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

825 the result is 25 equilibrium stages exercise 813

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Unformatted text preview: s given by the inverse lever arm rule as the ratio of line length MF to line length MS, which equals 1.67. Take a basis of 100 kg/h of feed. Then the solvent rate is 1.67(100) = 167 kg/h. The flow rates of raffinate and extract are obtained from an overall total material balance and an overall material balance for A: F + S = 100 + 167 = 267 = R + E xA F (1) F + xA S S = 0.45(100) + 0 = 45 = xA R R + xA E E = 0.025R + 0.20 E (2) Solving Eqs. (1) and (2) simultaneously, R = 48 kg/h and E = 219 kg/h. The resulting overall material balance is as follows, where raffinate and extract compositions are read from the phase boundary curve. Component Isopropyl alcohol (A) Diisopropyl ether (C) Water (S) Total Feed 45.0 50.0 5.0 100.0 Solvent 0.0 0.0 167.0 167.0 Raffinate 1.20 46.28 0.52 48.00 Extract 43.80 3.72 171.48 219.00 To determine the number of equilibrium stages for the above solvent rate and material balance, using the Varteressian-Fenske method with a McCabe-Thiele diagram, an equilibrium curve and an operating curve are needed in a McCabe-Thiele plot of mass fraction of alcohol in the extract against the mass fraction of alcohol in the raffinate, as illustrated for a different system in Fig. 8.25. The equilibrium curve is determined from the equilibrium tie-line data, shown in a right triangle diagram below and obtained directly from the statement of the exercise. These equilibria data are tabulated below. The operating curve is determined by first determining the operating point, P, as illustrated in Fig. 8.15, followed by drawing arbitrary straight lines through that point Analysis: (continued) Exercise 8.13 (continued) Exercise 8.13 (continued) Analysis: (continued) to give corresponding extract and raffinate compositions from intersections with the phase boundary curve, as illustrated in Fig. 8.21c. Several such operating lines are shown on the righttriangle diagram below. The operating points used to draw the operating curve on the McCabeThiele are tabulated below. The points for the equilibrium curve are: Mass fraction alcohol in Extract 0.081 0.102 0.117 0.175 0.217 0.268 Mass fraction alcohol in Raffinate 0.024 0.050 0.093 0.249 0.380 0.452 Analysis: (continued) Exercise 8.13 (continued) The determined points for the operating curve are: Mass fraction alcohol in Extract 0.000 0.009 0.044 0.095 0.150 0.200 Mass fraction alcohol in Raffinate 0.025 0.050 0.150 0.250 0.350 0.450 These points are plotted in the McCabe-Thiele diagram below, where the stages are stepped off as illustrated in Fig. 8.25. The result is 2.5 equilibrium stages. Exercise 8.13 (continued) Analysis: (continued) To determine if an extract of 25 wt% A can be obtained, as shown in the diagram below, the feed and desired extract points are plotted as F and E, respectively, with a straight operating line drawn between the two points. Since that line appears to be coincident with an equilibrium tie line, the extract could be obtained, but only with an infinite number of stages. An extract with an alcohol...
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