Solution Manual for MWH's Water Treatment: Principles and Design, 3rd Edition

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MWH’S WATER TREATMENTPRINCIPLES AND DESIGN3rd Editionby Crittenden, Trussell, Hand, Howe, and TchobanoglousHOMEWORK SOLUTION MANUALFORChapter 2Physical and ChemicalQuality of Water

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Homework Solution ManualPage 2 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterPROBLEM2-1Problem Statement- Given the following test results, determine the mole fraction ofcalcium (Ca2+).Concentration,Concentration,Cationmg/LAnionmg/LCa2+40.0HCO3–91.5Mg2+12.2SO42–72Na+15.1Cl–22.9K+5.1NO3–5.0Solution1.By definition, using Eq. 2-2, the mole fraction of Ca2+is as follows.2+2+2+2+++-2---343CaCaCaMgNaKHCOSOClNOnxnnnnnnnn=+++++++2.Substituting in the MW for each of the constituents yields the following:2+Ca40.08x40.0824.3122.9939.1619635.456240.080.11380.93=+++++++==PROBLEM2-2Problem Statement- Determine the mole fraction of magnesium (Mg2+) for the watergiven in Problem 2-1.Solution1.By definition, using Eq. 2-2, the mole fraction of Mg2+is as follows.2+2+2+2+++-2---343MgMgCaMgNaKHCOSOClNOnxnnnnnnnn=+++++++2.Substituting in the MW for each of the constituents yields the following:

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Homework Solution ManualPage 3 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of Water2+Mg24.31x40.0824.3122.9939.1619635.456224.310.06380.93=+++++++==PROBLEM2-3Problem Statement- Determine the mole fraction of sulfate (SO42–) for the water givenin Problem 2-1.Solution1.By definition, using Eq. 2-2, the mole fraction of2-4SOis as follows:2-42-42+2+++---33SOSOCaMgNaKHCOClNOnxnnnnnnn=++++++2.Substituting in the MW for each of the constituents yields the following:2-4SO96x40.0824.3122.9939.1619635.4562960.25280.93=+++++++==PROBLEM2-4Problem Statement- Commercial-grade sulfuric acid is about 95 percent H2SO4bymass. If the specific gravity is 1.85, determine the molarity, mole fraction, and normalityof the sulfuric acid. Use Eq. 2-4 to determine molarity.()()mass of solute, gM, mole/Lmolecular weight of solute, g/molevolume of solution, L=()()()( )0.951.85M, mole/L0.0179 mole/L981==1.Use Eq. 2-2 to determine the mole fraction.

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Homework Solution ManualPage 4 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of Water2424224H SOH SOH OH SOnxnn980.841898=+==+2.Use Eq. 2-6 to determine the normality.()()mass of solute, gN, eq/Lequivalent weight of solute, g/eqvolume of solution, L=()()()( )0.951.85N, eq/L0.036 eq/L491==PROBLEM2-5Problem Statement- If the UV intensity measured at the surface of a water sample is180 mW/cm2, estimate the average intensity in a petri dish with an average depthof 15 mm (used to study the inactivation of microorganisms after exposure to UVlight, as discussed in Chap. 13). Assume the absorptivity of the water,kA(λ) atλ=254 nm, is 0.10 cm–1and that the following form of the Beer–Lambert law applies:A0Iln2.303k( )xI = −λSolution1.The variables for the Beer-Lambert law are the following:I = unknownI0= 180 mJ/cm2kA(λ) = 0.10/cmx = 1.5 cm2.Rearrange the Beer-Lambert law given in the problem statement to solve for I asfollows:()()()A-2.303kλ x0-2.303 0.101.52II eI280eI174 mJ/cm===

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Homework Solution ManualPage 5 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterPROBLEM2-6Problem Statement- If the average UV intensity in a Petri dish containing water at adepth of 10 mm is 120 mW/cm2, what is the UV intensity at the surface of the watersample? Assume the absorptivity of the water,kA(λ) atλ= 254 nm, is 0.125 cm–1and that the equation given in Problem 2-5 applies.Solution1.The variables for the Beer-Lambert law are the following:I = 120 mJ/cm2I0= unknownkA(λ) = 0.125/cmx = 1.0 cm2.Rearrange the Beer-Lambert law given in the problem statement to solve for I0asfollows:()()()A0-2.303kλ x20-2.303 0.1251.0IIe120I160 mJ/cme===PROBLEM2-7Problem Statement- If the transmittance is 92 percent and a photo cell with a 12-mmpath length was used, what is the absorptivity?Solution1.Set up Eq. 2-12 to solve for T in terms of the variables which are given in theproblem statement; -kA(λ), and x.a.From Eq. 2-10, = -A(λ)b.T, % = 10-kA(λ)xx 100c.Variables for this equation are the following:T, % = 92-kA(λ) = unknown

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Homework Solution ManualPage 6 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of Waterx = 1.2 cm2.Rearrange Eq. 2-12 to solve for absorptivity, kA(λ), as follows:()()A-1ATlog 100kλx92log 1000.0301.2kλ0.030 cm−=== −=PROBLEM2-8Problem Statement- Given the following data obtained on two water supply sources,determine the constants in Eq. 2-16 and estimate the number of particles in thesize range between 2.1 and 5. Also, comment on the nature of the particle sizedistributions.Particle CountBin Size, mmWater AWater B5.1–10250011010.1–158508015.1–205005520.1–302503630.1–40802540.1–50602050.1–75281575.1–1001010Solution1.Calculate the information necessary to graphically represent the data. Thisinformation is shown in the following table.Bin size, mmGeometric meandiameter, (dp)gWater A particlenumberWater B particlenumber

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Homework Solution ManualPage 7 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of Water5.1 – 107.14a250011010.1 -1512.318508015.1 – 2017.385005520.1 - 3024.562503630.1 - 4034.70802540.1 - 5044.78602050.1 - 7561.30281575.1 – 10086.661010a7.145.110=x2.Prepare a plot of the geometric mean diameter for the bin size, (dp)g, versus thenumber of particles in the corresponding bin size.3.Determine A and β using Eq. 9-4 and the data plot from step 2.The value of A is determined when dp = 1mm for the best-fit regression linethrough the data, along withβ, which is the power law slope coefficient.a.Water A

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Homework Solution ManualPage 8 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterThe equation for the data regression line is shown on the graph in step 2, andthe values for A and β are:A = 222492 and β = 2.1953b.Water BThe equation for the data regression line is shown on the graph in step 2, andthe values for A and β are:A = 864.89 and β = 0.99034.Estimate the number of particles in the size range between 2.1 and 5.a.Calculate the geometric mean size.2.1 53.24×=b.Estimate the number of particles in water A using the equation for the dataregression line determined in step 2.()2.1953n 3.24222492 3.2416,842−=×=c.Estimate the number of particles in water B using the equation for the dataregression line determined in step 2.()0.9903n 3.24864.89 3.24270−=×=5.Comment on the nature of the particle size distributions.Parameter A is much greater for water A than for water B. As the value of Aincreases, the total number of particles in each size classification increases, thus,the total number of particles in water A in each size classification is greater than inwater B.The slopeβis a measure of the relative number of particles in each size range. Thus, ifβis less than one 1, the particle size distribution is dominated by large particles, ifβisequal to one all particle sizes are represented equally, and ifβis greater the one theparticle size distribution is dominated by small particles. The particle size distribution isdominated by small particles for water A because β (2.1953) is greater than one. Allparticle sizes are represented nearly equally for water B because β (0.9903) is close toone.

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Homework Solution ManualPage 9 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterPROBLEM2-9Problem Statement- The following particle size data were data obtained for theinfluent and effluent from a granular medium filter. Determine the constants in Eq.2-16 and assess the effect of the filter in removing particles.Particle CountBin Size, mmInfluentEffluent2.51–5200001015.1–1080003210.1–202000620.1–408003.240.1–804001.280.1–160850.34160.1–320400.12Solution1.Calculate the information necessary to graphically represent the data. Thisinformation is shown in the following table.Bin size, mmGeometric meandiameter, (dp)gInfluent particlenumberEffluent particlenumber2.51 - 53.54a200001015.1 – 107.1480003210.1 – 2014.212000620.1 - 4028.358003.240.1 – 8056.644001.280.1 - 160113.21850.34160.1 - 320226.34400.12a3.542.51 5=×2.Prepare a plot of the geometric mean diameter for the bin size, (dp)g, versus thenumber of particles in the corresponding bin size.

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Homework Solution ManualPage 10 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of Water3.Determine A and β:a.InfluentThe equation for the data regression line is shown on the graph in step 2,and the values for A and β are:A = 136189 and β = 1.514b.EffluentThe equation for the data regression line is shown on the graph in step 2,and the values for A and β are:A = 652.15 and β = 1.5954.Assess the effect of the filter in removing particles.a.Parameter A is much greater for the influent than it is for the effluent.Therefore, the effluent contains fewer particles than the influent, meaning thefilter successfully removed many particles.b.Both the influent and the effluent are dominated by small particlesbecause β (1.514 for influent and 1.595 for effluent) is greater than one. Thefilter is effective in removing all sizes of particles because β for the effluent isclose to the value for the influent.

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Homework Solution ManualPage 11 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterPROBLEM2-10Problem Statement- Determine the alkalinity and hardness in milligrams per liter asCaCO3for the water sample in Problem 2-1.Solution1.Determine alkalinity.a.Use Eq. 2-26 and determine the alkalinity in milliequivalents per liter.Alk, meq/L = (HCO3-) + (CO32-) + (OH-) - (H+)Alkalinity ionsConc., mg/Lmg/meqameq/LHCO3-91.561.021.50a Molecular weight/Z (see Eq. 2-7)b.Adding the conversion from meq/L to mg/L as CaCO3(see page xxx of the text)with Eq. 2-26 yields the following:33Alkalinity as CaCOmeq/L of substance x meq/L mass of CaCO , 50 mg/meq=()()3Alk1.505075 mg / L as CaCO==2.Determine hardness.a.Use Eq. 2-23 and determine the hardness in milliequivalents per liter.Hardness, meq/l = (Ca+2) + (Mg+2)Hardness ionsConc., mg/Lmg/meqameq/LCa2+4020.04b2.00Mg2+12.212.151.00∑hardness ions3.00a Molecular weight/Z (see Eq. 2-8)b.Adding the conversion from meq/L to mg/L as CaCO3(see page xxx of thetext) with Eq. 2-23 yields the following:33Hardnessmeq/L of substance x meq/L mass of CaCO , 50 mg/meqas CaCO=()()3Hardness350150 mg/L as CaCO==

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Homework Solution ManualPage 12 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterPROBLEM2-11Problem Statement- Given the following incomplete water analysis, determine theunknown values if the alkalinity and noncarbonate hardness are 50 and 150 mg/Las CaCO3, respectively:IonConcentration,mg/LCa2+42.0Mg2+?Na+?K+29.5HCO3–?SO42–96.0Cl–35.5NO3–4.0Solution1.Determine the concentration of HCO3-using Eq. 2-26.Alk, meq/L = (HCO3-) + (CO32-) + (OH-) - (H+)a.Convert alkalinity from mg/L as CaCO3to meq/L.3333mg/L alkalinity as CaCO:meq/L of alkalinity ions x meq/L mass of CaCO , 50 mg/meqmg/L alkalinity as CaCOmeq/L of alkalinity ionsmeq/L mass of CaCO50 mg/Lmeq/L of alkalinity ions1 meq/L50 mg/meq====b.Use Eq. 2-26 to determine the alkalinity ions that are present in the wateranalysis. The only alkalinity ion listed in the water analysis is-3HCO, which isunknown.c.Convert the concentration of-3HCOin meq/L to mg/L as follows:Alkalinity ionsConc., mg/Lmg/meqameq/L

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Homework Solution ManualPage 13 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterHCO3-61.0261.021.00a Molecular weight/Z (see Eq. 2-7)2.Determine the value of Mg2+using Eq. 2-23. (Non-carbonate hardness is theconcentration of Ca2+and Mg2+associated with nonalkalinity anions such as Cl-and SO42-. In this problem, 1.0 meq/L of Ca2+are associated with HCO3-analkalinity anion. The remaining concentration of Ca2+, 1.1 meq/L and all of theMg2+are associated with SO42-and Cl-, both nonalkalinity anions.)a.Convert hardness from mg/L as CaCO3to meq/L.3333mg/L hardnessmeq/L of hardness ions x meq/L mass of CaCO , 50 mg/meqas CaCOmg/L hardness as CaCOmeq/L of hardness ionsmeq/L mass of CaCO150 mg/Lmeq/L of hardness ions3 meq/L50 mg/meq====b.Use Eq. 2-23 and determine the hardness ions that are present in the wateranalysis.Hardness, meq/l = (Ca2+) + (Mg2+)The ions listed in the water analysis are Ca2+and Mg2+; the concentration ofMg2+unknown.c.Determine the concentration of Mg2+and convert the concentration of Mg2+inmeq/L to mg/L as follows:Hardness ionsConc., mg/Lmg/meqameq/LCa2+42.020.04b2.1Mg2+10.9412.150.9∑hardness ions3.0a Molecular weight/Z (see Eq. 2-8)3.Determine the concentration of Na+from an anion, cation balance.a.Prepare a cation-anion balance.CationConc.,mg/Lmg/meqameq/LAnionConc.,mg/Lmg/meqameq/L

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Homework Solution ManualPage 14 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterCa2+42.020.04b2.1HCO3-61.0261.021.00Mg2+10.9412.150.9SO42-96.048.032.00Na+23Cl-35.535.451.00K+29.539.10.75NO3-4.062.010.06∑cations3.75∑anions4.06a Molecular weight/Z (see Eq. 2-7)b For calcium, equivalent weight = 40.08/2 = 20.04 g/eq or 20.04 mg/meqb.Calculate the concentration of Na+ from the difference between the sum ofthe cations and the sum of the anions.++Nameq/LanionscationsNameq/L4.063.750.31=−=−=∑∑()()++++Namg/LNamg/meq x Nameq/LNamg/L230.317.13 mg/L===PROBLEM2-12Instructors Note:See Chap. 19 for a more thorough discussion of hardness, wheretotal, carbonate, and non-carbonate hardness are explained.Problem Statement- Given the following incomplete water analysis measured at 25ºC,determine the unknown values if the alkalinity and noncarbonate hardness are 40and 180 mg/L as CaCO3:IonConcentration, mg/LCa2+55.0Mg2+?Na+23.0K+?HCO3–?SO42–48.0Cl–?CO24.0

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Homework Solution ManualPage 15 of 18MWH’s Water Treatment: Principles and Design, 3rd ed.Version 1Chapter 2 - Physical and Chemical Quality of WaterSolution1.Determine the concentration of-3HCOusing Eq. 2-26.a.Convert alkalinity from mg/L as CaCO3to meq/L.3333mg/L alkalinity =meq/L of alkalinity ions x meq/L mass of CaCO , 50 mg/meqas CaCOmg/L alkalinity as CaCOmeq/L of alkalinity ionsmeq/L mass of CaCO40 mg/Lmeq/L of alkalinity ions0.8 meq/L50 mg/meq===b.Use Eq. 2-26 and determine the alkalinity ions that are present in the wateranalysis.-2--+33Alk, meq/L = (HCO ) + (CO) + (OH ) - (H )The only alkalinity ion listed in the water analysis is-3HCO, which is unknown.c.Convert the concentration of-3HCOin meq/L to mg/L as follows:Alkalinity ionsConc., mg/Lmg/meqameq/LHCO3-48.8261.020.80a Molecular weight/Z (see Eq. 2-8)2.Determine the value of Mg2+using Eq. 2-23, which is for non-carbonate hardness.a.Convert hardness from mg/L as CaCO3to meq/L.3333mg/L hardnessmeq/L of hardness ions x meq/L mass of CaCO , 50 mg/meqas CaCOmg/L hardness as CaCOmeq/L of hardness ionsmeq/L mass of CaCO180 mg/Lmeq/L of hardness ions3.6 meq/L50 mg/meq====b.Use Eq. 2-23 and determine the hardness ions that are present in the wateranalysis.Hardness, meq/l = (Ca2+) + (Mg2+)The hardness ions listed in the water analysis are Ca2+and Mg2+, and theconcentration of Mg2+is unknown.

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