- #1
James Brady
- 105
- 4
I am trying to solve the steady state heat equation with a heat source. I am starting out in 1 dimension (my book gives the solution in 2, but I'm just trying to get a feel right now) and I have a heat source Q, located at 0. It radiates heat through an infinite medium. So what would the steady state solution be?
##\frac{\partial U}{\partial t} = k_0\frac{\partial^2U}{\partial^2 x} + Q##
So since this is steady state, I figure I can set the left hand side to zero to get the equation:
##0 = k_0\frac{\partial^2U}{\partial^2 x} + Q##
In my book it gives the solution that looks something like (2d solution, radial coordinates): ##U(x) = \frac{Q}{2\pi}ln(x)## I know that that equation isn't exactly right, but the natural log term is in there. I am not sure how you can solve the heat equation to get the natural log term, I can only guess that it is by integrating 1/x at some point, but I'm not sure as to the process.
Also, the book is on geo-exchange systems and they are referring to a pipe underground radiating energy. They refer to this as a "line source" and they are modeling it in 2d at first.
##\frac{\partial U}{\partial t} = k_0\frac{\partial^2U}{\partial^2 x} + Q##
So since this is steady state, I figure I can set the left hand side to zero to get the equation:
##0 = k_0\frac{\partial^2U}{\partial^2 x} + Q##
In my book it gives the solution that looks something like (2d solution, radial coordinates): ##U(x) = \frac{Q}{2\pi}ln(x)## I know that that equation isn't exactly right, but the natural log term is in there. I am not sure how you can solve the heat equation to get the natural log term, I can only guess that it is by integrating 1/x at some point, but I'm not sure as to the process.
Also, the book is on geo-exchange systems and they are referring to a pipe underground radiating energy. They refer to this as a "line source" and they are modeling it in 2d at first.