Let’s solve a heat conduction problem. A region a ≤ r < ∞ in the cylindrical coordinate system is initially at a temperature F(r). For times t > 0 the boundary at r=a is kept insulated. Obtain an expression for the temperature distribution T(r, t) in the region for times t > 0.
Heat Conduction problem solving
I will solve the below problem:
A region a ≤ r < ∞ in the cylindrical coordinate system is initially at a temperature F(r). For times t > 0 the boundary at r=a is kept insulated. Obtain an expression for the temperature distribution T(r, t) in the region for times t > 0.
(sol)
Governing equation: $\frac{\partial^2 T}{\partial r } + \frac{1}{r} \frac{\partial T }{\partial r} = \frac{1}{\alpha} \frac{\partial T}{\partial t} $
BC: $\frac{\partial T}{\partial r } = 0 $ at $r=a$.
IC: $T(r,t=0)=F(r)$.
Let $T(r,t) = R(r) \Gamma(t)$.
Then, $$\frac{1}{R}\left( \frac{d^2 R}{ d r^2} + \frac{1}{r} \frac{d R}{d r} \right) = \frac{1}{\alpha \Gamma} \frac{ d \Gamma}{d t} $$
Let $\frac{ d \Gamma}{d t}=-\lambda^2$
$$ \Gamma(t) = \exp(-\alpha \lambda^2 t) $$
Then, $$ \frac{d^2 R}{ d r^2} + \frac{1}{r} \frac{d R}{d r} + R\lambda^2=0$$
Then, $R(r) = C_1 J_0 (\lambda r) + C_2 Y_0 (\lambda r)$. From BC, $C_1 = -C_2 \frac{Y_1(\lambda a)}{ J_1 (\lambda a)}$.
$$R(r) = – \frac{C_2}{J_1 (\lambda a) } \left[ J_0 (\lambda r ) Y_1 (\lambda a ) – Y_0 (\lambda r) J_1 (\lambda a) \right]$$
Let $R_0 (\lambda , r) = J_0 (\lambda r ) Y_1 (\lambda a ) – Y_0 (\lambda r) J_1 (\lambda a)$.
$$T(r,t) \int_0^\infty c(\lambda) R_0 (\lambda, r) d\lambda$$
$$F(r) = \int_a^\infty c(\lambda) R_0(\lambda, r) d\lambda$$
$$c(\lambda) = \frac{1}{N(\lambda)} \lambda \int_a^\infty r^\prime R_0(\lambda,r^\prime) F(r) d r^ \prime$$
$$N(\lambda) = J_1^2(\lambda a) + Y_1^2(\lambda a)$$
Reference: 6.7 in Hahn, D. W., & Özisik, M. N. (2012). Heat conduction. John Wiley & Sons.