---
# System prepended metadata

title: CH 4. Note
tags: [熱傳學]

---

###### 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

![](https://i.imgur.com/4Ou5Bxk.png)

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