Ch11. Fluids
Mass Density
Fluids are materials that can flow, and they include both
gases and liquids.
DEFINITION OF MASS DENSITY
The mass density
by its volume V:

is the mass m of a substance divided
m

V
SI Unit of Mass Density: kg/m3
1
Mass Densities of Common Substances
(Unit: kg/m3)
Solids
Aluminum
2 700
Ice
Brass
8 470
Iron (steel)
7 860
Concrete
2 200
Lead
11300
Copper
8 890
Quartz
2 660
Diamond
3 520
Silver
10 500
Gold
19 300
Wood (yellow pine)
917
550
2
Gases
Liquids
Blood (whole,
37°C)
Ethyl alcohol
Mercury
Air
1.29
Carbon dioxide
1.98
Helium
0.179
Hydrogen
0.0899
Nitrogen
1.25
Oxygen
1.43
1060
806
13600
Oil (hydraulic)
Water (4 °C)
800
1 × 103
Gases have the smallest densities.
3
Example 1.
Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about 5.2
× 10–3 m3 (5.5 qt) of blood. (a) Find the blood’s weight and
(b) express it as a percentage of the body weight.
(a)
(b)
4
A convenient way to compare densities: specific gravity
Specific gravity has no units.
5
Pressure
The SI unit for pressure is
a newton/meter2 (N/m2), a
combination that is
referred to as a pascal (Pa).
105 Pa = 1 bar
Another unit for pressure
is pounds per square inch
(lb/in.2), often abbreviated
as “psi.”
6
In general, a static fluid
cannot produce a force
parallel to a surface.
Pressure has no
directional characteristic.
The force generated by
the pressure of a static
fluid is always
perpendicular to the
surface that the fluid
contacts.
7
Example 2.
The Force on a Swimmer
Suppose the pressure acting on the back of a swimmer’s hand
is 1.2 × 105 Pa, a realistic value near the bottom of the diving
end of a pool. The surface area of the back of the hand is 8.4
× 10–3 m2 (a) Determine the magnitude of the force that acts
on it. (b) Discuss the direction of the force.
(a)
(b) The hand (palm downward) is oriented parallel to the bottom of the pool.
Since the water pushes perpendicularly against the back of the hand, the force
F is directed downward in the drawing. This downward-acting force is
balanced by an upward-acting force on the palm, so that the hand is in
equilibrium. If the hand were rotated by 90°, the directions of these forces
would also be rotated by 90°, always being perpendicular to the hand.
8
9
Pressure and Depth in a Static Fluid
10
11
Conceptual Example 3.
The Hoover Dam
Lake Mead is the largest wholly
artificial reservoir in the United
States and was formed after the
completion of the Hoover Dam in
1936. The water in the reservoir
backs up behind the dam for a
considerable distance (about 200
km or 120 miles). Suppose that all
the water were removed, except
for a relatively narrow vertical
column in contact with the dam.
12
A side view of this hypothetical situation, in which the
water against the dam has the same depth as in Figure
part a. Would the Hoover Dam still be needed to contain
the water in this hypothetical reservoir, or could a much
less massive structure do the job?
The dam for our imaginary reservoir would sustain the
same forces that the Hoover Dam sustains and would need
to be equally large.
13
Example 4. The Swimming Hole
The cross section of a swimming hole. Points A and B are
both located at a distance of h = 5.50 m below the surface of
the water. Find the pressure at each of these two points.
14
Since points A, B, C, and D are at the same distance
h beneath the liquid surface, the pressure at each of
them is the same.
15
Example 5. Blood Pressure
Blood in the arteries is
flowing, but as a first
approximation, the effects of
this flow can be ignored and
the blood can be treated as a
static fluid. Estimate the
amount by which the blood
pressure P2 in the anterior
tibial artery at the foot
exceeds the blood pressure P1
in the aorta at the heart when
a person is (a) reclining
horizontally and (b) standing.
16
(a)
h = 0 m,
(b)
17
Conceptual Example 6.
Pumping Water
Two methods for pumping water
from a well. In one method, the
pump is submerged in the water at
the bottom of the well, while in the
other, it is located at ground level. If
the well is shallow, either technique
can be used. However, if the well is
very deep, only one of the methods
works. Which is it?
A ground-level (on top of the ground)
pump can only cause water to rise to a
certain maximum height, so it cannot
be used for very deep wells.
18
Check Your Understanding 1
A scuba diver is swimming under water, and the graph
shows a plot of the water pressure acting on the diver as a
function of time. In each of the three regions, A B,
B
C, and C
D, does the depth of the diver increase,
decrease or remain constant?
A B: Increase
B
C: Decrease
C
D: Remain constant
19
Pressure Gauges
The atmospheric pressure
can be determined from
the height h of the
mercury in the tube, the
density  of mercury,
and the acceleration due
to gravity.
PA = PB
(PB is Atmospheric pressure.)
20
P1  hg  PA
If P1=0, then PA  hg
If liquid is mercury,   13.6 103 kg / m3
h  29.9 inches ( 30inches)
(760mm)
What if water
(   110 )
3
h  29.9 13.6  407 inches
 34 feet!!
21
The height h is
proportional to P2 – Patm,
which is called the gauge
pressure. The gauge
pressure is the amount
by which the container
pressure differs from
atmospheric pressure.
The actual value for P2
is called the absolute
pressure.
22
A sphygmomanometer is
used to measure blood
pressure.
23
Pascal's Principle
PASCAL’S PRINCIPLE
Any change in the pressure
applied to a completely
enclosed fluid is transmitted
undiminished to all parts of
the fluid and the enclosing
walls.
24
Example 7. A Car Lift
In a hydraulic car lift, the input piston has a radius of r1 =
0.0120 m and a negligible weight. The output plunger has a
radius of r2 = 0.150 m. The combined weight of the car and
the plunger is F2 = 20 500 N. The lift uses hydraulic oil that
has a density of 8.00 × 102 kg/m3. What input force F1 is
needed to support the car and the output plunger when the
bottom surfaces of the piston and plunger are at (a) the
same level and (b) the levels with h = 1.10 m?
(a)
25
(b)
P2 = P1 +  gh
P2 =F2/(pr22) and P1 =F1/(pr12):
F2
F1



gh
2
2
pr2 pr1
26
Archimedes' Principle
Buoyant force exists because fluid
pressure is larger at greater depths.
27
ARCHIMEDES’ PRINCIPLE
Any fluid applies a buoyant force to an object that is
partially or completely immersed in it; the magnitude of the
buoyant force equals the weight of the fluid that the object
displaces:
FB
Magnitude of
buoyant force
=
Wfluid
Weight of
displaced fluid
28
The effect that the buoyant force has depends on its
strength compared with the strengths of the other
forces that are acting.
29
Example 8. A Swimming Raft
A solid, square pinewood
raft measures 4.0 m on a
side and is 0.30 m thick. (a)
Determine whether the raft
floats in water, and (b) if so,
how much of the raft is
beneath the surface (see the
distance h in the figure).
30
(a)

3,V
3
=
550
kg/m
=
4.0
m
×
4.0
m
×
0.30
m
=
4.8
m
pine
pine
31
(b) F = 26 000 N = W
B
fluid ,
Vwater = 4.0 m × 4.0 m × h
32
Conceptual Example 9. How Much
Water Is Needed to Float a Ship?
A ship floating in the ocean is a familiar sight. But is all that
water really necessary? Can an ocean vessel float in the
amount of water that a swimming pool contains, for instance?
All that is needed, in
principle, is a thin
section of water that
separates the hull of
the floating ship
from the sides of the
canal.
33
A useful application of Archimedes’ principle can
be found in car batteries.
34
Example 10. A Goodyear Airship
Normally, a Goodyear
airship, contains about
5.40 × 103 m3 of
helium (He) whose
density is 0.179 kg/m3.
Find the weight of the
load WL that the
airship can carry in
equilibrium at an
altitude where the
density of air is 1.20
kg/m3.
35
36
Check Your Understanding 2
A glass is filled to the brim with water and has an ice
cube floating in it. When the ice cube melts, what
happens?
a. Water spills out of the glass.
b. The water level in the glass drops.
c. The water level in the glass does not change.
(c)
37
Fluids in Motion
In steady flow the velocity of the fluid particles at any
point is constant as time passes.
Unsteady flow exists whenever the velocity at a point in
the fluid changes as time passes.
Turbulent flow is an extreme
kind of unsteady flow and
occurs when there are sharp
obstacles or bends in the path
of a fast-moving fluid.
38
Fluid flow can be compressible or incompressible,
viscous or nonviscous.
An incompressible, nonviscous fluid is called an
ideal fluid.
A streamline is a line drawn in the fluid such that a
tangent to the streamline at any point is parallel to the
fluid velocity at that point.
39
Steady flow is often called streamline flow.
40
The Equation of Continuity
Equation of continuity: If a fluid
enters one end of a pipe at a
certain rate (e.g., 5 kilograms per
second), then fluid must also leave
at the same rate, assuming that
there are no places between the
entry and exit points to add or
remove fluid.
The mass of fluid per second (e.g.,
5 kg/s) that flows through a tube is
called the mass flow rate.
41
Mass flow rate
at position 2
Mass flow rate
at position 1
 m2
  2 A2 v2
t
m1
 1 A1v1
t
42
EQUATION OF CONTINUITY
The mass flow rate ( Av) has the same value at every
position along a tube that has a single entry and a single
exit point for fluid flow. For two positions along such a
tube
1 A1v1  2 A2 v 2

= fluid density (kg/m3)
A = cross-sectional area of tube (m2)
v = fluid speed (m2)
SI Unit of Mass Flow Rate: kg/s
43
Incompressible fluid
A1v1=A2v2
Q = Volume flow rate = Av
44
Example 11. A Garden Hose
A garden hose has an unobstructed opening with a crosssectional area of 2.85 × 10–4 m2, from which water fills a
bucket in 30.0 s. The volume of the bucket is 8.00 × 10–3
m3 (about two gallons). Find the speed of the water that
leaves the hose through (a) the unobstructed opening and
(b) an obstructed opening with half as much area.
45
Example 12. A Clogged Artery
In the condition known as atherosclerosis, a deposit or
atheroma forms on the arterial wall and reduces the opening
through which blood can flow. In the carotid artery in the neck,
blood flows three times faster through a partially blocked
region than it does through an unobstructed region. Determine
the ratio of the effective radii of the artery at the two places.
46
Check Your Understanding 3
Water flows from left to right through the five sections (A, B,
C, D, E) of the pipe shown in the drawing. In which section(s)
does the water speed increase, decrease, and remain constant?
Treat the water as an incompressible fluid.
47
Speed
Increases
Speed
Decreases
Speed is
Constant
a.
A, B
D, E
C
b.
D
B
A, C, E
c.
D, E
A, B
C
d.
B
D
A, C, E
e.
A, B
C, D
E
d
48
Bernoulli's Equation
49
(a)In this horizontal pipe, the pressure in region 2 is greater
than that in region 1. The difference in pressures leads to the
net force that accelerates the fluid to the right.
(b)When the fluid changes elevation, the pressure at the bottom
is greater than that at the top, assuming the cross-sectional
area of the pipe is constant.
50
51
BERNOULLI’S EQUATION
In the steady flow of a nonviscous, incompressible fluid of
density  , the pressure P, the fluid speed v, and the
elevation y at any two points (1 and 2) are related by
52
Applications of Bernoulli's Equation
When a moving fluid is contained in a horizontal
pipe, all parts of it have the same elevation (y1 = y2) .
53
Conceptual Example 13.
Tarpaulins and Bernoulli’s Equation
A tarpaulin is a piece of canvas that is used to cover a cargo,
like that pulled by the truck. When the truck is stationary the
tarpaulin lies flat, but it bulges outward when the truck is
speeding down the highway. Account for this behavior.
The greater inside pressure
generates a greater force on
the inner surface of the
canvas, and the tarpaulin
bulges outward.
54
Example 14.
An Enlarged Blood Vessel
An aneurysm is an abnormal enlargement of a blood vessel
such as the aorta. Suppose that, because of an aneurysm, the
cross-sectional area A1 of the aorta increases to a value A2 =
1.7A1. The speed of the blood (  = 1060 kg/m3 ) through a
normal portion of the aorta is v1 = 0.40 m/s. Assuming that
the aorta is horizontal (the person is lying down), determine
the amount by which the pressure P2 in the enlarged region
exceeds the pressure P1 in the normal region
55
56
57
58
Example 15. Efflux Speed
The tank is open to the atmosphere at
the top. Find an expression for the
speed of the liquid leaving the pipe at
the bottom.
59
Check Your Understanding 4
Fluid is flowing from left to right through a pipe (see the
drawing). Points A and B are at the same elevation, but the
cross-sectional areas of the pipe are different. Points B and
C are at different elevations, but the cross-sectional areas
are the same. Rank the pressures at the three points, highest
to lowest.
a. A and B (a tie), C
b. C, A and B (a tie)
c. B, C, A
e
d. C, B, A
e. A, B, C
60
Viscous Flow
(a) In ideal (nonviscous)
fluid flow, all fluid particles
across the pipe have the
same velocity.
(b) In viscous flow, the speed
of the fluid is zero at the
surface of the pipe and
increases to a maximum
along the center axis.
61
For a highly viscous fluid, like
thick honey, a large force is
needed; for a less viscous fluid,
like water, a smaller one will
do. As part b of the drawing
suggests, we may imagine the
fluid to be composed of many
thin horizontal layers. When
the top plate moves, the
intermediate fluid layers slide
over each other. The velocity
of each layer is different,
changing uniformly from v at
the top plate to zero at the
bottom plate. The resulting
flow is called laminar flow,
since a thin layer is often
referred to as a lamina.
62
FORCE NEEDED TO MOVE A LAYER OF VISCOUS
FLUID WITH A CONSTANT VELOCITY
The magnitude of the tangential force F required to move a
fluid layer at a constant speed v, when the layer has an area A
and is located a perpendicular distance y from an immobile
surface, is given by
F
where

Av
y
is the coefficient of viscosity.
SI Unit of Viscosity: Pa·s
Common Unit of Viscosity: poise (P)
63
Values of viscosity depend on the nature of the fluid. Under
ordinary conditions, the viscosities of liquids are significantly larger
than those of gases. Moreover, the viscosities of either liquids or
gases depend markedly on temperature. Usually, the viscosities of
liquids decrease as the temperature is increased. In contrast, the
viscosities of gases increase as the temperature is raised. An ideal
fluid has  = 0 P.
64
Factors that determine the volume flow rate Q (in m3/s) of the fluid.
First, a difference in pressures P2 – P1 must be maintained between
any two locations along the pipe for the fluid to flow. In fact, Q is
proportional to P2 – P1, a greater pressure difference leading to a
larger flow rate.
Second, a long pipe offers greater resistance to the flow than a
short pipe does, and Q is inversely proportional to the length L.
Because of this fact, long pipelines, such as the Alaskan pipeline,
have pumping stations at various places along the line to
compensate for a drop in pressure (see Figure 11.40).
Third, high-viscosity fluids flow less readily than low-viscosity
fluids, and Q is inversely proportional to the viscosity  .
Finally, the volume flow rate is larger in a pipe of larger radius,
other things being equal.
65
POISEUILLE’S LAW
A fluid whose viscosity is  , flowing through a pipe of
radius R and length L, has a volume flow rate Q given
by
pR P2  P1 
Q
8L
4
where P1 and P2 are the pressures at the ends of the pipe.
66
Example 16. Giving an Injection
A hypodermic syringe is filled
with a solution whose viscosity is
1.5 × 10–3 Pa·s. As the figure
shows, the plunger area of the
syringe is 8.0 × 10–5 m2, and the
length of the needle is 0.025 m.
The internal radius of the needle
is 4.0 × 10–4 m. The gauge
pressure in a vein is 1900 Pa (14
mm of mercury). What force
must be applied to the plunger, so
that 1.0 × 10–6 m3 of solution can
be injected in 3.0 s?
67
.
68
Concepts & Calculations Example 17.
Pressure and Force
69
The figure shows a rear view of a loaded two-wheeled
wheelbarrow on a horizontal surface. It has balloon tires and
a weight of W = 684 N, which is uniformly distributed. The
left tire has a contact area with the ground of AL = 6.6×10–4
m2, whereas the right tire is underinflated and has a contact
area of AR = 9.9×10–4 m2. Find the force and the pressure
that each tire applies to the ground.
70
71
Concepts & Calculations Example 18.
The Buoyant Force
72
A father (weight W = 830 N) and his daughter (weight W =
340 N) are spending the day at the lake. They are each sitting
on a beach ball that is just submerged beneath the water.
Ignoring the weight of the air within the balls and the parts of
their legs that are underwater, find the radius of each ball.
73
74
Conceptual Question 19
REASONING AND SOLUTION Consider a stream of water that falls
from a faucet. According to the equation of continuity, the mass flow rate
must be
Avthe same at every point along the stream ( = water density, A =
of the stream, v = speed of stream). As the water falls, it is accelerated due
to gravity; therefore, the speed of the water increases as it falls. Since the
density of water is uniform throughout the stream, when v increases, the
cross-sectional area A of the stream must decrease in order to maintain a
constant mass flow rate. Therefore, the cross-sectional area of the stream
becomes smaller as the water falls.
If the water is shot upward, as it is in a fountain, the velocity of the
stream is upward, while the acceleration due to gravity is directed
downward. Therefore, the speed of the stream decreases as the stream
rises. Since v decreases, the cross-sectional area A must increase in
order to maintain a constant mass flow rate Av. Therefore, the crosssectional area of the stream becomes larger as the water is shot upward.
75
Problem 25
REASONING AND SOLUTION The pump must generate
an upward force to counteract the weight of the column of
water above it. Therefore, F = mg = (  hA)g. The required
pressure is then
P = F/A =  gh = (1.00 *103 kg/m3)(9.80 m/s2)(71 m)
=
7.0  105 Pa
76
Problem 48
REASONING AND SOLUTION Only the weight of the
block compresses the spring. Applying Hooke's law gives W
= kx. The spring is stretched by the buoyant force acting on
the block minus the weight of the block. Hooke's law again
gives FB – W = 2kx. Eliminating kx gives FB = 3W. Now FB
= w gV, so that the volume of the block is
V = 3M /  w = 3(8.00 kg)/(1.00 * 103 kg/m3) = 2.40 * 10–2 m3
Vw = M/  b = (8.00 kg)/(840 kg/m3) = 9.52 * 10–3 m3
V – Vw = 1.45*10-2 m3
100(1.45* 10–2)/(2.40*10–2) =
60.3 %
77
Problem 52
REASONING
a. According to Equation 11.10, the volume flow rate Q is equal
to the product of the cross-sectional area A of the artery and the
speed v of the blood, Q = Av. Since Q and A are known, we can
determine v .
b. Since the volume flow rate Q2 through the constriction is the
same as the volume flow rate Q1 in the normal part of the artery,
Q2 = Q1. We can use this relation to find the blood speed in the
constricted region.
Solution
a. Since the artery is assumed to have a circular cross-section,
its cross-sectional area is A1 = p r 2 , where r1 is the radius.
1
Thus, the speed of the blood is
78
Q1 Q1
3.6 106 m3 / s
2
v1 
 2

4.2

10
m/s
2
A1 p r1 p 5.2 103 m


b. The volume flow rate is the same in the normal and
constricted parts of the artery, so Q2 = Q1. Since Q2  A2v2 ,
the blood speed is v2 = Q2/A2 = Q1/A2. We are given that the
radius of the constricted part of the artery is one-third that
of the normal artery, so r  1 r . Thus, the speed of the
2
3 1
blood at the constriction is
Q1
Q1
Q1
v2 
 2 
A2 p r2 p 1 r
3 1
 
2

3.6 106 m3 / s


p  13 5.2 103 m 

2
 0.38 m/s
79
Problem 57
REASONING AND SOLUTION
a. Using Equation 11.12, the form of Bernoulli's equation with
y1=y2 , we have


3
1.29
kg/m
2
1
2
2
2 

  150 Pa
P1  P2   v2  v1 
(15
m/s)

0
m/s


2
2
b. The pressure inside the roof is greater than the pressure on
the outside. Therefore, there is a net outward force on the roof.
If the wind speed is sufficiently high, some roofs are "blown
outward."
80
Problem 60
Reasoning
a. The drawing shows
two points, labeled 1
and 2, in the fluid.
Point 1 is at the top of
the water, and point 2
is where it flows out of
the dam at the bottom.
Bernoulli’s equation,
Equation 11.11, can be
used to determine the
speed v2 of the water
exiting the dam.
1
P1=1 atmosphere
v1 = 0 m/s
y1
2 P2=1 atmosphere
v2
y2
81
b. The number of cubic meters per second of water that
leaves the dam is the volume flow rate Q. According to
Equation 11.10, the volume flow rate is the product of the
cross-sectional area A2 of the crack and the speed v2 of the
water; Q = A2v2
Solutions
a. According to Bernoulli’s equation, as given in Equation
11.11, we have
P1 
1 v2
1
2
  gy1  P2 
1 v2
2
2
  gy2
82
Setting P1 = P2 , v1 = 0 m/s, and solving for v2, we obtain


v2  2g  y1  y2   2 9.80 m/s2 15.0 m  17.1 m/s
b. The volume flow rate of the water leaving the dam is

3
Q  A2v2  1.30 10 m
2
 17.1 m/s  2.22 10
2
3
m /s
83
Problem 77
REASONING The
drawing at the right
shows the situation. As
discussed in
Conceptual Example 6,
the job of the pump is
to draw air out of the
pipe that dips down
into the water. The
atmospheric pressure
in the well then pushes
the water upward into
the pipe.
84
In the drawing, the best the pump can do is to remove all of
the air, in which case, the pressure P1 at the top of the water
in the pipe is zero. The pressure P2 at the bottom of the pipe
at point A is the same as that at the point B, namely, it is equal
to atmospheric pressure (1.013  10 5 Pa ), because the two points
are at the same elevation, and point B is open to the
atmosphere. Equation 11.4, P2  P1  gh can be applied to
obtain the maximum depth h of the well.
SOLUTION Setting P1  0 Pa, and solving Equation
11.4 for h, we have
1.013  10 5 Pa
h

 10.3 m
3
3
2
g (1.000  10 kg / m ) (9.80 m / s )
P1
85
Problem 87
REASONING AND SOLUTION Let rh represent the inside
radius of the hose, and rp the radius of the plug, as suggested
by the figure below.
hose
plug
r
h
r
p
86
Then, from Equation 11.9, A1v1 = A2v2 , we have
p rh2v1  (p rh2  p rp2 )v2 or
v1
v2


rh2
 rp2
2
rh
  1   rp 
2
r 
 h
or
rp
rh
 1
v1
v2
According to the problem statement, v2 = 3v1 , or
rp
rh
 1
v1
3v1

2
 0.816
3
87