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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Transistor
    Jun 12, 2021 at 20:20
  • $\begingroup$ @Transistor Pretty much the case with every "How do I calculate the temperature?" question. $\endgroup$
    – DKNguyen
    Jun 13, 2021 at 1:38
  • $\begingroup$ Buy a book for ME's for the heat transfer class and work through it until you get to this problem. $\endgroup$
    – Tiger Guy
    Jun 13, 2021 at 6:42

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.