Standard Methods for Examination of Water & Wastewater_7 doc

TX249_frame_C08.fm Page 373 Friday, June 14, 2002 4:35 PM 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 of separating solids or particles from fluids A unit operation of filtration carried out using membranes as filter media is advanced filtration This chapter discusses advanced filtration using electrodialysis membranes and pressure membranes Filtration using pressure membranes include reverse osmosis, nanofiltration, microfiltration, and ultrafiltration 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 fibrous activated carbon to remove impurities from water and wastewater Carbon adsorption and filtration share some similar characteristics For example, head loss calculations and backwashing calculations are the same Carbon adsorption will be discussed as the last part of this chapter 8.1 ELECTRODIALYSIS MEMBRANES Figure 8.1a shows a cut section of an electrodialysis filtering membrane The filtering membranes are sheet-like barriers made out of high-capacity, highly cross-linked ion exchange resins that allow passage of ions but not of water Two types are used: cation membranes, which allow only cations to pass, and anion membranes, which 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 fixed onto the insoluble matrix, and mobile cations reside in the pore spaces occupied by water 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 if they are replaced by other cations that enter the structure If the entering cations came from water external to the membrane, then, the cations are removed from the water, thus filtering them out In anion membranes, the mechanics just described are reversed The mobile ions in the pore spaces are the anions; the ions fixed to the insoluble matrix are the cations The entering and replacing ions are anions from the water external to the membrane In this case, the anions are filtered out from the water Figure 8.1b portrays the process of filtering out the ions in solution Inside the tank, cation and anion membranes are installed alternate to each other Two electrodes are put on each side of the tank By impressing electricity on these electrodes, the positive 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 © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 374 Friday, June 14, 2002 4:35 PM 374 C A C A FIGURE 8.1 Cation filtering membrane (a); the electrodialysis process (b) 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 compartments are diluting compartments; the middle compartment is a concentrating compartment The water in the diluting compartments is withdrawn as the product water, and is the filtered water The concentrated solution in the concentrating compartment is discharged to waste 8.1.1 POWER REQUIREMENT OF ELECTRODIALYSIS UNITS 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 is deionized, while in compartment AC, the water is not deionized The number of deionizing compartments is equal to two Also, note that the membranes are always arranged 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 increase from two to three; if increased from six to eight, the number of deionizing membranes will 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 concentrating compartment in both directions; this pairing forms a cell For example, deionizing compartment CA pairs with concentrating compartment AC in the left direction and with 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 deionizing compartment Thus, the number of cells formed in an electrodialysis unit can be determined by counting only the number of deionizing compartments The number © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 375 Friday, June 14, 2002 4:35 PM 375 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 Qo The flow per deionizing compartment or cell is then equal to Qo /(m/2) If the influent ion concentration (positive or negative) is [Co] equivalents per unit volume, the total rate of inflow of ions is [Co]Qo /(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[Co]Qoη/(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[Co]Qoη/(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[Co]Qoη/(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, I M = 96,494 [ C o ]Q o η / ( m/2 ) 96,494 [ C o ]Q o η / ( m/2 ) I = M (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 E = IR If I is in amperes and R is in ohms, then E is in volts © 2003 by A P Sincero and G A Sincero (8.2) TX249_frame_C08.fm Page 376 Friday, June 14, 2002 4:35 PM From basic electricity, the power P is EI = I R Thus, [ C o ]Q o η 10 P = EI = I R = 3.72 ( 10 )  -  R  mM  (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 Qo must be in corresponding consistent units Example 8.1 A brackish water of 378.51 m /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 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: 4.0 geq geq -; [ C o ] = [ Nacl ] = - = = 0.068 - = 68 - N = 0.068 NaCl 23+35.45 L m Therefore, CD = 700 ( 0.068 ) = 47.6 mA/cm 47.6 ( 45.72 ) ( 50.8 ) I = = 122.84 A Ans 1000 ( 0.9 ) E = IR = 122.84 ( ) = 737 V Ans 378.51 96,494 ( 68 ) ( -) ) ( η )/ ( 400/2 ) 96,494 [ C o ]Q o η / ( m/2 ) 24 ( 60 ) ( 60 I = - = M 0.90 η = 0.77 Ans 2 P = EI = I R = 122.84 ( ) = 90,538 W Ans 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) © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 377 Friday, June 14, 2002 4:35 PM 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 the nature of its membrane, RO creates a barrier to this diffusion Figures 8.2 through 8.4 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.) © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 378 Friday, June 14, 2002 4:35 PM Porous tube Cellulose acetate membrane Perforated PVC baffle Membrane Feed flow Grooved phenolic support plate Paper substrate Permeate Plate and frame Large tube Roll to assemble Feed side spacer Feed water Feed flow Hollow fibers Permeate out Permeate flow (after passage through membrane) Permeate side backing material with membrane in each side and glued around edges and to center tube Product water Hollow fine fibers Spiral wound FIGURE 8.4 Reverse osmosis module designs The basics of a normal osmosis process are shown in Figure 8.5a A bag of semipermeable membrane is shown placed inside a bigger container full of pure water Inside the membrane bag is a solution of sucrose Because sucrose has osmotic pressure, it “sucks” water from outside the bag causing the water to pass through the membrane Introduction of the water into the membrane bag, in turn, causes the solution level to rise as indicated by the height π in the figure The height π is a measure of the osmotic pressure It follows that if sufficient pressure is applied to the tip of the tube in excess of that of the osmotic pressure, the height π will be suppressed and the flow of water through the membrane will be reversed (i.e., it would 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, theoretically, 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, to reverse 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., applying sufficient pressure to the tip of the tube to reverse the flow of water) is the fundamental description of the basic reverse osmosis process © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 379 Friday, June 14, 2002 4:35 PM p= Glass tube Osmotic pressure Pressure at A = P Dilute solution of sucrose Pressure at A' = P + Bag made of p semipermeable Passage of membrane Water solvent (water) (a) High-pressure Feedwater pump pump Pretreatment Reverse osmosis module Product pump Membrane Waste discharge Product water (b) FIGURE 8.5 (a) Osmosis process; (b) reverse osmosis system 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 process operated 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 the Sanibel–Captiva RO Plant in Florida This plant treats water for drinking purposes Take careful note of the pretreatment requirement indicated in Figure 8.5b As mentioned 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 The influent to a RO plant is already “clean,” only that it contains the ions, molecules, and molecular aggregates that need to be removed © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 380 Friday, June 14, 2002 4:35 PM After pretreatment the high-pressure pump forces the flow into the membrane module where the solutes are rejected The flow splits into two, one producing the product water and the other producing the waste discharge The waste discharge has one drawback in the use of RO filtration in that it may need to be treated separately before discharge 8.2.1 MEMBRANE MODULE DESIGNS 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 pass through 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 are rolled 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 is introduced into the influent spacer, thereby allowing it to permeate through the membrane into the spacer between the sealed membrane The permeate, now inside the sealed 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 effluent reject or effluent concentrate This concentrate, which may contain hazardous molecules, 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-coated steel 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 also exits in a separate one or more module ports, depending on the design All these module designs may be combined into banks of modules and may be connected in parallel or in series 8.2.2 FACTORS AFFECTING SOLUTE REJECTION AND BREAKTHROUGH The reason why the product or the permeate contains solute (that ought to be removed) is that the solute has broken through the membrane surface along with the product water It may be said that as long as the solute stays away from the membrane surface, only water will pass through into the product side and the permeate will © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 381 Friday, June 14, 2002 4:35 PM be solute-free; However, it is not possible to exclude the solute from contacting the membrane surface; hence, it is always liable to break through The efficiency at which solute is rejected is therefore a function of the interaction of the solute and the membrane surface As far as solute rejection and breakthrough are concerned, a review of literature revealed the following conclusions (Sincero, 1989): • Percentage removal is a function of functional groups present in the membrane • Percentage removal is a function of the nature of the membrane surface For example, solute and membrane may have the tendency to bond by hydrogen bonding Thus, the solute would easily permeate to the product side if the nature of the surface is such that it contains large amounts of hydrogen bonding sites • In a homologous series of compounds, percentage removal increases with molecular weight of solute • Percentage removal is a function of the size of the solute molecule • Percentage removal increases as the percent dissociation of the solute molecule 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 by the presence of other solutes For example, methyl formate experienced a drastic change in percentage removal when mixed with ethyl formate, methyl propionate, and ethyl propionate When alone, it was removed by only 14% but when mixed with the others, the removal increased to 66% Therefore, design of RO processes should be done by obtaining design criteria utilizing laboratory or pilot plant testing on the given influent 8.2.3 SOLUTE–WATER SEPARATION THEORY The sole purpose of using the membrane is to separate the solute from the water molecules 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 somewhat different way In addition to operating similar to conventional filtration, some other mechanisms operate during the process Several theories have been advanced as to how the separation in RO is effected Of these theories, the one suggested by Sourirajan with schematics 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 molecules for the surface of the membrane Because the membrane is an organic substance, several hydrogen bonding sites exist on its surface which preferentially bond water molecules to them (The hydrogen end of water molecules bonds by hydrogen bonding to other molecules.) As shown in Figure 8.6a, H2O molecules are shown layering over the membrane surface (preferential sorption), to the exclusion of the + − solute 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 © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 382 Friday, June 14, 2002 4:35 PM H20 Na+Cl- H20 Na+Cl- H20 H20 Na+Cl- H20 Na+Cl- H20 Bulk of the H Na+Cl- H Na+Cl- H 2 solutions H Na+Cl- H Na+Cl- H 2 Demineralized water H Na+Cl- H20 Na+Cl- H20 Pore water H20 Na+Cl- H20 Na+Cl- H20 interface Film Porous film surface H20 Porous film surface of appropriate H20 of appropriate chemical nature H20 chemical nature Film Critical pore diameter (b) (a) 100 Percent rejection Periodic cleaning of membrane Correction factor, Cf Flux 1.6 1.4 1.2 1.0 0.8 Na+ 10 20 30 Feedwater temperature (°C) (c) 90 0.6 Time Ca2+ 95 (d) SO2- Divalent rejection ions Monovalent rejection ions 85 1000 2000 3000 7000 Operating time (h) (e) 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 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 applying pressure, pushing H2O through the pores (capillary flow) As the process progresses, solutes build and line up near the membrane surface creating a concentration boundary layer This layer concentration is much larger than in the bulk solution and, also, much larger, of course, than the concentration in the permeate side This concentration difference creates a pressure for diffusive transport The membrane, however, creates a barrier to this diffusion, thus, retaining the solute and not allowing it to pass through easily Eventually, however, the solute will diffuse out and leak to the permeate side 8.2.4 TYPES OF MEMBRANES The first RO membrane put to practical use was the cellulose acetate membrane (CA membrane) The technique of preparation was developed by Sourirajan and Loeb and 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 © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 404 Friday, June 14, 2002 4:35 PM Assume diameter of actual column m; As = π (4 )/4 = 12.57 m Using columns, adsorbate retained in δ length of the columns = 2(12.57)(0.0051)(14.23) = 1.82 kg 2 Assuming carbon replacement is to be done every week, adsorbate retained in column length of length L − δ = 237(7) − 1.82 = 1657 kg Carbon required for the 1657 kg of adsorbate = 1657/0.02 = 82,850 kg = 41,425 kg/column Therefore, 41.425 - 721.58 L = + 0.0051 12.57 = 4.75, say 4.8 m including freeboard and other allowances Ans Diameter = m Ans Interval of carbon replacement = once every week Ans As soon as the bed is exhausted, as determined by the breakthrough of concentrations, the carbon may be replaced The replaced carbon may be reactivated again for reuse Up to 30 or more reactivations may be made without appreciable loss of adsorptive power of the reactivated carbon 8.3.6 RELATIVE VELOCITIES IN BED ADSORPTION As mentioned before, the unit operation of bed adsorption may be carried out in a moving-bed mode, either co-currently or countercurrently When the breakthrough experiment is carried out, the superficial velocity should also be recorded The reason is that adsorption is a function of the time of contact between the liquid phase containing the solute to be adsorbed and solid-phase carbon bed Thus, for the breakthrough data to be applicable to an actual prototype adsorption column, the relative velocities that transpired during the test must be maintained in the actual column When the relative velocities between the flowing water and the carbon bed are maintained, it is immaterial whether or not the bed is moving Consider first the co-current operation Let Vs be the superficial velocity of the flowing water relative to the stationary earth and Vb be the velocity of the bed also relative to the stationary earth Thus, the relative velocity of the flowing water Vs/b relative to the bed is V s/b = V s – V b V s = V s/b + V b (8.40) For the countercurrent operation, the formula is V s/b = V s – ( – V b ) V s = V s/b + ( – V b ) = V s/b – V b © 2003 by A P Sincero and G A Sincero (8.41) TX249_frame_C08.fm Page 405 Friday, June 14, 2002 4:35 PM In a breakthrough experiment, the superficial velocity may be obtained by dividing the volume V of water collected in t time by the superficial area of the experimental column Breakthrough experiments are invariably conducted in stationary beds Thus, from the previous equations this superficial velocity is actually the relative velocity of the flowing water with respect to the bed, with Vb equal to zero This relative velocity must be maintained in the actual column design, if the data collected in the breakthrough experiment are to be applicable Example 8.9 Design the column of the previous example if the feed is introduced at the bottom The carbon is continuously removed at the bottom and continuously added at the top Due to the countercurrent operation, assume the bed expands by 40% Solution: The design will include, in addition to those in the previous example, the determination of the carbon removal and addition rates at the bottom and top of column, respectively These rates are determined from the length of the packed carbon in the column and the interval of replacement as stated in the previous example 4.57 Removal rate = addition rate = - (12.57)(721.58) = 246.73 kg/h ( 24 ) Ans The superficial velocity, from the previous example = 0.0088 m/s, velocity relative to the stationary bed In the countercurrent operation, this relative velocity must be maintained if the breakthrough curve is to be applicable Because of the expansion, Vb is increased by 40% Thus, Vs in present design considering the expansion: 4.57 ( 1.4 ) V s = V s/b – V b = 0.0088 – - = 0.0088 – 0.00001 = 0.0088 m/s ( 24 ) ( 60 ) ( 60 ) Therefore, 0.11 A s = - = 12.5 m ⇒ diameter = m Ans 0.0088 L = 1.4 ( 4.57 ) = 6.4 m say 6.7 m, including freeboard and other allowances 8.3.7 HEAD LOSSES IN Ans BED ADSORPTION The operation of carbon bed adsorption units is similar to that of filtration units In fact, bed adsorption is partly filtration Countercurrent flow operation is analogous to filter backwashing, and co-current flow operation is analogous to normal downflow or gravity filtration Head loss calculations for bed adsorption are therefore the same as those with filtration Since head loss formulas through beds of solids have already been discussed under filtration, they will not be pursued here The important point to remember is that for filtration formulas to apply under moving-bed adsorption operations, the superficial velocity should now be considered relative velocity © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 406 Friday, June 14, 2002 4:35 PM GLOSSARY Anion membrane—An electrodialysis membrane that allows passage of anions Apolar membrane—An RO membrane in which the nonpolar regions exceed the polar regions Activation—The process of enhancing a particular characteristic Active zone—A small length of column where adsorption is taking place Activated carbon—Carbon with enhanced adsorption characteristic Adsorbate—The solute adsorbed onto the surface of a solid Adsorbent—The solid that adsorbs the adsorbate Adsorption—The process of concentrating solute at the surface of a solid as a result of the unbalanced attraction of atoms at the surface of the solid Absorption capacity—The X/M that produces the lowest possible residual concentration of the adsorbate Adsorption isotherm—An equation relating the amount of solute adsorbed onto the solid and the equilibrium concentration of the solute in solution at a given temperature Benzene ring— Cation membrane—An electrodialysis membrane that allows passage of cations Chemical activation—The process of activating carbon using chemicals Chemical adsorption or chemisorption—A type of adsorption aided by chemical bonding on the surfaces Coulomb efficiency—The fraction of the input number of equivalents of an ionized substance that is actually acted upon by an input of electricity Current density—The current in milliamperes that flows through a square centimeter of membrane perpendicular to the current direction Electrodialysis membrane—Sheet-like barriers made out of high-capacity, highly cross-linked ion exchange resins that allow passage of ions but not of water Flux—The rate of flow or mass flow per unit area per unit time Microfiltration—A reverse osmosis process that removes particles larger than those removed by ultrafiltration and operating at the lowest operating range of the reverse osmosis processes in the neighborhood of 70 kPag Nanofiltration—A reverse osmosis process that removes particles in the size range of the order of 0.001 µm at an operating pressure range of 500 to 1,400 kPag Peptide bond— O –C –NH– Physical adsorption—A type of adsorption brought about by weak van der Waals forces Polarization—A localized deficiency of ions in the vicinity of the membrane surfaces Polar membrane—An RO membrane in which the polar regions exceed the nonpolar regions Pressure membrane—A membrane that is used to separate materials from a fluid by the application of high pressure on the membrane © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 407 Friday, June 14, 2002 4:35 PM Product or permeate—The portion of the influent that passes (permeates) through the membrane Reverse osmosis—A process of removing minute particles such as ions, molecules, and molecular aggregates The process operating at the highest pressure range of 1,400 to 8,300 kPag removes particles in the size range of 0.0001 to 0.001 µm Steam activation—The process of activating carbon using steam Ultrafiltration—A reverse osmosis process that removes particles ranging in sizes from 0.001 to 10 µm at an operating range of 100 to 500 kPag Water gas reaction—C + H2O → H2 + CO SYMBOLS a As b c k [C] [Cci] [Co] [Coi] [Cpi] E F I K m M M n P Pf Pp Pfn Ppn ∆P Qo Qc Qp R Rm A constant in the Langmuir isotherm Cross-sectional area of carbon column A constant in the Langmuir isotherm Mass of cake per unit volume of filtrate A constant in the Freundlich isotherm Equilibrium concentration of adsorbate in solution Concentration of species i in concentrate of RO units Influent concentration to an active zone, column, or electrodialysis RO units Influent concentration of species i to an RO unit Concentration of species i in permeate of RO units Electromotive force across electrodes Product flux of RO units Electric current impressed on the electrodes of an electrodialysis unit A constant in the flux decline equation of RO units Number of membranes in an electrodialysis unit; a constant in the flux decline equation of RO units Amount of adsorbent that adsorbs X amount of adsorbate Current efficiency A constant in the Freundlich isotherm Power to electrodialysis unit Applied pressure on feed side of RO units Pressure on permeate side of RO units Net applied pressure on feed side of RO units Net pressure on permeate side of RO units Pressure drop Influent flow Concentrate flow Permeate flow Electric resistance across electrodialysis unit; percent rejection of RO units Resistance of filter or RO membrane © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 408 Friday, June 14, 2002 4:35 PM s So t Vb Vs Vs/b V Vb Vx X (X/M)ult α αm α mo δ η µ ρp πf πp Index of membrane and boundary layer compressibility Surficial area Time of operation Absolute velocity of bed Absolute superficial velocity of water Relative superficial velocity of flowing water relative to bed Volume of sample; volume of permeate Volume of effluent in column at breakthrough Volume of effluent in column at exhaustion Amount of adsorbate adsorbed to M amount of adsorbent Adsorption capacity Specific cake resistance Combined effect of compressibility, membrane resistance, and solute resistance in RO s A constant of proportionality in α m = α mo (−∆P) Length of an active zone Coulomb efficiency Absolute viscosity of water Bulk density of carbon Osmotic pressure on feed side of RO units Osmotic pressure on permeate side of RO units PROBLEMS 8.1 A wastewater containing a [Co] = 25 mg/L of phenol is to be treated using PAC to produce an effluent concentration [C]eff = 0.10 mg/L The PAC is simply added to the stream and the mixture subsequently settled in the following sedimentation tank The constants of the Langmuir equation are determined by running a jar test producing the results below The volume of waste subjected to each test is one liter If the flow rate Qo is 0.11 m /s, calculate the quantity of PAC needed for the operation What is the adsorption capacity of the PAC? Calculate the quantity of PAC needed to treat the influent phenol to the ultimate residual concentration Use the Langmuir isotherm Test [C] (mg/L) 8.2 8.3 PAC Added X (g) 0.25 0.5 1.0 2.6 6.0 0.25 0.09 0.06 Solve Problem 8.1 using the Freundlich isotherm A wastewater containing a [Co] = 25 mg/L of phenol is to be treated using PAC to produce an effluent concentration [C]eff = 0.10 mg/L The PAC is simply added to the stream and the mixture subsequently settled in the following sedimentation tank The constants of the Langmuir equation is © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 409 Friday, June 14, 2002 4:35 PM determined by running a jar test producing the results below The volume of waste subjected to each test is one liter If the flow rate Qo is 0.11 m /s, calculate the quantity of PAC needed for the operation What is the adsorption capacity of the PAC? Calculate the quantity of PAC needed to treat the influent phenol to the ultimate residual concentration Use the Langmuir isotherm Test 8.4 8.5 PAC Added X (g) [C ] (mg/L) 0.25 2.6 6.0 0.06 Solve Problem 8.3 using the Freundlich isotherm A wastewater containing a [Co] = 25 mg/L of phenol is to be treated using PAC to produce an effluent concentration [C]eff = 0.10 mg/L The PAC is simply added to the stream and the mixture subsequently settled in the following sedimentation tank The constants of the Langmuir equation are determined by running a jar test producing the results below The volume of waste subjected to each test is one liter If the flow rate Qo is 0.11 m /s, calculate the quantity of PAC needed for the operation What is the adsorption capacity of the PAC? Calculate the quantity of PAC needed to treat the influent phenol to the ultimate residual concentration Use the Langmuir isotherm Test [C] (mg/L) 8.6 8.7 PAC Added X (g) 0.25 6.0 Solve Problem 8.5 using the Freundlich isotherm A breakthrough experiment is conducted for phenol producing the results below The length of the active zone is calculated to be 5.1 mm The diameter of the column used is 2.5 cm, and the packed density of the bed is 721.58 kg/m (X/M)ult = 0.020 kg/kg Calculate the influent phenol concentration C (mg/L) 0.06 1.0 6.0 10 15 18 20 23 25 8.8 V (L) 1.0 1.24 1.31 1.43 1.48 1.58 1.72 1.83 2.00 A breakthrough experiment is conducted for phenol producing the results shown in Problem 8.7 The length of the active zone is calculated to be © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 410 Friday, June 14, 2002 4:35 PM 8.9 8.10 8.11 5.1 mm The influent phenol concentration is 25 mg/L The packed density of the bed is 721.58 kg/m ⋅(X/M)ult = 0.020 kg/kg Calculate the diameter of the column used in the experiment A breakthrough experiment is conducted for phenol producing the results shown in Problem 8.7 The length of the active zone is calculated to be 5.1 mm The influent phenol concentration is 25 mg/L (X/M)ult = 0.020 kg/kg The diameter of the column used in the experiment is 2.5 cm Calculate the packed density of the carbon used A breakthrough experiment is conducted for phenol producing the results shown in Problem 8.7 The length of the active zone is calculated to be 5.1 mm The influent phenol concentration is 25 mg/L The diameter of the column used in the experiment is 2.5 cm and the packed density of the bed is 721.58 kg/m Calculate the adsorption capacity of the bed A breakthrough experiment is conducted for phenol producing the results below Determine the length δ of the active zone The diameter of the column used is 2.5 cm, and the packed density of the bed is 721.58 kg/m [Co] is equal to 25 mg/L (X / M)ult = 0.020 kg/kg C (mg/L) 0.06 1.0 6.0 10 20 25 8.12 V (L) 1.0 1.24 1.31 1.43 1.72 2.00 A breakthrough experiment is conducted for phenol producing the results below Determine the length δ of the active zone The diameter of the column used is 2.5 cm, and the packed density of the bed is 721.58 kg/m [Co] is equal to 25 mg/L (X / M)ult = 0.020 kg/kg C (mg/L) 0.06 1.0 6.0 25 8.13 V (L) 1.0 1.24 1.31 2.00 A breakthrough experiment is conducted for phenol producing the results below Determine the length δ of the active zone The diameter of the column used is 2.5 cm, and the packed density of the bed is 721.58 kg/m © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 411 Friday, June 14, 2002 4:35 PM [Co] is equal to 25 mg/L (X/M)ult = 0.020 kg/kg C (mg/L) 0.06 25 8.14 8.15 8.16 8.17 8.18 V (L) 1.0 2.00 A wastewater containing 25 mg/L of phenol and having the characteristic breakthrough of Problem 8.11 is to be treated by adsorption onto an activated carbon bed Assume that the flow rate during the breakthrough experiment is 0.11 m /s The X/M ratio of the bed for the desired effluent of 0.1 mg/L is 0.029 kg solute per kg carbon If the flow rate for design is also 0.11 m /s, design the absorption column Assume the influent is introduced at the top of the bed The packed density of the carbon bed is 721.58 kg/m A wastewater containing 25 mg/L of phenol and having the characteristic breakthrough of Problem 8.12 is to be treated by adsorption onto an activated carbon bed Assume that the flow rate during the breakthrough experiment is 0.11 m /s The X/M ratio of the bed for the desired effluent of 0.1 mg/L is 0.029 kg solute per kg carbon If the flow rate for design is also 0.11 m /s, design the absorption column Assume the influent is introduced at the top of the bed The packed density of the carbon bed is 721.58 kg/m A wastewater containing 25 mg/L of phenol and having the characteristic breakthrough of Problem 8.13 is to be treated by adsorption onto an activated carbon bed Assume that the flow rate during the breakthrough experiment is 0.11 m /s The X/M ratio of the bed for the desired effluent of 0.1 mg/L is 0.029 kg solute per kg carbon If the flow rate for design is also 0.11 m /s, design the absorption column Assume the influent is introduced at the top of the bed The packed density of the carbon bed is 721.58 kg/m A wastewater containing 25 mg/L of phenol and having the characteristic breakthrough of Problem 8.12 is to be treated by adsorption onto an activated carbon bed Assume that the flow rate during the breakthrough experiment is 0.11 m /s The X/M ratio of the bed for the desired effluent of 0.1 mg/L is 0.020 kg/kg If the flow rate for design is also 0.11 m /s, design the absorption column Assume the influent is introduced at the top of the bed The packed density of the carbon bed is 721.58 kg/m A wastewater containing 25 mg/L of phenol and having the characteristic breakthrough of Problem 8.13 is to be treated by adsorption onto an activated carbon bed Assume that the flow rate during the breakthrough experiment is 0.11 m /s The X/M ratio of the bed for the desired effluent of 0.1 mg/L is 0.020 kg/kg If the flow rate for design is also 0.11 m /s, design the absorption column Assume the influent is introduced at the top of the bed The packed density of the carbon bed is 721.58 kg/m © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 412 Friday, June 14, 2002 4:35 PM 8.19 8.20 8.21 8.22 8.23 8.24 8.25 8.26 8.27 8.28 Design the column of Problem 8.14 if the feed is introduced at the bottom The carbon is continuously removed at the bottom and continuously added at the top Due to the countercurrent operation, assume the bed expands by 40% Design the column of Problem 8.15 if the feed is introduced at the bottom The carbon is continuously removed at the bottom and continuously added at the top Due to the countercurrent operation, assume the bed expands by 40% Design the column of Problem 8.16 if the feed is introduced at the bottom The carbon is continuously removed at the bottom and continuously added at the top Due to the countercurrent operation, assume the bed expands by 40% Design the column of Problem 8.17 if the feed is introduced at the bottom The carbon is continuously removed at the bottom and continuously added at the top Due to the countercurrent operation, assume the bed expands by 40% Design the column of Problem 8.18 if the feed is introduced at the bottom The carbon is continuously removed at the bottom and continuously added at the top Due to the countercurrent operation, assume the bed expands by 40% A brackish water of 379 m /day containing 4000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit The coulomb efficiency is 0.78 and there are 400 membranes in the unit each measuring 51 cm × 46 cm Resistance across the unit is ohms and the current efficiency is 90% Estimate the power requirement A brackish water of 379 m /day is to be deionized using an electrodialysis unit The coulomb efficiency is 0.78, and 400 membranes are in the unit each measuring 51 cm × 46 cm Resistance across the unit is ohms and the current efficiency is 90% The input power required to run the unit is 93.3 kW Estimate the concentration of ions to be removed expressed as NaCl A brackish water containing 4000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit The coulomb efficiency is 0.78, and 400 membranes are in the unit each measuring 51 cm × 46 cm Resistance across the unit is ohms and the current efficiency is 90% The input power required to run the unit is 93.3 kW Estimate the input flow to the unit A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit There are 400 membranes in the unit each measuring 51 cm × 46 cm Resistance across the unit is ohms and the current efficiency is 90% The input power required to run the unit is 93.3 kW If the inflow to the unit is 379 m /day, calculate the coulomb efficiency A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit Resistance across the unit is ohms and the current efficiency is 90% The input power required to run © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 413 Friday, June 14, 2002 4:35 PM 8.29 8.30 8.31 8.32 8.33 8.34 8.35 8.36 the unit is 93.3 kW If the inflow to the unit is 379 m /day and the coulomb efficiency is 0.78, estimate the number of membranes in the unit A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit Resistance across the unit is ohms The input power required to run the unit is 93.3 kW The inflow to the unit is 379 m /day, the coulomb efficiency is 0.78, and 400 membranes are in the unit each measuring 51 cm × 46 cm Calculate the current efficiency A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit The input power required to run the unit is 93.3 kW The inflow to the unit is 379 m /day, the coulomb efficiency is 0.78, and 400 membranes are in the unit each measuring 51 cm × 46 cm The current efficiency is 90% What is the electric resistance across the unit? A brackish water of 379 m /day containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit There are 400 membranes in the unit each measuring 51 cm × 46 cm inches Resistance across the unit is ohms, the current and coulomb efficiencies are, respectively, 90% and 78% Estimate the impressed current A brackish water of 379 m /day is to be deionized using an electrodialysis unit There are 400 membranes in the unit each measuring 51 cm × 46 cm inches Resistance across the unit is ohms, the current and coulomb efficiencies are, respectively, 90% and 78% If the impressed current is 124.76 amperes, what is the concentration of the ions in the raw water expressed as NaCl? A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit There are 400 membranes in the unit each measuring 51 cm × 46 cm inches Resistance across the unit is ohms, the current and coulomb efficiencies are, respectively, 90% and 78% If the impressed current is 124.76 amperes, what is the influent flow to the unit? A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit There are 400 membranes in the unit each measuring 51 cm × 46 cm inches Resistance across the unit is ohms; the current efficiency is 90% If the impressed current is 124.76 amperes and the influent flow to the unit is 379 m /day, what is the coulomb efficiency? A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit Resistance across the unit is ohms, the current and coulomb efficiencies are, respectively, 90% and 78% If the impressed current is 124.76 amperes and the influent flow to the unit is 379 m /day, calculate the number of membranes in the unit A brackish water containing 4,000 mg/L of ions expressed as NaCl is to be deionized using an electrodialysis unit Resistance across the unit is ohms; the coulomb efficiency is 78% If the impressed current is © 2003 by A P Sincero and G A Sincero TX249_frame_C08.fm Page 414 Friday, June 14, 2002 4:35 PM 8.37 124.76 amperes and the influent flow to the unit is 379 m /day, calculate the current efficiency A long term experiment for a CA membrane module operated at 2758 kPag using a feed of 2,000 mg/L of NaCl at 25°C produces the results below What is the expected flux at the end of one year of operation? What is the expected flux at the end of two years? How long does it take for the flux to decrease to 0.37 m /m ⋅ day? Time (h) Flux (m / m ⋅ day) 8.38 8.39 8.40 8.41 0.66 25,000 0.45 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 cellulose acetate Applying a pressure of 4826 kPag, the flux is found to be 0.203 m /m ⋅ day If s and α mo are, respectively, 0.5597 and 54.72 in the MKS system of units (meter-kilogram-second), at what temperature is the unit being operated? 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 cellulose acetate Applying a pressure of 4,826 kPag, the flux is found to be 0.203 m /m ⋅ day If s is equal to 0.5597 in the MKS system of units (meter-kilogram-second) and the temperature of operation is 25°C, what is the value of α mo ? 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 cellulose acetate The flux is 0.203 m /m ⋅ day If s and α mo are, respectively, 0.5597 and 54.72 in the MKS system of units (meter-kilogram-second) and the temperature is 25°C, what is the pressure applied to the membrane? 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 cellulose acetate Applying a pressure of 4,826 kPag, the flux is found to be 0.203 m /m ⋅ day If α mo is 54.72 in the MKS system of units (meter-kilogram-second) and the temperature is 25°C, what is the value of s? 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Index of membrane and boundary layer compressibility Surficial area Time of operation Absolute velocity of bed Absolute superficial velocity of water Relative superficial velocity of flowing water. .. Oxford England Bou-Hamad, S., et al (1997) Performance evaluation of three different pretreatment systems for seawater reverse osmosis technique, Desalination Int Symp Pretreatment of Feedwater... function of the size of the solute molecule • Percentage removal increases as the percent dissociation of the solute molecule increases The degree of dissociation of a molecule is a function of pH,