I'm not totally sure what you mean by "remapping" the 2D heat equation onto a cylindrical region. Could you give me some more specifics as to how to begin doing that?
Yes, the heat loss due to radiation will be negligible. This bellows is going to be used in a cryogenic setting (around 15K) in a vaccum of ##10^{-6}## Torr, so the radiation will be miniscule.
That actually ends up giving very nearly the same answer as my disk answer, except that it has the ##\sin{\theta}## in it and the argument of ##\ln{\frac{r_1}{r_2}}## is flipped. $$\frac{dT}{dr} = -\frac{\dot{Q}}{2 \pi rt \lambda \sin{\theta}} \frac{1}{r}$$ Integrating this gives $$\Delta T =...
This problem seems best treated in cylindrical coordinates. There is azimuthal symmetry, and there is no heat loss or generation within the cone, so our thermal conductivity equation reads:
$$\vec{q} = -k(\frac{\partial T}{\partial \rho} \hat{\rho} + \frac{\partial T}{\partial z} \hat{z})$$
We...
Thinking about it, it seems reasonable to replace the cone with a disk that has a hole in the center. That changes our cross-sectional area through which heat flows from approximately ##A = 2 \pi r t## to exactly ##A = 2 \pi r t##. We can then integrate over the radius itself rather than the...
The bellows are similar to this, but much taller. The drawing in the question statement is an exaggeration in terms of height of the cone, it is simply to show the sort of shape that I mean.
You are right that it is practically flat enough to be a disk (h approx. 1mm), but it is still a cone...
We can write our radius as a function of the height, z, of our cone: $$R(z) = \frac{R_2 - R_1}{h} z + R_1$$
Where h is the height of our cone, ##h = \frac{L}{40}##.
Our cross sectional area, $$A = 2 \pi R t$$ can then be written as $$A = 2 \pi t [\frac{R_2 - R_1}{h} z + R_1]$$
This I am all...