# 5.11: Untitled Page 109

## Chapter 5

We now use Eq. 5‐80 to eliminate (y ) so that Eq. 5‐87 provides the following A 3

result

( y )  ( y ) A 1  A 1  A   ( y ) (5‐88)

A 1

A 1



A 4

which can be solved for ( y ) as a function of ( y ) and the absorption A 1

A 4

coefficient.

( y )

Three Equilibrium Stages:

A 4

( y )

(5‐89)

A 1

2

3

1  A A A

Here we are certainly in a position to deduce that N equilibrium stages would produce the result given by

( y )

N Equilibrium Stages:

A N1

( y )

(5‐90)

A 1

2

3

1

N

A A A .... A

This type of sequential analysis can be used to analyze any multi‐stage process such as the one illustrated in Figure 5‐10; however, one must remember that there are three important simplifications associated with this result. These three simplifications are: (1) The total molar flow rates, M

 and M  , are treated as

constants, (2) a linear equilibrium relation is applicable, and (3) the mole fraction of species A entering the system in the  ‐phase is zero, i.e., ( x )  0 . This latter A o

constraint can be easily removed and that will be left as an exercise for the student.

If the absorption factor and the number of stages are specified for the system illustrated in Figure 5‐10, one can easily calculate the mole fraction of species A at the exit of the system, ( y ) , in terms of the mole fraction at the entrance, A 1

( y )

. If ( y ) is specified as some fraction of ( y ) A N1

A 1

A N and the number of

1

stages are specified, the required absorption factor can be determine using the implicit expression for A given by (see Appendix B for the solution procedure) ( y )

2

3

N

A N1

1  A A A .... A

(5‐91)

( y )

A 1

In general, it is convenient to use this result along with the macroscopic balance around the entire cascade illustrated in Figure 5‐10. This is given by ( x ) M

 ( y ) M

( x ) M

 ( y )

M

(5‐92)

A N

A 1

A o

A N





1 

molar flow of species A

molar flow of species A

out of the control volume

into the control volume

Two‐Phase Systems & Equilibrium Stages 199

and it is often convenient to arrange this result in the form ( y )

 ( y )

1

1

( x )

A N

A

A N

(5‐93)

M

M

Here we have continued to impose the condition that ( x )  0 in order to keep A o

this initial study as simple as possible.

There are several types of problems that can be explored using the analysis leading to Eq. 5‐91 and Eq. 5‐93, and we list three of the types as follows: Type I: Given the inlet mole fractions, ( x ) and ( y )

, the

A o

A N1

system parameters, and the desired value of ( y ) , we would like to A 1

determine the number of stages, N.

Type II: Given the inlet mole fractions, ( x ) and ( y )

, the

A o

A N1

system parameters, and the number of stages, N, we would like to determine the value of ( y ) .

A 1

Type III: Given the inlet mole fractions, ( x ) and ( y )

, the

A o

A N1

system parameters K

, M

 , the number of stages, N, and a

eq,A

desired value of ( y ) , we would like to determine the molar flow A 1

rate of the  ‐phase, M

  .

EXAMPLE 5.9. Absorption of acetone from air into water

In order to extract acetone from air using water, one can apply a contacting device such as the one illustrated in Figure 5‐7. The cascade of equilibrium stages can be represented by Figure 5‐10; however, the analogous system illustrated in Figure 5.9 is often used to more closely represent the gas‐liquid contacting process under consideration.

There are various problems associated with the design and operation of this unit and we begin with the first type of problem identified in the previous paragraph.

Type I: Given the inlet mole fractions, ( x ) and ( y )

, the

A o

A N1

system parameters, and the desired value of ( y ) , we would A 1

like to determine the number of stages, N.

200