Question

22.1-4. Equilibrium Stage Contact for Gas–Liquid System. A gas mixture at 2.026 × 10 5 Pa total pressure containing air and SO2 is brought into contact in a single-stage equilibrium mixer with pure water at 293 K. The partial pressure of SO2 in the original gas is 1.52 × 10 4 Pa. The inlet gas contains 5.70 total kg mol and the inlet water 2.20 total kg mol. The exit gas and liquid leaving are in equilibrium. Calculate the amounts and compositions of the outlet phases. Use equilibrium data from Fig. 22.2-1. Ans. xA1

Answer 1

Explanation:

Let the amount and composition of the exit streams is $L_{1}$ and $V_{1}$.

It is assumed that air is insoluble in $H_{2}O$ and $H_{2}O$ does not vaporize.

This means that there is no presence of air in $L_{1}$ stream and no presence of $H_{2}O$ in $V_{1}$ stream.

Hence, the system's overall balance will be as follows.

             $L_{0} + V_{2} = L_{1} + V_{1}$

                    2.2 + 5.7 = $L_{1} + V_{1}$

             $L_{1} + V_{1}$ = 7.9 ............ (1)

As it is assumed that water will not vaporize. Therefore, all of it will go into $L_{1}$. And, as air is not soluble in water so all of it will exit into stream $V_{1}$.

Hence, the inert water flow will be as follows.

                  L' = $L_{0}(1)$

                      = 2.2(1) = 2.2 kg mol

The inert air flow is as follows.

               V' = $V_{2} (1 - \frac{1.52 \times 10^{4}}{2.026 \times 10^{5}}$

                   = 5.27 kg mol

Making $SO_{2}$ (component A) balance:

    $L'(\frac{X_{A_{0}}}{1 - X_{A_{0}}}) + V'(\frac{Y_{A_{2}}}{1 - Y_{A_{2}}}) = L'(\frac{X_{A_{1}}}{1 - X_{A_{1}}}) + V'(\frac{Y_{A_{1}}}{1 - Y_{A_{1}}})$

       $2.2 (\frac{0}{1 - 0}) + 5.27 (\frac{0.075}{1 - 0.075}) = 2.2 (\frac{X_{A_{1}}}{1 - X_{A_{1}}}) + 5.27 (\frac{Y_{A_{1}}}{1 - Y_{A_{1}}})$ ......... (2)

Relationship between $X_{A_{1}}$ and $Y_{A_{1}}$ as follows. As $V_{1}$ is present in equilibrium with $L_{1}$, the relation between $Y_{A_{1}}$ and $X_{A_{1}}$ may assumed to follow Henry's Law.

                    $Y_{A_{1}} = H'X_{A_{1}}$        

where,    H' = $\frac{H}{P_{T}}$

where,    H = Henry's constant

           $P_{T}$ = total pressure

Value of H for $SO_{2}-H_{2}O$ plot at 293 K is 29.6 atm.

Therefore,      $P_{A} = Hx_{A} = 29.6x_{A}$

                            H' = $\frac{29.6}{2}$       (as $P_{T} = 2.026 \times 10^{5} Pa$ = 2 atm)

                                 = 14.8

So,    $Y_{A_{1}} = 14.8 \times x_{A_{1}}$      

Now, substitute the value of Henry's equilibrium relationship into equation (2) to make $X_{A_{1}}$ as your single unknown value.      

       $2.2(\frac{0}{1-0}) + 5.27(\frac{0.075}{1 - 0.075}) = 2.2(\frac{X_{A_{1}}}{1 - X_{A_{1}}}) + 5.27(\frac{14.8X_{A_{1}}}{1 - 14.8X_{A_{1}}})$

          0.4273 = $(\frac{2.2X_{A_{1}}}{1 - X_{A_{1}}}) + (\frac{77.996X_{A_{1}}}{1 - 14.8X_{A_{1}}})$

           $X_{A_{1}} = 4.947 \times 10^{-3}$

                         = 0.004947

          $Y_{A_{1}} = 14.8 \times X_{A_{1}}$

                          = 0.07322

Now, solving for the amount of existing fluid in $V_{1}$ and $L_{1}$ as follows.

               L' = $L_{1}(1 - X_{A_{1}})$

            $L_{1} = \frac{2.2}{1 - 0.004947}$

                   = 2.211 kg mol

                V' = $V_{1}(1 - Y_{A_{1}})$

            $V_{1} = \frac{5.27}{1 - 0.07322}$

                      = 5.686 kg mol

Hence, for the exiting stream $V_{1}$ with a total amount 5.686 kg mol has a composition of 7.32% $SO_{2}$ and the balance air. And, the exiting stream $L_{1}$ with a total amount 2.21 kg mol has a composition of 0.5% $SO_{2}$ and balance $H_{2}O$.