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05 Sedimentation

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    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.htm
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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.html
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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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