Separations analysis fundamentals - Principles of Chemical Separations with Environmental Applications - page 72

or, assuming α is constant,

x A
x B

x A
x B

b ,

N min
AB

d = α

(3.58)

where subscript d denotes the top of the cascade where component A is enriched and b
denotes the bottom of the cascade where component B is enriched. This equation can be
rearranged to solve for the minimum number of stages

ln ( x A / x B ) b

x b ) d / ( x A /

N min = . (3.59)

ln α AB

This equation is called the Fenske-Underwood equation.
Note that the minimum number of equilibrium stages increases non-linearly as the
separation required becomes more difficult [more extreme x A / x B ratio or α AB closer to

unity].
Equation (3.59) illustrates two important points: (1) the value of the minimum number
of stages is independent of the feed conditions and only depends on the separation
requirements; and (2) increasing the number of stages is usually more effective than
increasing flow to increase product purity for difficult separations.

Example 3.7

Problem:
Using the information in the previous example, calculate N min . The binary system
is H 2 S and N 2 . The water is essentially an immiscible liquid since it is in low
concentration in the gas phase and there is no net mass transfer of water.

10 4 x N 2 . y N 2 = 8 . 6 ×

Solution:
For equilibration of immiscible phases, one can write

m A
m B .
Since N 2 is the component being enriched,

α AB =

10 4 8 . 6 ×

α AB = = 172 ,

5 . 0 × 10 2

and

ln [(99 / 1) / (90 / 10)]

N min = = 0 . 47 .

ln (172)

Scaling up, one equilibrium stage would work as L infinity. This result empha-

sizes the point that it is better to increase the number of stages than increase the volumes
(or flowrates).

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