Standard Methods for Examination of Water & Wastewater_7 doc

8.1 ELECTRODIALYSIS MEMBRANES Figure 8.1a shows a cut section of an electrodialysis filtering membrane.. The number FIGURE 8.1 Cation filtering membrane a; the electrodialysis process b.

Advanced Filtration and Carbon Adsorption

This chapter continues the discussion on filtration started in Chapter 7, except that

it deals with advanced filtration We have defined filtration as a unit operation ofseparating solids or particles from fluids A unit operation of filtration carried outusing membranes as filter media is advanced filtration This chapter discusses advancedfiltration using electrodialysis membranes and pressure membranes Filtration usingpressure membranes include reverse osmosis, nanofiltration, microfiltration, andultrafiltration

In addition to advanced filtration, this chapter also discusses carbon adsorption.This is a unit operation that uses the active sites in powdered, granular, and fibrousactivated carbon to remove impurities from water and wastewater Carbon adsorptionand filtration share some similar characteristics For example, head loss calculationsand backwashing calculations are the same Carbon adsorption will be discussed asthe last part of this chapter

8.1 ELECTRODIALYSIS MEMBRANES

Figure 8.1a shows a cut section of an electrodialysis filtering membrane The filteringmembranes are sheet-like barriers made out of high-capacity, highly cross-linkedion exchange resins that allow passage of ions but not of water Two types are used:

allow only anions to pass The cut section in the figure is a cation membrane composed

of an insoluble matrix with water in the pore spaces Negative charges are fixedonto the insoluble matrix, and mobile cations reside in the pore spaces occupied bywater It is the residence of these mobile cations that gives the membrane the property

of allowing cations to pass through it These cations will go out of the structure ifthey are replaced by other cations that enter the structure If the entering cationscame from water external to the membrane, then, the cations are removed from thewater, thus filtering them out In anion membranes, the mechanics just describedare reversed The mobile ions in the pore spaces are the anions; the ions fixed tothe insoluble matrix are the cations The entering and replacing ions are anions fromthe water external to the membrane In this case, the anions are filtered out from thewater

Figure 8.1b portrays the process of filtering out the ions in solution Inside thetank, cation and anion membranes are installed alternate to each other Two electrodesare put on each side of the tank By impressing electricity on these electrodes, thepositive anode attracts negative ions and the negative cathode attracts positive ions.This impression of electricity is the reason why the respective ions replace their like

8

ions in the membranes As shown in the figure, two compartments become “cleaned”

of ions and one compartment (the middle) becomes “dirty” of ions The two partments are diluting compartments; the middle compartment is a concentratingcompartment The water in the diluting compartments is withdrawn as the productwater, and is the filtered water The concentrated solution in the concentratingcompartment is discharged to waste

com-8.1.1 P OWER R EQUIREMENT OF E LECTRODIALYSIS U NITS

The filtering membranes in Figure 8.1b are arranged as CACA from left to right,where C stands for cation and A stands for anion In compartments CA, the water isdeionized, while in compartment AC, the water is not deionized The number ofdeionizing compartments is equal to two Also, note that the membranes are alwaysarranged in pairs (i.e., cation membrane C is always paired with anion membrane A).Thus, the number of membranes in a unit is always even If the number of membranes

is increased from four to six, the number of deionizing compartments will increasefrom two to three; if increased from six to eight, the number of deionizing membraneswill increase from three to four; and so on Thus, if m is the number of membranes

in a unit, the number of deionizing compartments is equal to m/2

As shown in the figure, a deionizing compartment pairs with a concentratingcompartment in both directions; this pairing forms a cell For example, deionizingcompartment CA pairs with concentrating compartment AC in the left direction andwith the concentrating compartment AC in the right direction of CA In this paring(in both directions), however, only one cell is formed equal to the one deionizingcompartment Thus, the number of cells formed in an electrodialysis unit can bedetermined by counting only the number of deionizing compartments The number

FIGURE 8.1 Cation filtering membrane (a); the electrodialysis process (b)

of deionizing compartments in a unit is m/2, so the number of cells in a unit is also

equal to m/2

Because one equivalent of a substance is equal to one equivalent of electricity,

in electrodialysis calculations, concentrations are conveniently expressed in terms

of equivalents per unit volume Let the flow to the electrodialysis unit be Q o The

flow per deionizing compartment or cell is then equal to Q o/(m/2) If the influent

ion concentration (positive or negative) is [C o] equivalents per unit volume, the total

rate of inflow of ions is [C o]Q o/(m/2) equivalents per unit time per cell One

equivalent is also equal to one Faraday Because a Faraday or equivalent is equal to

96,494 coulombs, assuming a coulomb efficiency of η, the amount of electricity

needed to remove the ions in one cell is equal to 96,494[C o]Q oη/(m/2) coulombs

per unit time Coulomb efficiency is the fraction of the input number of equivalents

of an ionized substance that is actually acted upon by an input of electricity

If time is expressed in seconds, coulomb per second is amperes Therefore, for

time in seconds, 96,494[C o]Q oη/(m/2) amperes of current must be impressed upon

the membranes of the cell to effect the removal of the ions The cells are connected

in series, so the same current must pass through all of the cells in the electrodialysis

unit, and the same 96,494[C o]Q oη/(m/2) amperes of current would be responsible

for removing the ions in the whole unit To repeat, not only is the amperage impressed

in one cell but in all of the cells in the unit

In electrodialysis calculations, a term called current density (CD) is often used

Current density is the current in milliamperes that flows through a square centimeter

of membrane perpendicular to the current direction: CD = mA/Acm, where mA is

the milliamperes of electricity and Acm is the square centimeters of perpendicular

area A ratio called current density to normality (CD/N) is also used, where N is

the normality A high value of this ratio means that there is insufficient charge to

carry the current away When this occurs, a localized deficiency of ions on the

membrane surfaces may occur This occurrence is called polarization In commercial

electrodialysis units CD/N of up to 1,000 are utilized

The electric current I that is impressed at the electrodes is not necessarily the

same current that passes through the cells or deionizing compartments The actual

current that successfully passes through is a function of the current efficiency which

varies with the nature of the electrolyte, its concentration in solution, and the

membrane system Call M the current efficiency The amperes passing through the

solution is equal to the amperes required to remove the ions Thus,

(8.1)

The emf E across the electrodes is given by Ohm’s law as shown below, where

R is the resistance across the unit

From basic electricity, the power P is EI=I2R Thus,

(8.3)

If I is in amperes, E is in volts, and R is in ohms, P in is in watts Of course,

the combined units of N and Q o must be in corresponding consistent units

Example 8.1 A brackish water of 378.51 m3/day containing 4,000 mg/L of

ions expressed as Nacl is to be de-ionized using an electrodialysis unit There are

400 membranes in the unit each measuring 45.72 cm by 50.8 cm Resistance across

the unit is 6 ohms and the current efficiency is 90% CD/N to avoid polarization is

700 Estimate the impressed current and voltage, the coulomb efficiency, and the

power requirement

Solution:

Therefore,

8.2 PRESSURE MEMBRANES

Pressure membranes are membranes that are used to separate materials from a fluid

by the application of high pressure on the membrane Thus, pressure membrane

filtration is a high pressure filtration This contrasts with electrodialysis membranes

in which the separation is effected by the impression of electricity across electrodes

Filtration is carried out by impressing electricity, therefore, electrodialysis membrane

filtration may be called electrical filtration

According to Jacangelo (1989), three allied pressure-membrane processes are

used: ultrafiltration (UF), nanofiltration (NF), and reverse osmosis (RO) He states

that UF removes particles ranging in sizes from 0.001 to 10 µm, while RO can

remove particles ranging in sizes from 0.0001 to 0.001 µm As far as size removals

are concerned, NF stays between UF and RO, being able to remove particles in

the size range of the order of 0.001 µm UF is normally operated in the range of

100 to 500 kPag (kilopascal gage); NF, in the range of 500 to 1,400 kPag; and RO,

in the range of 1,400 to 8,300 kPag Microfiltration (MF) may added to this list

MF retains larger particles than UF and operates at a lesser pressure (70 kPag)

L

- 68geq

m3 -;

( )( )/ 400/2η ( )0.90

η = 0.77 Ans

P = EI = I2R = 122.842( )6 = 90,538 W Ans

Whereas the nature of membrane retention of particles in UF is molecular screening,the nature of membrane retention in MF is that of molecular-aggregate screening.

On the other hand, comparing RO and UF, RO presents a diffusive transport barrier.Diffusive transport refers to the diffusion of solute across the membrane Due to thenature of its membrane, RO creates a barrier to this diffusion Figures 8.2 through8.4 present example installations of reverse osmosis units

FIGURE 8.2 Bank of modules at the Sanibel–Captiva reverse osmosis plant, Florida.

FIGURE 8.3 Installation modules of various reverse osmosis units (Courtesy of Specific

Equipment Company, Houston, TX.)

The basics of a normal osmosis process are shown in Figure 8.5a A bag ofsemipermeable membrane is shown placed inside a bigger container full of purewater Inside the membrane bag is a solution of sucrose Because sucrose has osmoticpressure, it “sucks” water from outside the bag causing the water to pass throughthe membrane Introduction of the water into the membrane bag, in turn, causes thesolution level to rise as indicated by the height π in the figure The height π is ameasure of the osmotic pressure It follows that if sufficient pressure is applied tothe tip of the tube in excess of that of the osmotic pressure, the height π will besuppressed and the flow of water through the membrane will be reversed (i.e., itwould be from inside the bag toward the outside into the bigger container); thus,the term “reverse osmosis.”

Sucrose in a concentration of 1,000 mg/L has an osmotic pressure of 7.24 kNa(kiloNewtons absolute) Thus, the reverse pressure to be applied must be, theoreti-cally, in excess of 7.24 kNa for a sucrose concentration of 1,000 mg/L For NaCl,its osmotic pressure in a concentration of 35,000 mg/L is 2744.07 kNa Hence, toreverse the flow in a NaCl concentration of 35,000 mg/L, a reverse pressure in excess

of 2744.07 kNa should be applied The operation just described (i.e., applyingsufficient pressure to the tip of the tube to reverse the flow of water) is the funda-

mental description of the basic reverse osmosis process

FIGURE 8.4 Reverse osmosis module designs.

Perforated

PVC baffle

Spiral wound

Permeate side backing material with membrane in each side and glued around edges and to center tube

Permeate out

Permeate flow (after passage through membrane)

Feed flow Feed side

spacer

Roll to assemble

Permeate

Feed flow Membrane Porous tube

water

Plate and frame

Grooved phenolic support plate

Cellulose acetate membrane

Paper

substrate

UF, NF, MF, and RO and are all reverse osmosis filtration processes; however,when the term reverse osmosis or RO is used without qualification, it is the processoperated at the highest pressure range to which it is normally referred Figure 8.5b is

a schematic of an RO plant Figure 8.2 is a photograph of a bank of modules in theSanibel–Captiva RO Plant in Florida This plant treats water for drinking purposes.Take careful note of the pretreatment requirement indicated in Figure 8.5b Asmentioned before, the RO process is an advanced mode of filtration and its purpose

is to remove the very minute particles of molecules, ions, and dissolved solids Theinfluent to a RO plant is already “clean,” only that it contains the ions, molecules,and molecular aggregates that need to be removed

FIGURE 8.5 (a) Osmosis process; (b) reverse osmosis system.

(b)

Glass tube p = Osmotic pressure

Dilute solution

of sucrose Bag made of

semipermeable

membrane

Water (a)

Pressure at

A = P Pressure at A' = P + p Passage of solvent (water)

Feedwater pump

High-pressure pump Pretreatment

Reverse osmosis module Membrane

Product

pump

Product water Waste discharge

After pretreatment the high-pressure pump forces the flow into the membranemodule where the solutes are rejected The flow splits into two, one producing theproduct water and the other producing the waste discharge The waste discharge hasone drawback in the use of RO filtration in that it may need to be treated separatelybefore discharge.

8.2.1 M EMBRANE M ODULE D ESIGNS

Over the course of development of the membrane technology, RO module designs,

as shown in Figure 8.4, evolved They are tubular, plate-and-frame, spiral wound,

and hollow fine-fiber modules In the tubular design, the membrane is lined

inside the tube which is made of ordinary tubular material Water is allowed to passthrough the inside of the tube under excess pressure causing the water to permeate

through the membrane and to collect at the outside of the tube as the product or

permeate The portion of the influent that did not permeate becomes concentrated.

This is called the concentrate or the reject

The plate-and-frame design is similar to the plate-and-frame filter press discussed

in the previous chapter on conventional filtration In the case of RO, the semipermeable

membrane replaces the filter cloth The spiral-wound design consists of two flat sheets

of membranes separated by porous spacers The two sheets are sealed on three sides;the fourth side is attached to a central collector pipe; and the whole sealed sheets arerolled around the central collector pipe As the sheets are rolled around the pipe, a

second spacer, called influent spacer, is provided between the sealed sheets In the

final configuration, the spiral-wound sealed membrane looks like a cylinder Water isintroduced into the influent spacer, thereby allowing it to permeate through the mem-brane into the spacer between the sealed membrane The permeate, now inside thesealed membrane, flows toward the central pipe and exits through the fourth unsealed

side into the pipe The permeate is collected as the product water The concentrate or

the reject continues to flow along the influent spacer and is discharged as the effluentreject or effluent concentrate This concentrate, which may contain hazardous mole-cules, poses a problem for disposal

In the hollow fine-fiber design, the hollow fibers are a bundle of thousands of

parallel, self-supporting, hair-like fibers enclosed in a fiberglass or epoxy-coatedsteel vessel Water is introduced into the hollow bores of the fibers under pressure.The permeate water exits through one or more module ports The concentrate alsoexits in a separate one or more module ports, depending on the design All thesemodule designs may be combined into banks of modules and may be connected inparallel or in series

8.2.2 F ACTORS A FFECTING S OLUTE R EJECTION

AND B REAKTHROUGH

The reason why the product or the permeate contains solute (that ought to beremoved) is that the solute has broken through the membrane surface along withthe product water It may be said that as long as the solute stays away from the membranesurface, only water will pass through into the product side and the permeate will

be solute-free; However, it is not possible to exclude the solute from contacting themembrane surface; hence, it is always liable to break through The efficiency at whichsolute is rejected is therefore a function of the interaction of the solute and themembrane surface As far as solute rejection and breakthrough are concerned, areview of literature revealed the following conclusions (Sincero, 1989):

• Percentage removal is a function of functional groups present in themembrane

• Percentage removal is a function of the nature of the membrane surface.For example, solute and membrane may have the tendency to bond byhydrogen bonding Thus, the solute would easily permeate to the productside if the nature of the surface is such that it contains large amounts ofhydrogen bonding sites

• In a homologous series of compounds, percentage removal increases withmolecular weight of solute

• Percentage removal is a function of the size of the solute molecule

• Percentage removal increases as the percent dissociation of the solutemolecule increases The degree of dissociation of a molecule is a function

of pH, so percentage removal is also a function of pH

This review also found that the percentage removal of a solute is affected bythe presence of other solutes For example, methyl formate experienced a drasticchange in percentage removal when mixed with ethyl formate, methyl propionate,and ethyl propionate When alone, it was removed by only 14% but when mixedwith the others, the removal increased to 66% Therefore, design of RO processesshould be done by obtaining design criteria utilizing laboratory or pilot plant testing

on the given influent

8.2.3 S OLUTE –W ATER S EPARATION T HEORY

The sole purpose of using the membrane is to separate the solute from the watermolecules Whereas MF, UF, and NF may be viewed as similar to conventional filtration,only done in high-pressure modes, the RO process is thought to proceed in a somewhatdifferent way In addition to operating similar to conventional filtration, some othermechanisms operate during the process Several theories have been advanced as to howthe separation in RO is effected Of these theories, the one suggested by Sourirajan withschematics shown in Figures 8.6a and 8.6b is the most plausible

Sourirajan’s theory is called the preferential-sorption, capillary-flow theory This

theory asserts that there is a competition between the solute and the water moleculesfor the surface of the membrane Because the membrane is an organic substance,several hydrogen bonding sites exist on its surface which preferentially bond watermolecules to them (The hydrogen end of water molecules bonds by hydrogenbonding to other molecules.) As shown in Figure 8.6a, H2O molecules are shownlayering over the membrane surface (preferential sorption), to the exclusion of thesolute ions of Na+ and Cl− Thus, this exclusion brings about an initial separation

In Figure 8.6b, a pore through the membrane is postulated, accommodating two

diameters of water molecules This pore size designated as 2t, where t is the diameter

of the water molecule, is called the critical pore diameter With this configuration,

the final separation of the water molecules and the solutes materializes by applyingpressure, pushing H2O through the pores (capillary flow)

As the process progresses, solutes build and line up near the membrane surfacecreating a concentration boundary layer This layer concentration is much largerthan in the bulk solution and, also, much larger, of course, than the concentration

in the permeate side This concentration difference creates a pressure for diffusivetransport The membrane, however, creates a barrier to this diffusion, thus, retainingthe solute and not allowing it to pass through easily Eventually, however, the solutewill diffuse out and leak to the permeate side

8.2.4 T YPES OF M EMBRANES

The first RO membrane put to practical use was the cellulose acetate membrane (CAmembrane) The technique of preparation was developed by Sourirajan and Loeband consisted of casting step, evaporation step, gelation step, and shrinkage step

The casting step involves casting a solution of cellulose acetate in acetone containing

an additive into flat or tubular surfaces The additive (such as magnesium perchlorate)must be soluble in water so that it will easily leach out in the gelation step creating

a porous film After casting, the solvent acetone is evaporated The material is then subjected to the gelation step where it is immersed in cold water The film material sets to a gel and the additive leaches out Finally, the film is subjected to the shrinkage

step that determines the size of the pores, depending upon the temperature used in

shrinking High temperatures create smaller pores

FIGURE 8.6 (a) schematic representation of preferential sorption-capillary flow theory;

(b) critical pore diameter for separation; (c) flux decline with time; (d) correction factor for surface area of cellulose acetate; and (e) solute rejection as a function of operating time.

Bulk of the solutions Pore water interface

Porous film surface

of appropriate chemical nature

Ca 2+ SO

2-Na +

H20 Na + Cl - H20 Na + Cl - H20 H20 Na + Cl - H20 Na + Cl - H20 H20 Na + Cl - H20 Na + Cl - H20 H20 Na + Cl - H20 Na + Cl - H20 H20 Na + Cl - H20 Na + Cl - H20 H20 Na + Cl - H20 Na + Cl - H20

H 2 0

H 2 0

H 2 0

Monovalent rejection ions

Porous film surface

of appropriate chemical nature

After this first development of the CA membrane, different types of membranes

followed: CAB, CTA, PBIL, and PA membranes CAB is membrane of cellulose acetate

butyrate; CTA is cellulose triacetate The PBIL membrane is a polybenzimidazolone

polymer and PA are polyamide membranes The structure of a PBIL unit is as follows:

Polyethylene amine reacted with tolylene diisocyanate produces the NS-100

membrane (NS stands for nonpolysaccharide) The reaction is carried out as follows:

In this reaction, the H bonded to the N of the n repeating units of polyethylene

amine [(−CH2CH2NH)n−] moves to the N of tolylene diisocyanate [−N=C=O] ing the double bond between N and C The C of the carboxyl group [=C=O] of tolylenediisocyanate then bonds with the N of the amine The reaction above simply shows two

destroy-of the tolylene molecules participating in the reaction, but in reality, there will bemillions of them performing the reaction of H moving and the C of the carboxyl group

of the tolylene bonding with the N of the amine and so on The final structure is a mesh

of cross-linked assembly, thus creating molecular pores

As indicated in the NS-100 product, a closed loop structure is formed Theethylene repeating units [−CH2CH2−] form the backbone of the membrane, and thebenzene rings form the cross-linking mechanism that tie together the ethylenebackbones forming the closed loop The ethylene units and the benzene rings are

nonpolar regions, while the peptide bonds and the amines [−NH2] arepolar regions In the NS-100, nonpolar regions exceed the polar regions; hence, this

membrane is said to be apolar

The CA membrane contains the OH− and the acetyl groups The

OH− region exceeds the acetyl region in the membrane OH− is polar, while theacetyl group is nonpolar regions Since the OH− region exceeds the acetyl region,

CA membranes are polar The polarity or apolarity of any membrane is very tant in characterizing its property to reject solutes

impor-N C O NH

C O NH

(CH2CH2N)n–CH2–CH2–N

C=O (CH2CH2NH)nCH2CH2NH + 2

Polyethylene amine

Tolylene diisocyanate N=C=O N=C=O

O O

C=O NH

NS-100 structure

– C – N H – O

CH3CO2−

Polyethylene amine reacted with isophthaloyl chloride produces the PA-100

membrane as shown in the following reaction:

The chloride atom in isophthaloyl chloride is attached to the carboxyl group Inthis reaction, the H in the amine group of polyethylene amine reacts with the chloride

in the carboxyl group producing HCl, as shown by the product over the arrow, andthe PA-100 membrane to the right of the reaction equation As in the case of theNS-100, the reaction forms the closed loop resulting in cross-linked structure of thePA-membrane Thus, a molecular pore is again produced

Epiamine, a polyether amine, reacting with isophthaloyl chloride produces the

PA-300 membrane as shown in the following reaction:

Again, in this reaction, the H in the amine group of the amine, epiamine, reactswith the chloride of isophthaloyl chloride forming HCl The N, in turn, of the aminegroup, from which the H that reacts with the C chloride were taken, bonds with the

C of the carboxyl group of isophthaloyl chloride producing the PA-300 membrane

As shown in its structure, this membrane is also a cross-linked membrane

meta-Phenylene diamine reacting with trimesoyl chloride produces the FilmTec FT-30 membrane according to the following reaction:

C=O (CH 2 CH 2 NH) n CH 2 CH 2 NH + 2

(CH 2 CH 2 N) n CH 2 CH 2 N – C

(CH 2 CH 2 N) n CH 2 CH 2 N –

C O

O Cl

–4HCI

Cl Isophthaloyl chloride

C O O

– 3HCI C=O

C=O CI

H N–C=O

O C–N–

–NH

–NH

– C – N – H

diamine

O

2-Hydroxy-methyl furan when dehydrated using H2SO4 produces the NS-200

according to the following reaction:

Another membrane formed from 2-hydroxy-methyl furan is Toray PEC-1000

The PAs in the prefixes for the naming of the membranes stand for polyamide

Thus, the membranes referred to are polyamide membranes PA membranes contain

the amide group ; it is for this component that they are called polyamidemembranes The formulas for the NS-100, NS-300, and the FT-30 membranes containthe amide group, thus, they are polyamide membranes The NS-200 is not a poly-amide membrane

In the early applications of the reverse osmosis technique, the membranes able were the polysaccharide membranes such as the CA membrane As new mem-branes were developed, they were differentiated from the saccharide membranes bycalling them nonpolysaccharide membranes Thus, the NS-200 is a nonpolysaccharidemembrane

avail-8.2.5 M EMBRANE P ERFORMANCE C HARACTERIZATION

The performance of a given membrane may be characterized according to its productflux and purity of product Flux, which is a rate of flow per unit area of membrane,

is a function of membrane thickness, chemical composition of feed, membrane ity, time of operation, pressure across membrane, and feedwater temperature Productpurity, in turn, is a function of the rejection ability of the particular membrane

poros-Flux decline Figure 8.6c shows the decline of the flux with time of operation.This curve applies to a given membrane and membrane pressure differential Thelower solid curve is the actual decline without the effect of cleaning The saw-toothedconfiguration is the effect of periodic cleaning As shown, right after cleaning, theflux rate shows an improvement, but then, it begins to decline again with time Fromexperience, the general trend of the curve plots a straight line in a log–log paper.Thus, empirically, the following equation fits the curve:

(8.4)

where F is the flux, t is the time, m is the slope of the line, and K is a constant The previous equation is a straight-line equation between ln t and ln F The equation is that of a straight line, so only two data points for ln t and ln F are required to calculate the constants m and K Using the techniques of analytic

geometry as applied to straight-line equations, the following equations are produced

points, the data are then grouped into two groups Thus, l is the number of mental data in the first group in a total of n experimental data points The second group would consist of n − l data points.

experi-Equation (8.4) may be used to estimate the ultimate flux of a given membrane

at the end of its life (one to two years)

Example 8.2 A long term experiment for a CA membrane module operated

at 2757.89 kPag using a feed of 2,000 mg/L of NaCl at 25°C produces the resultsbelow What is the expected flux at the end of one year of operation? What is theexpected flux at the end of two years? How long does it take for the flux to decrease

Let l = 1

that is, if the membrane has not broken before this time! Ans

Flux through membrane as a function of pressure drop The flow of

perme-ate through a membrane may be considered as a “microscopic” form of cake tion, where the solute that polarizes at the feed side of the membrane may beconsidered as the cake In cake filtration (see Chapter 7), the volume of filtrate

filtra-that passes through the cake in time t can be solved from

(8.7)

where µ is the absolute viscosity of filtrate; c, the mass of cake per unit volume of

filtrate collected; , the specific cake resistance; −∆P, the pressure drop across the cake and filter; S o , the filter area; and R m , the filter resistance In RO, c is the solute

collected on the membrane (in the concentration boundary layer) per unit volume of

permeate; and R m, the resistance of the membrane All the other parameters have similarmeanings as explained earlier in Chapter 7

The volume flux F is Using this and solving the above equation for

(8.8)

Initially neglecting the resistance of the solute in the concentration boundary layer,

µc in the denominator of the first factor on the right-side of the equation may

be set to zero, producing

1 -

+

Now, considering the resistance of the solute, designate the combined effect of

compressibility, membrane resistance R m, and solute resistance as Analogous

to cake filtration, call as specific membrane resistance Hence,

(8.10)

(8.11)

where s is an index of membrane and boundary layer compressibility When s is

equal to zero, is equal to , the constant of proportionality of the equation

Calling the pressure in the feed side as P f , the net pressure P fn acting on themembrane in the feed side is

(8.12)where πf is the osmotic pressure in the feed side Also, calling the pressure in the

permeate side as P p , the net pressure P pn acting on the membrane in the permeateside is

(8.13)where πp is the osmotic pressure in the permeate side Thus,

of the osmotic pressure of the latter In solution for the same masses, NaCl yieldsabout 1.6 times more particles than Na2SO4 From this it may be concluded thatosmotic pressure is a function of the number of particles in solution Comparing the1,000 mg/L concentrations of Na2SO4 and MgSO4, the osmotic pressure of the former

is about to 1.4 times that of the latter In solution Na2SO4 yields about 1.3 moreparticles than MgSO4 The same conclusions will be drawn if other comparisons aremade; therefore, osmotic pressure depends on the number of particles in solution.From this finding, osmotic pressure is, therefore, additive

Determination of and s The straight-line form of Equation (8.15) is

This equation needs only two data points of ln(µF) and ln(−∆P) to determine the

constants and s Assuming there are a total of n experimental data points and

using the first l data points for the first equation, the last n − l data points for the

second equation, and using the techniques of analytic geometry, the followingequations are produced:

(8.17)

(8.18)

Example 8.3 The feedwater to an RO unit contains 3,000 mg/L of NaCl,

300 mg/L of CaCl2, and 400 mg/L of MgSO4 The membrane used is celluloseacetate and the results of a certain study are shown below What will the flux be

if the pressure applied is increased to 4826.31 kPag? Assume that for the givenconcentrations the osmotic pressures are NaCl = 235.80 kPa, CaCl2= 17.17 kPa,and MgSO4 = 9.93 kPa Also, assume the temperature during the experiment is

25°C

TABLE 8.1 Osmotic Pressures at 25 °°°°C

Compound

Concentration (mg/L)

Osmotic Pressure (kPa)

NaCl 35,000 2758 NaCl 1,000 76 NaHCO 3 1,000 90

Na2SO4 1,000 41 MgSO 4 1,000 28 MgCl2 1,000 69 CaCl 2 1,000 55 Sucrose 1,000 7 Dextrose 1,000 14

Applied Pressure (kPag) Flux (m 3 /m 2

⋅⋅⋅⋅ day)

1723.68 0.123 4136.84 0.187

αmo

s 1 l∑n l+1ln(µF)–(n–l)∑1l ln(µF)

l∑l n+1ln(–∆P) n–( –l)∑1l ln(–∆P) -–

Therefore,

Effect of temperature on permeation rate As shown in Equation (8.15), the

flux is a function of the dynamic viscosity µ Because µ is a function of temperature,the flux or permeation rate is therefore also a function of temperature As temper-ature increases, the viscosity of water decreases Thus, from the equation, the flux

) ln ( −∆P) Flux (m3 /m 2 ⋅⋅⋅⋅ day) µµµµ F ln( µµµµ F)

1723.68 1,460,780a 14.19 0.123 9.56 2.26 4136.84 3,873,940 15.17 0.187 14.54 2.68

αmo

1–0.57( )14.19 2.26–

1 -

0.20 m

3

m2⋅ day - Ans

is expected to increase with increase in temperature Correspondingly, it is also expectedthat the flux would decrease as the temperature decreases Figure 8.6d shows the

correction factor C f for membrane surface area (for CA membranes) as a function

of temperature relative to 25°C As shown, lower temperatures have larger correctionfactors This is due to the increase of µ as the temperature decreases The opposite

is true for the higher temperatures These correction factors are applied to themembrane surface area to produce the same flux relative to 25°C

Percent solute rejection or removal The other parameter important in the

design and operation of RO units is the percent rejection or removal of solutes Let

Q o be the feed inflow, [C o ] be the feed concentration of solutes, Q p be the permeate

outflow, [C p ] be the permeate concentration of solutes, Q c be the concentrate outflow,

and [C c] be the concentrate concentration of solutes By mass balance of solutes,

the percent rejection R is

(8.19)

The index i refers to the solute species i.

Figure 8.6e shows the effect of operating time on percent rejection As shown,this particular membrane rejects divalent ions better than it does the monovalentions Generally, percent rejection increases with the value of the ionic charge

Example 8.4 A laboratory RO unit 152.4 cm in length and 30.48 cm indiameter has an active surface area of 102.18 m2 It is used to treat a feedwater withthe following composition: NaCl = 3,000 mg/L, CaCl2= 300 mg/L, and MgSO4=

400 mg/L The product flow is 0.61 m3/m2⋅ day and contains 90 mg/L NaCl,

6 mg/L CaCl2, and 8 mg/L MgSO4 The feedwater inflow is 104.9 m3/day (a) What

is the percent rejection of NaCl? (b) What is the over-all percent rejection of ions?

Solution:

8.3 CARBON ADSORPTION

Solids are formed because of the attraction of the component atoms within the solidtoward each other In the interior of a solid, attractive forces are balanced among thevarious atoms making up the lattice At the surface, however, the atoms are subjected

to unbalanced forces—the ones toward the interior are attracted, but the ones at the

R ∑Q o[C oi ] ∑Q– p[C pi]

∑Q o[C oi] - 100( ) ∑Q c[C ci]

∑Q o[C oi] - 100( )

a

( ) R ∑Q o[C oi ] ∑Q– p[C pi]

∑Q o[C oi] - 100( ) 104.9 3,000 - 100( 104.9 3,000) 0.61 102.18–( ( ) ) 90 ( )

surface are not Because of this unbalanced nature, any particle that lands on the surface

may be attracted by the solid This is the phenomenon of adsorption, which is the

process of concentrating solute at the surface of a solid by virtue of this attraction

Adsorption may be physical or chemical Physical adsorption is also called van

der Waals adsorption, and chemical adsorption is also called chemisorption In the

former, the attraction on the surface is weak, being brought about by weak van derWaals forces In the latter, the attraction is stronger as a result of some chemicalbonding that occurs Adsorption is a surface-active phenomenon which means largersurface areas exposed to the solutes result in higher adsorption The solute is called

the adsorbate; the solid that adsorbs the solute is called the adsorbent The adsorbate

is said to be sorbed onto the adsorbent when it is adsorbed, and it is said to bedesorbed when it passes into solution

Adsorption capacity is enhanced by activating the surfaces In the process usingsteam, activation is accomplished by subjecting a prepared char of carbon material such

as coal to an oxidizing steam at high temperatures resulting in the water gas reaction:

C + H2O → H2 + CO The gases released leave behind in the char a very porous structure.The high porosity that results from activation increases the area for adsorption.One gram of char can produce about 1000 m2 of adsorption area After activation,the char is further processed into three types of finished product: powdered form

called powdered activated carbon (PAC), the granular form called granular activated

carbon (GAC), and activated carbon fiber (ACF) PAC is normally less than 200

mesh; GAC is normally greater than 0.1 mm in diameter ACF is a fibrous form ofactivated carbon Figure 8.7 shows a schematic of the transformation of raw carbon

to activated carbon, indicating the increase in surface area

8.3.1 A CTIVATION T ECHNIQUES

Activation is the process of enhancing a particular characteristic Carbon whose

adsorption characteristic is enhanced is called activated carbon The activation

techniques used in the manufacture of activated carbons are dependent on the natureand type of raw material available The activation techniques that are principally

used by commercial production operations are chemical activation and steam

acti-vation As the name suggests, chemical activation uses chemicals in the process and

FIGURE 8.7 Raw carbon material on the left transforms to the carbon on the right after

activation.

is generally used for the activation of peat- and wood-based raw materials The rawmaterial is impregnated with a strong dehydrating agent, typically phosphoric pen-toxide (P2O5) or zinc chloride (ZnCl2) mixed into a paste and then heated to tem-peratures of 500–800°C to activate the carbon The resultant activated carbon iswashed, dried, and ground to desired size Activated carbons produced by chemicalactivation generally exhibit a very open pore structure, ideal for the adsorption oflarge molecules.

Steam activation is generally used for the activation of coal and coconut shellraw materials Activation is carried out at temperatures of 800–1100°C in the pres-ence of superheated steam Gasification occurs as a result of the water–gas reaction:

(8.20)

This reaction is endothermic but the temperature is maintained by partial burning

of the CO and H2 produced:

(8.21)(8.22)The activated carbon produced is graded, screened, and de-dusted Activatedcarbons produced by steam activation generally exhibit a fine pore structure, idealfor the adsorption of compounds from both the liquid and vapor phases

8.3.2 A DSORPTION C APACITY

The adsorption capacity of activated carbon may be determined by the use of an

adsorption isotherm The adsorption isotherm is an equation relating the amount of

solute adsorbed onto the solid and the equilibrium concentration of the solute in solution

at a given temperature The following are isotherms that have been developed:Freundlich; Langmuir; and Brunauer, Emmet, and Teller (BET) The most commonlyused isotherm for the application of activated carbon in water and wastewatertreatment are the Freundlich and Langmuir isotherms The Freundlich isotherm is

an empirical equation; the Langmuir isotherm has a rational basis as will be shownbelow The respective isotherms are:

(8.23)

(8.24)

X is the mass of adsorbate adsorbed onto the mass of adsorbent M; [C] is the

concentration of adsorbate in solution in equilibrium with the adsorbate adsorbed;

n, k, a, and b are constants.

C+H2O→H2+CO+175,440kJ/kgmol

2CO+O2→2CO2–393,790 kJ/kgmol2H2+O2→2H2O–396,650 kJ/kgmol

X M

- = k C[ ]1/n

X M

- ab C[ ]

1+b C[ ] -

=

The Langmuir equation may be derived as follows Imagine a particular iment in which a quantity of carbon adsorbent is added to a beaker of samplecontaining pollutant Immediately, the solute will be sorbed onto the adsorbent untilequilibrium is reached One factor determining the amount of the sorbed materialshas to be the number of adsorption sites in the carbon The number of these sites

exper-may be quantified by the ratio X /M By the nature of equilibrium processes, some

of the solutes adsorbed will be desorbed back into solution While these solutes aredesorbing, some solutes will also be, again, adsorbed This process continues on,like a seesaw; this “seesaw behavior” is a characteristic of systems in equilibrium

The rate of adsorption r s is proportional to the concentration in solution, [C],

(at equilibrium in this case) and the amount of adsorption sites left vacant by thedesorbing solutes Now, let us determine these vacant adsorption sites On a giventrial of the experiment, the number of adsorption sites filled by the solute may be

quantified by the ratio X /M, as mentioned previously The greater the concentration

of the solute in solution, the greater this ratio will be For a given type of solute andtype of carbon adsorbent, there will be a characteristic one maximum value for this

ratio Call this (X /M) ult Now, we have two ratios: X /M, which is the ratio at any time and (X /M) ult, which is the greatest possible ratio The difference of these tworatios is proportional to the number of adsorption sites left vacant; consequently, the

rate of adsorption r s is therefore equal to k s [C][(X /M) ult − (X/M)], where k s is aproportionality constant

For the desorption process, as the ratio (X /M) forms on the adsorbent, it must become a driving force for desorption Thus, letting k d be the desorption proportion-

ality constant, r d = k d (X /M), where r d is the rate of desorption The process is inequilibrium, so the rate of adsorption is equal to the rate of desorption Therefore,

(8.25)

Solving for X /M produces Equation (8.24), where a = (X/M) ult and b = k s /k d.Note that in the derivation no mention is made of how many layers of moleculesare sorbed onto or desorbed from the activated carbon It is simply that solutes aresorbed and desorbed, irrespective of the counts of the layers of molecules; however,

it is conceivable that as the molecules are deposited and removed, the process occurslayer by layer

The straight-line forms of the Freundlich and Langmuir isotherms are, respectively,

=

To use the isotherms, constants are empirically determined by running anexperiment This is done by adding increasing amounts of the adsorbent to a sample

of adsorbate solution in a container For each amount of adsorbent added, M i, the

equilibrium concentration [C i] is determined The pairs of experiment trial valuescan then be used to obtain the desired parameter values from which the constantsare determined Once the constants are determined, the resulting model is used to

determine M, the amount of adsorbent (activated carbon) that is needed From the derivation, the adsorption capacity of activated carbon is a = (X/M) ult From thisratio, the absorption capacity of activated carbon is shown as the maximum value

of the X /M ratios This ratio corresponds to a concentration equal to the maximum

possible solute equilibrium concentration

The value of X is obtained as follows: Let [C o] be the concentration of solute

in a sample of volume before adsorption onto a mass of adsorbent M Then

(8.28)

8.3.3 D ETERMINATION OF THE F REUNDLICH C ONSTANTS

Using the techniques of analytic geometry, let us derive the Freundlich constants in alittle more detail than used in the derivation of the constants in the discussion of reverseosmosis treated previously As mentioned, the straight-line form of the equation requiresonly two experimental data points; however, experiments are normally conducted toproduce not just two pair of values but more Thus, the experimental results must bereduced to just the two pairs of values required for the determination of the parameters;

therefore, assuming there are m pairs of values, these m pairs must be reduced to just

two pairs Once the reduction to two pairs has been done, the isotherm equation may

be then be written to just the two pairs of derived values as follows: