Environmental Engineering Reference
In-Depth Information
5.11
More extraction-related examples
Example 5.6: countercurrent vs cross-flow extraction
Problem:
Suppose we have a feed stream which contains acetic acid (A) and water (W). Because
it would not be economically feasible to incinerate the water stream, we'd like to
extract the acetic acid into an isopropyl ether (I) phase. The isopropyl ether-acetic
acid stream could then be used as a fuel stream to another process. Note that the
isopropyl ether has a limited solubility in the water, and vice versa.
Graphically analyze two configurations of equilibrium stages (Figure 5.26): coun-
tercurrent and cross-flow. A small number of mass balance calculations are required
in addition to the graphical work. All compositions given are in mass fractions. Use
two equilibrium stages in each of the configurations.
Solution:
Given:
V
3
=
30 lb
/
hr
V
3
/
2
=
15 lb
/
hr
O
0
=
10 lb
/
hr
y
3
(A)
=
0
y
3
(I)
=
1
.
0
y
3
(
W
)
=
0
x
0
(A)
=
0
.
45
x
0
(I)
=
0
x
0
(
W
)
=
0
.
55
.
Equilibrium data can be found in Table 5.1, Example 5.1.
Additional information valid for countercurrent
only
:
y
1
(A)
=
0
.
125
x
2
(A)
=
0
.
13
.
The solution for the countercurrent cascade (Figure 5.27) is slightly different than in
Example 5.5, because now the number of stages
inste
ad of the product concentration
is specified. From the information given, the line
V
3
2
O
can be drawn.
V
1
and
O
2
can also be plotted from the given information because they both lie on
the solubilit
y env
elop
e (they
are leaving the column in equilibrium). The intersection
of the lines
O
0
V
1
and
V
3
2
point.
The mixing point is also shown on the graph, but it is not used to locate
V
1
as usual
since
V
1
is already specified. A tie-line through
V
1
locates the point
O
1
, and the line
O
2
locates the
O
1
crosses the solubility envelope at the point
V
2
.
Now all of the streams are located, and their compositions can be read from the
graph. Using the lever-arm rule to find the amounts of the streams:
O
0
V
1
14
.
75 cm
V
1
O
0
=
=
=
3
.
14 (using a ruler)
.
∴
V
1
=
.
/
.
31
4lb
hr
.
4
7cm
O
1
V
2
=
V
2
3
.
15 cm
O
1
=
85 cm
=
0
.
25
,
also:
O
1
+
V
1
=
O
0
+
V
2
.
12
.
Solving simultaneously gives
V
2
=
28.5 lb
/
hr and
O
1
=
7.1 lb
/
hr.
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