MEMBRANE SEPARATIONS The most rapidly advances area of filtration technology. There are three major types of membrane-based filtration techniques:

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.

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

Types of membrane filtration

Microfiltration

Retain particles as small as 0.1 m.

- 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.


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.


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.


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 themembrane.


A continuous countercurrent diafiltration system:

Types of membrane filtration: diafiltration (2/2)

1 2 3A + B A

waterB

Electrodialysis

Employ semi-permeable ion-exchange membranes that are impervious to water.

The separation is electrically driven instead of pressure-driven.

Cross-Flow Filtration

“Most of the pressure drop in conventional filtration comes from the cake.”

* Concentration polarization: accumulation of solute near the membrane surface

* We may wish that we could filter without a filter cake.

Use filtration where cross flow is dominant.

Cross-Flow Filtration

ULTRAFILTRATION

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

Because of the range of pore sizes in an ultrafiltration membrane, there is no absolute cutoff point.

ULTRAFILTRATION (2/6)

_____________

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.



For hollow fiber, feed streams generally need to be prefiltered.

______________________________________________________


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.


ANALYSIS OF ULTRAFILTRATION

In ultrafiltration, the species transported is the solvent and the chief force is the transmembrane pressure (P).

Solvent velocity force on solvent

Darcy's law:

Pk

v

ANALYSIS OF ULTRAFILTRATION (2/4)

Darcy's law:

Pk

v

where = thickness the membrane

= 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

PLRR

Pj p

pmv

)(

Taking the osmotic pressure into consideration,

)( PLj PvPLj pv


)( PLj Pv

where = the osmotic pressure

= a reflection coefficient

* = 1

The membrane rejects all solutes.

* = 0

The membrane freely passes both solvent and solute.


Estimation of Osmotic Pressure—the van’t Hoff equation (1885)

= c1RT

where c1 = molar concentration of solute (mol/L)

R = gas law constant (0.082 L atm mol1 K1)

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.

* Osmotic pressure of aqueous sucrose solutions at 30C:

The van’t Hoff equation (2/4)

______________________

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.


The van’t Hoff equation is a special case of the Gibbs equation for a binary system consisting of water and a solute.

22

ln :equation Gibbs xV

RT

where V2 is the molar volume of water, and x2 is the mole fraction of water.

When x2 >> x1

2

1

21

1112 )1ln(ln

n

n

nn

nxxx

RTcn

n

V

RT1

2

1

2

(Note: n2 V2 = total volume of water)


[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.

PLPLj pPv )(

)psi 130(day-ft

gal36

2 pL Lp = 0.28 gal/ft2-day-psi

(To be continued)

[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

s 360024

day

cm) (30.48

ft

gal

cm 3785

day-ft

gal36

2

23

2

= 0.0017 cm/s

#

* Molecular Weight of Yeast Cell

d = 8 m

= 1.05 g/cm3

6

3dV

= 2.7 1010 cm3

MW = (6.02 1023)(2.7 1010)(1.05)

= 1.7 1014

Estimation of Surface Concentration

Material balance:

Rate of solute accumulation

= (rate of solute flow in)

(rate of solute diffuse out)

dx

dcDcjv 0

B.C. 1: x = 0, c = c10

B.C. 2: x = , c = c1

1

10lnc

cDjv

[Example] We are carrying out the ultrafiltration of chymotrypsin in a spiral wound module at a rate of 1.3 105 cm/s. The solution concentration is 0.44 wt%, the protein’s diffusion coefficient is 9.5 107 cm2/s, and the boundary layer is about 0.018 cm thick. How high is the surface concentration?

Solution:

1

10lnc

cDjv

1

107

5 ln018.0

105.9103.1

c

c 3.1

1

10 c

c

#

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. = 1

P

RTcPALPALAj

dt

dVppv

11)(

(V = liquid volume in the system)

V

PRTnPAL

dt

dVp

/1 1

A Simplified Case of Batch Concentration (2/2)

V

PRTnPAL

dt

dVp

/1 1

I. C.: t = 0, V = V0

PRTnV

PRTnV

P

RTnVV

PALt

p /

/ln)(

1

1

1010

[Example] We want to ultrafilter 840 L of a solution containing 0.061 wt% of a protein used as a vaccine for herpes. This protein has a diffusion coefficient of 1.1 106 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 4C during the operation. The membrane in these cartridges gives an initial flux of 5.7 105 cm/s under a pressure drop of 31 atm. Assuming negligible concentration polarization, estimate the time to complete this filtration.

Solution:

PRTnV

PRTnV

P

RTnVV

PALt

p /

/ln)(

1

1

1010

(To be continued)

[Example] We want to ultrafilter 840 L of a solution containing 0.061 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 4C 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.


PRTnV

PRTnV

P

RTnVV

PALt

p /

/ln)(

1

1

1010

V0 = 840 L ; L62.25% 2

% 0.061 L840

V

mol 03.0g/mol 16900

g )10840(%061.0 3

1

n

L02.0atm 31

K)K)(277 atm/mol-- Lmol)(0.082 03.0(1

P

RTn

(To be continued)

Initial flux:

)( PLj Pv

)K 277( Kmol-

atm-L 082.0

L840

mol 03.010

RTc = 8.11 104 atm

5.7 10-5 = Lp (31 8.11 10-4)

Lp = 1.84 10-6 atm-s

cm

[Example] We want to ultrafilter 840 L of a solution containing 0.061 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 4C 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.


(To be continued)

PRTnV

PRTnV

P

RTnVV

PALt

p /

/ln)(

1

1

1010

PRTnV

PRTnV

P

RTnVV

PALt

p /

/ln)(

1

1

1010

ALpP = (8 1.2 104)(1.84 106)(31) = 5.48 cm3/s = 5.48 103 L/s

02.062.25

02.0840ln02.0)62.25840(

1048.5

13

= 1.486 105 s = 41.3 h

[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.


#07.0

02.062.25

02.0840ln02.0 :Note

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)

dx

dcDcjv

0

1

10

dxD

j

c

dc v

c

c

c

vv

k

j

D

j

c

c

1

10ln

c

vv

k

j

D

j

c

c

1

10ln

Correlation for flow inside pipes (for turbulent flow): 346.0

Sc913.0

ReSh 0096.0 NNN

NSh (Sherwood number) = (d: pipe diameter)

NRe (Reynolds number) =

NSc (Schmidt number) = D

vd

D

dkc

33.0Sc

69.0ReSh 082.0 NNN or

* Concentration polarization becomes severe when .101

10 c

c

* Equivalent diameter of the flow channel

perimeter wetted

area sectional-cross 4

* Prediction of liquid diffusivities—Stokes-Einstein equation (for particles or large spherical molecules, VA > 500 cm3/mol)

0

161032.7

r

TD

T = absolute temperature, K

r0 = radius of the particles, cm

= viscosity of solution, cP

VA = molar volume of solute as liquid at its normal boiling point, cm3/g mol

CONCENTRATION POLARIZATION IN ULTRAFILTRATION (3/3)

[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 9 107 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 m2 h1.

Solution: 33.0Sc

69.0ReSh 082.0 NNN

2750

s-cm

g 101.2

cm 1.0s

cm 300

cm

g 1.1

2-

3

Re

vd

N

c

v

k

j

c

c

1

10ln D

dkN cSh

(To be continued)

Example: Determination of the polarization modulus (c10/c1).


33.0Sc

69.0ReSh 082.0 NNN 2750Re N

4

27

3

2

Sc 1021.1

s

cm 109

cm

g 1.1

s-cm

g 102.1

DN

431)1021.1()2750(082.0 33.0469.0Sh

D

dkN c

s

cm 1088.3

cm 0.1s

cm 109

431 3

27

Sh

d

DNkc

;

(To be continued)

D

dkN cSh


s

cm 1088.3 3ck

c

v

k

j

c

c

1

10ln

38.1

s

cm 103.88

s 3600

h

cm 10

m

L

cm 1000

h-m

L 45

expexp3-

24

23

2

1

10

c

v

k

j

c

c

Example: Determination of the polarization modulus (c10/c1).

#

[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 4 107 cm2/s, and the osmotic pressure in atm is given by Jonsson’s equation:

= 4.4 103c 1.7 106c2 + 7.9 108c3

where c is the protein concentration in g/L. Calculate the applied pressure if the permeate flux is 103 cm/s. Assume the protein rejection is 100 percent and the bulk solution has the same density and viscosity as water.

Solution:

)( PLj Pv Need c10 to estimate .

c

v

k

j

c

c

1

10ln Need kc.

(To be continued)

[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; = 4.4 10-3c10 1.7 10-6c10

2 + 7.9 10-8c103 ;

= 1 g/cm3; = 1 cp; jv = 10-3 cm/s; P = ?


000,30s-g/cm 0.01

cm) cm/s)(2 150)(g/cm 1( 3

Re vd

N

000,25s)/cm 104)(g/cm (1

s-g/cm 01.0273Sc

D

N

D

dkN cSh = 0.0096NRe

0.913NSc0.346 = 3.9 103

cm/s 108.72

)104)(109.3( 473

Sh

d

DNkc

(To be continued)

cm/s 108.7 4ck

c

v

k

j

c

c

1

10ln4

310

108.7

10

10ln

c c10 = 36.04 g/L

= 4.4 10-3c 1.7 106c2 + 7.9 108c3 = 0.16 atm

Permeability, Lp

s 3600

h

cm 10000

m

L

cm 1000

atm-h-m

L 250

2

23

2

= 6.94 10-3 cm/s-atm

Permeate flux, )( PLj Pv

0.001 = 6.94 103 (P 0.16) P = 0.304 atm#

[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; = 4.4 10-3c10 1.7 10-6c10

2 + 7.9 10-8c103 ;

= 1 g/cm3; = 1 cp; jv = 10-3 cm/s; P = ?


CONCENTRATION POLARIZATION IN ULTRAFILTRATION WITH PARTIAL REJECTION OF SOLUTES

2cjdx

dcDcj vv

B.C.1: c = c10 at x = 0

B.C.2: c = c1 at x =

The solution is:

c

vv

k

j

D

j

cc

cc

21

210ln

0dx

dcDcjv

1

10lnc

cDjv

Recall:

c

vv

k

j

D

j

cc

cc

21

210ln

c10 c2 = (c1 c2)exp c

v

k

j

c10 = c2 + (c1 c2)exp c

v

k

j

c

v

c

v

k

jRRc

k

j

c

c

c

ccc exp1exp1 1

1

2

1

2110

, 1 where1

2

c

cR

CONCENTRATION POLARIZATION WITH PARTIAL REJECTION OF SOLUTES (2/4)

the fraction rejected

c

v

k

jRRcc exp1110

Assume the permeate concentration, c2, is in equilibrium with c10, i.e., c2 = Kc10.

c

v

k

jRR

Kc

c

c

cexp1

1

2

1

10 c

v

k

jRRR

Kexp1)1(

1

c

v

k

j

R

R

Kexp

11

1

c

v

k

j

R

R

K

K

Kexp

1

11

1

c

v

k

j

K

K

R

Rexp

1

1

1

21c

cR


c

v

k

j

K

K

R

Rexp

1

1

(1)The rejection, R, approaches (1 K) as the permeate flux, jv, approaches zero.

(2)The rejection decreases with increasing the permeate flux.

c2 = Kc10

K is a constant, and R varies with jv.


[Example] Ultrafiltration tests with a 1.5-cm tubular membrane at NRe = 25,000 gave a permeate flux of 40 L m2 h1 and 75 percent rejection for a 5 percent polymer solution. The polymer has an estimated diffusivity of 5 107 cm2/s. The bulk solution has the same density and viscosity as water. Predict the fraction rejected for a flux of 20 L m2 h1. What is the maximum rejection?

Solution:c

v

k

j

K

K

R

Rexp

1

1

Need K and kc.

000,20/s)cm 105)(g/cm 1(

s-g/cm 01.0273Sc

D

N

D

dkN cSh = 0.0096NRe

0.913NSc0.346 = 3060

cm/s 1002.15.1

)105)(3060( 37

d

DNk Shc

(To be continued)

[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 m2 h1. What is the maximum rejection?


c

v

k

j

K

K

R

Rexp

1

1

cm/s 1002.1 3ck

Permeate flux, jv = 40 L m2 h1 = 1.11 103 cm/s

c

v

k

j

K

K

R

Rexp

1

1

3

3

1002.1

1011.1exp

175.0

75.01

K

K

K = 0.101

;

(To be continued)

c

v

k

j

K

K

R

Rexp

1

1

cm/s 1002.1 3ck ; K = 0.101

If the permeate flux is 20 L m2 h1 or 0.556 103 cm/s,

3

3

1002.1

10556.0exp

101.01

101.01

R

R R = 0.84

The maximum rejection occurs as the flux approaches zero.

Rmax = 1 K = 0.899 #

;

[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 m2 h1. What is the maximum rejection?


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.

Gas-liquid two phase flow patterns:

Abatement of Concentration Polarization in Ultrafiltration (2/4)

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 (3.032 103 Pa s at 25C compared to 1.86 105 Pa s at 27C).

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.


Time (min)

0 30 60 90

Per

mea

te F

lux

(m3 m

-2h

-1)

0.00

0.02

0.04

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.


DIAFILTRATION

Batch concentration versus diafiltration

Batch concentration:

Diafiltration:

In diafiltration, the dialyzate (or wash solvent) is added at a rate equal to removal of ultrafiltrate.

Batch concentration versus diafiltration (2/2)

DIAFILTRATION (2/4)

For microspecies (i.e., solutes) that are freely permeable to the membrane,

CdVCVd )( 0 wf VC

C

dVVC

dC

00

1

0 0

0lnV

V

C

C w

f

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

DIAFILTRATION (3/4)

For microspecies (i.e., solutes) that are partially permeable to the membrane,

0

0 )1(lnV

VR

C

C w

f

where R = product retention (dimensionless).

Note:

* R = 0

Complete product passage through the membrane.

* R = 1 Rejected solute.

0

0lnV

V

C

C w

f

CdVCVd )( 0

DIAFILTRATION (4/4)

[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 m2 h1. To remove 99.99% of the salt, determine the time and the volume of water required.

Solution:

0

0lnV

V

C

C w

f

10000001.0

1ln wV Vw = 9210 L

Vw = Jv A t t

2

2m 100

h m

L 30 L9210

t = 3.07 h #

REVERSE OSMOSIS

Adding a soluble salt to water.

Reducing the chemical potential of the water.

Osmotic flow.

The osmotic pressure of a solution:

= 1.12(T + 273)mi

= osmotic pressure, psi

T = temperature, C

mi = summation of molalities (mol/1000 g of water) of all ionic and nonionic constituents in the solution

* = 15 psi, for a typical brackish water; = 350 psi, for seawater.

REVERSE OSMOSIS (2/19)

External pressure applied for reverse osmosis to occur:

400600 psig for brackish water

8001,000 psig for seawater

Reverse osmosis is used most extensively in the purification of water and the concentration of biological and food processing streams.


Performance variables of reverse osmosis:

A simple schematic of an RO system:


The rate of water passage through a semipermeable membrane is defined by:

Qw = kwA(P )/

Qw = water flow rate through the membrane

kw = membrane permeability coefficient for water

A = membrane area

P = hydraulic pressure differential across the membrane

= osmotic pressure differential across the membrane

= membrane thickness

Performance variables of reverse osmosis (2/5):


The rate of salt flow through the membrane is given by:

Qs = ksAc/

Qs = flow rate of salt through the membrane

c = salt concentration differential across the membrane

ks = membrane permeability coefficient for salt



Qw = kwA(P )/ ; Qs = ksAc/

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.



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



The membranes used in reverse osmosis are asymmetric.

The thin “skin” is supported by a porous substructure.


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.

The membranes used in reverse osmosis are asymmetric. (2/2)


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.


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.


Reverse Osmosis Devices—Spiral-Wound

* 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.

* This device is like a huge envelope made of membrane and containing a feed spacer.


Reverse Osmosis Devices—Hollow Fiber

Aramid membranes: commercialized by Du Pont in 1970, made from an aromatic polyamide polymer, operated at pH range of 411, not susceptible to biological attack and resist hydrolysis. (But, they are degraded by chlorine.)

________


* 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.

Reverse Osmosis Devices—Hollow Fiber

____________


RO system design:


Remarks:

* Cartridge filter: remove large-particle matter that could damage the high-pressure pump or cause device plugging.

___


* 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.

______

______


* High-pressure shutdown switch or pressure-relief device: prevent the permeator from over-pressurization.

* Flow-control valve on the brine: set conversion.

______

____


CONCENTRATION POLARIZATION IN REVERSE OSMOSIS

Concentration polarization has two deleterious effects:

(1) The increase in c10 increases .

Qw = kwA(P )/

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 = ksAc/

It engenders more salt leakage into the product water.

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.

CONCENTRATION POLARIZATION IN REVERSE OSMOSIS (2/3)

[Example] A reverse osmosis process is used for desalination of seawater. The volumetric flux of water through the membrane is 3 105 m/s (or m3 s1 m2), 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, = 1.12(T + 273)mi

2.1)(

)(

1

2

1

2

i

i

m

m

2= 1.21 = 1.2 2.5 = 3.0 MPa

(To be continued)

[Example] Reverse osmosis process for desalination of seawater.

Flux of water = 3 10-5 m/s (or m3 s-1 m-2); P = 8.0 MPa; = 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?


2 = 3.0 MPa

Rate of water passage through the membrane:

Qw = kwA(P )/ or Qw/A = jv = Lp(P )

91.05.20.8

0.30.8

)(

)(

1

2

1

2

P

P

j

j

v

v

jv2 = 0.91jv1 = 0.91 (3 105) = 2.73 105 m/s

#

Correlation of Dittus and Boelter (1930) for concentration polarization in turbulent flow:

67.08.0

47.02.0

1

10

023.0

)/(exp

Dv

dj

c

c hv

where jv = the volume of solvent per area per time

dh = equivalent hydraulic diameter

= viscosity of fluid

= density of fluid

v = average velocity of fluid

D = local solute diffusivity in solution

CONCENTRATION POLARIZATION IN REVERSE OSMOSIS (3/3)

[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 18.5C. 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:67.08.0

47.02.0

1

10

023.0

)/(exp

Dv

dj

c

c hv

(a) Lower temperature increases the viscosity and lowers the diffusivity.

Lower temperature makes concentration polarization more severe.

(To be continued)

[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:67.08.0

47.02.0

1

10

023.0

)/(exp

Dv

dj

c

c hv

(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).

#

* Production of Low Alcohol Beer Using Reverse Osmosis

_____________

Remarks

The annual replacement cost is up to 7% of the original capital cost.

* 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.

* Production of Low Alcohol Beer Using Reverse Osmosis (2/2)

DIALYSIS

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

The external fluid must be repeatedly changed to reach the required final composition.

DIALYSIS (2/3)

________________________________________

Place a stirrer of some kind in the external fluid (or inside the bag).

To increase the rate of movement.

DIALYSIS (3/3)

REVERSE DIALYSIS

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

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.)

REVERSE DIALYSIS (2/2)

GLASS FIBER DIALYSIS

Semipermeable glass fibers: hollow-bore fibers whose glass walls contain pores of controlled size

Two ways of using hollow fibers: dialysis and concentration.

GLASS FIBER DIALYSIS (2/2)

ANALYSIS OF DIALYZER

A shell-and-tube type of hollow-fiber dialyzer:

Q = flow rate, A = area of membrane, K = dialysis coefficient

For a counterflow dialyzer,

The rate of salt removed

lmDDDFFF CKACCQCCQ )()( 1221

)]/()ln[(

)()(

1221

1221

DFDF

DFDFlm CCCC

CCCCC

ANALYSIS OF DIALYZER (2/2)

[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 0.0415 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 )()( 1221 DDDFFF CCQCCQ

g/L 362 DC

)0(500)1001.0100(200 2 DC

(To be continued)

lmFFF CKACCQ )( 21

[Example] Shell-and-tube type of hollow-fiber dialyzer

CF1 = 100 g/L; K = 0.0415 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):g/L 362 DC

)]/()ln[(

)()(

1221

1221

DFDF

DFDFlm CCCC

CCCCC

g/L 1.29)]010/()36100ln[(

)010()36100(

221 cm 905,14)1.29)(0415.0(

)10100(200)(

lm

FFF

CK

CCQA

lmFFF CKACCQ )( 21

#

Documents

MEMBRANE SEPARATIONS The most rapidly advances area of filtration technology. There are three major types of membrane-based filtration techniques: