Environmental Engineering Reference
In-Depth Information
V
0
y
A
0
+
O
0
x
A
0
z
A
1
=
=
0
.
31
,
V
0
+
O
0
and it was plotted as a point on the line
O
0
V
0
(the lever-arm rule could also have been
used). A tie-line was then drawn through this mixing point; the ends of the tie-line
are the points
O
1
(0.73, 0.25) and
V
1
(0.02, 0.06), and their compositions were read
from the graph. Two mass balances (overall and one component) solved simultaneously
give the flowrates of the streams exiting the first stage.
Overall mass balance:
O
1
+
V
1
=
1000 kg
/
hr
+
1250 kg
/
hr
=
2250 kg
/
hr
.
Forwater
:
0
.
7(1000 kg
/
hr)
=
0
.
73
O
1
+
0
.
02
V
1
;
V
1
=
1327 kg
/
hr
,
O
1
=
923 kg
/
hr
.
In Stage 2, the stream
O
1
is mixed with a solvent stream identical in composition
and mass as the one in Stage 1, so the next line drawn connects
O
1
with
V
0
. The next
mixing point,
M
2
is found, and the analysis for Stage 1 is repeated for Stage 2. Note
that the
O
streams would not all be connected to
V
0
if different compositions of the
solvent phase were used for each stage. The compositions of
O
2
,
V
1
, and
V
2
can be
read directly from Figure 5.9. Expressed as fractions, they are:
x
A
2
=
0
.
78
x
B
2
=
0
.
205
y
A
1
=
0
.
02
y
B
1
=
0
.
06
y
A
2
=
0
.
015
y
B
2
=
0
.
045
.
The flowrate,
O
2
, can be calculated using the lever-arm rule:
M
2
V
2
O
2
V
2
×
2250
=
837 kg
/
hr
=
O
2
.
Table 5.1
Equilibrium data for Example 5.1.
Water layer, wt%
Isopropyl ether layer, wt%
Isopropyl ether
Water
Acetic acid
Isopropyl ether
Water
Acetic acid
1.2
98.1
0.69
99.3
0.5
0.18
1.5
97.1
1.41
98.9
0.7
0.37
1.6
95.5
2.89
98.4
0.8
0.79
1.9
91.7
6.42
97.1
1.0
1.93
2.3
84.4
13.30
93.3
1.9
4.82
3.4
71.1
25.50
84.7
3.9
11.40
4.4
58.9
36.70
71.5
6.9
21.60
10.6
45.1
44.30
58.1
10.8
31.10
16.5
37.1
46.40
48.7
15.1
36.20
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