1
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
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
πD2 = 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)(ρs) 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)(3.785liters/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/nEc where I= current in amperes, F=Faraday’s constant, Ec = current efficiency
as a fraction, Q = solution flow rate, N = normality of the solution, n = number of cells
Power = RI2 where R = resistance, I = current in amperes
Faraday Constant = 96,500 ampere-seconds per gram equivalent
2
Probability equation of True recurrence interval: Z = 1 – (1-1/TR)N , where Z is the
probability of event will be equal or exceeded during a specified design period; TR is the
recurrence interval of the event; N is the design period in years
Kf = 528Qlog(r2/r1) / m(h2-h1), for 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 – h1 = 10 ft – 1 ft. = 9 feet
Kf = 1055Q log(r2/r1) / h22 – h1
2 , for unconfined aquifer
TR = V/Q where TR is the theoretical residence time of a process reator, V = volume; Q = flow,
Surface Overflow rate = gpd / surface area of 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
G = (P/µV)1/2
where G = velocity gradient, fps/ft; P = power input, ft-lb/s; µ = absolute
viscosity, lb . s/ft
2; V = volume, ft
3
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 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 and 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:
h/l = Jv(1-ε)2VS
2 / gε
3d
2 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
v= kinematic viscosity (µ/p), ft2/s, (m
2/s)
g = acceleration of gravity, ft/s2
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.
Calculate 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.
Sodium alkalinity = total alkalinity – total hardness
NCH = Total Hardness – Total Alkalinity ( If Mg Alkalinity present then no Ca NCH)
3
Water
1. Groundwater is used for a water supply. It is taken from the ground at 25 degrees
C. The initial 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 CO2 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 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 (HCO3
-) form. Therefore, the equilibrium
expression for the first ionization of carbonic acid may be used.
CO2 + H2O → H2CO3 ↔ H+ + HCO3
-
K1 = [H+][HCO3
-] = 4.45 X 10
-7
[H2CO3]
The coefficients of CO2 and H2CO3 are 1. The number of moles of each compound is the same.
One mole of CO2 produces one mole of H2CO3.
[H2CO3- = [CO2] = 5.6 mg/L / (100 g/mol)(1000 mg/g) = 5.6 X 10
-5
[HCO3-] = 200 mg/L / 100 g/mol (1000 mg/g) = 1.246 X 10
-8
Solve for the hydrogen ion concentration.
[H+] = K1(H2CO3] = K1[H2CO3] = (4.45 X 10
-7)(5.6 X 10
-8) = 1.246 X 10
-8)
[HCO3-] [CO2]
pH = -log[H+] = - log(1.246 X 10
-8) = 7.9
2. The solids concentration of a stream water sample is to be determined. The total
solids concentration is determined by placing a portion 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 C. The following masses are recorded.
Mass of empty dish: 50.326 g
Mass of dish and 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 – 50.326 g / 1 g/mL = 68.1 mL
TS = (50.437 g – 50.326 g) (1000 mg/L) (1000 mg/L) = 1630 mg/L
68.1 ml
b). The total volatile solids concentration is:
TVS = (50.437 g – 50.383 g) X (1000 mg/g (1000 ml/L) = 793 mg/L
68.1 ml
c). The total fixed solids concentration is
TFS = 1630 mg/L – 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.
The following masses are recorded.
Volume of sample = 25 mL.
Mass of empty 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) – (51.494 g) X (1000 mg/g)(1000mg/L) = 1200 mg/L
25 ml
b. VDS = (51.524 g – 51.506 g) (1000 mg/L)(1000 mL/L = 720 mg/L
25 ml
c. FDS = 1200 mg/L – 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 rate of a rectangular clarifier that is 90 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.
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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 of the mixed reactor.
Solution: T10 = C(tracer, mg/L) / Co(initial dose, mg/L) = 0.10. The T10 will fall between 12
and 15 minutes. By interpolation, T10 = 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
gradient if µ = 2.735 X 10-5
lb-s/ft2.
Solution: G = (P/µV)1/2
where G = velocity gradient(power to the water), fps/ft; P = power
input, ft-lb/s; µ = absolute viscosity, lb . s/ft
2; V = volume, ft
3
6
G = (3960 / 2.735 X 10-5
X 16 X 64 X 100) ½
= 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 26 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.
h/l = Jv(1-ε)2VS
2 / gε
3d
2 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
v= kinematic viscosity (µ/p), ft2/s, (m
2/s)
g = acceleration of gravity, ft/s2
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.
h/l = 6 X 1.410 X 10-5
(1 – 0.42)20.00557 X (6.0)
2 / 32.2 X (0.42)
3 X (1.6 X 10
-3)2 = 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
Ca2+
40 20 2.00
Mg2+
10 12.2 0.82
Na+
11.7 23.0 0.51
K+
7.0 39.1 0.18
Total Cations = 3.51
HCO3-
110 61.0 1.80
SO42-
67.2 48.0 1.40
Cl-
11.0 35.5 0.31
Total Anions= 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 OH-
Alkalinity as CaCO3
CO3-
Alkalinity as CaCO3
HCO3-
Alkalinity as
CaCO3
P = 0 0 0 T
P<1/2 T 0 2P T-2P
P=1/2 T 0 2P 0
P>1/2 T 2P – T 2(T- P) 0
P = T T 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
OH- Alkalinity as
CaCO3
CO3- Alkalinity as
CaCO3
HCO3- Alkalinity as
CaCO3
P.Alk = 0, T. Alk. =100 0 0 100
P Alk = 40, T.Alk.=100 0 80 20
P Alk = 50, T.Alk. =100 0 100 0
P Alk=60, T Alk.=100 20 80 0
P = 100, T Alk.=100 100 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 = meq.
Meqs = (102 mg/L) / 23 mg/meq = 4.43 meqs/L
Ca+2
: 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+]
[H+] = 10
-7.6 moles /L
[H+][OH
-] = 10
-14
[10-7.6
][OH-] = 10
-14
[OH-] = 10
-6.4 moles/L
In water coagulation with alum (aluminum sulfate) and bicarbonate alkalinity, the
simplified reaction is: Al2(SO4)3 . 14 H2O + 6 HCO3 → 2Al(OH)3 ↓ 3SO4
-2 + 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
9
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 hardness
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 alkalinity, calcium
non-carbonate hardness, and the magnesium non-carbonate hardness from the given water
analyses.
AAnnaallyysseess WWaatteerr ##11 WWaatteerr ##22 WWaatteerr ##33
TToottaall HHaarrddnneessss 330000 330000 330000
CCaallcciiuumm HHaarrddnneessss 220000 220000 220000
MMgg HHaarrddnneessss 110000 110000 110000
TToottaall AAllkkaalliinniittyy 115500 225500 335500
IInntteerrpprreettaattiioonnss WWaatteerr ##11 WWaatteerr ##22 WWaatteerr ##33
CCaallcciiuumm AAllkkaalliinniittyy 115500 220000 220000
MMgg.. AAllkkaalliinniittyy NNoonnee 5500 110000
SSooddiiuumm AAllkkaalliinniittyy NNoonnee NNoonnee 5500
CCaa NN..CC.. HHaarrddnneessss 5500 NNoonnee NNoonnee
MMgg.. NN..CC.. HHaarrddnneessss 110000 5500 nnoonnee
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.
10
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 CaCO3
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, CO2: 19 mg/L X 100/44 or 2.27 = 43.13 mg/L as CaCO3
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 CaCO3 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 CaCO3 of non-carbonate
hardness.
11
29. If the town’s water supply has the following ionic concentrations 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
(0.5 lbm/1000 gr)(909.6 lbm/MG)(7000 gr/lbm) = 3.18 X 103 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%.
12
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
Month Inflow
MG
Draft
MG
Sum of
Inflows (MG)
Defieciency
MG
Cumulative
Deficiency(MG)
Jan 37.2 53.1 37.2 15.9 15.9
Feb 64.8 53.1 102 -11.7 0
Mar 108 53.1 210 -54.9 0
Apr 12 53.1 222 41.1 41.1
May 8.4 53.1 230.4 44.7 85.8
June 9.6 53.1 240 43.5 129.3
July 2.4 53.1 242.4 50.7 180
Aug 33.6 53.1 276 19.5 199.5
Sept 50.4 53.1 326.4 2.7 202.2
Oct 129.6 53.1 456 -76.5 0
Nov 117.6 53.1 573.6 -64.5 0
Dec. 26.4 53.1 600 26.7 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.
Answer: Z = 1 – (1-1/TR)N , where Z is the probability of event will be equal or exceeded
during a specified design period; TR is the recurrence interval of the event; N is the
design period in years
Example: Z = 1 – (1- 1/10)10
= 1 – (1-0.1)10
= 1 – (0.9)10
= 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 1.0 1.0 1.0 1.0
10 0.1 0.65 0.995 ~1.0
50 0.02 0.18 0.64 0.87
100 0.01 0.10 0.40 0.63
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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.
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 ac-
ft.
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:
14
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?
Equation: Kf = 1055Q log(r2/r1) / h22 – h1
2 , for unconfined aquifer
Solve for Q = Kf (h22 – h1
2) / 1055 log(r2/r1)
Log(r2/r1) = log (235/100) = 0.37107
h2 = 100 – 21 = 79 feet
h1 = 100 – 22,2 = 77.8 feet
Q = 1320 ( 792 – 77.8
2) / 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.
15
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
16
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.
17
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
HL = [(Vb2 – Va
2) / 2g] (1/0.7) = [(3.0)
2 – (2.31)
2/ (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.
18
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.
19
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?
20
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.
21
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
22
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
. Determine:
a). The ultimate first stage BOD in mg/L, Lo.
b). The ratio, y/Lo as a fraction and is a percent.
Solution:
a). y = Lo (1 – e-kt
)
10 = Lo (1- e-0.23x5
)
Lo = 14.6 mg/L
b). y/Lo = 10/14.6 = 68.5%
Label the diagr6am of population dynamics in anaerobic digestion.
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 ponds.
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
23
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 CnH2nOn.
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.
24
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 single-
stage trickling filter given the following information:
NRC formula = E20 = 100 / 1 + 0.0561 (w/VF)0.5
w/V = BOD loading, 23.5 lbs/1000 cu.ft./day
F, recirculation factor = 1.36
E20 = 100/ 1 + 0.0561(23.5/1.36)0.5
= 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%
25
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
26
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.
27
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.
28
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.
29
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.
Use equation: П D2/4 = area = 113 sq. ft : Solve diameter of tank.
П D2 = 452
D2 = 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.
30
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
31
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.
Equation: µN = 0.47e0.098(T-15)
where µN = maximum specific growth rate of Nitrosomonas, d-1
e = base of Napierian logarithms, 2.718
T = temperature, centigrade
Solution:
µN = 0.47(e)0.098(17-15)
= 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/ε
4) (Σ CDX/d). D=feet, Va = ft/sec, g = gravity or 32.17 ft/sec
2
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HL = (1.067/0.75)(2.5ft/32.17 ft)(0.0050133 ft/sec)2 X (1/0.41)
4 (35,700.0 ft
-1) = 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.
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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; QB = 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 carbon 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
BV = (π(3.75 in)2/4 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
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?
1. V = (π(4 in/12 in/ft)]2/4) (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, non-
degradable 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)(St) (0.236)(2.60)(134)
b. V = QӨ = (4,200,000 gals)(29.7 hrs.)(ft3/7.48gal) = 695,000 ft
3
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