# Laplace heat equation

An infinite straight metal pipe has annular cross-section $a \leq r \leq b$. The temperature of the inner surface of the pipe is equal to $\cos(\theta)$, and the outer surface is thermally insulted.

(i) Write down the differential equation and the boundary conditions that the temperature must satisfy

(ii) Find the steady state temperature in $a \leq r \leq b$ and show that the temperature of the outer surface is $\dfrac{2ab}{a^2+b^2}\cos(\theta)$

(iii) Find the heat flux across unit area of the inner surface, given that the thermal conductivity of the pipe is equal to $\kappa$.

For (i) it must satisfy Laplace's equation in polar coordinates, so

$\dfrac{\partial^2T}{\partial r^2} + \dfrac{1}{r}\dfrac{\partial T}{\partial r} + \dfrac{1}{r^2}\dfrac{\partial^2 T}{\partial\theta^2}=0$

and $T(a, \theta) = cos(\theta)$

How do I use the fact it's thermally insulated at $r=b$? In cartesians I know this would imply that $-\kappa \dfrac{\partial T}{\partial x} = 0$ so $T_x = 0$. Does this need to be solved via the Neumann problem in polar coordinates, so $\dfrac{\partial T}{\partial n}(b,\theta) = 0$

For (ii) I don't know what 'steady state' is (it's not mentioned in my notes at all). Is it the temperature independent of time? If so, how is it possible to determine this while working in polar coordinates?

Thanks for any help.

You are correct in that you use Neumann BC at the edge to express thermal insulation. Because you work in polars, this means that

$$\left[ \frac{\partial T}{\partial r}\right]_{r=b} = 0$$

Keep in mind that Laplaces's equation is for the steady-state temperature. For a changing temmperature, you would be solving the heat equation, which has a time derivative in it.

For steady state temp, you should be able to show that

$$T(r,\theta) = \left(\frac{A}{r}+ B r \right )\cos{\theta}$$

There are no higher-order terms because of the nature of your boundary conditions. Now $T(a,\theta) = \cos{\theta}$ so that

$$\frac{A}{a} + B a = 1$$

Your second boundary condition on thermal insulation implies that

$$-\frac{A}{b^2} + B = 0$$

For (iii), the heat flux is just the derivative with respect to $r$ at the inner surface, multiplied by $-\kappa$.

• Thanks, I'm not entirely sure how to show $T(r,\theta) = (A/r + Br)\cos(\theta)$. I can see how, when separating the variables, there's no $\sin(\theta)$ term, but how can we conclude $r^2F''(r) + rF'(r) = \lambda F(r)$ means we should be looking at a solution $F(r) = A/r + Br$? Thanks! Commented Apr 6, 2013 at 15:34
• Because of the fact that $n=1$ from the boundary condition, and that the allowed powers of $r$ are then $\pm n = \pm 1$ as a consequence of the radial equation. Commented Apr 6, 2013 at 15:44
• @Rob Thanks yeah, sorry for asking the blinding obvious. Thanks again! Commented Apr 6, 2013 at 15:54
• I've found the heat flux at $r=a$ but I'm unsure of how I deal with the "across unit area of the inner surface". Or does calculating the derivative with respect to $r$ at $r=a$ effectively give the required result? Thanks again. Commented Apr 6, 2013 at 17:13
• Computing that derivative gives you the result you need. Commented Apr 6, 2013 at 17:16