This is my point of view, I am not taking sides, nor using McIntosh/others info/views. This is my own thought process.
It's not that they're trash, it's all given by the amount of current flowing through the wire. It has to deal with the specific heat of the material, and the resistance of it.
Let's go with his specific situations that he pointed out: 300A@20' and 800A@5'
First we find the overall resistance of the wire:
0.09827ohms/kfeet
This divided by 1000 gives ohms per foot. (Or just add 3 0s after the decimal point.)
.00009827ohms/foot
This multiplied by 20 gives us the resistance of the wire.
0.0019654ohms
Now, plug this into ohms law to determine how many watts this wire dissipates, just the same as viewing it as a resistor. (I^2)*R=P
(300*300)0.0019654=176.886W
At the same time, lets plug this in to calculate the voltage drop across the resistor. I*R=E
300*0.0019654=0.58962V | This drops about .6V, it is not really relevant to the rest of what I'm saying, but it does drop ~.6V
Now that we know how many watts it dissipates, we still cannot calculate how much it will heat the wire. We have to find the mass of the wire.
We have to find the volume of the wire by multiplying it's area by it's height, given by (r^2)piH. H is determined by converting feet into cm, resulting in 609.6cm. r is .535cm
.535(609.6)=326.136cm^3
The density of copper is 8.96g/cm^3.
326.136*8.96=2,922.17856g | Now, we finally have how much mass of copper is in the wire. I'm assuming there's a minimal amount of air, but if there is more, it will heat up faster, since there is less copper mass, and will not evenly heat up the air since the thermal transfer is by convection instead of conduction.
The specific heat of copper is 0.385 J/g•C or K.
One watt dissipated is 1J for 1 second. We finally are able to determine how much it will heat the wire.
Q=mC(deltaT) | Delta T stands for change in temperature, which is (Tf-Ti). We are trying to find Tf.
Q=176.886J
m=2922.17856g
C=.385
Ti=25
With the variables stated, we can now start plugging things in and solve.
176.886=2922.17856*.385(Tf-25)
Multiply 2922.17856 by .385
176.886=1125.03875(Tf-25)
Divide 176.88600 by 1125.03875
0.157226584=Tf-25
Add 25 to 0.157226584
Tf= 25.157226584 degrees C
IF TL;DR:
In other words, 0.157226584 degrees C/second, or 0.2830079 degrees F/second. Five seconds, and it's slightly more than a degree hotter. :\ I don't know if I would run it for too long. Let's run this for five minutes, assuming that none of the energy is lost/absorbed by the environment.
0.2830079(60*5)=84.90237 degrees F. I am not about to go into the math for dissipation of this heat into the surrounding air/insulation. After consideration, I would say the wire has a temperature rise of about 50-60 degrees F after five minutes.
On second though, there is too much math involved, and it is a hell of a lot easier to do on paper, so I'm not going to do the 800A@5' unless someone asks.
Please ignore small grammatical errors, or me leaving s or other letters off words.
