Register to reply

Derivation of heat transfer equation for spherical coordinates

Share this thread:
eddysd
#1
May21-12, 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

qx=-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
Attached Thumbnails
Heat Transfer.png  
Phys.Org News Partner Science news on Phys.org
Mysterious source of ozone-depleting chemical baffles NASA
Water leads to chemical that gunks up biofuels production
How lizards regenerate their tails: Researchers discover genetic 'recipe'
rude man
#2
May22-12, 10:30 PM
HW Helper
Thanks
PF Gold
rude man's Avatar
P: 4,827
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 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".
eddysd
#3
May23-12, 07:47 AM
P: 39
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

rude man
#4
May23-12, 04:17 PM
HW Helper
Thanks
PF Gold
rude man's Avatar
P: 4,827
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/(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.


Register to reply

Related Discussions
Derivation of Heat Conduction in Spherical Co-Ordinates Advanced Physics Homework 1
Magnetic Field Equation in Spherical Coordinates to Cartesian Coordinates Advanced Physics Homework 26
Heat transfer derivation problem Engineering, Comp Sci, & Technology Homework 1
Transient heat transfer problem in cylinderical coordinates Advanced Physics Homework 6
Help for heat equations in spherical coordinates Calculus & Beyond Homework 1