Part $1$: Dynamic Modeling

The tank shown in Figure $1$ is used for continuous extraction of a solute from a liquid

solution to a solvent. We assume two perfectly mixed phases, the extract and the raffinate,

separated by an interface across which the solute diffuses at a rate given by

$n(t)=K_{a}V[c_1(t)-c_1^{**}(t)]$

where

$n(t)=$ the rate of solute mass transfer across the interface

$K_a,=$ the coefficient of mass transfer

$V,=$ is the contact volume

$c(t)=$ the solute concentration in the raffinate phase

$c_1^{**}(t)=$ the raffinate concentration that would be in equilibrium with the extract phase.

The equilibrium relationship can be expressed as a straight line, where $m$ is the slope of the

line, and $c_2(t)$ is the concentration of the solute in the extract phase

$c_1^{**}(t)=m{c}_2(t)$

For simplicity, it is assumed that the volume of each phase is half the total contact volume and

that the feed flow, $f_1,$ is constant. Furthermore, the variation of the densities of the streams with

concentration can be neglected.

The objective is to adjust the concentration of the solute in the extract phase, by

adjusting the feed concentration $c_i(t)$ subject to variations in the flow of pure solvent, $f_2(t).$

$1-$ Based on the unsteady$-$state mole balance for both of raffinate and extract phase$-$solute,

show that the dynamics of $c_1(t)$ and $c_2(t)$ can be expressed as follows and give the

expression of $K_1,K_2,K_3,K_4,{}_1,$ and ${}_2$ as a function of process parameters

${}_1\frac{d{c}_1(t)}{dt}+c_1(t)=K_1{c}_i(t)+K_2{c}_2(t)$

and

${}_2\frac{d{c}_2(t)}{dt}+c_2(t)=K_3{c}_1(t)-K_4{F}_2(t)$

Let ${}_1=2\mathrm{.}63,{}_2=4\mathrm{.}4,K_1=5\mathrm{.}62,K_2=0\mathrm{.}35,K_3=0\mathrm{.}82,$ and $K_4=4\mathrm{.}67$

$2-$ Draw the block diagram of the process