[Heat Conduction Problem Solving 7.4] A solid cylinder, 0 ≤ r ≤ b, is initially at zero temperature. For times t > 0, the boundary condition at r = b is the convection condition given as ∂T /∂x + H T = f (t), where f (t) is a function of time. Obtain an expression for the temperature distribution T(r,t) in the cylinder for times t > 0

Let’s solve a heat conduction problem.A solid cylinder, 0 ≤ r ≤ b, is initially at zero temperature. For times t > 0, the boundary condition at r = b is the convection condition given as ∂T /∂x + H T = f (t), where f (t) is a function of time. Obtain an expression for the temperature distribution T(r,t) in the cylinder for times t > 0

Heat Conduction problem solving

I will solve the below problem:

A solid cylinder, 0 ≤ r ≤ b, is initially at zero temperature. For times t > 0, the boundary condition at r = b is the convection condition given as ∂T /∂x + H T = f (t), where f (t) is a function of time. Obtain an expression for the temperature distribution T(r,t) in the cylinder for times t > 0

(sol)
$$ \frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial t} = \frac{1}{\alpha} \frac{\partial T}{\partial t} \text{ in } 0 \leq r \leq b, t>0$$
$$BC1: T(r\to 0) \text{ is finite}, BC2: \frac{\partial T}{\partial r} + HT = f(t) \text{ at } r=b, IC:T(t=0) =0$$
Auxiliary Problem is below as:
$$ \frac{\partial^2 \Phi}{\partial r^2} + \frac{1}{r} \frac{\partial \Phi}{\partial t} = \frac{1}{\alpha} \frac{\partial \Phi}{\partial t} \text{ in } 0 \leq r \leq b, t>0$$
$$BC1: \Phi(r\to 0) \text{ is finite}, BC2: \frac{\partial \Phi}{\partial r} + H\Phi = 1 \text{ at } r=b, IC:\Phi(t=0) =0$$
Let $\Phi(r,t) = \Phi_H(r,t) + \Phi_{SS}(r)$.
$$ \frac{\partial^2 \Phi_H}{\partial r^2} + \frac{1}{r} \frac{\partial \Phi_H}{\partial t} = \frac{1}{\alpha} \frac{\partial \Phi_H}{\partial t} \text{ in } 0 \leq r \leq b, t>0$$
$$BC1: \Phi_H(r\to 0) \text{ is finite}, BC2: \frac{\partial \Phi_H}{\partial r} + H\Phi_H = 0 \text{ at } r=b, IC:\Phi_H(t=0) =0$$
$$ \frac{\partial^2 \Phi_{SS}}{\partial r^2} + \frac{1}{r} \frac{\partial \Phi_{SS}}{\partial t} = 0 \text{ in } 0 \leq r \leq b$$
$$BC1: \Phi_{SS}(r\to 0) \text{ is finite}, BC2: \frac{\partial \Phi_{SS}}{\partial r} + H\Phi_{SS} = 1 \text{ at } r=b$$
Then, $\Phi_{SS}(r) = -\frac{1}{H}$.
Let $\Phi_{H}(r,t) = R(r)\Gamma(t)$.
$$ \frac{1}{R}\left( \frac{d^2 R}{dr^2} + \frac{1}{r} \frac{dR}{dr}\right) = \frac{1}{\alpha \Gamma} \frac{d \Gamma}{dt} = – \lambda^2$$
$$R(r) = C_1 J_0 ( \lambda r) + C_2 Y_0 (\lambda r)$$
$\Phi_H(r \to 0 )$ is finite means that $C_2=0$.
From $\frac{\partial \Phi_{SS}}{\partial r} + H\Phi_{SS} = 1$, $-\lambda J_1 (\lambda b) + H J_0 (\lambda b)=0$
and $\lambda_n (n=1,2,3,…)$ are eigenvalues and $\lambda_n \neq 0$.
$$\Phi_H (r,t) = \sum_{n=1}^\infty C_n J_0 (\lambda_n r) e^{-\alpha \lambda_n^2 t}$$
Since $\Phi_H(r,0) = – \frac{1}{H}$,
$$C_n = -\frac{1}{H} \frac{ \int_0^b r J_0 (\lambda_n r) dr}{ \int_0^b rJ_0^2 (\lambda_nr) dr}$$
$$\Phi(r,t) = \Phi_H(r,t) + \Phi_{SS}(r)$$
$$T(r,t) = \int_{\tau = 0 }^t f(\tau) \frac{ \partial \Phi (x,t-\tau)}{\partial \tau} d \tau$$