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

# Feed flow rate 150060359 2507 lbmolh acetone in the

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Unformatted text preview: 2 are combined. As is Design 1, P1 = 50 gal/min. With only 50% of the membrane area of Stage 1, P2 = 25 gal/min and R2 = 25 gal/min. Thus, the total permeate is 75 gal/min, giving a recovery of 75%. Design 3: The design is shown in a sketch below. Retentates R1 and R2 are feeds to Stages 2 and 3, respectively. Permeates from the 3 stages are combined to give the final permeate. Stage 1 consists of 4 membrane modules in parallel, each with 4,000 ft2 for a total of 16,000 ft2; Stage 2 consists of 2 membrane modules in parallel, each with 4,000 ft2 for a total of 8,000 ft2; and Stage 3 consists of a single unit of 4,000 ft2. The permeate from Stage 3 is only 12.5 gal/min because of the single unit. The total permeate = 50 + 25 + 12.5 = 87.5 gal/min for an 87.5% recovery. It is also of interest to compute the purity of the potable water produced. For example, for Design 2, the permeate from Stage 1 is 50 gal/min with 0.0001 lb salt/lb water. The feed to Stage 2 is 50 gal/min with 0.0199 lb salt/lb water. Using the results above from Design 1, Salt mass balance around Stage 2: 0 = ρ(R1*X1 - P2*Y2 - R2*X2 or 0 = 50(0.0199) - 25*X2 - 25*Y2 (1) Salt mass transfer rate in Stage 2: X1 + X2 0 = ρ∗P2*Y2 - PM salt AM − Y2 or 2 0.0199 + X2 − Y2 (2) 2 Solving these 2 equations with a nonlinear solver such as POLYMATH, the following results are obtained: X2 = 0.0396 and Y2 = 0.000199. Therefore, the final combined permeate contains a salt mass ratio of [50(0.0001) + 25(0.0001987)]/75 = 0.0001329 which is higher than the mass ratio of 0.0001 for Design 1. Thus, a higher recovery is counterbalanced by a less pure potable water. 8.33(25)*Y2 - 1.75 x 10-4(8,000) Analysis: (continued) Exercise 14.17 (continued) Exercise 14.18 Subject: Concentration of Kraft black liquor by a two-stage reverse osmosis process. Given: Fresh feed of 1,000 lb/h at 180oF and containing 15 wt% dissolved solids. Two-stage arrangement shown in Fig. 14.36, with recycle of retentate from Stage 2 to Stage 1. Combined feed to Stage 1 at 1,756 psia. Permeate from Stage 1 contains 0.4 wt% dissolved solids at a pressure of 15 psia. It is pumped to 518 psia to become feed to Stage 2. Permeate from Stage 2 contains 300 ppm dissolved solids at 15 psia. Retentate from Stage 2 contains 2.6 wt% dissolved solids and is recycled to Stage 1. Permeance for water = 0.0134 lb/ft2-h-psi. Osmotic pressure for 25 wt% dissolved solids is 1,700 psia. Other osmotic pressures are proportional to wt% dissolved solids. Assumptions: Arithmetic-mean osmotic pressure for plug flow on feed side of both stages. Perfect mixing on permeate side. No pressure drop on feed side. Find: Material balance and membrane areas. Analysis: Compute overall material balance: Let R1 = retentate flow rate from Stage 1, and P2 = permeate flow rate from Stage 2. Fresh feed in is 850 lb/h of H2O and 150 lb/h of dissolved solids. Total material balance: 1,000 = R1 + P2 (1) (2) Dissolved solids material balance: 150 = 0.25 R1 + [300/(1,000,000+300)] P2 Solving Eqs. (1) and (2) simultan...
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