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

Solution Manual for MWH's Water Treatment: Principles and Design, 3rd Edition simplifies the toughest textbook questions, providing easy-to-follow solutions for every chapter.

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    MWH’S WATER TREATMENT PRINCIPLES AND DESIGN 3rd Edition by Crittenden, Trussell, Hand, Howe, and Tchobanoglous HOMEWORK SOLUTION MANUAL FOR Chapter 2 Physical and Chemical Quality of Water

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    Homework Solution Manual Page 2 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water PROBLEM 2-1 Problem Statement - Given the following test results, determine the mole fraction of calcium (Ca 2+ ). Concentration, Concentration, Cation mg/L Anion mg/L Ca 2+ 40.0 HCO 3 – 91.5 Mg 2+ 12.2 SO 4 2– 72 Na + 15.1 Cl – 22.9 K + 5.1 NO 3 – 5.0 Solution 1. By definition, using Eq. 2-2, the mole fraction of Ca 2+ is as follows. 2+ 2+ 2+ 2+ + + - 2- - - 3 4 3 Ca Ca Ca Mg Na K HCO SO Cl NO n x n n n n n n n n = + + + + + + + 2. Substituting in the MW for each of the constituents yields the following: 2+ Ca 40.08 x 40.08 24.31 22.99 39.1 61 96 35.45 62 40.08 0.11 380.93 = + + + + + + + = = PROBLEM 2-2 Problem Statement - Determine the mole fraction of magnesium (Mg 2+ ) for the water given in Problem 2-1. Solution 1. By definition, using Eq. 2-2, the mole fraction of Mg 2+ is as follows. 2+ 2+ 2+ 2+ + + - 2- - - 3 4 3 Mg Mg Ca Mg Na K HCO SO Cl NO n x n n n n n n n n = + + + + + + + 2. Substituting in the MW for each of the constituents yields the following:

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    Homework Solution Manual Page 3 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water 2+ Mg 24.31 x 40.08 24.31 22.99 39.1 61 96 35.45 62 24.31 0.06 380.93 = + + + + + + + = = PROBLEM 2-3 Problem Statement - Determine the mole fraction of sulfate (SO 4 2– ) for the water given in Problem 2-1. Solution 1. By definition, using Eq. 2-2, the mole fraction of 2- 4 SO is as follows: 2- 4 2- 4 2+ 2+ + + - - - 3 3 SO SO Ca Mg Na K HCO Cl NO n x n n n n n n n = + + + + + + 2. Substituting in the MW for each of the constituents yields the following: 2- 4 SO 96 x 40.08 24.31 22.99 39.1 61 96 35.45 62 96 0.25 280.93 = + + + + + + + = = PROBLEM 2-4 Problem Statement - Commercial-grade sulfuric acid is about 95 percent H 2 SO 4 by mass. If the specific gravity is 1.85, determine the molarity, mole fraction, and normality of the sulfuric acid. Use Eq. 2-4 to determine molarity. ( )( ) mass of solute, g M, mole/L molecular weight of solute, g/mole volume of solution, L = ( )( ) ( )( ) 0.95 1.85 M, mole/L 0.0179 mole/L 98 1 = = 1. Use Eq. 2-2 to determine the mole fraction.

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    Homework Solution Manual Page 4 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water 2 4 2 4 2 2 4 H SO H SO H O H SO n x n n 98 0.84 18 98 = + = = + 2. Use Eq. 2-6 to determine the normality. ( )( ) mass of solute, g N, eq/L equivalent weight of solute, g/eq volume of solution, L = ( )( ) ( )( ) 0.95 1.85 N, eq/L 0.036 eq/L 49 1 = = PROBLEM 2-5 Problem Statement - If the UV intensity measured at the surface of a water sample is 180 mW/cm 2 , estimate the average intensity in a petri dish with an average depth of 15 mm (used to study the inactivation of microorganisms after exposure to UV light, as discussed in Chap. 13). Assume the absorptivity of the water, k A ( λ ) at λ = 254 nm, is 0.10 cm –1 and that the following form of the Beer–Lambert law applies: A 0 I ln 2.303k ( )x I   = − λ     Solution 1. The variables for the Beer-Lambert law are the following: I = unknown I 0 = 180 mJ/cm2 kA ( λ ) = 0.10/cm x = 1.5 cm 2. Rearrange the Beer-Lambert law given in the problem statement to solve for I as follows: ( ) ( )( ) A -2.303k λ x 0 -2.303 0.10 1.5 2 I I e I 280e I 174 mJ/cm = = =

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    Homework Solution Manual Page 5 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water PROBLEM 2-6 Problem Statement - If the average UV intensity in a Petri dish containing water at a depth of 10 mm is 120 mW/cm 2 , what is the UV intensity at the surface of the water sample? Assume the absorptivity of the water, k A ( λ ) at λ = 254 nm, is 0.125 cm –1 and that the equation given in Problem 2-5 applies. Solution 1. The variables for the Beer-Lambert law are the following: I = 120 mJ/cm 2 I 0 = unknown kA ( λ ) = 0.125/cm x = 1.0 cm 2. Rearrange the Beer-Lambert law given in the problem statement to solve for I 0 as follows: ( ) ( )( ) A 0 -2.303k λ x 2 0 -2.303 0.125 1.0 I I e 120 I 160 mJ/cm e = = = PROBLEM 2-7 Problem Statement - If the transmittance is 92 percent and a photo cell with a 12-mm path length was used, what is the absorptivity? Solution 1. Set up Eq. 2-12 to solve for T in terms of the variables which are given in the problem statement; -k A (λ), and x. a. From Eq. 2-10, = - A(λ) b. T, % = 10 - kA(λ)x x 100 c. Variables for this equation are the following: T, % = 92 -k A (λ) = unknown

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    Homework Solution Manual Page 6 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water x = 1.2 cm 2. Rearrange Eq. 2-12 to solve for absorptivity, k A (λ), as follows: ( ) ( ) A -1 A T log 100 k λ x 92 log 100 0.030 1.2 k λ 0.030 cm       − =       = = − = PROBLEM 2-8 Problem 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 the size range between 2.1 and 5. Also, comment on the nature of the particle size distributions. Particle Count Bin Size, mm Water A Water B 5.1–10 2500 110 10.1–15 850 80 15.1–20 500 55 20.1–30 250 36 30.1–40 80 25 40.1–50 60 20 50.1–75 28 15 75.1–100 10 10 Solution 1. Calculate the information necessary to graphically represent the data. This information is shown in the following table. Bin size, mm Geometric mean diameter, (dp)g Water A particle number Water B particle number

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    Homework Solution Manual Page 7 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water 5.1 – 10 7.14 a 2500 110 10.1 -15 12.31 850 80 15.1 – 20 17.38 500 55 20.1 - 30 24.56 250 36 30.1 - 40 34.70 80 25 40.1 - 50 44.78 60 20 50.1 - 75 61.30 28 15 75.1 – 100 86.66 10 10 a 7.14 5.1 10 = x 2. Prepare a plot of the geometric mean diameter for the bin size, (dp)g, versus the number 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 = 1 m m for the best-fit regression line through the data, along with β , which is the power law slope coefficient. a. Water A

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    Homework Solution Manual Page 8 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water The equation for the data regression line is shown on the graph in step 2, and the values for A and β are: A = 222492 and β = 2.1953 b. Water B The equation for the data regression line is shown on the graph in step 2, and the values for A and β are: A = 864.89 and β = 0.9903 4. Estimate the number of particles in the size range between 2.1 and 5. a. Calculate the geometric mean size. 2.1 5 3.24 × = b. Estimate the number of particles in water A using the equation for the data regression line determined in step 2. ( ) 2.1953 n 3.24 222492 3.24 16,842 − = × = c. Estimate the number of particles in water B using the equation for the data regression line determined in step 2. ( ) 0.9903 n 3.24 864.89 3.24 270 − = × = 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 A increases, 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 in water B. The slope β is a measure of the relative number of particles in each size range. Thus, if β is less than one 1, the particle size distribution is dominated by large particles, if β is equal to one all particle sizes are represented equally, and if β is greater the one the particle size distribution is dominated by small particles. The particle size distribution is dominated by small particles for water A because β (2.1953) is greater than one. All particle sizes are represented nearly equally for water B because β (0.9903) is close to one.

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    Homework Solution Manual Page 9 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water PROBLEM 2-9 Problem Statement - The following particle size data were data obtained for the influent 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 Count Bin Size, mm Influent Effluent 2.51–5 20000 101 5.1–10 8000 32 10.1–20 2000 6 20.1–40 800 3.2 40.1–80 400 1.2 80.1–160 85 0.34 160.1–320 40 0.12 Solution 1. Calculate the information necessary to graphically represent the data. This information is shown in the following table. Bin size, mm Geometric mean diameter, (dp)g Influent particle number Effluent particle number 2.51 - 5 3.54 a 20000 101 5.1 – 10 7.14 8000 32 10.1 – 20 14.21 2000 6 20.1 - 40 28.35 800 3.2 40.1 – 80 56.64 400 1.2 80.1 - 160 113.21 85 0.34 160.1 - 320 226.34 40 0.12 a 3.54 2.51 5 = × 2. Prepare a plot of the geometric mean diameter for the bin size, (dp)g, versus the number of particles in the corresponding bin size.

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    Homework Solution Manual Page 10 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water 3. Determine A and β: a. Influent The 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 = 1.514 b. Effluent The 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 = 1.595 4. 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 the filter successfully removed many particles. b. Both the influent and the effluent are dominated by small particles because β (1.514 for influent and 1.595 for effluent) is greater than one. The filter is effective in removing all sizes of particles because β for the effluent is close to the value for the influent.

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    Homework Solution Manual Page 11 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water PROBLEM 2-10 Problem Statement - Determine the alkalinity and hardness in milligrams per liter as CaCO 3 for the water sample in Problem 2-1. Solution 1. Determine alkalinity. a. Use Eq. 2-26 and determine the alkalinity in milliequivalents per liter. Alk, meq/L = (HCO 3 - ) + (CO 3 2- ) + (OH - ) - (H + ) Alkalinity ions Conc., mg/L mg/meq a meq/L HCO 3 - 91.5 61.02 1.50 a Molecular weight/Z (see Eq. 2-7) b. Adding the conversion from meq/L to mg/L as CaCO 3 (see page xxx of the text) with Eq. 2-26 yields the following: 3 3 Alkalinity as CaCO meq/L of substance x meq/L mass of CaCO , 50 mg/meq = ( )( ) 3 Alk 1.50 50 75 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 ions Conc., mg/L mg/meq a meq/L Ca 2+ 40 20.04 b 2.00 Mg 2+ 12.2 12.15 1.00 ∑ hardness ions 3.00 a Molecular weight/Z (see Eq. 2-8) b. Adding the conversion from meq/L to mg/L as CaCO 3 (see page xxx of the text) with Eq. 2-23 yields the following: 3 3 Hardness meq/L of substance x meq/L mass of CaCO , 50 mg/meq as CaCO = ( )( ) 3 Hardness 3 50 150 mg/L as CaCO = =

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    Homework Solution Manual Page 12 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water PROBLEM 2-11 Problem Statement - Given the following incomplete water analysis, determine the unknown values if the alkalinity and noncarbonate hardness are 50 and 150 mg/L as CaCO 3 , respectively: Ion Concentration, mg/L Ca 2+ 42.0 Mg 2+ ? Na + ? K + 29.5 HCO 3 – ? SO 4 2– 96.0 Cl – 35.5 NO 3 – 4.0 Solution 1. Determine the concentration of HCO 3 - using Eq. 2-26. Alk, meq/L = (HCO 3 - ) + (CO 3 2- ) + (OH - ) - (H + ) a. Convert alkalinity from mg/L as CaCO 3 to meq/L. 3 3 3 3 mg/L alkalinity as CaCO : meq/L of alkalinity ions x meq/L mass of CaCO , 50 mg/meq mg/L alkalinity as CaCO meq/L of alkalinity ions meq/L mass of CaCO 50 mg/L meq/L of alkalinity ions 1 meq/L 50 mg/meq = = = = b. Use Eq. 2-26 to determine the alkalinity ions that are present in the water analysis. The only alkalinity ion listed in the water analysis is - 3 HCO , which is unknown. c. Convert the concentration of - 3 HCO in meq/L to mg/L as follows: Alkalinity ions Conc., mg/L mg/meq a meq/L

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    Homework Solution Manual Page 13 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water HCO 3 - 61.02 61.02 1.00 a Molecular weight/Z (see Eq. 2-7) 2. Determine the value of Mg 2+ using Eq. 2-23. (Non-carbonate hardness is the concentration of Ca 2+ and Mg 2+ associated with nonalkalinity anions such as Cl - and SO 4 2- . In this problem, 1.0 meq/L of Ca 2+ are associated with HCO 3 - an alkalinity anion. The remaining concentration of Ca 2+ , 1.1 meq/L and all of the Mg 2+ are associated with SO 4 2- and Cl - , both nonalkalinity anions.) a. Convert hardness from mg/L as CaCO 3 to meq/L. 3 3 3 3 mg/L hardness meq/L of hardness ions x meq/L mass of CaCO , 50 mg/meq as CaCO mg/L hardness as CaCO meq/L of hardness ions meq/L mass of CaCO 150 mg/L meq/L of hardness ions 3 meq/L 50 mg/meq = = = = b. Use Eq. 2-23 and determine the hardness ions that are present in the water analysis. Hardness, meq/l = (Ca 2+ ) + (Mg 2+ ) The ions listed in the water analysis are Ca 2+ and Mg 2+ ; the concentration of Mg 2+ unknown. c. Determine the concentration of Mg 2+ and convert the concentration of Mg 2+ in meq/L to mg/L as follows: Hardness ions Conc., mg/L mg/meq a meq/L Ca 2+ 42.0 20.04 b 2.1 Mg 2+ 10.94 12.15 0.9 ∑ hardness ions 3.0 a Molecular weight/Z (see Eq. 2-8) 3. Determine the concentration of Na + from an anion, cation balance. a. Prepare a cation-anion balance. Cation Conc., mg/L mg/meq a meq/L Anion Conc., mg/L mg/meq a meq/L

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    Homework Solution Manual Page 14 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water Ca 2+ 42.0 20.04 b 2.1 HCO 3 - 61.02 61.02 1.00 Mg 2+ 10.94 12.15 0.9 SO 4 2- 96.0 48.03 2.00 Na + 23 Cl - 35.5 35.45 1.00 K + 29.5 39.1 0.75 NO 3 - 4.0 62.01 0.06 ∑ cations 3.75 ∑ anions 4.06 a Molecular weight/Z (see Eq. 2-7) b For calcium, equivalent weight = 40.08/2 = 20.04 g/eq or 20.04 mg/meq b. Calculate the concentration of Na+ from the difference between the sum of the cations and the sum of the anions. + + Na meq/L anions cations Na meq/L 4.06 3.75 0.31 = − = − = ∑ ∑ ( )( ) + + + + Na mg/L Na mg/meq x Na meq/L Na mg/L 23 0.31 7.13 mg/L = = = PROBLEM 2-12 Instructors Note: See Chap. 19 for a more thorough discussion of hardness, where total, 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 40 and 180 mg/L as CaCO 3 : Ion Concentration, mg/L Ca 2+ 55.0 Mg 2+ ? Na + 23.0 K + ? HCO 3 – ? SO 4 2– 48.0 Cl – ? CO 2 4.0

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    Homework Solution Manual Page 15 of 18 MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1 Chapter 2 - Physical and Chemical Quality of Water Solution 1. Determine the concentration of - 3 HCO using Eq. 2-26. a. Convert alkalinity from mg/L as CaCO 3 to meq/L. 3 3 3 3 mg/L alkalinity =meq/L of alkalinity ions x meq/L mass of CaCO , 50 mg/meq as CaCO mg/L alkalinity as CaCO meq/L of alkalinity ions meq/L mass of CaCO 40 mg/L meq/L of alkalinity ions 0.8 meq/L 50 mg/meq = = = b. Use Eq. 2-26 and determine the alkalinity ions that are present in the water analysis. - 2- - + 3 3 Alk, meq/L = (HCO ) + (CO ) + (OH ) - (H ) The only alkalinity ion listed in the water analysis is - 3 HCO , which is unknown. c. Convert the concentration of - 3 HCO in meq/L to mg/L as follows: Alkalinity ions Conc., mg/L mg/meq a meq/L HCO 3 - 48.82 61.02 0.80 a Molecular weight/Z (see Eq. 2-8) 2. Determine the value of Mg 2+ using Eq. 2-23, which is for non-carbonate hardness. a. Convert hardness from mg/L as CaCO 3 to meq/L. 3 3 3 3 mg/L hardness meq/L of hardness ions x meq/L mass of CaCO , 50 mg/meq as CaCO mg/L hardness as CaCO meq/L of hardness ions meq/L mass of CaCO 180 mg/L meq/L of hardness ions 3.6 meq/L 50 mg/meq = = = = b. Use Eq. 2-23 and determine the hardness ions that are present in the water analysis. Hardness, meq/l = (Ca 2+ ) + (Mg 2+ ) The hardness ions listed in the water analysis are Ca 2+ and Mg 2+ , and the concentration of Mg 2+ is unknown.

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