Skip to main content
Engineering LibreTexts

4.26: Untitled Page 81

  • Page ID
    18214
  • Chapter 4

    in which u is the column matrix of unknown mass flow rates and b is the column matrix of known mass flow rates.

      

    b

    2

    m

    1000 lbm / h

     2 

    u

    m

    ,

    b

    500 lb / h

    b

    (4‐143)

    3

    m

     3 

      

    b

     4

    m

     300 lbm / h 

     4 

    We can also make use of the unit matrix

    1 0 0

    I 

    0 1 0

    (4‐144)

    0 0 1

    to express Eq. 4‐142 in the form

    A u = I b

    (4‐145)

    Now we wish to repeat the Gaussian elimination used in Sec. 4.9.1, but in this case we will make use of Eq. 4‐145 rather than Eq. 4‐142 in order to retain the terms associated with Ib . Multiplying the first of Eqs. 4‐134 by ‐0.045 and adding the result to the second equation leads to

    (4‐146)

    Note that this set of equations is identical to Eqs. 4‐135 except that we have included the terms associated with I b . We now proceed with the Gaussian elimination represented by Eqs. 4‐136 through 4‐139 to arrive at (4‐147)

    This result is identical to Eq. 4‐139 except for the fact that we have retained the terms in the second column matrix that are associated with I b in Eq. 4‐145. In

    Multicomponent systems

    144

    Sec. 4.9.1, we used Eq. 4‐139 to carry out a backward elimination in order to solve for the mass flow rates given by Eq. 4‐140, and here we want to present that backward elimination explicitly in order to demonstrate that it leads to the inverse of the coefficient matrix, A . This procedure is known as the Gauss-Jordan algorithm and it consists of elementary row operations that reduce the left hand side of Eq. 4‐147 to a diagonal form.

    We begin to construct a diagonal form by multiplying the third equation by 37.917 and subtracting it from the second equation so that our set of equations takes the form

    m

      m  m 

    b

    0

    0

     1000 lb / h

    2

    3

    4

    2

    m

    0  m

     0   0 . 1152 b  0 . 0677 b  1 . 2326 b  288 . 42 lb / h (4‐148) 3

    2

    3

    4

    m

    0  0

    m   0 0464

    .

    b  1 . 0971 b  0 . 03251 b  492 . 39 lb / h 4

    2

    3

    4

    m

    We now multiply the third equation by 1

     and add the result to the first

    equation to obtain

    m

      m  0  1 0464

    .

    b  1 . 09711 b  0 03251

    .

    b  507 61

    .

    lb / h

    2

    3

    2

    3

    4

    m

    0  m

     0   0 . 1152 b  0 . 0677 b  1 . 2326 b  288 . 42 lb / h (4‐149) 3

    2

    3

    4

    m

    0  0  m

      0 0464

    .

    b  1 . 0971 b  0 03251

    .

    b  492 . 39 lb / h

    4

    2

    3

    4

    m

    Finally we multiply the second equation by 1

     and add the result to the first

    equation to obtain the desired diagonal form

    (4‐150)

    Here we see that the unknown mass flow rates are determined by (4‐151)

    where only three significant figures have been listed since it is unlikely that the input data are accurate to more than 1%. The coefficient matrix contained in the central term of Eqs. 4‐150 is the inverse of A and we express this result as

    145