Current capacity of wires

So this question stems from a previous question.
Here

Where from several tables one can see that the maximum current (Imax) is proportional to $1/R$ where $R$ is the resistance. (Or that $I_{\text{max}}R$ is a constant.)

Because it will be easier (mathematically) I'd rather work with the wire radius $r$ (And I'll have to be a bit careful not to mix up $R$ and $r$).

Now resistance goes as $1/r^2$. $$R = \frac{\rho l}{\pi r^2}$$ where $\rho$ is the resistivity, $l$ is the length). So saying that the max current goes as $1/R$ is the same as saying it goes as $r^2$.

From table $I_{\text{max}}$ ~ $r^2$.
Here is a re-posting of the wire table.

Now a naive guess for the current capacity, would be that all wires can carry the same power. $I_{\text{max}}^2R$ = constant. And with a little algebra one finds that $I_{\text{max}}$ ~ $r^1$.

Now I realized yesterday that bigger wires will also have a bigger area. And since the loss mechanisms (radiation, convection, or conduction) will all scale with the surface area of the wire, no matter what the mechanism I can add in the area effect. (This is the outside area of the wire ($2\pi r l$) and not it's cross section area)

So then I wrote, $I_{\text{max}}^2R$ ~ Area = $2\pi l r$.
(I'll leave out the algebra, but you can search for Preece's law and find it reproduced. as in the first few pages here.)

My result is that $I_{\text{max}}$ ~ $r^{3/2}$ in contradiction to the tables.

Does anyone have any idea what I'm missing? Does convection from a wire not scale with the surface area?

Edit: Adding a graph generated from Spehro's convection link. I set the length at 100m, Tc at 100 C and varied the diameter (5m to 100um) (Granted 5m diameter wire is a bit much :^) It seems to me that this goes the wrong way! (But I must admit I'm having a bit of trouble getting my head around all the different plots. I'll sit down tonight with a beer and mull it over.) The wire tables that show Imax ~ 1/R imply that the heat loss from the wire must go as $r^2$, where simple convection assumes loss ~ r. The plot shows that for thin wire the loss is even "slower" than r.

Edit2: I updated the graph of power loss to include data from the I-max. wire table. Data was scaled for 100m and the power was taken to be $Imax.^2 * R$ I also only plotted the relevant part of the graph for diameters form 50 um to 20 mm. The cable Imax is not too bad. (I've been using the blue data points to pick wire sizes.) • $r^2$ -> $r^2$ – Ignacio Vazquez-Abrams Oct 30 '14 at 17:04
• It looks like you've independently arrived at the "Preece Equation" I=a.$d^{3/2}$ ... – brhans Oct 30 '14 at 19:39
• In case of bare wire, the linearity depends of the conductor temperature (change of resistance due to temperature rise). So you can consider the wire as an energy source and heatsink. But in case of insulated wire, the temperature it is logarithmic since a part of the equation is the ln of the insulation thikness, which is not the same for all AWG sizes. – GR Tech Oct 31 '14 at 5:23
• @GRTech It is not at all clear to me if the insulation around the wire means it can carry more current or less. I could argue that if the insulation was a relatively good thermal conductor (compared to conv/ rad.) then there would not be much of a thermal drop across the insulation layer. and the insulation would then have a larger area for loss. (and perhaps better emissivity.) – George Herold Oct 31 '14 at 13:51
• @George Herold What you say above called critical thickness of insulation i.e a crosspoint between optimal thermal resistance caused by the insulation thickness and the increase of outer surface for convection. In any case the resistance of the cable changing as the temperature of the cable increase as well as the heat-dissipating ability of the cable related with heat-dissipation coefficient k which is changed because temperature of cable changing until the final temperature. So your plot it is not a straight line connecting the start temperature point and the final temperature point. – GR Tech Nov 1 '14 at 7:34

Convection does not scale linearly with surface area, it's a fluid dynamics problem rife with Nusselt numbers and Rayleigh coefficients. Since it's not my field, and since I'm more likely to use CFD than first principles in such a situation anyway, I'll point to a calculator that might allow some feel for the problem here (assuming it's actually implemented correctly, of course).

If I stick in reasonable numbers like 100°C for the wire 1, 2, 10mm for the wire diameter then I get total heat loss scaling with roughly the diameter (and thus the surface area) to the power of 2/3.

Preece was looking at fusing current, which would be at a much higher temperature than typical operation of an electrical wire.

• Thanks Spehro, unfortunately it looks like the corrections for small wires goes the wrong way. I don't know if the calculator is any good. (At least it got the radiative part correct.) – George Herold Oct 30 '14 at 20:19

It looks like you've independently arrived at the "Preece Equation" I=a.$d^{3/2}$.

Apparently W.H. Preece investigated the fusing current of wires and that's what he arrived at.

I is the fusing current, d is the diameter of the wire, a is a constant depending on the material of the wire.

• You are exactly right. (Preece did it back in the 19th century.) The fusing current will certainly have more radiative losses. (I'm still getting my head around how big the radiative losses are even at, say, 100 C.) – George Herold Oct 31 '14 at 0:11