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3.19: Untitled Page 45

  • Page ID
    18178
  • Chapter 3

    in Figure 3.5b. The next step in the bisection method is to bisect the distance between

    and

    to produce the value indicated by

    o

    x

    1

    x

    x  ( x x ) / 2 . Next one evaluates

    2

    1

    o

    H( x , )

    2

    in order to determine

    whether it is positive or negative. From Figure 3.5b we see that Figure 3.5b. Graphical iterative solution for x   h

    H( x  

    2 , )

    0 , thus the next estimate for the solution is given by x

    3

    ( x 2

    1

    x ) / 2 . This type of geometrical construction is not necessary to carry out the bi‐section method. Instead one only needs to evaluate H( x

    2 , ) H( 1

    x , )

     to determine whether it is positive or negative. If H( x

    2 , ) H( 1

    x , )

    0 the next estimate is given by x

    ( x

    x ) / 2 .

    3

    2

    1

    However, if H( x , )

    H x ,   0 the next estimate is given by

    2

    ( 1 )

    Single component systems

    75

    x

    3

    ( x 2

    o

    x ) / 2 . This procedure is repeated to achieve a converged value

    of x  0.3832 as indicated in Table 3.5. Given the solution for the dimensionless depth defined by Eq. 21, and given the specified dimensionless time defined by Eq. 22, we can determine that the fluid depth will be 0.435 meters at 11 hours and 9.3 minutes after the start time.

    The results tabulated in Table 3.5 can be extended to a range of dimensionless times in order to produce a curve of x versus  , and these results can be transformed to produce a curve of h versus t for any value of  and  .

    2

    Table 3.5. Converging Values for x   h  at    t 2  0 1

    . 0

    n

    xn

    H( xn)

    2

    0.5000

    0.09315

    3

    0.3000

    – 0.04333

    4

    0.4000

    0.01083

    5

    0.3500

    – 0.01922

    6

    0.3750

    – 0.00500

    7

    0.3875

    0.00271

    8

    0.3812

    – 0.00120

    9

    0.3844

    0.00074

    10

    0.3828

    – 0.00023

    11

    0.3836

    0.00026

    12

    0.3832

    0.00001

    13

    0.3830

    – 0.00011

    14

    0.3831

    – 0.00005

    15

    0.3832

    – 0.00002

    16

    0.3832

    0.00000

    In Examples 3.4 and 3.5 we have illustrated how one can develop solutions to transient macroscopic mass balances. For the system analyzed in Example 3.5 an iterative method of solution was required to find the root of an implicit equation for the fluid depth. More information about iterative methods is provided in Appendix B.

    index-85_1.png

    index-85_2.png

    76