5-Practice Problem.pdf

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ssm_ch5.pdf
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× $
447291216.010.png
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
447291216.011.png 447291216.012.png
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
'
447291216.013.png 447291216.001.png 447291216.002.png 447291216.003.png 447291216.004.png
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.
447291216.005.png 447291216.006.png
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
447291216.007.png 447291216.008.png 447291216.009.png

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