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CEE 633 Physical & Chemical TreatmentLecture 13. Ch. 9 Coagulation, Mixing and Flocculation
(cont’d)
Prof. Albert S. Kim
Civil and Environmental Engineering, University of Hawaii at Manoa
Tuesday, October 2, 2012
1 / 25
Stoichiometry of metal ion coagulants
Al2 (SO4)3 ⇒ Al3+ + 3H2O Al (OH)3 (am) ↓+ 3H+
FeCl3 ⇒ Fe3+ + 3H2O Fe (OH)3 (am) ↓+ 3H+
“am” = amorphous solid with much higher solubility than trueprecipitate (s).
Al2(SO4)3 · 14H2O+ 6(HCO3)− → 2Al(OH)3(am) ↓ +3SO2−
4 + 6CO2 + 14H2O
Fe2(SO4)3 · 9H2O+ 6(HCO3)− → 2Fe(OH)3(am) ↓ +3SO2−
4 + 6CO2 + 9H2O
FeCl3 · 6H2O+ 3(HCO3)− → 2Fe(OH)3(am) ↓ +3Cl− + 3CO2 + 6H2O
2 / 25
Alkalinity Control
Soda (NaOH), Lime (Ca(OH)2), or Soda ash (Na2CO3)
Al2(SO4)3 · 14H2O + 6NaOH→ 2Al(OH)3(am) ↓ +3Na2SO4 + 14H2O
Al2(SO4)3 · 14H2O + 3Ca(OH)2 → 2Al(OH)3(am) ↓ +3CaSO4 + 14H2O
See Example 9-3, p. 680
3 / 25
Problem 9-10
Added 60mg/L of [FeCl3 ·H2O]Question: produced Fe(OH)3 and CO2 & consumed alkalinity ?Using Eq. 9-17
FeCl3 · 6H2O + 3HCO−3 → Fe(OH)3(am) ↓ +3Cl− + 3CO2 + 6H2O
4 / 25
Cont.
1. Fe(OH)3 produced
= (60 mg/L of FeCl3 ·H2O)
×[
1 mole of FeCl3 · 6H2O
270.5 g of FeCl3 · 6H2O
]×[
1 mole of Fe(OH)31 mole of FeCl3 · 6H2O
]×[
107 g of Fe(OH)31 mole of Fe(OH)3
]=
(60
mg
Lof FeCl3 · 6H2O
)× 107 g of Fe(OH)3
270.5 g of FeCl3 · 6H2O
= 23.7mg/L of Fe(OH)3
5 / 25
Cont.2. CO2 generated
= (60mg/L FeCl3 · 6H2O)× 1mole
270.5g
×3mole CO2
1mole× 44g CO2
1mole CO2
= 60mg
L× 44× 3
270.5= 29.3mg/L CO2
3. Alkalinity consumed
= (60mg/L FeCl3 · 6H2O)
× 1mole
270.5g× 3mole HCO−
3
1mole
× 1eq Alk1mole of HCO−
3
× 50g CaCO3
1eq Alk= 33.3g/L as CaCO3
6 / 25
Basicity
is defined as the degree to which the hydrogen ions (that wouldbe released by hydrolysis) are preneutralized.• Generated acid (H+) is neutralized with base (OH−)
Basicity% =[OH−]
[M ]ZM× 100
where ZM = charge on metal ions, M = metal molarity• Commercial prehydrolyzed Alum salts:
Ala(OH)b(Cl)c(SO4)d
7 / 25
Problem 9-11(c)
Basicity of Al15O6(OH)24SO4
[OH]
M=
[OH] + 2[O]
[M]
=24 + 2× 6
15= 2.4
because each mole of oxygen will neutralize two moles ofhydrogen:
O6 ⇒ (H2O)6 ⇒ 12H+ → H12
Basicity:2.4
3= 80% bec
because Al+3
8 / 25
Effects of NOM on coagulation process
Dissolved (DOC)← 0.45µm(filtration)→ particulate
• Presence of NOM increases required coagulant dosages.• DOC remaining after coagulation:
1. adsorbable: adsorbed on metal coagulant, suspended insolution, [C]eq=Equilibrirum DOC concentration in water.(Eq. 9-24)
2. non-adsorbable: not adsorbed & present in solution. (Eq.9-22)
9 / 25
DOC concentration estimation• The nonadsorable DOC (mg/L)
DOC = [K1 (SUVA)rawwater +K2]×DOCinitial
where K1 and K2 are empirical constants,DOCinitial is theinitial concentration of DOC, and SUVA is the specific UVabsorbance of raw water. [L/mg ·m]
SUVA =
(102cm
1m
)(UV254/cm
DOC
)raw water
• The amount of DOC adsorbed using Langmuir isotherm:
q =QMb [C]eq1 + b [C]eq
where q is the equilibrium amount of DOC adsorbed,mg-DOC/g-adsorbent (adsorbent is floc), QM is themonolayer coverage, b is Langmuir equilibirum constant,and [C]eq is the equilibirum DOC concentration.
10 / 25
Mass BalanceThe mass of DOC (mg) adsorbed per meq of coagulant ot themount adsorbed is equated to the Langmuir isotherm:
adsorable − equilibrium
added coagulant
=[1− (SUVA)rawwaterK1 −K2] DOCinitial − [C]eq
M
=
(x3pH3 + x2pH
2 + x1pH)b [C]eq
1 + b [C]eq
where M is coagulant aded and metal hydorxide solid formed,[mM], pH is the coagulation pH, and b is Langmuir equilibriumconstant for adsorbable DOC to hydroxide surface, L/mg-DOC.
[C]eq =−(MB + 1−Ab) +
√(MB + 1−Ab)2 + 4bA
2b
11 / 25
Problem 9-14.
• DOC removal = ?• [C]eq = equilibrium DOC conc. in water• Alum = 0.05− 0.3mM (let’s take 0.05 mM)• DOC = 5mg/L
• pH = 7
• SUVA = 2.75L/mg ·m (measured)• M = coagulant dose [mM]
See Table 9-9 p. 692
12 / 25
Cont.• QM is the monolayer coverage equal to
= x3pH3 + x2pH2 + x1pH
[mg −DOC
g − adsorbent
]For Alum, x3 = 4.91, x2 = −74.2, x1 = 284.7 )
QM = 4.91× 73 − 74.2× 72 + 284.7
= 36.33[mg −DOC] / [g − adsorbent]
• b [L/mg −DOC] is Langmuir equilibrium constant foradsorbable DOC to hydroxyde surface (monolayer) = 0.147L/mg-DOC.
• B = QMb
B = 36.33mg −DOC
g − adsorbent× 0.147
L
mg −DOC
= 5.34L
g − adsorbent
13 / 25
Cont.
A = [1− {(SUVA)raw water ×K1 +K2}] DOCinitial
= (1− {(2.75L/mg ·m) · (−0.075) + 0.56}) (5mg/L)
= 3.23mg/L
(MB + 1−Ab) = 0.05× 5.34 + 1− 3.23× 0.147
= 0.81
[C]eq =−0.81 +
√(−0.81)2 + 4 · (0.147)(3.23)
2(0.147)
=0.7885
2× 0.147= 2.68 mg/L
14 / 25
Cont.
Non-adsorbable DOC = [K1 (SUV A)raw water +K2]DOCinitial
= [(−0.075)(2.75) + 0.56] · 5mg/L
= 1.77mg/L
DOC removed = initial− Equil. conc.− nonadsorbable
= 5− 2.68− 1.77
= 0.55mg/L
15 / 25
9-5 Mixing theory and practiceMixing purposes
1. to promote particle contacts (in flocculation)2. to uniformly blend chemicals in water
Virtually all water treatment processes take place in turbulentflow[Reynold’s number]
D vρ, µ
ν=µ/ρ
Re =Dvρ
µ=Dv
ν
where ν = µ/ρ is the kinematic viscosity with the same unit ofdiffusivity 16 / 25
Fraction factor
defined as
f =D
2ρv2|∆p|L
• Laminar flow (Re < 2100)
f =16
Re
• Turbulent flow (Re > 4000)
1√f
= 4.0 log10
(Re√f)− 0.4
called Von Karman-Nikuradse equation. Nonlinear w.r.t. f
17 / 25
Turbulent flow• Characteristics
1. consisting of a cascade of eddies• cascade of energy from large eddies to small eddies.
2. As large eddies becomes smaller, inertial force is overcomeby viscous force.
• Mixing and scale of turbulence• The smallest eddy (size) η
η =
(ν3
ε
)1/4
where ν is kinematic viscosity, ε is energy dissipation rateof a point of interest, J/kg · s, and ε is the overall averagerate of energy dissipation, which is
ε ≈ P
M=
power of mixing input to entire mixing vessel [J/s]mass of water [kg]
18 / 25
Camp-Stein Root Mean Square (RMS)
Global velocity gradient, G, i.e. shear rate generally defined as
γ =∂v
∂y
and
G =
√P
µV⇒ effective shear rate (widely used)
where V is the volume of the mixing vessel.
19 / 25
Uniformity and time series in mixing (p.702)
• At a given point in a stream, measure C n-times with acertain interval.
• Standard deviation below is normalized to the averageconcentration.
σ =
√√√√ 1
n− 1
n∑i=1
(Ci − C
)2where Ci is concentration of ith measurement and C is anaverage concentration.
• The coefficient of variation is defined as
COV =σ
C× 100%
20 / 25
Intensity of segregation Is by Danckwerts (1952)
Is =
(σmσu
)2
• σm = standard deviation of mixed stream• σu = standard deviation between two streams in unmixed
condition
Is =
{0 completely mixed1 completely unmixed
Note: A square of standard deviation is variance.
21 / 25
Standard deviation of unmixed streams1. When a chemical “a” is being added to a stream of water
Qa
Ca
Qw, C=0
The volume fraction of the chemical solution being added to theoriginal solution is
Xa =Qa
Qw +Qa
by definition the volume fraction of the water is Xw = 1−Xα,then the standard deviation of concentrations before mixing(expressed as volume fraction) is for a large number ofsampling
σu ∼=√Xa
(1−Xa
)22 / 25
2. When designing systems to dose chemicals
Qa
Ca
Qw, C=0
Cdose
Q
QaCa = QCdose
Xa =Qa
Qw +Qa=QaQwCdose
Ca
where Xa is the volume fraction of chemical “a” in an unmixedsystem.
23 / 25
Rapid mixing by Toor (1969)
• tk = reaction half-time• tm = time required to achieve COV < 5%
tktm
� 1 sloww 1 moderate� 1 fast
24 / 25
Time scale by Crank (1979)The molecular diffusion time scale for transport in or out ofsmall eddies
td =3R2
4Dl
where R = 12η is the radius of eddy, Dl is the liquid diffusivity of
chemical molecules (∼ 1× 10−9m2/s)
Gtd =
√P
µV· 3R2
4Dl
=
√P
µV· 3
4Dl
η2
4=
√P
µV· 3
16Dl
(ν3
ε
)12/4
=
√Pν3
µV ε· 3
16Dl=
1
4
(3ν
4Dl
)= O
(103)
25 / 25