How can the heat equation be derived for a long circular cylinder?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 3K views
BustedBreaks
Messages
62
Reaction score
0
Consider heat flow in a long circular cylinder where the temperature depends only on t and on the distance r to the axis of the cylinder. Here [tex]r=\sqrt{x^{2}+y^{2}}[/tex] is the cylindrical coordinate. From the three dimensional heat equation derive the equation [tex]u_{t}=k(u_{rr}+\frac{u_{r}}{r})[/tex].

My book describes how the general heat equation is derived, but I'm having trouble incorporating the equation for r and in general understanding these concepts.

The book starts out with:

[tex]H(t)=\int\int\int c\rho u dxdydz[/tex] which I'm somewhat confident I can change to

[tex]H(t)=\int\int\int c\rho u dxdydr[/tex] because of the equation with r

Then the book has
[tex]\frac{dH}{dt}=\int\int\int c\rho u_{t} dxdydr[/tex] which makes sense to me

then after this I am a bit confused. I don't really know what to do next in terms of answering the question.
 
Physics news on Phys.org
This can be done very easily, by rewriting the heat equation in spherical coordinates and taking advantage of the symmetry of the situation. A little bit of re-arranging gives you the equation.