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Monroe L. Weber-Shirk
School of Civil and Environmental Engineering
Water Treatment
Reflections
What are the two broad tasks of environmental engineers?
What is the connection between the broad tasks of environmental engineers and building a water treatment plant?
Why may the water need to be changed/treated?
Simple Sorting
Goal: clean waterSource: (contaminated) surface waterSolution: separate contaminants from waterHow?
Where are we going?
particlesdissolved chemicals
pathogens
Unit processes* designed to remove ___________ remove __________ ___________ inactivate __________
*Unit process: a process that is used in similar ways in many different applications sedimentation filtration ...
Unit Processes Designed to Remove Particulate Matter
ScreeningSedimentationCoagulation/flocculationFiltration
slow sand filtersrapid sand filtersdiatomaceous earth filtersmembrane filters
Conventional Surface Water Treatment
Screening
Coagulation
Flocculation
Sedimentation
Filtration
Disinfection
Storage
Distribution
Raw water
AlumPolymers Cl2
sludge
sludge
sludge
Screening
Removes large solids logsbranches rags fish
Simple processmay incorporate a mechanized trash
removal system Protects pumps and pipes in WTP
Sedimentation
the oldest form of water treatmentuses gravity to separate particles from wateroften follows coagulation and flocculationoccurs in NYC’s __________reservoirs
Sedimentation: Effect of the particle concentration
Dilute suspensionsParticles act independently
Concentrated suspensionsParticle-particle interactions are significantParticles may collide and stick together
(form flocs)Particle flocs may settle more quicklyParticle-particle forces may prevent further
consolidation
How fast do particles fall in dilute suspensions?
GravityFluid drag
What are the important parameters?Initial conditionsAfter falling for some time...
What are the important forces?___________________
projected
Sedimentation:Particle Terminal Fall Velocity
maF 0 WFF bd
p p g
2
2t
wPDdVACF W
dF
bF
p wgr" velocity terminalparticletcoefficien drag
gravity todueon acceleratidensitywater
density particle
area sectional cross particle
volumeparticle
t
D
w
p
p
p
VCgρ
ρ
A
_______W
________bF =
Identify forces
Particle Terminal Fall Velocity (continued)
bd FWF
gVAC wppt
wPD )(2
2
wPD
wppt
AC
gV
)(2 2
dAp
p
3
2
w
wp
D
tC
gdV
3
4 2 ( )4 3
p wt
D w
gdVC
r rr-=
Force balance (zero acceleration)
3
3
4 rp 2rAp
We haven’t yet assumed a shape
Assume a _______sphere
0.1
1
10
100
1000
Reynolds Number
Dra
g C
oeffi
cien
tDrag Coefficient on a Sphere
laminar
Re tV d
turbulentturbulent boundary
Stokes Law
Drag Coefficient:Equations
Laminar flow Re < 1
Transitional flow 1 < Re < 104
Fully turbulent flow Re > 104
24ReDC
Re tV d
18
2wp
t
gdV
w
wpt
gdV
3.0
w
wp
D
tC
gdV
3
4 General Equation
0.4DC
Use the graph
Example Calculation of Terminal Velocity
Determine the terminal settling velocity of a cryptosporidium oocyst having a diameter of 4 m and a density of 1.04 g/cm3 in water at 15°C [=1.14x10-3 kg/(s•m)].
Reynolds?
3
2
999 kg/m
9.81 m/swρ
g
==
Work in your teams.Use mks units (meters, kilograms, seconds).Convert your answer to some reasonable set of units that you understand.
Solution
0.001
0.01
0.1
0.1 1 10
floc diameter (mm)
floc
dens
ity
10
100
1000
floc
term
inal
vel
ocity
(m/d
ay)
floc densityVt (m/day)
Floc Density and Velocity (Approximate)
floc w
w
Water inlet
36 - 100 m/dayWater inlet
36 - 100 m/day
0.4 mm
______ kg/m3 floc 1030
Based on experimental data for Alum-clay flocs
Sedimentation Basin:Critical Path
Horizontal velocity
Vertical velocityL
H
captured getsbarely just that velocity terminal cV
A
QVh
18
2wp
t
gdV
Sludge zoneInle
t zo
ne Outle
t zo
ne
Sludge out
tVhV
A = WH
Q = flow rate
(property of the particle)
(property of the tank)
Sedimentation Basin:Importance of Tank Surface Area
cV
hV
L
H
W
Suppose water were flowing up through a sedimentation tank. What would be the velocity of a particle that is just barely removed?
Q
cs
H HQ Q QVLW Aq
= = = ="
s
residence timevolume of tank
A top surface area of tankWHL
Want a _____ Vc, ______ As, _______ H, _______ . small large
Time in tank
small large
cs
QVA
=
Conventional Sedimentation Basin
Settling zone
Sludge zoneInle
t zo
ne Outle
t zo
ne
Sludge out
long rectangular basins
4-6 hour retention time
3-4 m deepmax of 12 m
widemax of 48 m
long
We can’t do this in our laboratory scale plants!
What is Vc for this sedimentation tank?
3 24 18 /4c
H m hrV m dayhr day
Settling zone
Sludge zoneInle
t zo
ne Outle
t zo
neDesign Criteria for Sedimentation Tanks
Minimal turbulence (inlet baffles)
Uniform velocity (small dimensions normal to velocity)
No scour of settled particles
Slow moving particle collection system
Q/As must be small (to capture small particles)
This will be one of the ways you can improve the performance of your water treatment plant.
___________________________________________________________________________________________________________________________________________________________
Lamella
Sedimentation tanks are commonly divided into layers of shallow tanks (lamella)
The flow rate can be increased while still obtaining excellent particle removal
Lamella decrease distance particle has to fall in order to be removed
Lamella
Design needs improvement! Need method to transport particles to bottom of tank.
Lamella Closeup
Region of particle-free fluid above the suspension
SuspensionThin particle-free fluid layer beneath
the downward-facing surfaceConcentrated sediment
bL
cos sinlamella
cQv
wL wb
w = width of lamella
cs
QVA
=
cos sin
lamellac
Vv Lb
Lamella Design Strategy
Angle is approximately 60° to get solids to slide down the incline
Re must be less than 2000Shear doesn’t causing resuspension
if flow is laminarLamella spacing must be large
relative to floc size (flocs can be several mm in diameter)
Upflow velocity (Q/As) can be as large as 100 m/day
lamellalamella
QQN
Re lamellaV b
tan cossin
klamella
L LN
b
Sedimentation of Small Particles?
How could we increase the sedimentation rate of small particles?
18
2
wpt
gdV
Increase d (stick particles together)
Increase g (centrifuge)
Decrease viscosity (increase temperature)
Increase density difference(dissolved air flotation)
Particle/particle interactions
Electrostatic repulsion In most surface waters, colloidal surfaces are
negatively charged like charges repel __________________
van der Waals force an attractive forcedecays more rapidly with distance than the electrostatic
force is a stronger force at very close distances
stable suspension
Energy Barrier
Increase kinetic energy of particles
increase temperaturestir
Decrease magnitude of energy barrierchange the charge of the particlesintroduce positively charged
particles
+++++++++
+++++++++
Layer of counter ions
van der Waals
Electrostatic
Coagulation
Coagulation is a physical-chemical process whereby particles are destabilized
Several mechanismsadsorption of cations onto negatively charged
particlesdecrease the thickness of the layer of counter
ionssweep coagulationinterparticle bridging
Coagulation Chemistry
The standard coagulant for water supply is Alum [Al2(SO4)3*14.3H2O]
Typically 5 mg/L to 50 mg/L alum is usedThe chemistry is complex with many possible
species formed such as AlOH+2, Al(OH)2+, and
Al7(OH)17+4
The primary reaction produces Al(OH)3 Al2(SO4)3 + 6H2O2Al(OH)3 + 6H+ + 3SO4
-2
pH = -log[H+]
Coagulation Chemistry
Aluminum hydroxide [Al(OH)3] forms amorphous, gelatinous flocs that are heavier than water
The flocs look like snow in waterThese flocs entrap particles as the flocs
settle (sweep coagulation)
Coagulant introduction with rapid mixing
The coagulant must be mixed with the waterRetention times in the mixing zone are typically
between 1 and 10 seconds Types of rapid mix units
pumpshydraulic jumps flow-through basins with many baffles In-line blenders In-line static mixers
Flocculation
Coagulation has destabilized the particles by reducing the energy barrier
Now we want to get the particles to collideWe need relative motion between particles
________ ________ (effective for particles smaller than 1 m)
_________ _____________ (big particles hit smaller particles)
_______
Differential sedimentation
Shear
Brownian motion
Mechanical Flocculation
Shear provided by turbulence created by gentle stirring
Turbulence also keeps large flocs from settling so they can grow even larger!
Retention time of 10 - 30 minutesAdvantage is that amount of
shear can be varied independent of flow rate
Disadvantage is the tanks are far from plug flow
Hydraulic Flocculators
TypesHorizontal baffleVertical bafflePipe flow
Questions for designHow long must the suspension be in the “reactor”How should the geometry of the reactor be
determined?
Velocity Gradient Flocculation
With red particle as frame of reference
With fixed frame of reference
Increase Velocity Gradient
Velocity gradient!
duGdy
How much water is cleared of particles from stationary particle’s perspective?
Volume cleared is proportional to projected area of stationary particle
Volume cleared is proportional to timeVolume cleared is proportional to the
velocity gradientThe velocity of the water flowing past the
particle increases with the diameter of the particle
3cleared d Gt
2dtG
d
How much volume must be cleared before a collision occurs?
What is the average volume of water occupied by a particle?
Given C mg/L of particles in suspension…Need to know particle diameter (d)And density (particles)How many particles are in a volume of
water?3
6
particles
particles
CN
d
numbervolume
Volume occupied by a particle
3
6particles
occupiedparticles
d
C
Set volume occupied by a particle equal to volume cleared
3cleared d Gt
3
36particles
particles
dd Gt
C
particlescollision
particles
tG C
Collision Time
A measure of how long the particles must be in the velocity gradient to double in size
A series of collisions must occur for particles to grow large enough to be easily removed by sedimentation
particlescollision
particles
tG C
Flocculation Reactor Design
Critical design is when particle concentration is low
Higher velocity gradients would decrease the characteristic collision time
Why not design a tiny reactor with huge velocity gradients?
SHEAR
particlescollision
particles
tG C
Shear
The tangential force experienced by a fluid in a velocity gradient is proportional to the viscosity of the fluid
dudy
G
Fluid viscosity
2
N sm
Velocity gradient
1s
Shear
2
Nm
Too much shear?
Flocs can be broken by too much shearAmazingly, we haven’t been able to find good
information on the shear level that causes aluminum-clay flocs to breakup
fine grained cohesive sediments within estuarine waters were shown to produce smaller flocs when the shear exceeded 0.35 Pa (equivalent to a G of approximately 400/s)
dudy
Reaction time?
Low particle concentrations require longer flocculation
Goal is to get flocculation to work when turbidity is as low as 10 NTU (equivalent to approximately 20 mg/L of kaolin clay)
particlescollision
particles
tG C
2650
1400 0.020collision
gLt
gs L
331 seconds
Reaction time is more complex
Aluminum hydroxide polymers significantly increase the number of particles and the probability of collision (and hence decrease tcollision)
So for now we have to go with some empirical guidelines
G should be at least 20,000 where is the hydraulic residence time in the flocculation reactor
Q
Reactor volumeFlow rate
Laminar Flow Pipe Flocculation: for tiny flows!
The max value for G is approximately 50/s
These equations assume laminar flowLaminar flow requires that the
Reynolds number be less than 2000
See if you can figure out equations for the length of the pipe
3
643
QGd
3643
QdG
4Re Vd QD
max 3
32QGd
max 1.5G G
Given G, Q and d, Find Floc Tube Length
3
643
QGd
2
4d LQ
2
3
64 164 3 3d L Q LGQ d d
316dGL
True for laminar flow
Q
2
4d L
Q
Laminar Pipe Flow
displacement
velocityVelocity gradient
r r
Coagulation/Flocculation
Inject Coagulant in rapid mixerWater flows from rapid mix unit into
flocculation reactorWater flows from flocculation reactor into
sedimentation tankmake sure flocs don’t break!flocs settle and are removed
Jar Test
Mimics the rapid mix, flocculation, sedimentation treatment steps in a beaker
Allows operator to test the effect of different coagulant dosages or of different coagulants
Low tech water bottle test
Unit Processes in Conventional Surface Water Treatment
We’ve coveredSedimentationCoagulation/flocculation
Coming up!FiltrationDisinfectionRemoval of Dissolved Substances
Conventional Surface Water Treatment
Screening
Coagulation
Flocculation
Sedimentation
Filtration
Disinfection
Storage
Distribution
Raw water
AlumPolymers Cl2
sludge
sludge
sludge
Filtration
Slow sand filtersDiatomaceous earth filtersMembrane filtersRapid sand filters (Conventional Treatment)
Slow Sand Filtration
First filters to be used on a widespread basisFine sand with an effective size of 0.2 mmLow flow rates (10 - 40 cm/hr)Schmutzdecke (_____ ____) forms on top
of the filtercauses high head lossmust be removed periodically
Used without coagulation/flocculation!
filter cake
Diatomaceous Earth Filters
Diatomaceous earth (DE) is made of the silica skeletons of diatoms
DE is added to water and then fed to a special microscreen
The DE already on the microscreen strains particles and DE from the water
The continuous DE feed prevents the gradually thickening DE cake from developing excessive head loss
Was seriously considered for Croton Filtration Plant
Membrane Filters
Much like the membrane filters used to enumerate coliformsmuch greater surface area
Produce very high quality water (excellent particle removal)
Clog rapidly if the influent water is not of sufficiently high quality
More expensive than sand and DE filters
Rapid Sand Filter (Conventional US Treatment)
Sand
Gravel
Influent
DrainEffluent Wash water
Anthracite
Size(mm)0.70
0.45 - 0.55
5 - 60
SpecificGravity
1.6
2.65
2.65
Depth(cm)30
45
45
Particle Removal Mechanisms in Filters
Transport
Attachment
Molecular diffusionInertiaGravityInterception
StrainingSurface forces
Filter Design
Filter media silica sand and anthracite coalnon-uniform media will stratify with _______ particles
at the top Flow rates
2.5 - 10 m/hrBackwash rates
set to obtain a bed porosity of 0.65 to 0.70 typically 50 m/hr
smaller
Sand
Gravel
Influent
DrainEffluent Wash water
Anthracite
Backwash
Wash water is treated water!
WHY?Only clean water should ever be on bottom of filter!
Ways to Improve Filtration
Filter to wasteExtended Terminal Sub-fluidization WashAlum feed directly to filter?Potato starch?
Disinfection
Disinfection: operations aimed at killing or ____________ pathogenic microorganisms
Ideal disinfectant_______________ _____________________________________________ _______________
inactivating
Toxic to pathogensNot toxic to humans Fast rate of kill Residual protectionEconomical
Disinfection Options
Chlorine chlorine gas sodium hypochlorite (bleach)
Ozone Irradiation with Ultraviolet light SonificationElectric CurrentGamma-ray irradiation
Poisonous gas – risk of a leak
Chlorine
First large-scale chlorination was in 1908 at the Boonton Reservoir of the Jersey City Water Works in the United States
Widely used in the USTypical dosage (1-5 mg/L)
variable, based on the chlorine demandgoal of 0.2 mg/L residual
Trihalomethanes (EPA primary standard is 0.08 mg/L) Pathogen/carcinogen tradeoff
Chlorine oxidizes organic matter
Chlorine Reactions
Cl2 + H2O H+ + HOCl + Cl-
HOCl H+ + OCl-
The sum of HOCl and OCl- is called the ____ ______ _______
HOCl is the more effective disinfectantTherefore chlorine disinfection is more
effective at ________ pHHOCl and OCl- are in equilibrium at pH 7.5
free chlorine residual
low
+1 -2 +10Charges -1
Hypochlorous acid Hypochlorite ion
EPA Pathogen Inactivation Requirements
SDWA requires 99.9% inactivation for Giardia and 99.99% inactivation of viruses
Giardia is more difficult to kill with chlorine than viruses and thus Giardia inactivation determines the CT
Concentration x TimeEnumerating Giardia is difficult, time-consuming and costly.How would you ensure that water treatment plants meet this criteria?Where are Giardia removed/inactivated?
Safe Drinking Water Act
EPA Credits for Giardia Inactivation
Treatment type CreditConventional Filtration 99.7%Direct Filtration* 99%Disinfection f(time, conc., pH,
Temp.)
* No sedimentation tanks
Disinfection CT Credits
Contact time (min)chlorine pH 6.5 pH 7.5(mg/L) 2°C 10°C 2°C 10°C
0.5 300 178 430 2541 159 94 228 134
To get credit for 99.9% inactivation of Giardia:
Inactivation is a function of _______, __________________, and ___________.
concentrationtimepH temperature
NYC CT?
Kensico
Hillview
Delaware Pipeline21.75 km long5.94 m diameterpeak hourly flow = 33 m3/s
3.4 x 106 m3
volume =603,000 m3
5 hour residence time!
NYC CT Problem
Hillview Reservoir is an open reservoirShould the chlorine contact time prior to arrival at
Hillview count?
Giardia contamination from Upstate Reservoirs will be decreased, butrecontamination at Hillview is possible
Ozone
Widely used in EuropeO3 is chemically unstableMust be produced on siteMore expensive than chlorine (2 - 3 times)Typical dosages range from 1 to 5 mg/LOften followed by chlorination so that the
chlorine can provide a protective _______residual
Removal of Dissolved Substances (1)
Aeration (before filtration)oxidizes iron or manganese in groundwateroxidized forms are less soluble and thus
precipitate out of solutionremoves hydrogen sulfide (H2S)
Softening (before filtration)used to remove Ca+2 and Mg+2
usually not necessary with surface waters
Removal of Dissolved Substances (2)
Activated Carbon (between filtration and disinfection) extremely adsorbentused to remove organic contaminants spent activated carbon can be regenerated with superheated
steamReverse Osmosis
semi-permeable membrane allows water molecules to pass, but not the larger ions and molecules
primarily used for desalination also removes organic materials, bacteria, viruses, and
protozoa
Conventional Surface Water Treatment
Screening
Coagulation
Flocculation
Sedimentation
Filtration
Disinfection
Storage
Distribution
Raw water
AlumPolymers Cl2
sludge
sludge
sludge
Summary
Cryptosporidium Oocyst
ms
kg1.14x1018
kg/m 999kg/m 1040m/s 189.m 4x10 3
33226
tV
18
2wp
t
gdV
m 4x10
m/s 189.
kg/m 999
kg/m 1040
6
2
3
3
d
g
ρ
ρ
w
p
m/s1014.3 7 xVt
cm/day 7.2 tV
Reynolds Number Check
Re<<1 and therefore in Stokes Law range
Re Vd
7 6 3
3
3.14 10 m/s 4 10 m 999kg/mRe kg1.14x10
s m
x x
Re = 1.1 x 10Re = 1.1 x 10-6-6
Diatomaceous Earth
DE
Clay
lamella lamellalamella
QQ v wbN
cos sinlamella cLv vb
tan cossin
klamella
L LN
b
tan
sincos sin
cosck
QbLvb wbL wbL
tan
cossin
sincosc k
LbQ
v wL wL
cos sinlamella
cQv
wL wb
tan
sin
cos cos sinc
k
Qv
LwL wLb
1
cos sinc
lamella
Qv LwbNb