91Ó°ÊÓ

It has been given that a slab of thickness 10cm has two faces at ${10}^{°}$and${20}^{°}$.

Laplace’s equation in cylindrical coordinates is,

$\begin{array}{c}{\mathbf{∇}}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{∂}\mathbf{u}}{\mathbf{∂}\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{u}}{\mathbf{∂}{\mathbf{θ}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{u}}{\mathbf{∂}{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{Θ}\mathbf{\left(}\mathbf{θ}\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

The initial steady-state temperature $u_0$satisfies Laplace's equation, which in this one-dimensional case is$\raisebox{1ex}{$d^2{u}_0$}\!\left/ \!\raisebox{-1ex}{$d{x}^2$}\right.=0$.

The solution of this equation is $u_0=ax+b,$where a and b are constants that must be found to fit the given conditions.

It has been given that $u_0=10$at x = 0 and $u_0=20$at x = l.

${\text{u}}_{\text{0}}\text{=x+10}$ ….. (1)

The flow equation is mentioned below.

${∇}^{2}u=\frac{1}{{\mathrm{Î\pm }}^2}\frac{∂u}{∂t}$

Here, u is the temperature and ${\mathrm{Î\pm }}^2$is a constant characteristic of the material through which heat is flowing. Assume a solution of the form mentioned below.

$u=F\left(x\right)T\left(t\right)$

Replacing this into the differential equation.

$\frac{1}{F}{∇}^{2}F=\frac{1}{{\mathrm{Î\pm }}^2}\frac{1}{T}\frac{dT}{dt}$

The left side of this identity is a function only of the space variable x, and the right side is a function only of time. Therefore, both sides are the same constant.

$\frac{dT}{dt}=−k^2{\mathrm{Î\pm }}^{2}T$

The time equation can be integrated to give the equation mentioned below.

$T=e^{−k^2{\mathrm{Î\pm }}^{2}t}$

For the space equation.

$\frac{d^{2}F}{d{x}^2}+k^{2}F=0$

The above equation is the solutions of the equation mentioned below.

$F\left(x\right)=\left\{\begin{array}{l}\sin \left(kx\right)\\ \cos \left(kx\right)\end{array}\right.$

Given the boundary conditions of the problem. Discard the $\cos \left(kx\right)$solution. Thus eigenfunctions.

$u=e^{−{(n\mathrm{Ï€}\mathrm{Î\pm }/10)}^{2}t}\sin \left(\frac{n\mathrm{Ï€}x}{10}\right)$

The solution of our problem will be the series as given below.

$u=u_f+\underset{n=1}{\overset{\mathrm{∞}}{∑}}{b}_n{e}^{−{(n\mathrm{Ï€}\mathrm{Î\pm }/10)}^{2}t}\sin \frac{n\mathrm{Ï€}x}{10}$

At t = 0 there is a need of $u=u_0$.

$\begin{array}{c}u=\underset{n=1}{\overset{\mathrm{∞}}{∑}}{b}_n\sin \frac{n\mathrm{π}x}{l}\\ =u_0\\ =x+10−u_f\end{array}$

This means finding the Fourier sine series for $x+10$on $(0,\text{10})$.

$\begin{array}{c}{b}_n=\frac{2}{10}\underset{0}{\overset{10}{∫}}(x+10)\sin (n\mathrm{π}x/10)dx\\ =\frac{2}{10}[(x+10)\underset{0}{\overset{10}{∫}}\sin (n\mathrm{π}x/10)dx−\underset{0}{\overset{10}{∫}}\left[\frac{d}{dx}(x+10)\underset{0}{\overset{10}{∫}}\sin (n\mathrm{π}x/10)dx\right]dx]\\ =\frac{2}{10}{[−(x+10)\left[\frac{\cos (n\mathrm{π}x/10)}{n\mathrm{π}}\right]−\left[\frac{sin(n\mathrm{π}x/10)}{{\left(n\mathrm{π}\right)}^2}\right]]}_0^{10}\\ =−\frac{40}{\left(n\mathrm{π}\right)}\text{;forneven}\end{array}$

Now, replace $b_n$and $u_0$.

$u=20−x−\frac{40}{\mathrm{Ï€}}\underset{\begin{array}{c}2\\ \text{ven}n\end{array}}{\overset{\mathrm{∞}}{∑}}\frac{1}{n}{e}^{−{(n\mathrm{Ï€}\mathrm{Î\pm }/10)}^{2}t}\sin \frac{n\mathrm{Ï€}x}{10}$

Hence, the solution is found to be$u=20−x−\frac{40}{\mathrm{Ï€}}\underset{\begin{array}{c}2\\ \text{ven}n\end{array}}{\overset{\mathrm{∞}}{∑}}\frac{1}{n}{e}^{−{(n\mathrm{Ï€}\mathrm{Î\pm }/10)}^{2}t}\sin \frac{n\mathrm{Ï€}x}{10}$.