SirJaxx wrote:
This section applies to Feeders, not branch circuits as you had specified in your earlier question.
Whoops; I was lax in flipping through the NEC to find the section I was referencing. I should have referenced 210.19(A)(1) Exception, which uses the essentially the same text, and presumably is in the code for exactly the same reason.
In any case, while you are preaching to the choir that 'breakers trip on heat', are you in a position to explain the difference between an 80% rated breaker and a 100% rated breaker? Everyone here (including me, in the post that opened this thread) agrees that breakers trip on heat. But what is the way in which this heat affects an 80% breaker _different_ from the way it affects a 100% breaker.
Again, I have not been trying to argue this point (do breakers trip on heat or not), nor am I intentionally trying to be thick headed. I said in my initial question:
Will the (presumably 80% rated) circuit breaker to heat up, do the temperature compensation thing, and eventually simply (and safely) trip? Or will the circuit breaker not trip, but instead overheat in some component part? Or will the circuit breaker eventually trip, but suffer some accelerated wear in doing so?
I really do get that the breaker trips because of heating.
What I am trying to understand is why an 80% breaker is different from a 100% breaker. Both presumably trip on heat, and both will trip at a lower current if they are in a warmer ambient environment. Both will heat up when current flows through them, and both will heat up more when enclosed in a panel, rather than sitting in free air.
Remember, when you say that a 'breaker trips on heat', what you really mean is that the breaker trips when its thermal trip element reaches a certain temperature. When I run current through a breaker, all current carrying components will generate heat, including the thermal trip element. When an 80% breaker is operated at 100% of its trip rating, does the heat generated in the thermal trip element combined with the heat conducted to the thermal trip element, cause the circuit breaker to open _before_ or _after_ other parts of the circuit breaker are damaged or subjected to premature degradation?
The obvious, most sensible, intuitively correct answer is that 'of course a circuit breaker is designed so that it will open before any part of it gets too hot' (if the heating is caused by current flowing through the circuit breaker in a uniform ambient). I no longer trust this answer because I have seen enough references that suggest that an 80% rated breaker _will_ be damaged if used in this fashion.
I believe that at this point the only thing to do is follow PCBelarge's suggestion, and ask the manufacturers.
Regards,
Jonathan Edelson
