5-Practice Problem, fluid mech

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Chapter 5
Control Volume Approach
and Continuity Principle
Problem 5.1
A 10-cm-diameter pipe contains sea water that
ß
ows with a mean velocity of 5
m/s. Find the volume
ß
ow rate (discharge) and the mass
ß
ow rate.
Solution
The discharge is
! = "#
where
"
is the mean velocity. Thus
4
×0%1
2
= 0%0393
m
3
/s
From Table A.4, the density of sea water is 1026 kg/m
3
%
The mass
ß
ow rate is
ú& = '! = 1026×0%0393 =
40%3
kg/s
29
! = 5×
$
30CHAPTER5. CONTROLVOLUMEAPPROACHANDCONTINUITYPRINCIPLE
Problem 5.2
The velocity pro
Þ
le of a non-Newtonian
ß
uid in a circular conduit is given by
(
(
MAX
!
³
)
*
´
2
¸
1!2
=
1!
where
(
MAX
is the velocity at the centerline and
*
is the radius of the conduit. Find
the discharge (volume
ß
ow rate) in terms of
(
MAX
and
*%
Solution
The volume
ß
ow rate is
Z
! =
(+#
"
For an axisymmetric duct, this integral can be written as
Z
#
! = 2$
()+)
0
Substituting in the equation for the velocity distribution
Z
#
!
³
)
*
´
2
¸
1!2
! = 2$(
MAX
1!
)+)
0
Recognizing that
2)+) = +)
2
,
we can rewrite the integral as
Z
#
!
³
)
*
´
2
¸
1!2
! = $(
MAX
1!
+)
2
0
!
¸
Z
#
³
)
*
´
2
1!2
³
)
*
´
2
= $(
MAX
*
2
1!
+
0
or
Z
1
! = $(
MAX
*
2
[1!-]
1!2
+-
0
= !
2
3
$(
MAX
*
2
[1!-]
3!2
|
0
=
2
3
$(
MAX
*
2
31
Problem 5.3
A jet pump injects water at 120 ft/s through a 2-in. pipe into a secondary
ß
ow
in an 8-in. pipe where the velocity is 10 ft/s. Downstream the
ß
ows become fully
mixed with a uniform velocity pro
Þ
le. What is the magnitude of the velocity where
the
ß
ows are fully mixed?
Solution
Draw a control volume as shown in the sketch below.
Because the
ß
ow is steady
X
'V·A = 0
$%
Assuming the water is incompressible, the continuity equation becomes
X
V·A = 0
$%
The volume
ß
ow rate across station
.
is
X
V·A = !10×
$
4
Μ
8
12

2
&
where the minus sign occurs because the velocity and area vectors have the opposite
sense. The volume
ß
ow rate across station
B
is
X
V·A = !120×
$
4
Μ
2
12

2
'
32CHAPTER5. CONTROLVOLUMEAPPROACHANDCONTINUITYPRINCIPLE
and the volume
ß
ow rate across station
C
is
X
V·A = " ×
$
4
Μ
8
12

2
$
where
"
is the velocity. Substituting into the continuity equation
!120×
$
4
Μ
2
12

2
!10×
$
4
Μ
8
12

2
+" ×
$
4
Μ
8
12

2
= 0
¡
120×2
2
+10×8
2
¢
" =
8
2
=
17%5
ft/s
Problem 5.4
Water
ß
ows into a cylindrical tank at the rate of 1 m
3
/
min and out at the rate
of 1.2 m
3
/
min. The cross-sectional area of the tank is 2 m
2
%
Find the rate at
which the water level in the tank changes. The tank is open to the atmosphere.
Solution
Draw a control volume around the
ß
uid in the tank. Assume the control surface
moves with the free surface of the water.
33
The continuity equation is
+
+0
Z
X
'+"+
'V·A = 0
$(
$%
The density inside the control volume is constant so
+
+0
Z
X
+"+
V·A = 0
$(
$%
X
+"
+0
+
V·A = 0
$%
The volume of the
ß
uid in the tank is
" = 1#%
Mass crosses the control surface at
two locations. At the inlet
V·A = !!
)*
and at the outlet
V·A = !
+,-
Substituting into the continuity equation
#
+1
+0
+!
+,-
!!
)*
= 0
or
+0
=
!
)*
!!
+,-
#
=
1!1%2
2
=
!0%1
m/min
+1
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