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 8 TX249_frame_C08.fm Page 373 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 374 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 com- partments 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 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 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 FIGURE 8.1 Cation filtering membrane (a); the electrodialysis process (b). CCAA TX249_frame_C08.fm Page 374 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 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 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 o η / (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 o η / (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 o η / (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 / A cm , where mA is the milliamperes of electricity and A cm 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. (8.2) If I is in amperes and R is in ohms, then E is in volts. IM 96,494 C o []Q o η / m/2()= I 96,494 C o []Q o η / m/2() M = EIR= TX249_frame_C08.fm Page 375 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero From basic electricity, the power P is EI = I 2 R . 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 m 3 /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 µ m, while RO can remove particles ranging in sizes from 0.0001 to 0.001 µ m. 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 µ m. 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). PEII 2 R 3.72 10 10 () C o []Q o η mM   2 R== = C o [] Nacl[] 4.0 NaCl 4 23+35.45 0.068 geq L 68 geq m 3 ;=== = = N 0.068= CD 700 0.068()47.6 mA/cm 2 == I 47.6 45.72()50.8() 1000 0.9() 122.84 A Ans== EIR122.84 6() 737 V Ans== = I 96,494 C o []Q o η / m/2() M 96,494 68() 378.51 24 60()60() () η ()/400/2() 0.90 == η 0.77 Ans= PEII 2 R 122.84 2 6() 90,538 W Ans== = = TX249_frame_C08.fm Page 376 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 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.) TX249_frame_C08.fm Page 377 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 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, 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, 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 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 Large tube Hollow fine fibers Product water Hollow fibers Feed water Plate and frame Grooved phenolic support plate Cellulose acetate membrane Paper substrate TX249_frame_C08.fm Page 378 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 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. FIGURE 8.5 (a) Osmosis process; (b) reverse osmosis system. TX249_frame_C08.fm Page 379 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 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 mem- brane 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 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-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 TX249_frame_C08.fm Page 380 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 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, H 2 O 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 TX249_frame_C08.fm Page 381 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero 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 H 2 O 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. 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 Film Film Demineralized water Critical pore diameter (a) (b) (c) (d) (e) Flux Time Periodic cleaning of membrane Feedwater temperature (°C) 0102030 Correction factor, C f 1.6 1.4 1.2 1.0 0.8 0.6 1000 2000 3000 7000 100 95 90 85 Percent rejection Operating time (h) Divalent rejection ions Ca 2+ SO 2- Na + H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0 H 2 0 H 2 0 Monovalent rejection ions Porous film surface of appropriate chemical nature TX249_frame_C08.fm Page 382 Friday, June 14, 2002 4:35 PM © 2003 by A. P. Sincero and G. A. Sincero [...]... 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... The performance of a given membrane may be characterized according to its product flux 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 porosity, time of operation, pressure across membrane, and feedwater temperature Product purity, in turn, is a function of the rejection ability of the particular... + 2S o µ R m (8. 8) Initially neglecting the resistance of the solute in the concentration boundary layer, µc α V in the denominator of the first factor on the right-side of the equation may be set to zero, producing V 1 = F = - ( – ∆P ) µ Rm tS o © 2003 by A P Sincero and G A Sincero (8. 9) TX249_frame_C 08. fm Page 388 Friday, June 14, 2002 4:35 PM Now, considering the resistance of the solute,... ( – ∆P ) l (8. 17) l ( 1 – s )∑ 1 ln ( – ∆P ) – ∑ 1 ln ( µ F ) α mo = exp  -    l (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 cellulose acetate and the results of a certain study are shown below What will the flux be if the pressure applied is increased to 482 6.31 kPag?... that as the molecules are deposited and removed, the process occurs layer by layer The straight-line forms of the Freundlich and Langmuir isotherms are, respectively, 1 X ln = ln k + ln [ C ] n M straight-line form (8. 26) [C ] 1 1 - = - + [ C ] - X /M ab a straight-line form (8. 27) Because the equations are for straight lines, only two pairs of values of the respective parameters are required... [C ] (8. 33) m ∑ - ( m – l )   + ∑ [ C ]  ab a X /M 1 l +1 1 (8. 34) l+1 Solving for the constants a and b produces m l  l∑ l+1[ C ] – ( m – l )∑ 1[ C ]  a =   m [C ] l [C ]  l∑ l+1 - – ( m – l )∑ 1 -  X /M X /M (8. 35) l b = l l [C ] a∑ 1 - – ∑ 1[ C ] X /M (8. 36) Example 8. 5 A wastewater containing [Co] = 25 mg/L of phenol... 8. 3.5 BED ADSORPTION AND Ans ACTIVE ZONE In bed adsorption, the water to be treated is passed through a bed of activated carbon, in which the form of carbon is normally GAC The method of introduction of the influent may be made similar to sand filtration In addition, the bed may be moving countercurrent or co-current to the flow of influent Figure 8. 8 shows a cutaway view of a carbon-bed adsorption unit... N and C The C of the carboxyl group [=C=O] of tolylene diisocyanate then bonds with the N of the amine The reaction above simply shows two of the tolylene molecules participating in the reaction, but in reality, there will be millions 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. .. 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... 0.0051 m = 5.1 mm Ans Example 8. 8 A wastewater containing 25 mg/L of phenol and having the characteristic breakthrough of the previous example is to be treated by adsorption onto an activated carbon bed The flow rate during the breakthrough experiment is 3 0.11 m /s; this is equivalent to a surficial velocity of 0.0 088 m/s The X/M ratio of the bed for the desired effluent of 0.06 mg/L is 0.02 kg solute . rejection ions Ca 2+ SO 2- Na + H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0Na + Cl - H 2 0Na + Cl - H 2 0 H 2 0 H 2 0 H 2 0 Monovalent. 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 CaCl 2 , and 400 mg/L of MgSO 4 . The membrane used is cellulose acetate and the results of a. 102. 18( )90()– 104.9 3,000() 100()== 98. 2%= Ans b()R 104.9 3,000 300 400++()0.61 102. 18( )90 68+ +()– 104.9 3,000 300 400++() 100()= 388 ,130 6, 482 .30– 388 ,130 = 100() 98. 3%= Ans TX249_frame_C 08. fm