Transcript Fluids and Buoyant Force

Chapter 9
Fluids and Buoyant Force
In Physics, liquids and gases are
collectively called fluids.
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Fluids and Buoyant Force
A fluid is a nonsolid state of matter in
which the atoms or molecules are free
to move past each other, as in a gas
or a liquid.
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Fluids and Buoyant Force
Liquids differ from gases in that liquids
have a definite volume, gases do not.
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Fluids and Buoyant Force
Recall, volume is the amount of
space that an object occupies.
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Fluids and Buoyant Force
Of course, whether
an object floats or
sinks has little to
do with size. It is
the object’s mass
density that
matters.
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Fluids and Buoyant Force
Mass density (or just density) is the
mass per unit volume of a substance.
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Fluids and Buoyant Force
The Mass density of
fresh water at 4oC and 1
atm of pressure is 1.00
g/cm3 or 1.00X103kg/m3
Recall, 1 ml = 1 cm3
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Fluids and Buoyant Force
That means that one liter (1000 ml) of
fresh water at 4oC and 1 atm of
pressure has a mass of one kilogram
(1000 g).
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Fluids and Buoyant Force
Scuba divers,
submarines and
many types of fish
control their density
to sink or rise in the
water.
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Fluids and Buoyant Force
Formula for Mass Density
mass
density 
volume
m

V
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Fluids and Buoyant Force
Example 1. If gold has a mass density
of 19.3 g/cm3 at STP (standard
temperature and pressure), how large
would a 3.75 g sample of pure gold be?
Given:  = 19.3 g/cm3
Find: V
m = 3.75 g
3.75 g
3
Solution: V 

 0.194 cm
3
 19.3 g /cm
m
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Fluids and Buoyant Force
You may have
noticed that objects
feel lighter in the
water than they do
in the air.
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Fluids and Buoyant Force
This is because the
water exerts a
buoyant force in
the opposite
direction as gravity.
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Fluids and Buoyant Force
Buoyant Force is a force that acts
upward on an object submerged in a
liquid or floating on a liquid’s surface.
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Fluids and Buoyant Force
Do you know the story of Archimedes,
and how he discovered what has come
to be known as “Archimedes
Principle”?
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Fluids and Buoyant Force
Archimedes (287-212 BC) is considered
to be one of the greatest mathematicians
of all times, and he is sometimes called
the “father of mathematical physics.”
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Fluids and Buoyant Force
Hiero, the king of Syracuse,
commissioned a craftsman to fashion a
crown, and provided he with an exact
amount of gold to make it.
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Fluids and Buoyant Force
The crown that Heiro received weighed
the same amount as the gold he had
provided, but he suspected the
craftsman had used some silver instead
of pure gold, and he asked Archimedes
to prove it.
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Fluids and Buoyant Force
Archimedes was contemplating the
problem while in the bath. He realized
that by getting in the bath, he displaced
an amount of water that was proportional
to the part of his body that he
submerged.
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Fluids and Buoyant Force
He is said to have become so excited
that he ran naked through the streets,
shouting “Eureka! Eureka!” (this
translates to, “I have found it!”)
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Fluids and Buoyant Force
He formulated what is now known as
Archimedes principle, and a way to
test the composition of the crown.
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Fluids and Buoyant Force
Archimedes Principle – any object
that is completely or partially
submerged in a fluid experiences an
upward buoyant force equal in
magnitude to the weight of the fluid
displaced by the object.
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Fluids and Buoyant Force
Formula for Buoyant Force
FB  Fw (displaced fluid)  mf g
magnitude of buoyant force  weight of fluid displaced
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Fluids and Buoyant Force
Example 2. A piece of an unknown mineral
that is dropped into mercury displaces 0.057
m3 of Hg. Find the buoyant force on the
mineral. (note,  of Hg = 13.6 x 103 kg/m3 at
STP.)
3
3
V = 0.057 m3
Given:  = 13.6 x 10 kg/m
Find: FB
m
so m  V so FB  mf g  ρf Vf g
Hint:  
V
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Fluids and Buoyant Force
Example 2. A piece of an unknown mineral
that is dropped into mercury displaces 0.057
m3 of Hg. Find the buoyant force on the
mineral. (note,  of Hg = 13.6 x 103 kg/m3 at
STP.)
3
3
V = 0.057 m3
Given:  = 13.6 x 10 kg/m
Find: FB Solution:
FB  m f g  ρf Vf g
3
3
2
 (13.6 x 103 kg/ m
 )(0.057 m
 )(9.81 m/s )  7600 N
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Fluids and Buoyant Force
If the buoyant force on an object is
greater than the weight of the object, the
object will bob to the surface.
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Fluids and Buoyant Force
For an object that is
floating, the
buoyant force is
equal and opposite
to the weight of the
object.
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Fluids and Buoyant Force
In other words, if an
object is less dense
than the fluid that it
rests in, it will float.
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Fluids and Buoyant Force
The density of the
object determines
how much of the
object’s volume will
be submerged
below the surface.
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Fluids and Buoyant Force
Formula for Determining what
Portion of a Floating Object will
Submerge
Vsub  object

V
 fluid
Sub=submerged
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V=Volume of entire object
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Fluids and Buoyant Force
Example 3. A 2.00 m3 block of wood has a
density of 0.500 x 103 kg/m3. If this object is
placed in fresh water at 4oC and 1 atm of
pressure, what volume will be below the surface
of the water?
3
3
V = 2.00 m3
Given: object = 0.500 x 10 kg/m
fluid = 1. 00 x 103 kg/m3 Find: Vsub
Solution:
Vsub 
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V object
 fluid
(2.00 m3 )(0.500 x 103 kg/m 3 )
3


1.00
m
1.00 x 103 kg/m 3
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Fluids and Buoyant Force
Notice, if an object
is ½ as dense as
water, then ½ of it
will be under the
water.
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Fluids and Buoyant Force
If an object is 90%
as dense as water,
than 90% of the
object will be below
the surface (AS IN
AN ICEBERG).
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Fluids and Buoyant Force
If the density of the
object is greater
than the density of
the fluid, it will
continue to sink.
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Fluids and Buoyant Force
The apparent weight of a submerged
object depends upon its density.
This, so called “apparent
weight” is really just the
net force (Fnet) in the yaxis.
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Fluids and Buoyant Force
Formulas for Determining the FNet of a
Submerged Object
FNet  FB  FW
FNet  mf g  mo g
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FNet  f Vf g  o Vo g
Other
o

FB  f
Fg
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Fluids and Buoyant Force
Example 4. Gold has a density of 19.3 x 103
kg/m3. If a piece of gold with a volume of 0.450
m3 is completely submerged in water, how much
will it seem to weigh?
Hint: The Volume of the water is the same as the
volume of the object.
3 kg/m3 V = 0.450 m3

=
19.3
x
10
Given: object
o
fluid = 1.00 x 103 kg/m3 Vf = 0.450 m3
Find: FNet
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Fluids and Buoyant Force
3 kg/m3 V = 0.450 m3

=
19.3
x
10
object
o
Given:
fluid = 1.00 x 103 kg/m3 Vf = 0.450 m3
Find: FNet
Solution: Because Vo = Vf, we can rewrite our equation.
FNet  f Vf g  o Vo g  ( f - o )Vg
FNet  (  f -  o )Vg
 (1.00 x 103 kg/m 3 - 19.3 x10 3 kg/m 3 )(0.450 m3 )(9.81 m/s 2 )
 - 80785.35 N  - 8.08 x 104 N
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DOWN (it will sink, not float!)
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Another useful formula is
 Apparent weight = Fg – FB
Remember…FW and Fg are the same…

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Chapter 9 Section 9-1
Fluids and Buoyant Force
Summary of Formulas
 substance
m
sp. gr. 

 water
V
FB  Fw (displaced fluid)  mf g
Vsub  object

V
 fluid
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Chapter 9 Section 9-1
Fluids and Buoyant Force
Summary of Formulas
.

Apparent weight = Fg – FB
FNet  FB  FW
FNet  mf g  mo g
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FNet  f Vf g  o Vo g
o

FB  f
Fg
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Fluid Pressure and Temperature
An object that is
submerged in water, or
any other fluid,
experiences pressure
on all sides. You have
likely noticed the
effects of this pressure
while swimming.
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Fluid Pressure and Temperature
Gases also exert
pressure. Your body
feels the weight of the
atmosphere on it.
Your ears also “pop”
when you change
altitude quickly.
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Fluid Pressure and Temperature
Pressure is the magnitude of the force
on a surface per unit area.
Movie Clip 
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Fluid Pressure and Temperature
Formula for Pressure
F
p
A
force
pressure 
area
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Fluid Pressure and Temperature
Can you explain why a finger does not
pop a balloon even when you exert more
force on the balloon than you do with a
needle?
F
p
A
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Fluid Pressure and Temperature
The SI unit of pressure is the pascal (Pa),
which is equal to 1 N/m2.
F
p
A
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Conversion Factor
1 Pa = 1 N/m2
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Fluid Pressure and Temperature
Example 1. A cylinder-shaped column with a
height of 2.0 m and a base radius of 0.50 m
has a mass of 1250 kg. Find the pressure it
exerts on the ground.
Given: m = 1250 kg h = 2.0 m r = 0.50 m
Find: P
Solution:
F mg (1250 kg)(9.81 m/s 2 )
4
p  2 

1.6
x
10
Pa
2
A r
 (0.50 m)
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Fluid Pressure and Temperature
There are many other common units of
pressure, including the atmosphere (atm)
and millimeters of mercury (mm of Hg).
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Fluid Pressure and Temperature
Important Conversion Factors
1 atm = 760 mm of Hg = 760 torr = 1.013 x 105 Pa
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Fluid Pressure and Temperature
Example 2. Convert 2.30 atm to Pa.
1.013 x 105 Pa
2.30 atm x
= 2.33 x 105 Pa
1.00 atm
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Fluid Pressure and Temperature
A barometer is an instrument that is
used to measure atmospheric pressure.
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Fluid Pressure and Temperature
You have probably noticed that when you
squeeze a balloon, it will bulge outwards in
a different area. This can be explained by
Pascal’s Principle.
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Fluid Pressure and Temperature
Pascal’s Principle – Pressure applied
to a fluid in a closed container is
transmitted equally to every point of the
fluid and to the walls of the container.
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Fluid Pressure and Temperature
Formula for Pascal’s Principle
F1
F2
p increase 

A1 A 2
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Fluid Pressure and Temperature
Example 3. In a hydraulic car lift, the compressed air
exerts a force on a piston with a base radius of 5.00
cm. The pressure is transmitted to a second piston
with a base radius of 35.0 cm. How much force must
be exerted on the first piston to lift a car with a mass of
1.40 x 103 kg on the second piston?
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Fluid Pressure and Temperature
Example 4. In a hydraulic car lift, the compressed air
exerts a force on a piston with a base radius of 5.00
cm. The pressure is transmitted to a second piston
with a base radius of 35.0 cm. How much force must
be exerted on the first piston to lift a car with a mass of
1.40 x 103 kg on the second piston?
Given: r1 = 5.00 cm
r2 = 35.0 cm
m2 = 1.40 x 103 kg
Find: F1
A1
r1
Equation: F1  A x F2  r 2 x m2g
2
2
2
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Fluid Pressure and Temperature
Given: r1 = 5.00 cm
r2 = 35.0 cm
Find: F1
m2 = 1.40 x 103 kg
A1
r1
Equation: F1  x F2  2 x m1g
A2
r2
2
Solution:
A1
r
x F2  1 2 x m1g
A2
r2
2
F1 
 (5.00 cm) 2
3
2

x
(1.40
x
10
kg)(9.81
m/s
)  28 0 N
2
 (35.0 cm)
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Lesson 9-2
Fluid Pressure and Temperature
Pressure varies with depth in a fluid. That
is why your ears “pop” when you go up in
an airplane or down in the water.
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Lesson 9-2
Fluid Pressure and Temperature
Atmospheric pressure (patm) is the pressure
exerted by the atmosphere. It varies from day
to day and from location to location.
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Lesson 9-2
Fluid Pressure and Temperature
Standard atmospheric pressure is equal
to 1.0 atm or approximately 1.013 x 105 Pa.
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Lesson 9-2
Fluid Pressure and Temperature
Gauge pressure (pgauge = gh) is the pressure
exerted by just the water, or other fluid. When
calculating gauge pressure, leave the
atmospheric pressure out of the calculation.
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Lesson 9-2
Fluid Pressure and Temperature
Gauge Pressure as a Function of
Depth
p gauge  gh
Where  = density of fluid, g = acceleration due to gravity
and h = depth
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Lesson 9-2
Fluid Pressure and Temperature
Example 5. Calculate the gauge pressure at a
point 10.0 m below the surface of the ocean.
(note, sea water = 1.025 x 103 kg/m3)
Given:  = 1.025 x 103 kg/m3
h = 10.0 m
g = 9.81 m/s2
Find: pgauge
Solution:
pgauge  gh  (1.025 x 103 kg/m 3 )(9.81 m/s 2 )(10.0 m)  1.0 x 105 Pa
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Lesson 9-2
Fluid Pressure and Temperature
Absolute pressure (p) is the total pressure
exerted by the water (or other fluid) and the
atmosphere above the water.
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Lesson 9-2
Fluid Pressure and Temperature
Absolute Pressure as a Function of
Depth
p  p 0  gh
Where p = absolute pressure, p0 = pressure at
point of comparison (usually atmospheric
pressure),  = density of fluid, g = acceleration
due to gravity and h = depth
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Lesson 9-2
Fluid Pressure and Temperature
Example 6. Calculate the absolute pressure
experienced by a scuba diver 20.0 m below the
sea ( = 1.025 x 103 kg/m3). Assume patm = 1.0
x 105 Pa
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Lesson 9-2
Fluid Pressure and Temperature
Example 6. Calculate the absolute pressure
experienced by a scuba diver 20.0 m below the
sea ( = 1.025 x 103 kg/m3). Assume patm = 1.0
x 105 Pa
Given: p0 = 1.0 x 105 Pa h = 20.0 m  = 1.025 x 103 kg/m3
Find: p
p  p 0  gh
Solution:
 (1.0 x 105 Pa)  (1.025 x 103 kg/m 3 )(9.81 m/s 2 )(20.0 m)
 3.0 x 105 Pa
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Lesson 9-2
Fluid Pressure and Temperature
Summary of Formulas
F
p
A
F1
F2
p increase 

A1 A 2
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p  p 0  gh
p gauge  gh
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Lesson 9-2
Fluid Pressure and Temperature
Summary of Formulas
1 atm = 760 mm of Hg = 760 torr = 1.013 x 105 Pa
C  K  273 K
0
K  C  273 K
0
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Fluids in Motion
The motion of a fluid can be described as
either laminar (smooth) or turbulent
(irregular).
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Fluids in Motion
When studying fluids, is helps to consider
the behaviors of an ideal fluid.
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Fluids in Motion
An ideal fluid is an imaginary fluid that
has no internal friction or viscosity, and
is incompressible.
The viscosity of a fluid is a measure of
its internal resistance, or its resistance to
flow.
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Fluids in Motion
Fluids with high
viscosity, like
molasses, flow
slowly, due to
internal friction.
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Fluids in Motion
The law of
conservation of
mass explains that
the mass entering a
section of pipe or
tube must be the
same as the mass
that leaves.
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Fluids in Motion
The Continuity Equation
A1ν1  A2 ν 2
area x speed in region 1 = area x speed in region 2
The Continuity Equation is an expression of
the conservation of mass.
A ν  flowrate(m3 / sec)
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Fluids in Motion
A1 ν1  A 2 ν 2
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Fluids in Motion
Example 1. An ideal fluid is moving at 3.0 m/s
through a section of pipe with a radius of 0.35 m.
How fast will it flow through another section of the
pipe with a radius of 0.20 m?
HINT: A = r2
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Fluids in Motion
Example 1. An ideal fluid is moving at 3.0 m/s
through a section of pipe with a radius of 0.35 m.
How fast will it flow through another section of the
pipe with a radius of 0.20 m?
Given: n1 = 3.0 m/s
r1 = 0.35 m
r2 = 0.20 m
Find: n2
2
2
A
A
Formula:  1r1 ν 1   2r2 ν 2
2

r1 ν1  (0.35 m) 2 (3.0 m/s)
Solution: ν 2 

 9.2 m/s
2
2
 (0.20 m)
 r2
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Fluids in Motion
From example 1, we
see that as the area
of the pipe
decreases, the
velocity of the fluid
becomes much
greater.
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Fluids in Motion
Another property of fluids explains how
the wings of an airplane produce lift.
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Fluids in Motion
To demonstrate Bernoulli’s principle for
yourself, try blowing air over the top of a
piece of paper.
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Fluids in Motion
Bernoulli’s Principle – The pressure in
a fluid decreases as the fluid’s velocity
increases.
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Fluids in Motion
The wings of an airplane are designed to
allow air to flow quicker over the top than
the bottom, resulting in lower pressure
above the wings.
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Fluids in Motion
Bernoulli’s Equation
P  n  gh  constant
1
2
2
pressure + kinetic energy per unit volume + gravitational
potential energy per unit volume = constant along a given
streamline.
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Fluids in Motion
Bernoulli’s Equation for Comparing a
Fluid at Two Different Points
P1  n 1  gh 1  P2  n 2  gh 2
1
2
2
1
2
2
pressure + kinetic energy per unit volume + gravitational
potential energy per unit volume = pressure + kinetic energy
per unit volume + gravitational potential energy per unit
volume .
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Important to know!!!!!

IF there is no height difference, you can
eliminate the
gh 1 andgh 2
If there is no area change, velocity stays
the same
 IF there is no pressure change, P1 = P2

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Fluids in Motion
Example 2. Water is circulated through a system.
If the water is pumped with a speed of 0.45 m/s
under a pressure of 2.2 x 105 Pa from the first
floor through a 6.0-cm diameter pipe, what will the
pressure be on the next floor 4.0 m above?
HINT: The key to this
question is that the diameter
of the pipe doesn’t change
and, therefore, the velocity
of the water remains
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constant.
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Fluids in Motion
If n1 = n2 , then we can simplify our equation.
P1  n 1  gh 1  P2  n 2  gh 2
1
2
2
1
2
2
P1  gh 1  P2  gh 2
P2  P1  g(h 1 - h 2 )
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Fluids in Motion
Example 2. Water is circulated through a system.
If the water is pumped with a speed of 0.45 m/s
under a pressure of 2.2 x 105 Pa from the first
floor through a 6.0-cm diameter pipe, what will the
pressure be on the next floor 4.0 m above?
Given: P1 = 2.2 x 105 Pa h1- h2 = -4.0 m
water = 1000 kg/m3 g = 9.81 m/s2
Find: P2
Formula: P2  P1  g(h 1 - h 2 )
P2  P1  g(h 1 - h 2 )
 2.2 x 105 Pa  (1000 kg/m 3 )(9.81 m/s 2 )(-4.0 m)  1.9 x 105 Pa
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Fluids in Motion
Example 3. A water tank has a water level of 1.2 m.
The spigot is located 0.40 m above the ground. If
the spigot is opened fully, how fast will water come
out of the spigot?
HINT: The container is
open to the atmosphere, so
the pressure will be the
same in both spots. We
can assume that the velocity
of the water at the top (v2) is
essentially
zero.
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Fluids in Motion
The pressure of the fluid in both areas is the
same, P1 = P2 so they cancel out. Also, the
velocity at the top (n2) is zero, so we cross out
that expression.
P1  n 1  gh 1  P2  n 2  gh 2
2
1
2
1
2
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1
2
2
n 1  gh 1  gh 2
2
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Fluids in Motion
Now, we can divide all expressions by  (the fluid
doesn’t change, so density is the same
throughout) and continue to isolate n1, the
unknown.
1
2
n 1  gh 1  gh 2
2
n  gh 1  gh 2
2
1
2 1
2
1
2 1
2
1
2 1
n  gh 2 - gh 1
n  g(h 2 - h1 )
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n 1  2g(h 2 - h1 )
94
Fluids in Motion
Example 3. A water tank has a water level of 1.2
m. The spigot is located 0.40 m above the
ground. If the spigot is opened fully, how fast will
water come out of the spigot?
Given: h1 = 0.40 m
Find: n2
h2 = 1.2 m
Formula: n 1
g = 9.81 m/s2
 2g(h 2 - h1 )
Solution:
n 1  2(9.81 m/s 2 )(1.2 m - 0.40 m)  4.0 m/s
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95
Fluids in Motion
Summary of Formulas
A1 ν1  A 2 ν 2
P1  n 1  gh 1  P2  n 2  gh 2
1
2
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2
1
2
2
96
Lesson 9-2
Fluid Pressure and Temperature
The Kinetic Molecular Theory (KMT) of
Gases also relates the temperature of a
gas to the motion of its particles.
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97
Lesson 9-2
Fluid Pressure and Temperature
Temperature is a measure of the
average kinetic energy of the particles of
a substance.
The SI unit of
temperature is the
Kelvin (K).
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98
Lesson 9-2
Fluid Pressure and Temperature
Converting between Celsius and
Kelvin
K  C  273 K
0
C  K  273 K
0
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99
Lesson 9-2
Fluid Pressure and Temperature
Example 7. Convert 22.0 oC to Kelvin.
K  C  273 K  22.0  273 K  295 K
0
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100