Question

A liquid stream containing solute A and carrier liquid C enters a packed column. The solute is to be stripped using pure gas B. The liquid stream enters the column at a rate of 150 kmole/hr and this stream contains 7 mole % A. The stripping gas B enters at a rate of 500 kmole/hr. The design calls for 1 mole % A in the exiting liquid. For the phase equilibrium relationship use yA= 0.4XA. At the conditions of interest we have Kza = 75kmole/(hr *m3)
Determine the height of packing required for this separation if the column cross sectional area is 1m².

Answer 1

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 $(L_s)$ = 150 kmole/hr

Mole fraction of the solute $(x_2)$ = 7% = 0.07

(A+C)liquid = $x_1$ = 0.01

Pure gas $(G_s)$ = 500 kmole/hr

$y_1 = 0$

The delivery force in the gas phase & liquid phase is expressed as:

$(K_x)(x_2-x_1) = k_y(y_2-y_1)$

Given that $(K_x)_a$ = 75 kmole/hr m³ and equilibrium relationship $y_A = 0.4x_A$

Then :

$K_{xa}(x_2-x_1) = (k_y)_a(y_2-y_1)$

$75(0.07-0.01)=(k_y)_a(y_2-0)$

where ;

$y_2 = 0.4x_2$

= 0.4(0.07)

= 0.028

$75(0.06)=(k_y)_a(0.028)$

$(k_y)_a = \frac{4.5}{0.028}$

$(ky)_a = 160.71 \ kmol/hr.m^3$

NOW; the overall mass transfer coefficient on the liquid phase is :-

$\frac{1}{(K_{ox})_a} = \frac{1}{(K_x)_a}+ \frac{1}{m(K_y)_a}$

where m= slope = 0.4

$\frac{1}{(K_{ox})_a} = \frac{1}{75}+ \frac{1}{0.4(160.71)}$

${(K_{ox})_a} = 34.61 \ kmol/hr.m^3$

Finally; the height of the tower (z) is = $(HTU)_{oL}(NTU)_{oL}$

$(HTU)_{oL} = \frac{\frac{L_s}{s} }{(K_{ox})_a}$

where :

s = 1m²

$L_s$ = 150 kmol/hr

$(HTU)_{oL} = \frac{\frac{150}{1} }{34.61}$ = 4.33 m

$(NTU)_{oL} = e^x [\frac{(\frac{x_2-\frac{y_1}{m} }{x_1-\frac{y_1}{m} })(1-A)+A }{1-A} ]$

where A  = $\frac{L_s}{G_s *m}$ = $\frac{150}{500(0.4)}$

= 0.75

Then:

$(NTU)_{oL} = e^x [\frac{(\frac{0.07-0 }{0.01-0 })(1-0.75)+0.75 }{1-0.75} ]$

$(NTU)_{oL} =3.66 \ m$

the height of the tower (z) is = $(HTU)_{oL}(NTU)_{oL}$

= (4.33)(3.66)

=15.85 m

Thus, the height of packing required for this separation is 15.85 m