Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Derivation of heat transfer equation for spherical coordinates

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data


    where λ= thermal conductivity
    [itex]\dot{q}[/itex]= dissipation rate per volume

    2. Relevant equations


    3. The attempt at a solution

    I don't know where to start from to be honest, so any help would be greatly appreciated

    Attached Files:

    Last edited: May 21, 2012
  2. jcsd
  3. May 22, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    I've had to think on this one for some time; I hope what I write is correct:

    Start with the fundamental equation for heat transfer:

    dQ/dt = λAΔT/Δr
    dQ/dt = Qdot = rate of heat flow across area A;
    λ = conductivity;
    ΔT = temperature difference across volume element AΔr.

    What is ΔT/Δr in the limit as Δr → 0?

    Then: what is the volume element AΔr in spherical coordinates? (Heat flows thru the volume element from one side of area A to the other side, also of area A, the two sides separated by Δr. )

    Now for the big step: realize that Qdot need not be constant along Δr. In other words, Qdot can be different for the two end-sides of your elemental volume. So in the limit the derivative d(Qdot)/dr can be finite. So your last equation is to equate how Qdot changes along Δr to what the problem calls the "dissipation rate per volume".
  4. May 23, 2012 #3
    OK so this is what I got:

    -λ4r2[itex]\frac{dT}{dr}[/itex] + [itex]\dot{q}[/itex]4∏r2dr = ρc4∏r2[itex]\frac{dT}{dτ}[/itex]dr -4∏r2(λ[itex]\frac{dT}{dr}[/itex] + [itex]\frac{d}{dr}[/itex](λ[itex]\frac{dT}{dr}[/itex])dr)

    Is this correct?

    Since the flow is steady the time derivative [itex]\frac{dT}{dτ}[/itex]=0

    But then when I rearrange everything I get:

    r2[itex]\frac{d}{dr}[/itex](λ[itex]\frac{dT}{dr}[/itex]) + [itex]\dot{q}[/itex]r2 = 0

    can I just take the r2 inside the differential bracket?

    EDIT: missed out a dr in the rearranged equation:

    r2[itex]\frac{d}{dr}[/itex](λ[itex]\frac{dT}{dr}[/itex])dr + [itex]\dot{q}[/itex]r2 = 0
    Last edited: May 23, 2012
  5. May 23, 2012 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Your (edited) equation has incompatible terms: the first is infinitesimal, the second isn't. Plus, the terms' dimensions don't agree: the first one's are (using SI) J/sec whereas the second one's are J/(sec-m).

    Ironically, your unedited equation has matching dimensions but you can't smuggle the r2 into the d/dr bracket as you wondered. (That's just basic calculus: for example, r2d/dr(r2) = 2r3 whereas d/dr(r4) = 4r3.)

    Going back to my "first principles" equation , Q_dot = λAΔT/Δr, you seem to have correctly determined that, in spherical coordinates, A = 4πr2 and, of course, ΔT/Δr → dT/dr. So your remaining task, and it does take some thinking, is to somehow get rid of Q_dot and substitute for it an expression containing q_dot. (Sorry, I haven't learned the itex thing yet). So that you wind up with
    -d/dr{λr(dT/dr)} = r2q_dot. That is really the hard part about this problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook