91Ó°ÊÓ

It has been given that the rectangular plate is covering the area $T=0$for $x=0$, $x=1$, $y=0$, and $T=1−x$for$y=2$.

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 steady state temperature distribution.

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

First 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$x=1$.

$\begin{array}{c}T(1,y)=(c_1{e}^{−ky}+c_2{e}^{ky})\left(c_4\sin k\right)\\ =0\end{array}$

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 that for $y=2T=0$then need $c_1$and$c_2$.

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

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

$T=\sinh \left(n\mathrm{Ï€}y\right)\sin \left(n\mathrm{Ï€}x\right)$

It 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.

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

Find the expression for$B_n$.

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

Find the value of$B_n$.

$\begin{array}{c}{b}_n=2{∫}_0^1(1−x)\sin \left(n\mathrm{π}x\right)dx\\ =2{[(1−x)\left(\frac{−\cos \left(n\mathrm{π}x\right)}{n\mathrm{π}}\right)−\frac{\sin \left(n\mathrm{π}x\right)}{n^2{\mathrm{π}}^2}]}_0^1\\ =2[(1−1)\left(\frac{−\cos n\mathrm{π}}{n\mathrm{π}}\right)−\frac{\sin \left(n\mathrm{π}\right)}{n^2{\mathrm{π}}^2}−(1−0)\left(\frac{−\cos 0}{n\mathrm{π}}\right)+\frac{\sin \left(0\right)}{n^2{\mathrm{π}}^2}]\\ =2[0−0−\left(\frac{−1}{n\mathrm{π}}\right)+0]\end{array}$

$b_n=\frac{2}{n\mathrm{Ï€}}$

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$T(x,y)$.

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

Hence, this is the steady-state temperature distribution.