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Heat conduction in a sphere

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the equation governing the stead-state radial conduction of heat in a sphere or spherical shell with volumetric heat production is given by

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

    where k is thermal conductivity, H is the heat production per unit mass and [tex]\rho[/tex] is the density

    2. Relevant equations

    See above

    3. The attempt at a solution

    My first attempt was to start off with the 3-dimensional fourier's law equation with a steady-state case aka

    [tex]\nabla^2 T = -\frac{A}{k}[/tex]
    [tex]\frac{d^2T}{dr^2} = -\frac{4\pi r^2}{k}[/tex]
    [tex]\frac{d}{dr}(\frac{dT}{dr}) = -\frac{4\pi r^2}{k}[/tex]
    [tex]\int \frac{d}{dr}(\frac{dT}{dr}) dr = \int -\frac{4\pi r^2}{k} dr[/tex]
    [tex]\frac{dT}{dr} = -\frac{4\pi r^3}{3k}[/tex]

    At this point I'm totally lost as to how to proceed. Have I even got the right equation to start with?
     
  2. jcsd
  3. Oct 25, 2009 #2
    Doesn't look right to start there. Almost seems like you want to take the answer and find the starting point, which might not get you anywhere.

    Consider the shell having inner radius of [tex]r[/tex] and total thickness [tex]\delta r[/tex]. The heat flow out of the surface is given by
    [tex]
    4\pi\left(r+\delta r\right)^2q_r(r+\delta r)
    [/tex]

    while the heat flow into the surface is
    [tex]
    4\pi r^2q_r(r)
    [/tex]

    The total heat flow out will be the first minus the second equation. If you then Taylor expand the first equation about the variation [tex]\delta r[/tex], you should see the direction which you need to go to finish the problem
     
  4. Oct 25, 2009 #3

    gabbagabbahey

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    You seem to be claiming that [itex]\nabla^2T(r)=\frac{d^2T}{dr^2}[/itex]....that's not really what you get for the Laplacian in spherical coordinates is it?:wink:
     
  5. Oct 26, 2009 #4
    It is if there is no angular dependence, which he does seem to imply in the initial problem.
     
  6. Oct 26, 2009 #5

    gabbagabbahey

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    No, it isn't....look up the Laplacian in spherical coordinates.
     
  7. Oct 26, 2009 #6
    I think we may be disagreeing on different aspects. I was talking about the use of [tex]d[/tex] instead of [tex]\partial[/tex] and not the form while you were talking about the form and not the use of [tex]d[/tex].
     
  8. Oct 26, 2009 #7

    gabbagabbahey

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    Okay, but that wasn't very clear from your last post.
     
  9. Oct 28, 2009 #8
    Sorry about the late response guys, I got held up at work.

    So I followed your instructions jdwood, and I see how you had come up with the equation that you came up with. This is how far I've gotten.

    [tex]4\pi\left((r^2+\delta r^2)\left(-k\frac{dT}{dr}+\delta r \left(-k\frac{d^2T}{dr^2}\right)\right) + r^2 k\frac{dT}{dr}\right) = q_{total}[/tex]

    I am unsure of how to proceed from this part though.

    Is the [tex]q_{total}[/tex] = some constant value?

    Sorry if I'm asking the stupidest questions.
     
  10. Oct 28, 2009 #9

    gabbagabbahey

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    I don't see why you are deriving the heat conduction equation...surely you've already seen it done in your textbook/class? If so, just use it.

    [tex]\rho H=-k\nabla^2T[/tex]

    Have you not seen the above form before?
     
  11. Oct 28, 2009 #10
    I have seen the above, but the problem is that because this is spherical, we need to start from the beginning and render the equation proper. At least that's how the prof wants to see it on the homework.

    A HUGE thank you to jdwood for that massively helpful hint.

    I just set your equation to what I should have realized it was right from the beginning

    [tex]\rho H (4\pi r^2)\delta r [/tex]

    I was able to derive it using just basic algebra, taylor expansion, and assumption that [tex]\delta r^2 = 0[/tex].
     
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