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Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Do this first by constructing a right triangle

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Unformatted text preview: gram 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 just slightly greater than Nt = 5. Exercise 8.12 (continued) Analysis: (continued) (c) First compute the two product flow rates, using the following product compositions read from the plot: Component Feed Solvent Raffinate Extract Mass fractions: Acetone Water 0.450 0.550 0.000 0.000 0.100 0.895 0.509 0.040 1,1,2 TCE 0.000 1.000 0.005 0.451 1) By overall total material balance, F + S = 1,000 + 350 = 1,350 = RN + E1 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 right-triangle 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. A total balance for the first n-1 stages is: F + En= 1,000 + En = Rn-1 + E1 = Rn-1 + 770 Rn-1 - En = 1,000 - 770 = 230 or (3) By the inverse lever arm rule with the right-triangle diagram, Rn −1 EP =n En Rn −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 are obtained: Stage 1 2 3 4 5 Raffinate: xA xC 0.390 0.581 0.325 0.651 0.260 0.724 0.190 0.800 0.100 0.895 xS 0.029 0.024 0.016 0.010 0.005 Extract: yA 0.509 0.440 0.360 0.270 0.165 yC 0.040 0.033 0.024 0.017 0.011 yS 0.451 0.537 0.616 0.713 0.824 kg/h: R 935 822 738 656 580 E 770 706 592 508 426 Exercise 8.13 Subject: Extraction of isopropanol (A) from diisopropylether (C) by water (S) at 25oC. Given: Feed containing 45 wt% A, 50 wt% C, and 5 wt% S. Raffinate to contain 2.5 wt% A. Extract to contain 20 wt% A. Liquid-liquid equilibrium data. Find: Using Varteressian-Fenske method with a McCabe-Thiele diagram, find the number of theoretical stages. Can an extract of 25 wt% A be obtained? Analysis: Because the wt% of the solute, A, in both the raffinate and extract are given, the solvent rate must be determined. Do this first by constructing a right-triangle diagram from the equilibrium tie line and phase boundary data. As shown on the next page, for a diagram that does not show the tie lines, points on the diagram are placed for: the feed, F, (45 wt% A, 50 wt% C, and 5 wt% S); the solvent, S, (100% S); the raffinate, R, (2.5 wt% A on the phase boundary line); and the extract, E, (20 wt% A on the phase boundary line). The mixing point, M, is determined by the intersection of lines drawn from: (1) F to S and (2) R to E. The ratio of solvent to feed i...
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