Transcript Chapter 9

Chapter 9
Fluids
Chapter 9: Fluids
• Introduction to Fluids
• Pressure
• Measurement of Pressure
• Pascal’s Principle
• Gravity and Fluid Pressure
• Archimedes’ Principle
• Continuity Equation
• Bernoulli’s Equation
• Viscosity and Viscous Drag
• Surface Tension
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Pressure
Pressure arises from the collisions between the particles
of a fluid with another object (container walls for
example).
There is a momentum
change (impulse) that
is away from the
container walls. There
must be a force exerted
on the particle by the
wall.
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By Newton’s 3rd Law, there is a force on the wall due
to the particle.
F
Pressure is defined as P  .
A
The units of pressure are N/m2 and are called Pascals
(Pa).
Note: 1 atmosphere (atm) = 101.3 kPa
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Example (text problem 9.1): Someone steps on your toe,
exerting a force of 500 N on an area of 1.0 cm2. What is
the average pressure on that area in atmospheres?
2
 1m 
4
2
1.0 cm 
  1.0 10 m
 100 cm 
2
F
500 N
Pav  
A 1.0 10 4 m 2
1 atm

6
2  1 Pa 
 5.0 10 N/m 

2 
5
 1 N/m  1.013 10 Pa 
 49 atm
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Measuring Pressure
A manometer
is a U-shaped
tube that is
partially
filled with
liquid.
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Both ends of the
tube are open to
the atmosphere.
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A container of gas is connected to one end of the U-tube
If there is a pressure difference between the gas and the
atmosphere, a force will be exerted on the fluid in the U-tube.
This changes the equilibrium position of the fluid in the tube.
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From the
figure:
At point C Pc  Patm
Also
PB  PB'
The pressure at point B is the pressure of
the gas.
PB  PB '  PC  gd
PB  PC  PB  Patm  gd
Pgauge  gd
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A Barometer
The atmosphere pushes on the container of mercury
which forces mercury up the closed, inverted tube. The
distance
d is called the
barometric pressure.
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From the figure
and
PA  PB  Patm
PA  gd
Atmospheric pressure is equivalent to a column
of mercury 76.0 cm tall.
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Pascal’s Principle
A change in pressure at any point in a confined
fluid is transmitted everywhere throughout the
fluid. (This is useful in making a hydraulic lift.)
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The applied force is
transmitted to the
piston of crosssectional area A2 here.
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Apply a force F1
here to a piston of
cross-sectional area
A1.
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Mathematically,
P at point 1  P at point 2
F1
F
 2
A1 A 2
 A2 
 F1
F2  
 A1 
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Example: Assume that a force of 500 N (about 110 lbs) is
applied to the smaller piston in the previous figure. For
each case, compute the force on the larger piston if the
ratio of the piston areas (A2/A1) are 1, 10, and 100.
Using Pascal’s
Principle:
A2 A1
1
10
100
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F2
500 N
5000 N
50,000 N
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Gravity’s Effect on Fluid Pressure
FBD for the fluid cylinder
A cylinder
of fluid
y
P1A
x
w
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P2A
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Apply Newton’s 2nd Law to
the fluid cylinder:
F  P A PA w  0
2
1
P2 A  P1 A  Ad g  0
P2  P1  gd  0
 P2  P1  gd
or P2  P1  gd
If P1 (the pressure at the top of the cylinder) is known, then the
above expression can be used to find the variation of pressure
with depth in a fluid.
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If the top of the fluid column is placed at the surface of the fluid,
then P1 = Patm if the container is open.
P  Patm  gd
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Example (text problem 9.19): At the surface of a
freshwater lake, the pressure is 105 kPa. (a) What is the
pressure increase in going 35.0 m below the surface?
P  Patm  gd
P  P  Patm  gd



 1000 kg/m 3 9.8 m/s 2 35 m 
 343 kPa  3.4 atm
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Example: The surface pressure on the planet Venus is 95
atm. How far below the surface of the ocean on Earth do
you need to be to experience the same pressure? The
density of seawater is 1025 kg/m3.
P  Patm  gd
95 atm  1 atm  gd
gd  94 atm  9.5 106 N/m 2
1025 kg/m 9.8 m/s d  9.5 10
3
2
6
N/m 2
d  950 m
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Archimedes’ Principle
y
F
1
An FBD for an object
floating submerged in a
fluid.
x
w
F
2
The total force on the block
due to the fluid is called the
buoyant force.
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FB  F2  F1
where F2  F1
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The magnitude of the buoyant force
is:
FB  F2  F1
 P2 A  P1 A
 P2  P1 A
From before: P2  P1  gd
The result is FB  gdA  gV
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Archimedes’ Principle: A fluid exerts an upward
buoyant force on a submerged object equal in
magnitude to the weight of the volume of fluid
displaced by the object.
FB  gV
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Example (text problem 9.28): A flat-bottomed barge
loaded with coal has a mass of 3.0105 kg. The barge is
20.0 m long and 10.0 m wide. It floats in fresh water.
What is the depth of the barge below the waterline?
y
FBD for
the barge
Apply Newton’s 2nd Law to the barge:
F
F  F
B
FB  w
B
mw g   wVw g  mb g
x
w
w0
 wVw  mb
 w  Ad   mb
mb
3.0 105 kg
d

 1.5 m
3
 w A 1000 kg/m 20.0 m *10.0 m 

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
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Example (text problem 9.40): A piece of metal is released
under water. The volume of the metal is 50.0 cm3 and its
specific gravity is 5.0. What is its initial acceleration?
(Note: when v = 0, there is no drag force.)
y
FBD
for the
metal
Apply Newton’s 2nd Law to
the piece of metal:
F
F  F
B
B
x
w
 w  ma
The magnitude of the buoyant force
equals the weight of the fluid displaced
by the metal.
F   Vg
B
Solve for a:
MFMcGraw
water
 ρ waterV

ρ waterVg
FB

a
g 
g g
 1
ρ V

m
ρ objectVobject
 object object 
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Example continued:
Since the object is completely submerged V=Vobject.
specific gravity 

 water
where water = 1000 kg/m3 is
the density of water at 4 °C.
Given
 object
specific gravity 
 5.0
 water
 ρ waterV

 1

 1

a  g
 1  g 
 1  g 
 1  7.8 m/s 2
ρ V

 S .G. 
 5.0 
object
object


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Fluid Flow
A moving fluid will exert forces parallel to the surface
over which it moves, unlike a static fluid. This gives rise
to a viscous force that impedes the forward motion of the
fluid.
A steady flow is one where the velocity at a given
point in a fluid is constant.
V1 =
constant
MFMcGraw
v1v2
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V2 =
constant
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Steady flow is laminar; the fluid flows in layers.
The path that the fluid in these layers takes is
called a streamline. Streamlines do not cross.
An ideal fluid is incompressible, undergoes
laminar flow, and has no viscosity.
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The continuity equation—Conservation of mass.
The amount of mass that flows though the crosssectional area A1 is the same as the mass that flows
through cross-sectional area A2.
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m
 Av
t
V
 Av
t
is the mass flow rate (units kg/s)
is the volume flow rate (units m3/s)
The continuity equation is
1 A1v1  2 A2v2
If the fluid is incompressible, then 1= 2.
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Example (text problem 9.41): A garden hose of inner
radius 1.0 cm carries water at 2.0 m/s. The nozzle at the
end has radius 0.20 cm. How fast does the water move
through the constriction?
A1v1  A2 v2
 A1 
 r12 
v2   v1   2 v1
 A2 
 r2 
2
 1.0 cm 

 2.0 m/s   50 m/s
 0.20 cm 
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Bernoulli’s Equation
Bernoulli’s equation is a statement of energy
conservation.
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1 2
1 2
P1  gy1  v1  P2  gy2  v2
2
2
Work per unit
volume done
by the fluid
MFMcGraw
Potential
energy
per unit
volume
Kinetic
energy
per unit
volume
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Points 1 and 2
must be on the
same streamline
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Example (text problem 9.49): A nozzle is connected to a
horizontal hose. The nozzle shoots out water moving at
25 m/s. What is the gauge pressure of the water in the
hose? Neglect viscosity and assume that the diameter of
the nozzle is much smaller than the inner diameter of the
hose.
Let point 1 be inside the hose and point 2 be
outside the nozzle.
1 2
1 2
P1  gy1  v1  P2  gy2  v2
2
2
The hose is horizontal so y1 = y2. Also P2 = Patm.
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Example continued:
Substituting:
1 2
1 2
P1  v1  Patm  v2
2
2
1 2 1 2
P1  Patm  v2  v1
2
2
v2 = 25 m/s and v1 is unknown. Use the continuity
equation.
2


   d 2 
 A2 
2

v1   v2    2
 A1 
   d1 

 2

2
 d2 

v2   d  v2
 1



Since d2<<d1 it is true that v1<<v2.
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Example continued:
P1  Patm
1 2 1 2
 v2  v1
2
2
1
1 2
2
2
  v2  v1  v2
2
2
1
2
 1000 kg/m 3 25 m/s 
2
 3.1 105 Pa


MFMcGraw


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Viscosity
A real fluid has viscosity (fluid friction). This implies a
pressure difference needs to be maintained across the
ends of a pipe for fluid to flow.
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Viscosity also causes the existence of a velocity
gradient across a pipe. A fluid flows more rapidly in the
center of the pipe and more slowly closer to the walls of
the pipe.
The volume flow rate for laminar flow of a viscous
fluid is given by Poiseuille’s Law.
V  P L 4

r
t 8 
where  is the viscosity
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Example (text problem 9.55): A hypodermic syringe attached to a
needle has an internal radius of 0.300 mm and a length of 3.00 cm.
The needle is filled with a solution of viscosity 2.0010-3 Pa sec; it is
injected into a vein at a gauge pressure of 16.0 mm Hg. Neglect the
extra pressure required to accelerate the fluid from the syringe into the
entrance needle.
(a) What must the pressure of the fluid in the syringe be in
order to inject the solution at a rate of 0.250 mL/sec?
Solve Poiseuille’s Law for the pressure difference:


8L V 8 2.00 10 3 Pa sec 3.00 cm 
3
P  4


0
.
250
cm
sec
4

1
r t
 0.3 10 cm


 4716 Pa
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Example continued:
This pressure difference is between the fluid in the
syringe and the fluid in the vein. The pressure in
the syringe is
P  Ps  Pv
Ps  Pv  P
 2140 Pa  4720 Pa  6860 Pa
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Example continued:
(b) What force must be applied to the plunger, which
has an area of 1.00 cm2?
The result of (a) gives the force per unit area on
the plunger so the force is just F = PA = 0.686 N.
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Viscous Drag
The viscous drag force on a sphere is given by Stokes’
law.
FD  6rv
Where  is the viscosity of the fluid, r is the radius of
the sphere, and v is the velocity of the sphere.
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Example (text problem 9.62): A sphere of radius 1.0 cm
is dropped into a glass cylinder filled with a viscous
liquid. The mass of the sphere is 12.0 g and the density
of the liquid is 1200 kg/m3. The sphere reaches a
terminal speed of 0.15 m/s. What is the viscosity of the
liquid?
y
Apply Newton’s Second Law
to the sphere
FBD
for
sphere
F FD
B
x
F  F
D
 FB  w  ma
w
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Example continued:
When v = vterminal, a = 0 and
FD  FB  w  0
6rvt  ml g  ms g  0
6rvt  lVl g  ms g  0
6rvt  lVs g  ms g  0
Solving for 
MFMcGraw

ms g  lVs g
 2.4 Pa sec
6rvt
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Surface Tension
The surface of a fluid acts like a a stretched membrane
(imagine standing on a trampoline). There is a force
along the surface of the fluid.
The surface tension is a force per unit length.
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Example (text problem 9.70): Assume a water strider has
a roughly circular foot of radius 0.02 mm. The water
strider has 6 legs.
(a) What is the maximum possible upward force on
the foot due to the surface tension of the water?
The water strider will be able to walk on water
if the net upward force exerted by the water
equals the weight of the insect. The upward
force is supplied by the water’s surface
tension.
 2  2
F  PA   r  9 10 6 N
 r 
MFMcGraw
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Example continued:
(b) What is the maximum mass of this water strider so
that it can keep from breaking through the water
surface?
To be in equilibrium, each leg must support
one-sixth the weight of the insect.
1
6F
F  w or m 
 5 10 6 kg
6
g
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Summary
•Pressure and its Variation with Depth
•Pascal’s Principle
•Archimedes Principle
•Continuity Equation (conservation of mass)
•Bernoulli’s Equation (conservation of energy)
•Viscosity and Viscous Drag
•Surface Tension
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Quick Questions
Consider a boat loaded with scrap iron in a swimming pool. If the iron is
thrown overboard into the pool, will the water level at the edge of the
pool rise, fall, or remain unchanged?
1. Rise
2. Fall
3. Remain unchanged
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Consider a boat loaded with scrap iron in a swimming pool. If the iron is
thrown overboard into the pool, will the water level at the edge of the
pool rise, fall, or remain unchanged?
1. Rise
2. Fall
3. Remain unchanged
MFMcGraw
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MFMcGraw
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In the presence of air, the small iron ball and large plastic ball balance
each other. When air is evacuated from the container, the larger ball
1. rises.
2. falls.
3. remains in place.
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MFMcGraw
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The weight of the stand and suspended solid iron ball is equal to the weight
of the container of water as shown above. When the ball is lowered into the
water the balance is upset. The amount of weight that must be added to the
left side to restore balance, compared to the weight of water displaced by
the ball, would be
1. half.
3. twice.
MFMcGraw
2. the same.
4. more than twice.
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MFMcGraw
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A pair of identical
balloons are inflated
with air and
suspended on the
ends of a stick
that is horizontally
balanced. When the
balloon on the left is punctured,
the balance of the stick is
1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.
MFMcGraw
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MFMcGraw
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Consider an air-filled balloon weighted so that it is on the verge of
sinking—that is, its overall density just equals that of water.
Now if you push it beneath the surface, it will
1. sink.
2. return to the surface.
3. stay at the depth to
which it is pushed.
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MFMcGraw
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Extras
The density of the block of
wood floating in water is
1. greater than the density of water.
2. equal to the density of water.
3. less than half that of water.
4. more than half the density of water.
5. … not enough information is given.
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The density of the block of
wood floating in water is
1. greater than the density of water.
2. equal to the density of water.
3. less than half that of water.
4. more than half the density of water.
5. … not enough information is given.
MFMcGraw
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In the presence of air, the small iron ball and large plastic ball balance
each other. When air is evacuated from the container, the larger ball
1. rises.
2. falls.
3. remains in place.
MFMcGraw
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The weight of the stand and suspended solid iron ball is equal to the weight
of the container of water as shown above. When the ball is lowered into the
water the balance is upset. The amount of weight that must be added to the
left side to restore balance, compared to the weight of water displaced by
the ball, would be
1. half.
3. twice.
MFMcGraw
2. the same.
4. more than twice.
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A pair of identical
balloons are inflated
with air and
suspended on the
ends of a stick
that is horizontally
balanced. When the
balloon on the left is punctured,
the balance of the stick is
1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.
MFMcGraw
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Consider an air-filled balloon weighted so that it is on the verge of
sinking—that is, its overall density just equals that of water.
Now if you push it beneath the surface, it will
1. sink.
2. return to the surface.
3. stay at the depth to
which it is pushed.
MFMcGraw
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