CEE 6330 HW-4 Solutions PDF

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CEE 6330 Physicochemical process:HOMEWORK #4: (Due April 11)Solution:1. (40 pts) Derive design equations for reverse osmosis (RO) membrane processes by combining the(a).(b).(c).2. (30 pts)a). (5 pts)b). (5 pts)Solution:c). (5 pts)d). (5 pts)e).(10 pts)3. (30pts). The MF membrane in Example 12-3 is u...

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CEE 6330 Physicochemical process: HOMEWORK #4: (Due April 11) 1. (40 pts) Derive design equations for reverse osmosis (RO) membrane processes by combining the

(a). (b). (c).

Solution:

2. (30 pts)

a). (5 pts) b). (5 pts)

c). (5 pts)

d). (5 pts) e).(10 pts)

Solution:

3. (30pts). The MF membrane in Example 12-3 is used under full-scale conditions in a water treatment facility, producing a flux of 84 L/m2 · h at 1.1 bar just before cleaning and 106 L/m2 · h at 0.52 bar immediately after cleaning, both at a standard temperature of 20º C. Calculate values for the membrane resistance coefficient, irreversible fouling resistance coefficient, and chemically reversible fouling resistance coefficient. Solution 1. The membrane resistance coefficient was calculated in Example 12-3 (under conditions when κir = 0 and κcr = 0) and found to be 3.81 × 1011 m–1. 2. Determine the irreversible fouling resistance coefficient. a. b.

The viscosity of water at 20º C from Table C-1 in Appendix C is 1.00 × 10–3 kg/m · s. Also recall that 1 bar = 100 kPa = 10 5 N/m2 = 105 kg/s2 · m. The reversible component of fouling is removed by cleaning, so immediately after cleaning the chemically reversible fouling resistance coefficient is zero (κcr = 0). The only factors that cause resistance to flow are the membrane resistance and the irreversible fouling resistance, so the resistance-in-series equation can be written

J= c.

Rearrange Eq. 1 to solve for κir:

∆P µ ( κ m + κ ir )

(1)

κir =

(

)

(

0.52 × 10 5 kg s 2 ⋅ m (3600 s h ) 1 ×10 3 L m 3 ∆P − κm = −3 2 µJ 1.00 ×10 kg m ⋅ s 106 L m ⋅ h

(

)(

)

)

−3.81× 1011 m −1 = 1.39 × 1012 m −1 3. Determine the chemically reversible fouling resistance coefficient. a.

J=

Prior to cleaning, three components of resistance are present:

∆P µ ( κm + κir + κ cr ) b.

Rearrange the above equation to solve for κcr:

(

)

(

1.1 ×105 kg s 2 ⋅ m ( 3600 s h ) 1 ×10 3 L m 3 ∆P κcr = − κm − κir = µJ 1.00 ×10 −3 kg m ⋅ s 84 L m 2 ⋅ h

(

)(

)

)

−3.81 ×1011 m −1 − 1.39 × 1012 m−1

= 2.94 ×1012 m −1 Comment In this example, the chemically reversible fouling resistance coefficient is the largest resistance and nearly an order of magnitude larger than the membrane resistance coefficient, which demonstrates the importance of fouling on overall membrane performance....