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.
- 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:
$$\kappa \frac{\partial u}{\partial x}=hu$$
($\kappa u_x=hu$)
- Assume that side of the cylinder loses heat through radiative losses only:
Use Stefan-Boltzmann:
$$\kappa\frac{\partial T(r,L,t)}{\partial x}=\epsilon \sigma[T(r,L,t)^4-T_{\infty}^4]$$
- Combine $1.$ and $2.$, by assuming the end loses heat through convection and radiation:
$$\kappa\frac{\partial T(r,L,t)}{\partial x}=h[T(r,L,t)-T_{\infty}]+\epsilon \sigma[T(r,L,t)^4-T_{\infty}^4]$$
- 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.
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