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.

$\left\{\begin{array}{l}\begin{array}{cc}\mathbf{100}\mathbf{°}\mathbf{,}& \mathbf{0}\mathbf{<}\mathbf{\theta }\mathbf{<}\mathbf{\pi }\mathbf{/}\mathbf{3}\mathbf{,}\end{array}\\ \begin{array}{cc}\mathbf{0}\mathbf{°}\mathbf{,}& \mathbf{\text{otherwise.}}\end{array}\end{array}\mathbf{Hint}\mathbf{:}\mathbf{See}\mathbf{Problem}\mathbf{9}\mathbf{.8}\mathbf{of}\mathbf{Chapter}\mathbf{12}\mathbf{.}\right.$

Short answer

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

$u=100(\frac{1}{4}{p}_0\left(\cos \theta \right)+\frac{9}{16}r{p}_1\left(\cos \theta \right)+\frac{15}{32}{r}^2{p}_2\left(\cos \theta \right)+\dots \dots )$

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 a steady-state temperature.

Calculate the steady-state temperature distribution function:

Consider the equation below:

$\begin{array}{c}{u}_{r=1}=\left\{\begin{array}{l}100,0<\theta <\frac{\pi }{3}\\ 0°,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}otherwise}\end{array}\right.\\ =\left\{\begin{array}{l}100,\frac{1}{2}<x<1\\ 0°,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{otherwise}\end{array}\right.\\ =100f\left(x\right)\end{array}$

Here,

role="math" localid="1664355824351" $f\left(x\right)=\left\{\begin{array}{l}1,\frac{1}{2}<x<1\\ 0°,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}otherwise}\end{array}\right.$

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{1/2}{\overset{1}{\int }}dx\\ =\frac{1}{2}{\left[x\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\\ =\frac{1}{2}[1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.]\\ =\frac{1}{4}\end{array}$

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

$\begin{array}{c}{c}_1=\frac{3}{2}\underset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{1}{\int }}xdx\\ =\frac{3}{2}{\left[\frac{x^2}{2}\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\\ =\frac{3}{2}[\frac{1}{2}-\frac{1}{8}]\\ =\frac{9}{16}\end{array}$

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

$\begin{array}{c}{c}_2=\frac{5}{4}\underset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{1}{\int }}(3x^2-1)dx\\ =\frac{5}{4}{(3\frac{x^3}{3}-x)}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^1\\ =\frac{5}{4}(\frac{3}{3}-1-\frac{3}{12}+\frac{1}{2})\\ =\frac{15}{32}\end{array}$

Use the equation

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

$u=100(\frac{1}{4}{p}_0\left(\cos \theta \right)+\frac{9}{16}r{p}_1\left(\cos \theta \right)+\frac{15}{32}{r}^2{p}_2\left(\cos \theta \right)+\dots \dots )$

Hence, this is the steady-state temperature distribution inside a sphere of radius 1.