Separations analysis fundamentals - Principles of Chemical Separations with Environmental Applications - page 79

Example 3.8

An air-stripping tower is to be designed to remove volatile organic compounds (VOCs)
from water droplets. The overall mass transfer coefficient in the liquid phase is
given by:

1
K L =

1
k L +

1
mk g .

(3.91)

m 3 The distribution coefficient, m , is 1.8 atm mg and the partial pressure of VOCs is

0.14 atm. Assume clean air is used such that C A , b =

· /

0. For a gas-phase mass transfer

10 − 9 cm 2

coefficient of 1.3 × mg / atm · · s, examine three designs for which the

×

10 − 3 , 6.2 10 − 2 , and 6.2 10 − 4 cm liquid-phase mass transfer coefficient is 6.2 × × / s,

respectively.
Solution:
For the liquid phase:

K L C A , b − C A , b

J A =

and

mC A , b .

P A , b =

So,

0 . 14 atm

C A , b = cm 3

mg =

0 . 078 mg /

1 . 8 atm · m 3 /

and

1 10 6 cm 3

m 3

1

m 3
mg

atm · mg

10 − 9 mk g = . 8 . 3 ×

atm · cm 2 · s

10 − 3 cm = 2 . 3 × / s .

10 − 3 cm Case I: k L = 6 . 2 × / s

1 1

1
K L =

10 − 3 cm

0062 +

K L = 1 . 7 × / s

. .

0 0 0023

s 0 0 cm 3 10 − 4 cm 2

J A = 0 . 0017 cm / . 078 mg / − = 1 . 33 × mg / · s .

10 − 2 Case II: k L = 6 . 2 × cm / s

1
K L =

1 1

10 − 3 cm

062 +

K L = 2 . 26 × / s

0 . 0 . 0023

s 0 0 = cm 3 cm 2 10 − 3 cm 10 − 4 mg J A = 2 . 26 × / . 078 mg / − 1 . 76 × / · s .

10 − 4 cm

Case III: k L = 6 . 2 × / s

10 − 4

K L = 4 . 9 × cm / s

cm 3 10 − 5 mg cm 2

J A = 0 . 00049 cm / s(0 . 078 mg / − 0) = 3 . 82 × / · s .