Absorption and stripping - Principles of Chemical Separations with Environmental Applications - page 165

/

G ) min =

.

68, also
starting from the point ( X 0 , Y N + 1 ). The result is four stages. Note that four stages
provides a better separation than required. This means that the actual Y 1 value will be
below the target value.

The actual operating line was plotted with a slope of 1.5( L 0

6.5 McCabe-Thiele analysis: stripping

Stripping is very similar in concept to absorption, but mass transfer occurs in the opposite
direction: it is the transfer of a component from a liquid stream into a gas stream mass-
separating agent. The mass balances and the operating line are derived in a similar fashion
to those for absorption. Referring to Figure 6.6, the operating line is

L
G

X n + 1 + Y 0 −

X 1

L
G

Y n = . (6.7)

The Y intercept is less than zero so the operating line is below the equilibrium line.
One useful limit for a stripping column is the maximum L / G ratio, which corresponds

to the minimum stripping gas flowrate (minimum G / L ratio) required for a desired sepa-

ration. The maximum L G is the slope of the line which begins at the point ( Y 0 , X 1 ) and

intersects the equilibrium line. As shown in Figure 6.7, the ( L

/

G ) max occurs as one rotates
the operating line clockwise around ( X 1 , Y 0 ) until it intersects the equilibrium line. This
can occur at a tangent pinch point and not necessarily at the end of the column ( X N , Y N ).

/

Example 6.3: stripping column

Problem :
Avolatile organic carbon (VOC) in a water stream is to be stripped out using an
air stream in a countercurrent staged stripper. Inlet air is pure, and flowrate is G

=

500 lb / hr. Inlet liquid stream has a mass ratio of X = 0.10 and a flowrate of pure

water of 500 lb 0.005. Assume that water is

non-volatile and air is insoluble. Find the number of equilibrium stages and the outlet
gas mass ratio. Equilibrium data can be represented by Y

/ hr. The desired outlet mass ratio is X =

= 1.5 X .

Solution:
Figure 6.8 is a schematic diagram of the column. The equilibrium data are simple to
plot (Figure 6.9) because everything is already given in terms of mass ratios. One
point on the operating line is ( X 1 , Y 0 ), and the slope is 1 since the inert liquid and gas
flowrates are equal. The separation will require five equilibrium stages, and the outlet
gas mass ratio ( Y N )is0.095.

6.5.1 Analytical methods

When the solute concentration in both the gas and liquid phase is very dilute, the total
flowrates can be considered constant. The mole or mass ratios reduce to the corresponding