4-Practice Problems, fluid mech

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Chapter 4
Flowing Fluids and Pressure
Variation
Problem 4.1
A
ß
ow moves in the
!
-direction with a velocity of 10 m/s from 0 to 0.1 second
and then reverses direction with the same speed from 0.1 to 0.2 second. Sketch
the pathline starting from
! = 0
and the streakline with dye introduced at
! = 0"
Show the streamlines for the
Þ
rst time interval and the second time interval.
Solution
The pathline is the line traced out by a
ß
uid particle released from the origin. The
ß
uid particle
Þ
rst goes to
! = 1"0
and then returns to the origin so the pathline is
The streakline is the con
Þ
guration of the dye at the end of 0.2 second. During
the
Þ
rst period, the dye forms a streak extending from the origin to
! = 1
m.
During the second period, the whole
Þ
eld moves to the left while dye continues to
be injected. The
Þ
nal con
Þ
guration is a line extending from the origin to
! = !1
m.
The streamlines are represented by
19
20
CHAPTER 4. FLOWING FLUIDS AND PRESSURE VARIATION
21
Problem 4.2
A piston is accelerating upward at a rate of 10 m/s
2
. A 50-cm-long water col-
umn is above the piston. Determine the pressure at a distance of 20 cm below the
water surface. Neglect viscous e
!
ects.
Solution
Since the water column is accelerating, Euler’s equation applies. Let the
#
-direction
be coincident with elevation, that is, the
$
-direction. Euler’s equation becomes
!
%
%$
(&+'$) = ()
!
(1)
Since pressure varies with
$
only, the left side of Euler’s equation becomes
%
%$
(&+'$) = !
Μ
*&
*$
+'

!
(2)
Combining Eqs. (1) and (2) gives
*&
*$
= !(()
!
+')
= !(1000
kg/m
3
×10
m/s
2
+9810
N/m
3
)
(3)
= !19+810
N/m
3
Integrating Eq. (3) from the water surface (
$
= 0 m
)
to a depth of 20 cm (
$
= -0.2
m
)
gives
"(!=!0#2)
Z
!=!0#2
*& =
!19"8 *$
"(!=0)
!=0
³
´
&($ = !0"2)!&($ = 0) =
!19"8
kN/m
3
(!0"2!0)
m
Since pressure at the water surface (
$ = 0)
is 0,
³
´
&
!=!0#2
m
=
!19"8
kN/m
3
(!0"2
m
)
=
3.96 kPa-gage
Z
22
CHAPTER 4. FLOWING FLUIDS AND PRESSURE VARIATION
Problem 4.3
A rectangular tank, initially at rest, is
Þ
lled with kerosene (
( = 1"58
slug/ft
3
)
to a depth of 4 ft. The space above the kerosene contains air that is at a pressure
of 0.8 atm. Later, the tank is set in motion with a constant acceleration of 1.2 g to
the right. Determine the maximum pressure in the tank after the onset of motion.
Solution
After initial sloshing is damped out, the con
Þ
guration of the kerosene is shown in
Fig. 1.
Figure 1 Con
Þ
guration of kerosene during acceleration
In Fig. 1,
,
is an unknown length, and the angle
-
is
TAN- =
)
$
.
= 1"2
So
- = 50"2
%
To
Þ
nd the length
,
, note that the volume of the air space before and after motion
remains constant.
(10)(1)(
width
) = (1/2)(1"2,)(,)(
width
)
So
P
, =
20/1"2 = 4"08
ft
23
The maximum pressure will occur at point
0
in Fig. 1. Before
Þ
nding this pressure,
Þ
nd the pressure at
1
by integrating Euler’s equation from point
)
to point
1
.
!
*&
*!
= ()
$
Z
&
Z
&
*& = !
()
$
*!
'
'
&
&
= &
'
+()
$
(10!,)
= (0"8)(2116"2)
lbf/ft
2
+(1.58)(1.2
×32"2)(10!4"08)
lbf/ft
2
= 2054
lbf/ft
2
absolute
To
Þ
nd the pressure at
0
, Euler’s equation may be integrated from
1
to
0
.
!
*(&+'$)
*$
= 0
*&
*$
= !'
Z
(
Z
(
*& =
!'*$
&
&
&
(
!&
&
= !'($
(
!$
&
)
so
&
(
= &
&
+'(5
ft)
= 2054
lbf/ft
2
+(1"58×32"2
lbf/ft
3
)(5
ft)
=
2310
lbf/ft
2
-absolute
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