120% rule clarification

[Smart $]

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In the proposed scenario the conductors could not physically be overloaded in any case. The only possible point of overloading is on the busbars, opposite the point where the utility connects, if not covered up by the solar breakers.

But you're only considering overload, or rather an overcurrent condition caused by a faulting condition of connected normal loads. The difference is the fault is remote to the feeding conductors of discussion. As I said earlier, it depends on where the fault is located.

 

[Smart $]

Which it probably would do anyway, not because the panel or conductor rating was too small.

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I think he was suggesting that the resistive fault could be in the conductors, but still, (as you say) although the fault itself could be fed by both sources, the conductors on either side of it couldn't be. Hence my question of what difference it would make what the rating of the conductors would be. A resistive fault capable of dissipating 750A is going to generate a lot of heat no matter what the conductors feeding it are rated. My opinion is that it would very quickly develop into a dead short and shut everything down.

Not [necessarily] true.

If the 120% rule was adhered to, the conductors and busbar between these OCPD's would be sized no less than 700A ? 120% = 584A. Though we likely would not, let's say we used single conductors. We'd be looking at 1250kcmil having a 590A rating at 75?C. Let's also say we use 90?C rated conductors, so at 665A our conductors would be 90?C [theoretically by Code]. Of course it depends on the details of the fault, but that in itself is typically not hot enough to cause a fire. It's not even hot enough to impart continued degradation of the insulation at the location of the fault.

 

[ggunn]

Not [necessarily] true.

If the 120% rule was adhered to, the conductors and busbar between these OCPD's would be sized no less than 700A ? 120% = 584A. Though we likely would not, let's say we used single conductors. We'd be looking at 1250kcmil having a 590A rating at 75?C. Let's also say we use 90?C rated conductors, so at 665A our conductors would be 90?C [theoretically by Code]. Of course it depends on the details of the fault, but that in itself is typically not hot enough to cause a fire. It's not even hot enough to impart continued degradation of the insulation at the location of the fault.

But where in the conductors would that much current be flowing? If the fault is in the middle of a conductor which is fed by 350A of breakers at one end and a 350A fuse at the other, the only portion which would see the total current from the two sources is at the fault itself, which has already failed, so the rating of the conductor at that point is irrelevant. I also don't see the viability of this sustainable resistive fault, anyway. If it's a failure of insulation on a conductor, it would avalanche into a dead short in very short order and shut the circuit down no matter what the rating of the conductors.

 

[Smart $]

But where in the conductors would that much current be flowing? If the fault is in the middle of a conductor which is fed by 350A of breakers at one end and a 350A fuse at the other, the only portion which would see the total current from the two sources is at the fault itself, which has already failed, so the rating of the conductor at that point is irrelevant. I also don't see the viability of this sustainable resistive fault, anyway. If it's a failure of insulation on a conductor, it would avalanche into a dead short in very short order and shut the circuit down no matter what the rating of the conductors.

Its not irrelevant in the sense that a lower rated conductor will have a higher temperature, especially at the fault. The premise here is to minimize the potential for causing a fire If the fault extends beyond the conductors and creates a fire, there's not much that can be done about it, from an electrical standpoint... but what we don't want is the conductors themselves causing a fire.

 

[ggunn]

Its not irrelevant in the sense that a lower rated conductor will have a higher temperature, especially at the fault. The premise here is to minimize the potential for causing a fire If the fault extends beyond the conductors and creates a fire, there's not much that can be done about it, from an electrical standpoint... but what we don't want is the conductors themselves causing a fire.

I understand that, but even in this scenario, unlikely as it is, there is no way that the full current from both sources can flow through the conductors. Through the fault, yes, but if the insulation is already breached at the point of the fault, the rating of the conductors is indeed irrelevant because the rating of a conductor is the rating of its insulation. The insulation has already failed, and the current in the conductor could not have caused it.

 

[jaggedben]

But you're only considering overload, or rather an overcurrent condition caused by a faulting condition of connected normal loads. The difference is the fault is remote to the feeding conductors of discussion. As I said earlier, it depends on where the fault is located.

I don't follow. I'm considering possible overload from your resistive fault.

As far as location, the only locations that could conduct a resistive fault (or a proper load) that could lead to an overcurrent in installed components (without tripping an OCPD) would be any exposed ends of the busbars that lie outside the path between the solar and the utility. (It's not even a given that such exposed locations will exist in the installation.) In other words, it could only happen inside the panelboard.

It seems to me that if you have some resistive material dissipating 168kW of heat inside a panelboard, that's going to damage the panelboard and likely cause a fire, regardless if the rating of the panelboard is 400A or 600A.

If the 120% rule was adhered to, the conductors and busbar between these OCPD's would be sized no less than 700A ? 120% = 584A. Though we likely would not, let's say we used single conductors. We'd be looking at 1250kcmil having a 590A rating at 75?C. Let's also say we use 90?C rated conductors, so at 665A our conductors would be 90?C [theoretically by Code]. Of course it depends on the details of the fault, but that in itself is typically not hot enough to cause a fire. It's not even hot enough to impart continued degradation of the insulation at the location of the fault.

It's not the temperature of the conductors that would cause the fire. What would cause the fire is whatever resistive material you are positing that would be dissipating over 168kW of heat somewhere outside the conductors.

As I mull it over, the most plausible scenario I could think of for such a resistive path (the only plausible scenario?) might be brackish water inside an underground conduit. Of course, as ggunn says, this implies that insulation is already compromised and thus current ratings are already fairly irrelevant. And as I try to think of various ways the fault could play out (water boiling, someone getting shocked), I still can't think how it would make any fire safety difference if the conductors were rated for 584A instead of 350A, especially when safety factors are considered.

It's worth noting that a resistive fault passing 350A at 480V implies a fault path with a resistance of 1.4 ohms. Consider that testing of insulation is done to look for resistive paths in the megaohm range. I take that to imply that the dangers of a resistive fault occur at much lower levels than where the 120% rule would come into play.

 

[Smart $]

I understand that, but even in this scenario, unlikely as it is, there is no way that the full current from both sources can flow through the conductors. Through the fault, yes, but if the insulation is already breached at the point of the fault, the rating of the conductors is indeed irrelevant because the rating of a conductor is the rating of its insulation. The insulation has already failed, and the current in the conductor could not have caused it.

I don't follow. I'm considering possible overload from your resistive fault.

As far as location, the only locations that could conduct a resistive fault (or a proper load) that could lead to an overcurrent in installed components (without tripping an OCPD) would be any exposed ends of the busbars that lie outside the path between the solar and the utility. (It's not even a given that such exposed locations will exist in the installation.) In other words, it could only happen inside the panelboard.

It seems to me that if you have some resistive material dissipating 168kW of heat inside a panelboard, that's going to damage the panelboard and likely cause a fire, regardless if the rating of the panelboard is 400A or 600A.



It's not the temperature of the conductors that would cause the fire. What would cause the fire is whatever resistive material you are positing that would be dissipating over 168kW of heat somewhere outside the conductors.

As I mull it over, the most plausible scenario I could think of for such a resistive path (the only plausible scenario?) might be brackish water inside an underground conduit. Of course, as ggunn says, this implies that insulation is already compromised and thus current ratings are already fairly irrelevant. And as I try to think of various ways the fault could play out (water boiling, someone getting shocked), I still can't think how it would make any fire safety difference if the conductors were rated for 584A instead of 350A, especially when safety factors are considered.

It's worth noting that a resistive fault passing 350A at 480V implies a fault path with a resistance of 1.4 ohms. Consider that testing of insulation is done to look for resistive paths in the megaohm range. I take that to imply that the dangers of a resistive fault occur at much lower levels than where the 120% rule would come into play.

At the moment, I can't think of any other more specific way to convince you guys of the "merits" of 120% sizing of conductors... but it seems you both question 120% sizing of conductors yet do not question 120% for the busbar rating. What makes the busbar rating any different? Ask yourselves, is it the insulation? Is a busbar's rating determined by its insulation? Why must a busbar be rated for not less than 84% the sum of OCPD's supplying it? ...and why is it required that when the sum of supply OCPD ratings exceed its 100% rating, the supply connections are required to be at opposite ends of the busbar?

 

[SolarPro]

Someone needs to come up with a one-way valve type of solution for the output of these inverter aggregation panels. There is no load, and yet we're treating downstream breakers as though they are "supplying power." They are not?unless there is a fault. And it takes some pretty creative thinking to come up with a fault that doesn't clear itself.

Assuming the intent of the CMP is to (basically) double the amount of copper in an ac collection system like this, then why not install a CB that trips on reverse current? Then the only OCPDs feeding the buss or conductor are from the inverters, and you design the system for normal current, like any other circuit.

Or maybe the CMP should have an allowance for an arc-fault detection device on that circuit? That device would presumably detect any imaginable resistive fault, and we could design for normal current.

Usually "because the CMP says so" is backed up by a logic that is self-evident to me. In this case, I do wind up scratching my head a lot. (Now where's the emoticon for that?)

 

[Smart $]

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Usually "because the CMP says so" is backed up by a logic that is self-evident to me. In this case, I do wind up scratching my head a lot. (Now where's the emoticon for that?)

I should note that I often feel compelled to play the devil's advocate... and I have done so here.

Consider the topic of discussion here and compare it to reasoning behind tap and transformer secondary conductor rules. The 120% rule then seems quite illogical.

 

[jaggedben]

At the moment, I can't think of any other more specific way to convince you guys of the "merits" of 120% sizing of conductors... but it seems you both question 120% sizing of conductors yet do not question 120% for the busbar rating. What makes the busbar rating any different? Ask yourselves, is it the insulation? Is a busbar's rating determined by its insulation? Why must a busbar be rated for not less than 84% the sum of OCPD's supplying it? ...and why is it required that when the sum of supply OCPD ratings exceed its 100% rating, the supply connections are required to be at opposite ends of the busbar?

You are now extrapolating the argument to questions outside of what it has been about. I was answering ggunn's narrow question about whether an AC combiner panel would be safe, not whether the 120% rule is a good idea in general.

The fact is that there could be lots of installations, besides AC combiner panels, that would violate 120% rule or the 'opposite ends' rule but would probably be perfectly safe. For example:

-Installations where the sum of the OCPDs for the loads is less than the largest supply OCPD.
-Installations where all the loads are connected in-between two supplies, and neither supply exceeds the rating of the busbar.
-Installations where the loads that are not connected in-between the utility and solar do not exceed the sum of those two OCPDs.
-Installations where all the OCPDs (loads and supplies), except the utility supply, do not exceed the rating of the busbar.

(In all of these cases, assume the utility supply OCPD does not exceed the rating of the panelboard.)

The last one has been proposed for the 2014 NEC, and it would deal with the AC combiner panels, and cover ggunn's example. We'll see if it makes it into the actual code.

For the CMP, I think there are three problems:
a) testing procedures for listing panelboards do not directly test any of these scenarios.
b) their fear that safe installations could be easily made unsafe by an electrician who comes in later and modifies something.
c) Coming up with sufficiently simple language that covers all the possible cases.

Hence the simple but very conservative 120% rule.

 

[Smart $]

You are now extrapolating the argument to questions outside of what it has been about. I was answering ggunn's narrow question about whether an AC combiner panel would be safe, not whether the 120% rule is a good idea in general.

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And I am trying to point out that safeness is relative to what you consider safe to be.

Analogous example: Do you generally feel safe driving your automobile? How many people are killed or injured as a result of traffic accidents? Isn't the ultimately safe thing to do is not put yourself in that position? Short of that, we make safety decisions on a basis of risk vs. reward.

Ultimately, the rule governs how we must do it, so we do it that way, whether the reasoning is sound or not. You must agree, compliance with the 120% rule is safer than ...well... anything less safe. There is an avenue to contest the rule. Given the number of propasals through time, the rule may eventually be less safe...

 

[ggunn]

At the moment, I can't think of any other more specific way to convince you guys of the "merits" of 120% sizing of conductors... but it seems you both question 120% sizing of conductors yet do not question 120% for the busbar rating.

There are situations where the 120% rule for busbars makes very good sense. IMO, this isn't one of them.

I talked about the conductor rating because you brought it up. It seemed to me that you were saying that the conductors would carry the total current from both sources in the event of a fault, and I do not agree. No matter where you put the fault, no portion of the conductors save at the spot of the fault itself will ever see more than the source feeding it from that side of the fault, and at the spot of the fault the insulation is already compromised so the conductor rating is irrelevant. Jaggedben pointed out that the busbars below the interconnection point in a very unlikely scenario might see the entire current, but that's why we put the last inverter feed point at the opposite end of the busbar from the lugs.

This resistive fault you postulate could cause a fire even in common household wiring when everything is firmly under code and no inverters are present. If someone were to staple a Romex cable to a rafter with a metallic staple and manage to nick both the hot and neutral with the same staple just enough to cause a resistive fault which dissipates just enough power not to blow the breaker, and it somehow managed to remain at that resistance without deteriorating into a dead short, it could cause a fire, and just like in our theoretical inverter scenario the rating of the conductors wouldn't matter.

 

[ggunn]

And I am trying to point out that safeness is relative to what you consider safe to be.

Analogous example: Do you generally feel safe driving your automobile? How many people are killed or injured as a result of traffic accidents? Isn't the ultimately safe thing to do is not put yourself in that position? Short of that, we make safety decisions on a basis of risk vs. reward.

IMO, applying the 120% rule to the conductors in this situation is like requiring all the rear safety belts in your automobile to be clicked shut even if there is no one sitting in the back seat.

 

[Smart $]

IMO, applying the 120% rule to the conductors in this situation is like requiring all the rear safety belts in your automobile to be clicked shut even if there is no one sitting in the back seat.

LOL

 

[Smart $]

There are situations where the 120% rule for busbars makes very good sense. IMO, this isn't one of them.

I agree... but requirements are requirements. We are not allowed to "misinterpret" at our discretion. When the requirements change to agree with our opinion, I may be less a PITA. :lol:

I talked about the conductor rating because you brought it up. It seemed to me that you were saying that the conductors would carry the total current from both sources in the event of a fault, and I do not agree. No matter where you put the fault, no portion of the conductors save at the spot of the fault itself will ever see more than the source feeding it from that side of the fault, and at the spot of the fault the insulation is already compromised so the conductor rating is irrelevant. ...

But the conductor rating is relevant. Not regarding the insulation though. As a matter of conductor temperature at the fault location. Getting back to the proposed scenario, compare conductor temperatures of 650A at fault location for 400A 90?C wire vs. 700A 90?C wire. The temperature at the fault location will be the same as the conductor carrying the full fault current.

This resistive fault you postulate could cause a fire even in common household wiring when everything is firmly under code and no inverters are present. ...

True... but here you are saying you think it is ok to feed the fault twice as much current using two sources.

 

[ggunn]

But the conductor rating is relevant. Not regarding the insulation though.

The conductor rating is ALL (and only) about the insulation.

 

[Smart $]

The conductor rating is ALL (and only) about the insulation.

So a #12 cu 90?C conductor can carry just as much current as a 500kcmil cu conductor with the same 90?C insulation.... :lol:


Kidding aside, that is not entirely correct. It is also about the temperature of the conductor. How would you size bare wire (if you were permitted) suspended on ceramic insulators? Wouldn't you still have to size it so the conductor did not get hot enough to cause a fire? Have you ever seen a conducting wire glow red hot?

 

[ggunn]

So a #12 cu 90?C conductor can carry just as much current as a 500kcmil cu conductor with the same 90?C insulation.... :lol:


Kidding aside, that is not entirely correct. It is also about the temperature of the conductor. How would you size bare wire (if you were permitted) suspended on ceramic insulators? Wouldn't you still have to size it so the conductor did not get hot enough to cause a fire? Have you ever seen a conducting wire glow red hot?

It's about the temperature of the conductor, sure, but the conductor temperature limit is there to protect the insulation. In the case of bare wire on ceramic insulators, the temperature tolerance and therefore the allowable current might be much higher depending on the environment. You wouldn't use the tables in that case.

I agree that if a conductor is fed from both ends by 350A OCPD's and develops a fault somewhere between them, there will be more current and therefore more heat in the fault than if it were only fed from one end by one 350A OCPD. And heat is bad. But in that situation, anywhere along the conductor other than at the fault itself the maximum current in the conductor is still only 350A, and at the fault the rating of the conductor doesn't matter because the rating of the conductor is set to protect the insulation, and the insulation has already failed - for whatever reason, but it's not due to the current/heat under normal operation. The current in the fault is still 700A and the amount of heat generated is the same irrespective of the rating of the conductor. Using a higher rated conductor would not prevent the fire.

 

[Smart $]

... Using a higher rated conductor would not prevent the fire.

But it may, and that IS the point. A higher ampacity conductor may prevent a fire because it will not get as hot at the fault location. It may also save the conductors such that the fault can be repaired and condutors spliced and reinsulated.

 

[jaggedben]

But it may, and that IS the point. A higher ampacity conductor may prevent a fire because it will not get as hot at the fault location. It may also save the conductors such that the fault can be repaired and condutors spliced and reinsulated.

Even if we accept this (and I don't think I do), what are the actual chances of the 120% rule making a difference?

Whatever the chances are, they are the product (mathematically) of the following:

-The chance that a resistive fault would remain stable, within the window that is between the utility breaker (350A) and what is required by the 120% rule (570A). Very small chance, I think, that the resistive fault would stay in that window rather than going high or occuring after and tripping a breaker or remaining lower than the conductor rating.
-The chances that the fault would occur during good sunlight; otherwise the utility breaker would trip. (Roughly 15%?).
-The chance that the material conducting the resistive fault would not itself ignite or cause damage. Very small chance, I think.
-The chance that the safety factor already included in the conductor rating would be inadequate. Already small enough to satisfy someone's cost benefit analysis.

As I said earlier, intuition leads me to the answer 'infinitesimal chance'. Lacking data that suggest otherwise I will probably stick with that.