이번에 풀어볼 열전달(heat transfer) 문제는 아래와 같다.

A 1-D slab, $0 ≤ x ≤ L$, is initially at a temperature of $F(x)$. For times t > 0, both of the boundary surfaces are perfectly insulated. Obtain an expression for the temperature $T(x,t)$ in the slab. Clearly show the steady-state temperature in your expression. Independently derive the steady-state temperature from simple conservation of energy and show that this agrees with your above expression.

## 열전달(heat transfer) 문제풀이

이번에 풀어볼 열전달(heat transfer) 문제는 아래와 같다.

A 1-D slab, $0 ≤ x ≤ L$, is initially at a temperature of $F(x)$. For times t > 0, both of the boundary surfaces are perfectly insulated. Obtain an expression for the temperature $T(x,t)$ in the slab. Clearly show the steady-state temperature in your expression. Independently derive the steady-state temperature from simple conservation of energy and show that this agrees with your above expression.

문제를 풀어보도록 하자.

Heat Diffusion Equation: $\frac{\partial^2 T}{\partial x^2}= \frac{1}{\alpha} \frac{\partial T}{\partial t}$

Boundary Conditions: $\frac{\partial T}{\partial x} = 0 $ at $x=0,L$

Initial Condition: $T(x,0)=F(x)$

Let $T(x,t) = X(x) \Gamma(t)$. From the heat diffusion equation,

$\begin{align*} \frac{1}{X} \frac{\partial ^2 X}{\partial x^2} = \frac{1}{\partial \Gamma} \frac{\partial \Gamma}{\partial t} = – \lambda^2 \end{align*} $

$X(x) = C_1 \cos \lambda x + C_2 \sin \lambda x $

$\frac{\partial X}{\partial x} = – \lambda C_1 \sin \lambda x + \lambda C_2 \cos \lambda x $

$\frac{\partial X}{\partial x} =C_2 \lambda=0 $ at $x = 0 $, then $C_2 =0$

$\frac{\partial X}{\partial x} = -C_1 \lambda \sin \lambda L =0 $ at $ x = L$, then $\lambda_n = \frac{n\pi}{L}, n=0,1,2,…$.

$X_n(x) = \cos \lambda_n x$

From $\frac{1}{\alpha \Gamma_n} \frac{ \partial \Gamma_n}{\partial t} = – \lambda_n^2$,

$$ \Gamma_n(t) = \exp(-\alpha \lambda_n^2 t) $$

$$T(x,t) = \sum_{n=0}^\infty C_n X_n(x) \Gamma_n(t) = C_0 + \sum_{n=0}^\infty X_n(x) \Gamma_n(t).$$

From $T(x,0)=F(x)$,

$$ C_n = \frac{\int_0^L \cos \lambda_n x F(x) dx }{\int_0^L \cos^2 \lambda_n x dx}$$

$$ C_0 = \frac{\int_0^L F(x) dx }{ L }$$

Therfore, $\lim_{t\to \infty} T(x,t) = C_0 = \frac{1}{L} \int_0^L F(x) dx$ is a steady temperature.

Since $T(x,0)=F(x)$ and the slab is insulated, steady temperature is a mean of $F(x)$.

출처- 문제 3.2 in Hahn, D. W., & Özisik, M. N. (2012). Heat conduction. John Wiley & Sons.

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