Unformatted text preview: + 622
∆παϖγ =
− 5.6 = 514.5 − 5.6 = 508.9 psi
2
(∆P  ∆π)avg = 1,950  508.9 = 1,441 psi
nH 2O
10, 000, 000
gal
From Eq. (1), PM H O =
=
= 3.47 × 10−3
2
AM ( ∆P − ∆π )avg 2, 000, 000(1, 441)
dayft 2 − psi
Now consider the actual crossflow case. The (P  π) on the permeate side is very small (less than
4% of that on the feedpermeate side). Therefore, the change in (∆P  ∆π) along the length of the
membrane is little affected by conditions on the permeate side. Therefore, the case computed
above well approximates crossflow.
From Eq. (1470), SP = ( csalt )Permeate / ( csalt )Feed = 0.00855 / 0.620 = 0.0138 Exercise 14.17
Subject: Reverse osmosis with multiple stages.
Given: 100 gal/min of feed.
Assumptions: All units of same size.
Find: Design 1  Drawing for single stage of 4 units in parallel to obtain 50% recovery.
Design 2  Drawing and % recovery for 2 stages in series with respect to retentate (4 units
in parallel followed by 2 units in parallel).
Design 3  Drawing and % recovery for 3 stages in series with respect to retentate (4 units
in parallel, then 2 units in parallel, then 1 unit). Analysis:
Design 1: The design is shown in a sketch below. Consider the following conditions for the
single stage of 4 units in parallel. To simplify calculations and comparison of designs, assume:
1. Water feed dilute in dissolved salts. Take the 100 gal/min of feed = F, containing 1 lb
salt/100 lb of water = XF1.
2. Density of all streams = 8.33 lb water/gal = ρ.
3. Driving force for water transfer across membrane = ∆P  ∆π = 1,000 psi for all units.
4. Water flux across all membrane units = 4.5 gal/dayft2.
5. Salt concentration in permeate from the stage = 0.01 lb salt/100 lb of water = Y1.
Therefore for the single stage of 4 units in parallel, the water transfer rate for 50% recovery of
potable water = 0.5(100) = 50 gal/min. Therefore, the total membrane area is,
AM = (50)(60)(24)/4.5 = 16,000 ft2 or 4,000 ft2 per unit.
Compute the permeance for water transport from a modification of Eq. (1469):
mwater
8.33(50)
=
= 1.666 × 10 −5 lb / min  ft 2  psi
AM ( ∆P − ∆π ) 16,000(1,000)
Compute the permeance for salt transport from a modification of Eq. (1452), where the transport
of salt across the membrane = 50(8.33)(0.0001) = 0.0416 lb/min, and a concentration difference
in terms of mass ratios is used, based on an arithmetic average on the feed side and perfect
mixing on the permeate side. Therefore,
PM water = PM salt =
AM msalt
0.0416
=
= 175 × 10−4 lb / min  ft 2  ( ∆ mass ratio)
.
0.01 + 0.0199
XF + XR
16,000
− 0.0001
− YP
2
2 Exercise 14.17 (continued)
Analysis: (continued)
Design 2: The design is shown in a sketch below. Retentate, R1 from Stage 1 is feed to
the Stage 2. Stage 1 consists of 4 membrane modules in parallel, each with 4,000 ft2 for a total of
16,000 ft2, and Stage 2 consists of 2 membrane modules in parallel, each with 4,000 ft2 for a total
of 8,000 ft2. Permeates from Stage 1 and...
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 Spring '11
 Levicky
 The Land

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