Environmental Engineering Reference
In-Depth Information
Solution:
(a) In Figure 3.24, the point
F
marks the initial solution: 0% F, 45% G, 55% W. As
furfural is added to this mixture, the path from point
F
to point
A
is followed.
When the point
M
has been r
eac
h
ed,
a mixture which is half furfural and half feed
has been obtained (because
AM
1). Notice that this is in the two-phase
region. It will separate into two phases: the extract phase at point
E
(28.5% G,
6.5% W, 65.0% F), and the raffinate phase at point
R
(8% G, 84% W, 8% F). These
points are found from the tie-line through point
M
. The compositions of the two
phases can also be solved simultaneously:
/
FM
=
1
/
R
+
E
=
200 lb
(overall balance)
0.65(lb F
/
lb extract)
E
+
0.08(lb F
/
lb raffinate)
R
=
100 lb
(furfural balance)
R
=
53 lb, and
E
=
147 lb.
Note that this answer could also have been obtained from the lever-arm rule:
E
=
200(
RM
/
RE
)
and
R
=
200(
EM
/
ER
)
.
(b) The path followed is from point
E
to point
P
on the diagram. The extract compo-
sition is 82.5% G, 17.5% W.
This problem can also be solved with the right triangle diagram in Figure 3.25
.
Note
that the diagonal axis represents different solutions which are all 0% furfural. Even
though the % furfural is difficult to pick off this graph because the horizontal and
vertical axes are not the same length, remember that it can be found by subtracting the
other two compositions from 100%
.
ETHYLENE
GLYCOL
0
100%
100%
0
P
20
80
GLYCOL
WATER
GLYCOL
60
40
FURFURAL
F
40
60
E
0
M
100%
20
80
100%
0
R
A
WATER
FURFURAL
80
60
40
20
100%
0
FURFURAL
0
WATER
100%
Figure 3.24
Furfural-ethylene glycol-water phase diagram [10]. Copyright
c
1998, J. Wiley and Sons, Inc.
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