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

# 826b ld vn sd d 93167 81521 3425 8221

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Unformatted text preview: The solvent in the final total raffinate = 724.9 - 657.5 = 67.4 lb/h. The total flow rate of the exiting solvent, SD , can not be found yet because the flow rate, VN , can not be determined until the extract reflux rate is known. However, for, YA = 0.83 in the extract leaving Stage N, the wt% C for an extract saturated with solvent = 87.5. Similarly, the entering fresh solvent rate, SB can not be determined yet. From the shape of the equilibrium curve on the previous page, at minimum extract reflux, the intersection of the operating lines for the extractor sections above and below the feed would appear to be at the intersection with a vertical line for the wt% A (solvent-free) in the feed, and with the equilibrium curve. Assume this is so. Then at minimum extract reflux, the pinch point will be at the feed stage. Accordingly, the composition of raffinate stream, LF+1, entering the feed stage F (in Fig. 8.26b) is identical to the composition of the feed. From the Y - X plot on the previous page, the solvent-free composition of the extract, VF , is that in equilibrium with the X = 0.35 of the feed, or Y = 0.56. From the figures on the next page, the wt% C in this extract is 91.3. Now, the entering and leaving flow rates can be calculated by material balances for the section of the extractor above the feed stage in Fig. 8.26b. An overall solvent-free material balance gives the following, where the prime superscript indicates a flow rate on a solvent-free basis: VF' = L'F +1 + D = L'F +1 + 342.5 (3) A similar balance on the solute, A, gives: YVF VF' = 0.56VF' = X LF +1 L'F +1 + YD D = 0.35 L'F +1 + 0.83(342.5) = 0.35 L'F +1 + 284.3 (4) Solving Eqs. (3) and (4), gives: VF' = 782.9 lb / h and L'F +1 = 440.4 lb / h The corresponding total flow rates (including solvent) are: VF = 782.9/(1 - 0.913) = 8,998.9 lb/h of which, 8,998.9 - 782.9 = 8,216.0 is solvent LF+1 = 440.4/(1 - 0.125) = 503.3 lb/h of which 503.3 - 440.4 = 62.9 is solvent Therefore, the exiting solvent rate, SD = 8,216.0 - 62.9 = 8,152.1 lb/h Analysis: (continued) Exercise 8.26 (continued) Exercise 8.26 (continued) Analysis: (continued) The extract, VN , contains 83 wt% A on a solvent-free basis. Therefore, From the equilibrium plots on a previous page, the corresponding wt% C in this extract is 87.5 wt%. Therefore, the total flow rate of VN = 8,152.1/(0.875) = 9,316.7 lb/h By total material balance around the solvent removal and stream divider above the top of the cascade (as shown in Fig. 8.26b), LD = VN - SD - D = 9,316.7 - 8,152.1 - 342.5 = 822.1 lb/h. Therefore, the minimum extract reflux rate = 822.1 lb/h, and the minimum extract reflux ratio = 822.1/342.5 = 2.40 To compute the number of equilibrium stages, take twice the minimum extract reflux flow rate or 2(822.1) = 1,644.2 lb/h with the same composition of 83 wt% A and 17 wt% B. The solvent-free extract is the same flow rate as before of D = 342.5 lb/h. Therefore, the solvent-free flow rate of V'N = 1,644.2 + 342.5 = 1,986.7 lb/h. Since, this flow is 87.5 wt% solvent, total flow rate VN = 1,986.7/(1 - 0.875) = 15,893.6 lb/h. The...
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