191
P6.29: A submarine releases a spherical flotation buoy containing a radio beacon. The
buoy has a diameter of 1 ft and weighs 22 lb. The coefficient of drag for the submerged
buoy is CD = 0.45. At what steady speed will the buoy rise to the surface?
Approach:
We will use an equilibrium force balance involving the drag force to determine the speed at
which the buoy will rise. At steady speed, the drag FD and weight w forces that act
downward in the free body diagram exactly balance the buoyancy force FB. The density
and viscosity of seawater are assumed to be constant and are listed in Table 6.1. The drag
force is given by Equation (6.14).
Solution:
The buoyancy force that develops is
3
ft
slug
Chapter 6: Fluids Engineering
192
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Following Equation (6.14) for the drag force, the velocity is found from:
lb55112
2
.
v
Discussion:
This speed seems quite reasonable assuming the density and viscosity of sea water are
194
P6.31: (a) A luxury sports car has a frontal area of 22.4 ft2 and a 0.29 coefficient of drag at
60 mph. What is the drag force on the vehicle at this speed? (b) A sport-utility vehicle
has CD = 0.45 at 60 mph and a slightly larger frontal area of 29.1 ft2. What is the drag
force in this case?
Approach:
Apply Equation (6.14) using the density of air ρ = 2.33× 10–3 slug/ft3 listed in Table 6.1.
Convert velocity to consistent units when calculating the drag force.
Solution:
(a) Luxury sports car
195
P6.32: A certain parachute has a drag coefficient of CD = 1.5. If the parachute and skydiver
together weigh 225 lb, what should the frontal area of the parachute be so that the
skydiver’s terminal velocity is 15 mph when approaching the ground? Is it reasonable
to neglect the buoyant force that is present?
Approach:
Solution:
Terminal velocity:
ft
22
hr
107782
ft
5280
mi
15 4
.v
196
P6.33: Submarines dive by opening vents that allow air to escape from ballast tanks and
water to flow in and fill them. In addition, diving planes located at the bow are angled
downward to help push the boat below the surface. Calculate the diving force produced
by a 20 ft2 hydroplane having a lift coefficient of 0.11 as the boat cruises at 15 knots (1
knot = 1.152 mph).
Approach:
Apply Equation (6.18) using consistent units for velocity. Use the density of sea water 1.99
slug/ ft3 listed in Table 6.1 and assume it stays constant.
Solution:
Velocity:
ft
hr
ft
hr
mi
Discussion:
198
time
2
= length
2
2
2
Discussion:
P6.35: A device called a venturi flowmeter can be used to determine the velocity of a
flowing fluid by measuring the change in its pressure between two points. Water flows
through a pipe, and, at the constriction, the cross-sectional area of the pipe reduces
from A1 to the smaller value A2. Two pressure sensors are located just before and after
the constriction, and the change p1 – p2 that they measure is enough information to
determine the water’s velocity. By using Equations (6.13) and (6.17), show that the
velocity downstream of the constriction is given by
2
12
21
21
2
AAρ
pp
v
This so-called venturi effect is the principle behind the operation of such hardware as
automotive carburetors and the aspirators that deliver pharmaceutical products to
medical patients by inhalation.
Approach:
With the two cross-sectional areas known, the velocities at the two points are related by
Equation (6.13).
Solution:
Chapter 6: Fluids Engineering
200
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1 2
2
2
2 1
2
1 /
p p
v
A A
Discussion:
202
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2
8.232 ft
lb
= 232.8 psf
By using the conversion factor from Table 6.2:
psi
Discussion:
If the constriction were even tighter, the downstream velocity and pressure drop would
both be larger. Also, if the pipe had an expansion instead of a constriction, the downstream
)(
3
1105.8
102.1
sm
kg
Stage 2:
s
m
s
m
m
v
d
d
v
A
v36.011.0
5.0
4
4
2
2
1
2
2
2
1
1
2
1
2
m
)(
102.1
sm
m
kg m
Stage 3:
s
m
s
m
m
v
d
d
v
A
v25.236.0
2.0
4
4
2
2
1
2
3
2
2
2
3
2
3
m
Chapter 6: Fluids Engineering
5
)(
3
3105.4
100.1
2.025.21000 3
sm
kg
s
m
m
kg m
Re
Stage 4:
s
m
s
m
m
v
d
d
v
A
v0.3625.2
05.0
4
4
2
2
1
2
4
2
3
3
4
3
4
s
m
v0.36
2
kg
Chapter 6: Fluids Engineering
206
For example, if glycerol was chosen ( 3
1260 m
kg
glycerol
, )(
4.1 sm
kg
glycerol
), the inlet
Chapter 6: Fluids Engineering
208
as a function of the outer diameter and the length of the pipe. A density of water of 1000
kg/m3 is assumed.
Using the force balance, we will develop an equation for the required length of the platform
(and tubes) necessary to support the given load and platform at the required half-way
L = Length of raft
g = Gravitational constant
Solution:
Buoyancy force:
pipesfluidB gVF
The volume of the pipes equals the collective volume of the water displaced when the four
2
Force balance:
The buoyancy force is exactly counteracted by the gravitational force since the system is in
equilibrium (neglected any movement due to waves, currents, etc.).
209
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
LODggmm
FF
fluidplatformpayload
BG
2
2
)(
LODggtwLm fluidplatformpayload
2
2
)(
Solving this force balance equation for L gives us:
twOD
L
platformfluid
2
2
Pipe and platform selection:
The dimensioning and pricing analysis is performed for each pipe option.
PIPE A Raft Length
(m)
Length of Piping
needed (m)
Cost of
Piping
Cost of
Platform Total Cost
Platform X
7.43
29.7
$186.12
$71.00
$257.12
Platform Y
9.92
39.7
$159.71
$99.40
$259.11
13.11
52.4
$129.13
$127.80
5.74
23.0
$143.79
$116.00
$259.79
7.12
28.5
$114.79
$145.00
$259.79
8.63
34.5
$85.01
$174.00
PIPE C Raft Length
(m)
Length of Piping
Needed (m)
Cost of
Piping
Cost of
Platform Total Cost