Derivation of heat transfer equation for spherical coordinates

[eddysd]

Homework Statement

where λ= thermal conductivity
$\dot{q}$= dissipation rate per volume

Homework Equations

q_x=-kA$\frac{dT}{dx}$

The Attempt at a Solution

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

 

[rude man]

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]

OK so this is what I got:

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

Is this correct?

Since the flow is steady the time derivative $\frac{dT}{dτ}$=0

But then when I rearrange everything I get:

r²$\frac{d}{dr}$(λ$\frac{dT}{dr}$) + $\dot{q}$r² = 0

can I just take the r² inside the differential bracket?

EDIT: missed out a dr in the rearranged equation:

r²$\frac{d}{dr}$(λ$\frac{dT}{dr}$)dr + $\dot{q}$r² = 0

 

[rude man]

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.

 

[DeeNos]

.


The heat transfer equation for spherical coordinates can be derived by using the conservation of energy principle and the Fourier's law of heat conduction. This equation is used to describe the transfer of heat through a spherical object or system.

First, we consider a spherical element of radius r, thickness dr and volume dV. The heat flow through this element can be expressed as qx=-kA\frac{dT}{dr}, where k is the thermal conductivity and A is the surface area of the element.

Next, we apply the conservation of energy principle, which states that the rate of change of internal energy of a system is equal to the net heat transfer into or out of the system. This can be expressed as:

\dot{q}dV = \frac{\partial u}{\partial t}dV - \nabla \cdot q dV

where \dot{q} is the dissipation rate per unit volume, u is the internal energy, t is time and \nabla \cdot q is the divergence of the heat flux vector.

Using the product rule for differentiation, we can expand the divergence term as:

\nabla \cdot q = \frac{\partial qx}{\partial x} + \frac{\partial qy}{\partial y} + \frac{\partial qz}{\partial z}

In spherical coordinates, the heat flux vector can be written as q=\hat{r}qx + \hat{\theta}q\theta + \hat{\phi}q\phi, where \hat{r}, \hat{\theta} and \hat{\phi} are unit vectors in the radial, polar and azimuthal directions respectively.

Substituting this into the divergence term and rearranging, we get:

\nabla \cdot q = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2qx) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta q\theta) + \frac{1}{r\sin\theta}\frac{\partial q\phi}{\partial \phi}

Now, we can substitute this into the conservation of energy equation and combine it with the expression for heat flow through the spherical element. This gives us:

\dot{q}dV = \frac{\partial u}{\partial t}dV - \frac{1}{r^2}\