Skip to main content
Engineering LibreTexts

3.18: Untitled Page 44

  • Page ID
    18176
  • Chapter 3

    where  and  are constants given by

    4 Q

    0 . 6  4  A

    2 g

    1

    o

     

    ,

     

    (11)

    2

    2

    D

    D

    The initial condition for this process takes the form

    IC.

    h  0 ,

    t  0

    (12)

    and we need to integrate Eq. 10 and impose this initial condition in order to determine the fluid depth as a function of time. Separating variables in Eq. 10 leads to

    dh

    dt

    (13)

       h

    and the integral of this result is given by

    2

        h  ln    h   t C

    2 

    (14)

    The constant of integration, C, can be evaluated by means of the initial condition given by Eq. 12 and the result is

    2

    C

      ln

    (15)

    2

    This allows us to express Eq. 14 in the form

    2

       h   ln 1   h    t

    (16)

    2 

    

    and this result can be used to predict the height of the water at any time.

    Since Eq. 16 represents an implicit expression for (

    h t) , some computation

    is necessary in order to produce a curve of the fluid depth versus time.

    Before we consider a numerical method that can be used to solve Eq. 16 for (

    h t) , we want determine if the tank will overflow. To explore this question, we note that Eq. 16 provides the result

    1   h   0 , t

     

    (17)

    and this can be used to determine whether (

    h t) is greater than H as time

    tends to infinity. One can also obtain Eq. 17 directly from Eq. 10 by

    Single component systems

    73

    imposing the steady‐state condition that dh / dt  0 . We can arrange Eq. 17

    in the form

       2

    h

    ,

    t  

    (18)

    and in terms of Eqs. 11 we obtain

    h

    Q 0 6 . A 2

    t

     

    (19)

    1

    2

    g

    ,

    o

    from the values given in the problem statement, we find that h   2 2

    . 6 m

    at steady state, thus the tank will not overflow.

    At this point we would like to use Eq. 16 to develop a general solution for the fluid depth in the tank as a function of time, i.e., we wish to know (

    h t) for the specific set of parameters given in this example. It is convenient to represent this problem in dimensionless variables so that H( x, )

       x  ln(1 x)  

    (20)

    in which x is the dimensionless dependent variable representing the depth and defined by

    x   h

    (21)

    In Eq. 20 we have used  to represent the dimensionless time defined by 2

       t 2

    (22)

    and our objective is to find the root of the equation

    H( x, )

     0 ,

    x x

    (23)

    Here we have identified the solution as x and it is clear from Eq. 20 that

    the solution will require that x  1. In Appendix B we have described several methods for solving implicit equations, such as Eq. 23, and in this example we will use the simplest of these methods.

    Bisection method

    Here we wish to find the solution to Eq. 23 when the parameter  is equal to 0.10. A sketch of H(x, ) is shown in Figure 3.5b where we see

    that x is located between zero and one. The bisection method begins by locating values of x that produce positive and negative values of the function H( x, ) and these values are identified as

    x

    o

    x

    0 . 90 and

    0 1

    .

    1

    0

    index-83_1.png

    index-83_2.png

    index-83_3.png

    index-83_4.png

    index-83_5.png

    74