水的基本化学性质.ppt

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1、1,水的基本化学性质,BASIC WATER CHEMISTRY,2,目的, 过程, 获益,目的: 了解水的基本化学性质 过程: By studying basic elements of water chemistry such as units, equivalent weight, CaCO3 equivalents, acidity, alkalinity(P, M and BaCl2), pH, SiO2 control in boilers, corrosion and scaling indices, and by testing knowledge with a quiz,3,

2、Purpose, Process, Pay-off, contd,Pay-off: A knowledge of water chemistry is the basis of our business; without it we cannot claim to be a water treatment “on-site expert”,4,Water, the Universal Solvent,The polar charges on the water molecule make it an exceptional solvent:,O-2,H+,H+,d+,d-,5,HYDROLOG

3、ICAL CYCLE,Ca. Mg, SiO2, Cl, SO4,SS,CO2, O2,organisms,colour (organics),Micro-,6,Expression of Results,% : used for concentrated solutions ppm: a weight relationship(one part/106 parts) ppb = ppm x 1000 ppm as CaCO3: an equivalent weight relationship mg/l = ppm in dilute waters epm = equivalents per

4、 million = mequiv/l = ppm/equivalent weight,7,Equivalent Weight,Equivalent weight = weight required to combine with or replace 1 gm hydrogen or = gm molecular weight divided by the total charge on cations or anions when substance dissociates in solution,8,Example,Calcium phosphate: Ca3(PO4)2 3Ca2+ +

5、 2PO4-3 or. Ca3(PO4)2 + 6H+ 3Ca2+ + 2H3PO4 Molecular Weight = (3)(Ca) + (2)(PO4) = (3)(40) + (2)(95) = 310 Equivalent weight = 310/6 = 51.7,9,Equivalent in Terms of CaCO3,CaCO3 selected as the “unit of exchange” because it has a MW= 100 and an equivalent weight = 50. This allows direct subtraction a

6、nd addition, e.g., Total Hardness(TH) = 20 ppm as CaCO3(by titrn) Calcium Hardness(CaH) = 8 ppm as CaCO3(by titrn), or 3.2 ppm as Ca(by AA, for example) Hence MgH = 20 - 8 = 12 ppm as CaCO3(simple) but MgH = ?! if Ca expressed as Ca,10,Equivalent in Terms of CaCO3, contd,CaCO3 equivalent = weight Ca

7、CO3 chemically equivalent to the amount of material present equiv. wt. CaCO3(= 50) = (ppm as substance) x - equiv. wt. substance,11,Example,Al2O3Al2O3 + 3H2O - 2Al3+ + 6OH- MW = (2)(27) + (3)(16) = 102 Equiv. wt. = 102/6 = 17 CaCO3 equiv. = ppm as Al2O3 x (50/17) = ppm as Al2O3 x 2.94 so 100 ppm as

8、Al2O3 = 294 ppm of Al2O3 as CaCO3,12,Workshop Exercise,A solution contains 50 ppm of Al2(SO4)3 . 14H2O What is the concentration expressed as CaCO3? Molecular weights are in Appendix 1. (Groups of 2/Time = 5 mins),13,Definition and Units of Conductivity,Conductivity or specific conductance 1 = - mho

9、s/m Resistivity in ohm . m where mho = (ohm)-1 The SI unit for mho = Siemens(S),14,Commonly Used Units of Conductivity and Their Relationships,- mmhos/cm - mmhos(= mmhos/cm) - mS/cm(= mmhos/cm) - mS/m(= 100 x mS/cm) - mS/m(SI standard) = 0.001 x mS/m = 0.1 x mS/cm - TDS(ppm) (0.7)(conductivity in mm

10、hos/cm), at 20oC and neutral pH for typical raw, boiler and cooling waters,15,Effect of Temperature on Conductivity,ct c25 = - 1 + 0.02(t - 25) where c25 = conductivity at 25oC ct = conductivity at toC Therefore, c20 = 0.9c25 and TDS 0.63c25,16,Definition of Acidity,Free Mineral Acidity(FMA): exists

11、 at pH 4.3 (due to strong acids like H2SO4, HNO3, HCl). Total Acidity: exists up to a pH = 8.3,17,Definition of Alkalinity,Alkalinity is the capacity to neutralize acid and is equal to the total amount of HCO3-, CO32-, and OH- Alkalinity exists above a pH = 4.3, and total alkalinity = the alkalinity

12、 that exists down to the methyl orange or screened indicator change point, i.e., = HCO3- + CO32- + OH- (see next slide). For this reason, the total alkalinity is also called the “M”(for methyl orange) alkalinity.,18,Effect of pH on Alkalinity and Acidity,pH,MO ENDPOINT,PHENOLPHTH. END POINT,HCO3-,-,

13、0 2 4 6 8 10 12 14,19,P Alkalinity,OH- + H+ - H2O CO32- + H+- HCO3- (60)(1.67) + (1)(50) -(61)(0.82) expressed as CaCO3 or 100 + 50 - 50 Hence, P = all OH- + 1/2 CO32-,20,M Alkalinity or Total Alkalinity,As for P alkalinity plus. HCO3- + H+ - H2O + CO2 i.e. M = OH- + HCO3- + CO32- and 2P - M = 2(OH-

14、 + 1/2CO32-) - (OH- + HCO3- + CO32-) = OH- - HCO3- which becomes.OH- (2P - M 0) andHCO3- (2P - M 0),21,P(BaCl2) Alkalinity in Boiler Water,BaCl2 +,3MgO.2SiO2.2H2O,(Serpentine),Ca3(PO4)2,CO3 Cl SO4,SiO3 PO4 OH,TP polymer,Ba(OH)2,BaCl2,TP polymer,Free caustic alkalinity by acid-base titn,BaCO3 BaSO4 B

15、aSiO3 Ba3(PO4)2 Serpentine Ca3(PO4)2,*,*NB: Do not filter, as CO2 pick-up will occur!,22,P(BaCl2), contd,TP polymer has COO- groups which will titrate as M alkalinity, reducing OH- alkalinity, but this is small enough to ignore in most cases( 10%). P(BaCl2) = free caustic alkalinity(C.A.) = OHt- - O

16、Hs- where OHt- = total C.A. before serpentine production OHs- = C.A. consumed in producing serpentine (This concept is analogous to free and total chlorine),23,pH - Its Estimation in CW Systems,pH = - log10(H+ concentration, moles/litre) (1) Yun(Pacific Corrosion 87) pH = 0.094c + 0.204pHmu + 0.0018

17、Mmu + 6.78 where c = cycles(1.3 - 4.9) M = alkalinity(CaCO3) mu= make-up water for CaH = 7 - 60, pH = 6.3 - 8.3 in make-up water,24,Estimation of pH, contd,(2) Nalco DT Manual(1982) M = cycled up M-alk = (c)(Mmu) M 200 ppm: pH = 7.5 + log10(0.3M - 60) * NB: 30/300 Rule - CW: Palk = 0.3M - 60.for M 2

18、00,25,Estimation of pH, contd,(3) Boroughs et al: IWC Paper, 1981: Discussion by D.J.Goldstein et al pH = 4.4 + 1.6log10(M).Kunz, Yen & Hess pH = 4.5 + 1.43log10(M).Ionac Chemical Co. The Nalco T equations on the previous slide were an improvement on these equations.,26,Estimation of pH, contd,Adjus

19、tment of pH with Acid Acid addition actually drops the M alkalinity, as follows: 1 ppm 100% H2SO4 = 1.02 ppm as CaCO3 (refer to Appendix 1 in your notes) 1 ppm 98% H2SO4 = (0.98)(1.02) = 1.0 ppm as CaCO3 Hence 1 ppm 98% H2SO4 will drop M alk. by 1 ppm as CaCO3 Then use the Yun or DT models to predic

20、t the new pH.,27,Estimation of pH, contd- Workshop Exercise,A make-up water for an open cooling system has the following analysis: M = 60 ppm as CaCO3 pH = 7.8 CaH = 20 ppm as CaCO3 Compare the cycled up pH at 4 cycles using the Yun and DT methods. (Groups of 2/Time = 5 mins),28,Alkalinity for Silic

21、a Control in Boilers,3Mg2+ + 6OH- - 3Mg(OH)2 (3)(24.3) + (6)(17) 3Mg(OH)2 + 2SiO2 - 3MgO.2SiO2.2H2O + H2O (2)(60) Serpentine(insol.) 2Na+ + 2OH- - 2NaOH (2)(17) 2NaOH + SiO2 - Na2O.SiO2 + H2O* (60) Sodium Silicate(sol.) (*For softened or demin FW or for SiO2 left after serpentine formation),29,Alkal

22、inity for Silica Control, contd,(1) Sodium Silicate Formation OH- (as OH) (2)(17) OH- (as CaCO3) (2)(17)(2.94) - = - or - = - SiO2 (as SiO2) 60 SiO2(as SiO2) 60 = 1.67 (2) Magnesium Silicate(Serpentine) Formation OH- (as OH) (6)(17) OH- (as CaCO3) (6)(17)(2.94) - = - or - = - SiO2(as SiO2) (2)(60) S

23、iO2(as SiO2) (2)(60) = 2.50,30,P(BaCl2) for SiO2 Control in Boilers,(1) Serpentine Formation Mg(as CaCO3) (3)(24.3)(4.10) - = - = 2.5 SiO2(as SiO2) (2)(60) or SiO2 reacted to serpentine = Mg/2.5 (2) Sodium Silicate Formation SiO2 to be controlled as sodium silicate = SiO2 - (Mg/2.5) or OHt- - OHs- =

24、 P(BaCl2) 1.67SiO2 - (Mg/2.5),31,P(BaCl2) for SiO2 Control, contd,(3) Design Caustic Alkalinity(Literature) = 1.67n SiO2 - (Mg/3) + 150 where 3 is the “round-off” of 2.5? and n = cycles of concentration Design range = + 50 on the above, i.e. + 100 - 200,32,“Rule of Thumb” for P(BaCl2) in Terms of TD

25、S - SiO2 Control in Boilers,(1) Untreated Feedwater Typical E.Aust. raw water: SiO2/TDS = 0.10 Mg/TDS = 0.15 P(BaCl2) 1.67n(0.10TDS) - n(0.15TDS/3) where n = cycles 1.670.10TDSBW - 0.05TDSBW 0.084TDSBW,33,“Rule of Thumb”, contd,Require 0.025 ppm OH- / ppm TP polymer i.e. 15 ppm max. or 0.006TDSBW He

26、nce P(BaCl2) 0.09TDSBW or as a rule of thumb: 0.1TDSBW for SiO2 control and TP polymer reaction,34,Rule of Thumb, contd,(2) Softened Feedwater As before, SiO2/TDS = 0.10 P(BaCl2) 1.67n(0.10TDS) 1.670.10TDSBW 0.167TDSBW Allowing for TP polymer. P(BaCl2) 0.17TDSBW,35,Example,An untreated boiler feedwa

27、ter contains 5 ppm Mg(as CaCO3) and 10 ppm SiO2. What min. caustic alk. will be required at 30 cycles? Answer: P(BaCl2) 1.67300 - (150/3) 1.67(250) 417 ppm,36,Workshop Exercise,The boiler feedwater in the above example is treated through a softener unit. What is the minimum caustic alkalinity requir

28、ed now? (Groups of 2/Time = 2 mins),37,Indices to Predict Corrosion and Scaling Potential,1. Corrosion Potential Corrosion Loading Index(CLI) = (Cl + SO4)/M where all concentrations expressed in ppm as CaCO3 The CLI is an indicator of the corrosivity of any given water. As a rough guide, a 10-fold i

29、ncrease in CLI may give a 4-fold increase in corrosivity, requiring a 4-fold increase in inhibitor concentration.,38,Indices, contd,Generally: CLI 10, corrosion control critical, water very aggressive, high potential for SCC of stainless steel For chromate inhibitor: CrO4 demand(ppm) = 2(CLI)0.6,39,

30、Indices, contd,2. Scaling Potential Ryznar Stability Index(RSI) Langelier Saturation Index(LSI) Puckorius calculation of RSI, LSI Miyamoto and Silbert LSI for pH 10.5 * etc * especially applicable to ash handling systems,40,Indices, contd,RSI and LSI Both based on difference between water pH and a c

31、alculated pH of saturation of CaCO3 (pHs), and indicate whether or not the water is capable of dissolving more CaCO3.,RSI = 2pHs - pH,41,Indices, contd,Table of Scaling Severity Related to Indices LSI RSI Condition 3.0 3.0 Extremely severe 1.0 5.0 Severe 0.2 5.8 Slight 0.0 6.0 Stable water -1.0 8.0

32、Moderate tendency to dissolve scale -3.0 10.0 v. strong tendency to dissolve scale *note: dissolve scale = corrosive,LSI = pH - pHs,42,Indices, contd,Practical Applications of LSI (1) H. Feitler, in a paper published in the Jan. 1972 issue of MP, and discussed at length in a Canadian Electrical Asso

33、ciation report in 1986, found that the LSI can reach values between 1.7 and 2.0 before scaling actually occurs, and I am recommending that we use this guideline as a quick check of the potential for a scaling problem.,43,Indices, contd,Practical Applications of LSI(2) The MicroCAPE program, “Mineral

34、 Solubility” uses an MIT model to more accurately predict scaling potentials. Make sure you are using Version 2.3.,44,Indices, contd,Puckorius Method to Calculate pHs pHs = 9.30 + A + B - (C + D) where A,B,C, and D are factors to allow for TDS, temperature, calcium and alkalinity, respectively. Tabl

35、es for these factors are in Appendix 2 of your notes.,45,Workshop Exercise,A cooling water at 38 oC has the following analysis: CaH = 100 ppm as CaCO3 TDS = 2500 ppm M-alk = 75 ppm as CaCO3 pH = 8.2 What is the RSI and LSI, and will the water be scale forming or corrosive? Compare the predictions of

36、 the Puckorius Method and MicroCAPE “Min. Sol.” (Groups of 2/Time = 10 mins),46,Quiz,22 questions worth 25 points - all worth 1 pt each, except (20)-(22) each worth 2 pts 16 points are required to pass the course(answers to be sent to K.Gehan for marking and issuing of certificates) Time allowed will be 60 mins.,47,Quiz, contd,48,Quiz, contd,49,Quiz, contd,