Binary Distillation

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Binary Distillation SHR Chapter 7 1 BinaryDistillation.key - March 28, 2014

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operaciones unitarias.

Transcript of Binary Distillation

  • Binary Distillation

    SHR Chapter 7

    1 BinaryDistillation.key - March 28, 2014

  • IntroductionDates back to 1

    st century AD

    first used in a batch mode (distillate changes in time)

    Goal: separate heavy key (less volatile) from light key (more volatile) by exploiting 1.

    for 1 or 1, this can be done very effectively

    unless an azeotrope exists (where =1).

    then we recover the azeotrope and the light or heavy key,

    depending on which side the feed lies.

    By 16th century, multiple stages were in use to

    improve separation.

    By 1976, distillation accounted for nearly 3% of the US energy consumption!

    mostly in petroleum refineries

    Binary distillation is simplest & most well-understood.

    we will limit our discussion to binary distillation

    Third Law of Thermo - typically low thermodynamic efficiency.

    SHR 7.1

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  • Design ConsiderationsOperating pressure - knob 1

    below ambient pressure requires vacuum operation

    many things may influence choice of operating pressure

    Thermo: azeotrope formation, , etc.

    Column operating temperature range (avoiding reactions,

    corrosion, etc.)

    most analyses do not account for pressure variation through the column

    Operating temperature - knob 2

    Reboiler & Condenser:

    Bottoms above ambient requires additional energy input to the

    reboiler

    Distillate below ambient requires energy removal from the condenser.

    Thermodynamics: critical points of fluids

    F

    D

    B

    V

    L

    L

    V

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  • Batch Distillation

    Conceptually, follows the T-x-y diagram.

    More rigorous analysis in SHR chapter 13

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  • The McCabe-Thiele Graphical Method

    1925

    Continuous (staged) distillation

    SHR 7.2

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  • NomenclatureSpecifications

    F total (molar) feed ratez LK mole fraction in feedP Column operating pressurex LK mole fraction in distillatex LK mole fraction in bottoms

    R/R reflux ratioFeed phase condition

    VLE data (y/x plot)Type of condenser (partial/total)

    Type of reboiler (partial/total)

    ResultsD Distillate (molar) flow rateB Bottoms (molar) flow rate

    N minimum number of stagesR minimum reflux flow rateV Boilup ratioN Number of equilibrium stages

    Feed stage location

    Stage compositions (

    Light Key (LK) - more volatile component

    Heavy Key (HK) - less volatile component

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  • Preliminaries

    Rectifying section - like an absorber

    Feed & reboiler supply vapor

    Condenser supplies liquid

    Stripping section - like a stripper

    Feed & condenser supply liquid

    Reboiler supplies vapor

    FzF = xDD + xBBF = D +B

    D = F

    zF xBxD xB

    light-key mole balance:overall mole balance:

    combine to eliminate B & solve for D:

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  • Rectifying Section Operating LineLight key mole balance:

    relates light-key compositions in passing streams

    (streams on a stage are assumed to be in equilibrium)

    If L and V are constant, then this is a straight line.

    Both components have equal and constant molar enthalpies of vaporization (latent heats).

    Sensible heat, CpT, is negligible compared to latent heat.

    Column is insulated (no heat loss on each stage).

    Column pressure is constant (thermodynamics can be

    done at a single pressure).

    Big assumptions, but allow for simple analysis, since L and V are constant under these assumptions.

    The McCabe-Thiele Assumptions

    L

    V=

    L

    L+D=

    L/DL/D + D/D

    =R

    R+ 1D

    V=

    1

    R+ 1

    Overall mole balance: V = L+D

    y =

    R

    R+ 1

    x+

    1

    R+ 1

    xD

    R LD

    reflux ratio

    operat

    ing lin

    e

    SHR 7.2.1

    V yn+1 = Lxn +DxD

    yn+1 =L

    Vxn +

    D

    VxD

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  • Range of Reflux Ratiosy =

    L

    Vx+

    D

    VxD

    0 L

    V=

    R

    R+ 1

    1 because 0 R 1

    What happens at R = 0?

    What happens at R = ?

    What is the minimum R that allows separation?

    (We will answer this question shortly)

    R =L

    D

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  • relates light-key compositions in passing streams

    (streams on a stage are assumed to be in equilibrium)

    Stripping Section Operating Line

    Light key mole balance:

    Overall mole balance:

    Lxm = V ym+1 +BxBMcCabe-Thiele assumptions

    have been applied.

    ym+1 =L

    Vxm B

    VxB

    L VThe feed stage material balance relates L and V to and

    y =

    VB + 1

    VB

    x

    1

    VB

    xB

    VB VB

    boilup ratio

    oper

    atin

    g lin

    e

    SHR 7.2.2

    L = V +D

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  • Feed Stage & the q-Line

    subcooled liquid saturated liquid partially vaporized saturated vapor superheated vapor

    rectifying section

    stripping section

    liquid flow increase across feed rate normalized by feed rate.q =

    L LF

    = 1 +V VF

    q > 1 q = 1 q = LF/F q = 0 q < 0

    Operating lines & q-line must intersect at

    a single point.

    cannot specify q, VB and R independently.

    SHR 7.2.3

    q =hsat. vaporF hF

    hsat. vaporF hsat. liquidF

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  • More on the q-liney =

    L

    Vx+

    D

    VxD

    rectifying section:

    y =L

    Vx B

    VxB

    stripping section:

    subtract

    q =L LF

    = 1 +V VF

    y

    V VF

    =

    L LF

    x+ zF

    y (1 q) = qx+ zFy =

    q

    q 1x zF

    q 1

    Typically the feed condition is known (specifying q).

    Then we can choose VB or R.Note: specifying R implies VB.

    yV V = x L L+DxD +BxB| {z }

    FzF

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  • Feed Stage & Number of Stagestoo low

    too high

    just right

    Locate feed stage nearest to the intersection of the operating lines & q-line as possible (just after

    the horizontal line on the staircase passes P

    SHR 7.2.4

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  • Partial Reboilers & Condensers

    F

    D

    B

    V

    L

    L

    VTotal Reboiler:

    all liquid is turned back to vapor

    Partial reboiler:

    bottoms product is liquid, boilup is vapor

    This is another equilibrium stage! very common...

    Total condenser:

    all vapor is condensed back to liquid

    Partial condenser:

    distillate is vapor, reflux is liquid

    This is another equilibrium stage!

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  • Limiting Cases: Rmin, Nmin.eq

    uilibr

    ium cu

    rve

    x

    y

    x = y

    0 1.0

    1.0

    x = xDx = xB

    equil

    ibrium

    curve

    xy

    x = y

    0 1.0

    1.0

    x = zFx = xDx = xB

    Total reflux

    R = , VB = .

    L = V, D = B = F = 0.

    F = 0, N = Nmin.

    y = x is operating line.

    No product...

    Minimum reflux

    N = .

    higher capital costshigher operating costs

    x

    y

    x = y

    0 1.0

    1.0

    x = xDx = xBx = zF

    1

    2

    3

    pinch point

    (L/V )min =Rmin

    Rmin + 1

    Rmin =(L/V )min

    1 (L/V )min(VB)min =

    L/V

    max

    11F

    D

    B

    V

    L

    L

    V

    SHR 7.2.4

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  • Perfect Separation - Another Limiting Case

    x

    yx =

    y

    0 1.0

    1.0

    x = zF

    Perfect separation: xB = 0, xD = 1.

    Pinch points form in each section of the column.

    Theoretical value for minimum reflux ratio and boilup to achieve perfect separation.

    To find this:

    Obtain x-y data from thermo.

    Determine q-line

    Determine slope of rectifying

    operating line = Rmin/(Rmin+1).

    Rmin =1

    zF ( 1)For saturated liquid feed,

    SHR 7.2.4

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  • ExampleWe want to separate a mixture of n-heptane and n-octane using distillation at

    atmospheric pressure.

    If the feed is 40 mole% n-heptane as a saturated vapor, determine the minimum reflux ratio and minimum number of stages required to obtain

    product streams with 95% and 5% n-heptane.

    F

    D

    B

    V

    L

    L

    V Known:

    xD = 0.95

    xB = 0.05

    zF = 0.4, saturated vapor

    Needed:

    K-values (equilibrium curve)

    q-line

    operating lines

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