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.

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 = 150 kmole/hr

Mole fraction of the solute = 7% = 0.07

(A+C)liquid = = 0.01

Pure gas = 500 kmole/hr

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

Given that = 75 kmole/hr m³ and equilibrium relationship

Then :

where ;

= 0.4(0.07)

= 0.028

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

where m= slope = 0.4

Finally; the height of the tower (z) is =

where :

s = 1m²

= 150 kmol/hr

= 4.33 m

where A  = =

= 0.75

Then:

the height of the tower (z) is =

= (4.33)(3.66)

=15.85 m

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

answer:

true

explanation: