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4.17: Untitled Page 72

  • Page ID
    18205
  • Chapter 4

    Table 4‐3. Specified and unknown conditions

    Stream #1

    Stream #2

    Stream #3

    M 1 = 1200 mol/h

    ?

    ?

    x A = 0.3

    x A = 0.6

    x A = 0.1

    x B = 0.2

    x B = 0.3

    ?

    ?

    ?

    ?

    When the spaces identified by question marks have been filled with results, our solution will be complete. We begin our analysis with the simplest calculations and make use of the constraints given by Eqs. 4‐90. These can be used to express Table 4‐3 as follows:

    Table 4‐4. Unknowns to be determined

    Stream #1

    Stream #2

    Stream #3

    M 1 = 1200 mol/h

    ?

    ?

    x A = 0.3

    x A = 0.6

    x A = 0.1

    x B = 0.2

    x B = 0.3

    x B = 0.9 ‐ x C

    x C = 0.5

    x C = 0.1

    ?

    This table indicates that we have three unknowns to be determined on the basis of the three species balance equations. Use of the results given in Table 4‐4 allows us to express the balance equations given by Eqs. 4‐89 as

    Species A:

    0 6

    . M

     0 1

    . M

    0

     360 mol/h

    (4‐91a)

    2

    3

    Species B:

    0 3

    . M

     0 9

    . M

    M  x

     240 mol/h

    (4‐91b)

    2

    3

    3

    C 3

    

    bi‐linear

    Species C:

    0 1

    . M

    0 

    M

      x

     600 mol/h

    (4‐91c)

    2

    3

    C 3

    

    bi‐linear

    in which the product of unknowns,

    and 

    C

    x 3 , has been identified as a

    bi‐linear form. This is different from a linear form in which and 

     would

    C

    x 3

    appear separately, or a non‐linear form such as or and

    2

    x  . In this

    C 3

    problem, we are confronted with three unknowns,

    ,

    and 

     , and three

    C

    x 3

    equations that can easily be solved to yield

    Multicomponent systems

    126

    (4‐92)

    This information can be summarized in the same form as the input data in order to obtain

    Table 4‐5. Solution for molar flows and mole fractions

    Stream #1

    Stream #2

    Stream #3

    M 1 = 1200 mol/h

    M 2 = 480 mol/h

    M 3 = 720 mol/h

    x A = 0.3

    x A = 0.6

    x A = 0.100

    x B = 0.2

    x B = 0.3

    x B = 0.133

    x C = 0.5

    x C = 0.1

    x C = 0.767

    The structure of this ternary distillation process is typical of macroscopic mass balance problems for multicomponent systems. These problems become increasing complex (in the algebraic sense) as the number of components increases and as chemical reactions are included, thus it is important to understand the general structure. Macroscopic mass balance problems are always linear in terms of the compositions and flow rates even though these quantities may appear in bi‐linear forms. This is the case in Eqs. 4‐91 where an unknown flow rate is multiplied by an unknown composition; however, these equations are still linear in

    and 

     , thus a unique solution is possible. When

    C

    x 3

    chemical reactions occur, and the reaction rate expressions (see Chapter 9) are non‐linear in the composition, numerical methods are generally necessary and one must be aware that nonlinear problems may have more than one solution or no solution.

    4.7.3 Solution of sets of equations

    To illustrate a classic procedure for solving sets of algebraic equations, we direct our attention to Eqs. 4‐91. We begin by eliminating the term,

    ,

    from Eq. 4‐91b to obtain the following pair of linear equations: Species A:

    (4‐93a)

    Species B:

    (4‐93b)

    To solve this set of linear equations, we make use of a simple scheme known as Gaussian elimination. We begin by dividing Eq. 4‐93a by the coefficient 0.6 in order to obtain

    127