One-dimensional transient heat conduction in cylinder

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A long cylinder with radius of $r o$ and a uniform initial temperature of $T i$ is exposed to a fluid with temperature of ${T_\infty }$ ( ${T_\infty } < {T_i}$ ). The convective heat transfer coefficient between the fluid and cylinder is $h$ . Assuming that there is no internal heat generation and constant thermophysical properties, obtain the transient temperature distribution in the cylinder.

Since the temperature changes along the r-direction only, the energy equation is

$\frac{{{\partial ^2}T}}{{\partial {r^2}}} + \frac{1}{r}\frac{{\partial T}}{{\partial r}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}},{\rm{ }}0 < x < {r_o},{\rm{ }}t > 0 \qquad \qquad(1)$

subject to the following boundary and initial conditions

$\frac{{\partial T}}{{\partial r}} = 0,{\rm{ }}r = 0{\rm{ (axisymmetric)}} \qquad \qquad(2)$ $- k\frac{{\partial T}}{{\partial r}} = h(T - {T_\infty }),{\rm{ }}r = {r_o} \qquad \qquad(3)$ $T = {T_i},{\rm{ }}0 < r < {r_o},{\rm{ }}t = 0 \qquad \qquad(4)$

Defining the following dimensionless variables

$\theta = \frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}},{\rm{ }}R = \frac{r}{{{r_o}}},{\rm{ Fo}} = \frac{{\alpha t}}{{r_o^2}},{\rm{ Bi}} = \frac{{h{r_o}}}{k} \qquad \qquad(5)$

eqs. (1) – (4) will be nondimensionalized as

$\frac{{{\partial ^2}\theta }}{{\partial {R^2}}} + \frac{1}{R}\frac{{\partial \theta }}{{\partial R}} = \frac{{\partial \theta }}{{\partial {\rm{Fo}}}},{\rm{ }}0 < R < 1,{\rm{ Fo}} > 0 \qquad \qquad(6)$ $\frac{{\partial \theta }}{{\partial R}} = 0,{\rm{ }}R = 0 \qquad \qquad(7)$ $- \frac{{\partial \theta }}{{\partial R}} = {\rm{Bi}}\theta ,{\rm{ R}} = 1 \qquad \qquad(8)$ $\theta = 1,{\rm{ }}0 < R < 1,{\rm{ Fo}} = 0 \qquad \qquad(9)$

Assuming that the temperature can be expressed as

$\theta (R,{\rm{Fo}}) = \Theta (R)\Gamma ({\rm{Fo}}) \qquad \qquad(10)$

and substituting eq. (10) into eq. (6), one obtains

$\frac{1}{\Theta }\left[ {\Theta '' + \frac{1}{R}\Theta '} \right] = \frac{{\Gamma '}}{\Gamma } = - {\lambda ^2} \qquad \qquad(11)$

which can be rewritten as the following two equations

$\Theta '' + \frac{1}{R}\Theta ' + {\lambda ^2}\Theta = 0 \qquad \qquad(12)$ $\Gamma ' + {\lambda ^2}\Gamma = 0 \qquad \qquad(13)$

Equation $\frac{{{d^2}\vartheta }}{{d{r^2}}} + \frac{1}{r}\frac{{d\vartheta }}{{dr}} - {m^2}\vartheta = 0,{\rm{ }}0 < r < {r_o}$ is a Bessel’s equation of zero order and has the following general solution

$\Theta (R) = {C_1}{J_0}(\lambda R) + {C_2}{Y_0}(\lambda R) \qquad \qquad(14)$

where $J 0$ and $Y 0$ are Bessel functions of the first and second kind, respectively. The general solution of eq. (13) is

$\Gamma = {C_3}{e^{ - {\lambda ^2}{\rm{Fo}}}} \qquad \qquad(15)$

where $C 1, C 2,$ and $C 3$ are integral constants. The boundary conditions for eq. (12) can be obtained by substituting eq. (10) into eqs. (7) and (8), i.e.,

$\Theta '(0) = 0 \qquad \qquad(16)$ $- \Theta '(1) = {\rm{Bi}}\Theta (1) \qquad \qquad(17)$

The derivative of $Θ$ is

$\Theta '(R) = - {C_1}\lambda {J_1}(\lambda R) - {C_2}\lambda {Y_1}(\lambda R) \qquad \qquad(18)$

Since ${J_1}(0) = 0{\rm{ and }}{Y_1}(0) = - \infty$ , $C 2$ must be zero. Substituting eqs. (14) and (18) into eq. (17) and considering $C 2 = 0,$ we have

$- {\lambda _n}{J_1}({\lambda _n}) + {\rm{Bi}}{J_0}({\lambda _n}) = 0 \qquad \qquad(19)$

where $n$ is an integer. The eigenvalue $λ n$ can be obtained by solving eq. (19) using an iterative procedure. The dimensionless temperature with eigenvalue $λ n$ is

${\theta _n} = {C_n}{J_0}\left( {{\lambda _n}R} \right){e^{ - \lambda _n^2{\rm{Fo}}}} \qquad \qquad(20)$

where $C n = C 1 C 3$ . Equation (20) is a solution that satisfies eqs. (6) – (8) but not eq. (9). For a linear problem, the sum of different $θ n$ for each value of $n$ also satisfies eqs. (6) – (8).

$\theta = \sum\limits_{n = 1}^\infty {{C_n}{J_0}\left( {{\lambda _n}R} \right){e^{ - \lambda _n^2{\rm{Fo}}}}} \qquad \qquad(21)$

Substituting eq. (21) into eq. (9) yields

$1 = \sum\limits_{n = 1}^\infty {{C_n}{J_0}\left( {{\lambda _n}R} \right)}$

Multiplying the above equation by $R{J_0}\left( {{\lambda _m}R} \right)$ and integrating the resulting equation in the interval of (0, 1), one obtains

$\int_0^1 {R{J_0}({\lambda _m}R)dR} = \sum\limits_{n = 1}^\infty {{C_n}\int_0^1 {R{J_0}({\lambda _m}R){J_0}({\lambda _n}R)dR} }$

According to the orthogonal property of Bessel’s function, the integral on the right-hand side equals zero if $m \ne n$ but it is not zero if $m = n .$ Therefore, we have

${C_m} = \frac{{\int_0^1 {R{J_0}({\lambda _m}R)dR} }}{{\int_0^1 {RJ_0^2({\lambda _m}R)dR} }} = \frac{2}{{{\lambda _m}}}\frac{{{J_1}({\lambda _m})}}{{J_0^2({\lambda _m}) + J_1^2({\lambda _m})}}$

Changing notation from $m$ to $n$ , we get

${C_n} = \frac{2}{{{\lambda _n}}}\frac{{{J_1}({\lambda _n})}}{{J_0^2({\lambda _n}) + J_1^2({\lambda _n})}} \qquad \qquad(22)$

thus, the dimensionless temperature becomes

$\theta = \sum\limits_{n = 1}^\infty {\frac{2}{{{\lambda _n}}}\frac{{{J_1}({\lambda _n}){J_0}\left( {{\lambda _n}R} \right)}}{{J_0^2({\lambda _n}) + J_1^2({\lambda _n})}}{e^{ - \lambda _n^2{\rm{Fo}}}}} \qquad \qquad(23)$

References

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.

External Links

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One-dimensional transient heat conduction in cylinder

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