# surface temperature due to constant heat flux

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?

• Temperature will stabilise when power in is equal to the heat lost to the surroundings. I think you will find that calculation is quite complex. Jun 12 at 20:20
• @Transistor Pretty much the case with every "How do I calculate the temperature?" question. Jun 13 at 1:38
• Buy a book for ME's for the heat transfer class and work through it until you get to this problem. Jun 13 at 6:42

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 $$T(r)=-\frac{Q}{2\pi k}\ln\left(\frac{r}{r_2}\right)+\frac{Q}{2\pi r_2 h}+T_\infty.$$