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Chapter 11
Fluids
11.1 Mass Density
DEFINITION OF MASS DENSITY
The mass density of a substance is the mass of a
substance divided by its volume:
m

V
SI Unit of Mass Density: kg/m3
11.1 Mass Density
11.1 Mass Density
Example 1 Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about
5.2x10-3 m3 of blood.
(a) Find the blood’s weight and (b) express it as a
percentage of the body weight.



m  V  5.2 103 m3 1060 kg m3  5.5 kg


a) W  mg  5.5 kg  9.80 m s 2  54 N
54 N
b) Percentage 
100%  7.8%
690 N
11.2 Pressure
DEFINITION OF PRESSURE:
Pressure is a measure of the
component of force acting
perpendicular to the surface
divided by the area of the
surface.
F
P
A
SI Unit of Pressure: 1 N/m2 = 1Pa
Pascal
11.2 Pressure
Example 2 The Force on a Swimmer
Suppose the pressure acting on the back of
a swimmer’s hand is 1.2x105 Pa. The
surface area of the back of the hand is
8.4x10-3m2.
(a) Determine the magnitude of the force
that acts on it.
(b) Discuss the direction of the force.


a) F  PA  1.2 105 N m 2 8.4 10 3 m 2

F  1.0 103 N
b) Since the water pushes perpendicularly against the back of the hand,
the force is directed downward in the drawing.
11.2 Pressure
While you may not feel it, the force of the atmosphere constantly acts on you.
Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere
11.3 Pressure and Depth in a Static Fluid
Anyone who has gone
swimming has experienced
the increase in pressure that
occurs as you descend to the
bottom of a pool or lake.
We can use Newton’s Second
Law of motion to summarize
the forces acting on an object.
F
y
 P2 A  P1 A  mg  0
P2 A  P1 A  mg
m  V
11.3 Pressure and Depth in a Static Fluid
V  Ah
Since we know that the volume of a
cube can be determined by
multiplying the area of the base times
the height (V = A  h).
P2 A  P1 A   Vg
P2 A  P1 A   Ahg
P2  P1   hg
This relationship works for liquids
since they are considered
incompressible and their densities will
not change with depth.
11.3 Pressure and Depth in a Static Fluid
Conceptual Example 3 The Hoover
Dam
Lake Mead is the largest wholly artificial
reservoir in the United States. The water
in the reservoir backs up behind the dam
for a considerable distance (120 miles).
Suppose that all the water in Lake Mead
were removed except a relatively narrow
vertical column.
Would the Hoover Dam still be needed to
contain the water, or could a much less
massive structure do the job?
The answer is that the dam would need
to be just as massive since the
pressures at the greatest depths would
be the same.
11.3 Pressure and Depth in a Static Fluid
Conceptual Example 3 The Hoover
Dam
Why would the dam need to be just as
massive if it is holding back much less
water?
The reason has to do with the relationship:
P2  P1   hg
This formula only deals with pressure
changes caused by changes in height in
the vertical direction. Pressure increases
as the depth increases, not due to
horizontal changes in distance.
Pressure increases in depth are
primarily dependent upon gh only.
11.3 Pressure and Depth in a Static Fluid
Example 4 The Swimming Hole
Points A and B are located a
distance of 5.50 m beneath the
surface of the water. Find the
pressure at each of these two
locations.
P2  P1   gh
atmospheric pressure




P2  1.01105 Pa  1.00 103 kg m 3 9.80 m s 2 5.50 m 

 


P2  1.55 105 Pa
Since points A and B are at the same depth, the pressures will
the same at both points.
Note that P1 is often the barometric pressure.
11.4 Pressure Gauges (The Closed-Tube Mercury Manometer)
The closed-tube manometer consists of a tube full of mercury inverted such that its
open end is underneath a pool of mercury.
The pressure at the top of the tube is zero Pa as shown.
The pressure at points A and B are equal since they are at the same elevation, and
are equal to the atmospheric pressure as shown.
What is h?
P2  P1   gh
  gh
PPatm
2 


Patm
1.01105 Pa
h

 g 13.6 103 kg m 3 9.80 m s 2

h  0.760 m  760 mm


11.4 Pressure Gauges (Open-Tube Manometer)
The open-tube manometer consists of a tube, often full of mercury, that is open to
atmospheric pressure as shown below.
The pressure, P2 is greater than the atmospheric pressure.
The pressure at points A and B, as well as P2 are all the same.
The pressure at P1 is equal to the atmospheric pressure.
P2  P1   gh
P2  Patm   gh



gauge pressure
Absolute vs. Gauge Pressure
• The difference between P2 and the
atmospheric pressure is called the
gauge pressure. This is the pressure
often read from a gauge such as one on
your water tank or boiler.
• The absolute pressure is the actual
value of P2.
( P2  PB  PA )
11.5 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
enclosing walls.
In the diagram to the right, force F1 acting on
A1 will cause an increase in pressure by the
amount F1 /A1 throughout the whole enclosure.
If the heights of A1 and A2 are the same, then
gh = 0.
The significance of this relationship is that
a small force acting on a small area will
result in a large force acting on a much
larger area.
P2  P1   g 0 m
F2 F1

A2 A1
 A2 
F2  F1  
 A1 
11.5 Pascal’s Principle
Example 5 A Car Lift
The input piston has a radius of 0.0120 m and the
output plunger has a radius of 0.150 m.
The combined weight of the car and the plunger is
20500 N. Suppose that the input piston has a
negligible weight and the bottom surfaces of the
piston and plunger are at the same level. What is
the required input force?
Since the pistons are cylindrical, we can simply
calculate their areas using A = r2.
 A2 
F2  F1  
 A1 
 0.0120 m 2
F1  20500 N 
 131 N
2
 0.150 m 
11.5 Pascal’s Principle
Example 6 A Car Elevated on a Lift
Using the same lift as in the previous
example, the vehicle is now raised such
that the difference in height between the
two pistons is 1.5 m. The density of the
hydraulic oil is 830 kg/m3. Everything
else is the same as before.
Start with: P2 = P1 + gh and substitute F/A for P.
F2 F1
   gh
A2 A1
F1 F2

  gh
A1 A2
 F2

F1     gh  A1
 A2

 20,500 N

3
2
2


F1  

(
830
kg
/
m
)(
9
.
81
m
/
s
)(
1
.
50
m
)

(
0
.
012
m
)
2

  (0.150m1 )

F1  126 N
11.6 Archimedes’ Principle
When an object is immersed in a fluid, it will experience a net force in the
upward direction (toward the surface) called the buoyant force (FB).
Consider the cylinder immersed in the diagram to the left. The buoyant
force results from the difference between the force acting on the bottom
of the cylinder and the one on top.
F  PA
FB  F2  F1
P2  P1   gh
FB  P2 A  P1 A  P2  P1 A
V  hA
FB   ghA
FB  
V g
mass of
displaced
fluid
11.6 Archimedes’ Principle
Since V = mass, and mg = Fg, or the weight of the fluid, we get:
FB

 Wfluid

Magnitudeof
buoyantforce
Weight of
displacedfluid
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:
• Buoyancy
• Eureka!
11.6 Archimedes’ Principle
• If the weight of the object is less than the weight of the fluid displaced,
the object will float, and the magnitude of the buoyant force is equal to
the magnitude of its weight (object < fluid).
• If the weight of the object is the same as the weight of the fluid
displaced, then the object will neither sink or float (object = fluid).
• If the weight of the object is greater than the weight of the fluid
displaced, the object will sink (object > fluid).
1N
2N
8N
11.6 Archimedes’ Principle
Example 7 A Swimming Raft
The raft is made of solid square
pinewood. Determine whether the raft
floats in water and if so, how much of
the raft is beneath the surface.
The density of water is 1000 kg/m3
and the density of pine is 550 kg/m3.
11.6 Archimedes’ Principle
Start by finding the buoyant force that the raft would experience if it were
to be completely submerged.
In order to do this, we will need to start by finding the volume of the raft.
Vraft  4.0 m4.0 m0.30 m  4.8 m3
FBmax   Vg   waterVwater g



FBmax  1000 kg m3 4.8m3 9.80 m s 2
FBmax  47000 N

11.6 Archimedes’ Principle
We will now find the weight of the raft, given a known density for pine of
550 kg/m3.
We will then compare this weight with that of the maximum buoyant
force of the water.
Wraft  mraft g   pineVraft g



 550 kg m 3 4.8m 3 9.80 m s 2
 26000 N  47000 N
Hence, the raft floats!

11.6 Archimedes’ Principle
Now that we know the raft is floating, how
much of it is submerged? i.e. what is h?
Wraft  FB
26000 N   waterVwater g



26000 N  1000 kg m3 4.0 m4.0 mh 9.80 m s 2

26000 N
h
 0.17 m
3
2
1000 kg m 4.0 m 4.0 m  9.80 m s




11.6 Archimedes’ Principle
Conceptual Example 8 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 than a swimming pool contains?
Yes, as long as the water maintains a gap between the
ship and the walls of the canal. Remember, the pressure
increases with depth.
11.7 Fluids in Motion
Types of Fluid Flow:
In steady flow the velocity of the fluid particles at any point is constant
as time passes.
Unsteady flow exists whenever the velocity of the fluid particles at a
point changes as time passes.
Turbulent flow is an extreme kind of unsteady flow in which the velocity
of the fluid particles at a point change erratically in both magnitude and
direction.
11.7 Fluids in Motion
Types of Fluid Flow (cont.):
Fluid flow can be compressible or incompressible,
though most liquids are nearly incompressible.
(Incompressible means that you cannot squeeze the
fluid into a smaller volume)
Viscous fluids are those types of fluids where drag forces
exist. Examples of such fluids include honey, syrup and
motor oil products.
An incompressible, nonviscous fluid is called an ideal
fluid. Water, alcohol and gasoline are a few examples of
ideal fluids.
11.7 Fluids in Motion
When the flow is steady, streamlines are often used to represent
the trajectories of the fluid particles. Note that the velocity vector is
always tangent to the streamline.
11.8 The Equation of Continuity
The mass of fluid per second that flows through a tube is called the
mass flow rate.
Whether the thumb is over the end to create a sharp stream, or the
fluid is allowed to flow freely out of the end of the hose, the mass
flow rates are equal.
11.8 The Equation of Continuity
The figure below shows fluid flowing through a tube with two
different cross-sectional areas. How much mass passes a given
point during a period of time, t?
m  V   A v
t
distance
m2
  2 A2 v2
t
m1
 1 A1v1
t
If 1 kg of liquid passes by point 2 in time t, then 1 kg of fluid will pass
by point 1 in the same amount of time.
11.8 The Equation of Continuity
EQUATION OF CONTINUITY
The mass flow rate has the same value at every position along a
tube that has a single entry and a single exit for fluid flow.
1 A1v1  2 A2v2
SI Unit of Mass Flow Rate: kg/s
11.8 The Equation of Continuity
Incompressible fluid:
Volume flow rate Q:
A1v1  A2v2
Q  Av
1  2
11.8 The Equation of Continuity
Example 9 A Garden Hose
A garden hose has an unobstructed
opening with a cross sectional area of
2.85x10-4m2. It fills a bucket with a
volume of 8.00x10-3m3 in 30 seconds.
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.
a)
Q  Av


Q 8.00 103 m3 30.0 s 
v 
 0.936 m s
-4
2
A
2.85 10 m
b)
A1v1  A2v2
A1
v2 
v1  20.936 m s   1.87 m s
A2
11.9 Bernoulli’s Equation
Using the Equation of Continuity,
we know that where the velocity is
higher, the pressure is lower. The
fluid experiences a net unbalanced
force that causes it to accelerate
from region 2 to region 1.
According to the pressure-depth
relationship, the pressure is also
lower at higher levels, provided the
cross-sectional area of the pipe
does not change.
11.9 Bernoulli’s Equation (Conservation of Energy & the Work-Energy Theorem)
If you think of the blue fluid element of mass m as an object moving in
the pipe from region 2 to region 1, we can analyze it from the
perspective of Conservation of Energy and the Work-Energy
Theorem where we will consider only changes in mechanical energy
(kinetic and potential energy).
The work done on the fluid as it moves from region 2 to region 1 can
be written as:
Wnc = E1 – E2
Non-Conservative Work
11.9 Bernoulli’s Equation (Work done on the fluid element)
Since the total mechanical energy of the fluid element in each region
can be defined by its potential and kinetic energies:
Wnc = (KE1 + PE1) - (KE2 + PE2)
Wnc = (½ mv12 + mgy1)  (½ mv22 + mgy2)
Total mechanical
Energy in region 1
Total mechanical
Energy in region 2
11.9 Bernoulli’s Equation (Work done on the fluid element)
We will now focus on the left side of the expression, Wnc. The pressure in
region 2 is greater than it is in region 1, which gives rise to a net force, F,
that causes the fluid to flow through a displacement, s, as seen in (b)
below.
On the top surface:
the pressure is: PT
the force is:
FT = PTA
On the bottom surface: the pressure is: PB = PT + P
the force is:
FB = FT + F = (PT + P)A
The net force that pushes the fluid element up the pipe is: F = (P)A
Knowing that Work = Fd, it follows that:
Wnc = Fs = (P)As = (P2 – P1)V
11.9 Bernoulli’s Equation
By combining the two relationships, we get:
Wnc = (P2 – P1)V = (½ mv12 + mgy1)  (½ mv22 + mgy2)
Knowing that  = m/V, we will divide both side by V to get:
Wnc = (P2 – P1) = (½ v12 + gy1)  (½ v22 + gy2)
A rearrangement results in BERNOULLI’S EQUATION
P1 + ½ v12 + gy1 = P2 + ½ v22 + gy2
Bernoulli’s Equation is an alternative representation for the Law
of Conservation of Energy. It is true in cases where there is a
steady flow of a nonviscous, incompressible fluid.
11.10 Applications of Bernoulli’s Equation
Conceptual Example 10 Tarpaulins and Bernoulli’s Equation
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 air is stationary both inside and
outside the cargo hold while the
vehicle is not moving. However,
as it drives down the highway, the
air flowing over the top of the tarp
will result in lower pressure outside
the cargo hold than inside, causing
the tarp to bulge outward.
11.10 Applications of Bernoulli’s Equation
Lift in aircraft is a result of pressure differences caused by differences
in the speed of the air flowing over and under the wings. As per
Bernoulli’s Equation, as velocity increases, pressure decreases. An
aircraft wing is designed such that the air flow will be faster over the
top of the wing rather than the bottom.
11.10 Applications of Bernoulli’s Equation
A curveball in baseball is a pitch where the pitcher causes the ball to
spin as it travels towards the batter and catcher. As a result of this
rotation, the ball will follow a curved path. The deflection occurs in the
direction where the airflow over the surface of the ball is the greatest.
11.10 Applications of Bernoulli’s Equation
Example 11 Efflux Speed
A large storage tank in a commercial winery
is cylindrical in shape. The height, h, of
the fluid is two meters above the spigot.
The opening of the spigot has a radius
of 0.5 cm.
a) Determine the speed of the fluid leaving
the tank.
b) Determine how long it will take to fill a 1
liter bottle of wine.
11.10 Applications of Bernoulli’s Equation
We will start by making a few assumptions:
1. The fluid at the top of the liquid and at
that leaving the spigot will experience
the same pressure (Patm).
2. The velocity at the top of the liquid can
be considered to be zero.
3. We will also consider wine to be an ideal
incompressible fluid with negligible
viscosity.
P1  12 v12  gy1  P2  12 v22  gy2
11.10 Applications of Bernoulli’s Equation (a)
P1  P2  Patm
P1  12 v12  gy1  P2  12 v22  gy2
1
2
v12  gh
y2  y1  h
v1  2 gh
v1  2(9.81m / s 2 )( 2.0m)  6.4m / s
v2  0
11.10 Applications of Bernoulli’s Equation (b)
A    r2
s  v1  t
V  A s
V    r 2  v1  t
V
t
  r 2  v1
1000cm3
t
 2.0s
2
( )(0.5cm) (640cm / s)