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5-Vapor Liquid Equilibrium._depriester Chart

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SKF 2213 Chemical Engineering Thermodynamics Chapter 10: Vapor/Liquid Equilibrium
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  • SKF 2213 Chemical Engineering Thermodynamics

    Chapter 10: Vapor/Liquid Equilibrium

  • It is expected that students have the ability to: Describe the behaviour of VLE and how to simplify the VLE problem. Derive and simplify equations of VLE. Apply simplified VLE equations to obtain data for P-XY, T-XY and X-

    Y diagrams. Apply Raoults law and Henrys law to solve simple thermodynamic

    problems. Carry out bubble and dew point calculations for a given mixture Carry out flash calculation in order to determine the vapor/liquid

    fraction as well as the mixture composition of each phase at specified conditions using available K-Values etc.

  • Binary mixturefor species 1 and 2.

    RKAC1 is vaporpressure vs. T curve for species 1

    UBHC2 is vaporpressure vs. T curve for species 2

    C1 and C2 are critical points for species 1 & 2

    Species 1 is more volatile.

  • Under surface issaturated-vapor (P,T,y1 surface)

    Below that is gasphase.

    Top surface issaturated-liquid(P,T,x1 surface)

    Above that isliquid phase.

  • AEDBLA givesFigure 10.2 (a)Pxy phase diagram at constant T

    KJIHLK givesFigure 10.2 (b)Txy phase diagram at constant P

    F (liquid solution)L (bubble point)W (dew point)G (Gas mixture)

    LV = tie lineL = bubble pointW = dew point

    P

  • Note: species 1 has higher vapor pressure or lower boiling point than species 2

    Most chemical processes operate at this P, T range

    Gas

    Gas

    Liquid

    Liquid

    Tieline

    1 0x2y2

  • Commonly found shape of Pxy

    diagrams at constant T

    This is similar shape toFigure 10.2 (a)

  • Commonly found shape of Txy

    diagrams at constant P

    This diagram is of practical interest as

    most VLE applications occur at

    constant P(example in

    distillation column)

    This is similar shape toFigure 10.2 (b)

  • y1-x1 diagrams at constant P

    Azeotrope

    Azeotropex1=y1 x2=y2Liquid composition= Gas composition

  • For Raoults law, needs PisatTherefore requires T
  • K-value Using DePriester Chart

    For light hydrocarbon mixture where Kiis essentially function of T and P only, the valuecould be simply tabulated as in DePriester chart.

  • Read value of K-valueat given T and Pe.g. 1000 kPa and 0 oC

    for methane

  • Bubblepoint & DewpointCalculations

  • BUBL P: Calculate {yi} and P, given {xi} and T

    To calculate the P when the 1st bubble appear as a result of decrease in P at constant T

    To calculate the P when the last bubble disappear as a result of increase in P at constant T

  • DEW P: Calculate {xi} and P, given {yi} and T

    To calculate the P when the 1st dew (drop of liquid) appear as a result of increase in P at constant T

    To calculate the P when the last dew disappear as a result of decrease in P at constant T

  • BUBL T: Calculate {yi} and T, given {xi} and P

    To calculate the T when the 1st bubble appear as a result of increase in T at constant P

    To calculate the T when the last bubble disappear as a result of decrease in T at constant P

  • DEW T: Calculate {xi} and T, given {yi} and P

    To calculate the T when the 1st dew (drop of liquid) appear as a result of decrease in T at constant P

    To calculate the T when the last dew disappear as a result of increase in T at constant P.

  • Overall mole balanceT L V

    Component mole balanceTzi Lxi Vyi

    Let T=1 mol, so V and L are mole fractionszi Lxi Vyi zi (1V )xi Vyi (A)

    Derivation

  • Substitute yi Kixi into (A)zi (1V )xi KixiV xi (1V VKi ) xi (1V (Ki 1))xi

    zi1V (Ki 1)

    Substitute xi yiKi

    into (A)

    zi (1V )yiKi

    yiV zi Ki (1V )yi yiVKi

    yi zi Ki

    1V (Ki 1) (10.16)

  • xi yi 0zi

    1V(Ki 1) ziKi

    1V(Ki 1) 0

    zi ziKi1V(Ki 1) 0

  • Bubblepoint Calculation

    So at bubble point, L=1, V=0 and zi xi

    (xi xiKi) 0xi xiKi xiKi 1 (10.13)

    zi ziKi1V(Ki 1) 0

  • Bubblepoint Calculation

    xiKi 1 (10.13)If Raoult's Law valid,

    xiPi

    sat

    P 1 P xiPisat (10.2) see example 10.1

    If Modified Raoult's Law valid,

    xi iPisat

    P 1 P xi iPisat (10.6) see example 10.3

  • Dewpoint Calculation

    So at dewpoint, L=0, V=1 and zi yi

    yi yiKiKi

    0yiKi yi 0

    yiKi

    =1 (10.14)

    zi ziKi1V(Ki 1) 0

  • Dewpoint CalculationyiKi

    =1 (10.14) If Raoult's Law valid,

    yiPi

    sat

    P

    =1 P 1yi

    Pisat

    (10.3), see example 10.1

    If Modified Raoult's Law valid,yi

    iPisatP

    =1 P 1yi

    iPisat (10.7), see example 10.3

  • Example 10.1

  • Example 10.1 (cont)

  • Plot Px1y1 at T=75oC

    lnP1sat / kPa14.2724 2945.47

    T / oC224.00lnP2

    sat / kPa14.2043 2972.64T / oC209.00

    At 75C,P1

    sat 83.21kPa P2sat 41.98kPa

    Acetonitrile(1)/Nitromethane(2)Antoine Eqn,

    Acetonitrile(1) is more volatile

    Example 10.1 (xtra1)

  • Plot Px1y1 at T=75oC

    Calculate P and y1, given a set of x1 and T=75oCThis is BUBL P calculation.

    xiKi 1 (10.13) . For Raoult's Law, P xiPisat (10.2) P x1P1sat x2P2sat x1P1sat (1 x1)P2satP (P1sat P2sat )x1 P2sat note: a linear line (y=mx+c)

    Calculate P for a set of x1 and then calculate y1using,

    y1 x1P1

    sat

    P

    Example 10.1 (xtra2)

  • Plot Px1y1 at T=75oC y1

    x1P1sat

    P

    183.211750.888074.960.8750.748366.720.6750.569258.470.4750.331350.230.275041.98075y1P(kPa)x1T

    P (P1sat P2sat )x1 P2sat

    P1sat

    P2sat

    So now plot Px1y1!!

    Point b

    Example 10.1 (xtra3)

  • a liquid solutionor subcooled

    liquid mixture

    b sat liquid solution or bubblepointPb is by BUBL P

    c sat vapor mixtureor dewpointPd is by DEW P

    d gas mixtureor superheated

    vapor mixtureData from previous slide

    Example 10.1 (xtra4)

  • DEW P calculationCalculate Pd and x1, given y1 and T

    So lets calculate Pd at z1=0.6 and T=75oC.This is point c in previous Px1y1 diagram. Note: z1=y1

    Then calculate x1 using,

    x1 y1PdP1

    sat 0.6(59.74)83.21 0.43

    yiKi

    =1 (10.14) If Raoult's Law valid, Pd 1yiPi

    sat (10.3)

    Pd 10.683.21

    0.441.98

    59.74kPa

    Example 10.1 (xtra5)

  • Plot Tx1y1 at P=70kPa

    T1sat / oC 2945.47

    14.2724 lnP / kPa 224.00

    T2sat / oC 2972.64

    14.2043 lnP / kPa 209.00

    so at 70kPa,T1

    sat 69.84C T2sat 89.58C

    Acetonitrile(1)/Nitromethane(2)

    Antoine Eqn,

    Acetonitrile(1) is more volatile

    Example 10.1 (xtra6)

  • Plot Tx1y1 at P=70kPaChoose T between T1sat and T2sat , calculate P1sat and P2sat and use these to calculate x1 by the following eqn.

    xiKi 1 (10.13) . For Raoult's Law, P xiPisat (10.2) P x1P1sat x2P2sat x1P1sat (1 x1)P2sat

    x1 P P2sat

    P1sat P2sat

    Calculate x1 for a set of T and then calculate y1 using,

    y1 x1P1

    sat

    P

    Example 10.1 (xtra7)

  • Plot Tx1y1 at P=70kPa y1

    x1P1sat

    P

    0 (x2=1)0.14240.31840.51560.73781 (x2=0)x1

    0 (y2=1)89.58 T2sat700.240186700.474282700.675978700.848474701 (y2=0)69.84 T1sat70y1T(oC)P (kPa)

    So now plot Tx1y1!!

    x1 PP2sat

    P1sat P2sat

    Example 10.1 (xtra8)

  • c sat vapor mixtureor dewpointTd is by DEW T

    d gas mixtureor superheated

    vapor mixture

    a liquid solutionor subcooled

    liquid mixture

    b sat liquid solution or bubblepointTb is by BUBL T

    Data from previous slide

    Example 10.1 (xtra9)

  • BUBL T calculationCalculate Tb and y1, given x1 and P.

    So lets calculate Tb at z1=0.6 and P=70kPa.This is point b in previous Tx1y1 diagram. Note: z1=x1

    xiKi 1 (10.13), For Raoult's Law, Pb xiPisat (10.2) Pb xiPisat Pksat xi Pi

    sat

    Pksat = Pksat xiik

    Pksat Pb

    xiik (A) , where ik Pi

    sat

    Pksat Relative volatility

    lnik ln Pisat

    Pksat =lnPi

    sat-lnPksat = Ai

    BiT Ci

    - Ak Bk

    T Ck

    The solution is not straightforward as T is unknown.

    Example 10.1 (xtra10)

  • BUBL T calculationSolution is through iteration,

    1. Start with an initial guess of T

    T xi Tisat2. Arbitrarily pick a component, e.g. k=2

    3. Calculate ik, (note: Number of ik is equal to number of component)

    lnik Ai BiT Ci

    - Ak Bk

    T Ck

    T 0.6(69.84)0.4(89.58)77.74oC

    12 1.9611 22 1

    Example 10.1 (xtra11)

  • BUBL T calculation

    5. Calculate new value of T,

    6. Stop if this is equal or close to earlier value of T, else use this value as a new guess. Repeat step 3,4 & 5 until converge.

    T Bk

    Ak lnPksatCk T

    2972.6414.2043 ln44.3977 20976.53

    oC

    4. Calculate Pksat using eqn A, Pksat P

    xiik P2

    sat Px112 x222

    = 700.6(1.9611)+0.4(1)

    =44.3977kPa

    Example 10.1 (xtra12)

  • BUBL T calculation

    7. Finally, calculate yi using Raoults law (Use Antoine Eqn for Pisat)

    y1=0.7472 (point b)

    76.4244.221.971776.4376.4344.241.970376.53TP2sat12T

    Answer(point b)

    Example 10.1 (xtra13)

  • Example 10.1 (xtra14)

  • DEW T calculation

    Calculate Td and x1, given y1 and P.

    Calculate Td and x1 for z1=0.6 and P=70kPaSee example 10.1 page 356 for the solution (by iteration)DEW T calculation.

    Answer: Td =79.58oC (point c)x1=0.4351 (point c)

    Example 10.1 (xtra15)

  • Example 10.1 (xtra16)

  • Example 10.2 (Henrys law)

  • Example 10.3 (Modified Raoults law)

  • Example 10.3 (cont)

  • Example 10.3 (cont)

  • Example 10.3 (cont)

  • Example 10.4 (K-value correlations)

  • Example 10.5 (Flash calculations)


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