Skip to main content
Engineering LibreTexts

3.16: Untitled Page 42

  • Page ID
    18175
  • Chapter 3

    is used as a water storage tank. The water is used as a cooling fluid in a batch distillation unit and then sent to waste. The distillation unit runs for six hours and consumes 5 gal/min of water. At the beginning of the process the tank is full of water, and since we do not want to empty the tank during the distillation process, we need to know the liquid level in the tank at the end of the process. The system under consideration is illustrated in Figure 3.4 and we have constructed the control volume on the following basis:

    I. A cut has been made at the exit of the tank where information is given.

    II. There is no required information that would generate another cut.

    III. The surface of the control volume is located where (v w ) n is known.

    IV. The control volume encloses the region for which information is required.

    For the moving control volume illustrate in Figure 3.4, the macroscopic mass balance takes the form

    d

    dV

    ( 

    ) dA

    v w n

    0

    (1)

    dt V( t)

    A( t)

    in which

    ( ) has been replaced with V t since the control volume is no a

    V t

    ( )

    Figure 3.4 Cooling water storage tank

    Single component systems

    69

    longer arbitrary. A little thought will indicate that (

    v w)n is zero

    everywhere on the surface of the control volume except at the exit of the tank. This leads to the representation given by

    v

    (

    w)n dA

    vn dA   Q

    (2)

    exit

    A( t)

    A exit

    in which we have assumed that the density can be treated as a constant.

    Use of Eq. 2 in Eq. 1 provides

    d

    dV   Q

    (3)

    dt

    0

    exit

    V ( t)

    and we again impose the condition of a constant density to obtain dV ( t)

      Q

     0

    exit

    (4)

    dt

    The control volume is computed as the product of the cross sectional area of the tank multiplied by the level of water in the tank, i.e., V( t) = AT h( t).

    Use of this relation in Eq. 4 and canceling the density leads to dh

    A

     

    T

    Qe

    dt

    xit

    (5)

    and this equation is easily integrated to give

    Qexit t

    h( t)  h t  0 

    (6)

    AT

    For consistency, all the variables are computed in SI units.

     2

    D

    A

    2

    3 . 14 m

    T

    4

    (7)

    (5gal / min)(3 . 78 

    3

    3

    10

    m / gal)

    Q

    4

    3

    3 . 1510

    m / s

    exit

    60 s / min

    At the end of the process,

    4

    t  6 h  2.16  10 s , the water level in the

    tank will be

    h( t  6 h)  3 m  2 . 167 m  0 8

    . 33 m

    index-79_1.png

    index-79_2.png

    index-79_3.png

    70