###### tags: `熱傳學`
# CH 4. Note
## 4.1 General Considerations and Solution Techniques
- Determine the temperature distribution in the medium

For two-dimensiona, steady-state conditions with no generation and constant thermal conducitivity, from Eq. 2.22 we obtain

$\dfrac{\partial ^2 T}{\partial x^2}+\dfrac{\partial ^2 T}{\partial y^2}=0$

- Using this equation to solve for $T(x,y)$, then determine the heat flux components $q_x''$ and $q_y''$

## 4.2 The Method of Separation of Variables


Assuming negligible heat transfer from the surfaces of the plate of the ends of the rod, temperature gradients normal to the x-y plane may be neglected $(\partial ^2 T/\partial z^2\approx 0)$

- Transformation $\theta \equiv \dfrac{T-T_1}{T_2-T_1}\quad\rightarrow\quad\dfrac{\partial^2\theta}{\partial x^2}+\dfrac{\partial^2\theta}{\partial y^2}=0$

- Boundary conditions

$\begin{aligned}
\theta(0,y)&=0\ \ and\quad\theta(x,0)=0\\
\theta(L,y)&=0\ \ and\quad\theta(x,W)=1
\end{aligned}$

- Assume the existence of a solution of the form

$\theta(x,y)=X(x)\bullet Y(y)$$\qquad\rightarrow\ \ \dfrac{1}{X}\dfrac{d^2X}{dx^2}=\dfrac{1}{Y}\dfrac{d^2Y}{dy^2}$

The equality can apply only if both sides are equal to the same constant

$\begin{aligned}
\dfrac{d^2 X}{dx^2}+\lambda^2X=0\\
\dfrac{d^2 Y}{dy^2}-\lambda^2Y=0
\end{aligned}$

the general form of the two-dimensional solution

$\theta=(C_1\ cos\lambda x+C_2\ sin\lambda x)\ (C_3e^{-\lambda y}+C_4e^{\lambda y})$

### Applying the conditions
- $\theta(0,y)=0$ and $\theta(x,0)=0$

we have $C_1=0$ and $C_3=-C_4$

- Applying $\theta(L,y)=0$, we obtain

$C_2C_4\ sin\lambda L\ (e^{\lambda y}-e^{-\lambda y})=0$

the condition is satisfied by requiring that $sin\ \lambda L=0$

$\lambda=\dfrac{n\pi}{L}\ \ n=1, 2, 3, ...$

- The desired solution may be expressed as

$\theta =C_2C_4\ sin\dfrac{n\pi x}{L}\ (e^{n\pi y/L}-e^{-n\pi y/L})$