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Varying thermal conductivity with length

  1. Feb 21, 2014 #1
    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?

  2. jcsd
  3. Feb 22, 2014 #2
    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?

  4. Feb 22, 2014 #3
    That's not the correct equation. It should be

    [tex]C \rho\frac{\partial}{\partial t}T =\frac{\partial}{\partial x} k(x,T)\frac{\partial}{\partial x} T,[/tex]
    where [itex]C, \rho,[/itex] and [itex] k [/itex] are the heat capacity, density, and conductivity of the material
  5. Feb 22, 2014 #4
    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)?
  6. Feb 22, 2014 #5
    So the equation you are solving is:
    [tex]\frac{d}{dx} \left(k(x)\frac{dT}{d x}\right)=0[/tex]
    Integrating this once with respect to x gives:
    [tex]k(x)\frac{dT}{d x}=C[/tex]
    where C is minus the (constant) heat flux.
    Integrating again gives:
    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.

  7. Feb 22, 2014 #6
    For a steady state the equation becomes
    [tex]\frac{\partial}{\partial x} k(x,T)\frac{\partial}{\partial x} T = 0[/tex]
    [tex]k(x,T)\frac{\partial}{\partial x} T = A[/tex]
    Where A is some constant.
    If k = x then
    [tex]x\frac{\partial}{\partial x} T = A[/tex]
    [tex]\frac{\partial}{\partial x} T = \frac{A}{x}[/tex]
    [tex]T = A ln(x) + B[/tex]
  8. Feb 22, 2014 #7
    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.
  9. Feb 22, 2014 #8
    Thanks! Much appreciated!!
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