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5.24: Untitled Page 121

  • Page ID
    18254
  • Chapter 5

    5‐31. In the previous problem we made use of a Taylor series expansion to obtain a simplified expression for the chemical potential as a function of the mole fraction. A Taylor series expansion is a powerful tool for predicting the value of a function at some position x b when information about the function is only available at x a . If we think about the definite integral given by x b

    f ( b) 

    f ( a) 

    df dx

    dx

    (1)

    x a

    we see that we can determine f ( b) , given f ( a) , only if we know the derivative, df / dx everywhere between a and b. To see how we can use Eq. 1 to determine f ( b) using only information at x a , we first make use of the change of variable, or transformation, defined by

      x b

    (2)

    This leads to the relations

    dx d,

    df dx df d

    (3)

    which can be used to express Eq. 1 in the form

      0

    f ( b) 

    f ( a) 

    df d

    d

    (4)

      a b

    Remember that the rule for differentiating a product

    d

    dV

    dU

    UV  U

    V

    d

    d

    d

    (5)

    is the basis for the technique known as integration by parts where it is employed in the form

    dV

    d

    dU

    U

    UV 

    d

    d

    V d

    (6)

    If we make use of the two representations

    df

    U

    ,

    V

     

    d

    (7)

    Eq. 6 provides the following identity for the derivative of f with respect to :

    index-232_1.png

    index-232_2.png

    Two‐Phase Systems & Equilibrium Stages

    223

    df

    d df

    d df

      

    (8)

    d

    d  d 

    d

    d

    Use this result in Eq. 4 to produce the second term in the Taylor series expansion for f ( b) about f ( a) . Repeat this entire procedure to extend the representation for f ( b) to obtain the third term in the Taylor series expansion and thus illustrate how one obtains a representation for f ( b) that involves only the function and its derivatives evaluated at x a . Keep in mind that the only mathematical tools used in this derivation are the definition of the definite integral and the rule for differentiating a product (Stein and Barcellos, 1992, pages 134 and 266).

    5‐32. A gas mixture leaves a solvent recovery unit as illustrated in Figure 5.32.

    Figure 5.32. Recovery‐condenser system

    The partial pressure of benzene in this stream is 80 mm Hg and the total pressure is 750 mm Hg. The volumetric analysis of the gas, on a benzene‐free basis, is 15%

    CO , 4% O and the remainder is nitrogen. This gas is compressed to 5 atm and 2

    2

    cooled to 100 F. Calculate the percentage of benzene condensed in the process.

    Assume that CO , O and N are insoluble in benzene, thus the liquid phase is 2

    2

    2

    pure benzene.

    224