AVP Extraction

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    Chemical Engineering Operations

    Solvent extraction (liquid-liquid extraction)

    Dr. Anand V. Patwardhan

    Professor of Chemical EngineeringInstitute of Chemical Technology

    Nathalal M. Parikh Road

    Matunga (East), Mumbai-400019

    [email protected]; [email protected]; [email protected]

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    Liquid-Liquid Extraction / Solvent Extraction

    Separation of constituents of liquid solution by contact

    ,

    original solution distribute differently between the two

    .

    EXTRACTION:

    1. Carrier phase is a liquid, not a solid, so the physical

    se aration techni ues chan e

    2. Two distinct li uid hases develo hence non-

    2uniformity of resulting solutions.

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    Driving force: chemical differences, not the vapor

    pressure erences, an ence can e use w endistillation is impractical.

    For example: to separate materials with similar boiling

    ,

    compounds.

    Common applications: separation and purification of

    broth, etc.

    3

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    Distillation and evaporation generally produce

    n s e pro uc s; qu ex rac on genera y oes no .

    products.

    Secondary separation: by distillation or evaporation.

    The overall process cost thus must be considered

    Extraction may become economical for dilute aqueous

    solutions when eva oration would re uirevaporisation of very large amounts of water.

    4

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    Terminology

    Feed: solution to be treated.

    Solvent: liquid used in contacting.

    Extract: enriched solvent product.

    Raffinate: depleted feed.

    Process: single stage, multistage crosscurrent, or

    countercurrent.

    5

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    A typical ternary diagram

    6

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    Plait point:

    Located near the top of the

    two- hase envelo e at the

    inflection point.

    represen s a con on

    where the 3-component

    phases, but the phases have

    .

    (analogous to azeotropicmixture of liquid and

    vapor.)

    7

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    2 main classes of liquid-liquid equilibrium:

    Class I system: 1 immiscible pair of compounds.

    ,

    triangular diagram.

    Solvent Selection:

    .

    Distribution coefficients: y/x at equilibrium; large

    values pre erable.

    Insolubilit : should not be soluble in carrier feedliquid.

    8

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    Recoverability: constraints such as azeotropes.

    Density: must be different phases can be separated by

    settling.

    Interfacial tension: if too high, liquids will be difficult

    .

    Chemical reactivity: solvent should be inert and stable.

    Viscosity, Vapor pressure, Freezing point: low values

    Safety: toxicity, flammability.

    Cost:

    9

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    Calculations:

    1. Number of stages needed to make a separation.

    2. Amount of solvent needed to make a separation.

    3. Liquid-Liquid equilibrium is not available as

    , .

    Choice of graphical approaches:

    i. McCabe-Thiele approach: if y versus x data is

    available mass fraction of solute in E- hase versusmass fraction in R-phase). The curve begins at the

    ori in and ends with the lait oint com osition.

    10

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    ii. Hunter-Nash method: Equilateral triangle

    agram: cons ruc on can e one rec y on etriangle.

    iii. Rectangular equilibrium diagrams.

    11

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    Solvent-to-Feed Ratio:

    For a given feed mixture, required degree of

    , ,

    choice of solvent, there exists a minimum solvent-to-

    .

    ,

    extract phase in equilibrium with the entering feed.

    A theoretical upper limit or maximum solvent-to-feed

    ratio also can be determined. The maximum solvent-to-feed ratio is thus that which puts the mixture on the

    hase boundar or sin le hase.

    12

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    Extraction equipment selection

    Design constraints:

    (1) Maximise area of mass transfer,

    (2) Adjust flow feeds for maximum solute recovery.

    3 main types of extractors:

    Mixer-settlers: when only one equilibrium stage is

    needed. The two hases are added and mixed.Separation based on density differences.

    Disadvanta e: re uires lar e-volume vessel and a hi h

    13solvent demand.

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    Contacting columns: packed / tray / spray column. In

    case o pac e co umn, e pac ng ma er a s ou ewetted by the continuous phase. The flow in a column

    .

    less than 4%. Also, it can offer multiple stages.

    14

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    Advantages and disadvantages of the various extractor types

    Unit Advantages Disadvantages Efficient

    Low head room

    Mixer-

    Settler

    Induces good

    contacting

    Can handle an

    arge oor

    High setup costs

    High operation costs

    number of stages

    Columns Small investment costs

    High head room

    w ou

    Agitation) Low operating costs

    cu o sca e up rom a

    Less efficient than mixer-settler

    Good dispersion

    (with

    Agitation)

    Low investment costs

    Can handle any

    number of stages

    differences

    Does not tolerate high flow ratios

    Centrifugal

    Can separate small

    density differences

    High setup cost

    High operating and maintenance

    15

    Small liquid inventory

    Cannot handle many stages

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    16

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    Liquid-Liquid Equilibrium

    Chemical potential of both liquid phases must be equal,

    L L L L L L1 2 1 1 2 2

    x xi i i i i i

    For a multi-component system, the UNIQUAC equation

    combinatorialln ln l residualn

    17

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    Activity coefficient models for finding mole fractions:

    UNIFAC (UNIquac Functional group Activity

    -

    Combinatorial and residual activities: based on

    statistical mechanical theory and compositions are

    computed from the size and energy differencesbetween the molecules in the mixture. The

    relationships for these two activities are available.

    18

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    Theory Ternary Phase Diagram

    General principles:

    within the triangle to the three sides equals the altitude

    .

    Each apex of triangle represents one pure component.

    Any point of a side of the triangle represents a binary

    Lines may be drawn parallel to the sides of the

    equ a era r ang e or e p o ng o e

    compositions.

    19

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    Phase diagram for a three component system

    20

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    Ternary phase diagram: directly from experimental

    a a.

    ,

    experimentally by a cloud point titration.

    For example, a solution containing A and C with

    ,

    is added until the onset of cloudiness (haze) due to

    Then the composition can be plotted onto the

    ternar hase dia ram.

    21

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    Tie lines: Join points (equilibrium) on miscibility

    oun ary.

    from an experiment

    For example, A mixture may be prepared with

    , ,

    B). If we allow it to equilibrate, then we can

    raffinate (R) phase.

    22

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    Point F: feed; Point S: solvent.

    Point H: composition of feed and solvent at

    .

    (S) compositions for each component

    Points Rand E: compositions of Raffinate and

    Point P: lait oint: At this oint onl one li uidphase exists and the compositions of two effluents are

    e ual.

    23

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    Curve JRDPEK: equilibrium between all three

    componen s.

    will exist.

    Area above JRDPEK: only one liquid phase.

    Operating line: FE and SR.

    Equilibrium constraint: chemical potential

    x xA A A A A A

    24

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    Separation factory

    E

    R Solute fluxes in Raffinate and Extract:

    Extract:N K A x xE E E

    Raffinate:N K A x xR R R

    ,

    KE = overall mass transfer coefficient on Extract side

    R

    A = total available mass transfer areai =E

    xiR= solute concentration at interface on Raffinate side

    25

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    A = carrier solvent B = solvent used for extraction C = component to be extracted

    26

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    The figure gives a general ternary diagram for a

    es re so u e , an ex rac ng so ven an acarrier solvent (A) from which the solute is to be

    .

    .

    extract C from feed.

    The Raffinate composition (R) is specified with respect

    to the recover ofC.

    27

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    Composition Flow rate

    g ven g venF A and C YES YES

    S B

    YES, usually pure

    OR

    NO, obtained

    fromre a ve y pure ca cu a on

    NO, obtainedNO, obtained

    E

    large C fromcalculation

    component

    balance

    A with

    YES, recovery

    amount

    NO, obtained

    fromsmall C of solute C

    needed

    component

    balance

    28

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    ,determination of the

    minimum solvent-to-feedra o min

    29

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    Step 1: Determine the minimum solvent-to-feed ratio

    min. s s nee e o n ou ex raccomposition)

    Procedure:

    1. Draw an operating line from S to Rthat extends

    beyond the boundaries of triangular diagram.

    2. Each tie line is considered to be a pinch point, and line

    drawn rom each t e l ne to operat ng l ne s named P ,

    , Pn.

    3. The Pinch Point farthest away from Ris called Pmin.

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    4. After Pmin is established, a line is drawn from Pmin,

    roug ee compos on , o e o er s e oequilibrium curve. This point will represent E1.

    5. After E1 is known, a mass balance around the system

    = =.

    31

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    Solving for (Smin/F), we get,

    x xS A Amin F M

    F x xA AM S

    Let (S/F)actual = z (S/F)min (1 < z < 2, generally).

    The new mixing point (M) is determined by movingalong the FS line until the new ratio point is reached.

    (FM )

    M S

    l

    l

    32

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    33

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    Step 2: Determine E (extract composition)

    Procedure:

    1. A line is drawn from the raffinate composition (R),

    side of the equilibrium line.

    2. This is the extract (solute-rich solvent) composition

    34

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    Step 3: Determine Operating Point P

    Procedure:

    1. Operating point is a graphical point that represents

    .

    .

    (R) points on diagram (operating line)

    3. Draw a line connecting extract (E) and feed (F). The

    oint at which these two lines intersect P is o eratinpoint P.

    35

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    extract (E) composition

    36

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    Step 4: Number of stages

    Procedure:

    1. Follow tie line from E to the other side of equilibrium

    stage.

    2. Another operating line drawn from operating point P,

    is an equilibrium stage of the system.

    3. This procedure is repeated until stages have been

    constructed to R the raffinate com osition.

    37

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    Step 5: Calculation of unknown flow rates

    Procedure:

    . .

    x F x S x R x EAF AS AR AEx F x S x R x E

    BF BS BR BE

    xAs and xBs = fractions ofA and B for the specifiedstreams.

    F, S, R, and E = flow rates of feed, solvent, raffinate,

    .

    Rand E = unknowns.

    38

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    Step 6: Determination of Extraction Column Diameter

    Procedure:

    1. Diameter of extraction column must be enough to

    flooding.

    2. Estimation of column diameter for L-L contacting-

    contactors due to larger number of important

    variables.

    39

    3 V i bl f l l i l di

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    3. Variables necessary for calculating column diameter

    nc u e:

    Viscosit and densit of continuous hase

    Geometr of internals

    40

    4 C l di t b b t d t i d th h

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    4. Column diameter may be best determined through

    sca e up o a ora ory es runs:

    . P .

    . .

    .

    constant for larger scaled up commercial units.

    iv. The superficial velocity data will be used to

    calculate the column diameter throu h thefollowing correlation derivation:

    41

    5 A t l l it f di d (d l t)

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    5. uD = Actual average velocity of dispersed (droplet)

    p ase.

    uC = Actual average velocity of the continuous phase.

    UD = Superficial velocity of dispersed phase.

    U = Su erficial velocit of continuous hase.

    D= Volume fraction of dispersed phase in column

    ur = verage rop e r se ve oc y re a ve o con nuous

    phase

    = apac ty Parameter or the extract on column

    CD = Drag CoefficientD, C = Densities of dispersed and continuous phases

    =

    42

    {1 } F t hi h t f hi d d i i

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    {1D} = Factor which accounts for hindered rising

    e ec o o er rop e s

    u0 = Characteristic rise velocity for a single droplet

    = Viscosity (subscript will determine component)

    = Interfacial tension subscri t will determinecomponent)

    =

    DT = Column diameter

    g = ccelerat on due to grav ty

    MD, MC = Mass flow rates of dispersed and continuousphases

    43

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    Countercurrent flows ofdispersed and continuous

    liquid phases in a column

    Diameter Calculation Procedure:

    Step A: Determination of Column Total Capacity

    relative to the column wall are:

    UU CDu ; u1D C

    44

    The average droplet rise velocity relative to the

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    The average droplet rise velocity relative to the

    con nuous p ase s e sum o ese ve oc es or

    counter-current operation):

    CDu1

    D D

    This relative velocity is also expressed in terms of

    forces acting upon droplet including drag forces,gravitational forces, and buoyancy forces. These

    variables are combined into one parameter called C:

    dCapacity Paramet P

    3 Cer C

    45

    If the droplet diameter dp is not known C may be

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    If the droplet diameter dp is not known C may be

    o a ne roug a corre a on ea er, . .; en ey,

    E.J.Separation Process Principles, John Wiley and

    , , . ,

    experimental data from operating equipment.

    2 1C Du C 1 12 fr

    C From ex erimental data Ga ler et al. found that the

    right-hand-side of the above equation may be

    ex ressed as:0 D

    46

    Eliminating the relative velocity (u ) by combining the

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    Eliminating the relative velocity (ur) by combining the

    a ove equa ons:UU

    CD1 0 D

    D D

    ...

    D.

    value ofUC/uo.

    This graph represents the holdup curve for the liquid-

    li uid extraction column.

    A typical value ofU /u may be assumed 0.1.

    47

    Typical holdup curve for liquid-liquid extraction

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    Typical holdup curve for liquid-liquid extraction

    Floodin oint

    UD

    UC 0.1

    u0

    u

    0

    48

    At fixed UC an increase in UD results in a increased

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    At fixed UC, an increase in UD results in a increased

    va ue o o upD

    un e oo ng po n s reac e

    at the maximum:

    D D U

    C On the other hand, with UD fixed, UC may be increased

    until the floodin oint is achieved at:

    U 0C D U

    D

    49

    Inserting these derivatives into equation (1) results in

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    Inserting these derivatives into equation (1) results in

    e o ow ng express on orD

    a oo ng con ons.

    The subscript fdenotes flooding:

    1 2U U

    C CU UD fD D

    50

    Apply derivatives of Equation (1) into Equation (2), the

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    Apply derivatives of Equation (1) into Equation (2), the

    express on so ve s mu aneous y resu ng n e

    following Figure for the variation of total capacity as a

    Total Capacity versus Phase flow ratio

    U U

    u

    Asymptotic limit = 0.25

    51

    U U

    D C

    Step B: Characteristic Rise Velocity calculation

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    Step B: Characteristic Rise Velocity calculation

    Dimensionless quantity [(u0 C C)/( )] may be

    . , ,

    experimentally.

    Hence, the characteristic rise velocity for a single

    0.01u

    0C C

    52

    Step C: Calculation of the superficial velocities at

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    p p

    velocity for best performance.

    The sum of superficial velocities is found by reading the

    ratio graph., and multiplying by the characteristic risevelocit then dividin the uantit b 2.

    53

    Step D: Determination of the Total Volumetric Flow rate

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    p

    e o a vo ume r c ow ra e s a unc on o e mass

    flow rates:

    UUCDQ

    D C

    -

    The cross-sectional area is the total volumetric flowrate divided b the sum of the su erficial velocities at

    50% of flooding: Q

    TotalC U UC D 50% floodin

    54

    Step F: Determination of Column Diameter

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    p

    The column diameter may be found from the cross-

    1 2

    D 4A

    55

    Step 7: Determine the Height of the Column

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    Procedure:

    HETS (Height Equivalent to a Theoretical Stage) gives

    .

    ,

    experimental data suggest that the dominant physical

    a. Interfacial tension

    b. Phase viscosities

    56

    c. ens y erence e ween p ases

    HETS is estimated by conducting laboratory

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    exper men s o e erm ne e co umn ame er as

    discussed in step 6.

    These values are scaled to commercial size column by

    diameter raised to an exponent, which may vary from

    . . , .

    following figure shows the HETS for columns and

    rotar contactor.

    57

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    58

    Find Value of (HETS/DT1/3)

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    From Figure, determine the value of (HETS/DT1/3) at a

    .

    the column diameter:

    HETS = (HETS/DT1/3) . DT1/3

    Total height of the column is derived from the number

    of e uilibrium sta es Ste 4 :

    Total Hei ht = HETS Number of E uilibrium Sta es

    59

    Performance of Several Types of Column Extractors

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    Extractor Type 1/HETS,(m1)

    UD + UC,(m/hr)

    Packed Column 1.5 - 2.5 12 - 30

    . .

    Rotating Disk2.5 - 3.5 15 - 30

    Karr Column 3.5 - 7.0 30 - 40

    60

    Important Properties in Liquid-Liquid Extraction:

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    1. Temperature: smaller role in extraction than in other

    .

    streams fed into the column.

    There is no heating requirement for the process and

    For these reasons extraction can be re arded as anisothermal process.

    61

    2. Pressure: small role in extraction.

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    When combined with the temperature considerations,

    -

    phase liquid region.

    Isothermal and isobaric condition is beneficial to

    .

    - -

    dependent.

    62

    3. Activity coefficients: most important; determine the

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    m sc y an ence e par on ac or o e so u e.

    -

    are the most accurate in predicting the activities of the

    ,equations.

    Once a predictive model has been plotted on a

    equilibrium line experimentally for the most accurate

    data.

    63

    4. Viscosity: affects flooding and choice of equipment.

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    Flooding of extraction columns is analogous to

    .

    spray or packed columns.

    64

    COSTS INVOLVED

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    Economic tradeoff exists for the design:

    1. At fixed solvent feed ratio, amount of solute extracted

    . ,value of the unextracted solute is balanced against the

    .

    required decreases as the solvent rate increases. The

    ca acit of the e ui ment necessar for handlin thelarger solvent flow increases with higher solvent rate.

    Thus the cost of the e ui ment asses throu h a

    65minimum.

    3. As solvent rate increases the extract solutions become

    more u e ere ore e cos o so ven remova s

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    more u e. ere ore, e cos o so ven remova s

    increased as well as the operating cost for increased

    .

    .annualized cost (investment and operating costs) must

    reflux rate.

    5. Cost models have been developed for the various types

    of extractor desi n such as column t e extractormixer-settler, and continuous centrifugal extractor.

    66

    REFERENCES

    1 S d J D H l E J S ti P P i i l J h Wil

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    1. Seader, J.D.; Henley, E.J. Separation Process Principles. John Wiley

    and sons, New York, 1999.

    . , . . .

    Company, Houston, 1994.

    3. Sandler, S.I. Chemical and Engineering Thermodynamics. John Wileyand sons, New York, 1998.

    4. Treybal, R.E. Mass Transfer Operations. McGraw-Hill, New York,

    .

    5. Douglas, J.M. Conceptual Design of Chemical Processes. McGraw-Hill, New York, 1988.

    6. Hanson, C.; Baird, M.H.I.; Lo, T.C. Handbook of Solvent Extraction.

    John Wiley and sons, New York, 1983.

    7. Reid, R.C.; Prausnitz, J.; Poling, B. The Properties of Gases and

    Liquids, 4th edition. McGraw-Hill, New York, 1987.

    67