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Derivation of Heat Conduction in Spherical Co-Ordinates

  1. Feb 20, 2012 #1
    I have two questions. I believe I have solved the first question and would like confirmation of this answer; the second question I'm a little bit lost on so any help there would be greatly appreciated!

    I am working on a problem set in which I must derive the equation for heat conduction in spherical co-ordinates. I have completed part a, in which I derived the heat flux equation:

    [itex]\frac{dq_{r}}{dr} + \frac{2}{r}q_{r} - \rho H = 0[/itex]

    I have used Fourier's Law (to rewrite this equation for teperature), which states

    [itex]\textbf{q}[/itex] = -k[itex]\nabla T[/itex], where [itex]\textbf{q}[/itex] is the heat flux and it is a vector (I couldn't find a vector symbol, so it is simply bolded) and [itex]\nabla [/itex] is the gradient.

    Using the gradiant for speherical co-ordinates, and considering only changes in the radial direction (so that [itex] \frac{\delta T}{\delta \phi}[/itex] and [itex]\frac{\delta T}{\delta \theta}[/itex] are 0) we can write:

    [itex]\nabla T = \frac{dT}{dr} \widehat{r}[/itex]

    So we can write:

    [itex]\textbf{q} = q_{r} = -k \nabla T = -k \frac{dT}{dr} \widehat{r}[/itex]

    [itex]\frac{dq_{r}}{dr} + \frac{2}{r}q_{r} - \rho H = 0[/itex]

    [itex]\frac{d}{dr} ( -k \frac{dT}{dr}) - \frac{2k}{r} \frac{dT}{dr} - \rho H = 0[/itex]

    Assuming k is a constant:

    [itex]-k \frac{d^{2}T}{dr^{2}} - \frac{2k}{r} \frac{dT}{dr} - \rho H = 0[/itex]

    [itex] (\frac{d^{2}T}{dr^{2}} + \frac{2}{r} \frac{dT}{dr}) + \frac{\rho H}{k} = 0[/itex]

    So I have written this as an equation for temperature instead of heat flux. Is this correct?

    ____________________________________________________________________________

    Next I am asked to solve the above equation for T(r) subject to the boundary conditions T(R) = [itex]T_{s}[/itex] (where R is the radius of the planet) and T(0) must be finite.

    [itex] (\frac{d^{2}T}{dr^{2}} + \frac{2}{r} \frac{dT}{dr}) + \frac{\rho H}{k} = 0[/itex]

    Using a hint that is provided I can write:

    [itex] \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dT}{dr}) = - \frac{\rho H}{k} [/itex]


    Now, I'm not too sure how to simplify this. My E&M textbook says that the radial component of the Laplacian in spherical co-ordinates is equal to what I have on the left side of that equation so I could write (because we are assuming, as above that [itex] \frac{\delta T}{\delta \phi}[/itex] and [itex]\frac{\delta T}{\delta \theta}[/itex] are 0):

    [itex] \nabla^{2}T + \frac{\rho H}{k} = 0 [/itex]

    What should I do next? Is there any way for me to simplify this further? Should I have not written it in terms of the Laplacian? How can I solve for an expression T(r)?

    Thanks in advance! I appreciate any insight you can give me as to what the next step might be! :)
     
  2. jcsd
  3. Feb 21, 2012 #2
    So I've been thinking about this a bit more and I know that this expression must be correct because it agrees with the expression derived in class using Cartesian co-ordinates.

    However, I need an expression for T in terms of r and this is where I get stuck.

    [itex] \nabla^{2}T + \frac{\rho H}{k} = 0 [/itex]

    [itex] \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dT}{dr}) = - \frac{\rho H}{k} [/itex]

    [itex] \frac{d}{dr} (r^{2} \frac{dT}{dr}) = - \frac{\rho Hr^{2}}{k} [/itex]

    We can then integrate both sides with respect to r:

    [itex] r^{2} \frac{dT}{dr} = - \frac{\rho Hr^{3}}{3k} + c_{1}[/itex]

    [itex] \frac{dT}{dr} = - \frac{\rho Hr}{3k} + c_{1}[/itex]

    Integrating again we can write:

    [itex] T = - \frac{\rho Hr^{2}}{6k} + c_{1}r + c_{2}[/itex]

    Is this correct? I'm concerned that I did something incorrect while integrating the first time; I feel like the 6 shouldn't be in the denominator. In Cartesian co-ordinates we have a 2 in the denominator. Where am I going wrong?

    I believe there must be a 2 and not a 6 because if r is solely in the y direction (x and z = 0) the equation should reduce to what we found in the Cartesian case.
     
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