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I tried to solve this problem:

I have a tube of inner and outer radius $R_1$ and $R_2$ (like coppers tube for fridges). The temperature at the exterior of the tube ($\rho = R_2$) is maintained to $T_e$. We denote $T_1$ the temperature inside the tube and $T_2$ the temperature between $R_1$ and $R_2$. The diffusivity coefficients of the material are $k_1$ (like the fluid inside the tube) and $k_2$ (copper coefficient). Copper tube, fluid inside, maintained temperature to <span class=$T_e$ outside the tube" />

So the heat equations are:

$$ \left[ k_i \frac{1}{\rho} \frac{\partial}{\partial \rho} \left( \rho \frac{\partial}{\partial \rho} \right) - \frac{\partial}{\partial t} \right] T_i = 0 $$

then boundary conditions are:

$$ T_1(R_1,t) = T_2(R_1,t) $$ $$ l_1 \frac{\partial}{\partial \rho} T_1(\rho = R_1,t) = l_2 \frac{\partial}{\partial \rho} T_2(\rho = R_1,t) $$ $$ T_2(R_2,t) = T_e $$

with $l_i$ the thermal conductivity, and initial condition is:

$$T_1(\rho,0) = T_2(\rho,0) = T_0$$

By separation of variables we find general solutions:

$$ T_1 = A + \sum_{n = 1} a_n e^{- \lambda_n t} J_0 \left( \sqrt{\lambda_n/k_1} \rho \right) $$ $$ T_2 = B + \sum_{n = 1} b_n e^{- \mu_n t} J_0 \left( \sqrt{\mu_n/k_2} \rho \right) + c_n e^{- \nu_n t} Y_0 \left( \sqrt{\nu_n/k_2} \rho \right) $$

as $t \rightarrow + \infty $ the steady state is reached so logically we have: $ A = B = T_e $ but I have difficulties to find the eigenvalues $\lambda_n$, $\mu_n$ and $\nu_n$ and the coefficients of the series. Does anyone know the solution of this problem?

What I know

For a simpler case, a disk of radius $R$, with initial temperature inside the disk $T_0$ and boundary temperature $T(R,t) = T_e$ we find the solution:

$$ T(\rho,t) = T_e + \sum_{n=1}^{+ \infty} a_n J_0 \left( \sqrt{\lambda_n/k_1} \rho \right) e^{- \lambda_n t} $$

with the $\lambda_n$ solving (linked to the zeros of $J_0$ Bessel function):

$$ J_0 \left( \sqrt{\lambda_n/k_1} R \right) = 0 $$

and coefficients $a_n$ are (using weighted scalar product):

$$ a_n = (T_0 - T_e) \frac{ \int_{0}^{R_1} \rho J_0 \left( \sqrt{\lambda_n/k_1} \rho \right) d \rho}{ \int_{0}^{R_1} \rho J_0^2 \left( \sqrt{\lambda_n/k_1} \rho \right) d \rho } $$

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  • $\begingroup$ I have on idea, using the 3 B.C and assuming $\nu_n = \lambda_n = \mu_n$ to get rid of exponential time terms, I found a determinental equation on $\lambda_n$ which gives the $\lambda_n$ (involving of course numerical calculations). To have the temperature inside, $T_1$ (so the coefficients $a_n$), I can integrate on radius $R_n$, which correspond to $J_0(R_n \sqrt(\lambda_n/k_1)) = 0$ and I get rid of the infinite sum because of orthogonality. What do you think folks? $\endgroup$
    – Fefetltl
    Dec 31, 2021 at 15:17

1 Answer 1

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I found it finally using this (complicated) paper:

http://verl.npre.illinois.edu/Documents/J-08-01.pdf

As I said in comment we need to apply BC and we find a matrix equality to zero with coefficients, which gives a zero determinant equation, able to find the $\lambda_n$ eigenvalues.

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