{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

# The stages are arbitrarily numbered from the feed end

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. (8-10), 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 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 as. A total balance for the first n-1 stages is: or F + En= 1,000 + En = Rn-1 + E1 = Rn-1 + 770 Rn-1 - En = 1,000 - 770 = 230 (3) By the inverse lever arm rule with the right-triangle diagram, Rn −1 EP =n En Rn...
View Full Document

{[ snackBarMessage ]}