Useful information, Formulas and Conversion Factors 8.34 pounds = 1 gallon of water 8.34 pounds/Million gallons = 1 mg/L 7.48 gallons = 1 cubic foot of water wa ter 325,851 gallons = 1 acre-foot of water 1440 minutes = one day 17.12 mg/liter = 1 grain per gallon 7000 grains = 1 pound 142.8 pounds/MG = 1 grain per gallon 2 πD = Area 4 πD = circumference of circle MW of carbon dioxide = 44 MW of calcium oxide = 56 MW of magnesium = 24.1 MW of calcium carbonate = 100 MW of CaCO3 = 100 MW of Ca = 40 MW of Mg = 24.1 MW of Na = 23 MW of S = 32 MW of O = 16 MW of F = 19 MW of Si = 28 Sodium Fluoride, NaF Sodium silicofluoride, Na2SiF6 Fluosilicic acid, H2SiF6 BV, cu. ft = Q/QB , where design flow rate = Q; QB = Breakthrough volume from an experimental pilot column M, lb = (BV)(ρ s) where BV, cu. ft and ρs = mass of carbon in lb/cu. ft. Mt = Q/Vb,, Mt = pounds exhausted per hour; Q = flow rate; Vb = gallons per pound treated BV, = area X length X (cu ft/ 1728 cu.in.)(7.48 gals/cu.ft)(3.785liter gals/cu.ft)(3.785liters/gal) s/gal) = liters Filter loading rate = gpm/sq.ft filter area Filter backwash rate = gpm/sq.ft.area BV = Q/Qb V = area X length Qb = (gal/hr)(BV/V)(cu.ft/7.48)=BV/hr. I=FQNEr /nE /nEc where I= current in amperes, F=Faraday’s constant, E c = current efficiency as a fraction, Q = solution flow rate, N = normality of the solution, n = nu mber of cells Power = RI2 where R = resistance, I = current in amperes Faraday Constant = 96,500 96, 500 ampere-seconds per gram equivalent
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N
Probability equation of True recurrence recurrence interval: interval: Z = 1 – (1-1/TR) , where Z is the probability of event will be equal or exceeded during a specified design period; T R is the recurrence interval of the event; N is the design period in years K 528Qlog(r2/r1) / m(h2-h1), for artesian artesian well where K is coefficient of permeability in
N
Probability equation of True recurrence recurrence interval: interval: Z = 1 – (1-1/TR) , where Z is the probability of event will be equal or exceeded during a specified design period; T R is the recurrence interval of the event; N is the design period in years Kf = 528Qlog(r2/r1) / m(h2-h1), for artesian artesian well where Kf is coefficient of permeability in gpd/sq. ft.; Q = gallons per minute steady-state pumping rate; r2/r1 is drawdown in feet; m = thickness of aquifer; h2 – h2 – h1 h1 = 10 ft – ft – 1 1 ft. = 9 feet 2 2 Kf = 1055Q log(r2 /r1) / h2 – h1 , for unconfined aquifer TR = V/Q where TR is the theoretical residence time of a process reator, V = volume; Q = flow, Surface Overflow Overflow rate = gpd / surface surface area area of tank, tank, where gpd is gallons per day and surface area is feet or gpd/ sq.ft. Detention time = Volume of tank (in gallons) / flow rate (gallons/ time) Weir loading, gpd/ft = flow, gpd / linear weir length, feet T10 = C(tracer, mg/L) / Co(initial dose, mg/L) = 0.10 0.1 0 1/2 G = (P/µV) where G = velocity gradient, fps/ft; P = power input, ft-lb/s; µ = absolute . viscosity, lb s/ft 2; V = volume, ft3 Calculation for Volume of Grit Chamber or other reaction vessel: Volume, cu. ft. = discharge rate, cfs X detention time, seconds Calculation of length of Grit chamber or other o ther reaction vessel: Length, feet = Volume, cu. ft. / depth, feet Calculation of cross sectional area of grit chamber or other reaction vessel: Cross sectional area = cfs/ fps, where cfs is cubic feet per second an d fps is feet per second Filtration rate, gpm/sq. foot = gpm/area, sq. feet Kozeny Equation for rate of head loss thru clean granular medium: 2 2 3 2 h/l = Jv(1-ε) Jv(1-ε) VS / gε d where h/l = head loss per unit depth of filter bed, gt H2O/ft J = constant, approximately 6 for filtration in the laminar flow region, dimensionless 2 2 v= kinematic viscosity viscosity (µ /p), ft /s, (m /s) 2 g = acceleration of gravity, ft/s ε = porosity of the stationary filter bed, dimensionless dimensionless V = superficial (approach) velocity of water above the bed, ft/s S = shape factor ranging from 6.0 for spherical grains to 7.5 for angular grains, dimensionless d = mean grain diameter, ft. Calculate Chemical Chemical Dosage: Million gallons X 8.34lbs/million gallons X ppm = pounds. Calculate the meq/l of water anions and cations of a water analysis: Meq/L = (Valence / MW) X mg/L or Meq/L = (1 / Eq.Wt.) X mg/L T. Hardness, as CaCO3 = (Ca as Ca, mg/L X 2.5) + (Mg as Mg, mg/L X 4.12) The calcium alkalinity = Calcium hardness or total alkalinity whichever is smaller Magnesium alkalinity = magnesium hardness if total alkalinity ≥ than total hardness Magnesium Alkalinity = Total Alkalinity – calcium hardness if total alkalinity is > than calcium hardness but less than total hardness. ha rdness. Sodium alkalinity = total alkalinity – total hardness NCH = Total Hardness – Total Alkalinity ( If Mg Alkalinity present then no Ca NCH)
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Water 1. Groundwater is used for a water supply. It is taken from the the ground at 25 degrees C. The initial properties properties of the water are as follows.
N
Probability equation of True recurrence recurrence interval: interval: Z = 1 – (1-1/TR) , where Z is the probability of event will be equal or exceeded during a specified design period; T R is the recurrence interval of the event; N is the design period in years Kf = 528Qlog(r2/r1) / m(h2-h1), for artesian artesian well where Kf is coefficient of permeability in gpd/sq. ft.; Q = gallons per minute steady-state pumping rate; r2/r1 is drawdown in feet; m = thickness of aquifer; h2 – h2 – h1 h1 = 10 ft – ft – 1 1 ft. = 9 feet 2 2 Kf = 1055Q log(r2 /r1) / h2 – h1 , for unconfined aquifer TR = V/Q where TR is the theoretical residence time of a process reator, V = volume; Q = flow, Surface Overflow Overflow rate = gpd / surface surface area area of tank, tank, where gpd is gallons per day and surface area is feet or gpd/ sq.ft. Detention time = Volume of tank (in gallons) / flow rate (gallons/ time) Weir loading, gpd/ft = flow, gpd / linear weir length, feet T10 = C(tracer, mg/L) / Co(initial dose, mg/L) = 0.10 0.1 0 1/2 G = (P/µV) where G = velocity gradient, fps/ft; P = power input, ft-lb/s; µ = absolute . viscosity, lb s/ft 2; V = volume, ft3 Calculation for Volume of Grit Chamber or other reaction vessel: Volume, cu. ft. = discharge rate, cfs X detention time, seconds Calculation of length of Grit chamber or other o ther reaction vessel: Length, feet = Volume, cu. ft. / depth, feet Calculation of cross sectional area of grit chamber or other reaction vessel: Cross sectional area = cfs/ fps, where cfs is cubic feet per second an d fps is feet per second Filtration rate, gpm/sq. foot = gpm/area, sq. feet Kozeny Equation for rate of head loss thru clean granular medium: 2 2 3 2 h/l = Jv(1-ε) Jv(1-ε) VS / gε d where h/l = head loss per unit depth of filter bed, gt H2O/ft J = constant, approximately 6 for filtration in the laminar flow region, dimensionless 2 2 v= kinematic viscosity viscosity (µ /p), ft /s, (m /s) 2 g = acceleration of gravity, ft/s ε = porosity of the stationary filter bed, dimensionless dimensionless V = superficial (approach) velocity of water above the bed, ft/s S = shape factor ranging from 6.0 for spherical grains to 7.5 for angular grains, dimensionless d = mean grain diameter, ft. Calculate Chemical Chemical Dosage: Million gallons X 8.34lbs/million gallons X ppm = pounds. Calculate the meq/l of water anions and cations of a water analysis: Meq/L = (Valence / MW) X mg/L or Meq/L = (1 / Eq.Wt.) X mg/L T. Hardness, as CaCO3 = (Ca as Ca, mg/L X 2.5) + (Mg as Mg, mg/L X 4.12) The calcium alkalinity = Calcium hardness or total alkalinity whichever is smaller Magnesium alkalinity = magnesium hardness if total alkalinity ≥ than total hardness Magnesium Alkalinity = Total Alkalinity – calcium hardness if total alkalinity is > than calcium hardness but less than total hardness. ha rdness. Sodium alkalinity = total alkalinity – total hardness NCH = Total Hardness – Total Alkalinity ( If Mg Alkalinity present then no Ca NCH)
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Water 1. Groundwater is used for a water supply. It is taken from the the ground at 25 degrees C. The initial properties properties of the water are as follows.
Water 1. Groundwater is used for a water supply. It is taken from the the ground at 25 degrees C. The initial properties properties of the water are as follows. CO2 = 60 mg/L as CaCO3 Alkalinity = 200 mg/L as CaCO3 pH = 7.1 The water is treated for CO 2 removal by spraying into the atmosphere through a nozzle. The final CO2 concentration is 5.6 mg/L as CaCO3 at 25 degrees C. The first ionization constant of carbonic acid is 4.45 X 10-7. What is the the final pH of the water after spraying and recovery, assuming the alkalinity is unchanged in the process? Since pH < 8.3, all alkalinity is in the bicarbonate (HCO 3 ) form. Therefore, the equilibrium expression for the first ionization of carbonic acid may be used. + CO2 + H2O → H2CO3 ↔ H + HCO3 + -7 K1 = [H ][HCO3 ] = 4.45 X 10 [H2CO3] 2 The coefficients of CO and H2CO3 are 1. The number of moles of each compound is the same. One mole of CO 2 produces one mole of H 2CO3. -5 [H2CO3 = [CO2] = 5.6 mg/L / (100 g/mol)(1000 mg/g) = 5.6 X 10 -8 [HCO3 ] = 200 mg/L / 100 g/mol (1000 mg/g) = 1.246 X 10 Solve for the hydrogen ion concentration. + -7 -8 -8 [H ] = K1(H2CO3] = K1[H2CO3] = (4.45 X 10 )(5.6 X 10 ) = 1.246 X 10 ) [HCO3 ] [CO2] + -8 pH = -log[H ] = - log(1.246 X 10 ) = 7.9 2. The solids concentration concentration of a stream water sample sample is to be determined. determined. The total solids concentration is determined by placing a portion po rtion of the sample into a porcelain evaporating dish, drying the sample at 105 degrees C, and igniting the residue by placing the dried sample in a muffle furnace at 550 degrees degrees C. The following masses are recorded. recorded. Mass of empty dish: dish: 50.326 g Mass of dish and sample: sample: 118.400 g Mass of dish and dry solids: 50.437 g Mass of dish and ignited solids: 50.383 g a). The total solids concentration is The density of water is 1 g/mL. The volume of the tested sample is 118.4 g – g – 50.326 50.326 g / 1 g/mL = 68.1 mL TS = (50.437 g – g – 50.326 50.326 g) (1000 mg/L) (1000 mg/L) = 1630 mg/L 68.1 ml b). The total volatile solids concentration is: – 50.383 g) X (1000 mg/g (1000 ml/L) = 793 mg/L TVS = (50.437 g – 50.383 68.1 ml c). The total fixed solids concentration is TFS = 1630 mg/L – mg/L – 793 793 mg/L = 837 mg/L
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3. The dissolved concentration is determined by filtering a portion of the sample through a glass-fiber disk into a porcelain evaporating dish, drying the sample at 105 degrees C, and igniting the residue by placing the dried sample in a muffle furnace at 550 degrees C. The following masses are recorded. recorded.
3. The dissolved concentration is determined by filtering a portion of the sample through a glass-fiber disk into a porcelain evaporating dish, drying the sample at 105 degrees C, and igniting the residue by placing the dried sample in a muffle furnace at 550 degrees C. The following masses are recorded. recorded. The following masses are recorded. Volume of sample = 25 mL. Mass of empty dish: dish: 51.494 g Mass of dish and dry solids = 51.524 g Mass of dish and ignited solids = 51.506 g a.
The total dissolved solids concentration is: TDS = (51.524 g) – g) – (51.494 (51.494 g) X (1000 mg/g)(1000mg/L) = 1200 mg/L 25 ml b. VDS = (51.524 g – g – 51.506 51.506 g) (1000 mg/L)(1000 mg/L)(1000 mL/L = 720 mg/L 25 ml c. FDS = 1200 mg/L – mg/L – 720 720 mg/L = 480 mg/L
Clarifiers
4. What is the theoretical mean residence time of a process reactor if the volume of the reactor is 100,000 gallons and the rate of flow through the reactor is 1000 gpm. Solution: TR = V/Q where V = volume; Q = flow TR = 100000 gallons / 1000 gpm = 100 minutes or 1.67 hours 5. Calculate the overflow overflow rate of a rectangular clarifier clarifier that is 90 feet feet long, 16 feet wide, and 12 feet deep with a flow rate of 1.5 mgd. Solution: Overflow rate = gpd / surface area of tank = 1,500,000 gallons / 90 feet long X 16 feet wide = 1041 gpd/sq.ft. gpd/sq.ft. 4
6. Calculate the detention time in hours of a rectangular clarifier that is 90 feet long, 16 feet wide, and 12 feet deep with a flow rate of 1.5 mgd. Detention time = Volume of tank (in gallons) / flow rate (gallons/ time) Detention time, hours = 90 feet X 16 feet X 12 feet X 7.48gals/cu. Ft / 1500000 gpd / 24 hours or
6. Calculate the detention time in hours of a rectangular clarifier that is 90 feet long, 16 feet wide, and 12 feet deep with a flow rate of 1.5 mgd. Detention time = Volume of tank (in gallons) / flow rate (gallons/ time) Detention time, hours = 90 feet X 16 feet X 12 feet X 7.48gals/cu. Ft / 1500000 gpd / 24 hours or 62500 gallons per hour = 2.1 hours. 7. Calculate the weir loading for a rectangular clarifier that is 90 feet long 16 feet wide, and 12 feet deep that treats 500, 000 gallons per day with a weir length equal to three tank widths. Solution: Weir loading, gpd/ft = flow, gpd / linear weir length, feet Weir loading, gpd/ft = 500000 gpd / 3 X 16 ft = 10416 gpd/foot 8. A completely mixed reactor showed the following results of a step-dose input tracer test using fluoride chemical. Calculate the T10 in minutes. The T10 time in minutes is when 10% of the tracer chemical passes through the effluent o f the mixed reactor.
Solution: T10 = C(tracer, mg/L) / Co(initial dose, mg/L) = 0.10. The T 10 will fall between 12 and 15 minutes. By interpolation, T 10 = 0.045/0.1 X 3 minutes difference + 12 minutes = 13.35 minutes
9. A water plant has a flocculation tank 64 feet long and 100 feet wide with a water depth of 16 feet and a power input of 3960 foot lbs/ s or 5.36 horsepower. Calculate the velocity -5 2 gradient if µ = 2.735 X 10 lb-s/ft . 1/2 Solution: G = (P/µV) where G = velocity gradient(power to the water), fps/ft; P = power . 2 3 input, ft-lb/s; µ = absolute viscosity, lb s/ft ; V = volume, ft 5
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½
G = (3960 / 2.735 X 10 X 16 X 64 X 100) = 38 fps/ft 10. Determine the volume of a hopper-bottom grit removal tank if the detention time is 1 minute at 1 cfs peak flow. Solution: Volume, cu. ft. = discharge rate, cfs X detention time, seconds
-5
½
G = (3960 / 2.735 X 10 X 16 X 64 X 100) = 38 fps/ft 10. Determine the volume of a hopper-bottom grit removal tank if the detention time is 1 minute at 1 cfs peak flow. Solution: Volume, cu. ft. = discharge rate, cfs X detention time, seconds Volume = 1 cfs X 60 seconds = 60 cubic feet. 11. Determine the length of a grit chamber if the detention time is 1 minute at peak flow of 1 cfs with flow velocity of 1 fps and depth of chamber is 2 feet. Solution: Q, cfs X detention time, seconds = Volume, cubic feet 1 cfs X 60s = 60 cubic foot Length, feet = Volume, cu. ft. / depth, feet = 60 cu.ft. / 2 foot depth = 30 feet long 12. Determine the cross-sectional area of a grit chamber if the detention time is 1 minute at peak flow of 1 cfs with flow velocity of 1 fps. Solution: The cross sectional area is equal to the discharge rate divided by the velocity. Cross sectional area = cfs/ fps = 1 cfs / 1 fps = 1 sq. foot 13. Determine the gpm/square foot filtration rate of a filter 26 feet by 2 6 feet that treats 4 mgd. Solution: Filtration rate, gpm/sq. foot = gpm/area, sq. feet = 4,000,000/1440 min/day / 26 X 26 = 4.1 gpm/sq.ft. 14. Calculate the initial head loss through a filter with 18 inches of uniform sand having a porosity of 0.42 and a grain diameter of 1.6 X 10-3 ft. Assume spherical particles and a water temperature of 50 degrees F. The filtration rate is 2.5 gpm/sq.ft. Use the Kozeny equation. 2 2 3 2 h/l = Jv(1-ε) VS / gε d where h/l = head loss per unit depth of filter bed, gt H2O/ft J = constant, approximately 6 for filtration in the laminar flow region, dimensionless 2 2 v= kinematic viscosity (µ /p), ft /s, (m /s) 2 g = acceleration of gravity, ft/s ε = porosity of the stationary filter bed, dimensionless V = superficial (approach) velocity of water above the bed, ft/s S = shape factor ranging from 6.0 for spherical grains to 7.5 for angular grains, dimensionless d = mean grain diameter, ft. V = 2.5 gpm/sq.ft / 7.48 gal/ft X 60 sec/min = 0.00557 ft/sec, 1 = 1.5 ft. -5 2 2 3 -3 2 h/l = 6 X 1.410 X 10 (1 – 0.42) 0.00557 X (6.0) / 32.2 X (0.42) X (1.6 X 10 ) = 0.93 feet h = 1.5 ft depth X 0.933 = 1.4 feet loss of head. Based on a given chemical dosage in ppm, calculate the pounds required per million gallons. Solution: Million gallons X 8.34lbs/million gallons X ppm = pounds. Example: Calculate how many pounds per day of chlorine if the dosage is 3.5 ppm and the amount of water to be treated is 10 million gallons per day. Answer: 10 MGD X 8.34 X 3.5 ppm = 292 pounds per day.
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15. Calculate the meq/l of water anions and cations of a water analysis. Meq/L = (Valence / MW) X mg/L or Meq/L = (1 / Eq.Wt.) X mg/L Table 11.3 - Example of Water Analysis Data Ion Mg/L Equivalent weight
Meq/l
15. Calculate the meq/l of water anions and cations of a water analysis. Meq/L = (Valence / MW) X mg/L or Meq/L = (1 / Eq.Wt.) X mg/L Table 11.3 - Example of Water Analysis Data Ion Mg/L Equivalent weight + Ca 40 20 + Mg 10 12.2 + Na 11.7 23.0 + K 7.0 39.1 Total Cations = HCO 110 61.0 SO4 67.2 48.0 Cl 11.0 35.5 Total Anions=
Meq/l 2.00 0.82 0.51 0.18 3.51 1.80 1.40 0.31 3.51
Name the three major classes of alkalinity The three major classes of alkalinity are defined by their pH range. a. Bicarbonates, HCO3- >4.5 to ≤ pH 8.3 b. Carbonates, CO3 - >8.3 < pH 10 c. Hydroxide, OH- >pH10 What is the titration endpoint for phenolthalein alkalinity and the total alkalinity respectively. P. Alkalinity = pH 8.3 T. Alkalinity = 4.5 16. Determine the alkalinity relationships from a water analysis giving the phenol alkalinity and total alkalinity concentrations. Titration Result
OHAlkalinity as CaCO3
CO3Alkalinity as CaCO3
0 0 0 2P – T T
0 2P 2P 2(T- P) 0
P=0 P<1/2 T P=1/2 T P>1/2 T P=T
HCO3Alkalinity as CaCO3 T T-2P 0 0 0
17. Fill in the correct amount in mg/L for each blank based on alkalinity results in the first column. Alkalinity Tests Results P.Alk = 0, T. Alk. =100 P Alk = 40, T.Alk.=100 P Alk = 50, T.Alk. =100 P Alk=60, T Alk.=100 P = 100, T Alk.=100
OH- Alkalinity as CaCO3 0 0 0 20 100
CO3- Alkalinity as CaCO3 0 80 100 80 0
HCO3- Alkalinity as CaCO3 100 20 0 0 0
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18. What are the atomic weight and the equivalent weight of Na+1 and Ca+2 if a water sample has 102 mg/l of Na+1 and 68 mg/L of Ca+2, how many milliequivalents of each are present? + Na AW = 23, Eq. Wt. = 23/1 = 23 grams, 23 mg =
18. What are the atomic weight and the equivalent weight of Na+1 and Ca+2 if a water sample has 102 mg/l of Na+1 and 68 mg/L of Ca+2, how many milliequivalents of each are present? + Na : AW = 23, Eq. Wt. = 23/1 = 23 grams, 23 mg = meq. Meqs = (102 mg/L) / 23 mg/meq = 4.43 meqs/L +2 Ca : AW = 40, Eq. Wt. = 40/2 = 20 grams, 20 mg = meq. Meqs = (68 mg/L) / 20 mg/meq. = 3.4 meqs/L 19. What are the molecular weight and the equivalent weight of CaCl2 if a water sample contains 168 mg/L of CaCl2, how many milliequivalents are present? CaCl2 MW = 40 + 2(35.45) = 110.45 Meq. Wt.= 110/2 = 55 mg Meqs = (168 mg/L) / 55 mg/meq = 3.05 meg/L 20. If a water sample contains 134 ppm of Na+1, how many mg/L are present? 1 ppm = 1 mg/L if specific gravity = 1.00, therefore 134 ppm = 134 mg/L. 21. If a water sample has a pH of 7.6, how many moles per liter of hydrogen ion, H+, and hydroxyl ion, OH-, are present? pH = 7.6 + pH = Log 1/[H ] + -7.6 [H ] = 10 moles /L + -14 [H ][OH ] = 10 -7.6 -14 [10 ][OH ] = 10 -6.4 [OH ] = 10 moles/L In water coagulation with alum (aluminum sulfate) and bicarbonate alkalinity, the . -2 simplified reaction is: Al2(SO4)3 14 H2O + 6 HCO3 → 2Al(OH)3 ↓ 3SO4 + 14 H2O + 6CO2, why does this reaction go to completion? It goes to completion because an insoluble product is formed. 22. Determine the total hardness as calcium carbonate if given the calcium and magnesium concentrations. Solution: T. Hardness, as CaCO3 = (Ca as Ca, mg/L X 2.5) + (Mg as Mg, mg/L X 4.12) Example: Ca = 30 mg/L as Ca X 2.5 = 75 mg/L as CaCO3 Mg = 10 mg/L as Mg X 4.12 = 41.2 mg/L as CaCO3 Total Hardness as CaCO3 = 116.2 mg/L as CaCO3 Name three ways carbonate hardness can be reduced in water. Boiling, lime, and ion exchange Name two ways noncarbonated hardness can be reduced in water. Soda ash and ion exchange. Write the chemical reactions in lime treatment of carbon dioxide. CO2 + Ca(OH)2 CaCO3 + H2O Write the chemical reactions in lime treatment of calcium and magnesium bicarbonate. Ca + 2HCO3 + Ca(OH)2 2CaCO3 + 2H2O(pH 8.3-9.) Mg + 2HCO3 + Ca(OH)2 CaCO3 + Mg + CO3 + 2H2O(pH >10.8) Write the chemical reactions in lime treatment of magnesium carbonate and sulfate. Mg + CO3 + Ca(OH)2 → CaCO3 + Mg(OH)2 Mg + SO4 + Ca(OH)2 Ca + SO4 + Mg(OH)2
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Write the chemical reactions of the second state treatment with soda ash of non-carbonate hardness. Ca + SO4 + Na2CO3 Na2SO4 + CaCO3 Ca + Cl2 + Na2CO3 CaCO3 + 2NaCl
Write the chemical reactions of the second state treatment with soda ash of non-carbonate hardness. Ca + SO4 + Na2CO3 Na2SO4 + CaCO3 Ca + Cl2 + Na2CO3 CaCO3 + 2NaCl 23. A water has an hardness of 185 m/L as CaCO3, what is the hardness expressed as meq/L? CaCO3 Meq wt = [40 +12+3(16)] / 2 = 50 Meqs/L = (185 mg/L) / 50 mg/meq. = 3.70 meq/L 25. A water has an alkalinity of 225 mg/L as CaCO3. What is the alkalinity expressed as meq/L? CaCO3 meq wt. = [40+12+3(26)] / 2 = 50 Meq/L = (225 mg/L) / 50 mg/meq = 4.5 meq. How can the calcium alkalinity be determined? The calcium alkalinity = Calcium hardness or total alkalinity whichever is smaller Name two ways the magnesium alkalinity be determined? Magnesium alkalinity = magnesium hardness if total alkalinity ≥ than total ha rdness Magnesium Alkalinity = Total Alkalinity – calcium hardness if total alkalinity is > than calcium hardness but less than total hardness. How can the sodium alkalinity be determined? Sodium alkalinity = total alkalinity – total hardness How can the noncarbonated hardness be determined? NCH = Total Hardness – Total Alkalinity ( If Mg Alkalinity present then no Ca NCH) 26. Determine the calcium alkalinity, magnesium alkalinity, sodium alka linity, calcium non-carbonate hardness, and the magnesium non-carbonate hardness from the given water analyses. Analyses Water #1 Water #2 Water #3 Total Hardness 300 300 300 Calcium Hardness 200 200 200 Mg Hardness 100 100 100 Total Alkalinity 150 250 350 Interpretations Water #1 Water #2 Water #3 Calcium Alkalinity 150 200 200 Mg. Alkalinity None 50 100 Sodium Alkalinity None None 50 Ca N.C. Hardness 50 None None Mg. N.C. Hardness 100 50 none What is the practical minimum total hardness level of the lime-soda ash treatment method? The practical minimum hardness level using the lime-soda ash treatment method is 50 mg/L.
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27. Calculate the hydrated lime (100%), soda ash, and carbon dioxide requirements to reduce the hardness of a water with the following analysis to about 50 to 80 mg/L by the excess Lime-soda ash process based on the following water analyses. Analyses: Total Hardness 280 mg/L CaCO3
27. Calculate the hydrated lime (100%), soda ash, and carbon dioxide requirements to reduce the hardness of a water with the following analysis to about 50 to 80 mg/L by the excess Lime-soda ash process based on the following water analyses. Analyses: Total Hardness = 280 mg/L as CaCO3 Mg++ = 21 mg/L Alkalinity = 170 mg/L as CaCO3 Carbon Dioxide = 6 mg/L Lime Requirement: Carbon Dioxide = (6) (56) / (44) = 8 Alkalinity = (170) (56) / (100) = 95 Mg ++ = (21) (56) / (24.3) = 48 Excess Lime = = 35 Total CaO required = 186 mg/L Soda Ash Requirement: NCH = 280 – 170 = 110 mg/L Soda Ash (Na2CO3) = (110) (106) / (100) = 117 mg/L 28. A town’s water supply has the following ionic concentrations. +++ Al = 0.5 mg/L ++ Ca = 80.2 mg/L Cl = 85.9 mg/L CO2 = 19 mg/L -CO3 = 0 ++ Fe = 1.0 mg/L Fl = 0 HCO3 = 185 mg/L ++ Mg = 24.3 mg/L + Na = 46.0 mg/L NO3 = 0 -SO4 = 125 mg/L a). What is the total hardness? ++ Ca = 80.2 mg/L X 2.5 = 200.5 mg/L as CaCO 3 ++ Mg = 24.3 mg/L X 4.12 = 100 mg/L Fe ++ = 1.0 mg/L X 1.79 = 1.79 mg/L +++ Al = 0.5 mg/L X 5.56 = 2.78 mg/L Total Hardness ------------ 305 mg/L b). How much slaked lime is required to combine with the carbonate hardness? To remove the carbonate hardness, CO 2: 19 mg/L X 100/44 or 2.27 = 43.13 mg/L as CaCO 3 HCO3 : 185 mg/L X 50mg/Las 1 meq/L/61mg/L as 1meq/L or 0.82 = 151.7 mg/L Total equivalents to be neutralized are: 43.13 mg/L + 151.7 mg/L = 194.83 mg/L as Ca(OH) 2 Convert Ca(OH) 2 to CaCO3 = 194.83 /100/74 or 1.35 = 144.3 mg/L as CaCO 3 as substance c). How much soda ash is required to react with the non-carbonate hardness? NONE, because soda ash only reacts with non-carbonate hardness and the problem is only asking for the carbonate hardness removal which is only removed by lime. However, there is non-carbonate hardness present in the water. It is calculated by the difference in the Total hardness and the total alkalinity or 305 mg/l – 152 mg/L = 153 mg/L as CaCO 3 of non-carbonate hardness.
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29. If the town’s water supply has the following ionic concentra tions and it is to be softened using a zeolite process with the following characteristics: exchange capacity, 10,000 grains/cu.ft., and salt requirement , 0.5 lbm per 1000 grains hardness removed. How much salt is required to soften the water to 100 mg/L hardness?
29. If the town’s water supply has the following ionic concentra tions and it is to be softened using a zeolite process with the following characteristics: exchange capacity, 10,000 grains/cu.ft., and salt requirement , 0.5 lbm per 1000 grains hardness removed. How much salt is required to soften the water to 100 mg/L hardness? Ca(HCO3)2 = 137 mg/L as CaCO3 MgSO4 = 72 mg/L as CaCO3 CO2 = 0 There are 7000 grains in a pound. The hardness removed is: 137 mg/L + 72 mg/L – 100 mg/L = 109 mg/L (109 mg/L) (8.345 lbm-L/mg-MG) = 909.6 lbm hardness/MG 3 (0.5 lbm/1000 gr)(909.6 lbm/MG)(7000 gr/lbm) = 3.18 X 10 lbm/MG or 3200 lbm/MG 30. Determine the % hypochlorous acid (HOCl) available for disinfection from chlorine at 7.0, 8.0, and 9.0 pH values at 20 degrees C.
Solution: At pH 7.0 and 20 degrees C, the % hypochlorous acid is 80%. At pH 8.0 and 20 degrees C, the % hypochlorous acid is 30%. At pH 9.0 and 20 degrees C, the % hypochlorous acid is 5%.
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31. Determine the storage requirement and critical design period of a reservoir based on the given mean inflow and draft flows for a year period. Col 1
Col 2
Col 3
Col 4
Col. 5
Col. 6
31. Determine the storage requirement and critical design period of a reservoir based on the given mean inflow and draft flows for a year period. Col 1 Month Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec.
Col 2 Inflow MG 37.2 64.8 108 12 8.4 9.6 2.4 33.6 50.4 129.6 117.6 26.4
Col 3 Draft MG 53.1 53.1 53.1 53.1 53.1 53.1 53.1 53.1 53.1 53.1 53.1 53.1
Col 4 Sum of Inflows (MG) 37.2 102 210 222 230.4 240 242.4 276 326.4 456 573.6 600
Col. 5 Defieciency MG 15.9 -11.7 -54.9 41.1 44.7 43.5 50.7 19.5 2.7 -76.5 -64.5 26.7
Col. 6 Cumulative Deficiency(MG) 15.9 0 0 41.1 85.8 129.3 180 199.5 202.2 0 0 26.7
Solution: Col 4 = Sum of Col 2; Col 5 = Col 3 – Col 1; Col 6 = Sum of Col 5 when > 0 Answer: Storage Requirement = 202.2 MG; Critical Design Period = 202.2/col 3 or 53.1=3.8 months. 32. Write the probability equation of a true recurrence interval and develop a probability table based on the recurrence intervals of 1 year, 10 years, 50 years and 100 years for design periods of 1 year, 10 years 50 years and 100 years. N
Answer: Z = 1 – (1-1/TR) , where Z is the probability of event will be equal or exceeded during a specified design period; T R is the recurrence interval of the event; N is the design period in years 10 10 10 Example: Z = 1 – (1- 1/10) = 1 – (1-0.1) = 1 – (0.9) = 1 – (1 – 0.35) = 0.65
Table : Probability That an Event Having a Prescribed Interval Will Be Equaled or Exceeded During a Specified Design Period 33. Design Period (Years) TR (year) 1 10 50 100 1.0 1.0 1.0 1.0 1 0.1 0.995 ~1.0 10 0.65 0.02 0.18 0.64 0.87 50 0.01 0.10 0.40 0.63 100
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33. A flow of 100 mgd is to be developed from a 190 sq. mile watershed. At the flow line the area’s reservoir is estimated to cover 3900 acres. The annual rainfall is 40 inches, the annual runoff is 14 inches, and the annual evaporation is 49 inches. Find the net gain or loss in storage this represents. Calculate the volume of water evaporated in acre-ft.
33. A flow of 100 mgd is to be developed from a 190 sq. mile watershed. At the flow line the area’s reservoir is estimated to cover 3900 acres. The annual rainfall is 40 inches, the annual runoff is 14 inches, and the annual evaporation is 49 inches. Find the net gain or loss in storage this represents. Calculate the volume of water evaporated in acre-ft. Solution: Reservoir area, square miles = 3900acres / 640 acres per square mile = 6.1 square miles Annual runoff, ac-ft. = (14inches annual runoff / 12inches per foot)(190 sq. miles watershed – 6.1 sq. miles for reservoir area) (640 acres per sq. mile reservoir area) = 137,704 ac-ft. Annual evaporation, ac-ft. = (49 inches per year/12 inches per foot)(3900 acres reservoir area) = 15,925 ac-ft. Draft, ac-ft. = (100 MGD X 365 days X 1000000 gallons per Million) / (7.48 gallons per cubic foot X 43,560 square feet per acre) = 112,022 ac-ft. Precipitation on Lake, ac-ft. = (40 inches per year / 12 inches per foot) (3900 acres in Reservoir area) = 13,000 ac-ft. Gain in Storage, ac-ft. = 137,704 ac-ft runoff + 13,000 ac-ft precipitation in lake = 150,704 acft. Loss in Storage, ac-ft. = 112,022 ac-ft Draft + 15925 ac-ft annual evaporation = 127,947 ac-ft loss in storage. Net gain in Storage, ac.-ft. = 150,704 ac-ft – 127,947 ac-ft = 22,757 ac-ft. gain in storage 34. If a constant annual yield of 1500 gpm was required, what reservoir capacity would be needed to sustain it? Fine the capacity in acre-ft per year. Solution: Constant annual yield, gpm = 1500 gpm Time of operation without recharge = 1 year Reservoir Capacity Required, ac.-ft. = 1500 gpm X365 days per year X24 hours per day X 60 minues per hour /325851gallons / ac.ft. = 788,000,000 gallons/325851 gallons per ac-ft = 2418 ac.ft 35. Determine the permeability of an artesian aquifer being pumped by a fully penetrating well. The aquifer is composed of medium sand and is 90 feet thick. The steady-state pumping rate is 850 gpm. The drawdown of an observation well 50 feet away is 10 feet, and the drawdown in a second observation well 500 feet away is 1 foot.
Solution: Use Coefficient of Permeability Equation below: 13
Kf = 528Qlog(r2/r1) / m(h2-h1) = 528 X 850 X log (10/1) / 90 X (10 – 1) = 554 gpd/sq. foot where Kf is coefficient of permeability in gpd/sq. ft.; Q = gallons per minute steady-state pumping rate; r2/r1 is drawdown in feet = 500/50 = 10 feet; m = 90 feet thickness of aquifer; h2 h1 = 10 ft 1 ft. = 9 feet
Kf = 528Qlog(r2/r1) / m(h2-h1) = 528 X 850 X log (10/1) / 90 X (10 – 1) = 554 gpd/sq. foot where Kf is coefficient of permeability in gpd/sq. ft.; Q = gallons per minute steady-state pumping rate; r2/r1 is drawdown in feet = 500/50 = 10 feet; m = 90 feet thickness of aquifer; h2 – h1 = 10 ft – 1 ft. = 9 feet 36. An 18-inch well fully penetrates an unconfined aquifer of 100 feet depth. Two observation wells located 100 and 235 feet from the pumped well have drawdowns of 22.2 and 21 feet, respectively. If the flow is steady and Kf = 1320 gpd per square feet, what would be the discharge? 2 2 Equation: Kf = 1055Q log(r2 /r1) / h2 – h1 , for unconfined aquifer 2 2 Solve for Q = K f (h2 – h1 ) / 1055 log(r 2 /r1) Log(r2 /r1) = log (235/100) = 0.37107 h2 = 100 – 21 = 79 feet h1 = 100 – 22,2 = 77.8 feet 2 2 Q = 1320 ( 79 – 77.8 ) / 1055 X 0.37107 = 634.44 gpm 37. Compare the annual water requirements of an 1500-acre irrigated farm and a city of 130000 population. Assume an irrigation requirement of 3 acre-feet/acre/year and a per capita water use rate of 180 gpcd. City Use, gpd = 130000 population X 180 gpcd = 2,340,000 gallons per day Irrigation use, gpd = 43560 sq. ft per acre X 7.48 gallons / cu.ft. X 3 acre-ft X 1500 acres / 365 days per year = 4,017,067 gpd 2,340,000 gpd / 4,017,067 gpd = to the Irrigation use of 1.72 times the city’s use. 38. Consider a 1000-acre residential area with a housing density of four dwellings per acre. Estimate the peak hourly water use requirement.
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Number of dwelling units = 1000 X 4 = 4000 units Use Figure 4.4 and read up from y-axis = 4000 units Find curve line for 4.0 dwelling units per acre that crosses the 4000 unit line at 5000 gpm as peak hourly flow.
Number of dwelling units = 1000 X 4 = 4000 units Use Figure 4.4 and read up from y-axis = 4000 units Find curve line for 4.0 dwelling units per acre that crosses the 4000 unit line at 5000 gpm as peak hourly flow. 39. A community had a population of 200,000 in 2000 and it is expected that this will increase to 260,000 by 2015. The water treatment capacity in 2000 was 43 mgd. A survey showed that the average per capita water use rate was 180 gpcd. Estimate the community’s water requirements in 2015 assuming (a) no change in use rate and (b) a reduced rate of 160 gpcd. Will expanded treatment facilities be needed by 2015 for condition (a)? For condition (b)? Solution: In 1995 the population is 200,000 with water treatment capacity of 43 mgd with 180 gpcd water use. In 2010 the population will be 260,000. (a) At no change in use rate of 180 gpcd, the water treatment capacity will be how much at 260000 population: 260000 X 180 = 46.8 mgd, expansion is required (b) If the per capita use decreases to 160 gpcd by 2015, how much water treatment capacity will be needed? 260000 X 160 = 41.6 mgd, less capacity required
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Wastewater Wastewater Terminolgy Define the following:
Wastewater Wastewater Terminolgy Define the following: Bacteria – are the simplest forms of microorganisms that use soluble food and are capable of self-reproduction. Heterotrophic bacteria – use organic compounds as an energy and carbon source for synthesis. Saprophte – refers to an organism that lives on dead or decaying organic matter. Aerobes – require free dissolved oxygen to live and multiply. Anaerobes – oxidize organic matter in the complete absence of dissolved oxygen. Facultative bacteria – are a class of bacteria that use free dissolved oxygen when available but can also respire and multiply in its absence. Escherichia coli – is a fecal coliform that is a facultative bacterium. Autotrophic bacteria – use carbon dioxide as a carbon source and oxidize inorganic compounds for energy. Thiobacillus – are autotrophic sulfur bacteria that convert hydrogen sulfide to sulfuric acid. Leptothrix and Crenothrix – are iron-accumulating bacteria and thrive in water pipes containing dissolved iron and form yellow-or reddish-colored slimes. Fungi – refer to microscopic non-photosynthetic plants, including yeasts and molds. Algae – are autotrophic microscopic photosynthetic plants that uses carbon dioxide or bicarbonates in solution as a carbon source with phosphorous and nitrogen necessary for growth. Photosynthesis – is illustrated by the equation: sunlight CO2 + 2H2O new cell tissue + O2 +H2O) Dark reaction Protozoans – are single-celled aerobes that reproduce by binary fission and ingest bacteria and algae. Rotifers – are the simplest multicelled animals (aerobes) found in low pollutional streams and lakes. Daphnia and Cyclops - are crustaceans which are strict aerobes that ingest microscopic plants and serve as food for fishes. Metabolism – is the biochemical process performed by living organisms to yield energy for synthesis, motility and respiration to remain viable. Oxidation – is the addition of oxygen, removal of hydrogen, or removal of electrons. Reduction – is the removal of oxygen, addition of hydrogen, or addition of electrons. Exponential growth phase – is limited only by the microorganisms ability to process the substrate, i.e., excess food and maximum rate of metabolism. Declining growth phase – is caused by an increasing shortage of substrate. Stationary Phase – The rate of reproduction of microorganisms equals the rate of death of the microorganisms. Endogenous Growth Phase – of microorganisms occurs when the rate of metabolism is decreasing at an increasing rate, resulting in a rapid decrease in the number of viable cells. Food to microorganism ratio (F/M) – is the balance between the food supply and mass of microorganisms.
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1. Compare the average daily sewage flow from an apartment building of 15 floors having 10 apartments per floor using 200 gpd/unit with the sewage flow from a 10 -acre residential area having four houses per acre with an average of 4 persons per house using 90 gpcd.
1. Compare the average daily sewage flow from an apartment building of 15 floors having 10 apartments per floor using 200 gpd/unit with the sewage flow from a 10 -acre residential area having four houses per acre with an average of 4 persons per house using 90 gpcd. For individual houses there will be: 10 acre residential area X 4 houses per acre = 40 houses per 10 acres 4 persons per house = 40 houses per 10 acres X 4 persons per house = 160 persons 160 persons X 90 gpcd = 14,400 gpd For the apartment there will be: 10 apartments per floor X 15 floors = 150 units 150 units @ 200 gpd/unit = 200 gpd/unit X 150 units = 30,000 gpd. This problem illustrates the impact of high rise units versus land space. 2. Find the peak hourly flow in mgd for an 800-acre urban area having the following features: Domestic flows = 90 gpcd, commercial flows of 15 gpcd, infiltration = 600 gpd/acre, population density of 20 persons per acre and peak hour to average day ratio of 3.0. Solution: Calculate the average daily flows for each contributor: Domestic flows: 90 gpcd X 20 persons per acre X 800-acre = 1.44 mgd Commercial: 15gpcd X 20 persons per acre X 800-acre = 0.24 mgd Infiltration: 600 gpd per acre X 800-acre = 0.48 mgd 3. Given a 300 acre housing development with 850 houses (4 persons per house), and an average annual rainfall of 28 inches, calculate the yearly volume of precipitation over the area and compare it with the annual sewage flow of 100 gpcd. Solution: Yearly volume of 28 inches precipitation, acre-feet per year = 300 acres X 28 inches per year/12 inches per foot = 700 acre-feet per year Given Sewage flow = 100 gpcd Sewage volume, gallons per year would be 850 houses X 4 persons per house X 100 gpcd X 365 days per year = 124,100,000 gallons per year 124,100,000gallons per year / 7.48 gallons per cu.ft. = 16,590,909.09 cu. ft. = 16,590,909.09 cu. ft. / 43560 sq.ft per acre = 380 acre-feet. The peak hourly flow = 3.0 peak hour to average day ratio X (1.44+0.24+.48) = 6.48 mgd 4. A mechanically cleaned bar screen has bars 3/8” (5 mm) thick and 1 1/4” clear spaces between the bars. If the velocity through the bars is 3 feet/sec., determine the approach velocity, in ft/sec., and the head loss through the screen, in feet. Solution: Assume depth = D Va Aa = VbAb Va = (Ab/Aa) Vb = (1.25”)(D) / (1.625”) (D) x 3.0 fps = 2.31 fps 2 2 2 2 HL = [(Vb – Va ) / 2g] (1/0.7) = [(3.0) – (2.31) / (2)(32.17) x (1/0.7) = 0.08 feet. 5. A municipal wastewater plant treats 12.3 MGD. The screenings amount to 1.5 cubic feet per million gallons treated, and the grit amounts to 4.0 cubic feet per million gallons treated. a. Determine the volume in cubic feet of screenings removed per day. b. Determine the volume in cubic feet of grit removed per day. 17
c. Determine the time in days required to fill a 4’ X 6’ X 6’ solid waste container. Solution: a. Volume of screenings = (12.3 MG/Day)(1.5 cu.ft/MG) = 18.5 cu.ft.per day b. Volume of grit = (12.3MG/day)(4.0cu.ft./MG) = 49.2 cu.ft. per day.
c. Determine the time in days required to fill a 4’ X 6’ X 6’ solid waste container. Solution: a. Volume of screenings = (12.3 MG/Day)(1.5 cu.ft/MG) = 18.5 cu.ft.per day b. Volume of grit = (12.3MG/day)(4.0cu.ft./MG) = 49.2 cu.ft. per day. c. Total screenings and grit = 18.5 + 49.2 = 67.7 cu.ft./day Time = (4’ X 6’ X 6’)(day/67.7 cu.ft.) = 2.13 days Be able to draw and label the carbon cycle.
Be able to draw and label the nitrogen cycle.
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Be able to draw and label the sulfur cycle.
Be able to draw and label the sulfur cycle.
Name 6 factors that affect the growth of microorganisms. 1. temperature 2. pH 3. nutrients 4. oxygen supply 5. toxins 6. substrate types 6. A water at elevation 1500 feet(457 mm Hq) is at 20 degrees C. The atmosphere pressure at Elevation of 1000 feet is 733 mm Hq and at elevation 2200 feet is 706 mm Hq. Determine the saturation of dissolved oxygen concentration in mg/L? . Solution: Cs = 43.8 mg/L atmospheres Elevation 1000 – 733 mm Hq Elevation 1500 - X Elevation 2000 – 706 mm Hq. X = (733 + 706) (0.5) = 719.5 mm Atmosphere = 719.5 / 760 = 0.9467 . Cs = (43.8 mg/L atm) (0.9467 atm)(0.209) = 8.67 mg/L 7. The high purity oxygen process used in some activated sludge wastewater plants uses almost pure oxygen gas as an oxygen source instead of air. If the gas is 80% oxygen and 20% nitrogen and is at 1 atm pressure, what is the equilibrium oxygen concentration at 20 degrees C.? – Cs = 43.8 mg/L atm X Pg Pg = 0.80 atmospheres Cs = (43.8 mg/L – atm) (0.80 atm) = 35.0 mg/L 7. An activated sludge plant has anaerobic digesters with sludge gas utilization. The digesters produce 110,000 cubic feet per day of sludge gas at 10 inches of water column pressure at 95 degrees F. The gas is to be compressed and stored in a gas dome and used to fuel internal combustion engines that drive generators to produce electricity. If the compressed gas is at 30 psig pressure and 95 degrees F, how many cubic feet per day will it occupy?
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P1V1 = P2V2 T1 T2 T1 = T2 , so P V =PV
P1V1 = P2V2 T1 T2 T1 = T2 , so P1V1 = P2V2 For USCS units: V1= 110,000 cubic feet P1 = 14.7 psi + [(10”/12” ft) / 2.31 ft/psi] = 15.06 psi P2 = 14.7 + 30 psi = 44.7 psi V2 = P1 = [15.06] (110,000) = 37,100 cubic feet P2 [ 44,7 ] Identify the 3 major bacteria types and their temperature ranges used in anaerobic digestion. 1. Thermophilic bacteria – 50 – 55 degrees C. 2. Mesophilic bacteria – 20 – 40 degrees C 3. Psychrophilic bacteria – 4 – 10 degrees C Identify the growth phases of a pure culture of bacteria
What is the ideal growth phase for good settling in an activated sludge plant? The ideal growth phase for good settling in an activated sludge plant is during the endogenous phase.
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Label the schematic of a continuous-flow activated-sludge process.
Label the schematic of a continuous-flow activated-sludge process.
Identify the range of operation for most activated sludge treatment systems in the figure below.
8. An aerated laboratory vessel is filled with a mixture of substrate and microbial cells. At time zero, the substrate concentration (S) is 150 mg/L and the cell concentration (X) is 1500 mg/L. The biochemical reaction is pseudo-first order, and the rate constant K is 0.40 L/ (gm cells)(hr). After 6 hours of aeration, the substrate concentration (S) is 4 mg/L, and the 21
cell concentration (X) is 1590 mg/L. Using the specific rate of substrate utilization equation, (1/X)(dS/dt) = KS, determine: a. The rate of substrate utilization, dS/dt, at time zero in mg/(L)(hr). b. The rate of substrate utilization, dS/dt, at time equals 6 hours in mg?(L)(hr).
cell concentration (X) is 1590 mg/L. Using the specific rate of substrate utilization equation, (1/X)(dS/dt) = KS, determine: a. The rate of substrate utilization, dS/dt, at time zero in mg/(L)(hr). b. The rate of substrate utilization, dS/dt, at time equals 6 hours in mg?(L)(hr). 1 dS = KS where S is substrate concentration, t is time, X is cell concentration X dt Or dS = XKS dt a.
dS = (1500 mg ) ( grams ) (0.40L) (150 mg) = 90 mg/L- hr dt L 1000 mg) (gm-hr) ( L) b. ds = [1591 mg] (grams ) (0.40 L) (4 mg) = 2.5 mg/L – hr. dt L 1000 mg) (gm-hr) ( L) 9. A municipal effluent has a BOD5 of 10 mg/L and K1(base e) is 0.23 day-1. a). The ultimate first stage BOD in mg/L, Lo. b). The ratio, y/Lo as a fraction and is a percent. Solution: -kt a). y = Lo (1 – e ) -0.23x5 10 = Lo (1- e ) Lo = 14.6 mg/L b). y/Lo = 10/14.6 = 68.5% Label the diagr6am of population dynamics in anaerobic digestion.
Determine:
What causes the pH drop in anaerobic digestion. An excess of organic matter is fed to a digester will cause acid formers to rapidly process the food resulting in excessive organic acids while the methane formers are unable to metabolize the organic acids fast enough resulting in a pH drop. Explain symbiosis as it applies to bacteria and algae in oxidation po nds. Symbiosis is the relationship of two or more species that live together for mutual benefit such as bacteria that metabolize organic matter releasing nitrogen and phosphorous and carbon
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dioxide for the benefit of algae that use these compounds with energy from sunlight for synthesis, releasing oxygen for the benefit of the bacteria. Define Facultative Stabilization Ponds. Facultative stabilization ponds are anaerobic at the bottom of the ponds while the surface is
dioxide for the benefit of algae that use these compounds with energy from sunlight for synthesis, releasing oxygen for the benefit of the bacteria. Define Facultative Stabilization Ponds. Facultative stabilization ponds are anaerobic at the bottom of the ponds while the surface is aerobic. 10. Calculate the pounds of BOD contributed to a sewage plant each day from a population of 100000 with an average of 0.20 pounds of BOD per person per day. Solution: 100000 X 0.20 = 20,000 pounds of BOD per day. 11. Calculate the pounds of suspended solids contributed to a sewage plant each day from a population of 100000 with an average of 0.24 pounds of S.S. per person per day. Solution: 100000 X 0.24 = 24,000 pounds of Suspended solids 12. Given the flow of 10 mgd with a raw BOD or suspended solids level of 250 mg/L, determine the pounds of BOD or suspended solids level contributed to the sewage plant. Solution: 10 mgd X 8.34#/MG X 250 ppm = 20850 pounds 13. If domestic wastewater contains 0.24 pounds of suspended solids and 0.20 pounds of BOD per 120 gallons per day per capita, what is the BOD equivalent population for an industry that discharges 0.10 mgd of wastewater with an average BOD of 450 mg/L? What is the hydraulic equivalent population of this wastewater? Solution: BOD equivalent population = 0.10 MG X 450 mg/L X 8.34 lb/MG / 0.20 lb/person = 1900 Hydraulic equivalent population = 100000 gal/day/120 gal/person = 830 Define the following terms: Carbohydrates – are hydrates of carbon with the empirical formula C nH2nOn. Monosaccharide – simplest carbohydrate unit such as glucose. Disaccharides – are composed of two monosaccharide units such as sucrose. Polysaccharides – are long chains of monosaccharides such as cellulose, starch and glycogen. Cellulose – is the common polysaccharide in wood, cotton, paper, and plant tissues. Starches – are primary nutrient polysaccharides for plant growth and are abundant in potatoes, rice, wheat, corn and other plant forms. Proteins – in simple form are long-chain molecules composed of amino acids connected by peptide bonds and are important in both the structure ( muscle tissue) and dynamic aspects (enzymes) of living matter. Lipids – form the bulk of organic matter of living cells which are soluble in varying degrees in organic solvents while being only sparingly soluble in water.
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Be able to label the parts of a trickling filter
Be able to label the parts of a trickling filter
Label the parts of the schematic diagram of the trickling filter process.
14. Calculate the total volume of media required for a trickling filter if 35 lbs/1000 cu. Ft/day of BOD is applied at a flow of 1.2 mgd at 315 mg/L BOD and 35% BOD removal is expected through the filter. Solution: 1.2 MGD X 8.34 lbs/gallon X 315 mg/L X .65 BOD remaining / 0.035 lbs/cu ft. = 58,500 cu. Ft. 15. Calculate the % BOD removal at 20 degrees C using the NRC formula for a singlestage trickling filter given the following information: 0.5 NRC formula = E20 = 100 / 1 + 0.0561 (w/VF) w/V = BOD loading, 23.5 lbs/1000 cu.ft./day F, recirculation factor = 1.36 0.5 E20 = 100/ 1 + 0.0561(23.5/1.36) = 81.1% 16. Calculate the plant efficiency if the primary sedimentation removes 35% BOD, first stage trickling filter removes 70% BOD, and the second stage trickling filter removes 60% BOD. Solution: E = 100 – 100[(1-0.35)(1-0.70)(1-0.60)] = 92.2%
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Label the parts of the continuous-flow activated sludge process.
Label the parts of the continuous-flow activated sludge process.
Name five activation-sludge processes. 1. Step Aeration 2. Conventional 3. Contact Stabilization 4. Extended Aeration 5. High-Purity Oxygen Name two ways of expressing BOD loadings in an activated-sludge process. 1. BOD/day/1000 cu. Ft 2. BOD/day/lb of MLSS (mixed liquor suspended solids) or MLVSS (mixed liquor volatile suspended solids) 17. Convert the average BOD concentration of 200 mg/L into units of pounds per 1000 cu. ft/day. Solution: MGD X 8.34#/gal X mg/l = pounds/day 1 MGD X 8.34 X 200 = 1668 pounds/ one million gallons/day or 133689 cu. Ft/day 1668/133689 = .0124767 pounds/cu.ft/day 0.01247 X 1000 cu. Ft. = 12.47 pounds/1000 cubic feet/day 18. If the BOD loading of 12.5 pounds/1000 cu.ft is aerated for a period 6 hours, determine what the BOD loading becomes in terms of pounds/1000 cubic feet/day? 12.5lb/1000 cu. Feet X 24/6.0 hr = 50 lb/1000 cu.ft./day. 19. Calculate the BOD loading in terms of lb/day/lb MLSS of an activated sludge if the settled wastewater flow = 3.67 mgd, the influent BOD = 128 mg/L, the MLSS in aeration tank = 2350 mg/L, the aeration tank volume = 0.898 million gallons. Solution: BOD load = 3.67 X 128 mg/L X 8.34 = 3920 lb/day MLSS in aeration tank = 0.898 MG X 2350 mg/L X 8.34 = 17,600 lb. BOD loading = 3920/17600 = 0.22 lb/day/lb of MLSS
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20. Determine the sludge age in days when given the volume and suspended solids in aeration tank, volume and suspended solids in effluent, and volume and suspended solids in the waste sludge. Solution:
20. Determine the sludge age in days when given the volume and suspended solids in aeration tank, volume and suspended solids in effluent, and volume and suspended solids in the waste sludge. Solution: Sluge age = MLSS X V/SSe X Qe + SSw X Qw where MLSS = mixed liquor suspended solids, mg/L V = volume of aeration tank, million gallons SSe = suspended solids in effluent, mg/L SSw = suspended solids in waste sludge, mg/l Qe = quantity of effluent wastewater, mgd Qw = quantity of waste sludge, mgd Example: MLSS = 2350 mg/L V = 120,000 cu.ft = 0.898 million gallons SSe = 26 mg/L Qe = 3.67 mgd SSw = 11,000 mg/L Qw = 0.0189 mgd Sludge age = mean cell residence time = 2350 X 0.898 / 26 X 3.67 + 11,000 X 0.0189 = 7.0 days. How does step-aeration and tapered aeration differ from the conventional activated sludge process? In step aeration the influent load is introduced at several points along the tank length while the tapered aeration attempts to supply air to match oxygen demand along the length of the tank. Step loading provides more uniform oxygen demand for an evenly distributed air supply. Identify the step-aeration activated sludge process diagram.
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Identify the Contact Stabilization Process without primary sedimentation.
Identify the Contact Stabilization Process without primary sedimentation.
Identify the Extended Aeration Process without primary sedimentation.
Identify 6 biological factors that can adversely affect settleability of activated sludge. 1. Species of dominant microorganisms (filamentous) 2. Ineffective biological flocculation 3. Denitrification in final clarifier (floating solids) 4. Excessive volumetric and food/microorganisms loadings 5. Mixed-liquor suspended-solids concentration 6. Unsteady-state conditions (nonuniform feed rate and discontinuous wasting of excess activated sludge). Identify 5 chemical factors that can adversely affect settleability of activated sludge. 1. Lack of nutrients 2. Presence of toxins 3. Kinds of organic matter 4. Insufficient aeration 5. Low temperature Identify 2 physical factors that can adversely affect settleability of activated sludge. 1. Excessive agitation during aeration resulting in shearing of floc. 2. Ineffective final clarification: inadequate rate of return sludge, excessive overflow rate or solids loading, or hydraulic turbulence.
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21. Calculate the specific gravity of solid matter in a sludge based on equation 13.1 as follows: Eq. 13.1: Ws/Ssy = Wf/Sfy + Wv/Svy, where Ws = weight of dry solids, lb., Ss = specific gravity of solids, y = unit weight of water, lb/cu.ft, Wf = weight of fixed solids (nonvolatile), lb.,
21. Calculate the specific gravity of solid matter in a sludge based on equation 13.1 as follows: Eq. 13.1: Ws/Ssy = Wf/Sfy + Wv/Svy, where Ws = weight of dry solids, lb., Ss = specific gravity of solids, y = unit weight of water, lb/cu.ft, Wf = weight of fixed solids (nonvolatile), lb., Sf = specific gravity of fixed solids, Wv = weight of volatile solids, lb., Sv=specific gravity of volatile solids. Example: Consider a waste biological sludge of 10% solids with a volatile fraction of 70%. Their specific gravity can be estimated using the above equation by assuming values of 2.5 for the fixed matter and 1.0 for the volatile residue. Solution: 1.00/Ss = 0.30/2.5 + 0.70/1.0 = 0.82 Ss = 1/0.82 = 1.22 22. Calculate the specific gravity of the wet sludge by Eq. 13.2 as follows: S = Ww + Ws/(Ww/1.00) + (Ws/Ss), where S = specific gravity of wet sludge; Ww = weight of water, lb; Ws = weight of dry solids, lb.; Ss = specific gravity of dry solids Example: Determine the specific gravity of a wet sludge of 10% solids. S = ___90 + 10________ = 1.02 (90/1.00) + (10/1.22) 23. Estimate the quantity of sludge produced by a trickling filter plant treating 1.0 mgd of domestic wastewater. Assume a suspended solids concentration of 220 mg/L in the raw wastewater, a solids content in the sludge equivalent to 90% removal and a sludge of 5.0% concentration withdrawn from the settling tanks. Solution: Solids in the sludge = 1.0 X 8.34 X 220 X 0.90 = 1650 lbs/day Volume of Sludge (using Eq. 13.3) = 1650/0.05 X 62.4 = 530 cu.ft/day 24. Calculate the total dry solids in lbs/day in primary settling with 50% settling of the solids for a wastewater plant treating 1 mgd of 200 mg/L suspended solids in unsettled wastewater. Solution: Wsp = f X SS X Q X 8.34, where Wsp = primary solids, lb of dry weight/day; f = fraction of suspended solids removed in primary settling; SS = suspended solids in unsettled wastewater, mg/L; Q = daily wastewater, mgd; 8.23 = conversion factor, lb/mg per mg/L Ws = 0.50 X 200 X 1.0 X 8.34 = 834 lbs/day 25. Calculate the total dry solids from an activated sludge process without primary sedimentation, in lbs/day of dry weight from a wastewater plant treating 1 mgd with a FM ratio of 0.05 and an influent BOD of 250 mg/L.
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Was, total dry solids from activated-sludge processes without primary sedimentation, lb/dayof dry weight = 2.0K X BOD X Q X 8.34, where K is the factor determined from figure 13.1 multiplied by 2.0. Was = 2.0(0.3) X 250 mg/L X 1.0 MGD X 8.34#/gallons = 1251 lbs dry weight solids/day
Was, total dry solids from activated-sludge processes without primary sedimentation, lb/dayof dry weight = 2.0K X BOD X Q X 8.34, where K is the factor determined from figure 13.1 multiplied by 2.0. Was = 2.0(0.3) X 250 mg/L X 1.0 MGD X 8.34#/gallons = 1251 lbs dry weight solids/day 26. A municipal wastewater plant effluent sample was filtered through a crucible and filter mat. The crucible plus the mat had a dry (tare) weight of 17.8216 grams. After filtration, the crucible mat and residue weighed 17.8374 gm. After weighing the crucible, mat, and residue were ignited at 600 degrees C. The crucible, mat, and ash weighed 17.8258 grams. If the sample was 50 ml. determine the suspended solids, the volatile solids, the fixed solids, and the percent volatile suspended solids. Solution: 17.8374 -17.8216 0.0158 gm = 15.8 mg TSS = (15.8 mg/50 ml)(1000 ml/L) = 316 mg/L 17,8258 -17.8216 0.0042 gm = 4.2 mg FSS = (4.2/50 ml) (1000 mg/L) = 84 mg/L VSS = TSS – FSS = 316 – 84 = 232 mg/L Percent VSS (232/316) 100% = 73.4% Label the parts of a gravity sludge thickener
27. Determine the diameter of a gravity thickener based on a solids loading of 10 lb/sq.ft/day and 1130 pounds of solids per day. Solution: Tank area required = 1130 lbs. per day / 10 lbs. per sq ft per day = 113 sq.ft. 2 Use equation: П D /4 = area = 113 sq. ft : Solve diameter of tank. 2 П D = 452 2 D = 143.87 D = 12.0 ft. 28. Determine the gallons per day of 1130 pounds per day of applied sludge at a concentration of 4.5%. Solution: Volume of applied sludge = 1130 lbs. per day / (0.045 sludge conc.X 62.4 lbs. per cu. ft of water) = 402 cu.ft of wet sludge. 402 cu. ft of wet sludge X 7.48 gallons per cu. ft. = 3007 gallons per day of sludge. 29
29. Determine the gpd of dilution water required to obtain an overflow rate of 400 gpd/sq.ft if 3007 gpd of applied sludge is sent to the gravity thickener and the area of the thickener is 113 square feet. Solution: Overflow rate of applied sludge = 3007 gpd/113 sq.ft. = 27 gpd/sq.ft.
29. Determine the gpd of dilution water required to obtain an overflow rate of 400 gpd/sq.ft if 3007 gpd of applied sludge is sent to the gravity thickener and the area of the thickener is 113 square feet. Solution: Overflow rate of applied sludge = 3007 gpd/113 sq.ft. = 27 gpd/sq.ft. Dilution flow required to attain 400 gpd/sq.ft = (400-27) 113 = 42,149 gpd ~ 42,000 gpd. 30. Determine the solids retention time in a gravity thickener with a surface area of 113 sq. feet from a trickling filter plant that contains 1130 pounds per day of solids assuming an underflow of 8.0% solids concentration and 95% solids capture at a sludge depth of 3.0 feet. Solution: Volume of thickened sludge = 1130 lbs per day X 0.95 / (8.0/100) 62.4 lbs. per cu. ft of water = 215 cu. Ft/ day Solids retention time = 3 feet depth X 113 sq. ft. area X 24 hours per day / 215 cu. ft per day = 38 hours. 31. A single-stage anaerobic digester has a capacity of 13,800 cu.ft., of which 10,600 cu.ft is below the landing brackets. The average raw-waste sludge solids fed to the digesters are 580 pounds of solids/day. Calculate the digester loading in pounds of volatile solids fed per cubic feet capacity below the landing brackets/day. Assume that 70% of the solids are volatile. Solution: Volatile solids loading = 0.70 volatile solids X 580 lbs. of volatile solids per day / 10,600 cu.ft digester volume = 0.038 lb of volatile solids fed / cu.ft./day 32. Calculate digester capacity using the following data based on the following equation and data: Vtotal = [(V1 + V2)/2 X T1] + [V2 X T2] where V = total digester capacity, cu.ft; V1 = volume of average daily raw-sludge feed, cuft/day; V2 = volume of daily digested sludge accumulation in tank, cu.ft/day; T1 = period required for digestion, days (approx. 25 days at temperature of 85 – 95 degrees F; T2 = period of digested sludge storage, days (normally 30-120 days). Problem Data: Average daily raw-sludge solids = 580 pounds Raw-sludge moisture content = 96% Digestion period = 30 days Solids reduction during digestion = 45% Digested-sludge moisture content = 94% Digested-sludge storage required = 90 days Solution: V1, volume of avg daily feed solids = 580avg daily raw solids/0.04 dry solids X62.4 lbs. per cu. ft = 232 cu.ft. V2, volume of daily digested sludge, cu. ft per day = 0.55 X 580/ 0.06 X 62.4 = 85 cu.ft per day Solution: Vtotal = [(232 cu ft + 85 cu ft)/ 2 (30 days digestion)] + [(85 cu. ft X 90 days digestion)] = 12,400 cu.ft. of digester volume Name three important parameters in nitrification kinetics. Temperature, pH, and dissolved-oxygen concentration
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What is the minimum dissolved oxygen level recommended in practice for nitrification to prevent reduced nitrification through the aeration tank? 2.0 mg/L is recommended in practice. Why is a long sludge age required in nitrification to prevent excessive loss of viable
What is the minimum dissolved oxygen level recommended in practice for nitrification to prevent reduced nitrification through the aeration tank? 2.0 mg/L is recommended in practice. Why is a long sludge age required in nitrification to prevent excessive loss of viable bacteria in continuous flow aeration systems? A long sludge age is required because the growth rate of viable bacteria must be rapid enough to replace microbes lost through sludge wasting and washout in the plant effluent. 33. Calculate the Mean Cell Residence Time in activated-sludge operation using equation: Equation: MLVSS X V/SSe X Qe + SSw X Qw Given: MLVSS = 1400 V = Volume of aeration tank = 1.5 million gallons SSe = Effluent suspended solids = 20 mg/L Qe = design flow = 5.0 mgd SSw = suspended solids in waste sludge, mg/l =8200 mg/L Qw =quantity of waste sludge, mgd= .026 mgd Solution: (1400 mlvss X 1.5 MG) / [(20 mg/L X 5.0 MGD) + 8200 mg/L X 0.026 MGD)] = 6.7 days 34. Calculate the Mean Cell Risidence Time for Nitrification given equation at 17 degrees C. 0.098(T-15) -1 Equation: µN = 0.47e where µN = maximum specific growth rate of Nitrosomonas, d e = base of Napierian logarithms, 2.718 T = temperature, centigrade Solution: 0.098(17-15) µN = 0.47(e) = 0.572 days 35. A rapid sand filter has a sand bed 30 inches in depth. Pertinent data are: specific gravity of the sand = 2.65, shape factor (Ø) = 0.75, porosity (ε) = 0.41, filtration rate = 2.25 gpm/sq.ft, and operating temperature = 50 degrees F. The sieve analysis of the sand is given in attached table. Determine the head loss for the clean filter bed in a stratified condition using the Rose equation: HL= (1.067/Ø) (D/g) (Va2/ε4) (Σ CDX/d). D=feet, Va = ft/sec, g = gravity or 32.17 ft/sec2
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2
4
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HL = (1.067/0.75)(2.5ft/32.17 ft)(0.0050133 ft/sec) X (1/0.41) (35,700.0 ft ) = 3.51 feet
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4
-1
HL = (1.067/0.75)(2.5ft/32.17 ft)(0.0050133 ft/sec) X (1/0.41) (35,700.0 ft ) = 3.51 feet
36. Filter Loading Rate = gpm/sq ft filter area Example 1: A rapid sand filter is 10 feet wide and 15 feet long. If the flow through the filter is 450,000 gpd, what is the filter loading rate in gallons per minute per square foot? Filter loading rate = gpm/sq.ft filter area Gpm/sq.ft = 450000gallons/1440 minutes/ (10 ft X 15 ft) = 2.08 gpm/sq.ft. 37. Filter loading rate = Inches of water fall / minute Example 2: How many inches drop per minute corresponds to a filter loading rate of 2.6 gpm/sq. ft? Inches drop/minute = 2.6 gpm/7.48 gals/ cu.ft. (12 inches/foot) = 4.17 inches/minute 38. What is the filter loading rate in gpm/sq. ft. of a sand filter in which the water level dropped 15 inches in 3minutes after the influent valve was closed? 15 inches/3 minutes = 5 inches per minute 5 in/min / 12 (7.48 gals/cu.ft) = 3.12 gpm/sq.ft. 39. A rapid sand filter is 15 feet wide and 40 feet long. If the flow through the filter is 1.6 mgd, what is the filter loading rate in gpm/sq. ft? 1,600,000/1440 minutes/day = 1111 gpm Filter loading rate = gpm/sq ft area = 1111 gpm / 15 X 40 = 1.85 gpm/sq.ft. 40. Filter backwash rate = gpm/sq.ft.area A rapid sand filter is 10 feet wide and 16 feet long. If backwash water is flowing upward at a rate of 4,950,000 gpd, what is the backwash rate in gallons per minute per square foot? Filter backwash rate = gpm/sq.ft area = 4,950,000 / 1440 minutes/day / 10 X 16 = 21.48 gpm/sq.ft. 41. A mixed media filter is 25 feet wide and 32 feet long. If the filter receives a backwash flow of 17,300,000 gpd, what is the filter backwash flow rate in gpm/sq.ft. Filter backwash rate = gpm/sq.ft area = 17,300,000 / 1440 min/day / 25 X 32 = 15.02 gpm/sq.ft. 42. How many inches rise per minute corresponds to a filter backwash rate of 18 gpm/sq.ft.? 18 gpm/sq.ft /7.48gal/cu.ft (12 inches/foot) = 28.9 inches per minute. 32
43. A rapid sand filter is 15 ft wide and 30 feet long. If the backwash water flow rate is 12 mgd, determine the filter backwash rate in gpm/sq.ft. Filter backwash rate, gpm/sq.ft = 12,000,000/1440 minutes/day / 15 ft X 30 ft = 18.52 gpm/sq.ft. 44. During backwashing of a rapid sand filter, the operator measured the backwash flow
43. A rapid sand filter is 15 ft wide and 30 feet long. If the backwash water flow rate is 12 mgd, determine the filter backwash rate in gpm/sq.ft. Filter backwash rate, gpm/sq.ft = 12,000,000/1440 minutes/day / 15 ft X 30 ft = 18.52 gpm/sq.ft. 44. During backwashing of a rapid sand filter, the operator measured the backwash flow rate to be 27 inches/minute. Express this filter backwash rate in gallons per minute per square foot. Gpm/sq. ft. = 27 inches/min / 12 inches/foot X 7.48 gal/cu.ft = 16.88 gpm/sq.ft. 45. Determine the design bed volume , cu.ft., for a flow rate of 40,000 gpd with an experimental pilot column result of 1.67 BV/hour. BV, cu. ft = Q/QB , where design flow rate = Q; Q B = Breakthrough volume from an experimental pilot column Design BV, cu. ft. = 40000 X hour X _cu.ft.___ = 133 cu. ft. 24 hrs 1.67 7.48gal 46. If the mass of carbon = 25 lb/cu.ft, how much is required for the design bed volume of 133 cu. ft? M, lb = (BV)(ρs) where BV, cu. ft and ρs = mass of carbon in lb/cu. ft. M, lb = (133)(25) = 3330 lb. 47. If the volume treated at the allowable breakthrough was 549.5 gallons, what was the number of gallons per pound treated if the pilot column’s mass of carbo n was 6.564 pounds? Vb = Vb/M = 549.5/ 6.564 = 83.71 gal/lb. 48. How many pounds of carbon per hour were exhausted in treating 40,000 gallons per day if each pound of carbon required 83.71 gallons? Mt = Q/Vb, Mt = pounds exhausted per hour; Q = flow rate; Vb = gallons per pound treated Mt = 40000/24 hr (lb/83.71) = 19.91 lb/hour 49. How many days would it take for the breakthrough time if the bed contained 3330 pounds of carbon and the bed was exhausted at the rate of 19.91 pounds per hour? T = M/Mt = (3330) (hr/19.91 lb) = 167 hours or 6.96 days. 50. A pilot column breakthrough test has been performed. Pertinent design data are inside diameter = 3.75 inches, length = 41 inches, mass of carbon = 2980 grams or 6.564 lbs., liquid flow rate = 17.42 liters per hour or 1.00 gpm/sq.ft., and packed carbon density = 25 lbs/cu.ft (401 gm/liter. The breakthrough of the carbon occurred after 295 gallons had passed through the column. Determine: 1). the liquid flowrate in bed volumes per hour and 2. the volume of liquid treated per unit mass of carbon ----in other words, the gal/lb at an allowable breakthrough amount. 1. BV, = area X length X (cu ft/ 1728 cu.in.)(7.48 gals/cu.ft)(3.785liters/gal) = liters 2 BV = (π(3.75 in) /4 X 41)(cu.ft/1728 cu.in)(7.48/cu.ft)(3.785 liters/gal) = 7.4193 liters BV = Q/Qb or Qb = Q/BV = 17.42liter/hr/7.4193 liters = 2.35 BV/hour 2. Vb = 295 gal/6.564 lb = 44.9 gal/lb.
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51. A water is to be softened for an industrial water supply. The calcium content is 107 mg/L and the magnesium content is 18 mg/L. The allowable breakthrough, Ca, is 5%, where Co is the hardness of the untreated water. A pilot column 4 inches in diameter and containing 3 feet of resin has been operated at a flowrate of 0.59 gal/hour to obtain
51. A water is to be softened for an industrial water supply. The calcium content is 107 mg/L and the magnesium content is 18 mg/L. The allowable breakthrough, Ca, is 5%, where Co is the hardness of the untreated water. A pilot column 4 inches in diameter and containing 3 feet of resin has been operated at a flowrate of 0.59 gal/hour to obtain breakthrough data. The resin has a specific weight of 44lb/cu.ft., and the design flowrate is 150,000 gal/day. Using the scale-up design approach, determine: 1. The pilot column volume(cu.ft) 2. The pilot column mass(lb) 3. The pilot column BV/hour 4. The full scale up volume(cu.ft) 5. The full scale up mass of resin(lb) 6. The amount of millequivalents of total hardness 7. If 740 gallons is obtained from the breakthrough curve, how many gallons per pound are obtained for the pilot column? 8. How many pounds per hour would the full scale column unit operate based on the pilot scale unit’s gallons per pound rate. 9. What is the design life of the column? 2 1. V = (π(4 in/12 in/ft)] /4) (3 feet) = 0.2618 cu. ft. 2. M = (0.2618 cu.ft)[44 lb/cu.ft.] = 11.519 lb. 3. Qb = (0.59 gal/hr)(BV/0.2618 cu.ft.)(cu.ft/7.48 gal) = 0.301 BV/hr. 4. BV = Q/Qb = (150000 gal/day)(1 day/24 hrs)(1 hr/0.301 BV)(cu.ft/7.48 gal) = 2780 cu. ft. 5. lb resin = (2780 cu. ft.)(44 lb/cu.ft) = 122000 lb. 6. Total hardness = 107/20 + 18/12 = 6.85 meq/l 7. Vb = 740 gal/11.519 lb = 64.24 gal/lb. 8. Mt = (150000 gal/day) (1 day/24 hours)/ Vb = 97.3 lb/hour 9. t = 122000 lb/ 97.3 lb/hour = 1250 hours, or 52.1 days 52. A soluble organic industrial wastewater with a COD of 2590 mg/L is to be treated by completely mixed activated sludge plant. A treatability study has been made, and the following data obtained: Design influent flow = 4.2 MGD, MLSS = 3000 mg/L, MLVSS = 86.7% of the MLSS, K = 0.236 liter/ (gm MLVss-hr), reaction is pseudo-first order, nondegradable COD = 16 mg/L, and effluent COD = 150 mg/L, Determine: a. Required reaction time, Ө, hours b. Required Reactor basin volume, cubic feet. Solution: a. Si = 2590 – 16 = 2574 mg/L St = 150 – 16 = 134 mg/l MLVSS = (0.867)(3000) = 2600 mg/L: = 2.60 gm/L Ө= Si – St = 2574 - 134 = 28.7 hours (Required Reaction Time) K(MLVSS)(S t) (0.236)(2.60)(134) 3 3 b. V = QӨ = (4,200,000 gals)(29.7 hrs.)(ft /7.48gal) = 695,000 ft
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