# One-dimensional transient heat conduction in cylinder

A long cylinder with radius of ro and a uniform initial temperature of Ti 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 J0 and Y0 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 C1,C2, and C3 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 ${J_1}(0) = 0{\rm{ and }}{Y_1}(0) = - \infty$, C2 must be zero. Substituting eqs. (14) and (18) into eq. (17) and considering C2 = 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 λn can be obtained by solving eq. (19) using an iterative procedure. The dimensionless temperature with eigenvalue λn is ${\theta _n} = {C_n}{J_0}\left( {{\lambda _n}R} \right){e^{ - \lambda _n^2{\rm{Fo}}}} \qquad \qquad(20)$

where Cn = C1C3. Equation (20) is a solution that satisfies eqs. (6) – (8) but not eq. (9). For a linear problem, the sum of different θn 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.