Current capacity of wires

Question

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.)

Answer 1

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.

Answer 2

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.