Varying thermal conductivity with length

[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.

 

[Chestermiller]

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.

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

 

[dauto]

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.

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

 

[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)?

 

[Chestermiller]

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

 

[dauto]

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)?

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

 

[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.

 

[j_phillips]

Thanks! Much appreciated!