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I was wondering if anyone can show me what approach to finding the steady state temperature distribution in this problem. The image is in the link below.

I know that this requires Laplace's Equation in polar coordinates, but the boundary conditions are kind of tricky with 3 different temperatures. I know the inside of the pipe owuld be T(r,theta) = 10, but how does one approach the problem with the outside boundary conditions?

Any help would be appreciated. Thanks!

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I'll provide you with a general approach to this problem. The general solution to the Dirichlet problem in an annulus (this problem) is

$$u(r,\theta) = a_0 + b_0 \log{r} + \sum_{k=1}^{\infty} (a_k r^k + b_k r^{-k}) \cos{k \theta} + \sum_{k=1}^{\infty} (c_k r^k + d_k r^{-k}) \sin{k \theta}$$

where $\theta$ is the angle from the positive $x$ axis in a ccw sense. Your inner boundary condition is easy:

$$u(3,\theta) = T_i = 10^{\circ}$$

The outer boundary condition is more complicated. The geometry has the outer surface being held to one of two temperatures, depending on the angle. You can deduce from the geometry that

$$u(5,\theta) = \begin{cases} \\ T_u = -20^{\circ} & \theta \in [\theta_0,\pi-\theta_0] \\ T_l = 2^{\circ} & \text{otherwise} \end{cases}$$


$$\sin{\theta_0} = \frac{5-0.5}{5} = \frac{9}{10}$$

We then represent this function $u(5,\theta)$ as a Fourier series:

$$u(5,\theta) = \frac12 A_0 + \sum_{k=1}^{\infty}( A_k \cos{k \theta} + B_k \sin{k \theta})$$

$$A_k = \frac{1}{2 \pi} \int_0^{2 \pi} d\theta\, u(5,\theta) \cos{k \theta}$$ $$B_k = \frac{1}{2 \pi} \int_0^{2 \pi} d\theta\, u(5,\theta) \sin{k \theta}$$

You do this calculation and get the $A_k$ and $B_k$. You then use this and the inner BC to get the $A_k$, $b_k$, $c_k$, and $d_k$:

$$\frac12 A_0 = a_0 + b_0 \log{5}$$ $$T_i = a_0 + b_0 \log{3}$$

$$A_k = a_k 5^k + b_k 5^{-k}$$ $$0 = a_k 3^k + b_k 3^{-k}$$

$$B_k = c_k 5^k + d_k 5^{-k}$$ $$0 = c_k 3^k + d_k 3^{-k}$$

This completely solves the problem - it is up to you to carry out the computations.

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