Environmental Engineering Reference
In-Depth Information
50
40
O
N
+1
30
V
N
,min
20
O
1
M
min
10
Conjugate
line
tie-
line
0
0
20
40
60
80
100
V
0
∆
used
wt% water (component
A
)
∆
min
Figure 5.18
Minimum solvent flowrate calculation: water-acetic acid-isopropyl
ether, Example 5.3.
Example 5.3: minimum solvent flowrate
Problem:
Find the minimum solvent flowrate for Example 5.2.
Solution:
The previous steps were used to create Figure 5.18.
V
0
O
N
+
1
O
N
+
1
M
min
V
0
M
min
min
=
=
1
.
6
.
Since
O
N
+
1
=
1000 kg
/
hr,
V
0min
=
1600 kg
/
hr. The actual solvent flowrate in the
previous example was 2500 kg
6 for that extraction process. Using
even more solvent would decrease the required number of equilibrium stages.
/
hr, so
V
0
/
V
0min
=
1
.
5.8
Countercurrent extraction with feed at intermediate stage
The concentration of solute (
A
)inthe outlet extract stream is limited to a relatively
low value because the outlet extract stream is a passing stream with feed
O
N
+
1
.So
the maximum concentration of
A
in
V
N
will occur when the column is operated at the
minimum solvent flowrate. As with distillation, this limitation can be overcome by using
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