# Heat transfer in a finite cylinder

This question has more to do with experimental physics than theoretical physics. I have a copper cylinder of radius $a$ and length $L$. On one side of the cylinder, let's say $x=0$, I have a source with a temperature that varies with time. Now, on the other side, $x=L$, the copper cylinder is exposed to air at atmospheric pressure and temperature.

Now, my problem is what happens at $x=L$. I know I can't consider that side as a source because it doesn't have a constant temperature. But if a wanted to solve the heat equation, which bonundary condition should I use for $x=L$?

## 1 Answer

Now, my problem is what happens at $x=L$. I know I can't consider that side as a source because it doesn't have a constant temperature. But if a wanted to solve the heat equation, which boundary condition should I use for $x=L$?

Your boundary condition must express what is physically happening.

There are various possibilities.

1. Assume that side of the cylinder loses heat through convection only:

If $T(r,x,t)$ is the temperature of the cylinder, then in that case:

$$\kappa\frac{\partial T(r,L,t)}{\partial x}=h[T(r,L,t)-T_{\infty}]$$ ($\kappa T_x(r,L,t)=h[T(r,L,t)-T_{\infty}]$)

$T_{\infty}$ is the temperature of air at that end. This boundary condition means that the heat conducted though the end bit is equal to what is lost through convection at that end. $\kappa$ is the thermal diffusivity and $h$ the the convection coefficient.

In that case it's easier to define a new variable:

$$u(r,x,t)=T(r,x,t)-T_{\infty}$$

So:

$$\kappa \frac{\partial u}{\partial x}=hu$$ ($\kappa u_x=hu$)

1. Assume that side of the cylinder loses heat through radiative losses only:

Use Stefan-Boltzmann:

$$\kappa\frac{\partial T(r,L,t)}{\partial x}=\epsilon \sigma[T(r,L,t)^4-T_{\infty}^4]$$

1. Combine $1.$ and $2.$, by assuming the end loses heat through convection and radiation:

$$\kappa\frac{\partial T(r,L,t)}{\partial x}=h[T(r,L,t)-T_{\infty}]+\epsilon \sigma[T(r,L,t)^4-T_{\infty}^4]$$

1. Assume the tip $x=L$ is insulated:

$$\kappa\frac{\partial T(r,L,t)}{\partial x}=0$$

($u_x(r,L,t)=0$)

The latter is of course by far the easiest case to solve.

The choice is yours.

If you assume $\frac{\partial T}{\partial r}=0$, then $T$ reduces to $T(x,t)$ but the principle remains the same.

Your question is closely related to this question

• Okay, wow, thanks. I'll see what I can do with that boundary conditions. Do you know any book with a complete discussion about this particular subject?
– user120474
Oct 21, 2016 at 1:50
• Have a look at the related question I just edited in.
– Gert
Oct 21, 2016 at 1:52
• math.ubc.ca/~peirce/HeatProblems.pdf These are all eigenvalue problems, requiring separation of variables method. Your case is quite simple.
– Gert
Oct 21, 2016 at 2:00