1. MEMBRANE SEPARATIONS
 The most rapidly advances area of filtration technology.
 There are three major types of membrane-based filtration techniques:
(1) microfiltration,
(2) ultrafiltration, and
(3) reverse osmosis
 They are classed according to the particle size they commonly remove from solutions.

2. Types of membrane filtration
 There are three major types of membrane-based filtration techniques: microfiltration, ultrafiltration, and reverse osmosis.

3. Microfiltration  Retain particles as small as 0.1 mm.
Types of membrane filtration
Microfiltration
 Retain particles as small as 0.1 mm.
 - They retain even the smallest bacteria.
- They will not retain dissolved proteins.
 Used extensively in bioprocesses as sterile filter for both liquid and gas streams.
 They are used to filter-sterilize heat-sensitive media.

4. Types of membrane filtration
Ultrafiltration
 Being able to retain dissolved proteins with molecular weights as low as a few thousand.
 Rated in terms of their molecular-weight cutoff.
 Widely used in the separation of biological products.

5. Types of membrane filtration
Reverse Osmosis
 Retain not only proteins but also dissolved ionic salts and small organic molecules with MW in the hundreds.
 Used most extensively in the purification of water and the concentration of biological and food processing streams.

6. Diafiltration  An alternative method of operating an ultrafilter.
Types of membrane filtration
Diafiltration
 An alternative method of operating an ultrafilter.
 Repeated or continuous addition of fresh solvent (usually water) in an ultrafiltration system.
 To wash out any contaminants not retained by the
membrane.

7.  A continuous countercurrent diafiltration system:
Types of membrane filtration: diafiltration (2/2)
 A continuous countercurrent diafiltration system:

8. Electrodialysis
 Employ semi-permeable ion-exchange membranes that are impervious to water.
 The separation is electrically driven instead of pressure-driven.

9. Cross-Flow Filtration
“Most of the pressure drop in conventional filtration comes from the cake.”
* Concentration polarization: accumulation of solute near the membrane surface

10. * We may wish that we could filter without a filter cake.
Cross-Flow Filtration
* We may wish that we could filter without a filter cake.
 Use filtration where cross flow is dominant.

12. ULTRAFILTRATION
 Ultrafiltration is a membrane process; it involves solvent transport under pressure.

13. ULTRAFILTRATION (2/6)
 Because of the range of pore sizes in an ultrafiltration membrane, there is no absolute cutoff point.
_____________

14.  Ultrafiltration processes have three distinctive characteristics:
(1) Use of high cross flow
 Reduce cake formation or concentration polarization.
(2) Dominated by the membrane
 In conventional filtration, the choice of filter medium usually has little effect on flow through the cake.
(3) Filtration performance depends on the membrane
geometry in actual equipment.

15. ULTRAFILTRATION (4/6)

16. ______________________________________________________
ULTRAFILTRATION (5/6)
______________________________________________________
 For hollow fiber, feed streams generally need to be prefiltered.

17.  Operation of a batch ultrafiltration:
* Ultrafiltration slows down as the solute concentration increases.
 Re-dilution once or twice may be useful before finally collecting the retentate.

18. Solvent velocity  force on solvent
ANALYSIS OF ULTRAFILTRATION
In ultrafiltration, the species transported is the solvent and the chief force is the transmembrane pressure (DP).
Solvent velocity  force on solvent
Darcy's law:

19. where q = thickness the membrane m = viscosity of the permeate
ANALYSIS OF ULTRAFILTRATION (2/4)
Darcy's law: 
where q = thickness the membrane
m = viscosity of the permeate
jv = the volume of solvent per area per time (or the solvent velocity)
Rm = membrane resistance
Rp = resistance of the polarized boundary layer
Lp = the permeability

20. Taking the osmotic pressure into consideration,
ANALYSIS OF ULTRAFILTRATION (3/4)
Taking the osmotic pressure into consideration, 

21. where p = the osmotic pressure s = a reflection coefficient
ANALYSIS OF ULTRAFILTRATION (4/4)
where p = the osmotic pressure
s = a reflection coefficient
* s = 1
 The membrane rejects all solutes.
* s = 0
 The membrane freely passes both solvent and solute.

22. Estimation of Osmotic Pressure—the van’t Hoff equation (1885)
p = c1RT
where c1 = molar concentration of solute (mol/L)
R = gas law constant (0.082 L atm mol-1 K-1)
T = absolute temperature
 The van’t Hoff equation should only be used at low molar concentrations.
 At higher concentrations, the van’t Hoff equation significantly underpredicts the osmotic pressure.

23. * Osmotic pressure of aqueous sucrose solutions at 30C:
The van’t Hoff equation (2/4)
* Osmotic pressure of aqueous sucrose solutions at 30C:
___________
___________

24.  Their maximum concentrations are often much less than 1 molar.
The van’t Hoff equation (3/4)
 The van’t Hoff equation can be readily used for many types of biotechnology products, particularly macromolecules.
 Their maximum concentrations are often much less than 1 molar.

25. (Note: n2  V2 = total volume of water)
The van’t Hoff equation (4/4)
 The van’t Hoff equation is a special case of the Gibbs equation for a binary system consisting of water and a solute.
where V2 is the molar volume of water, and x2 is the mole fraction of water.
When x2 >> x1  
(Note: n2  V2 = total volume of water)

26. [Example] Ultrafiltration of a well-stirred suspension containing 0
[Example] Ultrafiltration of a well-stirred suspension containing 0.1 vol% yeast suspension gives a flux of 36 gal/ft2-day under a pressure difference of 130 psi. (a) What is the value of Lp? (b) What is the water velocity through the membrane?
Solution:
(a) The yeast cells have a very high molecular weight, so that their molar concentration and the resulting osmotic pressure will be small.
 Lp = 0.28 gal/ft2-day-psi
(To be continued)

27. (b) The water velocity through the membrane = jv
[Example] Ultrafiltration of a well-stirred suspension containing 0.1 vol% yeast suspension gives a flux of 36 gal/ft2-day under a pressure difference of 130 psi. (a) What is the value of Lp? (b) What is the water velocity through the membrane?
Solution (cont’d):
(b) The water velocity through the membrane = jv
= cm/s #

28. * Molecular Weight of Yeast Cell
d = 8 mm
r = 1.05 g/cm3
= 2.7  cm3
 MW = (6.02  1023)(2.7  10-10)(1.05)
= 1.7  1014

29. Estimation of Surface Concentration
Material balance:
Rate of solute accumulation
= (rate of solute flow in)
- (rate of solute diffuse out) 
B.C. 1: x = 0, c = c10
B.C. 2: x = , c = c1 

30. [Example] We are carrying out the ultrafiltration of chymotrypsin in a spiral wound module at a rate of 1.3  10-5 cm/s. The solution concentration is 0.44 wt%, the protein’s diffusion coefficient is 9.5  10-7 cm2/s, and the boundary layer is about cm thick. How high is the surface concentration?
Solution:  #

31. (V = liquid volume in the system)
A Simplified Case of Batch Concentration
Assumptions:
(1) There is little concentration polarization.  c10 = c1
 When (c10 – c1)/c1 < 0.1, the effect of concentration polarization can be neglected.
(2) The membrane rejects all the solute.  s = 1
(V = liquid volume in the system) 

32. A Simplified Case of Batch Concentration (2/2)
I. C.: t = 0, V = V0

33. [Example] We want to ultrafilter 840 L of a solution containing 0
[Example] We want to ultrafilter 840 L of a solution containing wt% of a protein used as a vaccine for herpes. This protein has a diffusion coefficient of 1.1  10-6 cm2/s and a molecular weight of 16,900. We would like to get the concentration up to about 2% by weight. The ultrafilter, which we hope to use, has eight hollow fiber cartridges, each of which has a surface area of 1.20 m2. It is cooled to 4C during the operation. The membrane in these cartridges gives an initial flux of 5.7  10-5 cm/s under a pressure drop of 31 atm. Assuming negligible concentration polarization, estimate the time to complete this filtration.
Solution:
(To be continued)

34. Solution (cont’d): V0 = 840 L ;
[Example] We want to ultrafilter 840 L of a solution containing wt% of a protein used as a vaccine for herpes. This protein has a diffusion coefficient of 1.1  10-6 cm2/s and a molecular weight of 16,900. We would like to get the concentration up to about 2% by weight. The ultrafilter, which we hope to use, has eight hollow fiber cartridges, each of which has a surface area of 1.20 m2. It is cooled to 4C during the operation. The membrane in these cartridges gives an initial flux of 5.7  10-5 cm/s under a pressure drop of 31 atm. Assuming negligible concentration polarization, estimate the time to complete this filtration.
Solution (cont’d):
V0 = 840 L ;
(To be continued)

35. Solution (cont’d): Initial flux: 5.7  10-5 = Lp (31 - 8.11  10-4)
[Example] We want to ultrafilter 840 L of a solution containing wt% of a protein used as a vaccine for herpes. This protein has a diffusion coefficient of 1.1  10-6 cm2/s and a molecular weight of 16,900. We would like to get the concentration up to about 2% by weight. The ultrafilter, which we hope to use, has eight hollow fiber cartridges, each of which has a surface area of 1.20 m2. It is cooled to 4C during the operation. The membrane in these cartridges gives an initial flux of 5.7  10-5 cm/s under a pressure drop of 31 atm. Assuming negligible concentration polarization, estimate the time to complete this filtration.
Solution (cont’d):
Initial flux:
= 8.11  10-4 atm
5.7  10-5 = Lp (  10-4)
 Lp = 1.84  10-6
(To be continued)

36. [Example] Ultrafiltration of a vaccine for herpes.
The ultrafilter, which we hope to use, has eight hollow fiber cartridges, each of which has a surface area of 1.20 m2.
Assuming negligible concentration polarization, estimate the time to complete this filtration.
Solution (cont’d):
ALpDP = (8  1.2  104)(1.84  10-6)(31) = 5.48 cm3/s = 5.48  10-3 L/s
=  105 s = 41.3 h #

37. CONCENTRATION POLARIZATION IN ULTRAFILTRATION
Material balance for solute in the boundary layer:
(Flux of solute in due to convection)
= (flux of solute out due to
diffusion)   

38. Correlation for flow inside pipes (for turbulent flow):
NSh (Sherwood number) = (d: pipe diameter)
NRe (Reynolds number) =
NSc (Schmidt number) =
* Equivalent diameter of the flow channel
* Concentration polarization becomes severe when

39. T = absolute temperature, K r0 = radius of the particles, cm
CONCENTRATION POLARIZATION IN ULTRAFILTRATION (3/3)
* Prediction of liquid diffusivities—Stokes-Einstein equation (for particles or large spherical molecules, VA > 500 cm3/mol)
T = absolute temperature, K
r0 = radius of the particles, cm
m = viscosity of solution, cP
VA = molar volume of solute as liquid at its normal boiling point, cm3/g mol

40. [Example] Equipment is available for ultrafiltration of a protein solution at constant volume to remove low molecular weight species (achieved by the addition of water or buffer to the feed in an operation called diafiltration). The flow channels for this system are tubes 0.1 cm in diameter and 100 cm long. The protein has a diffusion coefficient of 9  10-7 cm2/s. The solution has a viscosity of 1.2 cp and a density of 1.1 g/cm3. The system is capable of operating at bulk stream velocity of 300 cm/s. At this velocity, determine the polarization modulus (c10/c1) for a transmembrane flux of 45 L m-2 h-1.
Solution:  
(To be continued)

41. Example: Determination of the polarization modulus (c10/c1).
Solution (cont’d): ; 
(To be continued)

42. Example: Determination of the polarization modulus (c10/c1).
Solution (cont’d):  #

43. [Example] A tubular membrane with a diameter of 2 cm and a water permeability of 250 L/m2-h-atm is used for ultrafiltration of cheese whey. The solution velocity is 1.5 m/s and the protein concentration is 10 g/L. The whey proteins have an average diffusivity of 4  10-7 cm2/s, and the osmotic pressure in atm is given by Jonsson’s equation:
p = 4.4  10-3c  10-6c  10-8c3
where c is the protein concentration in g/L. Calculate the applied pressure if the permeate flux is 10-3 cm/s. Assume the protein rejection is 100 percent and the bulk solution has the same density and viscosity as water.
Solution:
 Need c10 to estimate p.
 Need kc.
(To be continued)

44. Solution (cont’d): = 0.0096NRe0.913NSc0.346 = 3.9  103 
[Example] d = 2 cm; Lp = 250 L/m2-h-atm; v = 1.5 m/s; c1 = 10 g/L; D = 4  10-7 cm2/s; p = 4.4  10-3c  10-6c  10-8c103 ;
r = 1 g/cm3; m = 1 cp; jv = 10-3 cm/s; DP = ?
Solution (cont’d):
= NRe0.913NSc0.346 = 3.9  103 
(To be continued)

45. Solution (cont’d):   c10 = 36.04 g/L
[Example] d = 2 cm; Lp = 250 L/m2-h-atm; v = 1.5 m/s; c1 = 10 g/L; D = 4  10-7 cm2/s; p = 4.4  10-3c  10-6c  10-8c103 ;
r = 1 g/cm3; m = 1 cp; jv = 10-3 cm/s; DP = ?
Solution (cont’d): 
 c10 = g/L
p = 4.4  10-3c  10-6c  10-8c3 = 0.16 atm
Permeability, Lp
= 6.94  10-3 cm/s-atm
Permeate flux,
 = 6.94  10-3 (DP )  DP = atm #

46. CONCENTRATION POLARIZATION IN ULTRAFILTRATION WITH PARTIAL REJECTION OF SOLUTES
B.C.1: c = c10 at x = 0
B.C.2: c = c1 at x =
The solution is:
Recall: 

47.  c10 - c2 = (c1 - c2)exp  c10 = c2 + (c1 - c2)exp 
CONCENTRATION POLARIZATION WITH PARTIAL REJECTION OF SOLUTES (2/4)
 c10 - c2 = (c1 - c2)exp
 c10 = c2 + (c1 - c2)exp 
the fraction rejected

48. CONCENTRATION POLARIZATION WITH PARTIAL REJECTION OF SOLUTES (3/4)
Assume the permeate concentration, c2, is in equilibrium with c10, i.e., c2 = Kc10.    \

49.  K is a constant, and R varies with jv.
CONCENTRATION POLARIZATION WITH PARTIAL REJECTION OF SOLUTES (4/4)
c2 = Kc10
 K is a constant, and R varies with jv.
 (1) The rejection, R, approaches (1 - K) as the permeate flux, jv, approaches zero.
(2) The rejection decreases with increasing the permeate flux.

50. [Example] Ultrafiltration tests with a 1
[Example] Ultrafiltration tests with a 1.5-cm tubular membrane at NRe = 25,000 gave a permeate flux of 40 L m-2 h-1 and 75 percent rejection for a 5 percent polymer solution. The polymer has an estimated diffusivity of 5  10-7 cm2/s. The bulk solution has the same density and viscosity as water. Predict the fraction rejected for a flux of 20 L m-2 h-1. What is the maximum rejection?
Solution:
 Need K and kc.
= NRe0.913NSc0.346 = 3060 
(To be continued)

51. Permeate flux, jv = 40 L m-2 h-1 = 1.11  10-3 cm/s
[Example] d = 1.5 cm; NRe = 25,000; jv = 40 L m-2 h-1; R = 0.75; D = 5  10-7 cm2/s. Predict the fraction rejected for a flux of 20 L m-2 h-1. What is the maximum rejection?
Solution (cont’d): ;
Permeate flux, jv = 40 L m-2 h-1 = 1.11  10-3 cm/s 
 K = 0.101
(To be continued)

52. If the permeate flux is 20 L m-2 h-1 or 0.556  10-3 cm/s,
[Example] d = 1.5 cm; NRe = 25,000; jv = 40 L m-2 h-1; R = 0.75; D = 5  10-7 cm2/s. Predict the fraction rejected for a flux of 20 L m-2 h-1. What is the maximum rejection?
Solution (cont’d):
; K = 0.101 ;
If the permeate flux is 20 L m-2 h-1 or  10-3 cm/s,
 R = 0.84
The maximum rejection occurs as the flux approaches zero.
 Rmax = 1 - K = 0.899 #

53. Abatement of Concentration Polarization in Ultrafiltration
“Concentration polarization increases the osmotic pressure, hampers the permeate flux, and adds to the severity of membrane fouling.”
 Air sparging (injecting air into the feed stream) has been the most popular technique proposed for the reduction of concentration polarization.
 The injected air induces hydrodynamic disturbances in the filtration module, which destabilizes the concentration layer over the membrane surface.

54. Gas-liquid two phase flow patterns:
Abatement of Concentration Polarization in Ultrafiltration (2/4)
Gas-liquid two phase flow patterns:

55. Abatement of Concentration Polarization in Ultrafiltration (3/4)
 An alternative idea: using n-hexadecane in place of air to generate two-phase flow.
 n-Hexadecane has a viscosity two order of magnitude higher than air (3.032  10-3 Pa s at 25C compared to 1.86  10-5 Pa s at 27C).
 When the oil droplets rub the surface of the membrane, the shear force and hence the disturbances generated by n-hexadecane droplets could be much more effective than that by air bubbles.

56. Abatement of Concentration Polarization in Ultrafiltration (4/4)
Comparison of permeate fluxes for the conventional, air/water, and n-hexadecane/water ultrafiltrations of a lipase solution. Symbols: (○) conventional, (▲) air/water, and (■) n-hexadecane/water.

57. DIAFILTRATION  Batch concentration versus diafiltration

58. DIAFILTRATION (2/4)
 Batch concentration versus diafiltration (2/2)
Diafiltration:
 In diafiltration, the dialyzate (or wash solvent) is added at a rate equal to removal of ultrafiltrate.

59. V0 = volume of initial preparation added to the cell (being constant)
DIAFILTRATION (3/4)
For microspecies (i.e., solutes) that are freely permeable to the membrane,  
where
V0 = volume of initial preparation added to the cell (being constant)
C0 = initial microsolute concentration in the reservoir
Cf = microsolute concentration after volume Vw of wash solution has passed through the cell

60. where R = product retention (dimensionless). Note: * R = 0
DIAFILTRATION (4/4) 
For microspecies (i.e., solutes) that are partially permeable to the membrane,
where R = product retention (dimensionless).
Note:
* R = 0
 Complete product passage through the membrane.
* R = 1  Rejected solute.

61. [Example] It is desired to use a crossflow filtration system to desalt 1000 L of a protein solution containing NaCl. The system has a membrane area of 100 m2 and is capable of operating at a transmembrane flux of 30 L m-2 h-1. To remove 99.99% of the salt, determine the time and the volume of water required.
Solution: 
 Vw = 9210 L
Vw = Jv  A  t 
 t = 3.07 h #

62. REVERSE OSMOSIS Adding a soluble salt to water.
 Reducing the chemical potential of the water.
 Osmotic flow.

63.  The osmotic pressure p of a solution: p = 1.12(T + 273)Smi
REVERSE OSMOSIS (2/19)
 The osmotic pressure p of a solution:
p = 1.12(T + 273)Smi
p = osmotic pressure, psi
T = temperature, C
Smi = summation of molalities (mol/1000 g of water) of all ionic and nonionic constituents in the solution
* p = 15 psi, for a typical brackish water; p = 350 psi, for seawater.

64. 400-600 psig for brackish water
REVERSE OSMOSIS (3/19)
 External pressure applied for reverse osmosis to occur:
psig for brackish water
800-1,000 psig for seawater
 Reverse osmosis is used most extensively in the purification of water and the concentration of biological and food processing streams.

65.  Performance variables of reverse osmosis:
A simple schematic of an RO system:

66. Qw = water flow rate through the membrane
REVERSE OSMOSIS (5/19)
 Performance variables of reverse osmosis (2/5):
The rate of water passage through a semipermeable membrane is defined by:
Qw = kwA(DP - p)/q
Qw = water flow rate through the membrane
kw = membrane permeability coefficient for water
A = membrane area
DP = hydraulic pressure differential across the membrane
p = osmotic pressure differential across the membrane
q = membrane thickness

67. The rate of salt flow through the membrane is given by: Qs = ksADc/q
REVERSE OSMOSIS (6/19)
 Performance variables of reverse osmosis (3/5):
The rate of salt flow through the membrane is given by:
Qs = ksADc/q
Qs = flow rate of salt through the membrane
Dc = salt concentration differential across the membrane
ks = membrane permeability coefficient for salt

68. Qw = kwA(DP - p)/q ; Qs = ksADc/q
REVERSE OSMOSIS (7/19)
 Performance variables of reverse osmosis (4/5):
Qw = kwA(DP - p)/q ; Qs = ksADc/q
  The rate of water flow through the membrane is proportional to the pressure differential across the membrane.
 The rate of salt flow is proportional to the concentration differential, and is independent of the applied pressure.
 An increase in operating pressure will increase the water flow without changing the salt flow.

69. Recovery (or conversion) is defined by: Y = 100Qp/Qf Y = % recovery
REVERSE OSMOSIS (8/19)
 Performance variables of reverse osmosis (5/5):
Recovery (or conversion) is defined by:
Y = 100Qp/Qf
Y = % recovery
Qp = product water flow rate
Qf = feed water flow rate
Salt passage is defined by:
SP = 100cp/cf
SP = % salt passage
cp = salt concentration in the product stream
cf = salt concentration in the feed stream

70.  The membranes used in reverse osmosis are asymmetric.
 The thin “skin” is supported by a porous substructure.

71.  The rate-determining step for water transport is across the skin.
REVERSE OSMOSIS (10/19)
 The membranes used in reverse osmosis are asymmetric. (2/2)
 The rate-determining step for water transport is across the skin.
 The flow rate through a membrane is inversely proportional to the membrane thickness.
 These membranes provide high water transport while still maintaining the important ability to reject salts.

72. (1) They are susceptible to degradation from biological attack.
REVERSE OSMOSIS (11/19)
 Cellulose acetate membranes (discovered by Loeb and Sourirajan at UCLA in the early 1960s) have two major limitations.
(1) They are susceptible to degradation from biological attack.
 Using chlorinated feed water to prevent such
attack.
(2) They hydrolyze back to cellulose under acidic and particularly basic conditions.
 Control the pH of the system at 4.5 to 7.5.

73. Reverse Osmosis Devices—Tubular
* The membrane is either inserted into, or coated onto, the surface of a porous tube designed to withstand the operating pressure.

74. Reverse Osmosis Devices—Spiral-Wound
* This device is like a huge envelope made of membrane and containing a feed spacer.
* Feed flowing around the envelope at high pressure goes across the membrane and is collected inside the envelope. The envelope is wound spirally about a plastic tube that receives the permeate.

75. Reverse Osmosis Devices—Hollow Fiber
Aramid membranes: commercialized by Du Pont in 1970, made from an aromatic polyamide polymer, operated at pH range of 4-11, not susceptible to biological attack and resist hydrolysis. (But, they are degraded by chlorine.)
________

76. REVERSE OSMOSIS (15/19)
Reverse Osmosis Devices—Hollow Fiber
____________
* Pressurized water passes through the fiber wall into the fiber bore. The salts and other impurities remain in the brine, which flows to the outer perimeter of the fiber bundle.

77. REVERSE OSMOSIS (16/19)
 RO system design:

78. REVERSE OSMOSIS (17/19)
___
Remarks:
* Cartridge filter: remove large-particle matter that could damage the high-pressure pump or cause device plugging.

79. REVERSE OSMOSIS (18/19)
______
______
* Low-pressure shutdown switch: prevent pump operation at inadequate flow rate.
* Valve on the pump discharge: control the pressure of the feed water.
* Temperature switch: protect the permeator.

80. * Flow-control valve on the brine: set conversion.
REVERSE OSMOSIS (19/19)
______
____
* High-pressure shutdown switch or pressure-relief device: prevent the permeator from over-pressurization.
* Flow-control valve on the brine: set conversion.

81. CONCENTRATION POLARIZATION IN REVERSE OSMOSIS
 Concentration polarization has two deleterious effects:
(1) The increase in c10 increases p.
Qw = kwA(DP - p)/q
 It necessitates a greater applied total pressure to produce a given water flux across the membrane.
(2) The increase in c10 serves to increase the driving
force for salt transport through the membrane.
Qs = ksADc/q
 It engenders more salt leakage into the product water.

82.  It is referred to the phenomenon as “fouling” of the membrane.
CONCENTRATION POLARIZATION IN REVERSE OSMOSIS (2/3)
 When the concentration of retained solutes at the membrane surface, c10, exceeds the solubility limit, it forms a thixotropic gel.
 It is referred to the phenomenon as “fouling” of the
membrane.

83. Osmotic pressure of seawater, p = 1.12(T + 273)Smi
[Example] A reverse osmosis process is used for desalination of seawater. The volumetric flux of water through the membrane is 3  10-5 m/s (or m3 s-1 m-2), and the applied feed pressure is 8.0 MPa greater than the product-water pressure. For seawater, the osmotic pressure is 2.5 MPa. What is the water velocity through the membrane if the polarization modulus (c10/c1) rises to 1.2-fold of the original?
Solution:
Osmotic pressure of seawater, p = 1.12(T + 273)Smi 
 p2= 1.2p1 = 1.2  2.5 = 3.0 MPa
(To be continued)

84. Qw = kwA(DP - p)/q or Qw/A = jv = Lp(DP - p)
[Example] Reverse osmosis process for desalination of seawater.
Flux of water = 3  10-5 m/s (or m3 s-1 m-2); DP = 8.0 MPa; p = 2.5 MPa. What is the water velocity through the membrane if the polarization modulus (c10/c1) rises to 1.2-fold of the original?
Solution (cont’d):
p2 = 3.0 MPa
Rate of water passage through the membrane:
Qw = kwA(DP - p)/q or Qw/A = jv = Lp(DP - p)
\ jv2 = 0.91jv1 = 0.91  (3  10-5) = 2.73  10-5 m/s #

85. where jv = the volume of solvent per area per time
CONCENTRATION POLARIZATION IN REVERSE OSMOSIS (3/3)
 Correlation of Dittus and Boelter (1930) for concentration polarization in turbulent flow:
where jv = the volume of solvent per area per time
dh = equivalent hydraulic diameter
m = viscosity of fluid
r = density of fluid
v = average velocity of fluid
D = local solute diffusivity in solution

86.  Lower temperature makes concentration polarization more severe.
[Example] A reverse osmosis desalting process is carried out using turbulent flow through a tubular 1.0-cm-diameter membrane with a system temperature of 18.5C. Which of the following factors would be effective in reducing the degree of concentration polarization if the water flux is held constant? (a) Reduced temperature; (b) reduced tube diameter with the same mass flow rate of seawater; and (c) recirculation of the seawater with the same size and length.
Solution:
(a) Lower temperature increases the viscosity and lowers the diffusivity.
 Lower temperature makes concentration polarization more severe.
(To be continued)

87.  Concentration polarization will be reduced.
[Example (cont’d)] Which of the following factors would be effective in reducing the degree of concentration polarization if the water flux is held constant? (b) Reduced tube diameter with the same mass flow rate of seawater; and (c) recirculation of the seawater with the same size and length.
Solution:
(b) Reducing tube diameter with the same mass flow rate of water will raise v.
 Concentration polarization will be reduced.
(c) Recirculation of the seawater will increase v.
 It alleviates concentration polarization but does so at the expense of much more pumping power (more flow and more pressure drop). #

88. * Production of Low Alcohol Beer Using Reverse Osmosis
_____________

89.  Some flavor losses occur.
* Production of Low Alcohol Beer Using Reverse Osmosis (2/2)
Remarks
* Some flavor components that have a molecular weight or size similar to ethanol also pass through the membrane.
 Some flavor losses occur.
* The membrane cost is high.
 The annual replacement cost is up to 7% of the original capital cost.

90. DIALYSIS
 At equilibrium, the concentration of small molecules is the same inside and outside the membrane.

91. _____________________________ ___________
DIALYSIS (2/3)
 The external fluid must be repeatedly changed to reach the required final composition.
_____________________________
___________

92.  To increase the rate of movement.
DIALYSIS (3/3)
 Place a stirrer of some kind in the external fluid (or inside the bag).
 To increase the rate of movement.

93. REVERSE DIALYSIS
 The filled bag is packed in a dry, water-soluble polymer which cannot enter the membrane.

94. Remarks of reverse dialysis:
 It is used to concentrate the material in the bag.
 Equilibrium is never reached.
 Water and salts are continuously removed until the sample is totally dry.
 Most macromolecules become irreversibly
bound to the dialysis tubing. (They are lost.)

95. GLASS FIBER DIALYSIS
 Semipermeable glass fibers: hollow-bore fibers whose glass walls contain pores of controlled size

96.  Two ways of using hollow fibers: dialysis and concentration.
GLASS FIBER DIALYSIS (2/2)
 Two ways of using hollow fibers: dialysis and concentration.

97. ANALYSIS OF DIALYZER A shell-and-tube type of hollow-fiber dialyzer:
Q = flow rate, A = area of membrane, K = dialysis coefficient

98. For a counterflow dialyzer,
ANALYSIS OF DIALYZER (2/2)
For a counterflow dialyzer,
The rate of salt removed

99. The rate of salt removed
[Example] A solution of raffinose containing 100 g/L of NaCl is to be dialysed in a shell-and-tube type of hollow-fiber dialyzer operating countercurrently. With a dialyzer having 1000 cm2 area of membranes the dialysis coefficient for NaCl was determined to be cm/min, when the feed rate was 200 cm3/min, and the flow rate of pure water was 500 cm3/min. If 90% of the salt is to be removed, what area of the hollow-fiber membranes will be needed, if the same flow rates for feed and water are used?
Solution:
The rate of salt removed 
(To be continued)

100. [Example] Shell-and-tube type of hollow-fiber dialyzer
CF1 = 100 g/L; K = cm/min; QF = 200 cm3/min; QD = 500 cm3/min. If 90% of the salt is to be removed, what area of the hollow-fiber membranes will be needed?
Solution (cont’d):  #

101. The End of
MEMBRANE
SEPARATIONS