Analytical properties of Graetz modes in parallel and concentric configurations

Abstract

The generalized Graetz problem refers to stationary convection–diffusion in uni-directional flows. In this contribution we demonstrate the analyticity of generalized Graetz solutions associated with layered domains: either cylindrical (possibly concentric) or parallel. Such configurations are considered as prototypes for heat exchangers devices and appear in numerous applications involving heat or mass transfer. The established framework of Graetz modes allows to recast the 3D resolution of the heat transfer into a 2D or even 1D spectral problem. The associated eigenfunctions (called Graetz modes) are obtained with the help of a sequence of closure functions that are recursively computed. The spectrum is given by the zeros of an explicit analytical series, the truncation of which allows to approximate the eigenvalues by solving a polynomial equation. Graetz mode computation is henceforth made explicit and can be performed using standard software of formal calculus. It permits a direct and mesh-less computation of the resulting solutions for a broad range of configurations. Some solutions are illustrated to showcase the interest of mesh-less analytical derivation of the Graetz solutions, useful to validate other numerical approaches.

Appendix

1.1 Cylindrical heated pipe case $Q\ne 0$

The solution provided by (20) is

$$\begin{aligned} T(r,z) = \frac{8}{\mathrm{Pe}}G(z)+\sum _{i\in \mathbb {Z}^\star } \alpha _i c_i(z) T_i(r)\, e^{\lambda _i z}, \end{aligned}$$

(33)

Rewritting (28) as

$$\begin{aligned} g(z)=H(z)H(1-z)\left( 1-\cos \left( 2\pi \left( z+\frac{1}{2}\right) \right) \right) \end{aligned}$$

with H(z) the Heaviside function, and using integration by parts leads to the primitive $G(z)=\int _{-\infty }^{z} g(z')dz'$ equals to

$$\begin{aligned} G(z)=H(z)H(1-z)\left( z-\frac{1}{2 \pi }\sin (2\pi \left( z+\frac{1}{2}\right) \right) . \end{aligned}$$

The function $c_{i}(z)$ in (33) are given by (23), the integration by part of which gives

$$\begin{aligned} c_{+i}(z)= g(z)\frac{e^{-\lambda _{i} z}}{\lambda _i}+\frac{H(1-z)}{\lambda _i^2+4\pi ^2}\left( \frac{\lambda }{2\pi }\sin (2\pi z) +\cos (2\pi z) \right) . \end{aligned}$$

The eigenfunctions $T_i$ are provided by the $\lambda$-analytical decomposition (10) upon functions $t_p(r)$ such that

$$\begin{aligned} T_i(r) = \sum _{p=0}^{N_p} t_p(r) \lambda _i^p, \end{aligned}$$

(34)

where each eigenvalue $\lambda _i$ of the discrete spectrum sets its eigenfunctions $T_i$ from (34). Here, Neumann adiabatic boundary condition at $R=2$, combined with—truncated—decomposition (34) provide a—hence finite—polynomial condition for $\lambda _i$ whose zeros are the approximated discrete spectrum. We hereby provide the first three elements of both downstream and upstream spectrum computed with finite truncation $N_p=20$ in (34) and parameter $Pe=1$, with a formal calculus Maple software: $\lambda _1 = 0.674240$, $\lambda _2 =3.306258$, $\lambda _3 =4.936416$, $\lambda _{-1} =0$, $\lambda _{-2} =-1.027741$, $\lambda _{-3} =-2.35726$. Function $t_p(r)$, $p\in \{0,5\}$ are also hereby given by the following piece-wise continuous analytic functions of r along the fluid-solid domains $r \in [0,1] \cup [1,2]$

$$\begin{aligned}&\left\{ \begin{array}{ll} {r \in [0,1]} &{}\quad t_{0}= 1\\ {r \in [1,2]} &{}\quad t_{0}= 1 \end{array} \right. \nonumber \\&\left\{ \begin{array}{ll} {r \in [0,1]} &{}\quad t_{1}= -\frac{5}{8} \,{r}^{4}+5/2\,{r}^{2} \\ {r \in [1,2]} &{}\quad t_{1}= \frac{15}{8}+\frac{5}{2}\,\ln \left( r \right) \end{array} \right. \nonumber \\&\left\{ \begin{array}{ll} {r \in [0,1]} &{}\quad t_{2}= \,-\frac{1}{4}\,{r}^{2}+\frac{25\,{r}^{4}}{16}-\frac{125\,{r}^{6}}{144}+\frac{25\,{r}^{8}}{256} \\ {r \in [1,2]} &{}\quad t_{2}= \frac{1825}{2304}-\frac{\,{r}^{2}}{4}+{\frac{175\,\ln \left( r \right) }{96}} \end{array} \right. \nonumber \\&\left\{ \begin{array}{ll} {r \in [0,1]} &{}\quad {t_{3}} = {-\frac{5\,{r}^{4}}{16}+\frac{25\,{r}^{6}}{48}-\frac{875\,{r}^{8}}{2304} +\frac{445\,{r}^{10}}{4608}-\frac{125\,{r}^{12}}{18432} } \\ {r \in [1,2]} &{}\quad {t_{3} } = {- \frac{4385}{18432}+\frac{5\,{r}^{2}}{32}+\frac{155\,\ln \left( r \right) }{4608} -\frac{5}{8}\,{r}^{2}\ln \left( r \right) } \end{array} \right. \nonumber \\&\left\{ \begin{array}{ll} {r \in [0,1]} &{}\quad {t_{4} } ={\frac{{r}^{4}}{64}-\frac{25\,{r}^{6}}{192} +\frac{1325\,{r}^{8}}{9216}-\frac{839\,{r}^{10}}{9216}+\frac{10975\,{r}^{12}}{331776}-\frac{3175\,{r}^{14}}{602112}+\frac{625\,{r}^{16}}{2359296}}\\ {r \in [1,2]} &{}\quad {t_{4} } ={-\frac{319528919}{1040449536} +\frac{2375\,{r}^{2}}{9216}-\frac{847715\,\ln \left( r \right) }{3096576}+\frac{{r}^{4}}{64}-\frac{175\,{r}^{2}\ln \left( r \right) }{384} } \end{array} \right. \nonumber \\&\left\{ \begin{array}{ll} {r \in [0,1]} &{}\quad {t_{5}} = {\frac{5\,{r}^{6}}{384} -\frac{95\,{r}^{8}}{3072}+\frac{575\,{r}^{10}}{18432} -\frac{3755\,{r}^{12}}{221184}+\frac{51755\,{r}^{14}}{8128512}-\frac{779375\,{r}^{16}}{520224768}+\frac{3201125\,{r}^{18}}{18728091648}-\frac{125\,{r}^{20}}{18874368}} \\ {r \in [1,2]} &{}\quad {t_{5} } = {-\frac{2789680345}{74912366592} +\frac{5005\,{r}^{2}}{73728}-\frac{9747175\,\ln \left( r \right) }{231211008}-\frac{15\,{r}^{4}}{512}-\frac{155\,{r}^{2}\ln \left( r \right) }{18432}+\frac{5\,{r}^{4}\ln \left( r \right) }{128} } \end{array} \right. \end{aligned}$$

(35)

Finally, each parameter $\alpha _i$ of (33) is given by (22) using the closure function $T_i(R=2)$ and its corresponding eigenvalue $\lambda _i$.

1.2 Parallel configuration and balanced case $Q=0$

The theoretical solution detailed in Sect. 4.1.3 is hereby detailed. From (25) we recall the temperature solution

$$\begin{aligned} T(r,z)= a G(z)+g(z)aT_0+\sum _{i\in \mathbb {Z}^\star } \alpha _i c_i(z) T_i(r)\, e^{\lambda _i z}, \end{aligned}$$

(36)

involving the constant a given in (32) and the function g(z) given in (28). Rewritting (28) as

$$\begin{aligned} g(z)=H(z)H(1-z)\left( 1-\cos \left( 2\pi \left( z+\frac{1}{2}\right) \right) \right) , \end{aligned}$$

with H(z) the Heaviside function, and using integration by parts leads to a primitive $G(z)=\int _{-\infty }^{z} g(z')dz'$ equals to

$$\begin{aligned} G(z)=H(z)H(1-z)\left( z-\frac{1}{2 \pi }\sin \left(2\pi \left(z+\frac{1}{2}\right)\right)\right) . \end{aligned}$$

Again, the functions $c_{i}(z)$ are given by (23), the integration by part of which gives

$$\begin{aligned} c_{+i}(z)= g(z)\frac{e^{-\lambda _{i} z}}{\lambda _i}+\frac{H(1-z)}{\lambda _i^2+4\pi ^2}\left( \frac{\lambda }{2\pi }\sin \left( 2\pi z\right) +\cos (2\pi z) \right) \end{aligned}$$

The eigenfunctions $T_i$ are provided by the $\lambda$-analytical decomposition (10) upon functions $t_p(r)$ such that

$$\begin{aligned} T_i(r) = \sum _{p=0}^{N_p} t_p(r) \lambda _i^p, \end{aligned}$$

(37)

where each eigenvalue $\lambda _i$ of the discrete spectrum sets its eigenfunctions $T_i$ from (37). Here again, Neumann boundary condition at $R=\pm 2$, combined with—truncated—decomposition (37) provide a polynomial condition for $\lambda _i$ the zeros of which are the approximated discrete spectrum. We hereby provide the five first elements of this spectrum computed with finite truncation $N_p=20$ in (37) and parameter $Pe=50$, computed with a formal calculus Maple software. $\lambda _1 = -1.738793$, $\lambda _2 = -1.738793$, $\lambda _3 = -1.585275$, $\lambda _4 =-1.3093020$, $\lambda _5 = -1.011529$. Function $t_p(r)$, $p\in \{0,5\}$ are also hereby given by the following piece-wise continuous polynomial functions of r along the various solid-fluid domains $[-2,-1] \cup [-1,0] \cup [0,1] \cup [1,2]$. Starting with $t_0=1$ identically equal to 1, we recursively compute the following (polynomial) functions $t_p(r)$

$$\begin{aligned}&\left\{ \begin{array}{llll} {r \in [-2,-1]} &{}\quad t_{1}= 0 \\ {r \in [-1,0]} &{}\quad t_{1}= 25\, \left( r-1 \right) \left( 1+r \right) ^{3} \\ {r \in [0,1]} &{}\quad t_{1}= -25\,{r}^{4}+50\,{r}^{3}-50\,r-25 \\ {r \in [1,2]} &{}\quad t_{1}= -50 \end{array} \right. \end{aligned}$$

(38)

$$\begin{aligned}&\left\{ \begin{array}{lllll} {r \in [-2,-1]} &{}\quad t_{2}= -\frac{1}{2}\, \left( r+2 \right) ^{2} \\ {r \in [-1,0]} &{}\quad t_{2}= {\frac{1347}{14}}+{\frac{2236\,r}{7}}-\frac{{r}^{2}}{2}-1250\,{r}^{3} -1875\,{r}^{4}-750\,{r}^{5}+500\,{r}^{6}+{\frac{3750\,{r}^{7}}{7}}+ {\frac{1875\,{r}^{8}}{14}}\\ {r \in [0,1]} &{}\quad t_{2} = {\frac{1347}{14}}+{\frac{2236\,r}{7}}-\frac{{r}^{2}}{2}-1250\,{r}^{3}-625\,{r}^{4}+750\,{r}^{5}+500\,{r}^{6}-{\frac{3750\,{r}^{7}}{7}}+ {\frac{1875\,{r}^{8}}{14}} \\ {r \in [1,2]} &{}\quad t_{2}= 1248-{\frac{13014\,r}{7}}-\frac{{r}^{2}}{2} \end{array} \right. \end{aligned}$$

(39)

$$\begin{aligned}&\left\{ \begin{array}{llll} {r \in [-2,-1] } &{}\quad {t_{3}= 0}\\ {r \in [-1,0]} &{}\quad {t_{3}= } {\frac{5(1+r)^{3} \left( -11661 -20261\,r+96921\,{r}^{2}+226961\,{r}^{3}+8750\,{r}^{4}-362250\,{r}^{5}-322000\,{r}^{6}-18500\,{r}^{7}+84375\,{r}^{8}+28125\,{r}^{9}\right) }{462} }\\ {r \in [0,1]} &{}\quad {t_{3} =} {-\frac{19435}{154}-\frac{19730\,r}{33}+\frac{25 r^2}{2}+\frac{101200{r}^{3}}{21}+\frac{78125{r}^{4}}{14}-\frac{33610{r}^{5}}{7}-\frac{74965{r}^{6}}{6}+\frac{31250{r}^{7}}{7}+\frac{103125{r}^{8}}{14}-\frac{3125{r}^{9}}{3} -\frac{72500{r}^{10}}{21}}\\ &{}\qquad {+\,\frac{140625{r}^{11}}{77}-\frac{46875{r}^{12}}{154} } \\ {r \in [1,2]} &{}\quad {t_{3}=} {\frac{14580}{11} -100\,r+25\,{r}^{2}} \end{array} \right. \end{aligned}$$

(40)

$$\begin{aligned}&\left\{ \begin{array}{llll} {r \in [-2,-1] } &{}\quad {t_{4}=} {\frac{1}{24}\, \left( r+2 \right) ^{4}}\\ {r \in [-1,0]} &{}\quad {t_{4}= } {\frac{21071161}{504504}+\frac{3288262\,r}{9009}-\frac{1347\,{r}^{2}}{28} -\frac{209989\,{r}^{3}}{33}-\frac{33452423\,{r}^{4}}{1848}-\frac{95900\,{r}^{5}}{11}+\frac{2031875\,{r}^{6}}{42}+\frac{15973250\,{r}^{7}}{147}+\frac{2275625\,{r}^{8}}{28}-\frac{2028125\,{r}^{9}}{63}}\\ &{}\qquad {-\,\frac{4179325\,{r}^{10}}{36}-\frac{7640625\,{r}^{11}}{77}-\frac{2265625\,{r}^{12}}{66}+\frac{421875\,{r}^{13}}{91}+\frac{60968750\,{r}^{14}}{7007} +\frac{234375\,{r}^{15}}{77}+ \frac{234375\,{r}^{16}}{616} } \\ {r \in [0,1]} &{}\quad {t_{4} =} {+\frac{21071161}{504504}+\frac{3288262\,r}{9009}-\frac{1347\,{r}^{2}}{28}-\frac{209989\,{r}^{3}}{33}-\frac{21791423\,{r}^{4}}{1848}+\frac{101400\,{r}^{5}}{11}+\frac{2019625\,{r}^{6}}{42}+\frac{796750\,{r}^{7}}{147}-\frac{10902625\,{r}^{8}}{196}-\frac{2018875\,{r}^{9}}{63} }\\ &{}\qquad {+\,\frac{14244725\,{r}^{10}}{252}+\frac{609375\,{r}^{11}}{77}-\frac{8828125\,{r}^{12}}{462}-\frac{421875\,{r}^{13}}{91}+\frac{60968750\,{r}^{14}}{7007}-\frac{234375\,{r}^{15}}{77}+\frac{234375\,r^{16}}{616} } \\ {r \in [1,2]} &{} \quad {t_{4}=} {-\frac{22888172}{1617}+\frac{1164110834\,r}{63063} -624\,{r}^{2}+\frac{2169\,{r}^{3}}{7} } \end{array} \right. \end{aligned}$$

(41)

$$\begin{aligned}&\left\{ \begin{array}{llll} {r \in [-2,-1] } &{}\quad t_{5}= 0\\ {r \in [-1,0]} &{}\quad t_{5}= {\frac{5\, \left( 1+r \right) ^{3}}{162954792} \left( 1594383325 -582590385\,r-978868182\,{r}^{2}+74397416190\,{r}^{3}+111713337741\,{r}^{4}+ 410289881841\,{r}^{5}-1279496305500\,{r}^{6} \right. }\\ &{}\qquad {-\,741396373500\,{r}^{7}+1797598889250\,{r}^{8}+3506620554250\,{r}^{9}+1894060853250\,{r}^{10}-1117168905750\,{r}^{11}} \\ &{}\qquad {\left. -2202785812500\,{r}^{12} -1228317562500\,{r}^{13}-156926250000\,{r}^{14}+140323125000\,{r}^{15}+68527265625\,{r}^{16}+ 9789609375\,{r}^{17} \right) } \\ {r \in [0,1]} &{}\quad t_{5} = {\frac{7971916625}{162954792}+\frac{3500466325\,r}{27159132}+ \frac{19435\,{r}^{2}}{308}+\frac{50174095\,{r}^{3}}{22932}+\frac{2038131125\,{r}^{4}}{252252} -\frac{25775835\,{r}^{5}}{4004}-\frac{19508095\,{r}^{6}}{308}-\frac{250061555\,{r}^{7}}{6468} }\\ &{}\qquad {+\,\frac{132625865\,{r}^{8}}{1176}+\frac{37396250\,{r}^{9}}{231} -\frac{251021875\,{r}^{10}}{1764}-\frac{269196250\,{r}^{11}}{1617} +\frac{1040385625\,{r}^{12}}{19404}+\frac{170490000\,{r}^{13}}{1001}-\frac{74039375\,{r}^{14}}{924} } \\ &{}\qquad {-\,\frac{62421875\,{r}^{15}}{1617}+\frac{434609375\,{r}^{16}}{24024}+\frac{7008984375\,{r}^{17}}{476476}-\frac{4114843750\,{r}^{18}}{357357}+\frac{17578125\,{r}^{19}}{5852}-\frac{3515625\,{r}^{20}}{11704} }\\ {r \in [1,2]} &{}\quad t_{5}= {-\frac{96144058760}{20369349}+\frac{83080\,r}{33} -\frac{7290\,{r}^{2}}{11}+\frac{50\,{r}^{3}}{3}-\frac{25\,{r}^{4}}{12}} \end{array} \right. \end{aligned}$$

(42)

Parameter $\alpha _i$ of (36) is given by (22) using closure function $T_i(R=2)$ and its corresponding eigenvalue $\lambda _i$.