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7/30/2019 05 Sedimentation
1/70
Monroe L. Weber-ShirkSchool ofCivil and
Environmental Engineering
Sedimentation
http://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cornell.edu/http://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/index.cfmhttp://www.cee.cornell.edu/faculty/info.cfm?abbrev=faculty&shorttitle=bio&netid=mw24http://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://ceeserver.cee.cornell.edu/mw24/Default.htmhttp://ceeserver.cee.cornell.edu/mw24/Default.htm7/30/2019 05 Sedimentation
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Simple Sorting
Goal: clean water
Source: (contaminated) surface water
Solution: separate contaminants from water
How?
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Unit processes* designed to
remove _________________________
remove __________ ___________
inactivate ____________
*Unit process: a process that is used in similarways in many different applications
Unit Processes Designed to Remove ParticulateMatter
Screening
Coagulation/flocculation
Sedimentation
Filtration
Where are we?
Particles and pathogensdissolved chemicals
pathogens
Empirical design
Theories developed later
Smaller particles
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Conventional Surface Water
Treatment
Screening
Rapid Mix
Flocculation
Sedimentation
Filtration
Disinfection
Storage
Distribution
Raw water
AlumPolymers
Cl2
sludge
sludge
sludge
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Screening
Removes large solids
logs
branches
rags
fish
Simple processmay incorporate a mechanized trash
removal system
Protects pumps and pipes in WTP
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Sedimentation
the oldest form of water treatment
uses gravity to separate particles from water
often follows coagulation and flocculation
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Sedimentation: Effect of the
particle concentration
Dilute suspensions
Particles act independently
Concentrated suspensionsParticle-particle interactions are significant
Particles may collide and stick together(form flocs)
Particle flocs may settle more quickly
At very high concentrations particle-particle forces may prevent further
consolidation
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projected
Sedimentation:
Particle Terminal Fall Velocity
maF 0 WFF bd
p pg
2
2
twPDd
VACF W
dF
bF
p w
gr"velocityterminalparticle
tcoefficiendrag
gravitytodueonaccelerati
densitywaterdensityparticle
areasectionalcrossparticle
volumeparticle
t
D
w
p
p
p
V
C
g
A
_______W
________bF=
Identify forces
( )4
3
p w
t
D w
gdV
C
r r
r
-
=
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Drag Coefficient on a Sphere
laminar
Re tV d
turbulentturbulent
boundary
0.1
1
10
100
1000
0.1 1 1
0100
100
0
100
00
100
000
100
0000
100
0000
0
Reynolds Number
DragCoeffici
Stokes Law
24
Red
C
18
2
wp
t
gdV
( )4
3
p w
t
D w
gdV
C
r r
r
-
=
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Floc Drag
C.Dtransition Re( )24
Re
3
Re 0.3
0.1 1 10 100 1 103
1 104
1 105
1 106
1 10
0.1
1
10
100
CDsphere
CDtransition Rek
Stokes Rek
Regraph Rek
Flocs created in the
water treatment
process can have
Re exceeding 1 and
thus their terminal
velocity must be
modeled using
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Sedimentation Basin:
Critical Path
Horizontal velocity
Vertical velocityL
HhV Sludge zoneIn
letzone
Outletzone
Sludge out
cV
hV
WH
flow rate
What is Vc for this sedimentation tank?
Vc = particle velocity that just barely ______________gets captured
Q
A
cV H
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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
tankofareasurfacetopA
tankofvolume
timeresidence
s
WHL
Want a _____ Vc, ______ As, _______ H, _______.small large
Time in tank
small large
c
s
QV A
=
HQ
QLW s
QA
c HV
Vc is a property of the
sedimentation tank!
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Conventional Sedimentation Basin
Settling zone
Sludge zoneInlet
zo
ne
Outlet
zone
Sludge out
long rectangular basins
4-6 hour retention time
3-4 m deepmax of 12 m wide
max of 48 m long
What is Vc forconventional design?
3 2418 /
4c
H m hrV m day
hr day
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_______________________________
_______________________________
_______________________________
_______________________________
_______________________________Vc of 20 to 60 m/day*
Residence time of 1.5 to 3 hours*
Settling zone
Sludge zoneInlet
zone
Outlet
zone
Design Criteria for
Horizontal Flow
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)
* Schulz and Okun And dont break flocs at inlet!
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Sedimentation Tank particle capture
What is the size of the smallest
floc that can be reliably captured
by a tank with critical velocity of60 m/day?
We need a measure of real water
treatment floc terminal velocitiesResearch
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Physical Characteristics of Floc:
The Floc Density Function
Tambo, N. and Y. Watanabe (1979). "Physicalcharacteristics of flocs--I. The floc densityfunction and aluminum floc." Water Research13(5): 409-419.
Measured floc density based on sedimentationvelocity (Our real interest!)
Flocs were prepared from kaolin clay and alum atneutral pH
Floc diameters were measured by projected area
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Floc Density Function:
Dimensional Analysis!
Floc density is a function of__________
Make the density dimensionless
Make the floc sizedimensionless
Write the functional
relationshipAfter looking at the data
conclude that a power lawrelationship is appropriate
floc w floc
w clay
df
d
floc
clay
dd
floc w
w
dn
floc w floc
w clay
da
d
floc size
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Model Results
For clay assume dclay was 3.5 m (based onTambo and Watanabe)
a is 10 and nd is -1.25 (obtained by fitting thedimensionless model to their data)
The coefficient of variation for predicteddimensionless density is0.2 for dfloc/dclay of 30 and
0.7 for dfloc/dclay of 1500The model is valid for __________flocs in the size
range 0.1 mm to 3 mm
dn
floc w floc
w clay
da
d
clay/alum
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Additional Model Limitation
This model is simplistic and doesnt include
Density of clay
Ratio of alum concentration to clay concentration
Method of floc formation
Data doesnt justify a more sophisticated model
Are big flocs formed from a few medium sized
flocs or directly from many clay particles?Flocs that are formed from smaller flocs may tend to be
less dense than flocs that are formed from accumulationof (alum coated) clay particles
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Model Results Terminal Velocity
dn
floc w floc
w clay
da
d
24 30.34
Re RedC
4
3
floc w
t
D w
gdV
C
= shape factor (1 for spheres)
Requires iterative solution for velocity
Ret flocV d
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Floc Sedimentation Velocity
1 103
0.01 0.1 1 10 100
1
10
100
1 103
Vt dfloci
dclay a nd
m
day
dfloci
mm
a: 10
nd: -1.25
dclay: 3.5 m: 45/24
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Floc density summary
Given a critical velocity for a sedimentationtank (Vc) we can estimate the smallest
particles that we will be able to captureThis is turn connects back to flocculator
design
We need flocculators that can reliablyproduce large flocs so the sedimentationtank can remove them
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Flocculation/Sedimentation:
Deep vs. Shallow
Compare the expected performance of shallow and deephorizontal flow sedimentation tanks assuming they havethe same critical velocity (same Q and same surface area)
More opportunities to
______ with otherparticles by _________
____________ or
________________
Expect the _______
tank to perform better!
deeper
collidedifferential
sedimentation
Brownian motion
But the deep tank is
expensive to make and
hard to get uniform flow!
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Flocculation/Sedimentation:
Batch vs. Upflow
Compare the expected performance of a batch
(bucket) and an upflow clarifier assuming they
have the same critical velocityHow could you improve the performance of
the batch flocculation/sedimentation tank?
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Lamella
Sedimentation tanks are commonly divided intolayers of shallow tanks (lamella)
The flow rate can be increased while still
obtaining excellent particle removal
Lamella
decrease
distanceparticle has
to fall in
order to be
removed
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Defining critical velocity for plate
and tube settlers
ab
L
cosL asin
b
a
Vup Va
How far must particle
settle to reach lower plate?
a
Path for critical particle?
cosc
b
ha
hc
cosc
bh
a
What is total vertical distancethat particle will travel?
sinh L a
h
What is net vertical velocity?
net up cV V V
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Compare times
Time to travel distance hc Time to travel distance h=
cosc
bh a
sinh L a
c
c up c
h h
V V V
sin
cosc up c
b L
V V V
a
a
sin cosup c cbV bV L V a a
sin cosup cbV L b V a a
sin cos
up
c
bVV
L ba a
ab
L
cosL asin
b
a
Vup Va
a
hc
h
1 cos sinup
c
V L
V b
a a
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Comparison with Q/As
ab
L
cosL a
sin
b
a
Vup Va
a
hc
h
Q V bwa
sinupV
Va a
cossin
bA L wa
a
1
sincos
sin
up
c
V bwQV
bAL w
aa
a
sin
upV bw
Q a
cos sinupc
V bV L ba a Same answer!
As is horizontal area over which particles can settle
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Performance ratio (conventional to
plate/tube settlers)
Compare the area on which
a particle can be removed
Use a single lamella tosimplify the comparison
a
bL
cosL asin
b
a
Conventional capture area
sinconventional
b
A w a
Plate/tube capture area
cos
sin
tube
bA w wL a
a
1 cos sinratio
L
A b a a
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Critical Velocity Debate?
cos sinc
VV
L
b
a
a a
cos1
sin
up
c
V L
V b
a
a
Schulz and Okun
Water Quality and Treatment (1999)
1 cos sinup
c
V L
V ba a Weber-Shirk
WQ&T shows this geometry
But has this equation
Assume that the geometry is
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9045
105
Check the extremes!
01020304050607080900
5
10
15
20
ratio ( )
ratioWS ( )
deg
1 cos sinup
c
V L
V ba a
cos1
sin
up
c
V L
V b
a
a
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Critical Velocity Guidelines
Based on tube settlers
1030 m/day
Based on Horizontal flow tanks20 to 60 m/day
Unclear why horizontal flow tanks have a higher
rating than tube settlers
Could be slow adoption of tube settler potential
Could be upflow velocity that prevents particle
sedimentation in the zone below the plate settlers
http://www.brentwoodprocess.com/tubesystems_main.html
Schulz and Okun
http://www.brentwoodprocess.com/tubesystems_main.htmlhttp://www.brentwoodprocess.com/tubesystems_main.html7/30/2019 05 Sedimentation
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Problems with Big Tanks
To approximate plug flow and to avoid shortcircuiting through a tank the hydraulic radiusshould be much smaller than the length of the tank
Long pipes work well!Vc performance of large scale sedimentation tanks
is expected to be 3 times less than obtained inlaboratory sedimentation tanks*
Plate and tube settlers should have much betterflow characteristics than big open horizontal flowsedimentation tanks
h
AR
P
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Goal of laminar flow to avoid floc
resuspension
4
RehV Ra
*
2 2h
b w bR
w
1cos
sinc
LV V
ba a
a
2Re
V ba
12 cossin
Re 390c LbV
ba
a
30 /cV m day
1L m5b cm60a
Re is laminar for typical designs, _____________________
Is Re a design constraint? sinupV
Vaa 1 cos sin
up
c
V L
V ba a
h
AreaR
Wet Perimeter
not a design constraint
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Mysterious Recommendations
Re must be less than 280 (Arboleda, 1983
as referenced in Schulz and Okun)
The entrance region should be discounteddue to possible turbulence (Yao, 1973 as
referenced in Schulz and Okun)
0.13Reuseful
L L
b b
At a Re of 280 we discard 36 and a typical L/b is 20
so this doesnt make sense
But this isnt about turbulence(see next slide)!!!
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Entrance Region Length
1
10
100
10
100
1000
10000
100000
100000
1000000
10000000
Re
le/D
1/ 6
4.4 Reel
D0.06Ree
l
D
laminar turbulent
Reel
fD
Distancefor
velocity
profile to
develop
0.12Reel
b
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Entrance region
The distance required to produce avelocity profile that then remainsunchanged
Laminar flow velocity profile isparabolic
Velocity profile begins as uniformflow
Tube and plate settlers are usuallynot long enough to get to theparabolic velocity profile
a
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Lamella Design Strategy
Angle is approximately 60 to getsolids to slide down the incline
Lamella spacing of 5 cm (b)L varies between 0.6 and 1.2 m
Vc of 10-30 m/day
Find Vup through active area of tankFind active area of sed tank
Add area of tank required for angled
plates: add L*cos(a) to tank length
tankactive
up
QA
V
1 cos sinup cL
V Vb
a a
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Sedimentation tank cross section
Effluent Launder (a manifold)
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Design starting with Vup
The value of the vertical velocity is
important in determining the effectiveness
of sludge blankets and thus it may beadvantageous to begin with a specified Vup
and a specified Vc and then solve for L/b
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Equations relating Velocities and
geometry
1 cos sinactiveup lamella
c
V L
V ba a
activeup total
up active
V L
V L
cosactive total lamellaL L L a
Continuity (Lengths are sed tank lengths)
Lamella gain
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Designing a plate settler
wplate
bplate
LplateQplant
Ntanks
a
Vertical space in the
sedimentation tank
divided between sludge storage and
collection
flow distribution
Plates flow collection
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AguaClara Plant Layout (draft)
Drain Valve
access holes
Chemicalstore room
Steps
Effluent launders
Sed tanks
Floc tank
To the distribution tank
Sed tank manifold
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Distributing flow between tanks
Which sedimentation tank will have the highestflow rate?
Where is the greatest head loss in the flow througha sedimentation tank?
Either precisely balance the amount of head loss
through each tankOr add an identical flow restriction in each flow
path
Where is the highest velocity?
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Will the flow be the same?
Dh
Long
Short
Head loss for long route = head loss for short route if KE is ignored
Q for long route< Q for short route
K=1
K=0.2K=0.5 K=1
2 2
1 1 2 21 2
2 2L
p V p Vz z h
g g g g
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Conservative estimate of effects of
manifold velocity
2
1 22 longportport
L
V
H H hg
long short
Control surface 1l
cs 3
Long orifice Short orifice2
1 32 shortportport L
VH H hg
cs 2
2
max
2 3 2 manifoldmanifold
L
V
H H hg
cs 4 cs 5
2 2
2 32 2longport shortport port port
L L
V VH h H h
g g
2
max
2 manifold longport shortport manifold
L L L
Vh h h
This neglects velocity head differences
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Modeling the flow
2
2
elong
long
longlong
ratio
short eshortshort
short
ghA
KQQ
Q ghA
K
Q.pipeminorD h.e K A.circle D( )2 g h.e
K
long short L Lh h
0.20.26
3
short
ratio
long
KQ
K
We are assuming that minor
losses dominate. It would be
easy to add a major loss term(fL/d). The dependence of the
friction factor on Q would
require iteration.
Since each point can have only one pressure
g y
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Design a robust system that gets the
same flow through both pipes
2
21
ratio long short
control
ratio
Q K KK
Q
short control
ratio
long control
K KQ
K K
2
2
0.95 3 0.225.7
1 0.95controlK
Add an identical minor head
loss to both paths
Solve for the control loss
coefficient
Design the orifice
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Piezometric head decrease in a
manifold assuming equal port flows
2 2
port port
2 4 2 41
8 8port
np
i
iQ C nQH
g d g d D
port
2
2 2
2 41
8
port
n
p
i
QH C i n
g d
D
2 2
1 2 3 16
n
i
n
i n n
Piezometric head decrease in a
manifold with n ports
d is the manifold diameterrepresents the head
loss coefficient in the
manifold at each port or
along the manifold as fL/dNote that we arent
using the total flow in
the manifold, we are
using Qport
port2
2 2
2 4
82 3 1
6portp
Q nH C n n n
g d
D
Head loss Kinetic energy
portpC
f l if ld
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Convert from port to total manifold
flow and pressure coefficient
total port Q nQ portp pC nC
port2
2 2
2 4
82 3 1
6portp
Q nH C n n n
g d
D
total
2
2 4 2
8 1 1 11
3 2 6p
QH C
g d n n
D
Loss coefficient Velocity head
manifold
p
manifold
LC f K
d Note approximation withf
These are losses in the manifold
C l l ddi i l h d l
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Calculate additional head loss
required to get uniform flow
2
1 1 11
3 2 6long pK C
n n
0shortK
Kcontrol is the minor losscoefficient we need
somewhere in the ports
connecting to the
manifold
Note that this Klong gives the correct head loss when using Qmaxmanifold
Long path Short path
2
2
1 1 11
3 2 6
11
p
control
ratio
Cn n
K
Q
2
21
ratio long short
control
ratio
Q K KK
Q
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Total Loss Coefficient
2
1 1 11
3 2 6long pK C
n n
2
21
ratio long
control
ratio
Q KK
Q
21
longtotal long control
ratio
KK K KQ
Excluding KE
Including KE (moreconservative)
2
2 2
11 1
1 1
long ratio long
total long control
ratio ratio
K Q KK K K
Q Q
We are calculating the total
loss coefficient so we can
get a relationship between
the total availablepiezometric head and the
diameter of the manifold
C l l h if ld di
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Calculate the manifold diameter
given a total manifold head loss
2
manifold
2 4
8 totall
manifold
Q Kh
g d
1
2 4
2
8 manifold totalmanifold
l
Q Kd
g h
Solve the minor
loss equation for D
We could use a total head loss of perhaps 5 to 20 cm to determine the
diameter of the manifold. After selecting a manifold diameter (a real
pipe size) find the required control head loss and the orifice size.
Minor loss equation
Ktotal is defined based on the total flow
through the manifold and includes KE .21
long
total
ratio
KK
Q
2
1 1 1
1 3 2 6long pK C n n
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Full Equation for Manifold Diameter
Cp is loss coefficient for entire length of manifold
12 4
2
8 manifoldmanifold total
l
Qd K
gh
1
4
2 2
2 2
1 1 11
8 3 2 6
1
pmanifold
manifold
l ratio
CQ n n
dgh Q
21
long
total
ratio
KK
Q
2
1 1 1
1 3 2 6long pK C n n
M if ld d i i i h j
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Manifold design equation with major
losses
n is number of ports
f is friction factor (okay to use f based on Qtotal)
Qratio is acceptable ratio of min port flow over max port flow
hl is total head loss through the ports and through the manifold
Qmanifold
is the total flow through the manifold from the n ports
K is the sum of the minor loss coefficients for the manifold (zero for a straight pipe)
Iteration is required!
1
4
22
2 2
1 1 11
3 2 68
1
manifold
manifoldmanifold
manifold
l ratio
Lf K
d n nQd
gh Q
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Head loss in a Manifold
2
4 2 2
8 1 1 1
3 2 6manifoldmanifoldtotal
l
manifold manifold
fLQh K
d g d n n
2
1 1 1
3 2 6manifoldl l
h hn n
The head loss in a manifold pipe can be obtained
by calculating the head loss with the maximum Q
through the pipe and then multiplying by a factor
that is dependent on the number of ports.
N fi d th ffl t l d ifi
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Now find the effluent launder orifice
area
2or control Q K A gh
2or control
QAK gh
Use the orifice equation to figure out what the area of the
flow must be to get the required control head loss. This
will be the total area of the orifices into the effluent launder
for one tank.
2
2
1 1 11
3 2 6
1
1
p
control
ratio
Cn n
K
Q
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Orifice flow (correction!)
2orQ K A g h D
2
2 2
1
2
or
or or
Qh
K gAD
Solve for h and substitute
area of a circle to obtain same
form as minor loss equation
Kor= 0.63
2.5 d 8 d
d
Dh
D
2
22
manifold
e
manifold
Qh K
gA
4 2
2 4 2
4
2 4 2
1
1
manifold or
or or manifold
or or manifold
manifold
or or or
d QK d Q
Q n Q
d
K d n
C l l ti th ifi di t b d
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Calculating the orifice diameter based
on uniform flow between orifices
2
2
1 1 11
3 2 6
11
p
control
ratio
Cn n
K
Q
4
2 4 2
1 manifoldcontrol
or or or
dK
K d n
1
4
2 2
1or manifold
or or control
d dK n K
manifold
p
manifold
LC f K
d
H ll t th ifi b ?
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How small must the orifice be?
Case of 1 orifice
4
2 4
pipe
or or
dK
K d
1
4
2
1or pipe
or
d dKK
For this case dorifice must be approximately 0.56dpipe.
We learned that we can obtain equal similarparallel flow by ensuring that the head loss
is similar all paths.We can compensate for small differences in
the paths by adding head loss that is largecompared with the small differences.
Effl t L d
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Effluent Launders:
Manifold Manifolds
Two Goals Extract water uniformly from the top of the sed tank so the flow
between all of the plates is the same
Create head loss that is much greater than any of the potentialdifferences in head loss through the sedimentation tanks toguarantee that the flow through the sedimentation tanks isdistributed equally
A pipe with orifices
Recommended orifice velocity is 0.46 to 0.76 m/s (WaterTreatment Plant Design 4th edition page 7.28) The corresponding head loss is 3 to 8 cm through the orifices
but it isnt necessarily this simple!
We need to get a low enough head loss in the rest of the system
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Effluent Launders and Manifold
We need to determine the required diameter of the effluent launder pipe
The number and the size of the orifices that control the flow ofwater into the effluent launder
The diameter of the manifold The head loss through the orifices will be designed to be
large relative to the differences in head loss for the variouspaths through the plant
We need an estimate of the head loss through the plant by
the different paths Eventually take into account what happens when one
sedimentation tank is taken off line.
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Head Loss in a sed tank?
Head loss through sed tank inlet pipes
and plate settlers is miniscule
The major difference in head loss
between sed tanks is due to thedifferent path lengths in the manifold
that collects the water from the sed
tanks.
We want equally divided flow twoplaces
Sed tanks
Plate settlers (orifices into launders)
Manifold head loss:
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Manifold head loss:
Sed tanks equal!
We will assume minor losses dominate to developthe equations. If major losses are important theycan be added or modeled as a minor loss.
The head loss coefficient from flowing straightthrough a PVC Tee is approximately 0.2
We make the assumption that the flow into eachport is the same
Eventually we will figure out the design criteria toget identical port flow
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Minor losses vs. Major losses
Compare by taking a ratio 2e
2
Vh K
g
2
ff
2
L Vh
D g
e
f f
Kh D
h L
f
e f
KhL
D h
11
0.02
L
D
Thus in a 10 cm diameter pipe, anelbow with a K of 1 gives as much
head loss as 5 m of pipe
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Now design the Effluent Launder
The effluent launder might be a smaller
diameter pipe than the sed tank manifold
(especially if there are many sedimentationtanks)
The orifice ports will be distributed along
both sides of the launder
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Now design the Effluent Launder
Port spacing should be less than the vertical distancebetween the ports and the top of the plate settlers (Im notsure about this constraint, but this should help minimizethe chance that the port will cause a local high flowthrough the plate settlers closest to the port)
The depth of water above the plate settlers should be
0.6 to 1 m with launders spaced at 1.5 m (Water Treatment PlantDesign 4th edition page 7.24)
This design guideline forces us to use a very deep tank. Deep tanksare expensive and so we need to figure out what the real constraintis.
It is possible that the constraint is the ratio of water depth tolaunder spacing.
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Effluent Launder
The solution technique is similar to the manifold
design
We know the control head lossthe head lossthrough the ports will ensure that the flow through
each port is almost the same
We need to find the difference in the head loss
between the extreme paths
Then solve for the diameter of the effluent launder
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Sedimentation Tank Appurtenances
Distributing the flow between parallel tanks
Effluent Launders
Sludge removal (manifold design similar toeffluent launders)
Isolating a tank for fill and drain: using only asingle drain valve per tank
Filling the tank with clean water
Not disturbing the water levels in the rest of the plant
Entrance manifolds: designed to not break up flocs
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Plate settlers
launder
Sludge
Sludge drain line thatdischarges into a
floor drain on the
platform
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