Question

Find the steady-state temperature distribution inside a sphere of radius 1 when the surface temperatures are as given in Problems 1 to 10.

$\left\{\begin{array}{l}\begin{array}{ccc}\mathbf{cos\theta }\mathbf{,}& \mathbf{0}\mathbf{<}\mathbf{\theta }\mathbf{<}\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{,}& \mathbf{\text{thatis,upperhemisphere,}}\end{array}\\ \begin{array}{ccc}\mathbf{0}\mathbf{,}& \mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{<}\mathbf{\theta }\mathbf{<}\mathbf{\pi }\mathbf{,}& \mathbf{\text{thatis,lowerhemisphere}}\end{array}\end{array}\right.$.

Short answer

The steady-state temperature distribution inside a sphere of radius 1:

$\frac{1}{4}{P}_0\left(\cos \theta \right)+\frac{9}{16}r{P}_1\left(\cos \theta \right)+\frac{15}{32}{r}^2{P}_2\left(\cos \theta \right)+\dots$.

Step-by-step solution

Given Information:

The radius of the sphere is 1.

Definition of steady-state temperature:

When a conductor reaches a point where no more heat can be absorbed by the rod, it is said to be at steady-state temperature.

Calculate the steady-state temperature distribution function:

The standard Legendre polynomials is$P_l\left(\cos \left(\theta \right)\right)$.

Consider the equation

$u_{r=1}=\left\{\begin{array}{l}\begin{array}{ll}0,& -1<x<0\end{array}\\ \begin{array}{ll}x,& 0<x<1\end{array}\end{array}\right.$

Take the equation

$c_m=\frac{2m+1}{2}\underset{-1}{\overset{1}{\int }}(5x^3+3x^2-3)p_m\left(x\right)dx$ ….. (1)

Simplify further:

Take m = 0 and put in equation (1).

$\begin{array}{c}{c}_0=\frac{1}{2}\underset{0}{\overset{1}{\int }}x{P}_0\left(x\right)dx\\ =\frac{1}{2}\underset{0}{\overset{1}{\int }}xdx\\ =\frac{1}{2}{\left[\frac{x^2}{2}\right]}_0^1\end{array}$

$c_0=\frac{1}{4}$

Take m = 1 and put in equation (1).

$\begin{array}{c}{c}_1=\frac{3}{2}\underset{0}{\overset{1}{\int }}(x\times x)dx\\ =\frac{3}{2}\underset{0}{\overset{1}{\int }}{\text{x}}^{2}dx\\ =\frac{3}{2}{\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{2}\end{array}$

Take m = 2 and put in equation (1).

$\begin{array}{c}{c}_2=\frac{5}{4}{\int }_0^{1}x(3x^2-1)dx\\ =\frac{5}{4}{\int }_0^1(3x^3-x)dx\\ =\frac{5}{4}{[\frac{3x^4}{4}-\frac{x^2}{2}]}_0^1\\ =\frac{5}{16}\end{array}$

Use the equation

$u=\sum _{l=0}^{\infty }{c}_l{r}^l{P}_l\left(\cos \theta \right)$

$u=\sum _{l=0}^{\infty }{c}_l{r}^l{P}_l\left(\cos \theta \right)=\frac{1}{4}{P}_0\left(\cos \theta \right)+\frac{9}{16}r{P}_1\left(\cos \theta \right)+\frac{15}{32}{r}^2{P}_2\left(\cos \theta \right)+\dots$

Hence the steady-state temperature distribution inside a sphere of radius 1:

$\frac{1}{4}{P}_0\left(\cos \theta \right)+\frac{9}{16}r{P}_1\left(\cos \theta \right)+\frac{15}{32}{r}^2{P}_2\left(\cos \theta \right)+\dots$.