Analysis of a Chromatographic Model with (Part 4)

Analysis of a Chromatographic Model with Irreversible and Reversible Reactions

Analysis of the EDM and LKM for Rectangular Pulse injection

The EDM and the LKM are two liquid chromatographic models that are examined in light of linear isotherms. And two different sets of BC’s at inlet and outlet. ( $N_c = 1$ ) for a rectangular single solute with limited length concentration pulses and Within linear limitations, a revolution of curves ( $\tau_{inj}\rightarrow \infty$ ) is infused. The Laplace transformation is used to determine the analytical solution. The temporal moments up to the fourth demands for studying the solute transport conduct. We analyzed the models using Laplace-transformed configurations. Our analysis included both exact and numerical results, which we compared for validation. For linear adsorption isotherms, we described the results of various experiments. We also determined the temporal moments to understand numerically and analytically.

LKM

The LKM ties combine internal and outside mass transport precautions to generate a mass exchange coefficient $k$ .

in these equations, $q^*$ represents the equilibrium liquid phase concentration. We use $c$ for liquid phase concentrations and $q$ for solid phase concentrations. The interstitial velocity is denoted by the letter $u$ , $\epsilon_t$ represents the porosity, $t$ is for time, $D_a$ denotes axial dispersion, $k$ denotes mass transfer coefficient, and $z$ is for axial coordinate.

The isotherm $q^*(c)$ describes the equilibrium relationship between mobile and stationary phase concentrations. This thermodynamic detail is crucial for accurately predicting the column’s concentration profile. We often model this relationship using the convex nonlinear Langmuir isotherm, expressed as follows:

Here, $b$ represents the isotherm’s nonlinearity, and $c$ denotes the Henry coefficient.The isotherm becomes linear. For low concentrations,

The dispersive coefficient, $D_a$ , is often expressed as a dimensionless Peclet number, Pe. This dimensionless parameter represents the ratio of advective transport to dispersive transport. The Pe can be calculated by,

$Pe = (u * L) / Da$

where L represents the length of the column. The IC’s for a column that is uniformly preequilibrated are:

EDM

The EDM is based on the assumption that mass transfer kinetics are infinitely quick, $k \rightarrow \infty$ . The visible dispersion coefficient $D_a$ is the aggregated of all band-expanding commitments. The mass equilibrium equation for a chromatographic column in the equilibrium dispersive model is,

Where the phase ratio $F =\frac{1-\epsilon}{\epsilon}$ depending on the porosity $\epsilon \in(0, 1)$ . We can express the apparent dispersion coefficient, D, in terms of the Peclet number, Pe, as follows:

When the column efficiency is high, the EDM accurately predicts chromatographic profiles, i.e. for huge peclet number Pe. The initial and BCs of the model are the same as LKM given by Eqs. 3.8-3.11.

Analytical solution of linear EDM and LKM

In this section we will discuss about exact solution for EDM and LKM for the two different pairs of BCs.

Analytical Solution of LKM

In this section, we’ll analyze the linear isotherm using the single-component LKM model. For the given boundary conditions, LT utilized to find analytical solution. We can express the LT as follows:

(1) $\begin{equation*} \dfrac{\partial c_1}{\partial t}+u\dfrac{\partial c_1}{\partial z}=D\dfrac{\partial^2 c_1}{\partial z^2}-\dfrac{k_1}{\epsilon}(q^1-q_1)-\mu_1c_1 \end{equation*}$

(2) $\begin{equation*} \dfrac{\partial c_2}{\partial t}+u\dfrac{\partial c_2}{\partial z}=D\dfrac{\partial^2 c_2}{\partial z^2}-\dfrac{k_2}{\epsilon}(q^_2-q_2)-\mu_1c_1\end{equation*}$

For solid phase, the governing equations are

(3) $\begin{equation*} \dfrac{\partial q_1}{\partial t}=\dfrac{k_1}{1-\epsilon}(q^_1-q_1)-\dfrac{v_1q_1}{1-\epsilon} \end{equation*}$

(4) $\begin{equation*} \dfrac{\partial q_2}{\partial t}=\dfrac{k_2}{1-\epsilon}(q^_2-q_2)-\dfrac{v_1q_1}{1-\epsilon}\end{equation*}$

In above equations, $c_i$ denotes the liquid phase concentration, $q_i$ represents the solid phase
concentration and $k_i$ is the mass transfer coefficients of component \textit{i} and $v_i$ denotes the solid phase reaction coefficient for first and second components of mixture , where \textit{i=1,2} and $\epsilon$ $\in$ $(0,1)$ . Now the initial conditions are given below,

(5) $\begin{equation*} c_i(0,z)=c{i,init} \qquad \qquadq_i(0,z)=q^{i,init}=a_i c{i,init}\end{equation*}$

Part 1: Analysis of a Chromatographic Model with Irreversible

where $q_{i,init}$ and $c_{i,init}$ represents the initial concentrations of components of the mixture in solid and liquid phase, respectively in column.
Let’s define some dimensionless variables to make the analysis easier.

(6) $\begin{equation*} x=\dfrac{z}{L} \qquad \tau=\dfrac{ut}{L} \qquad Pe=\dfrac{Lu}{D} \qquad \eta_1=\dfrac{L\mu_1}{u} \qquad \omega_i=\dfrac{Lk_i}{u} \quad (i=1,2),\end{equation*}$

Where L represents the length of column. Now use these variables in equations 1 to 4 , we get

$\dfrac{\partial c_1}{\partial t}=\dfrac{\partial c_1}{\partial \tau} \dfrac{u}{L}$

$\dfrac{\partial c_1}{\partial z}=\dfrac{\partial c_1}{\partial x} \dfrac{1}{L}$

and

$\dfrac{\partial^2 c_1}{\partial z^2}=\dfrac{\partial^2 c_1}{\partial x^2} \dfrac{1}{L^2}$

Use these above equations in eq 1, we get,

$\dfrac{\partial c_1}{\partial \tau}+\dfrac{L \tau}{u t}\dfrac{\partial c_1}{\partial x}=\dfrac{1}{Pe}\dfrac{\partial^2 c_1}{\partial x^2}-\dfrac{\omega_1}{\epsilon}(q^1-q_1)-\eta c_1$

From initial condition\

$q^_1=a_1 c_1$

$\qquad$ we obtain

(7) $\begin{equation*} \dfrac{\partial c_1}{\partial \tau}+\dfrac{\partial c_1}{\partial x}=\dfrac{1}{Pe}\dfrac{\partial^2 c_1}{\partial x^2}-\dfrac{\omega_1}{\epsilon}(a_1 c_1-q_1)-\eta_1 c_1 \end{equation*}$

Now putting

$\dfrac{\partial c_2}{\partial t}=\dfrac{\partial c_2}{\partial \tau} \dfrac{u}{L}$

$\dfrac{\partial c_2}{\partial z}=\dfrac{\partial c_2}{\partial x} \dfrac{1}{L}$

and

$\dfrac{\partial^2 c_2}{\partial z^2}=\dfrac{\partial^2 c_2}{\partial x^2} \dfrac{1}{L^2}$

in equation 2, we obtain,

(8) $\begin{equation*} \dfrac{\partial c_2}{\partial \tau}+\dfrac{\partial c_2}{\partial x}=\dfrac{1}{Pe}\dfrac{\partial^2 c_2}{\partial x^2}-\dfrac{\omega_2}{\epsilon}(a_2 c_2-q_2)-\eta_1 c_1 \end{equation*}$

Now we have

$\dfrac{\partial q_1}{\partial t}=\dfrac{u}{L}\dfrac{\partial q_1}{\partial \tau}$

put into equation 3 we get

$\dfrac{u}{L}\dfrac{\partial q_1}{\partial \tau}=\dfrac{\omega_1 u}{(1-\epsilon) L}(a_1 c_1-q_1)-\dfrac{v_1 q_1}{1- \epsilon}$

multiplying above equation with $\dfrac{L}{u}$ and simplify then the equation becomes

(9) $\begin{equation*} \dfrac{\partial q_1}{\partial \tau}=\dfrac{\omega_1}{1- \epsilon}\Big[a_1 c_1-(1+\dfrac{v_1}{k_1})q_1\Big] \end{equation*}$

Read Part 2: Analysis of a Chromatographic Model with Irreversible and

similarly we can obtain the equation

(10) $\begin{equation*} \dfrac{\partial q_2}{\partial \tau}=\dfrac{\omega_2}{1- \epsilon}\Big[a_2 c_2-q_2+\dfrac{v_1 q_1}{k_2}\Big] \end{equation*}$

ICs in non-dimensionalize form are given as

(11) $\begin{equation*} c_i(0,x)=c{i,init} \qquad q_i(0,x)=q^_{i,init}=a_i c{_i,init} \quad (i=1,2)\end{equation*}$

Apply the Laplace transformation in $\tau$ domain in equation 7 and 8 and eliminate the solid concentrations $q_i$

applying Laplace transformation on equation 7

$s \bar{c_1}-c_1(x,0)+\dfrac{d \bar{c_1}}{dx}=\dfrac{1}{Pe}\dfrac{d^2 \bar{c_1}}{dx^2}-\dfrac{\omega_1}{\epsilon}(a_1 \bar{c_1}-\bar{q_1})- \eta_1 \bar{c_1}$

multiplying above equation with $Pe$ and use ICs then equation becomes

(12) $\begin{equation*} \dfrac{d^2 \bar{c_1}}{dx^2}-Pe\dfrac{d \bar{c_1}}{dx}-Pe \bar{c_1}\Big(s+\dfrac{\omega_1 a_1}{\epsilon}+\eta_1\Big)+\dfrac{\omega_1 Pe \bar{q_1}}{\epsilon}=-Pe c_{1,init}\end{equation*}$

Now apply Laplace transformation to equation 9 then\

$s \bar{q_1}-q_1(x,0)=\dfrac{\omega_1}{1- \epsilon}[a_1 \bar{c_1}-h_1 \bar{q_1}]$

$\quad$ where $h_1=\Big(1+\dfrac{v_1}{k_1}\Big)$

use ICs and simplify then above expression becomes

$\bar{q_1}=\dfrac{\omega_1 a_1 \bar{c_1}+(1- \epsilon)a_1 c_{1,init}}{s(1-\epsilon)+\omega_1 h_1}$

so use this above value in equation 12

$\dfrac{d^2 \bar{c_1}}{dx^2}-Pe\dfrac{d \bar{c_1}}{dx}-Pe \bar{c_1}\Big(s+\dfrac{\omega_1 a_1}{\epsilon}+\eta_1\Big)+\dfrac{\omega_1 Pe[\omega_1 a_1 \bar{c_1}+(1-\epsilon)a_1 c_{1,init}]}{\epsilon s (1-\epsilon)+\omega_1 h_1 \epsilon}= -Pe c_{1,init}$

Part 3: Analysis of a Chromatographic Model with Irreversible and

simplify and then rearrange the above equation

$\dfrac{d^2\bar{c_1}}{dx^2}-Pe\dfrac{d \bar{c_1}}{dx}-\bar{c_1}\Big[r_1+Pe s \Big[ {1+\dfrac{\omega_1 a_1 F}{s (1-\epsilon)+\omega_1 h_1}+\dfrac{\omega^2_1 a_1 v_1}{k_1 s \epsilon [s(1-\epsilon)+\omega_1 h_1]}} \Big]\Big]= c_{1,init}\Big[-Pe\Big[1+\dfrac{\omega_1 a_1 F}{s(1-\epsilon)+\omega_1 h_1}+\dfrac{a_1 v_1 \omega^2_1}{s \epsilon k_1 [s(1-\epsilon)+\omega_1 h_1]}\Big]+\dfrac{Pe a_1 v_1 \omega^2_1}{s \epsilon k_1[s(1-\epsilon)+h_1 \omega_1}\Big]$

Part 1: Analysis of a Chromatographic Model with Irreversible

(13) $\begin{equation*} \dfrac{d^2\bar{c_1}}{dx^2}-Pe\dfrac{d \bar{c_1}}{dx}-\bar{c_1}(r_1+s \alpha_1)= c_{1,init}(Pe \beta_1 -\alpha_1)\end{equation*}$

Now equation 8 takes the form by appliyang Laplace transformation

$s \bar{c_2}-c_2(x,0)+\dfrac{d \bar{c_2}}{dx}=\dfrac{1}{Pe}\dfrac{d^2 \bar{c_2}}{dx^2}-\dfrac{\omega_2}{\epsilon}(a_1 \bar{c_2}-\bar{q_2})- \eta_1 \bar{c_1}$

from the ICs

(14) $\begin{equation*} \dfrac{d^2\bar{c_2}}{dx^2}-Pe\dfrac{d\bar{c_2}}{dx}-Pe\dfrac{\omega_2}{\epsilon}a_2 \bar{c_2}+Pe\dfrac{\omega_2}{\epsilon}\bar{q_2}+Pe\eta_1\bar{c_1}-Pe s\bar{c_2}=-Pe c_{2,init}\end{equation*}$

Now apply Laplace transformation on equation 10

$s\bar{q_2}-q_2(x,0)=\dfrac{\omega_2}{1-\epsilon}\Big(a_2\bar{c_2}-\bar{q_2}+\dfrac{v_1}{k_2}\bar{q_1}\Big)$

now use the ICs and putting the value of $\bar{q_1}$

$\bar{q_2}=\dfrac{a_2\omega_2\bar{c_2}}{s(1-\epsilon)+\omega_2}+\dfrac{(1-\epsilon)a_2 c_{2,init}}{s(1-\epsilon)+\omega_2}+\dfrac{v_1\omega_2}{k_2[s(1-\epsilon)+\omega_2]}\Big[\dfrac{\omega_1 a_1 \bar{c_1}+(1- \epsilon)a_1 c_{1,init}}{s(1-\epsilon)+\omega_1 h_1}\Big]$

using the value of $\bar{q_2}$ in equation 14

(15) $\begin{equation*} \dfrac{d^2 \bar{c_2}}{dx^2}-Pe \dfrac{d \bar{c_2}}{dx}+ r_2 \bar{c_1} - s \alpha_2 \bar{c_2} = - \alpha_2 c_{2,init}-s Pe \beta_2 c_{1,init}\end{equation*}$

Part 3: Analysis of a Chromatographic Model with Irreversible and

where

(16) $\begin{eqnarray*} \alpha_1=Pe\Big[1+\dfrac{F a_1 \omega_1}{s(1-\epsilon)+h_1 \omega_1}+\dfrac{a_1 \omega^2_1 v_1}{s k_1 \epsilon{s(1-\epsilon)+\omega_1 h_1}}\Big],\nonumber \\ \alpha_2=Pe\Big[1+\dfrac{F a_2 \omega_2}{s(1-\epsilon)+\omega_2} \Big ], \quad F=\dfrac{1-\epsilon}{\epsilon}, \qquad h_1=1+\dfrac{v_1}{k_1}, \nonumber \\ r_1= \eta_1 Pe, \quad r_2=Pe\Big[\eta_1+\dfrac{a_1 v_1\omega_1 \omega^2_2}{\epsilon k_2 (1-\epsilon)[s(1-\epsilon)+\omega_1 h_1]}\Big], \nonumber \\ \beta_i=\dfrac{\omega^2_i a_1 v_1}{s k_i \epsilon[s(1-\epsilon)+h_1 \omega_1]}, \qquad i=1,2\end{eqnarray*}$

The concentrations of mixture components in liquid phase are represented by $\bar{c_1}$ and $\bar{c_2}$ in the Laplace domain. Equations 13 and 15 takes the form in matrix form

(17) $\begin{equation*} \dfrac{d^2}{dx^2}\begin{Bmatrix}\bar{c_1} \ \bar{c_2}\end{Bmatrix} -Pe\dfrac{d}{dx} \begin{Bmatrix}\bar{c_1} \ \bar{c_2}\end{Bmatrix} + \begin{bmatrix} -r_1-s\alpha_1 & 0 \ r_2 & -s\alpha_2 \end{bmatrix}\begin{Bmatrix}\bar{c_1} \ \bar{c_2}\end{Bmatrix}= \begin{Bmatrix} (-\alpha_1+\beta_1 Pe)c_{1,init} \ -c_{2init} \alpha_2 - Pe s \beta_2 c_{1,init}\end{Bmatrix}\end{equation*}$

where the square matrix is represented by the $[ \quad ]$ and the column matrix is represented by the ${ \quad }$ . Thus, on the left hand side of equation 17, a combined reaction coefficient matrix [D] is given as