On Wednesday, March 11, 2015 at 3:14:01 PM UTC-4, dpb wrote:
On 03/11/2015 1:53 PM, trader_4 wrote:
On Wednesday, March 11, 2015 at 2:43:59 PM UTC-4, wrote:


Consider this. The Microwave (MW) develops a problem, causing a dead
short. There is no properly sized breaker to trip. Several things can
happen.

A dead short will trip the existing 40A or whatever breaker for
the stove instantly. It's an overload that exceeds the wiring to the
outlet, but not the breaker, ie the range of 15A or 20A to 40A, that would
be the fire hazard from the wiring overheating.
...

And, of course, while presuming 14ga/15A outlet, any load above that is
in violation of Code, the actual breaker thermal trip level is generally
about 5-7.5% _above_ the nominal breaker rating at which point the
bimetallic strip opens so the actual load might be as much as 43A or so
before the breaker itself saw the overload.

But, mitigating that is that the actual current-carrying capacity of
wiring or the outlet is quite a lot greater than that for which it is
rated for Code application so there's quite a margin of safety before a
real fire danger is present.

I don't have the data at hand on what the temperature rise in a 14ga Cu
would be in open air at 43A but it'll be warm...will it be fire-setting
hot I don't know for certain.


Copper 14g is 2.5 ohms per 1000 ft, or .0025 per ft. At 40 amps,
I2R is 1600 x .0025, or 4 watts per foot. You have two conductors,
so double that, 8W. 8W spread out over a foot of wire doesn't sound
like a lot of heat to me either. Like you say, warm, but I doubt
it would melt romex insulation. There is a lot of margin in the code.


See the previous comment above as well...it certainly should be fixed
but I don't believe it's an imminent hazard in the next 10 minutes.

--

I agree, especially now that he knows not to plug another big load
into that same receptacle with the microwave.