Derivation of Heat Conduction in Spherical CoOrdinatesby Tsunoyukami Tags: conduction, coordinates, derivation, heat, spherical 

#1
Feb2012, 08:56 PM

P: 201

I have two questions. I believe I have solved the first question and would like confirmation of this answer; the second question I'm a little bit lost on so any help there would be greatly appreciated!
I am working on a problem set in which I must derive the equation for heat conduction in spherical coordinates. I have completed part a, in which I derived the heat flux equation: [itex]\frac{dq_{r}}{dr} + \frac{2}{r}q_{r}  \rho H = 0[/itex] I have used Fourier's Law (to rewrite this equation for teperature), which states [itex]\textbf{q}[/itex] = k[itex]\nabla T[/itex], where [itex]\textbf{q}[/itex] is the heat flux and it is a vector (I couldn't find a vector symbol, so it is simply bolded) and [itex]\nabla [/itex] is the gradient. Using the gradiant for speherical coordinates, and considering only changes in the radial direction (so that [itex] \frac{\delta T}{\delta \phi}[/itex] and [itex]\frac{\delta T}{\delta \theta}[/itex] are 0) we can write: [itex]\nabla T = \frac{dT}{dr} \widehat{r}[/itex] So we can write: [itex]\textbf{q} = q_{r} = k \nabla T = k \frac{dT}{dr} \widehat{r}[/itex] [itex]\frac{dq_{r}}{dr} + \frac{2}{r}q_{r}  \rho H = 0[/itex] [itex]\frac{d}{dr} ( k \frac{dT}{dr})  \frac{2k}{r} \frac{dT}{dr}  \rho H = 0[/itex] Assuming k is a constant: [itex]k \frac{d^{2}T}{dr^{2}}  \frac{2k}{r} \frac{dT}{dr}  \rho H = 0[/itex] [itex] (\frac{d^{2}T}{dr^{2}} + \frac{2}{r} \frac{dT}{dr}) + \frac{\rho H}{k} = 0[/itex] So I have written this as an equation for temperature instead of heat flux. Is this correct? _______________________________________________________________________ _____ Next I am asked to solve the above equation for T(r) subject to the boundary conditions T(R) = [itex]T_{s}[/itex] (where R is the radius of the planet) and T(0) must be finite. [itex] (\frac{d^{2}T}{dr^{2}} + \frac{2}{r} \frac{dT}{dr}) + \frac{\rho H}{k} = 0[/itex] Using a hint that is provided I can write: [itex] \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dT}{dr}) =  \frac{\rho H}{k} [/itex] Now, I'm not too sure how to simplify this. My E&M textbook says that the radial component of the Laplacian in spherical coordinates is equal to what I have on the left side of that equation so I could write (because we are assuming, as above that [itex] \frac{\delta T}{\delta \phi}[/itex] and [itex]\frac{\delta T}{\delta \theta}[/itex] are 0): [itex] \nabla^{2}T + \frac{\rho H}{k} = 0 [/itex] What should I do next? Is there any way for me to simplify this further? Should I have not written it in terms of the Laplacian? How can I solve for an expression T(r)? Thanks in advance! I appreciate any insight you can give me as to what the next step might be! :) 



#2
Feb2112, 11:29 AM

P: 201

So I've been thinking about this a bit more and I know that this expression must be correct because it agrees with the expression derived in class using Cartesian coordinates.
However, I need an expression for T in terms of r and this is where I get stuck. [itex] \nabla^{2}T + \frac{\rho H}{k} = 0 [/itex] [itex] \frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dT}{dr}) =  \frac{\rho H}{k} [/itex] [itex] \frac{d}{dr} (r^{2} \frac{dT}{dr}) =  \frac{\rho Hr^{2}}{k} [/itex] We can then integrate both sides with respect to r: [itex] r^{2} \frac{dT}{dr} =  \frac{\rho Hr^{3}}{3k} + c_{1}[/itex] [itex] \frac{dT}{dr} =  \frac{\rho Hr}{3k} + c_{1}[/itex] Integrating again we can write: [itex] T =  \frac{\rho Hr^{2}}{6k} + c_{1}r + c_{2}[/itex] Is this correct? I'm concerned that I did something incorrect while integrating the first time; I feel like the 6 shouldn't be in the denominator. In Cartesian coordinates we have a 2 in the denominator. Where am I going wrong? I believe there must be a 2 and not a 6 because if r is solely in the y direction (x and z = 0) the equation should reduce to what we found in the Cartesian case. 


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