# Derivation of Heat Conduction in Spherical Co-Ordinates

by Tsunoyukami
Tags: conduction, coordinates, derivation, heat, spherical
 P: 209 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 co-ordinates. I have completed part a, in which I derived the heat flux equation: $\frac{dq_{r}}{dr} + \frac{2}{r}q_{r} - \rho H = 0$ I have used Fourier's Law (to rewrite this equation for teperature), which states $\textbf{q}$ = -k$\nabla T$, where $\textbf{q}$ is the heat flux and it is a vector (I couldn't find a vector symbol, so it is simply bolded) and $\nabla$ is the gradient. Using the gradiant for speherical co-ordinates, and considering only changes in the radial direction (so that $\frac{\delta T}{\delta \phi}$ and $\frac{\delta T}{\delta \theta}$ are 0) we can write: $\nabla T = \frac{dT}{dr} \widehat{r}$ So we can write: $\textbf{q} = q_{r} = -k \nabla T = -k \frac{dT}{dr} \widehat{r}$ $\frac{dq_{r}}{dr} + \frac{2}{r}q_{r} - \rho H = 0$ $\frac{d}{dr} ( -k \frac{dT}{dr}) - \frac{2k}{r} \frac{dT}{dr} - \rho H = 0$ Assuming k is a constant: $-k \frac{d^{2}T}{dr^{2}} - \frac{2k}{r} \frac{dT}{dr} - \rho H = 0$ $(\frac{d^{2}T}{dr^{2}} + \frac{2}{r} \frac{dT}{dr}) + \frac{\rho H}{k} = 0$ 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) = $T_{s}$ (where R is the radius of the planet) and T(0) must be finite. $(\frac{d^{2}T}{dr^{2}} + \frac{2}{r} \frac{dT}{dr}) + \frac{\rho H}{k} = 0$ Using a hint that is provided I can write: $\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dT}{dr}) = - \frac{\rho H}{k}$ Now, I'm not too sure how to simplify this. My E&M textbook says that the radial component of the Laplacian in spherical co-ordinates is equal to what I have on the left side of that equation so I could write (because we are assuming, as above that $\frac{\delta T}{\delta \phi}$ and $\frac{\delta T}{\delta \theta}$ are 0): $\nabla^{2}T + \frac{\rho H}{k} = 0$ 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! :)
 P: 209 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 co-ordinates. However, I need an expression for T in terms of r and this is where I get stuck. $\nabla^{2}T + \frac{\rho H}{k} = 0$ $\frac{1}{r^{2}} \frac{d}{dr} (r^{2} \frac{dT}{dr}) = - \frac{\rho H}{k}$ $\frac{d}{dr} (r^{2} \frac{dT}{dr}) = - \frac{\rho Hr^{2}}{k}$ We can then integrate both sides with respect to r: $r^{2} \frac{dT}{dr} = - \frac{\rho Hr^{3}}{3k} + c_{1}$ $\frac{dT}{dr} = - \frac{\rho Hr}{3k} + c_{1}$ Integrating again we can write: $T = - \frac{\rho Hr^{2}}{6k} + c_{1}r + c_{2}$ 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 co-ordinates 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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