열전달(Heat Transfer) 문제풀이해보겠습니다. 이번에 풀려고 하는 문제는 아래와 같습니다.

A 1-D slab,$0\leq x \leq L$, is initially at zero temperature. For times $t > 0$, the boundary at $x = 0$ is kept insulated, the boundary at x = L is kept at zero temperature, and there is internal energy generation within the solid at a constant rate of $g_0$ (W/m3). Obtain an expression for the temperature distribution $T(x, t)$ in the slab for times $t > 0.$

## 열전달(Heat Transfer) 문제풀이

다시 한번 remind 하면 제가 풀려고 하는 문제는 아래와 같습니다.

A 1-D slab,$0\leq x \leq L$, is initially at zero temperature. For times $t > 0$, the boundary at $x = 0$ is kept insulated, the boundary at x = L is kept at zero temperature, and there is internal energy generation within the solid at a constant rate of $g_0$ (W/m3). Obtain an expression for the temperature distribution $T(x, t)$ in the slab for times $t > 0.$

Heat diffusion equation:

$$ \frac{\partial^2 T}{\partial x^2} + \frac{1}{k} g_0 = \frac{1}{\alpha} \frac{\partial T}{ \partial t} $$

$$BC1: \frac{\partial T}{\partial x} = 0 \text{ at } x=0, BC2: T=0 \text{ at } x=L IC: T(x,0)=0 $$

Let $T(x,t) = T_H(x,t) + T_{SS}(x)$.

PDE for $T_H$:

$$\frac{\partial ^2 T_H}{\partial x^2} = \frac{1}{\alpha} \frac{\partial T_H}{\partial t}$$

$$BC1: \frac{\partial T_H}{\partial x} = 0 \text{ at }x=0, BC2: T_H=0 \text{ at } x=L, T_H(x,0)=-T_{SS}(x) $$

PDE for $T_{SS}$:

$$\frac{\partial ^2 T_{SS}}{\partial x^2} + \frac{1}{k}g_0 = 0 $$

$$BC1: \frac{\partial T_{SS}}{\partial x } = 0 \text{ at } x=0, BC2: T_{SS}=0 \text{ at } x=L $$

From the PDE about $T_{SS}$,

$$ T_{SS}(x) = -\frac{1}{2k}g_0 x^2 + C_1 x + C_2 $$

From the BC1, $C_1=0$.

From the BC2, $C_2 = \frac{1}{2k}g_0 L^2 $.

$$T_{SS}(x) = – \frac{1}{2k}g_0 x^2 + \frac{1}{2k}g_0 L^2 $$

Let $T_H(x,t) = X(x) \Gamma(t)$.

From the PDE about $T_H$,

$$ \frac{1}{X} \frac{\partial ^2 X}{\partial x^2 } = \frac{1}{ \alpha \Gamma} \frac{ \partial \Gamma}{\partial t} = – \lambda^2 $$

$$X(x) = C_3 \cos \lambda x + C_4 \sin \lambda x$$

From the BC1 about $T_H$, $C_4=0$.

From the BC2 about $T_H$, $\cos \lambda L = 0 \to \lambda_n L = \frac{\pi}{2} + n \pi, n=0,1,2,…$

$$T_H(x,t) = \sum_{n=0}^\infty C_n \cos \lambda_n x \exp (- \alpha \lambda_n^2 t) $$

From the IC about $T_H$,

$$-T_{SS}(x) = \sum_{n=0}^\infty C_n \cos \lambda_n x $$

$$C_n = \frac{-\int_0^L T_{SS}(x) \cos \lambda x dx }{ \int_{0}^L \cos^2\lambda_n x dx} = \frac{2g_0}{Lk\lambda_n^3} (-1)^{n+1} $$

$$T(x,t) = T_H(x,t) + T_{SS}(x) $$

$$T(x,t) = \sum_{n=0}^\infty C_n \cos \lambda_n x \exp(-\alpha \lambda_n^2 t ) -\frac{1}{2k} g_0 ( x^2 -L^2)$$

출처- 문제 3.10 in Hahn, D. W., & Özisik, M. N. (2012). Heat conduction. John Wiley & Sons.

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