Distillation - Principles of Chemical Separations with Environmental Applications - page 96

Example 4.2: enriching column

Problem:
Steam stripping is to be used to remove a solvent from contaminated soil. An enriching
column will be used to recover the solvent from the stream. A vapor feed of 40 mol hr

with a composition of 20 mol% solvent and 80 mol% water enters an enriching column.
The distillate stream is to have a flow rate of 5 mol

/

hr and a concentration of 90 mol%
solvent. The internal reflux ratio is 0.875 and constant molar overflow (CMO) may be
assumed. Graph the operating line to predict the number of equilibrium stages in this
enriching column.
Solution:
Assuming CMO means that L and V are constant, and the operating line will be straight.
The internal reflux ratio is L

/

V ,sothe easiest way to plot the mass balance for the
column (operating line) is to use the Equation (4.10):

/

L
V x n +

x D ( V − L )

y n + 1 = .

V

The slope of the operating line is the internal reflux ratio, 0.875. Now that we know
the slope, we need only one point to plot the line. It is possible to plot the y -intercept
of this line, but an easier point to find is the one where the operating line crosses the
diagonal, at x D = 9 (liquid and vapor compositions are equal since a total condenser

is used, i.e., all vapor is condensed to liquid).
Now we can solve the two mass balances simultaneously to find the composition of
the stream leaving the bottom of the column:

0 .

mol
hr =

mol
hr +

F = D + B ⇒ 40 5 B

40

moles solvent
moles feed

mol
hr

x F F = x D D + x B B ⇒ 0 . 20

5

moles solvent
moles distillate

mol
hr

= 0 . 90 + x B B .

So the bottoms have a concentration of 0.10 mol% solvent and a flowrate of 35 mol hr.

Once this concentration is found on the operating line, the stages can be “stepped off ”
(Figure 4.4). Starting at x D , draw a horizontal line to the equilibrium curve. This point
will give the mole fractions of the vapor and liquid streams on the first stage. From
this point, draw a vertical line down to the operating line. This point will give the mole
fractions of the passing streams ( x n and y n + 1 ,

/

1 for the passing streams below the
first stage). The first “step” is now complete, and the process is repeated until the passing
streams from the last stage meet or exceed the required value for x B . Remember that the
points on the operating line are passing stream mole fractions, while the points on the

n =