열전달(Heat Transfer) 문제풀이해보겠습니다. 이번에 풀려고 하는 문제는 아래와 같습니다.
A long, hollow cylinder, $a \leq r \leq b$, is initially at a temperature of $T = F(r)$. For times $t > 0$ the boundaries at $r = a$ and $r = b$ are kept insulated. Obtain an expression for temperature distribution $T(r, t)$ in the solid for times $t > 0.$
열전달(Heat Transfer) 문제풀이
다시 한번 remind 하면 제가 풀려고 하는 문제는 아래와 같습니다.
A long, hollow cylinder, $a \leq r \leq b$, is initially at a temperature of $T = F(r)$. For times $t > 0$ the boundaries at $r = a$ and $r = b$ are kept insulated. Obtain an expression for temperature distribution $T(r, t)$ in the solid for times $t > 0.$
풀이)
$$IC: T(r,0)=F(r), BC1: \frac{\partial T}{\partial r} = 0 \text{ at } r=a, BC2: \frac{\partial T}{\partial r} = 0 \text{ at } r=b $$
Heat diffusion equation:
$$\frac{\partial^2 T}{\partial r} + \frac{1}{r} \frac{\partial T}{\partial r} = \frac{1}{\alpha} \frac{ \partial T}{\partial t}$$
Let $T(r,t) = R(r) \Gamma(t) $.
$$ \frac{1}{R} \left[ \frac{\partial^2 R}{\partial r^2} + \frac{1}{r} \frac{\partial R}{\partial r} \right]= \frac{1}{\alpha \Gamma} \frac{ \partial \Gamma} { \partial t} = – \lambda^2$$
$$ \frac{\partial^2 R}{\partial r^2} + \frac{1}{r} \frac{\partial R}{\partial r}+\lambda^2 R =0 $$
$$R(r) = C_1 J_0 ( \lambda r) + C_2 Y_0 ( \lambda r) $$
$$\frac{dR}{dr} = -C_1 \lambda J_1 (\lambda r) – C_2 \lambda Y_1 (\lambda r) $$
From BC1, BC2
$$-C_1 \lambda J_1 (\lambda r) – C_2 \lambda Y_1 (\lambda r) = 0 \text{ for } r=a,b$$
$$-\frac{C_1}{C_2} = \frac{Y_1(\lambda a)}{J_1(\lambda a)} = \frac{Y_1 (\lambda b)}{J_1 (\lambda b)} $$
$$ \lambda_n > 0 (n=1,2,…) \text{ is an eigen value iff } J_1(\lambda_n b) Y_1 (\lambda_na) -J_1(\lambda_n a) Y_1 (\lambda_n b) = 0 $$
$$R_n(r) = C_n(Y_1(\lambda_n a) J_0 (\lambda_n r) – J_1 (\lambda_n a) Y_0 (\lambda_n r)$$
From $\frac{1}{\alpha \Gamma} \frac{\partial \Gamma}{\partial t} = -\lambda^2$,
$$\Gamma_n(t) = \exp (-\alpha \lambda_n^2 t) $$
$$T(r,t) = \sum_{n=1}^\infty R_n(r)\Gamma_n(t)$$
From the IC
$$F(r) = \sum_{n=1}^\infty R_n(r) $$
$$C_n =\frac{\int_a^b rF(r)(Y_1(\lambda_n a) J_0 (\lambda_n r)-J_1 (\lambda_n a) Y_0(\lambda_n r)) dr}{\int_a^b r(Y_1(\lambda_n a) J_0 (\lambda_n r)-J_1 (\lambda_n a) Y_0(\lambda_n r))^2 dr}$$
출처- 문제 4.1 in Hahn, D. W., & Özisik, M. N. (2012). Heat conduction. John Wiley & Sons.
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