# 5.9: Untitled Page 107

## Chapter 5

5.6.1 Sequential analysis‐algebraic

There are a variety of problem statements associated with a system of the type illustrated in Figure 5‐10. For example, one might be given the following: (1) The number of stages, N, (2) the mole fractions for the inlet conditions, ( x ) and A o

( y )

, (3) the equilibrium relation applicable to each stage, and (4) the molar A N1

flow rates of the two phases, M

 and M  . Given this information one would then be asked to determine the outlet conditions, ( y ) and ( x ) .

A 1

A N

Alternatively, one might be given: (1) The mole fractions for the inlet streams, ( x ) and ( y )

, (2) the equilibrium relation, and (3) the molar flow rates of A o

A N1

the two phases, M

 and M  . In this case one would be asked to determine the number of stages, N, that would be required to achieve a desired composition in one of the outlet streams.

In the following paragraphs we develop an algebraic solution for the cascade of N equilibrium stages illustrated in Figure 5‐10, and we illustrate how this result can be applied to several different problem statements. We begin our analysis of the cascade of equilibrium stages with the single unit illustrated in Figure 5‐11 in which the obvious control volume has been identified.

Figure 5‐11. Single unit extraction process

For this case the mole balance for species A takes the form ( x ) M

 ( y ) M

( x ) M

 ( y ) M

(5‐66)

A 1

A 1

A o

A





2 

molar flow of species A

molar flow of species A

out of the control volume

into the control volume

and we impose the special condition given by

Special Condition:

( x )

 0

(5‐67)

A o

Two‐Phase Systems & Equilibrium Stages 195

For this condition Eq. 5‐66 takes the form

( x ) M

 ( y ) M

 ( y ) M

(5‐68)

A 1

A 1

A 2

At this point our objective is to develop a relation between ( y ) and ( y ) , and A 1

A 2

we begin by arranging Eq. 5‐68 in the form

( y )  ( x )

M

M

 ( y )

(5‐69)

A 1

A 1 

 

A 2

Here we note that the process equilibrium relation is given by Process equilibrium relation:

( y )

K

( x )

(5‐70)

A 1

eq,A

A 1

and use of this result allows us to simplify the mole balance for species A to obtain

( y )  ( y )

M

M

K

 ( y )

(5‐71)

A 1

A 1 

eq,A

A 2

We now define the absorption factor according to

A   MMK

(5‐72)

eq,A

in order to develop the following relation between ( y ) and ( y ) A 1

A 2

( y )

One Equilibrium Stage:

A 2

( y )

A 1

1 

(5‐73)

A

Here we have chosen to arrange the macroscopic mole balance for species A in terms of the mole fraction in the  ‐phase. However, the choice is arbitrary and we could just as well have set up the analysis in terms of x instead of y .

A

A

Having developed an expression for the exit mole fraction, ( y ) , in terms of A 1

the entrance mole fraction, ( y ) , for a single equilibrium stage, we now wish to A 2

relate ( y ) to ( y ) . To accomplish this we examine the two equilibrium stages A 1

A 3

illustrated in Figure 5‐12. In this case the molar balance for species A can be expressed as

( x ) M

 ( y ) M

( x ) M

 ( y ) M

(5‐74)

A 2

A 1

A o

A







3 



molar flow of species A

molar flow of species A

out of the control volume

into the control volume

and we continue to impose the special condition

196