# 5.16: Untitled Page 113

## Chapter 5

To continue this type of comparison, we recall Figure 5‐12 in the form given below by Figure 5‐17. There we have clearly identified Points #3, #4 and #5 in the graphical analysis with those same pairs of values in the sequential analysis. In Figure 5‐15 and Figure 5‐17 we see that Point #3 represents a point on the operating line, that Point #4 represents a point on the equilibrium line, and that Point #5 represents a second point on the operating line.

Figure 5‐17. Mole balance around Units #1 and #2

The sequential analysis presented in Sec. 5.6.1 has the advantage of providing a clear set of equations that describe the mass transfer process occurring in a cascade of equilibrium stages. The graphical analysis is certainly less accurate, but it can illustrate quite effectively how the system responds to changes in the operating conditions. As an example, we can reduce the molar flow rate of the

 ‐phase (water) and examine the effect that will have on the separation process.

For the case in which species A is transferred from the  ‐phase to the  ‐phase, the reduction of M

  will surely make the separation process less efficient. This change is clearly indicated in Figure 5‐18 where the effect of reducing M

  by

21% diminishes the efficiency of the cascade significantly. The graphical representation of a cascade of equilibrium stages is explored more thoroughly in Example 5.10.

Two‐Phase Systems & Equilibrium Stages

207

Figure 5‐18. Reduced molar flow rate for the  ‐phase Example 5.10. Graphical analysis of the absorption process

Here we extend Example 5.9 to explore the use of graphical methods to analyze the absorption of acetone from air into water. The molar flow rate of the gas mixture (air‐acetone) entering the cascade (at stage N) is M

30 kmol/h and the mole fraction of acetone is given by

 

( y )

0 010

.

The molar flow rate of the liquid (water) entering the A N 

1

cascade (at stage #1) is M

90 kmol/h and the mole fraction of acetone

 

in this entering liquid stream is zero, i.e., ( x )  0 . In this case the slope A o

of the operating line and the equilibrium coefficient are given by M

M

3 0

. ,

K

2 5

. 3

(1)

eq, A

and we can use these values to determine an operating line and an equilibrium line. Following the development given in Sec. 5.6.2 we construct the graph illustrated in Figure 5.10a. There we find that 5 stages

208