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

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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

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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

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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

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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

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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

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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

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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

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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)

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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.

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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

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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

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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

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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

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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

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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

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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]

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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.

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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%

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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.

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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.

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Rapid mixing by Toor (1969)

• tk = reaction half-time• tm = time required to achieve COV < 5%

tktm

� 1 sloww 1 moderate� 1 fast

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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)

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