Unformatted text preview: −1 P (4) where the lines are measured from the right-triangle diagram.
Eqs. (3) and (4) are solved simultaneously for Rn-1 and En for n = 2 to 6. The following results
0.100 0.895 xS
0.824 580 E
426 Exercise 8.12
Subject: Extraction of acetone (A) from water (C) by 1,1,2-trichloroethane (S) at 25oC
Given: 1,000 kg/h of 45 wt% A in C. Liquid-liquid equilibrium data. 10 wt% A in raffinate.
Find: Using a right triangle diagram.
(a) Minimum flow rate of S.
(b) Number of equilibrium stages for solvent rate of 1.5 times minimum.
(c) Flow rate and composition of each stream leaving each stage.
Analysis: Using the given equilibrium data in weight fractions, the right-triangle diagram is
shown on the next page, where a solid line is used for the equilibrium curve and dashed lines are
used for the tie lines, only three of which are given. Additional tie-line locations can be made
using either of the two techniques illustrated in Fig. 8.16.
(a) The minimum solvent flow rate corresponds to an infinite number of equilibrium
stages. To determine this minimum: (1) Points are plotted on the triangular diagram for the
given compositions of the solvent, S (pure S), feed, F (45 wt% A, 55 wt% C), and raffinate, RN
(10 wt% A on the equilibrium curve), followed by drawing an operating line through the points S
and RN and extending it only to the right because the tie lines slope down from left to right; (2)
Because the pinch region, where equilibrium stages crowd together, may occur anywhere,
additional potential operating lines are drawn through the tie lines extended to the right until they
cross the operating line at the solvent end, drawn in (1), where these intersections for four tie
lines (the three given and one other that extends through the feed point) are denoted P1, P2, P3,
and P4; (3) The intersection point farthest from the triangular diagram is the one used to
determine the minimum solvent rate, where here it is P1 = Pmin , corresponding to a pinch point at
the feed end of the cascade; (4) An operating line is drawn from P1 through F to the determine
the extract E1, followed by drawing lines from E! to RN and from S to F, with the intersection of
these two lines determining the mixing point, M. The wt% A at the mixing point is 36.5%.
From Eq. (8-10),
xA F − xA M 0.45 − 0.365
xA M − xA S
0.365 − 0.0
Therefore, Smin = 0.233F = 0.233(1,000) = 233 kg/h
(b) Solvent rate = 1.5 Smin = 1.5(233) = 350 kg/h
With this rate, the total (feed + solvent) component flow rates becomes: A = 450 kg/h, C
= 550 kg/h, and S = 350 kg/h, giving a total of 1,350 kg/h. Thus, the mixing point is at (xA)M =
450/1,350 = 0.333, (xC)M = 550/1,350 = 0.408, and (xS)M = 350/1,350 = 0.259. This point is
plotted as M on the second triangular dia...
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