fluid_mechanics

194 Chapter 5 ■ Finite Control Volume Analysis
or
u max 2U
(Ans)
V 2 = u max /2
(parabolic)
(b) Since this flow is incompressible, we conclude from Eq.
5.7 that U is the average velocity at all sections of the control volume.
Thus, the average velocity at section (2), V 2 , is one-half the
maximum velocity, u max , there or
V 2 u max
2
(Ans)
u max
COMMENT The relationship between the maximum velocity
at section (2) and the average velocity is a function of the
“shape” of the velocity profile. For the parabolic profile assumed
in this example, the average velocity, u max /2, is the actual
“average” of the maximum velocity at section (2), u 2 u max ,
and the minimum velocity at that section, u 2 0. However, as
shown in Fig. E5.4c, if the velocity profile is a different shape
(non-parabolic), the average velocity is not necessarily one half
of the maximum velocity.
V 2 = u max /2
(non-parabolic)
F I G U R E E5.4c
E XAMPLE 5.5
Conservation of Mass—Unsteady Flow
GIVEN A bathtub is being filled with water from a faucet. The
rate of flow from the faucet is steady at 9 gal/min. The tub volume
is approximated by a rectangular space as indicated in Fig. E5.5a.
V j
A j
Control volume
FIND Estimate the time rate of change of the depth of water in
the tub, ∂h/∂t, in inches per minute at any instant.
SOLUTION
We use the fixed, nondeforming control volume outlined with a
dashed line in Fig. E5.5a. This control volume includes in it, at
any instant, the water accumulated in the tub, some of the water
flowing from the faucet into the tub, and some air. Application of
Eqs. 5.4 and 5.5 to these contents of the control volume results in
0
0t
air air dV air 0
0t
water water dV water
volume
volume
m # water m # air 0 (1)
Recall that the mass, dm, of fluid contained in a small volume
dV is dm dV. Hence, the two integrals in Eq. 1 represent the
total amount of air and water in the control volume, and the sum
of the first two terms is the time rate of change of mass within
the control volume.
Note that the time rate of change of air mass and water mass
are each not zero. Recognizing, however, that the air mass must
be conserved, we know that the time rate of change of the mass of
air in the control volume must be equal to the rate of air mass flow
out of the control volume. For simplicity, we disregard any water
evaporation that occurs. Thus, applying Eqs. 5.4 and 5.5 to the air
only and to the water only, we obtain
0
0t
air air dV air m # air 0
volume
1.5 ft
5 ft
F I G U R E E5.5a
for air, and
0
(2)
0t
water water dV water m # water
volume
for water. The volume of water in the control volume is given by
where A j is the cross-sectional area of the water flowing from the
faucet into the tub. Combining Eqs. 2 and 3, we obtain
and, thus, since
water water dV water water 3h12 ft215 ft2
volume
water 110 ft 2 A j 2 0h
0t m# water
m # Q,
h
11.5 ft h2A j 4
0h
0t Q water
110 ft 2 A j 2
2 ft
(3)