We could continue to speculate until the cows come home (or choose your own adage)....
As I said earlier, intuition leads me to the answer 'infinitesimal chance'. Lacking data that suggest otherwise I will probably stick with that.
We could continue to speculate until the cows come home (or choose your own adage)....
As I said earlier, intuition leads me to the answer 'infinitesimal chance'. Lacking data that suggest otherwise I will probably stick with that.
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.
The only way the conductor does not carry 700A (or less) of fault current at the point of the fault (if it is a single point fault rather than a distributed fault) is if the conductor is cut by an interposing conductive material. If the conductor is not cut, then the entire 700A (or less) exits the side of the conductor at a single point. Call the conductor to either side of the fault a separate conductor. Now "T" cadweld those two 400A conductors to a third 400A conductor. All 700A of fault current is carried by the third 400A conductor. What would be the temperature of that third conductor?How so? The fault is not actually in the conductor but orthogonal to it. At no point within the conductor itself in this scenario is the current more than 350A, and the conductor is rated to be less than 90 degrees C at 400A. Why do we care if the temperature of the conductor is lower at points removed from the fault as long as it stays below the maximum rated temperature for its insulation?
What would be the temperature of that third conductor?
The temperature of the third conductor would be not be dependent on the size/rating of the conductor(s) from which it sprang. In the primary conductor the only direction you could have a resultant current of 700A is out from the conductor at the point of your cadweld because the currents from the two supplying OCPD's are in opposite directions. There is no cross section of that conductor which will have 700A passing through it. Kirchoff's Law.The only way the conductor does not carry 700A (or less) of fault current at the point of the fault (if it is a single point fault rather than a distributed fault) is if the conductor is cut by an interposing conductive material. If the conductor is not cut, then the entire 700A (or less) exits the side of the conductor at a single point. Call the conductor to either side of the fault a separate conductor. Now "T" cadweld those two 400A conductors to a third 400A conductor. All 700A of fault current is carried by the third 400A conductor. What would be the temperature of that third conductor?
Now we are getting much closer to my true beliefs on the matter :happyyes: so I'm not going to continue fueling the discussion as devil's advocate.As I see it, the 'conductor' here would really be the resistive fault path that you are postulating. The temperature of our 400A wire conductor at the point of the fault would depend in part on the heat transfer between the wire conductor and the resistive fault path and other surrounding materials. Hard to say what that would be without knowing the specific heat of those materials. (Hard for me to say regardless, not knowing really anything about heat transfer.)
You could probably only determine the answer with a lab test anyway. Using NEC tables is just not appropriate, given safety factors and probably a few other things.
ggunn is really right, I think. Once the insulation is compromised, all bets are off anyway. The NEC ampacity ratings are designed to prevent this fault, not prevent it from doing further damage.
Here's another point: the 120% rule only applies to a load side connection. This same 400A conductor would be allowed by the code if it were, say, a service entrance conductor that was tapped somewhere for the solar. In that case your resistive fault, which could also be fed by both the utility and the solar, could be as big as allowed by the sum of the circuit breakers on the solar and the utility transformer. Which would be way more than 120% of its ampacity. And not only that, but the full fault current could flow through some length of the conductor, not just at a single exit point.
In other words, your resistive fault condition would be just as bad (worse actually) on an allowed line-side tap than on this disallowed example of ggunn's. So whatever the reason that the CMP picked 120% (and I wonder sometimes if they pulled it out of a hat), it wasn't due to any consistent logic concerning a fault of this type.
A larger conductor would be a lower temperature because of lower impedance. It would also act as a better heat sink....I have only submitted that upsizing the conductor won't make it get any less so.
Let me get this straight. If this were a supply side tap, I would not be allowed to put OCP on the disco for the conductors between the tap and the inverter breakers? That sounds like what Smart $ was saying earlier in this thread, and it makes no sense to me. Why would the code dictate that conductors be unprotected? I must be misunderstanding something, but if not, please cite the section of the code that addresses this. It's not that I don't believe you guys, but this is something I need to understand.Here's another point: the 120% rule only applies to a load side connection. This same 400A conductor would be allowed by the code if it were, say, a service entrance conductor that was tapped somewhere for the solar. In that case your resistive fault, which could also be fed by both the utility and the solar, could be as big as allowed by the sum of the circuit breakers on the solar and the utility transformer. Which would be way more than 120% of its ampacity. And not only that, but the full fault current could flow through some length of the conductor, not just at a single exit point.
Let me get this straight. If this were a supply side tap, I would not be allowed to put OCP on the disco for the conductors between the tap and the inverter breakers?
Now we are getting much closer to my true beliefs on the matter :happyyes: so I'm not going to continue fueling the discussion as devil's advocate.![]()
Could be either or both... I'll neither confirm nor deny...
1) You've just been string us along for nothin'. In which case I please gullible. :ashamed1:
2) You were wrong about the danger of a resistive fault.
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