Analysis of Two-Component Non-Equilibrium Model of Linear Reactive Chromatography

Abstract

This article presents semi-analytical solutions and analytical temporal moments of a two-component linear reactive lumped kinetic model incorporating irreversible ($A\rightarrow B$) and reversible ($A\leftrightarrows B$) reactions in a fixed-bed liquid chromatographic column. Both solid and liquid phase reactions and two sets of boundary conditions are considered. The current model equations contain a coupled system of two partial differential equations (PDEs) and two ordinary differential equations (ODEs). The solution methodology successively employs the Laplace transform and linear transformation steps to uncouple the governing set of coupled differential equations. The resulting system of uncoupled ODEs is solved by applying an elementary solution technique. The numerical Laplace inversion is employed to transform back the solutions in the actual time domain. To further analyze the effects of different kinetic parameters, statistical temporal moments are derived from the Laplace-transformed solutions. The current solutions extend and generalize our recent solutions for single-solute transport models of non-reactive liquid chromatography. For verification, the analytical results are compared with the numerical solutions of a high-resolution finite volume scheme. Several case studies of practical interest are considered. Good agreements in the results validate the correctness of semi-analytical solutions and the accuracy of proposed numerical algorithm.

Appendices

Appendix

Analytical Moments

Here, the analytical temporal moments are presented for two different sets of boundary conditions. For the derivation of the moments, $c_{i,\mathrm init}=0\,\mathrm{g}/\mathrm{l}$ (for $i=1,2$), $\eta _i=0$, and $c_{2,\mathrm inj}=0\,\mathrm{g}/\mathrm{l}$ are considered, i.e. we are considering an empty column initially and injecting only component 1 into the reactor. Moreover, only the effect of solid-phase reaction is taken into account, while the liquid-phase reaction is neglected.

Irreversible Reaction and Dirichlet BCs

Here, we neglect the liquid-phase reaction, i.e. $\eta _1=0\,\mathrm{min}^{-1}$. Eqs. (60) and (61) are used to derived the moments $\mu _n^{(i)}$ of Laplace-transformed solutions given in Eqs. (32) and (35) for $i=1,2$ and $n=0,1,2,3$. Let us define

$$\begin{aligned} r&=\,a_1F\widetilde{\nu }_1 \mathrm{Pe},\quad \gamma =\sqrt{\mathrm{Pe}(\mathrm{Pe}+4a_1F\nu ) },\quad \delta _{1,2}=\mathrm{Pe}\mp \gamma ,\nonumber \\ \chi _{1,2}&=\frac{\gamma \mathrm{Pe}^4}{64}+\frac{3\gamma ^3 \mathrm{Pe}^2}{32}+\frac{\gamma ^5}{64}\mp \ r^2\mathrm{Pe}\mp \frac{3rPe^3}{4}\mp \frac{ \mathrm{Pe}^5}{8},\nonumber \\ \chi _{3,4}&=\frac{\gamma ^3\mathrm{Pe}}{8}\mp \ \frac{\gamma ^2(\mathrm{Pe}^2+2r)}{8},\,\,\, \chi _5=\frac{\gamma ^3(\mathrm{Pe}^2-4r)-\gamma \mathrm{Pe}^4}{32},\,\,\,\chi _6:=F(a_1-a_2)\mathrm{Pe}+(1+a_2F)r,\nonumber \\ \chi _{7,8}&=\frac{\mathrm{Pe}^2\ \,\mathrm{e}^{\mathrm{Pe}}}{2}\biggr (\biggr [\frac{F(a_1-a_2)\mathrm{Pe}}{4}+\frac{r(1+a_1F)}{8}\biggr ]\gamma ^3+\frac{rPe^2(1+a_1F) \gamma }{8}\mp \frac{F(a_1-a_2)\mathrm{Pe}^4}{4}\nonumber \\&\quad \mp \,\frac{3Fr(a_1-a_2)\mathrm{Pe}^2}{2}\mp \frac{r(1+a_1F)^3\mathrm{Pe}}{4}\mp \ r^2(1+a_1F)\mathrm{Pe}\mp \ r^2[1+(3a_1-2a_2)F]\biggr ). \end{aligned}$$

(64)

Using Eq. (60), the zeroth moments are given as

$$\begin{aligned} \mu ^{(1)}_{0}=C_{1,\mathrm inj}\tau _\mathrm{inj}\,\mathrm{e}^{\frac{\delta _1}{2}},\quad \mu ^{(2)}_{0}=C_{1,\mathrm inj}\tau _\mathrm{inj}\left( 1-\,\mathrm{e}^{\frac{\delta _1}{2}}\right) . \end{aligned}$$

(65)

From Eq. (65), we get $\mu ^{(1)}_0+\mu ^{(2)}_0=C_{1,\mathrm inj}\tau _\mathrm{inj}$, as $C_{2,\mathrm inj}=0$ is considered.

The first moments are calculated using the Eq. (61) for $n=1$:

$$\begin{aligned} \mu ^{(1)}_{1}&=\frac{\tau _\mathrm{inj}}{2}+\frac{\mathrm{Pe}(1+a_1F)}{\gamma } ,\end{aligned}$$

(66)

$$\begin{aligned} \mu ^{(2)}_1&=\frac{\tau _\mathrm{inj}}{2}+\frac{\mathrm{Pe}[\gamma F(a_2-a_1)-r(1+a_1F)]e^\frac{\delta _1}{2}-\gamma [F(a_2- a_1) \mathrm{Pe}-r(1+a_2F)]}{r\gamma \left( 1-e^\frac{\delta _1}{2}\right) }. \end{aligned}$$

(67)

The second moments are expressed as

$$\begin{aligned} \mu ^{(1)}_{2}=\frac{\tau ^2_\mathrm{inj}}{3}+\frac{\mathrm{Pe}(1+a_1F)\tau _\mathrm{inj}}{\gamma }+\frac{\mathrm{Pe}^2(1+a_1F)(\gamma +2)}{\gamma ^3}+\frac{2Pea_1F (1-\epsilon )}{\gamma \widetilde{k}_1}, \end{aligned}$$

(68)

$$\begin{aligned} \mu ^{(2)}_2&=\frac{\tau ^2_\mathrm{inj}}{3}+\left[ \frac{\mathrm{Pe}\Bigr (\gamma F(a_2-a_1)-r(1+a_1F)\Bigl )e^\frac{\delta _1}{2}-\gamma \Bigr (F(a_2- a_1) \mathrm{Pe}-r(1+a_2F)\Bigl )}{r\gamma \Bigr (1-e^\frac{\delta _1}{2}\Bigl )}\right] \tau _\mathrm{inj}\nonumber \\&\quad +\,\frac{1}{1-\,\mathrm{e}^{\frac{\delta _1}{2}}}\left( e^\frac{\delta _1}{2} \left[ -\frac{2Pea_1F(1-\epsilon )}{\gamma \widetilde{k}_1}-\frac{2PeF(1-\epsilon )(a_1\widetilde{k}_2-a_2\widetilde{k}_1)}{r\widetilde{k}_1\widetilde{k}_2}-\frac{2\mathrm{Pe}^2(1+ a_1F)^2}{\gamma ^3}-\,\frac{\mathrm{Pe}^2F^2(a_2-a_1)^2}{r^2}\right] \right. \nonumber \\&\quad \left. +\,\frac{2PeF(1-\epsilon )(a_1\widetilde{k}_2-a_2\widetilde{k}_1)}{r\widetilde{k}_1\widetilde{k}_2}+\frac{\mathrm{Pe}^2F^2(a_2- a_1)^2}{r^2} +\frac{2(1+a_2F)^2}{\mathrm{Pe}}+\frac{2a_2F(1-\epsilon )}{\widetilde{k}_2}\right) \nonumber \\&\quad -\,\frac{\mathrm{Pe}^2[ \gamma F(a_2-a_1)-r(1+a_1F)]^2e^\frac{\delta _1}{2}-\gamma ^2[F (a_2-a_1) \mathrm{Pe}-r(1+a_2F)]^2}{r^2\gamma ^2\left( 1-e^\frac{\delta _1}{2}\right) } . \end{aligned}$$

(69)

The above equations are helpful to calculate the second central moments using the relations:

$$\begin{aligned} {\mu '}^{(i)}_2=\mu ^{(i)}_2-\left( \mu ^{(i)}_1\right) ^2,\quad i=1,2. \end{aligned}$$

(70)

Thus, the second central moments are given as

$$\begin{aligned} {\mu '}^{(1)}_2&=\frac{\tau ^2_\mathrm{inj}}{12}+\frac{2\mathrm{Pe}\left[ \gamma ^2a_1 F(1-\epsilon )+\mathrm{Pe}\widetilde{k}_1(1+a_1F)^2\right] }{\gamma ^3 \widetilde{k}_1},\end{aligned}$$

(71)

$$\begin{aligned} {\mu '}^{(2)}_2&=\frac{\tau ^2_\mathrm{inj}}{12}+\,\frac{1}{1-\,\mathrm{e}^{\frac{\delta _1}{2}}}\left( e^\frac{\delta _1}{2} \bigg [-\frac{2Pea_1F(1-\epsilon )}{\gamma \widetilde{k}_1}-\frac{2PeF(1-\epsilon )(a_1\widetilde{k}_2-a_2 \widetilde{k}_1)}{r\widetilde{k}_1\widetilde{k}_2}-\frac{2\mathrm{Pe}^2 (1+a_1F)^2}{\gamma ^3}\right. \nonumber \\&\quad -\,\left. \frac{\mathrm{Pe}^2F^2(a_2-a_1)^2}{r^2}\bigg ] +\,\frac{2PeF(1-\epsilon )(a_1\widetilde{k}_2-a_2\widetilde{k}_1)}{r\widetilde{k}_1\widetilde{k}_2}+\frac{\mathrm{Pe}^2F^2 (a_2-a_1)^2}{r^2} +\frac{2(1+a_2F)^2}{\mathrm{Pe}}+\frac{2a_2F(1-\epsilon )}{\widetilde{k}_2}\right) \nonumber \\&\quad -\,\left[ \frac{\mathrm{Pe}^2 [\gamma F(a_2-a_1)-r(1+a_1F)]^2e^\frac{\delta _1}{2}-\gamma ^2[ F(a_2-a_1) \mathrm{Pe}-r(1+a_2F)]^2}{r^2\gamma ^2\left( 1-e^\frac{\delta _1}{2}\right) }\right] \nonumber \\&\quad -\,\left[ \frac{\mathrm{Pe}[\gamma F(a_2-a_1)-r(1+a_1F)]e^\frac{\delta _1}{2}-\gamma [F(a_2- a_1) \mathrm{Pe}-r(1+a_2F)]}{r\gamma \left( 1-e^\frac{\delta _1}{2}\right) }\right] ^2. \end{aligned}$$

(72)

The third moments are given as

$$\begin{aligned} \mu ^{(1)}_3&=\frac{\tau ^3_\mathrm{inj}}{4}+\frac{\mathrm{Pe}(1+a_1F)\tau ^2_\mathrm{inj}}{\gamma } +\left[ \frac{\mathrm{Pe}^2(1+a_1F)^2(\gamma +2)}{\gamma ^3}+\frac{2Pea_1F(1- \epsilon )}{\gamma \widetilde{k}_1}\right] \frac{3\tau _\mathrm{inj}}{2}\nonumber \ \\&\quad +\,\frac{\mathrm{Pe}^3(1+a_1F)^3(\gamma ^2+6\gamma +12)}{\gamma ^5}+\frac{6\mathrm{Pe}^2 a_1F(1+a_1F)(\gamma +2)(1-\epsilon )}{\gamma ^3\widetilde{k}_1} + \frac{6Pea_1F(1-\epsilon )^2}{\gamma \widetilde{k}^2_1}, \end{aligned}$$

(73)

$$\begin{aligned} \mu ^{(2)}_3&=\frac{\tau ^3_\mathrm{inj}}{4}\,+\,\Biggr (\frac{\mathrm{Pe}[\gamma F(a_2-a_1)-r(1+a_1F)]e^\frac{\delta _1}{2}-\gamma [F(a_2- a_1) \mathrm{Pe}-r(1+a_2F)]}{r\gamma \left( 1-e^\frac{\delta _1}{2}\right) }\Biggr )\tau ^2_\mathrm{inj}\nonumber \\&\quad +\,\frac{3\tau _\mathrm{inj}}{2\left( 1-\,\mathrm{e}^{\frac{\delta _1}{2}}\right) }\biggr (\,\mathrm{e}^{ \frac{\delta _1}{2}}\Bigg [ \frac{-2\mathrm{Pe}^2(1+a_1F)^2}{\gamma ^3}-\frac{\mathrm{Pe}^2F^2(a_2-a_1)^2}{r^2}- \frac{2Pea_1F(1-\epsilon )}{\gamma \widetilde{k}_1}+\nonumber \\&\quad -\,\frac{2PeF(1-\epsilon )(a_1\widetilde{k}_2-a_2\widetilde{k}_1)}{r\widetilde{k}_1\widetilde{k}_2}\Biggr ] +\frac{\mathrm{Pe}^2F^2(a_2-a_1)^2}{r^2}+\frac{2PeF(1-\epsilon )(a_1 \widetilde{k}_2-a_2\widetilde{k}_1)}{r\widetilde{k}_1\widetilde{k}_2} \nonumber \\&\quad +\,\frac{2(1+a_2F)^2}{\mathrm{Pe}}+\frac{2a_2F(1-\epsilon )}{\widetilde{k}_2} \biggr )\nonumber \ \\&\quad -\,\left( \frac{\mathrm{Pe}^2[\gamma F(a_2-a_1)-r(1+a_1F)]^2e^\frac{\delta _1}{2}-\gamma ^2[F (a_2-a_1) \mathrm{Pe}-r(1+a_2F)]^2}{2r^2\gamma ^2\left( 1-e^\frac{\delta _1}{2}\right) }\right) 3\tau _\mathrm{inj}\nonumber \\&\quad -\,\frac{6\mathrm{Pe}^3F^3(a_2-a_1)^3}{r^3}-\frac{1}{\left( 1-\,\mathrm{e}^{\frac{\delta _1}{2}} \right) }\left[ \frac{\mathrm{Pe}^3\,\mathrm{e}^{\frac{\delta _1}{2}}(1+a_1F)^3(\gamma ^2+ 6\gamma +12)}{\gamma ^5}\right. \nonumber \\&\quad \left. \,-\,\frac{3FPe^3\,\mathrm{e}^{\frac{\delta _1}{2}}(1+a_1F)^2(a_2-a_1)(\gamma +2)}{r\gamma ^3}+\frac{6F^2\mathrm{Pe}^3(1+a_1F)(a_2-a_1)^2}{r^2\gamma }\right. \nonumber \\&\quad \left. +\frac{6F(1+a_2F)^2(a_2-a_1)}{r}-\frac{(\mathrm{Pe}^2+12)(1+a_2F)^3}{\mathrm{Pe}^2}\right] \nonumber \\&\quad +\,\frac{12\mathrm{Pe}^2F^2(a_2-a_1) (1-\epsilon )(a_2\widetilde{k}_1-a_1\widetilde{k}_2)}{r^2\widetilde{k}_1\widetilde{k}_2} -\frac{\,\mathrm{e}^{\frac{\delta _1}{2}}\mathrm{Pe}^2(1-\epsilon )}{(1-\,\mathrm{e}^{\frac{\delta _1}{2}})}\left[ \frac{12a_1F(1+a_1F)^2}{\gamma ^3\widetilde{k}_1}\right. \nonumber \\ &\quad \left. -\frac{6a_1F^2 (a_2-a_1)}{r\gamma \widetilde{k}_1}+\frac{6a_1F(1+a_1F)}{\gamma ^2\widetilde{k}_1} -\frac{6F(1+a_1F)(a_2\widetilde{k}_1-a_1\widetilde{k}_2)}{r\gamma \widetilde{k}_1\widetilde{k}_2}\right] \nonumber \\&\quad -\,\frac{1}{(1-\,\mathrm{e}^{\frac{\delta _1}{2}})}\left[ \frac{-12a_2F(1+a_2F)(1-\epsilon )}{\mathrm{Pe}\widetilde{k}_2}-\frac{6\mathrm{Pe}^2F^2(1+a_2F)(a_2-a_1)^2}{r^2}\right] \nonumber \\&\quad -\,\frac{6PeF(1- \epsilon )^2(a_2\widetilde{k}^2_1-a_1\widetilde{k}^2_2) }{r\widetilde{k}^2_1\widetilde{k}^2_2}-\frac{1}{(1-\,\mathrm{e}^{\frac{\delta _1}{2}})}\left[ \frac{6Pea_1Fe^{\frac{\delta _1}{2}}(1-\epsilon )^2}{\gamma \widetilde{k}^2_1}+\frac{6Pea_2F^2(1-\epsilon )(a_2-a_1)}{r \widetilde{k}_2}\right. \nonumber \\&\quad \left. +\,\frac{3PeF(1+a_2F)^2(a_2-a_1)}{r}+\frac{6PeF(1- \epsilon )(1+a_2F)(a_2\widetilde{k}_1-a_1\widetilde{k}_2)}{r \widetilde{k}_1 \widetilde{k}_2}\right] \nonumber \\&\quad -\,\frac{1}{\left( 1-\,\mathrm{e}^{\frac{\delta _1}{2}}\right) }\left[ \frac{-6(1+a_2F)^3}{\mathrm{Pe}}-\frac{6a_2F(1-\epsilon )^2}{\widetilde{k}^2_2} -\frac{6a_2F(1-\epsilon )(1+a_2F)}{\widetilde{k}_2}\right] . \end{aligned}$$

(74)

Finally, the third central moments can be deduced from the given relations as

$$\begin{aligned} {\mu '}^{(i)}_3=\mu ^{(i)}_3-3\mu ^{(i)}_1\mu ^{(i)}_2+2\left( \mu ^{(i)}_1\right) ^3,\quad i=1,2. \end{aligned}$$

(75)

The expressions of third central moments were very lengthy. Therefore, only plots of these moments are shown in the test problems.

Irreversible Reaction with Danckwerts BCs

Here, the moments are derived of the solutions given in Eqs. (40) and (41).

The zeroth moments are given as

$$\begin{aligned} \mu ^{(1)}_{0}=\frac{-4C_{1,\mathrm inj}\tau _\mathrm{inj}\gamma \mathrm{Pe}\, \,\mathrm{e}^{\mathrm{Pe}}}{\delta _1^2 \,\mathrm{e}^{\frac{\delta _1}{2}}-\delta _2^2 \,\mathrm{e}^{\frac{\delta _2}{2}}},\quad \mu ^{(2)}_{0}=C_{1,\mathrm inj} \tau _\mathrm{inj}\left[ 1+\frac{4\gamma \mathrm{Pe}\, \,\mathrm{e}^{\mathrm{Pe}}}{\delta _1^2e^\frac{\delta _1}{2}-\delta _2^2e^\frac{\delta _2}{2}} \right] . \end{aligned}$$

(76)

From Eq. (76), it follows that $\mu ^{(1)}_0+\mu ^{(2)}_0=C_{1,\mathrm inj}\tau _\mathrm{inj}$, as $C_{2,\mathrm inj}=0$ is considered.

The first moments take the form

$$ \mu ^{(1)}_{1}=\frac{\tau _\mathrm{inj}}{2}-\frac{2\mathrm{Pe}(1+a_1F)}{\gamma ^2\left( \delta _1^2 \,\mathrm{e}^{\frac{\delta _1}{2}}-\delta _2^2 \,\mathrm{e}^{\frac{\delta _2}{2}}\right) }\Biggl [\gamma (2r+\mathrm{Pe}^2) \left( \,\mathrm{e}^{\frac{\delta _1}{2}}+\,\mathrm{e}^{\frac{\delta _2}{2}}\right) -(4r(1+\mathrm{Pe})+\mathrm{Pe}^3)\left( \,\mathrm{e}^{\frac{\delta _1}{2}}-\,\mathrm{e}^{\frac{\delta _2}{2}}\right) \Biggr ]. $$

(77)

$$\begin{aligned} \mu ^{(2)}_1&=\frac{32\tau _\mathrm{inj}\left( \chi _1\,\mathrm{e}^{\delta _1}+[\chi _5e^\frac{\delta _2}{2}-\mathrm{Pe}\, \,\mathrm{e}^{\mathrm{Pe}}\chi _3]e^\frac{\delta _1}{2} +[\chi _2e^\frac{\delta _2}{2}-\mathrm{Pe}\, \,\mathrm{e}^{\mathrm{Pe}}\chi _4]e^\frac{\delta _2}{2} \right) }{\gamma \left( \delta _1^2 \,\mathrm{e}^{\frac{\delta _1}{2}}-\delta _2^2 \,\mathrm{e}^{\frac{\delta _2}{2}}\right) \left( \delta _1^2 \,\mathrm{e}^{\frac{\delta _1}{2}}-\delta _2^2 \,\mathrm{e}^{\frac{\delta _2}{2}}+4\gamma \mathrm{Pe}\, \,\mathrm{e}^{\mathrm{Pe}}\right) }\nonumber \ \\&\quad +\,\frac{64\left[ \chi _1\chi _6\,\mathrm{e}^{\delta _1}+\left[ \chi _5\chi _6e^\frac{\delta _2}{2}-\chi _7\right] e^\frac{\delta _1}{2} +\left[ \chi _2\chi _6e^\frac{\delta _2}{2}-\chi _8\right] e^\frac{\delta _2}{2} \right] }{r\gamma \left( \delta _1^2 \,\mathrm{e}^{\frac{\delta _1}{2}}-\delta _2^2 \,\mathrm{e}^{\frac{\delta _2}{2}}\right) \left( \delta _1^2 \,\mathrm{e}^{\frac{\delta _1}{2}}-\delta _2^2 \,\mathrm{e}^{\frac{\delta _2}{2}}+4\gamma \mathrm{Pe}\, \,\mathrm{e}^{\mathrm{Pe}}\right) }. \end{aligned}$$

(78)

The second and third central moments are not given here due to their lengthy expressions.

Reversible Reaction with Dirichlet BCs

The Eqs. (60) and (61) are used to derive the moments $\mu _n^{(i)}$ of Laplace-transformed solutions given in Eqs. (53) and (54) for $i=1,2$ and $n=0,1,2,3$. Let us define

$$\begin{aligned} R_1&=(a_1\widetilde{\nu }_1+ a_2\widetilde{\nu }_2)\mathrm{FPe},\quad \gamma _{1,2}=\sqrt{\mathrm{Pe}^2+2R_1\mp 2\sqrt{P+R_1^2}},\quad \delta _{1,2}=\mathrm{Pe}\mp \gamma _1,\nonumber \\ \delta _{3,4}&=\mathrm{Pe}\mp \sqrt{\mathrm{Pe}^2+2R_1+2\sqrt{P+R_1^2}},\quad \chi _{1,2}:=W\mp \sqrt{P+R_1^2} ,\nonumber \\ \chi _3&=2(P-W^2+R_1^2)\{(a_1F+a_2)\epsilon -a_2\}\gamma _{1}+\{(2+a_1F -a_2)\epsilon +a_2\}(P+R_1^2)^\frac{3}{2}\nonumber \\&\quad +\,W\left[ W\{(a_1F+a_2)\epsilon - a_2\}\sqrt{P+R_1^2}-2\epsilon (P+R_1^2)(1+a_1F)\right] ,\nonumber \\ \chi _4&=-2(P-W^2+R_1^2)\{(a_1F+a_2)\epsilon -a_2\}\gamma _{2}+\{(2+a_1F -a_2)\epsilon +a_2\}(P+R_1^2)^\frac{3}{2}\nonumber \\&\quad +\,W\left[ W\{(a_1F+a_2)\epsilon -a_2\}\sqrt{P+R_1^2}+2\epsilon (P+R_1^2)(1+ a_1F)\right] ,\nonumber \\ \chi _5&=-2W\{(a_1F+a_2)\epsilon -a_2\}\gamma _{1}+W\{(a_2+a_1F) \epsilon -a_2\}\sqrt{P+R_1^2}\nonumber \\&\quad -\,(P+R_1^2)\{(2-a_2+a_1F)\epsilon +a_2\},\nonumber \\ \chi _6&=2W\{(a_1F+a_2)\epsilon -a_2\}\gamma _{2}+W\{(a_2+a_1F) \epsilon -a_2\}\sqrt{P+R_1^2}\nonumber \\&\quad +\,(P+R_1^2)\{(2-a_2+a_1F)\epsilon +a_2\}. \end{aligned}$$

(79)

Using Eq. (60), the zeroth moments can be calculated as

$$\begin{aligned} \mu ^{(1)}_0=\frac{C_{1,\mathrm inj}\tau _\mathrm{inj}\left( \chi _2\,\mathrm{e}^{\frac{\delta _3}{2}}-\chi _1\,\mathrm{e}^{\frac{\delta _1}{2}}\right) }{2\sqrt{P+R_1^2}},\quad \mu ^{(2)}_0=\frac{r_3C_{1,\mathrm inj}\tau _ \mathrm{inj}}{\sqrt{P+R_1^2}}\,\left( \,\mathrm{e}^{\frac{\delta _1}{2}}-\,\mathrm{e}^{\frac{\delta _3}{2}}\right) . \end{aligned}$$

(80)

The first moments are derived using Eq. (61) for $n=1$

$$\begin{aligned} \mu ^{(1)}_1=&\frac{\tau _\mathrm{inj}}{2}+\frac{\mathrm{Pe}\left( \gamma _2\chi _3\,\mathrm{e}^{\frac{\delta _1}{2}}+\gamma _1\chi _4\,\mathrm{e}^{\frac{\delta _3}{2}}\right) }{2\epsilon \gamma _1\gamma _2(P+R_1^2)\left( \chi _2\,\mathrm{e}^{\frac{\delta _3}{2}} -\chi _1\,\mathrm{e}^{\frac{\delta _1}{2}}\right) },\end{aligned}$$

(81)

$$\begin{aligned} \mu ^{(2)}_1=&\frac{\tau _\mathrm{inj}}{2}+\frac{\mathrm{Pe}\left( \gamma _2\chi _5\,\mathrm{e}^{\frac{\delta _1}{2}}+\gamma _1\chi _6\,\mathrm{e}^{\frac{\delta _3}{2}}\right) }{2\epsilon \gamma _1\gamma _2(P+R_1^2)\left( \mathrm{e}^{\frac{\delta _3}{2}} -\,\mathrm{e}^{\frac{\delta _1}{2}}\right) }. \end{aligned}$$

(82)

Moreover, the expressions of analytical ${\mu }_2^{(i)}$ and ${\mu }_3^{(i)}$ were very lengthy. Therefore, the plots of second and third central moments are shown in Fig. 6.

Reversible Reaction with Danckwerts BCs

Here, the moments are derived for the solutions given in Eqs. (58) and (59). The zeroth moments are given as

$$\begin{aligned} \mu ^{(1)}_0=&\frac{2C_{1,\mathrm inj}\tau _\mathrm{inj}\mathrm{Pe} \,\mathrm{e}^{\mathrm{Pe}}}{\sqrt{P+R_1^2}}\left[ \frac{\chi _1\gamma _1}{\delta ^2_1 \,\mathrm{e}^{\frac{\delta _1}{2}} -\delta ^2_{2}\,\mathrm{e}^{\frac{\delta _2}{2}}}-\frac{\chi _2\gamma _2}{\delta ^2 _3 \,\mathrm{e}^{\frac{\delta _3}{2}} -\delta ^2_{4}\,\mathrm{e}^{\frac{\delta _4}{2}}}\right] ,\end{aligned}$$

(83)

$$\begin{aligned} \mu ^{(2)}_0=&\frac{-4r_3C_{1,\mathrm inj}\tau _\mathrm{inj}\mathrm{Pe} \,\mathrm{e}^{\mathrm{Pe}}}{\sqrt{P+R_1^2}}\left[ \frac{\gamma _1}{\delta ^2_1 \,\mathrm{e}^{\frac{\delta _1}{2}} -\delta ^2_{2}\,\mathrm{e}^{\frac{\delta _2}{2}}}-\frac{\gamma _2}{\delta ^2 _3 \,\mathrm{e}^{\frac{\delta _3}{2}} -\delta ^2_{4}\,\mathrm{e}^{\frac{\delta _4}{2}}}\right] . \end{aligned}$$

(84)

Other moments are not presented here due to their lengthy expressions.