Heat equation and stainless-steel wire

[Physgeek64]

Homework Statement

A stainless-steel wire is 0.1 mm in diameter and 1 m long. If the outside of the
wire is held fixed at 20◦C, estimate the steady-state current passing through the wire
when the stainless steel at the centre of the wire begins to melt.

Homework Equations

The Attempt at a Solution

The solution to the heat equation will be $T=-\frac{Hr^2}{4\kappa} +c$ The ln term disappears because the temperature must be finite at the origin.

H is the source term $H=\frac{I^2R}{Al}= \frac{I^2\rho}{A^2}$ where A is the c.s.a of the wire

Since the outer surface of the wire is held at 20 degrees
$c=293+\frac{Ha^2}{4\kappa}$ where a is the radius of the wire

$T=293+\frac{Ha^2}{4\kappa}-\frac{Hr^2}{4\kappa}$

Therefore the temperature at the centre of the wire is

$T=293+\frac{Ha^2}{4\kappa}$

setting $T=1400$ which we are told is the melting point of the wire we get

$1380 = \frac{I^2\rho a^2}{\pi^2 a^4 4\kappa}$

$I=\sqrt{\frac{4(1380) \pi^2 a^2 \kappa}{\rho}}$
However this isn't right but i can't see where I've gone wrong

Many thanks

 

[Charles Link]

I obtained a similar expression for the current $I$, (with a 2 instead of a 4), when I used a much simpler heat transfer model, where I assumed the $\Delta T$ occurred in a planar type geometry in a distance $a$ and the surface area was $2 \pi a L$. Thermal conductivity $\kappa \approx 50$ W(/m K) , but I think at elevated temperatures $\rho$ is likely to increase from $10^{-7 }$ to perhaps $10^{-6}$ or larger. The radius $a=.5 \cdot 10^{-4} \, m$, so that your answer may be in the right ballpark. It is also a somewhat extreme case where there is a heat sink at the surface that can keep the temperature at 20 degrees Centigrade.

 

[Chestermiller]

@Physgeek64 I confirm your equation. What value did you use for a?

 

[Physgeek64]

@Physgeek64 I confirm your equation. What value did you use for a?

$0.5 X 10^{-3}$

 

[Charles Link]

$0.5 X 10^{-3}$

10{-4}?

 

[Chestermiller]

$0.5 X 10^{-3}$

Sounds OK overall.

 

[Physgeek64]

-4?

sorry, yes!

 

[Physgeek64]

Sounds OK overall.

I wonder how I've got the wrong answer then

 

[Charles Link]

I wonder how I've got the wrong answer then

You get an answer where $I \approx 100 \, A$ or more if you use $\rho=10^{-7}$. I think $\rho$ is somewhat larger than that. See also post 2.

 

[haruspex]

If you are not given the values of κ and ρ to use then I am not sure how you are supposed to solve this. From what I read, over the temperature range involved, κ increases from 15 to 30 W/mK, while ρ increases from 7x10^-7 to 12x10^-7 Ωm.
To take that into account you would need either to use numerical methods or to approximate these temperature dependencies with analytical functions and solve the resulting diffusion equation.