the increase in size not related to length. I believe the intent of the rule would still be met if the run were short and a GEC sized to the OCPD were used, but I don't think there are any provisions for this.

I believe table 250.122 does that.

Again, this is just my 'common sense' explanation. In my feeble little mind it just makes sense to up size all the conductors, not just the CCC's. I admit that I would not be able to easily mathematically prove that I am correct, but I don't need to so long as the NEC requires it.

The CCC's were increased because of voltage drop. The egc may have needed to be increased, but that is a design decision. The NEC required the EGC to be increased 4 conductor sizes.

 

[don_resqcapt19]

Don, I do seem recall a debate of the sort you mention, but I don't believe this was the subject matter. I already posted the original Proposal; the only Comment was to note that deleted text was indicated by italics.

If my fuzzy memory is correct, the debate was whether a wire-type EGC should be included with all circuits. It was accepted, and later rejected after MUCH Comment.

Bob,
Actually I was commenting on something that I had read into Bob Wilson's post. I was thinking of the issue caused by 250.122(F) when cables are run in parallel. Prior to the 93 code, the part of 250-95 that addressed the sizing of EGCs for parallel circuits only applied to parallel circuits that were installed in raceways. I had thought this was a conduit v cable fight, but in checking the 93 code ROP and ROC, there was only a single accepted proposal and no comments for the change that added cables to this rule.

 

[hurk27]

Another way to look at it......

Let's say you have a 1200 foot long driveway and want to feed a light array at the end that draws 12 amps. It is on a single 20 amp circuit.

Let's also say you wanted to achieve a goal of a 3 percent voltage drop.

You find that you need a 1/0 hot and neutral to get 3.1 percent and go for it.

So.....do you depend on 1200 feet of 12 AWG to be your ECG?

Hopefully this extreme condition illustrates the need for the ECGs to be up-sized with the other conductors.

At 1200' If you leave the 12 for the EGC you would still have over 50 amps as a bolted fault with the 1/0 as a ungrounded conductor, if you leave both the ungrounded conductor and the EGC at a#12 it would drop to less then 29 amps bolted fault at 1200' which might not clear it. At least fast enough

But as Bob stated most faults are not bolted, and contain high resistance, so even without the voltage drop issue we can have a high resistance fault that would not clear the OPCD.
this is one of the reasons for the Arc fault breaker, even though a GFCI would do the same job:roll:

 

[Smart $]

The origional conductor was to be 3/0 cu. It was increased to 300 kcm. I would hope that you would agree that the fault current would be higher with the 300 kcm. If you look at table 250.122 , the egc for a 200 amp breaker is #6 cu. If #6 is ok for a 3/0 ckt it certainly good enough for a 300 kcm ckt because the fault current is higher and will trip the breaker quicker.

...

The CCC's were increased because of voltage drop. The egc may have needed to be increased, but that is a design decision. The NEC required the EGC to be increased 4 conductor sizes.

I think you are leaving out a critical piece of information in the drawing your conclusion. You are talking as if the circuit length using 3/0 is the same as when using 300 kcmil. The fact is, they are not. And available fault current at the load end of a circuit is affect by path resistance/impedence, and thus by voltage drop.

Let's exaggerate for a minute. Using the same system as the source, you have two 150A circuits. One uses 3/0 Cu and is 10 ft to the load. The other uses 300 kcmil Cu and is ten miles to the load (I did say we were exaggerating ). At the load end of each, which circuit has more available fault current???

When you upsize for voltage drop, the available fault current at the load end of the circuit is reduced approximately by the same percentage as the voltage drop.

 

[bob]

I think you are leaving out a critical piece of information in the drawing your conclusion. You are talking as if the circuit length using 3/0 is the same as when using 300 kcmil.

In this example are the lenghts not the same?

The fact is, they are not. And available fault current at the load end of a circuit is affect by path resistance/impedence, and thus by voltage drop.

The lengths are not the same? Then we have a different problem.

Let's exaggerate for a minute. Using the same system as the source, you have two 150A circuits. One uses 3/0 Cu and is 10 ft to the load. The other uses 300 kcmil Cu and is ten miles to the load (I did say we were exaggerating . At the load end of each, which circuit has more available fault current???

Whats your point?

When you upsize for voltage drop, the available fault current at the load end of the circuit is reduced approximately by the same percentage as the voltage drop.

Are you saying that the fault current is less with the 300 kcm than with the 3/0?

 

[Smart $]

In this example are the lenghts not the same?


The lengths are not the same? Then we have a different problem.


Whats your point?


Are you saying that the fault current is less with the 300 kcm than with the 3/0?

I'm saying that I believe the available fault current at the load end of a 300kcmil circuit at 400' in length is the same as a much shorter 3/0 circuit that has the same percentage of voltage drop for the same load.

 

[bob]

I'm saying that I believe the available fault current at the load end of a 300kcmil circuit at 400' in length is the same as a much shorter 3/0 circuit that has the same percentage of voltage drop for the same load.

I would agree but I don't see your point.

 

[Smart $]

I would agree but I don't see your point.

Aren't ground faults just an abnormal circuit?

To maintain a nominal level of circuit operation, wouldn't a larger grounding conductor be necessary for longer cirucit runs?

 

[macmikeman]

There should have been an exception to that code section for upsized phase conductors that are in metal raceways(installed the right way of course, all wrench tight that has an optional egc (while not required, but still provided) and is bonded to the raceway. In that case I see the only real need to upsize the egc would be code compliance as presently written. Any available fault current would be safely carried by the coresponding increased size metal conduit. Since I cannot spell well, I was kinda hoping one of you respected friends might submit this for me next cycle.

 

[bob]

Aren't ground faults just an abnormal circuit?

To maintain a nominal level of circuit operation, wouldn't a larger grounding conductor be necessary for longer cirucit runs?

I'm not sure what you mean by "nominal level of circuit operation".
When you have a ground fault the only operation required is to trip the breaker. That does not necessarly require a larger EGC for a longer ckt.

 

[George Stolz]

When you have a ground fault the only operation required is to trip the breaker.

...quickly.

 

[bob]

...quickly.

Thank you George

 

[Smart $]

I'm not sure what you mean by "nominal level of circuit operation".
When you have a ground fault the only operation required is to trip the breaker. That does not necessarly require a larger EGC for a longer ckt.

...quickly.

Thank you George

And that is the point. If you do not upsize the grounding conductor, it will take longer to trip the breaker because fault current at the end of the circuit is subject to the resistance of the grounding conductor. An upsized grounding conductor would permit more fault current to flow, but not more than the not-upsized grounding conductor at a shorter circuit length... hence a nominal level of circuit operation.

 

[hurk27]

And that is the point. If you do not upsize the grounding conductor, it will take longer to trip the breaker because fault current at the end of the circuit is subject to the resistance of the grounding conductor. An upsized grounding conductor would permit more fault current to flow, but not more than the not-upsized grounding conductor at a shorter circuit length... hence a nominal level of circuit operation.

The problem is in many cases the EGC does not need to be up sized as much as the ungrounded conductors, and before "2002" a PE could do a fault calculation and many times install EGC sized for the fault that would safely clear the breaker that would be much lower cost that what the code requires now, Bob (rbalex) touched on this in post 8.

If you have up sized the ungrounded conductors, you have already reduced half of the fault clearing path, so many times the EGC doesn't require to be much larger to to reach the fault clearing ability to open the OCPD's if any at all.
a set of #12 sol will produce a fault of about 29 amps at 1,000' maybe not enough to open a 20 amp breaker fast enough, but increase the ungrounded conductor to a 3/0 with the #12 EGC, and it rises to over 50 amps of fault, which should clear the same 20 amp breaker in less than a couple seconds, increase the EGC to a #10 and now you have almost 90 amps of fault current. (the above is just for reference as no other factors was used)

Fault current and maintaining voltage drop do not run hand in hand, they are two different calculations.

So the change in 2002 might have made it easier to inspect but went a long way in making it more costly in adjusting for voltage drop, which It might keep some from adjusting for voltage drop because of this.

 

[Smart $]

The problem is in many cases the EGC does not need to be up sized as much as the ungrounded conductors, and before "2002" a PE could do a fault calculation and many times install EGC sized for the fault that would safely clear the breaker that would be much lower cost that what the code requires now, Bob (rbalex) touched on this in post 8.

I'm not saying the grounding conductor needs to be upsized by linear ratio as it is in current code. Aside from that, I would like to know the formula or method is used to determine such. The short-circuit calculations I have seen to date are regarding circuit conductors (service, feeder, and branch), not grounding conductors.

If you have up sized the ungrounded conductors, you have already reduced half of the fault clearing path, so many times the EGC doesn't require to be much larger to to reach the fault clearing ability to open the OCPD's if any at all.

That's a rather "programmed response"... and a misleading one at that. Just because the conductor is upsized does not necessarily increase the fault current at the end of the circuit. The very reason you increased the size of the circuit conductors (distance>voltage drop) is the same reason by which you have not facilitated fault current. Is there some magic involved that a different voltage drives the fault current? I think not. SO we are talking about the same voltage driving the current. Then we must consider that the fault may occur when the circuit conductor has been subjected to the load for a duration which has already brought it up to some temperature above ambient, but lower than it's rating. Not that I'm fully versed in such calculations, but I've yet to see any method which accounts for "already warm" conductors.

... a set of #12 sol will produce a fault of about 29 amps at 1,000' maybe not enough to open a 20 amp breaker fast enough, but increase the ungrounded conductor to a 3/0 with the #12 EGC, and it rises to over 50 amps of fault, which should clear the same 20 amp breaker in less than a couple seconds, increase the EGC to a #10 and now you have almost 90 amps of fault current. (the above is just for reference as no other factors was used)

The current values you are providing are an rms measurement of the current between the occurence of the fault and the point of being interrupted by the ocpd of a properly designed circuit. This is not the same as the available fault current. Additionally, your premise has the same fault as bob's in that it does not take into account the resistance of longer circuit conductors... especially ones that are "already warm".

Fault current and maintaining voltage drop do not run hand in hand, they are two different calculations.

Perhaps different calculations, but the fault current calculation should take into account the voltage drop (which in this sense would actually be current drop) of the circuit.

So the change in 2002 might have made it easier to inspect but went a long way in making it more costly in adjusting for voltage drop, which It might keep some from adjusting for voltage drop because of this.

True... but easily overcome with feeders by oversizing the source end ocpd, and using a closer-to-calculated-load ocpd at the load end (i.e. by making your conductors not upsized; using the loophole in the code ). So the real problem is at the branch circuit level, where there is no ocpd at the load end.

 

[bob]

That's a rather "programmed response"... and a misleading one at that. Just because the conductor is upsized does not necessarily increase the fault current at the end of the circuit.

I do not know where you are getting this information but it is just plain wrong

The very reason you increased the size of the circuit conductors (distance>voltage drop) is the same reason by which you have not facilitated fault current.

Wrong

Is there some magic involved that a different voltage drives the fault current? I think not. SO we are talking about the same voltage driving the current. Then we must consider that the fault may occur when the circuit conductor has been subjected to the load for a duration which has already brought it up to some temperature above ambient, but lower than it's rating. Not that I'm fully versed in such calculations, but I've yet to see any method which accounts for "already warm" conductors.

The characteristics of the conductors in table 9 are listed at 75C.

The current values you are providing are an rms measurement of the current between the occurence of the fault and the point of being interrupted by the ocpd of a properly designed circuit. This is not the same as the available fault current.

What else would you use in place of the RMS value?
No it is not the same as the available fault current. The available fault current is taken at the OCPD and has direct affect on what the fault current magnitude is at the fault point.

Additionally, your premise has the same fault as bob's in that it does not take into account the resistance of longer circuit conductors... especially ones that are "already warm".

See table 9

Perhaps different calculations, but the fault current calculation should take into account the voltage drop (which in this sense would actually be current drop) of the circuit.

not true.

True... but easily overcome with feeders by oversizing the source end ocpd, and using a closer-to-calculated-load ocpd at the load end (i.e. by making your conductors not upsized; using the loophole in the code ). So the real problem is at the branch circuit level, where there is no ocpd at the load end.

not sure why this is here.

 

[Smart $]

I do not know where you are getting this information but it is just plain wrong


Wrong



The characteristics of the conductors in table 9 are listed at 75C.


What else would you use in place of the RMS value?
No it is not the same as the available fault current. The available fault current is taken at the OCPD and has direct affect on what the fault current magnitude is at the fault point.

See table 9



not true.



not sure why this is here.

Let me reciprocate the sentiment I get reading your comments... "You're wrong."

I often have unintelligent people (no reference to you intended) tell me I'm wrong. If they knew any better, they wouldn't be able to say that with accuracy.

So please, when you say I'm wrong, at least provide some commentary to establish the how, why, where, when, etc. to my error. I am not beyond learning, but at least point me in the right direction. "Table 9" is a start, but I'm don't see how you are relating it to the issue under discussion. Enlighten me...

 

[hurk27]

ok lets do it this way.

I did the calculations of the OP in this thread, with the given information

400' of 300Kcmil
400' of #6 EGC

going from the resistance table 8 in chapter 9 at 75?C resistance

300 Kcmil has a resistance of .01784 @400 feet
# 6 has a resistance of .204 @400 feet
this gives us a total fault circuit resistance of .22184

If the voltage from the ungrounded to the X0 is 120 volts you will have about 541 amps of fault current.

if the voltage from the ungrounded conductor to the X0 is 277 volts you will have about 1248.6 amps of fault current

if at 240 volts expect around 1081.8 amps


now every one of these will open a 150 amp breaker.

so why would we need a larger EGC other than the code says so.

The only time would be, if the panel with the breaker in it is also fed by a very long run and has feeders conductors close to the size of the circuit breaker feeding this circuit, like when we see a 200 amp panel feeding a 150 amp circuit, it might be a problem if the 200 amp has long feeders? this is why we need to have a PE stamp to say the #6 is ok

I just don't see it, and as Bob said in post 8 this code change was not made in the realm of safety. It was for the convenience of inspectors.

If I'm wrong in my caculations for a fault let me know