10-Practice Problems, fluid mech
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Chapter 10
Flow in Conduits
Problem 10.1
Water at 20
o
C (
! = 10
!3
N
·
s/m
2
,
" = 1000
kg/m
3
) ß
ows through a 0.5-mm tube
connected to the bottom of a reservoir. The length of the tube is 1.0 m, and the
depth of water in the reservoir is 20 cm. Find the
ß
ow rate in the tube. Neglect
the entrance loss at the junction of the tube and reservoir.
Solution
Applying the energy equation between the top of the water in the reservoir (1) and
the end of the tube (2) gives
#
1
$
+%
1
&
2'
+(
1
+)
!
=
#
2
$
+%
2
&
2'
+(
2
+)
"
+)
#
Thepressureatpoints1and2isthesame(atmospheric),thevelocityinthereservoir
is zero, and there is no pump or turbine in the system. Also, the only losses are
87
88
CHAPTER 10. FLOW IN CONDUITS
friction losses in the tube. The energy equation simpli
Þ
es to
(
1
= %
2
&
2'
+(
2
+)
$
We will assume the
ß
ow is laminar, so
%
2
= 2*
The head loss due to friction in a
laminar
ß
ow is
)
$
= 32
!+&
$,
2
Substituting into the energy equation gives
(
1
= %
2
&
2'
+(
2
+32
!+&
2
$,
2
and replacing the variables with values
1*2 =
&
2
9*81
+32
10
!3
×1×&
2
9810×0*0005
2
0*102&
2
+13*05&
2
!1*2 = 0
Solving
&
2
= 0*092
m/s
The volume
ß
ow rate is
4
×0*0005
2
×0*092 = 1*8×10
!8
m
3
/s
=
1*8×10
!2
ml/s
To determine whether the
ß
ow is laminar, calculate the Reynolds number.
RE =
"&,
!
=
10
3
×0*092×0*0005
10
!3
= 46
Since the Reynolds number is less than 2000, the laminar
ß
ow assumption is justi-
Þ
ed.
- = .& =
/
89
Problem 10.2
An oil supply line for a bearing is being designed. The supply line is a tube
with an internal diameter of
16
in. and 10 feet long. It is to transport SAE 30W
oil (
! = 2×10
!3
lbf
·
s/ft
2
0 " = 1*71
slugs/ft
3
) at the rate of 0.01 gpm. Find the
pressure drop across the line.
Solution
First determine the Reynolds number to establish if the
ß
ow is laminar or turbulent.
The velocity in the line is
& =
-
.
=
0*01
gpm
×0*00223
cfs/gpm
%
4
×
¡
16
×
12
¢
2
ft
2
= 1*05
ft/s
The Reynolds number is
RE =
"&,
!
=
1*71×1*05×
16
×
12
2×10
!3
= 4*68
The
ß
ow is laminar, so the head loss is
)
$
= 32
!+&
$,
2
= 32
2×10
!3
×10×1*05
1*71×32*2×
¡
16
×
12
¢
2
= 450
ft
The pressure drop is
!# = $)
$
= 1*71×32*2×450
= 24*8×10
3
psf =
172
psi
1
1
90
CHAPTER 10. FLOW IN CONDUITS
Problem 10.3
Kerosene (
! = 1*9 × 10
!3
N
·
s/m
2
,
1
=0.81)
ß
ows in a 2-cm diameter commer-
cial steel pipe (
2
&
= 0*046
mm). A mercury manometer (
1 = 13*6)
is connected
between a 2-m section of pipe as shown, and there is a 5-cm de
ß
ection in the
manometer. The elevation di
!
erence between the two taps is 0.5 mm. Find the
direction and velocity of the
ß
ow in the pipe.
Solution
First
Þ
nd the di
!
erence in piezometric pressure between the twopressure taps inthe
pipe. Flow is always in the direction of decreasing piezometric head. Take station
1 on the left and station 2 on the right. De
Þ
ne the distance
3
as the distance from
the center of the pipe at station 2 and the top of the mercury in the manometer.
Using the manometer equation from 1 to 2 gives
#
2
= #
1
+$
k
((
1
!(
2
)+$
k
3+$
Hg
!)!$
k
!)!$
k
3
Thus we can write
#
2
+$
k
(
2
!(#
1
+$
k
(
1
) = ($
Hg
!$
k
)!)
#
'(2
!#
'(1
= ($
Hg
!$
k
)!)
Since
$
Hg
4 $
k
0 #
'(2
4 #
'(1
0
the
ß
ow must be from right to left (uphill).
The energy equation from 2 to 1 is
#
'(2
+%
2
"
k
&
2
+$
k
)
!
= #
'(1
+%
1
"
k
&
2
+$
k
)
"
+$
k
)
#
Since the pipe has a constant area,
&
1
= &
2
0
and there are no turbines or pumps in
the system, the equation reduces to
#
'(2
!#
'(1
= $
k
)
#
= ($
Hg
!$
k
)!)
91
The head loss can be expressed using the Darcy-Weisbach relation
)
#
= 5
+
,
&
2
2'
so
Μ
¶
5
+
,
&
2
2'
=
$
Hg
$
k
!1
!)
Substituting in the values
Μ
13*6
0*81
!1
1
2×9*81
5&
2
= 0*155
m
2
6
s
2
¶
= 5&
2
2
0*02
0*05×
Since
5
depends on the Reynolds number (and velocity), this equation has to be
solved by iteration. The relative roughness for the pipe is
,
=
0*046
= 0*0023
20
From the Moody diagram (Fig. 10.8), the friction factor for a fully rough pipe
would be about 0.025. This would give a velocity of
R
0*155
0*025
= 2*49
m/s
& =
The corresponding Reynolds number is
RE =
"&,
!
=
1000×0*81×2*49×0*02
1*9×10
!3
= 2*12×10
4
From the Moody diagram, the friction factor for this Reynolds number is 0.0305.
The velocity is corrected to
R
0*155
0*0305
= 2*25
m/s
& =
ThenewReynoldsnumberis
1*92×10
4
*
Thefrictionfactoris0.030, givingavelocity
of 2.27 m/s. Further iterations would not signi
Þ
cantly change the value so
& =
2*27
m/s
and the
ß
ow is from right to left.
2
&
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