- #1
Tsunoyukami
- 215
- 11
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:
[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 co-ordinates, 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 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 [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! :)
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:
[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 co-ordinates, 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 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 [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! :)