Unformatted text preview: 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. (810), xA F − xA M 0.45 − 0.365
S min
=
=
= 0.233
F
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 diagram on a following page. A line from RN through M to
an intersection with the equilibrium curve determines the extract point E1, as illustrated in Fig.
8.14. The operating point P is now determined, as in Fig. 8.17, by finding the intersection of
lines drawn through S and RN, and through F and E1. The number of equilibrium stages is now
stepped off, as in Fig. 8.17, giving a value of Nt slightly greater than 5. Exercise 8.11 (continued)
Analysis: (continued)
(c) First compute the two product flow rates, using the following product compositions
read from the plot:
Mass fractions:
Component Acetone Water
Feed
0.450
0.550
Solvent
0.000
0.000
Raffinate
0.100
0.895
Extract
0.509
0.040 1,1,2 TCE
0.000
1.000
0.005
0.451 By overall total material balance, F + S = 1,000 + 350 = 1,350 = RN + E1
1)
By overall acetone balance, 0.45F = 450 = 0.100RN + 0.509E1
(2)
Solving Eqs. (1) and (2) simultaneously, E1 = 770 kg/h and RN = 580 kg/h
The composition of each stream leaving each stage can be read from the righttriangle plot
prepared in part (b). Let x be mass fraction in the raffinate and y be mass fraction in the extract.
The stages are arbitrarily numbered from the feed end. The flow rate of each stream leaving each
stage is best obtained by total material balances around groups of stages and the inverse lever
arm rule, using the operating lines as. A total balance for the first n1 stages is:
or F + En= 1,000 + En = Rn1 + E1 = Rn1 + 770
Rn1  En = 1,000  770 = 230
(3) By the inverse lever arm rule with the righttriangle diagram,
Rn −1
EP
=n
En
Rn...
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 Spring '11
 Levicky
 The Land

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