Page 170 - Separation process principles 2

Water
(A)
Mass fraction furfural
Figure 4.15 Solution to Example 4.7a.

SOLUTION phase at point E (8.5% B, 4.5% A, and 87.0% C) and the raffi-
nate at point R (34.0% B, 56.0% A, and 10.0% C).
Assume a basis of 100 g of 45% glycol-water feed. Thus, in Fig- Step 5. The inverse-lever-arm rule applies to points E, M, and
--
ure 4.13b, the feed (F) is 55 g of A and 45 g of B. The solvent (S) is R, so E = M(RM/ER). Because M = 100 + 200 = 300 g, and
200 g of C. Let E = the extract, and R = the raffinate. from measurements of the line segments, E = 300(147/200) =
(a) By an equilateral-triangular diagram, Figure 4.15: 220gandR = M- E =300-220=80g.
(b) By a right-triangular diagram, Figure 4.16:
Step 1. Locate the feed and solvent compositions at points F
and S, respectively. Step 1. Locate the points F and S for the two feed streams.
Step 2. Define M, the mixing point, as M = F + S = E + R Step 2. Define the mixing point M = F + S.
Step 3. Apply the inverse-lever-arm rule to the equilateral- Step3. The inverse-lever-arm rule also applies to right-
triangular phase-equilibrium diagram. Let wi(') be the mass triangular diagrams, so MF/MS = ;.
fraction of species i in the extract, w!') be the mass fraction of Step 4. Points R and E are on the ends of the interpolated dash-
species i in the raffinate, and wjMi be the mass fraction of dot tie line passing through point M.
species i in the combined feed and solvent phases.
The numerical results of part (b) are identical to those of
From a balance on the solvent, C: (F + s)w&*) =
FW~) + SW:). Part (a).
(c) By an equilibrium solute diagram, Figure 4.14~. A material
Therefore,
balance on glycol, B,

Thus, points S, M, and F lie on a straight line, and, by the
must be solved simultaneously with a phase-equilibrium relation-
inverse lever arm rule,
ship. It is not possible to do this graphically using Figure 4.14~ in
any straightforward manner unless the solvent (C) and carrier (A)
are mutually insoluble. The outlet-stream composition can be
found, however, by the following iterative procedure.
The composition at point M is 18.3% A, 15.0% B, and 66.7% C.
Step 4. Since M lies in the two-phase region, the mixture must Step 1. Guess a value for wr) and read the equilibrium value,
separate along an interpolated dash-dot tie line into the extract wr), from Figure 4.14~.