This is based on a problem that came up today. During the course of this problem I realized that I wasn't so sure I understood the relationship between wattage and heat produced.
In the past we did a test in the lab using .305 Ω /ft wire. The jacket is rated for 150C. We were able to get about 6.5A (at 3.66V) out of it at 24C ambient, with out exceeding the jacket rating. I want to estimate what the ampacity of 0.027 Ω/ft wire is. So I am wondering if I did it correctly, because this amperage seems a little high for the wire to handle, then again most copper wire is only rated at 90C.
So the math I did on it was this
So you do .305 * 2 = .61 Ω /ft 6A^2 * .61 Ω = 21.62W (I^2*r = W) 21.96W * / 2ft = 10.98 W/ ft
So would it be safe to assume that if I did the same with a 0.027w/ft wire with the same jacked I would arrive at this amperage?
If you start with 11W/ft * 2ft = 22W 0.027 Ω * 2ft = .054 Ω Sqrt(22W/(0.054 Ω)) = 20.18A
ETA: we are planning on testing this tomorrow when we get some wire in. So we shall find out.
Answer
I don't quite follow your calculations. But if you are doing I^2 *R as the power, and assuming the same max power for each wire.. then that's what I would have done.
However I went here and it looks like I*R is about constant. (?)
(I had to plot it.) Still looks linear.
Maybe someone can tell us both why.
Edit: I*R dependence. (Thanks Spehro, it was suddenly obvious on the drive home.) No matter the thermal loss mechanism (convection, radiation..) It will go as the area of the wire. 2 * pi *r * l (r - radius and l - length), so bigger wire will need more heat to get to a given temperature. (more later)
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