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Engineering LibreTexts

4.24: Untitled Page 79

  • Page ID
    18212
  • Chapter 4

    We suppose that A is invertible and that the inverse, along with u and b , are given by

     11

    a

    12

    a

    13

    a

    14

    a

     1

    u

     1

    b

     

     

    a

    a

    a

    a

    u

    b

    1

    21

    22

    23

    24

    2

    2

    A

     

    ,

    u

       ,

    b

      

     

     

    (4‐127)

    31

    a

    32

    a

    33

    a

    34

    a

    3

    u

    3

    b

     

     

     41

    a

    42

    a

    43

    a

    44

    a

    n

    u

    n

    b

    We can multiply Eq. 4‐126 by

    1

    A to obtain

    1

    1

    A

    Au

    I u

    u

    A

    b

    (4‐128)

    and the details are given by

    1 0 0 0  1

    u

     1

    a 1

    1

    a 2

    1

    a 3

    1

    a 4   1

    b

      

      

    0 1 0 0 u

    a

    a

    a

    a

    b

    2

    21

    22

    23

    24

    2

        

      

    (4‐129)

    0 0 1 0 

      

    3

    u

    3

    a 1

    3

    a 2

    3

    a 3

    3

    a 4

    3

    b

      

      

    0 0 0 1  n

    u

     41

    a

    42

    a

    4

    a 3

    44

    a   n

    b

    Carrying out the matrix multiplication on the left hand side leads to

     1

    u

     1

    a 1

    1

    a 2

    1

    a 3

    1

    a 4   1

    b

     

      

    u

    a

    a

    a

    a

    b

    2

    21

    22

    23

    24

    2

       

      

     

      

    (4‐130)

    3

    u

    3

    a 1

    3

    a 2

    3

    a 3

    3

    a 4

    3

    b

     

      

    n

    u

     41

    a

    42

    a

    43

    a

    44

    a   n

    b

    while the more complex multiplication on the right hand side provides

     1

    u

     11

    a

    1

    b

    12

    a

    2

    b

    13

    a

    3

    b

    14

    a

    4

    b

     

    u

    a b

    a b

    a b

    a b

    2

    21 1

    22 2

    23 3

    24 4

       

     

    (4‐131)

    3

    u

    31

    a

    1

    b

    32

    a

    2

    b

    33

    a

    3

    b

    34

    a

    4

    b

     

    n

    u

     41

    a

    1

    b

    42

    a

    2

    b

    43

    a

    3

    b

    44

    a

    4

    b

    Each element of the column matrix on the left hand side is equal to the corresponding element on the right hand side and this leads to the solution for the unknowns given by

    Multicomponent systems

    140

    1

    u

    11

    a

    1

    b

    12

    a

    2

    b

    13

    a

    3

    b

    14

    a

    4

    b

    2

    u

    21

    a

    1

    b

    22

    a

    2

    b

    23

    a

    3

    b

    24

    a

    4

    b

    (4‐132)

    3

    u

    31

    a

    1

    b

    32

    a

    2

    b

    33

    a

    3

    b

    34

    a

    4

    b

    4

    u

    41

    a

    1

    b

    42

    a

    2

    b

    43

    a

    3

    b

    44

    a

    4

    b

    It should be clear that there are two main problems associated with the use of Eqs. 4‐125 to obtain the solution for the unknowns given by Eq. 4‐132. The first problem is the correct interpretation of a physical process to arrive at the original set of equations, and the second problem is the determination of the inverse of the matrix A .

    There are a variety of methods for developing the inverse for a matrix; however, sets of equations are usually solved numerically without calculating the inverse of a matrix. One of the classic methods is known as Gaussian elimination and we can illustrate this technique with the problem that was studied in Sec. 4.8. We can make use of the data provided in Figure 4‐10 to express Eqs. 4‐105 in the form

    species A:

    (4‐133a)

    species B:

    (4‐133b)

    Total:

    (4‐133c)

    and our objective is to determine the three mass flow rates,

    ,

    and

    . The

    solution is obtained by making use of the following three rules which are referred to as elementary row operations:

    I. Any equation in the set can be modified by multiplying or

    dividing by a non‐zero scalar without affecting the

    solution.

    II. Any equation can be added or subtracted from the set without affecting the solution.

    III. Any two equations can be interchanged without affecting

    the solution.

    For this particular problem, it is convenient to arrange the three equations in the form

    141