1. Membrane Separation Process

2. Objectives Estimate the extent of concentration polarization in crossflow filtration Select filtration unit operations to meet product requirements, consistent with product properties

3. WHAT IS A MEMBRANE? Membranes are materials which have voids in them letting some molecules pass more conveniently than some other molecules. A semi-permeable membrane is a VERY THIN film that allows some types of matter to pass through while leaving others behind

4. WHAT IS A MEMBRANE?

5. HOW SEPARATION OCCURS Difference in permeabilities through a membrane: Difference in size, Affinity to the membrane, Charge, etc.

6. DRIVING FORCES Pressure difference, Concentration difference, Voltage difference, etc.

7. ADVANTAGES Continuous separation Low energy requirement Meet various separation demands

8. DISADVANTAGES Fouling Short service periods

9. TYPES OF PROCESSES Classification according to pore size Microfiltration Ultrafiltration Nanofiltration Reverse Osmosis

10. REVERSE OSMOSIS

11. only remove some suspended materials larger than 1 micron the process eliminates the dissolved solids, bacteria, viruses and other germs contained in the water only water molecules allowed to pass via very big pressure assymmetric type membranes (decrease the driving pressure of the flux) almost all membranes are made polymers, cellulosic acetate and matic polyamide types rated at 96%-99+% NaCl rejection

12. extensive applications: –potable water from sea or brackish water –ultrapure water for food processing and electronic industries –harmaceutical grade water –water for chemical, pulp & paper industry –waste treatment REVERSE OSMOSIS

13. future directions: –municipal and industrial waste treatment –process water for boilers –de-watering of feed streams –processing high temperature feed- streams REVERSE OSMOSIS

14. MICROFILTRATION

15. MICROFILTRATION largest pores a sterile filtration with pores 0.1-10.0 microns micro-organisms cannot pass through them operated at low pressure differences used to filter particles. may or may not be assymmetric

16. wide array of applications: –parenterals and sterile water for pharmaceutical industry –food & beverages –chemical industry –microelectronics industry –fermentation –laboratory/analytical uses MICROFILTRATION

17. MICROFILTRATION applications in the near future: –biotechnology (concentration of biomass) –diatomaceous earth displacement –non-sewage waste treatment (removing intractable particles in oily fluids) –paints (separating solvents from pigments)

18. ULTRAFILTRATION

19. ULTRAFILTRATION to separate a solution; mixture of desirable and undesirable components has smaller pores than microfiltration membranes driving force → pressure differential (2-10 bars to 25-30 bars) used to separate species with pore sizes 10-1000 Å (103-0.1 microns) Can be obtained down to a a molecular weight cutoff (MWCO) level of 1000 Daltons (Da) and up yo as high as 1 000 000 Da. assymmetric; the pores are small

20. wide range of applications : –oil emulsion waste treatment –treatment of whey in dairy industries –concentration of biological macromolecules –electrocoat paint recovery –concentration of textile sizing –concentration of heat sensitive proteins for food additives –many more … ULTRAFILTRATION

21. wide range of applications in the near future: –ultraflitration of milk –bioprocessing (separation and concentration of biologically active components) –protein harvesting –refining of oils ULTRAFILTRATION

22. NANOFILTRATION

23. NANOFILTRATION less pore sizes than ultrafiltration membranes the mass transfer mechanism is diffusion & separate small molecules from the solution (assymmetric) cellulosic acetate and aromatic polyamide type membranes (salt rejections; 95% for divalent salts to 40% for monovalent salts) can typically operate at higher recoveries; conserving total water usage due to a lower concentrate stream flow rate (advantage over reverse osmosis) not effective on small molecular weight organics (e.g.methanol)

24. typical applications: –desalination of food, dairy and beverage products or byproducts –partial desalination of whey, UF permeate or retentate as required –desalination of dyes and optical brighteners –purification of spent clean-in-place (CIP) chemicals –color reduction or manipulation of food products –concentration of food, dairy and beverage products or byproducts –fermentation byproduct concentration NANOFILTRATION

26. 0.00010.0010.010.1110 100  m hair Crypto- sporidium smallest micro- organis m polio virus Suspended solids Parasites Bacteria Org. macro. molecules Viruses Colloids Dissolved salts Sand filtration Microfiltration Ultrafiltration Nanofiltration Reverse Osmosis ZW500: 0.04 um ZW1000: 0.02 um

27. SUMMARY

28. Membrane Configurations

30. 1. Flat membranes Mainly used to fabricate and used In some cases modules are stacked together like a multilayer sandwich or plate-and-frame filter press Three basic structures are commonly uses for membranes: homogeneous (no significant variation in pore diameter from the filtering surface to the other side), asymmetric (has a thin layer next to the filtering surface that has very pores), and composite (containing very small pores next to the filtering surface; however, the thin and thick layers of this membrane are made of two different types of material)

31. 2. Spiral wound membranes Constructed from flat sheet membranes separated by spacer screens Consists of a sandwich of four sheets wrapped around a central core of a perforated collecting tube The four sheets consist of a top sheet of an open separator grid for feed channel, a membrane, a porous felt backing for the permeate channel, and another membrane The feed solution is fed into one end of the module and flows through the separator screens along the surface of the membranes The retentate is the collected in the other end of the module The permeate spirals radially inward, eventually to be collected through a central tube The flow in these systems tend to be turbulent The main advantage of spiral-wound modules is that they are susceptible to fouling by particulates because of the narrow and irregular flow through spacers Membrane Configurations

32. Spiral-wound elements and assembly Local solution flow paths for spiral-wound separator.

33. 3. Hollow-fiber membranes The membranes are in the shape of very-small-diameter hollow fibers Typically, the high-pressure feed enters the shell side at one end and leaves at the other end The hollow fibers are closed at one end of the tube bundles The permeate solution inside the fibers flows countercurrent to the shell-side flow and is collected in a chamber where the open ends of the fibers terminate Then the permeate exits the device Membrane Configurations Hollow-fiber separator assembly.

34. 1. Osmotic pressure of solutions Experimental data shows that the osmotic pressure π of a solution is proportional to the concentration of the solute and temperature T Van’t Hoff originally showed that the relationship is similar to that for pressure of an ideal gas For example, for dilute water solutions, If a solute exists as two or more ions in solution, n represents the total number of ions For more concentrated solutions, Eq. (1) is modified using the osmotic coefficient Φ, which is the ratio of the actual osmotic pressure π to the ideal π calculated from the equation Flux Equation for Reverse Osmosis (1) n: number of kg mol of solute, V m : volume of pure solvent water in m 3 associated with n kg mol of solute, R: gas law constant 82.057 x 10 -3 m 3 ·atm/kg mol·K, T: temperature in K.

35. Calculate the osmotic pressure of a solution containing 0.10 g mol NaCl/1000 g H 2 0 at 25°C. Solution: The density of water = 997.0 kg/m 3 and then, n = 2 x 0.10 x 10 -3 = 2.00 x 10 -4 kg mol (NaCl gives two ions). Also, the volume of the pure solvent water V m = 1.00 kg/(997.0 kg/m3). Substituting into Eq. (1), This compares with the experimental value in Table 1 of 4.56 atm. Example 1

36. TABLE 1. Osmotic Pressure of Various Aqueous Solutions at 25°C

37. 2. Diffusion-type models For diffusion type membranes, the steady-state equations governing the transport of solvent and solute are to a first approximation. For the diffusion of the solvent through the membrane, where N w is the solvent (water) flux in kg/s·m 2 ; P w the solvent membrane permeability, kg solvent/s·m·atm; L m the membrane thickness, m; A w the solvent permeability constant, kg solvent/s·m2·atm; ΔP = P 1 - P 2 (hydrostatic pressure difference with P 1 pressure exerted on feed and P 2 on product solution), atm; and Δπ = π 1 - π 2 (osmotic pressure of feed solution - osmotic pressure of product solution), atm. Flux Equation for Reverse Osmosis (2) (3)

38. Flux Equation for Reverse Osmosis For the diffusion of the solute through a membrane, an approximation for the flux of the solute is where N s is the solute (salt) flux in kg solute/s·m 2 ; D s the diffusivity of solute in membrane, m 2 /s; K s = c m /c (distribution coefficient), concentration of solute in membrane/concentration of solute in solution; A s is the solute permeability constant, m/s; c 1 the solute concentration in upstream or feed (concentrate) solution, kg solute/m 3 ; and c 2 the solute concentration in downstream or product (permeate) solution, kg solute/m 3. The distribution coefficient K s is approximately constant over the membrane. (4) (5)

39. Making a material balance at steady state for the solute, the solute diffusing through the membrane must equal the amount of solute is leaving in the downstream or product ( permeate) solution: where c w2 is the concentration of solvent in stream 2 (permeate), kg solvent/m 3. If the stream 2 is dilute in solute, c w2 is approximately the density of the solvent. In reverse osmosis, the solute rejection R is defined as the ration concentration difference across the membrane divided by the bulk concentration on the feed or concentrate side (fraction of solute remaining in the feed stream): Flux Equation for Reverse Osmosis (6) (7)

40. Flux Equation for Reverse Osmosis This can be related to the flux equation as follows, by first substituting Eq. (2) and (4) into (6) to eliminate N w and N s in Eq. (6) Then, solving for c 2 /c 1 and substituting this result to Eq. (7) (8) (9) where B is in atm -1

41. Example 2 Experiments at 25°C were performed to determine the permeabilities of a cellulose acetate membrane. The laboratory test section shown in Fig. 3 has membrane area A = 2.00 x 10 -3 m 2. The inlet feed solution concentration of NaCl is c 1 = 10.0kg NaCl/m 3 solution (10.0 g NaCI/L, ρ 1 = 1004 kg solution/m 3 ). The water recovery is assumed low so that the concentration c 1 in the entering feed solution flowing past the membrane and the concentration of the exit feed solution are essentially equal. The product solution contains c 2 = 0.39 kg NaCl/m 3 solution (ρ 2 = 997 kg solution/m 3 ) and its measured flow rate is 1.92 x 10 -8 m 3 solution/s. A pressure differential of 5514 kPa (54.42 atm) is used. Calculate the permeability constants of the membrane and the solute rejection R. FIGURE. Process flow diagram of experimental reverse-osmosis laboratory unit.

42. Example 2 Solution: Since c 2 is very low (dilute solution), the value of c 2 can be assumed as the density of water (Table 1), or c w2 = 997 kg solvent/m 3. To convert the product flow rate to water flux, N w, using an area of 2.00 x 10 -3 m 2, Substituting into Eq. (6),

43. To determine the osmotic pressures from Table 1, the concentrations are converted as follows. For c 1, 10 kg NaCl is in 1004 kg solution/m 3 (ρ 1 = 1004). Then, 1004 – 10 = 994 kg H 2 O in 1 m 3 solution. Hence, in the feed solution, where the molecular weight of NaCl = 58.45, (10.00 X 1000)/ (994 x 58.45) = 0.1721 g mol NaC1/kg H 2 O. From Table 1, π 1 = 7.80 atm by linear interpolation. Substituting into Eq. (1), the predicted = 8.39 atm, which is higher than the experimental value. For the product solution, 997 - 0.39 = 996.6 kg H 2 O. Hence, (0.39 x 1000)/(996.6 x 58.45) = 0.00670 g mol NaCl/kg H 2 O. From Table 1, π 2 = 0.32 atm. Then, Δπ = π 1 - π 2 =7.80 - 0.32 = 7.48 atm and ΔP = 54.42 atm. Substituting into Eq. (2), Example 2

44. Solving, (P w /L m ) = A = 2.039 x 10 -4 kg solvent/s m 2 atm. Substituting into Eq. (4), Solving, (D s K s /L m ) = A s = 3.896 x 10 -7 m/s. To calculate the solute rejection R by substituting into Eq. (7), Also, substituting into Eq. (9) and then Eq. (8), Example 2

45. Concentration Polarization in Reverse- Osmosis Diffusion Model In desalination, localized concentrations of solute build up at the point where the solvent leaves the solution and enters the membrane The solute accumulates in a relatively stable boundary layer next to the membrane Concentration polarization, β, is defined as the ratio of the salt concentration at the membrane surface to the salt concentration in the bulk feed stream c 1 Concentration polarization causes the water flux to decrease, since the osmotic pressure π 1 increases as the boundary layer concentration increases and the overall driving force (ΔP-Δπ) Also, the solute increases, since the solute concentration increases at the boundary Hence, often the ΔP must be increased to compensate, which results in higher power costs

46. Concentration Polarization in Reverse- Osmosis Diffusion Model The effect of the concentration polarization β can be included approximately by modifying the value of Δπ in Eq. (2) and (8) as follows: It is assumed that the osmotic pressure π 1 is directly proportional to the concentration, which is approximately correct. Also, Eq. (4) can be modified as: The usual concentration polarization ratio is 1.2 to 2.0, that is, the concentration in the boundary layer is 1.2-2.0 times c 1 in the bulk feed solution. This ratio is often difficult to predict (10) (11)

47. Flux Equation for Ultrafiltration The flux equation for diffusion of solvent through the membrane is the same as Eq. (2) for reverse-osmosis: In ultrafiltration the membrane does not allow passage of the solute, which is generally a macromolecule, The concentration in moles/liter of the large molecules is usually small. Hence, the osmotic pressure is very low and can be neglected Then Eq. (2) becomes Ultrafiltration units operate about 5-100 psi pressure drop, compared to 400-2000 for reverse osmosis (12) (13)

48. Flux Equation for Ultrafiltration Since the solute is rejected by the membrane, it accumulates and starts to build up at the surface of the membrane As pressure drop is increased and/or concentration of the solute is increased, concentration polarization occurs, which is much more severe than in reverse osmosis As the pressure drop increases, this increases the solvent flux N w to and through the membrane The concentration c s increases and gives a larger back molecular diffusion of solute from the membrane to the bulk solution At steady state, the convective flux equals the diffusion flux: where N w c/ρ = [kg solvent/(s·m 2 )](kg solute/m 3 )/(kg solvent/m 3 ) = kg solute/s m 2 ; D AB is diffusivity of solute in solvent, m2/s; and x is distance, m. (14)

49. Integrating this equation between the limits of x = 0, and c = c s and x = δ and c = c 1, where k c is the mass-transfer coefficient, m/s. Further increases in pressure drop increase the value of c s to a limiting concentration, at which the accumulated solute forms a semisolid gel where c s = c g, as shown in Fig. b. For the usual case of almost-complete solute retentation, c p = 0 and Eq. (15) becomes Still further increases in pressure drops do not change cg and the membrane is said to be “gel polarized”. Then Eq. (15) becomes Flux Equation for Ultrafiltration (15) (16) (17)

50. With increases in pressure drop, the gel later increases in thickness, causing the solvent flux to decrease of the added gel-layer resistance Finally, the net flux of solute by convective transfer becomes equal to the back diffusion of solute into the bulk solution because of the polarized concentration gradient, as given by Eq. (15) The added gel-layer resistance next to the membrane causes an increases resistance to solvent flux, as given by Where 1/A w is the membrane resistance and R g is the variable gel-layer resistance, (s·m 2· atm)/kg solvent. The solvent flux in this gel-polarized regime is independent of pressure difference and is determined by Eq. (15) for back diffusion Flux Equation for Ultrafiltration (18)

51. Flux Equation for Ultrafiltration FIGURE. Concentration polarization in ultrafiltration: (a) concentration profile before gel formation, (b) concentration profile with a gel layer formed at membrane surface.

52. Models for Microfiltration 1. Dead-end microfiltration flow model The particles build up with time as a cake and the clarified permeate is forced through the membrane. The permeate flux equation is 2. Cross-flow microfiltration flow model Similar to reverse osmosis and ultrafiltration in that the flow of bulk solution is parallel to the membrane surface and not through it The concentration-polarization model of convection of particles to the cake layer is balanced by particle diffusion by Brownian diffusion away from the cake surface. This model is similar to the flux Eq. (13) for ultrafiltration (19)