# Varying thermal conductivity with length

1. Feb 21, 2014

### j_phillips

I'm interested in modeling a system where the material varies along its length, thus the conductivity coefficient would be a function of both T, and x. k(T,x). For starters, if I assume negligible change w.r.t T, then he heat diffusion equation would be d/dt(k(x)dT/dx)=0. Correct? What if k just equals x (ie. linear)

I'm a little rusty with this solution. Any help?

Thanks.

2. Feb 22, 2014

### Staff: Mentor

Hi j_phillips. Welcome to physics forums!!!
In your equation, what does the lower case t represent? Is this supposed to be a steady state problem, or an unsteady state problem?

Chet

3. Feb 22, 2014

### dauto

That's not the correct equation. It should be

$$C \rho\frac{\partial}{\partial t}T =\frac{\partial}{\partial x} k(x,T)\frac{\partial}{\partial x} T,$$
where $C, \rho,$ and $k$ are the heat capacity, density, and conductivity of the material

4. Feb 22, 2014

### j_phillips

I appreciate the welcome!

This would be steady-state so the right side of the equation would go to zero. If k=x, then d/dx(k) would equal a constant K, but then there is t d/dx*T term which would be unaccounted for?

Instead of a constant (K), let's say k(x,T)=100+cos(2x) for 0<x<(pi/2) (I'm just trying to liven up the problem.) then d'K=-2*sin(2x) then 0=2*sin(2x)*d/dx(T)?

5. Feb 22, 2014

### Staff: Mentor

So the equation you are solving is:
$$\frac{d}{dx} \left(k(x)\frac{dT}{d x}\right)=0$$
Integrating this once with respect to x gives:
$$k(x)\frac{dT}{d x}=C$$
where C is minus the (constant) heat flux.
Integrating again gives:
$$T=C\int_0^x{\frac{dζ}{k(ζ)}}+T(0)$$
where ζ is a dummy variable of integration. If the temperature is specified at x = 0 and x = L, that is enough information to get the heat flux C.

Chet

6. Feb 22, 2014

### dauto

For a steady state the equation becomes
$$\frac{\partial}{\partial x} k(x,T)\frac{\partial}{\partial x} T = 0$$
$$k(x,T)\frac{\partial}{\partial x} T = A$$
Where A is some constant.
If k = x then
$$x\frac{\partial}{\partial x} T = A$$
$$\frac{\partial}{\partial x} T = \frac{A}{x}$$
$$T = A ln(x) + B$$

7. Feb 22, 2014

### dauto

Chestermiller beat me to the punch, except that his solution assumes x=0 as the reference point. That doesn't work for that specific problem because k(x) would be zero there.

8. Feb 22, 2014

### j_phillips

Thanks! Much appreciated!!