91Ó°ÊÓ

It has been given that the rectangular plate is covering the area $0<x<1$, $0<y<2$, if $T=0$for $x=0$, $x=1$ and $y=2$and $\text{T=1-x}$for$y=0$.

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$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{Θ}\mathbf{\left(}\mathbf{θ}\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Assume that the plate is so long compared to its width that the mathematical approximation can be made that it extends to infinity in the y direction, the temperature T satisfies Laplace's equation inside the plate that is${∇}^{2}T=0$.

To solve this equation, try a solution of the form mentioned below.

$\begin{array}{l}T(x,y)=X\left(x\right)Y\left(y\right)\\ Y\frac{d^{2}X}{d{x}^2}+X\frac{d^{2}Y}{d{y}^2}=0\end{array}$

Divide the above equation by XY.

$\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}=0$

Now, following the process of separation of variables it can be written

$\begin{array}{c}\frac{1}{X}\frac{d^{2}X}{d{x}^2}=−\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}\\ =\text{constant}\\ =−k^2,\text{ â¶Ä‰â¶Ä‰}kâ‰\yen 0\end{array}$

$X^{\text{'}\text{'}}=−k^{2}X$and $Y^{\text{'}\text{'}}=k^{2}Y$

Here, the constant $k^2$is called the separation constant. The solutions for these equations.

$\begin{array}{l}X=\left\{\begin{array}{l}\sin \left(kx\right)\\ \cos \left(kx\right)\end{array}\right.\\ Y=\left\{\begin{array}{l}{e}^{ky}\\ e^{−ky}\end{array}\right.\end{array}$

Since none of the four solutions satisfies the given boundary temperatures, take a combination of the previous solutions.

Write the steady state temperature distribution.

$T=(c_1{e}^{−ky}+c_2{e}^{ky})(c_3\cos kx+c_4\sin kx)$

Set the value of the constants $c_1,c_2,c_3$and $c_4$. So, if $T=0$for $x=0$.

$\begin{array}{c}T(0,y)=(c_1{e}^{−ky}+c_2{e}^{ky})\left(c_3\right)\\ =0\end{array}$

Since an exponential can't be zero, $c_3=0$. If $T=0$for$x=1$.

$T(1,y)=(c_1{e}^{−ky}+c_2{e}^{ky})\left(c_4\sin k\right)$

It is known that $\sin kx=0$if $kx=n\mathrm{Ï€},$where $n=0,1,2..,$therefore for $x=1$there is $k=n\mathrm{Ï€}.$Finally, there is for $y=2T=0$ then $c_1$and $c_2$are needed.

$\begin{array}{c}(c_1{e}^{−ky}+c_2{e}^{ky})=\frac{1}{2}(e^{k(y−2)}−e^{−k(2−y)})\\ =\sinh k(2−y)\end{array}$

Thus $e^{ky}$, there is $c_1=\frac{1}{2}{e}^{2k}$and $c_2=−\frac{1}{2}{e}^{−2k}.$Thus for any integral n, the solution.

$c_2=−\frac{1}{2}{e}^{−2k}â‹\dots T=\sinh n\mathrm{Ï€}(2−y)\sin n\mathrm{Ï€}x$.

The above equation satisfies the given boundary conditions on the three $T=0$sides.

The $y=0$condition is not satisfied by any value of n. But a linear combination of solutions is a solution. Thus, write an infinite series of T.

$T(x,y)=\underset{n=1}{\overset{\mathrm{∞}}{∑}}{B}_n\sinh n\mathrm{π}(2−y)\sin n\mathrm{π}x$

Find the expression for$B_n$.

$\begin{array}{c}T(x,0)=1−x=\underset{n=1}{\overset{\mathrm{∞}}{∑}}{B}_n\sinh \left(2n\mathrm{Ï€}\right)\sin \left(n\mathrm{Ï€}x\right)\\ =\underset{n=1}{\overset{\mathrm{∞}}{∑}}{b}_n\sin \left(n\mathrm{Ï€}x\right)\text{ â¶Ä‰â¶Ä‰}(b_n=B_n\sinh 2n\mathrm{Ï€})\end{array}$

Find the value of$B_n$.

$\begin{array}{c}{b}_n=2{∫}_0^1(1−x)\sin \left(n\mathrm{π}x\right)dx\\ =2{[(1−x)\left(\frac{−\cos n\mathrm{π}x}{n\mathrm{π}}\right)−\frac{\sin \left(n\mathrm{π}x\right)}{n^2{\mathrm{π}}^2}]}_0^1\\ =\frac{2}{n\mathrm{π}}\end{array}$

Hence $B_n$is derived as mentioned below.

$\begin{array}{c}{B}_n=\frac{b_n}{\sinh \left(2n\mathrm{Ï€}\right)}\\ =\frac{2}{n\mathrm{Ï€}\sinh \left(2n\mathrm{Ï€}\right)}\end{array}$

Finally, substitute $B_n$into the distribution $T(x,y)$.

$T(x,y)=\underset{n=1}{\overset{\mathrm{∞}}{∑}}\frac{2}{n\mathrm{π}\sinh 2n\mathrm{π}}\sinh n\mathrm{π}(2−y)\sin \left(n\mathrm{π}x\right)$.

Hence, the steady-state temperature distribution is obtained as below.

$T(x,y)=\underset{n=1}{\overset{\mathrm{∞}}{∑}}\frac{2}{n\mathrm{π}\sinh 2n\mathrm{π}}\sinh n\mathrm{π}(2−y)\sin \left(n\mathrm{π}x\right)$