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

Feed flow rate 150060359 2507 lbmolh acetone in the

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online