Standard Methods for Examination of Water & Wastewater_9 potx

TX249_frame_C10.fm Page 465 Friday, June 14, 2002 4:40 PM Part III Unit Processes of Water and Wastewater Treatment Part III covers the various unit processes employed in water and wastewater treatment including water softening; water stabilization; coagulation; removal of iron and manganese by chemical precipitation; removal of nitrogen by nitrification–denitrification; removal of phosphorus by chemical precipitation; ion exchange; and disinfection © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 467 Friday, June 14, 2002 4:40 PM 10 Water Softening Softening is the term given to the process of removing ions that interfere with the use of soap These ions are called hardness ions due to the presence of multivalent cations, mostly calcium and magnesium In natural waters, other ions that may be 2+ present to cause hardness but not in significant amounts are iron (Fe ), manganese 2+ 2+ 3+ (Mn ), strontium (Sr ), and aluminum (Al ) In the process of cleansing using soap, lather is formed causing the surface tension of water to decrease This decrease in surface tension makes water molecules partially lose their mutual attraction toward each other, allowing them to wet ‘‘foreign” solids, thereby, suspending the solids in water As the water is rinsed out, the solids are removed from the soiled material In the presence of hardness ions, however, soap does not form the lather immediately but reacts with the ions, preventing the formation of lather and forming scum Lather will only form when all the hardness ions are consumed This means that hard waters are hard to lather Hard waters are those waters that contain these hardness ions in excessive amounts Softening using chemicals is discussed in this chapter Other topics related to softening are discussed as necessary 10.1 HARD WATERS The following lists the general classification of hard waters: Soft Moderately hard Hard Very hard < 50 mg/L as CaCO3 50–150 mg/L as CaCO3 150–300 mg/L as CaCO3 > 300 mg/L as CaCO3 A very soft water has a slimy feel For example, rainwater, which is exceedingly soft, is slimy when used with soap For this reason, hardness in water used for domestic purposes is not completely removed Hardness is normally removed to the level of 75 to 120 mg/L as CaCO3 10.2 TYPES OF HARDNESS Two basic types of hardness are associated with the ions causing hardness: carbonate and noncarbonate hardness When the hardness ions are associated with the − HCO ions in water, the type of hardness is called carbonate hardness; otherwise, the type of hardness is called noncarbonate hardness An example of carbonate hardness is Ca(HCO3)2, and an example of noncarbonate hardness is MgCl2 © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 468 Friday, June 14, 2002 4:40 PM 468 In practice, when one addresses hardness removal, it means the removal of the calcium and magnesium ions associated with the two types of hardness Our discussion of hardness removal therefore is divided into the categories of calcium and magnesium Corresponding chemical reactions are developed If other specific hardness ions are also present and to be removed, such as iron, manganese, strontium, and aluminum, corresponding specific reactions have to be developed for them The removal of iron and manganese will be discussed in a separate chapter In general, water is softened in three ways: chemical precipitation, ion exchange, and reverse osmosis Only the chemical precipitation method is discussed in this chapter 10.3 PLANT TYPES FOR HARDNESS REMOVAL In practice, two types of plants are generally used for chemical precipitation hardness removal: One type uses a sludge blanket contact mechanism to facilitate the precipitation reaction The second type consists of a flash mix, a flocculation basin, and a sedimentation basin The former is called a solids-contact clarifier The latter arrangement of flash mix, flocculation, and sedimentation were discussed in previous chapters on unit operations A solids-contact clarifier is shown in Figure 10.1 The chemicals are introduced into the primary mixing and reaction zone Here, the fresh reactants are mixed by the swirling action generated by the rotor impeller and also mixed with a return sludge that are introduced under the hood from the clarification zone The purpose of the return sludge is to provide nuclei that are important for the initiation of the chemical reaction The mixture then flows up through the sludge blanket where secondary reaction and mixing occur The reaction products then overflow into the clarification zone, where the clarified water is separated out by sedimentation of the reaction product solids The clarified water finally overflows into the effluent discharge The settled sludge from the clarification is drawn off through the sludge discharge pipe Motor drive Chemical feed cat walk Influent Secondary mixing and Clarified water reaction zone Effluent Clarified water Impeller Concentrated sludge Sludge blanket Hood Hood Return sludge Bleed off and drain Primary mixing and reaction zone Sludge discharge Return sludge FIGURE 10.1 Solids-contact clarifier (Courtesy of Infilco Degremont, Inc.) © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 469 Friday, June 14, 2002 4:40 PM 469 10.4 THE EQUIVALENT CaCO3 CONCENTRATION In the literature, hardness is frequently expressed in terms of CaCO3 Expressing concentrations in terms of CaCO3 can be confusing For example, when the hardness of water is 60 mg/L as CaCO3, what does this really mean? Related to this question is a second question: Given the concentration of a hardness substance, how is this converted to the equivalent CaCO3 hardness concentration? To obtain the mass of any substance, the number of equivalents of that substance is multiplied by its equivalent mass Because the number of equivalents of all substances participating in a given chemical reaction are equal, what differentiates the various species in this chemical reaction are their respective equivalent masses Thus, to obtain the concentration of 60 mg/L as CaCO3, this equal number of equivalents must have been multiplied by the equivalent mass of calcium carbonate This section will determine this equivalent mass This is actually 50 and converting the concentration of any hardness substance to the equivalent CaCO3 hardness concentration is obtained by multiplying the number of equivalents of the substance by 50 Let [CT,hard]eq represent the total concentration of hardness in equivalents, where the symbol [ ] is read as “the concentration of ” and the subscript eq means that the concentration is expressed in terms of equivalents If the only hardness ions present are calcium and magnesium, 2+ 2+ [CT,hard]eq = [Ca ]eq + [Mg ]eq (10.1) The number of equivalents of one substance in a given chemical reaction is equal to the same number of equivalents of any other substance in this reaction, so it is entirely correct to arbitrarily express the total concentration of hardness in terms of only one of the ions that participates in the chemical reaction The concentrations of the other ions must then be subsumed in the concentration of this one ion being chosen For example, if the total hardness is to be expressed in terms of the magnesium ion only, the previous equation will become [CT,hard]eq = [ Mg ] ′ eq 2+ (10.2) In this equation, the part of the total hardness contained in calcium is subsumed 2+ 2+ ′ ′ in magnesium, [ Mg ] eq Note the prime The term [ Mg ] eq means that it is a concentration in terms of magnesium but that the calcium ion concentration is subsumed in it and expressed in terms of magnesium equivalents Moreover, since an equivalent of one is equal to the equivalent of another, it is really immaterial under what substance the total equivalents of hardness is expressed Then, in a similar manner, the total concentration of hardness may also be expressed in terms of the calcium ion alone as follows ′ [CT,hard]eq = [ Ca ] eq 2+ © 2003 by A P Sincero and G A Sincero (10.3) TX249_frame_C10.fm Page 470 Friday, June 14, 2002 4:40 PM 470 In this equation, the part of the total hardness contained in magnesium is now 2+ 2+ ′ ′ subsumed in calcium, [ Ca ] eq Again, note the prime Similar to the term [ Mg ] eq, 2+ ′ [ Ca ] eq means that it is a concentration in terms of calcium but the magnesium ion is subsumed in it and expressed in terms of calcium equivalents If other hardness ions are present as well, then the concentrations of these ions will also be subsumed in the one particular ion chosen to express the hardness When any ion participates in a chemical reaction, it will react to the satisfaction of its ionic charge For example, when the calcium and magnesium ions participate in a softening reaction, they will react to the satisfaction of their ionic charges of two Thus, if the chemical reaction is written out, the number of reference species 2+ 2+ for Ca and Mg will be found to be two and the respective equivalent masses are 2+ then Ca/2 and Mg/2 In terms of molar concentrations, [Ca ]eq is then equal to 2+ 2+ 2+ 2+ 2+ ′ [Ca ](Ca/Ca/2) = 2[Ca ] By analogy, [ Ca ] eq = [ Ca ]′ , where [Ca ]′ is now an equivalent molar concentration in terms of the calcium ion that subsumes all 2+ 2+ 2+ concentrations Also, [Mg ]eq = [Mg ](Mg/Mg/2) = 2[Mg ] and, again, by analogy, 2+ 2+ 2+ [ Mg ]′ eq = 2[Mg ]′ [Mg ]′ is the equivalent molar concentration in terms of the magnesium ion that also subsumes all concentrations Considering Eqs (10.2) and (10.3) simultaneously and expressing the molar concentrations of the calcium and magnesium hardness in terms of calcium only, ′ ′ [CT,hard]eq = [ Mg ] eq = [ Ca ] eq = [ Ca ]′ 2+ 2+ 2+ (10.4) 2+ The term 2[Ca ]′ can be converted to [CaCO3]′ This is done as follows: In 2+ 2+ CaCO3, one mole of [Ca ] is equal to one mole of [CaCO3] Hence, [ Ca ]′ = 2[CaCO3]′ and Equation (10.4) becomes ′ ′ [CT,hard]eq = [ Mg ] eq = [ Ca ] eq = [ Ca ]′ = 2[CaCo3]′ or (10.5) 2+ 2+ [ C T,hard ] eq ′ ′ [ Mg ] eq [ Ca ] eq [CaCO3]′ = = = -2 2 (10.6) 2+ 2+ 2+ [CaCO3]′ is an equivalent molar concentration expressed in terms of moles of CaCO3 Thus, to get the equivalent calcium carbonate mass concentration, it must be multiplied by CaCO3 = 100 Therefore, [ C T,hard ] eq [ CaCO ]′ ( 100 ) =   ( 100 ) = 50 [ C T,hard ] eq    [ Mg 2+ ] eq ′ 2+ =   ( 100 ) = 50 [ Mg ] ′ eq    [ Ca 2+ ] eq ′ 2+ =   ( 100 ) = 50 [ Ca ] ′ eq   © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 471 Friday, June 14, 2002 4:40 PM Or, ′ ′ [ C T,hard ] asCaCO = 50[CT,hard]eq = 50 [ Mg ] eq = 50 [ Ca ] eq 2+ 2+ (10.7) Thus, the concentration of hardness expressed in terms of the mass of CaCO3 2+ 2+ is equal to the number of equivalents of hardness ([CT,hard]eq, [ Mg ] ′ , or [ Ca ] ′ ) eq eq divided times the molecular mass of calcium carbonate Or, simply put, the concentration of hardness expressed in terms of the mass of CaCO3 is equal to the 2+ 2+ number of equivalents of hardness ([CT,hard]eq, [ Mg ] ′ , or [ Ca ] ′ ) times 50 eq eq Note that Equation (10.7) merely states the concentration of hardness in terms of the mass of CaCO3 Although the symbol for calcium carbonate is being used, it does not state anything about the actual concentration of the CaCO3 species present; it is even possible that no species of calcium carbonate exists, but MgCO3 or any other species where the concentrations are simply being expressed in terms of CaCO3 2+ To apply Equation (10.7), suppose that the concentration of Mg in a sample 2+ 2+ of water is given as 60 mg/L as CaCO3 The meq/L of Mg is then equal to [Mg ]eq = 2+ [ Mg ] ′ = 60/50 = 1.2 and the concentration of magnesium in mg/L is 1.2(Mg/2) = eq 1.2(24.3/2) = 14.58 mg/L Take the following reaction and the example of the magnesium ion: Mg ( HCO ) + 2Ca ( OH ) → Mg (OH ) ↓ + 2CaCO ↓ + 2HOH In this reaction, the equivalent mass of CaCO3 is 2(CaCO3)/2 = 100 Again, in any given chemical reaction, the number of equivalents of all the participating species are equal Thus, if the number of equivalents of the magnesium ion is 1.2 meq/L, the number of equivalents of CaCO3 also must be 1.2 meq/L Thus, from this reaction, 2+ the calcium carbonate concentration corresponding to the 1.2 meq/L of Mg would be 1.2(100) = 120 mg/L as CaCO3 Or, because the species is really calcium carbonate, it is simply 120 mg/L CaCO3––no more ‘‘as.” How is the concentration of 120 mg/L CaCO3 related to the concentration of 60 mg/L as CaCO3 for the magnesium ion? The 60 mg/L is not a concentration of calcium carbonate but a concentration of the magnesium expressed as CaCO3 These two are very different In the “120,” there is really calcium carbonate present, while in the “60,” there is none Again, expressing the magnesium concentration as 60 mg/L CaCO3 does not mean that there are 60 mg/L of the CaCO3 but that there are 60 mg/L of the ion of magnesium expressed as CaCO3 Expressing the concentration of one substance in terms of another is a normalization This is analogous to expressing other currencies in terms of the dollar Go to the Philippines and you can make purchases with the dollar, because there is a normalization (conversion) between the dollar and the peso Example 10.1 The concentration of total hardness in a given raw water is found to be 300 mg/L as CaCO3 Calculate the concentration in milligram equivalents per liter © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 472 Friday, June 14, 2002 4:40 PM Solution: 300 [ C T,hard ] meq = = 6.0 meq/L Ans 50 10.5 SOFTENING OF CALCIUM HARDNESS Calcium hardness may either be carbonate or noncarbonate The solubility product 2+ −9 2− constant of CaCO3 is Ksp = [Ca ][ CO ] = 5(10 ) at 25°C A low value of the Ksp −9 means that the substance has a low solubility; a value of 5(10 ) is very low Because of this very low solubility, calcium hardness is removed through precipitation of CaCO3 Because there are two types of calcium hardness, there corresponds two general methods of removing it When calcium is associated with the bicarbonate ion, the hardness metal ion can be easily removed by providing the hydroxide radical + − The H of the bicarbonate becomes neutralized by the OH provided forming water 2− and the CO ion necessary to precipitate calcium carbonate The softening reaction is as follows: − 2− Ca ( HCO ) + 2OH → CaCO ↓ + CO + 2HOH (10.8) The precipitate, CaCO3, is indicated by a downward pointing arrow,↓ As shown, the two bicarbonate species in the reactant side are converted to the two carbonate species in the product side The carbonate ion shown unpaired will pair with whatever − cation the OH was with in the reactant side of the reaction − Once all the bicarbonate ions have been destroyed by the provision of OH , they − convert to the carbonate ion, as indicated If the OH source contains the calcium ion, the carbonates all precipitate as CaCO3 according to the following softening reaction: 2− 2+ Ca + CO → CaCO ↓ (10.9) In this case, no carbonate ion is left unpaired, because the calcium ion is present to cause precipitation The other method of calcium removal involves the case when the hardness is in the form of the noncarbonate, such as in the form of CaCl2 In this case, a carbonate ion must be provided, whereupon Equation (10.9) will apply, precipitating calcium as calcium carbonate The usual source of the carbonate ion is soda ash 10.6 SOFTENING OF MAGNESIUM HARDNESS As in the case of calcium hardness, magnesium can also be present in the form of −12 carbonate and noncarbonate hardness The Ksp of Mg(OH)2 is a low value of 9(10 ) Thus, the hardness is removed in the form of Mg(OH)2 To remove the carbonate − hardness of magnesium, a source of the OH ion is therefore added to precipitate the Mg(OH)2 as shown in the following softening chemical reaction: − 2− Mg ( HCO ) + 4OH → Mg ( OH ) ↓ + 2CO + 2HOH © 2003 by A P Sincero and G A Sincero (10.10) TX249_frame_C10.fm Page 473 Friday, June 14, 2002 4:40 PM − The carbonate ions in the product side will pair with whatever cation the OH was with in the reactant side of the reaction If this cation is calcium, in the form of Ca(OH)2, then the product will again be the calcium carbonate precipitate In natural waters, the form of noncarbonate hardness normally encountered is the one associated with the sulfate anion; although, occasionally, large quantities of the chloride and nitrate anions may also be found The softening reactions for the removal of the noncarbonate hardness of magnesium associated with the possible anions are as follows: − 2− MgSO + 2OH → Mg ( OH ) ↓ + SO − MgCl + 2OH → Mg ( OH ) ↓ + 2Cl − Mg ( NO ) + 2OH → Mg ( OH ) ↓ + 2NO (10.11) (10.12) (10.13) As shown in all the previous reactions, the removal of the magnesium hardness, both carbonate and noncarbonate, can use just one chemical This chemical is normally lime, CaO 10.7 LIME–SODA PROCESS As shown by the various chemical reactions above, the chemicals soda ash and lime may be used for the removal of hardness caused by calcium and magnesium Thus, the lime–soda process is used This process, as mentioned, uses lime (CaO) and soda ash (Na2CO3) As the name of the process implies, two possible sets of chemical reactions are involved: the reactions of lime and the reactions of soda ash To understand more fully what really is happening in the process, it is important to discuss these chemical reactions Let us begin by discussing the lime reactions CaO first reacts with water to form slaked lime, before reacting with the bicarbonate The slaking reaction is CaO + HOH → Ca ( OH ) (10.14) After slaking, the bicarbonates are neutralized according to the following reactions: Ca ( HCO ) + Ca ( OH ) → 2CaCO ↓ + 2HOH (10.15) Mg ( HCO ) + 2Ca ( OH ) → Mg ( OH ) ↓ + 2CaCO ↓ + HOH (10.16) Note that in Equation (10.16) two types of solids are produced: Mg(OH)2 and CaCO3 and that the added calcium ion from the lime that would have produced an added hardness to the water has been removed as CaCO3 Although the hardness ions have been precipitated out, the resulting solids, however, pose a problem of disposal in water softening plants © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 474 Friday, June 14, 2002 4:40 PM Magnesium, whether in the form of the carbonate or noncarbonate hardness, is always removed in the form of the hydroxide Thus, to remove the total magnesium hardness, more lime is added to satisfy the overall stoichiometric requirements for both the carbonates and noncarbonates (Later, we will also discuss the requirement of adding more lime to raise the pH.) The pertinent softening reactions for the removal of the noncarbonate hardness of magnesium follow MgSO + Ca ( OH ) → Mg ( OH ) ↓ + CaSO (10.17) MgCl + Ca ( OH ) → Mg ( OH ) ↓ + CaCl (10.18) Mg ( NO ) + Ca ( OH ) → Mg ( OH ) ↓ + Ca ( NO ) (10.19) As shown from the previous reactions, there is really no net mole removal of hardness that results from the addition of lime For every mole of magnesium hardness removed [MgSO4, MgCl2, or Mg(NO3)2], there is a corresponding mole of by-product noncarbonate calcium hardness produced [CaSO4, CaCl2, or Ca(NO3)2] Despite the fact that these reactions are useless, they still must be considered, because they will always transpire if the noncarbonate hardness is removed using lime For this reason, the implementation of the lime-soda process should be such that as much magnesium as possible is left unremoved and rely only on the removal of calcium to meet the desired treated water hardness If the desired hardness level is not met by the removal of only the calcium ions, then removal of the magnesium may be initiated This will entail the use of lime followed by the possible addition of soda ash to remove the resulting noncarbonate hardness of calcium As shown by Eqs (10.17), (10.18), and (10.19), soda ash is needed to remove the by-product calcium hardness formed from the use of the lime As noted before, the calcium ion is removed in the form of CaCO3 This is the reason for the use of the second chemical known as soda ash for the removal of the noncarbonate hardness of calcium Thus, another set of chemical reactions involving soda ash and calcium would have to written The pertinent softening reactions are as follows: CaSO + Na CO → CaCO ↓ + Na SO (10.20) CaCl + Na CO → CaCO ↓ + 2NaCl (10.21) Ca ( NO ) + Na CO → CaCO ↓ + 2NaNO (10.22) Some of the calcium hardness of these reactions would be coming from the by-product noncarbonate hardness of calcium that results if lime were added to remove the noncarbonate hardness of magnesium It is worth repeating that soda ash is used for two purposes only: to remove the original calcium noncarbonate hardness in the raw water and to remove the by-product calcium noncarbonate hardness that results from the precipitation of the noncarbonate © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 475 Friday, June 14, 2002 4:40 PM hardness of magnesium It is also important to remember that by using lime the carbonate hardness of magnesium does not produce any calcium noncarbonate hardness It is only the noncarbonate magnesium that is capable of producing the byproduct calcium noncarbonate hardness when lime is used 10.7.1 CALCULATION OF STOICHIOMETRIC LIME REQUIRED IN THE LIME–SODA PROCESS Because the raw water is exposed to the atmosphere, there will always be some CO2 dissolved in it As will be shown later, carbon dioxide also consumes lime We will, however, ignore this requirement for the moment, and discuss it in the latter part of this chapter Ignoring carbon dioxide, the amount of lime needed comes from the requirement to remove the carbonate hardness of calcium and the requirements to remove both the carbonate and the noncarbonate hardness of magnesium We will first calculate the amount of lime required for the removal of the carbonate hardness of calcium Let M CaHCO be the mass of Ca ( HCO ) in the carbonate hardness of calcium to be removed From Equation (10.15), taking the positive oxidation state, the number of reference species is moles of positive charges Thus, the equivalent mass of calcium bicarbonate is Ca ( HCO ) /2 and the number of equivalents of the M CaHCO mass of calcium bicarbonate is M CaHCO /( Ca ( HCO ) /2) This is also the same number of equivalents of CaO (or slaked lime) used to neutralize the bicarbonate From Eqs (10.14) and (10.15), the equivalent mass of CaO is CaO/2 Thus, the mass of CaO, M CaOCaHCO 3, needed to neutralize the calcium bicarbonate is M CaHCO / (Ca(HCO ) /2)(CaO/2) Let M T CaHCO be the total calcium bicarbonate in the raw water and f CaHCO be its fractional removal Thus, M CaHCO = f CaHCO M T CaHCO and f CaHCO M T CaHCO M CaOCaHCO = ( CaO/2 ) = 0.35 f CaHCO M T CaHCO (10.23) Ca ( HCO ) /2 Now, calculate the amount of lime needed to remove Mg ( HCO ) Let M MgHCO be the mass of Mg ( HCO ) in the carbonate hardness of magnesium to be removed From Equation (10.16), again taking the positive oxidation state, the number of reference species is moles of positive charges Thus, the equivalent mass of magnesium bicarbonate is Mg ( HCO ) /2 and the number of equivalents of the M MgHCO mass of magnesium bicarbonate is M MgHCO /( Mg(HCO ) /2) This is also the same number of equivalents of CaO (or slaked lime) used to neutralize the magnesium bicarbonate From Eqs (10.14) and (10.16), the equivalent mass of CaO is 2CaO/2 Thus, the mass of CaO, M CaOMgHCO , needed to neutralize the magnesium bicarbonate is M MgHCO /( Mg ( HCO ) /2)CaO Let M T MgHCO be the total magnesium bicarbonate in the raw water and f MgHCO be its fractional removal Thus, M MgHCO = f MgHCO M T MgHCO and f MgHCO M T MgHCO M CaOMgHCO = - CaO = 0.77 ( f MgHCO M T MgHCO ) (10.24) Mg ( HCO ) /2 © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 497 Friday, June 14, 2002 4:40 PM Therefore, Msolids = 5483.04 + 3201.56 + 1248.5 = 9933.1 kg/d Ans V sl = M solids - ; f solids S sl ( 1000 ) Therefore, V sl = assume fsolids = 0.99 9933.1 -0.99 ( 1.04 ) ( 1000 ) = 9.65 m /d Example 10.3 In the previous example, (a) determine the ionic composition of the finished water and (b) show that the cations and anions are balanced Assume −10.33 the temperature is 25°C such that K sp,HCO3 = 10 Also, assume that the effluent is recarbonated Solution: (a) M T CaHCO ( – f CaHCO ) M T Ca ( – f Ca ) 2+ [ Ca ] meq = 12.3 + 50.0 V V Recarbonation will transform the 15.0 mg/L as CaCO3 of carbonate ions from the limit of technology to 15.0 mg/L as CaCO3 of bicarbonate ions From the order of removal, these carbonate ions will be coming solely from the removal of the raw calcium bicarbonate The carbonate ions coming from the use of soda ash to form calcium carbonate will not be accounted for in this limit of technology consideration, since by the time they are formed, the water is already saturated with the carbonate coming from the removal of the raw calcium bicarbonate Therefore, 15 f CaHCO = – = 0.90 f Ca = 152.5 M T Ca = 745.75 kg/d M T CaHCO = 5065.63 kg/d V = 25,000 m /d Therefore, 5065.63 ( – 0.90 ) 745.75 ( – ) 2+ [ Ca ] meq = 12.3 - + 50.0 = 0.25 meq/L Ans 25,000 25,000 M T MgHCO ( – f MgHCO ) M T Mg ( – f MgCa ) 2+ [ Mg ] meq = 13.7 - + 82.3 - : V V From the limit of technology, Mg(OH)2 = 16 mg/L as CaCO3 will dissolve The given concentration of 1.0 meq/L of magnesium is equal to 50 mg/L as CaCO3 Therefore, 16 fMgCa = − - = 0.68; MT Mg = 303.75 kg/d; 50 © 2003 by A P Sincero and G A Sincero M T MgHCO = TX249_frame_C10.fm Page 498 Friday, June 14, 2002 4:40 PM Therefore, ( – f MgHCO ) 303.75 ( – 0.68 ) 2+ [Mg ]meq = 13.7 - + 82.3 = 0.32 meq/L Ans 25,000 25,000 M sodAsh + + [Na ]meq = [Na ]meq,inf + 18.9 - : V + [Na ]meq,inf = meq/L MsodAsh = 3300.59 kg/d Therefore, 3300.59 + [Na ]meq = + 18.9   = 4.50 meq/L Ans  25,000  M T CaHCO ( – f CaHCO ) − [ HCO ] meq =  [ HCO ] OHCO meq + 12.3  V M T MgHCO ( – f MgHCO ) + 13.7 -  { A }  V – K sp,HCO3 + K sp,HCO3 + 4K sp,HCO   A =  – 3   – 10.33 – 10.33 – 10.33   – 10 + ( 10 ) + ( 10 ) =  – -  Ӎ1.0   [HCO3]OHmeq = 0.32 meq/L; [ HCO ] CO3 meq = Therefore, 5065.63 ( – 0.90 ) − [ HCO ] meq =  0.32 + 12.3  25,000 ( – f MgHCO ) + 13.7  { } = 0.57 meq/L Ans  25,000  – K sp,HCO3 + K sp,HCO3 + 4K sp,HCO  − [ CO ] meq =  -3    – 10 –10.33 + ( 10 –10.33 ) +4 ( 10 –10.33 ) =  - Ӎ0   © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 499 Friday, June 14, 2002 4:40 PM 2− 2− [ SO ] meq = [ SO ] meq,in f = 4.49 meq/L Ans − − [ Cl ] meq = [ Cl ] meq,in f = − − [ NO ] meq = [ NO ] meq,in f = (b) No of Equiv (meq/L) Ions 2+ Ca 2+ Mg + Na − HCO 2− SO 0.25 0.32 4.50 ∑ = 5.07 0.57 4.49 ∑ = 5.06 As shown in the table, the ions in the finished water are balanced Ans 10.16 TYPICAL DESIGN PARAMETERS AND CRITERIA Table 10.5 shows typical design criteria for softening systems 10.17 SPLIT TREATMENT Water with a high concentration of magnesium is often softened by a process called split treatment Water softening may be done in either a single- or two-stage treatment In a single-stage treatment, all the chemicals are added in just one basin, whereas, in a two-stage treatment, chemicals are added in two stages In split treatment, the operation is in two stages Part of the raw water is bypassed from the first stage (split) Excess lime to facilitate the precipitation of magnesium hydroxide to the limit of technology is added in the first stage but, instead, of neutralizing this excess, it is used in the second stage to react with the calcium hardness of the bypassed flow that is introduced into the second stage Referring to Figure 10.3, let Q be the rate of flow of water treated, Mgr be the concentration of Mg in the raw water, and Mg f be the concentration of magnesium TABLE 10.5 Design Parameters and Criteria for Softening Systems Parameter Mixer Flocculator Settling Basin Solids-Contact Basin Detention time Velocity gradient, 1/sec Flow-through velocity, m/s Overflow rate, m /min ⋅ m 800 — — 30–60 10–90 0.15–0.5 — 2–4 h — 0.15–0.5 0.8–1.8 2–4 h — — 4.0 © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 500 Friday, June 14, 2002 4:40 PM FIGURE 10.3 Schematic of the split-treatment mode of softening in the finished water The amount of magnesium removed is then equal to QMgr − QMg f If qb is the bypassed flow, the amount of magnesium introduced to the second stage from the bypassed flow is qbMgr The flow coming out of the first stage and flowing into the second stage is Q − qb; the corresponding amount of magnesium introduced into the second stage is (Q − qb)Mg1, where Mg1 is the concentration of magnesium in the effluent of the first stage The total amount of magnesium introduced into the second stage is then the sum of those coming from the bypassed flow and those coming from the first stage and is equal to qbMgr + (Q − qb) Mg1 No removal of magnesium is effected in the second stage, so this is equal to QMg f Expressing in terms of an equation, QMg f = q b Mgr + ( Q – q b )Mg1 (10.70) Solving for the fraction of bypassed flow fb, qb Mg f – Mg1 f b = = -Q Mgr – Mg1 (10.71) Example 10.4 For the raw water of Example 10.2, the lime–soda process in the split treatment mode is used to remove the total hardness to 120 mg/L as CaCO3 containing magnesium hardness of 30 mg/L as CaCO3 Calculate the chemical requirements in the first stage Solution: Ions Conc (mg/L) 80 12.0 46.0 2+ Ca 2+ Mg + Na − HCO 2− SO a b Equiv Mass 20.05 12.15 23 152.5 216.0 61 b 48.05 1.008+12.0+3 ( 16 ) - = 61 32.1+4 ( 16 ) = 48.05 © 2003 by A P Sincero and G A Sincero a No of Equiv (meq/L) 3.99 1.0 ∑ = 6.99 2.5 4.49 ∑ = 6.99 Conc as CaCO3 (mg/L) 199.5 50 100 349.5 125 224.5 349.5 TX249_frame_C10.fm Page 501 Friday, June 14, 2002 4:40 PM qb Mg f – Mg1 f b = = Q Mgr – Mg1 Mg1 = 16 mg/L as CaCO Mg f = 30 mg/L as CaCO Mgr = 50 mg/L as CaCO qb 30 – 16 f b = = - = 0.41 Q 50 – 16 Therefore, bypassed flow to the second stage = 0.41(25,000) = 10,250 m d and flow to the first stage = 14,750 m d M CaO = M CaOCaHCO + M CaOMgHCO + M CaOMgCa + M CaOCO + M CaOExcess M CaOCaHCO = 0.35 f CaHCO M T CaHCO : Removal of calcium bicarbonate takes precedence Therefore, meq/L of Ca(HCO3)2 = 2.5 2.5 – 0.3 f CaHCO = = 0.88 2.5 M T CaHCO = 2.5  - ( 14,750 ) = 36.88 kgeq/d  1000 Ca ( HCO ) 40.1 + ( 61 ) = 36.88 ( 20.05 ) -2 = 36.88 ( 20.05 )     Ca 40.1 = 2989.12 kg/d Therefore, M CaOCaHCO = 0.35 ( 0.88 ) ( 2989.12 ) = 920.65 kg/d M CaOMgHCO = 0.77 f MgHCO M T MgHCO = M CaOMgCa = 2.30 f MgCa M T Mg : – 0.32 f MgCa = = 0.68; 1 M T Mg = 1.0  - ( 14,750 ) = 14.75 kgeq/d  1000 = 14.75 ( 12.15 ) = 179.21 kg/d Therefore, M CaOMgCa = 2.30 ( 0.68 ) ( 179.21 ) = 280.28 kg/d M CaOCO = 1.27 ( M CO2 ) = 1.27 ( 22.0 )  - ( 14,750 ) = 412.12 kg/d  1000 M CaOExcess = 0.028 V kg = 0.028 ( 14,750 ) = 413 kg/d © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 502 Friday, June 14, 2002 4:40 PM Therefore, M CaO = 920.65 + + 280.28 + 412.12 + 413 = 2026.05 kg/d of pure lime 2026.05 = = 2251.17 kg/d of lime Ans 0.90 Soda ash will not be used in the first stage Example 10.5 In Example 10.4, calculate the chemical requirements in the second stage Solution: M CaO = M CaOCaHCO + M CaOMgHCO + M CaOMgCa + M CaOCO + M CaOExcess First, we have to find the influent hardness concentrations to the second stage Calcium introduced to the second stage from the first stage is equal to the limit of technology of 15.0 mg/L CaCO3, plus the calcium from the lime used to remove the noncarbonate hardness of magnesium, which is 50 mg/L CaCO3, plus the noncarbonate calcium from the raw water, which is 199.5 − 125 = 74.5 mg/L CaCO3 This will produce a total outflow of calcium from the first stage of 50 + 15 + 74.5 = 139.5 mg/L as CaCO3 from a flow of 14,750 m /d As this outflow mixes with the bypass flow, the influent calcium concentration to the second stage then becomes (139.5(14,750) + 199.5(10,250))/25,000 = 164.1 mg/L as CaCO3 = 3.28 meq/L Calcium to remain in the effluent is 120 − 30 = 90 mg/L as CaCO3 = 1.8 meq/L Because of the limit of technology of 0.3 meq/L, pseudo calcium concentration to remain in effluent is 1.8 − 0.3 = 1.5 meq/L Thus, calcium to be removed is 3.28 − 1.5 = 1.78 meq/L − Considering the bypass, the influent HCO concentration to the second stage is = (0(14,750) + 2.5(10,250))/25,000 = 1.025 meq/L Because we have 3.28 meq/L of influent calcium but only 1.025 meq/L of bicarbonate, by the order of removal, influent Ca(HCO3)2 = 1.025 meq/L An amount of 1.78 meq/L of calcium is to be removed Thus, noncarbonate calcium to be removed = 1.78 − 1.025 = 0.755 meq/L Noncarbonate calcium = 3.28 − 1.025 = 2.255 meq/L The limit of technology for magnesium hydroxide is 16 mg/L as CaCO3 = 0.32 meq/L Considering the bypass flow, Mg influent concentration to the second stage = (0.32(14,750) + 1.0(10,250))/25,000 = 0.60 meq/L M CaOCaHCO = 0.35 f CaHCO M T CaHCO 1.025 – 0.3 f CaHCO = = 0.71 1.025   M T CaHCO = 1.025   ( 25,000 ) = 25.625 keq/d  ( 1000 )  Ca ( HCO ) 40.1+2 ( 61 ) = 25.625 ( 20.05 ) -2 = 25.625 ( 20.05 )  -    Ca 40.1 = 2076.91 kg/d © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 503 Friday, June 14, 2002 4:40 PM Therefore, M CaOCaHCO = 0.35 ( 0.71 ) ( 2076.91 ) = 514.16 kg/d M CaOMgHCO = 0; + M CaOMgCa = 0; M CaOCO = 1.27 ( M CO2 ) = 1.27 ( 22.0 )  - ( 10,250 ) = 286.39 kg/d  1000 M CaOExcess = – 413 kg/d Therefore, MCaO = 514.16 + + + 286.39 − 413 = 387.55 kg/d of pure lime 387.55 = - = 430.61 kg/d of lime Ans 0.9 MsodAsh = MsodAshCa + MsodAshCaMg MsodAshCa = 2.65 fCaMT Ca:   MT Ca = 0.755   (25,000) = 18.875 keq/d  ( 1000 )  = 18.875(20.05) = 378.44 kg/d Therefore, MsodAshCa = 2.65(1)(378.44) = 1002.88 kg/d of pure soda ash MsodAshCaMg = Therefore, MsodAsh = 1002.88 + = 1002.88 kg/d 1002.88 = = 1179.85 kg/d of soda ash Ans 0.85 Example 10.6 In Example 10.4, calculate the ionic concentrations of the finished water assuming recarbonation is done and show that the ions are balanced Solution: The solution will pertain only to the second stage M T CaHCO ( – f CaHCO ) M T Ca ( – f Ca ) 2+ [ Ca ] meq = 12.3 + 50.0 - : V V [ Ca ( HCO ) ] = 1.025 meq/L   M T CaHCO = 1.025   ( 25,000 ) = 25.625 keq/d  ( 1000 )  © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 504 Friday, June 14, 2002 4:40 PM Ca ( HCO ) 40.1 + ( 61 ) = 25.625 ( 20.05 ) -2 = 25.625 ( 20.05 )     Ca 40.1 f CaHCO = 2076.91 kg/d 1.025 – 0.3 = = 0.71 1.025 V = 25,000 m /d M T Ca = 2.255  - ( 25,000 ) = 56.375 kgeq/d  1000 f Ca = 56.375 ( 20.05 ) = 1130.32 kg/d 0.755 = = 0.33 2.255 Therefore, 2076.91 ( – 0.71 ) 1130.32 ( – 0.33 ) 2+ [ Ca ] meq = 12.3 - + 50.0 25,000 25,000 2+ [ Mg ] meq = 0.296 + 1.51 = 1.8 meq/L Ans M T MgHCO ( – f MgHCO ) M T Mg ( – f MgCa ) = 13.7 - + 82.3 -V V M T MgHCO = 0; M T Mg = 0.60  - ( 12.15 ) ( 25,000 ) = 182.25 kg/d;  1000 f MgCa = Therefore, ( – f MgHCO ) 182.25 ( – ) 2+ [ Mg ] meq = 13.7 - + 82.3 = 0.6 meq/L 25,000 25,000 M sodAsh 1002.88 + + [ Na ] meq = [ Na ] meq,inf + 18.9  -  = + 18.9    V   25,000  = 2.758 meq/L Ans M T CaHCO ( – f CaHCO )  − [ HCO ] meq =  [ HCO ] OHCO meq + 12.3 V  M T MgHCO ( – f MgHCO )  + 13.7 - { A }: V  [ HCO ] OHmeq = 0.32 meq/L; [ HCO ] CO3 meq = – K sp,HCO3 + K sp,HCO3 + 4K sp,HCO   A =  – 3   © 2003 by A P Sincero and G A Sincero Ans TX249_frame_C10.fm Page 505 Friday, June 14, 2002 4:40 PM Assuming temperature = 25°C, K sp,HCO3 = 10 −10.33 −10.33 −10.33 −10.33   – 10 + ( 10 ) + ( 10 ) A =  – - Ӎ1.0   Therefore,  2076.91 ( – 0.71 ) 0(1 – 0)  − [ HCO ] meq =  0.32 + 12.3 - + 13.7 - { } 25,000 25,000   − [ HCO ] meq = 0.32 + 0.296 + = 0.62 meq/L Ans  – K sp,HCO3 + K sp,HCO3 + 4K sp,HCO  − [ CO ] meq =  3    – 10 −10.33 + ( 10 −10.33 ) + ( 10 −10.33 ) =  - Ӎ   2− 2− [ SO ] meq = [ SO ] meq,inf = 4.49 meq/L Ions Ans No of Equiv (meq/L) 2+ Ca 2+ Mg + Na − HCO 2− SO 1.80 0.60 2.758 ∑ = 5.15 0.62 4.49 ∑ = 5.11 The ions may be considered balanced Ans 10.18 USE OF ALKALINITY IN WATER SOFTENING CALCULATIONS It will be noted that we have avoided the use of alkalinity in any of the softening equations, yet, in the literature, this parameter is used in softening calculations The reason is that the use of this parameter in softening is often misleading It will be recalled that the carbonate ion and the bicarbonate ion play very different roles in the unit process reactions They are on the opposite sides, the carbonate being produced from the bicarbonate How then is the literature able to use alkalinity in calculations? The answer is that the literature is conceptually wrong Alkalinity values cannot be used in any softening © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 506 Friday, June 14, 2002 4:40 PM calculations, unless it is assumed that the alkalinity is mostly bicarbonate If this is assumed, then the calculation can proceed, because as we have seen, this bicarbonate will then produce the carbonate Although this assumption is, of course, accurate, it does not make its use conceptually correct To repeat, alkalinity values should not be used in any softening calculations, without prior knowledge of its constituent species Its various constituents have very different roles to play in the process GLOSSARY − Carbonate hardness—Hardness ions associated with the HCO ions in water Equivalent mass—Mass of any substance participating in a reaction per unit of reference species Equivalents or number of equivalents—Mass of a substance divided by the equivalent mass Hardness ions—Multivalent cations mostly calcium and magnesium that interfere in the use of soap Lime–soda process—A process of removing hardness ions that uses lime and soda ash Noncarbonate hardness—Hardness ions are associated with anions other than the bicarbonate ion Recarbonation—A method of stabilizing water that uses carbon dioxide Reference species—A number that represents the combining or reacting capacity of the substances participating in a given chemical reaction Softening—Term given to the process of removing ions that interfere with the use of soap Solids-contact clarifier—A clarifier that uses a sludge blanket contact mechanism to facilitate the precipitation reaction SYMBOLS 2+ [Ca ]eq 2+ ′ [ Ca ] eq 2+ [Ca ]meq − [Cl ]meq − [Cl ]meq,inf [CT,hard] [CT,hard]eq f fCa f1 f2 f CaHCO Concentration of the calcium ion in equivalents per unit volume Concentration of equivalent calcium ions per unit volume Milligramequivalent per liter of the calcium ion in the treated water Milligramequivalent per liter of the chloride ion in treated water Milligramequivalent per liter of the chloride ion in influent to the softening plant Concentration of total hardness Concentration of total hardness in equivalents per unit volume Overall removal of total hardness Fraction of noncarbonate hardness of calcium removed Overall removal of the calcium hardness Overall removal of the magnesium hardness Fraction of Ca(HCO3)2 removed © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 507 Friday, June 14, 2002 4:40 PM fMgCa f MgHCO − [ HCO ] meq K Ksp MCa M CaCO Ca M CaCO CaHCO M CaCO CO2 M CaCO MgCa M CaCO MgHCO M CaHCO M CaOCaHCO MCaO M CaOCO MCaOExcess MCaOMgCa M CaOMgHCO M CO2 [Mg 2+]eq 2+ [Mg ]meq 2+ ′ [ Mg ] eq MMgCa M MgHCO M MgOHMgHCO MMgOHMgCa moleMass MsodAsh MsodAshCa MsodAshMgCa MsolidsCarb MsolidsNonCarb Fraction of mass of magnesium noncarbonate hardness removed Fraction of Mg(HCO3)2 removed Milligramequivalent per liter of the bicarbonate ion in the treated water Equilibrium constant Solubility product constant Mass of noncarbonate hardness of calcium removed Mass of calcium carbonate solids produced from the precipitation of the noncarbonate hardness of calcium Mass of calcium carbonate solid produced from the calcium bicarbonate removed Mass of calcium carbonate solids produced from reaction of lime with dissolved carbon dioxide Mass of calcium carbonate solids produced from the precipitation of the calcium that results from the precipitation of the noncarbonate hardness of magnesium Mass of calcium carbonate produced from the precipitation of the magnesium bicarbonate removed Mass of Ca(HCO3)2 removed Mass of lime needed to precipitate the Ca(HCO3)2 removed Mass of lime used in the lime-soda process in kg Mass of lime needed for the dissolved carbon dioxide Mass of excess lime in kg needed to raise pH to 10.4 Mass of lime needed to precipitate the noncarbonate hardness of magnesium removed Mass of lime needed to precipitate the Mg(HCO3)2 removed Mass of dissolve carbon dioxide Concentration of the magnesium ion in equivalents per unit volume Milligramequivalent per liter of the magnesium ion in the treated water Concentration of equivalent magnesium ions Mass of noncarbonate hardness of magnesium removed Mass of magnesium bicarbonate removed Mass of magnesium hydroxide solid produced from the magnesium bicarbonate removed Mass of magnesium solids produced from the removal of the noncarbonate hardness of magnesium Molecular mass Amount of soda ash required in the lime-soda process Mass of soda ash needed to precipitate the noncarbonate calcium hardness removed Mass of soda ash needed to precipitate the calcium produced from the removal of the noncarbonate hardness of magnesium Mass of solids produced from the removal of carbonate hardness Mass of solids produced from the removal of noncarbonate hardness © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 508 Friday, June 14, 2002 4:40 PM Msolids MTCa M T CaHCO MTMg M T MgHCO + [Na ]meq − [ NO ] meq − [ NO ] meq,inf qb Q Mg1 Mg f Mgr ratioChem 2− [ SO ] meq 2− [ SO ] meq,inf V V sl Mass of solids produced in the lime-soda process Total mass of noncarbonate hardness of calcium Total mass of Ca(HCO3)2 Total mass of noncarbonate hardness of magnesium Total mass of Mg(HCO3)2 Milligramequivalent per liter of the sodium ion in treated water Milligramequivalent per liter of the nitrate ion in treated water Milligramequivalent per liter of the nitrate ion in influent to softening plant Bypass flow in the split-treatment mode Inflow to softening plant Concentration of magnesium in influent from first stage of splittreatment softening plant Concentration of magnesium in treated water Concentration of raw water to softening plant Chemical ratio Milligramequivalent per liter of the sulfate ion in treated water Milligramequivalent per liter of the sulfate ion in influent to softening plant Volume of water softened in cubic meters Volume of sludge in cubic meters PROBLEMS 10.1 Write the balanced chemical reaction in the softening of calcium bicarbonate using lime 10.2 Write the balanced chemical reaction in the softening of calcium sulfate using soda ash 10.3 The total mass of calcium carbonate solids produced from softening a bicarbonate of calcium using lime is 500 g Calculate the original number of equivalents of the bicarbonate Calculate the original number of moles of the bicarbonate 10.4 The total mass of calcium carbonate solids produced from softening of calcium sulfate using soda ash is 150 g Calculate the original number of equivalents of the sulfate Calculate the original number of moles of the sulfate 10.5 The number of moles of a particular substance is 2.5 and its mass is 500 g What is its molecular mass? 10.6 The reaction in the softening of magnesium bicarbonate using lime is Mg(HCO3)2 + 2Ca(OH)2 → Mg(OH)2↓ + 2CaCO3↓ + 2HOH Calculate the chemical ratio between CaCO3 and Mg(HCO3)2 10.7 From the chemical reaction in Problem 10.6, if the mass of CaCO3 is 100 grams, how many grams of magnesium carbonate were removed? 10.8 A raw water to be treated by the lime-soda process to the minimum hardness +2 possible has the following characteristics: CO2 = 22.0 mg/L, Ca = 80 mg/L, © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 509 Friday, June 14, 2002 4:40 PM +2 10.9 10.10 10.11 10.12 10.13 10.14 10.15 10.16 10.17 10.18 10.19 10.20 + − –2 Mg = 12.0 mg/L, Na = 46.0 mg/L, HCO = 152.5 mg/L, and SO = 216.0 mg/L Check if the number of equivalents of positive and negative ions are balanced The flow is 25,000 m /d In Problem 10.8 for a complete removal of hardness, calculate the lime requirement Assume that the lime used is 90% pure In Problem 10.8 for a complete removal of hardness, calculate the soda ash requirement Assume that the soda ash used is 85% pure In Problem 10.8 for a complete removal of hardness, calculate the mass of solids and the volume of sludge produced Assume that the lime used is 90% pure and the soda ash used is 85% pure Also, assume that the specific gravity of the sludge is 1.04 Taking Problems 10.8 through 10.11 together, (a) determine the ionic composition of the finished water and (b) show that the cations and anions are balanced For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total lime needed For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total soda ash needed Taking Problems 10.13 and 10.14 together, calculate the ionic concentrations of the finished water and show whether or not the ions are balanced Assume that recarbonation is done For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of lime used to precipitate the calcium bicarbonate For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of lime used to precipitate the magnesium bicarbonate For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of lime needed to precipitate the magnesium in the noncarbonate hardness of magnesium removed For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of lime needed to neutralize the dissolved carbon dioxide For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of lime used to raise the pH to 10.4 © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 510 Friday, June 14, 2002 4:40 PM 10.21 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of soda ash needed to precipitate the calcium in the noncarbonate hardness of calcium 10.22 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of soda ash needed to precipitate the calcium ion produced from the precipitation of the magnesium hardness in the noncarbonate hardness of magnesium 10.23 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of soda ash needed to precipitate the noncarbonate hardness 10.24 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of calcium carbonate solids that precipitated from the removal of the carbonate hardness of calcium 10.25 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of magnesium hydroxide solids that precipitated from the removal of the magnesium bicarbonate of the carbonate hardness of magnesium 10.26 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of calcium carbonate solids precipitated from the removal of the magnesium bicarbonate of the carbonate hardness of magnesium 10.27 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of solids precipitated from the removal of the carbonate hardness 10.28 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of magnesium hydroxide solids produced from the precipitation of magnesium in the noncarbonate hardness of magnesium 10.29 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of calcium carbonate solids produced from the precipitation of calcium from the noncarbonate hardness of calcium 10.30 For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass © 2003 by A P Sincero and G A Sincero TX249_frame_C10.fm Page 511 Friday, June 14, 2002 4:40 PM 10.31 10.32 10.33 10.34 10.35 10.36 10.37 10.38 of calcium carbonate solids produced from the precipitation of magnesium in the magnesium of the noncarbonate hardness of magnesium For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of solids produced from the removal of the noncarbonate hardness For the raw water of Example 10.8, using the lime-soda process in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3, calculate the total mass of calcium carbonate solids produced from neutralizing the dissolved carbon dioxide Calculate the total amount of solids produced from applying the lime-soda process to the raw water of Example 10.8 in the split treatment mode to remove the total hardness to 130 mg/L as CaCO3 containing magnesium hardness of 40 mg/L as CaCO3 What is the anion mostly associated with the hardness ions in natural waters? Write the chemical formula of EDTA What are the two principal cations that cause hardness in natural waters? The hardness of a given sample of water is 120 mg/L as CaCO3 Calculate the hardness in terms of the calcium ion and the magnesium How does soap aid in cleansing? BIBLIOGRAPHY Davis, L M and D A Cornwell (1991) Introduction to Environmental Engineering MacGraw-Hill, New York Doran, G F., et al (1998) Evaluation of technologies to treat oil field produced water to drinking water or reuse quality Proc SPE Annu Western Regional Meeting, May 11–15, Bakersfield, CA, Soc Pet Eng (SPE) Richardson, TX Kedem, O and G Zalmon (1997) Compact accelerated precipitation softening (CAPS) as a pretreatment for membrane desalination I softening by NaOH, Desalination, 113, 1, 65–71 Malakhov, I A., et al (1995) Small-waste technology of water softening and decarbonization for heat network feeding Teploenergetika, 12, 61–63 Rich, L G (1963) Unit Process of Sanitary Engineering John Wiley & Sons, New York Sincero, A P and G A Sincero (1996) Environmental Engineering: A Design Approach Prentice Hall, Upper Saddle River, NJ Sladeckova, A (1994) Biofilm and periphyton formation in storage tanks, Water Supply Proc 19th Int Water Supply Cong Exhibition Oct 2–8 1993, Budapest, Hungary, 12, 1–2, Blackwell Scientific Publishers, Oxford Snoeyink, V L (1980) Water Chemistry John Wiley & Sons, New York Tchobanoglous, G and E D Schroeder (1985) Water Quality Addison-Wesley, Reading, MA © 2003 by A P Sincero and G A Sincero ... Inflow to softening plant Concentration of magnesium in influent from first stage of splittreatment softening plant Concentration of magnesium in treated water Concentration of raw water to softening... the number of reference species is moles of positive charges for each of these reactions Thus, the equivalent mass of calcium is Ca/2 and the number of equivalents of the MCa mass of calcium... typical design criteria for softening systems 10.17 SPLIT TREATMENT Water with a high concentration of magnesium is often softened by a process called split treatment Water softening may be done