Respuesta :
Explanation:
Let the amount and composition of the exit streams is [tex]L_{1}[/tex] and [tex]V_{1}[/tex].
It is assumed that air is insoluble in [tex]H_{2}O[/tex] and [tex]H_{2}O[/tex] does not vaporize.
This means that there is no presence of air in [tex]L_{1}[/tex] stream and no presence of [tex]H_{2}O[/tex] in [tex]V_{1}[/tex] stream.
Hence, the system's overall balance will be as follows.
[tex]L_{0} + V_{2} = L_{1} + V_{1}[/tex]
2.2 + 5.7 = [tex]L_{1} + V_{1}[/tex]
[tex]L_{1} + V_{1}[/tex] = 7.9 ............ (1)
As it is assumed that water will not vaporize. Therefore, all of it will go into [tex]L_{1}[/tex]. And, as air is not soluble in water so all of it will exit into stream [tex]V_{1}[/tex].
Hence, the inert water flow will be as follows.
L' = [tex]L_{0}(1)[/tex]
= 2.2(1) = 2.2 kg mol
The inert air flow is as follows.
V' = [tex]V_{2} (1 - \frac{1.52 \times 10^{4}}{2.026 \times 10^{5}}[/tex]
= 5.27 kg mol
Making [tex]SO_{2}[/tex] (component A) balance:
[tex]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}}})[/tex]
[tex]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}}})[/tex] ......... (2)
Relationship between [tex]X_{A_{1}}[/tex] and [tex]Y_{A_{1}}[/tex] as follows. As [tex]V_{1}[/tex] is present in equilibrium with [tex]L_{1}[/tex], the relation between [tex]Y_{A_{1}}[/tex] and [tex]X_{A_{1}}[/tex] may assumed to follow Henry's Law.
[tex]Y_{A_{1}} = H'X_{A_{1}}[/tex]
where, H' = [tex]\frac{H}{P_{T}}[/tex]
where, H = Henry's constant
[tex]P_{T}[/tex] = total pressure
Value of H for [tex]SO_{2}-H_{2}O[/tex] plot at 293 K is 29.6 atm.
Therefore, [tex]P_{A} = Hx_{A} = 29.6x_{A}[/tex]
H' = [tex]\frac{29.6}{2}[/tex] (as [tex]P_{T} = 2.026 \times 10^{5} Pa[/tex] = 2 atm)
= 14.8
So, [tex]Y_{A_{1}} = 14.8 \times x_{A_{1}}[/tex]
Now, substitute the value of Henry's equilibrium relationship into equation (2) to make [tex]X_{A_{1}}[/tex] as your single unknown value.
[tex]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}}})[/tex]
0.4273 = [tex](\frac{2.2X_{A_{1}}}{1 - X_{A_{1}}}) + (\frac{77.996X_{A_{1}}}{1 - 14.8X_{A_{1}}})[/tex]
[tex]X_{A_{1}} = 4.947 \times 10^{-3}[/tex]
= 0.004947
[tex]Y_{A_{1}} = 14.8 \times X_{A_{1}}[/tex]
= 0.07322
Now, solving for the amount of existing fluid in [tex]V_{1}[/tex] and [tex]L_{1}[/tex] as follows.
L' = [tex]L_{1}(1 - X_{A_{1}})[/tex]
[tex]L_{1} = \frac{2.2}{1 - 0.004947}[/tex]
= 2.211 kg mol
V' = [tex]V_{1}(1 - Y_{A_{1}})[/tex]
[tex]V_{1} = \frac{5.27}{1 - 0.07322}[/tex]
= 5.686 kg mol
Hence, for the exiting stream [tex]V_{1}[/tex] with a total amount 5.686 kg mol has a composition of 7.32% [tex]SO_{2}[/tex] and the balance air. And, the exiting stream [tex]L_{1}[/tex] with a total amount 2.21 kg mol has a composition of 0.5% [tex]SO_{2}[/tex] and balance [tex]H_{2}O[/tex].
