Answer:
the height of packing required for this separation is 15.85 m
Explanation:
Given that :
liquid stream (A+C) enters a packed column which the solute is stripped by using a pure gas B
The solute flowrate [tex](L_s)[/tex] = 150 kmole/hr
Mole fraction of the solute [tex](x_2)[/tex] = 7% = 0.07
(A+C)liquid = [tex]x_1[/tex] = 0.01
Pure gas [tex](G_s)[/tex] = 500 kmole/hr
[tex]y_1 = 0[/tex]
The delivery force in the gas phase & liquid phase is expressed as:
[tex](K_x)(x_2-x_1) = k_y(y_2-y_1)[/tex]
Given that [tex](K_x)_a[/tex] = 75 kmole/hr mÂł and equilibrium relationship [tex]y_A = 0.4x_A[/tex]
Then :
[tex]K_{xa}(x_2-x_1) = (k_y)_a(y_2-y_1)[/tex]
[tex]75(0.07-0.01)=(k_y)_a(y_2-0)[/tex]
where ;
[tex]y_2 = 0.4x_2[/tex]
= 0.4(0.07)
= 0.028
[tex]75(0.06)=(k_y)_a(0.028)[/tex]
[tex](k_y)_a = \frac{4.5}{0.028}[/tex]
[tex](ky)_a = 160.71 \ kmol/hr.m^3[/tex]
NOW; the overall mass transfer coefficient on the liquid phase is :-
[tex]\frac{1}{(K_{ox})_a} = \frac{1}{(K_x)_a}+ \frac{1}{m(K_y)_a}[/tex]
where m= slope = 0.4
[tex]\frac{1}{(K_{ox})_a} = \frac{1}{75}+ \frac{1}{0.4(160.71)}[/tex]
[tex]{(K_{ox})_a} = 34.61 \ kmol/hr.m^3[/tex]
Finally; the height of the tower (z) is = [tex](HTU)_{oL}(NTU)_{oL}[/tex]
[tex](HTU)_{oL} = \frac{\frac{L_s}{s} }{(K_{ox})_a}[/tex]
where :
s = 1m²
[tex]L_s[/tex] = 150 kmol/hr
[tex](HTU)_{oL} = \frac{\frac{150}{1} }{34.61}[/tex] = 4.33 m
[tex](NTU)_{oL} = e^x [\frac{(\frac{x_2-\frac{y_1}{m} }{x_1-\frac{y_1}{m} })(1-A)+A }{1-A} ][/tex]
where A Â = [tex]\frac{L_s}{G_s *m}[/tex] = [tex]\frac{150}{500(0.4)}[/tex]
= 0.75
Then:
[tex](NTU)_{oL} = e^x [\frac{(\frac{0.07-0 }{0.01-0 })(1-0.75)+0.75 }{1-0.75} ][/tex]
[tex](NTU)_{oL} =3.66 \ m[/tex]
the height of the tower (z) is = [tex](HTU)_{oL}(NTU)_{oL}[/tex]
= (4.33)(3.66)
=15.85 m
Thus, the height of packing required for this separation is 15.85 m
