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It has been given that the rectangular plate is covering the area $T=30掳$for $x=0$and $y=3$; $T=20掳$for $y=0$and$x=5$.

Laplace鈥檚 equation in cylindrical coordinates is,

$\begin{array}{c}{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{鈭�}\mathbf{u}}{\mathbf{鈭�}\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{u}}{\mathbf{鈭�}{\mathbf{胃}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{u}}{\mathbf{鈭�}{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

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

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${鈭�}^{2}T=0$.

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

$\begin{array}{c}\frac{1}{X}\frac{d^{2}X}{d{x}^2}=鈭�\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}\\ =\text{constant}\\ =鈭�k^2,\text{鈥夆赌夆赌�}k鈮�0\end{array}$

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

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

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

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

Subtract $20掳$so you end up with the sum of two cases. Set the first case where $T=0$but for $x=0$, and another case where $T=0$but for$y=3$.

Solution 1:

Thus, the possible solution for Y(x).

$Y\left(y\right)=\sinh \left(ky\right)$

Thus, the possible solutions for X(x).

$X\left(x\right)=\sin \left(kx\right)$

The boundary condition $T=10掳$for $y=3$gives us the equation mentioned below.

$\begin{array}{c}T(x,y)=\underset{n=1}{\overset{\mathrm{鈭�}}{鈭�}}{c}_n\sin (n蟺x/5)\sin (n蟺3/5)\\ =10\end{array}$

The Fourier coefficient.

$c_n=\frac{40}{蟺}\frac{1}{\sinh (3蟺n/5)}$

Thus, the solution is as mentioned below.

$T_1(x,y)=\frac{40}{蟺}\underset{\text{oddn}}{\overset{\mathrm{鈭�}}{鈭�}}\frac{1}{\sinh (3蟺n/5)}\sin (n蟺y/5)\cos (n蟺x/5)$

Solution 2:

Thus, the possible solution for Y(x).

$Y\left(y\right)=\sin \left(ky\right)$

Thus, the possible solutions for X(x).

$X\left(x\right)=\sinh k(5鈭�x)$

The boundary condition $T=10掳$for $x=0$gives us the equation mentioned below.

$T(x,y)=\underset{n=1}{\overset{\mathrm{鈭�}}{鈭�}}{c}_n\sinh (n蟺5/3)\sin (n蟺y/3)=10$

The Fourier coefficient.

$c_n=\frac{40}{蟺}\frac{1}{\sinh (5蟺n/3)}$

Thus, the solution is as mentioned below.

$T_2(x,y)=\frac{40}{蟺}\underset{oddn}{\overset{\mathrm{鈭�}}{鈭�}}\frac{1}{\sinh (蟺n5/3)}\sinh (蟺n(5鈭�x)/3)\sin (n蟺y/3)$

Combine the result and the solution becomes as mentioned below.

$\begin{array}{c}T(x,y)=20+\frac{40}{蟺}\underset{oddn}{\overset{\mathrm{鈭�}}{鈭�}}\frac{1}{\sinh (3蟺n/5)}\sin (n蟺y/5)\cos (n蟺x/5)\\ +\frac{40}{蟺}\underset{\text{oddn}}{\overset{\mathrm{鈭�}}{鈭�}}\frac{1}{\sinh (蟺n5/3)}\sinh (蟺n(5鈭�x)/3)\sin (n蟺y/3)\end{array}$

Hence, this is the steady-state temperature distribution.