surface temperature due to constant heat flux

Question

Imagine a hollow cylinder with the outside surface exposed to air (convection). If there is a constant heat flux on the inside surface how would I calculate both the inside and outside surface temperature?

Answer 1

The steady-state heat equation is $\nabla^2 T=0$, where $\nabla^2$ is the Laplacian. In the axisymmetric 2-D case, $T=T(r)$ and $\nabla^2 T=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T(r)}{\partial r}\right)$, where $r$ is the radial distance. From $$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T(r)}{\partial r}\right)=0,$$ multiply by $r$ and integrate to obtain $$r\frac{\partial T(r)}{\partial r}=C_1\Longrightarrow \frac{\partial T(r)}{\partial r}=-\frac{Q}{2\pi rk},$$ where I've plugged in the heat-flux boundary condition of $Q$ (in watts per cylinder length in meters, corresponding to a heat flux of $q=Q/2\pi r_1$) and denoted the annulus thermal conductivity by $k$. Integrate again: $$T(r)=-\frac{Q}{2\pi k}\ln r+C_2.$$ The convective boundary condition at the outer radius $r_2$ for convective coefficient $h$ and ambient temperature $T_\infty$ is $$Q=2\pi r_2 h[T(r_2)-T_\infty].$$ Solve for $C_2$ to obtain $$T(r)=-\frac{Q}{2\pi k}\ln\left(\frac{r}{r_2}\right)+\frac{Q}{2\pi r_2 h}+T_\infty.$$