Derivation of heat transfer equation for spherical coordinatesby eddysd Tags: coordinates, derivation, equation, heat, spherical, transfer 

#1
May2112, 11:59 AM

P: 39

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 q_{x}=kA[itex]\frac{dT}{dx}[/itex] 3. The attempt at a solution I don't know where to start from to be honest, so any help would be greatly appreciated 



#2
May2212, 10:30 PM

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P: 4,406

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 where 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 endsides 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". 



#3
May2312, 07:47 AM

P: 39

OK so this is what I got:
λ4r^{2}[itex]\frac{dT}{dr}[/itex] + [itex]\dot{q}[/itex]4∏r^{2}dr = ρc4∏r^{2}[itex]\frac{dT}{dτ}[/itex]dr 4∏r^{2}(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: r^{2}[itex]\frac{d}{dr}[/itex](λ[itex]\frac{dT}{dr}[/itex]) + [itex]\dot{q}[/itex]r^{2} = 0 can I just take the r^{2} inside the differential bracket? EDIT: missed out a dr in the rearranged equation: r^{2}[itex]\frac{d}{dr}[/itex](λ[itex]\frac{dT}{dr}[/itex])dr + [itex]\dot{q}[/itex]r^{2} = 0 



#4
May2312, 04:17 PM

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Derivation of heat transfer equation for spherical coordinates
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/(secm).
Ironically, your unedited equation has matching dimensions but you can't smuggle the r^{2} into the d/dr bracket as you wondered. (That's just basic calculus: for example, r^{2}d/dr(r^{2}) = 2r^{3} whereas d/dr(r^{4}) = 4r^{3}.) Going back to my "first principles" equation , Q_dot = λAΔT/Δr, you seem to have correctly determined that, in spherical coordinates, A = 4πr^{2} 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)} = r^{2}q_dot. That is really the hard part about this problem. 


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