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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
Salt mass transfer rate in Stage 2:
X1 + X2
0 = ρ∗P2*Y2 - PM salt AM
− Y2 or
2 0.0199 + X2
Solving these 2 equations with a nonlinear solver such as POLYMATH, the following results are
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%
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
Dissolved solids material balance: 150 = 0.25 R1 + [300/(1,000,000+300)] P2
Solving Eqs. (1) and (2) simultan...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
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