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3.17: Untitled Page 43

  • Page ID
    18177
  • Chapter 3

    EXAMPLE 3.5. Water level in a storage tank with an inlet and outlet In Figure 3.5a we have shown a tank into which water enters at a volumetric rate

    and leaves at a rate

    that is given by

    1

    Q

    o

    Q

    o

    Q

    0 6

    . A o 2 gh

    (1)

    Here o

    A is the area of the orifice in the bottom of the tank. The tank is initially empty when water begins to flow into the tank, thus we have an initial condition of the form

    I.C.

    h  0 ,

    t  0

    (2)

    The parameters associated with this process are given by

    4

    3

    2

    Q

     10 m / s , A  0 . 354 cm , D  1.5 m , h  2.7

    8 m

    (3)

    1

    o

    o

    and in this example we want to derive an equation that can be used to predict the height of the water at any time. That equation requires a numerical solution and we illustrate one of the methods (see Appendix B) that can be used to obtain numerical results.

    The control volume used to analyze this process is illustrated in Figure 3.5a where we have shown a moving control volume for which the Figure 3.5a. Tank filling and overflow process

    Single component systems

    71

    “cuts” at the entrance and exit are joined by a su face

    r

    at which (v w)n

    is zero. If the water overflows, a second exit will be created; however, we are only concerned with the process that occurs prior to overflow. Our analysis is based on the macroscopic mass balance for a moving control volume given by

    d

    dV

    ( 

    ) dA

    v w n

    0

    (4)

    dt V( t)

    A( t)

    and the assumption that the density is constant leads to

    d V( t)

    (

    ) dA

    v w n

    0

    (5)

    dt

    A( t)

    The normal component of the relative velocity, (v w)n , is zero everywhere except at the entrance and exit, thus Eq. 5 simplifies to dV( t)

    o

    Q

    1

    Q

    0

    (6)

    dt

    The volume of the control volume is given by

    2

    D

    V( t) 

    (

    h t)  V ( t)

    (7)

    4

    jet

    and use of this expression in Eq. 6 provides

    2

    D dh

    dV

    jet

    o

    Q

    1

    Q

    0

    (8)

    4

    dt

    dt

    If the area of the jet is much, much smaller than the cross sectional area of the tank, we can impose the following constraint on Eq. 8

    2

    dV

    D

    dh

    jet

    

    (9)

    dt

    4

    dt

    however, one must keep in mind that there are problems for which this inequality might not be valid. When Eq. 9 is valid, we can simplify Eq. 8

    and arrange it in the compact form

    dh

        h

    (10)

    dt

    72