- #1
alex_amvdor
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- Homework Statement
- This is not a homework problem but rather a problem I need to solve for my research, though it seems like something that would be assigned to an undergraduate. There is a similar question asked on this forum, but that question stipulates that each end of our cone is at a constant temperature, while in this problem that is not true of the larger end.
We have a truncated cone made of copper with a small radius ##\rho_1##, a large radius ##\rho_2##, and a height h (thus, the outside radius can be written as a function of z, $$\rho(z) = \frac{\rho_2-\rho_1}{h}z + \rho_1$$). The small face of the cone has a circular heater of radius ##\rho_1## attached to it which is inputting a constant heat flow rate ##\dot{Q}## which is homogeneous over the entire small surface (##\vec{q}## constant over small end), and the small end is at some temperature ##T_1##. All of the heat flows out of the large end of the cone, but the large end does not necessarily have a homogeneous heat flux through it.
The system is in thermal equilibrium. The cone is surrounded otherwise by vacuum, so no heat flows out of the sides. The goal is to find either exactly how homogeneous the heat flow and/or temperature through the large end of the cone is, or at least to find an upper bound on the homogeneity.
- Relevant Equations
- $$\vec{q} = -k \nabla T$$
This problem seems best treated in cylindrical coordinates. There is azimuthal symmetry, and there is no heat loss or generation within the cone, so our thermal conductivity equation reads:
$$\vec{q} = -k(\frac{\partial T}{\partial \rho} \hat{\rho} + \frac{\partial T}{\partial z} \hat{z})$$
We know that the total ##\dot{Q}## flowing through each end is constant, and at the small end ##\vec{q}## is constant as well, so: $$|\vec{q}| = \frac{\dot{Q}}{\pi \rho_1^2}$$
The fact that ##\dot{Q}## must also flow through through the large end means that: $$\dot{Q} = \oint_{z = h} \vec{q}(\rho, z=h) \cdot dA\hat{z}$$ meaning the integral of the heat flux over the entire surface of the large end will give the heat input through the small end. This is also true for any cross-sectional area taken with respect to the z-axis.
The only other boundary condition is that there must be no heat flow through the side walls, so at that boundary all heat flux must be perpendicular to the wall.
These are all the boundary conditions I was able to come up with. I have tried to work through this problem a few different times, but I get stuck very early on every time. The only thing I know is right is that, for z = 0 (over the entire small end):$$\vec{q} = -k\frac{\partial T}{\partial z} \hat{z}$$ as the heat flux is only in the z direction at this end. We also know ##|\vec{q}| = \frac{Q}{\pi \rho_1^2}##, so it seems we can solve this differential equation for some function of##z##, with an unknown function of ##\rho## in there as well.
Expressing the side-wall boundary condition in more mathematical terms:
The side-wall makes an angle ##\theta## with the ##z##-axis, with ##\tan{\theta}=\frac{\rho_2-\rho_1}{h}## so we can define the direction perpendicular to the sidewall as ##\hat{r}_\bot = (\rho, \phi, z) = (\cos{\theta}, \phi, -\sin{\theta})##, for any ##\phi##. The boundary condition is then that $$\vec{q} \cdot \hat{r}_\bot = 0$$ for ##\rho = \rho(z)##. This gives us $$\frac{\partial T}{\partial \rho} \cos{\theta} - \frac{\partial T}{\partial z}\sin{\theta} = 0$$ but as ##\tan{\theta} = \frac{\rho_2-\rho_1}{h}##, we find that $$\frac{\partial T}{\partial \rho} = \frac{\partial T}{\partial z} \cdot \frac{\rho_2-\rho_1}{h}$$ for ##\rho = \rho(z)##
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