[Heat Conduction Problem Solving 5.1] A hollow sphere a ≤ r ≤ b is initially at temperature T = F (r). For times t > 0, the boundary surface at r = a is kept insulated, and the boundary at r = b dissipates heat by convection with convection coefficient h into a medium at zero temperature. Obtain an expression for the temperature distribution T(r, t) in the sphere for times t > 0.

Let’s solve a heat conduction problem.A hollow sphere a ≤ r ≤ b is initially at temperature T = F (r). For times t > 0, the boundary surface at r = a is kept insulated, and the boundary at r = b dissipates heat by convection with convection coefficient h into a medium at zero temperature. Obtain an expression for the temperature distribution T(r, t) in the sphere for times t > 0.

Heat Conduction problem solving

I will solve the below problem:

A hollow sphere a ≤ r ≤ b is initially at temperature T = F (r). For times t > 0, the boundary surface at r = a is kept insulated, and the boundary at r = b dissipates heat by convection with convection coefficient h into a medium at zero temperature. Obtain an expression for the temperature distribution T(r, t) in the sphere for times t > 0.

(sol)
$$\frac{\partial^2 T}{\partial r^2} + \frac{2}{r} \frac{\partial T}{\partial r} = \frac{1}{\alpha} \frac{\partial T}{\partial t}$$
BC1: $\frac{\partial T}{\partial r} = 0 $ at $r=a$, BC2: $-k \frac{\partial T}{\partial r} = h T $ at $r=b$, IC: $T(t=0)=F(r)$
$U(r,t) = rT(r,t)$
$$\frac{\partial^2 U}{\partial r^2} = \frac{1}{\alpha } \frac{\partial U}{\partial t}$$
BC1: $\frac{\partial U}{\partial r}- \frac{1}{\alpha} U =0$, BC2: $\frac{\partial U}{\partial r} + K U = 0 $ at $r=b$ where $K= (\frac{h}{k}-\frac{1}{b})$, IC: $U(t=0)=rF(r)$.
Let $U(r,t)=R(r)\Gamma(t)$.
$$ \frac{1}{R} \frac{\partial^2 R}{\partial r^2} = \frac{1}{\alpha \Gamma} \frac{\partial \Gamma}{\partial t} = – \lambda^2$$
$$\Gamma(t) = C_1 e^{-\alpha \lambda^2 t}$$
$$R(r) = C_2 \cos \lambda r + C_3 \sin \lambda r$$
From BC1,
$$\left(-\lambda \sin \lambda a – \frac{1}{a} \cos \lambda a\right) C_2 + \left(\lambda \cos \lambda a – \frac{1}{a} \sin \lambda a\right) C_3 = 0$$
$$C_2 = C_3 \tan (\theta_\lambda – \lambda a)$$
Where $\theta_\lambda = \arctan (\lambda a)$.
From BC2, eigenvalue $\lambda_n$ is obtained. If $\lambda_n=0$, then $C_2=C_3=0$. Therefore $\lambda_n \neq 0$.
$$U(r,t) = \sum_{n=1}^\infty C_n \left[ \cos (\lambda_n r) \sin(\theta_{\lambda_n}-\lambda_n a)+\sin(\lambda_nr) \cos(\theta_{\lambda_n}-\lambda_n a)\right]e^{-\alpha \lambda_n^2 t}$$
Then, $$U(r,t) = \sum_{n=1}^\infty C_n \sin (\lambda_n r – \lambda_n a + \theta_{\lambda_n})e^{-\alpha \lambda_n^2 t}$$
From IC, $U(r,0)=rF(r)$ and
$$ C_n = \frac{\int_a^b rF(r)\sin({\lambda_n}(r-a) + \theta_{\lambda_n})dr}{\int_a^b \sin^2({\lambda_n}(r-a) + \theta_{\lambda_n})dr}$$
$$T(r,t) = \frac{1}{r} U(r,t)$$