KremserEquation (1)

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    The Kremser Equation

    Tore Haug-Warberg

    Dept. of Chemical Engineering

    October 1st, 2004

    The Kremser Equation represents an analytical solution to a classical sepa-

    ration problem ofN ideal equilibrium stages connected with countercurrent

    gas and liquid flows:

    V1

    1

    V2

    Vn

    n

    Vn+1

    VN

    N

    VN+1

    L0

    L1

    Ln1

    Ln

    LN1

    LN

    It is assumed that one (single) chemical component takes part in the separa-

    tion, or, if the system is multicomponent, that all the components are being

    separated independently. Moreover, the equilibrium relation is assumed to

    be linear (Henrys law). The component balance is written:

    Lnxn + VN+1yN+1 = LNxN+ Vn+1yn+1

    Constant molar flow assumption:

    Ln = Ln+1 = = LN = LConstant

    and Vn+1 = Vn+2 = = VN+1 = VConstant

    Combine the equations above into a simplified component balance:

    L (xn xN) = V(yn+1 yN+1) , n [0,N]

    The compositions (xn+1,yn+1) for n [0,N are assumed to represent true

    equilibrium pairs whereyn+1 = mxn+1 is a linear equilibrium relation (Henrys

    law). The mass balance becomes L (xn xN) = V(mxn+1 yN+1), or on a

    slightly modified form (operating line):

    xn+1 L

    mVxn =

    yN+1

    m

    L

    mVxN

    A L/mV is the so-called absorption factor (problem dependent). The

    final balance equation reads

    xn+1 Axn =yN+1

    mAxN

    Constant c

    , n [0,N]

    These are the N+ 1 equations that must be solved simultaneously to fix

    the concentration profile of the column. The equations constitute a set of

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    coupled algebraic equations ideally suited for matrix algebra. Let us start

    by visualizing the coupling in matrix terms:

    AA...A

    1 0 0

    A 1 0

    0 A 1. . .

    A 1

    x0x1x2...

    xN

    =

    x0c

    c...

    c

    The first (and only) stage in the elimination process is indicated to the left.

    The outcome of the elimination is

    1 0 0

    0 1 00 0 1

    . . .

    0 1

    x0

    x1x2...

    xN

    =

    x0

    c + Ax0c + A(c + Ax0) = c + cA + A

    2x0...

    cN

    k=1 Ak1

    +ANx0

    The vector elements on the right hand side are recognized as members of

    the geometric seriesNk=1

    Ak1 =1 AN

    1 A

    The solution ofxN can thereby be written

    xN = cNk=1

    Ak1 + ANx0 = c1 AN

    1 A+ ANx0

    This equation constitutes The Solution to the original problem, and the

    remaining part of the note shows simply(?) how to tweak it into something

    recognizable from Geankoplis. Substitutes the definition ofc yN+1m

    AxN

    xN =

    yN+1

    m AxN

    1 AN

    1 A+ ANx0

    and isolates an expression for xN (the right hand side is now in terms of

    column input parameters only)

    xN =

    yN+1m

    1AN1A

    +ANx0

    1 + A

    1AN

    1A

    Multiplication on both sides by 1A

    1Agives

    xN = xN 1 A

    1 A1

    =

    yN+1m

    1 AN

    + ANx0 (1 A)

    (1 A) + A (1 AN)

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    Add and subtract the term1 AN

    x0 in the nominator

    xN =

    yN+1m

    1 AN

    + ANx0 (1 A) +

    0

    1 AN

    x0

    1 AN

    x0

    1 AN+1

    =

    yN+1m

    x0

    1 AN+ x0

    1 AN+1

    1 AN+1

    =

    1 AN

    1 AN+1

    yN+1

    m x0

    + x0

    Collect all the concentration variables on the left hand side and the absorp-

    tion parameters to the right

    x0 xN

    x0 yN+1m

    =1 AN

    1 AN+1=

    1 AN

    1 AN+1A(N+1)

    A(N+1)1

    =

    1A

    N+1

    1A

    1A

    N+1 1

    Eq. 10.3-21

    =S N+1 S

    S N+1 1

    Here, S A1 mV/L is the so-called stripping factor. Rather than

    asking for xN we can ask for y1. The same procedure repeated leads to:

    yN+1 y1

    yN+1 mx0= A

    N+

    1 AAN+1 1

    Eq. 10.3-24

    The equations above express the concentration differences in a column of

    fixed size (number of trays N). We may occasionally want to invert the

    relationship and calculate Ngiven a separation specification :

    x0 xN

    x0 yN+1m

    =1 AN

    1 AN+1=

    1 AN

    1 A AN

    Rewrites the last equation into 1 AN

    = AAN

    which yields

    AN =1

    1 A= N=

    ln

    11A

    lnA