Transcript Chapter 9

Chapter 9
Fluids
Chapter 9: Fluids
MFMcGraw-PHY 1401
•
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

2 1m
4
2
1.0 cm 
  1.0 10 m
 100 cm 
A 500N person
weighs about
113 lbs.
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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Gravity’s Effect on Fluid Pressure
FBD for the fluid cylinder
An imaginary
cylinder of
fluid
y
x
Imaginary cylinder
can be any size
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P1A
w
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P2A
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Apply Newton’s 2nd Law to the fluid cylinder. Since the
fluids isn’t moving the net force is zero.
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
You noticed on the previous slide that the areas canceled out.
Only the height matters since that is the direction of gravity.
Think of the pressure as a force density in N/m2
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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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Measuring Pressure
A manometer is a
U-shaped tube that
is partially filled
with liquid,
usually Mercury
(Hg).
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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
Also
Pc  Patm
PB  PB'
The pressure at point B is the pressure of the gas.
PB  PB '  PC  gd
PB  PC  PB  Patm  gd
Pgauge  gd
Pgauge = Pmeas - Patm
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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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The Many Units of Pressure
1 ATM equals
1.013x105 N/m2
14.7 lbs/in2
1.013 bar
76 cm Hg
760 mm Hg
760 Torr
34 ft H2O
29.9 in Hg
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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 cross-sectional area
A2 here.
Apply a force F1 here
to a piston of crosssectional area A1.
In these problems neglect
pressure due to columns
of fluid.
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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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Archimedes’ Principle
y
F1
An FBD for an object floating
submerged in a fluid.
x
w
F2
The total force on the block due to
the fluid is called the buoyant force.
FB  F2  F1
where F2  F1
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Buoyant force = the weight of the fluid displaced
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?
Apply Newton’s 2nd Law to the barge:
y
FBD for
the barge
F  F
B
FB
FB  w
x
w
w0
mw g   wVw g  mb g
 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.)
Apply Newton’s 2nd Law to
y
the piece of metal:
FBD for
FB
F  FB  w  ma
the metal

x
w
Solve for a:
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The magnitude of the buoyant force
equals the weight of the fluid displaced
by the metal.
FB   waterVg
 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.
 object
 5.0
Given specific gravity 
 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


The sign is minus because gravity acts down. BF causes a < g.
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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
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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.
Crossing streamlines would indicate a volume of fluid with
two different velocities at the same time.
An ideal fluid is incompressible, undergoes laminar
flow, and has no viscosity.
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The Continuity Equation—Conservation of Mass
Faster
Slower
The amount of mass that flows though the cross-sectional 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?
Simple ratios
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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This is the most general equation
1 2
1 2
P1  gy1  v1  P2  gy2  v2
2
2
Work per unit
volume done
by the fluid
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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.
  d 2 2 
  
2
A 
d 
2 

v1   2 v2    2 v2   2  v2
 A1 
 d1 
   d1  


2




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
Since v1  0
2
2
1
2
 1000 kg/m 3 25 m/s 
2
 3.1 105 Pa


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

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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 
4th power
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
Conversion:
1.013x10 5 Pa
16 (mm Hg)×
= 2132 Pa
760 mm Hg
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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 that the sphere is falling
through, 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?
Apply Newton’s Second Law to
the sphere
y
FBD for
sphere
FB F
D
F  F
D
 FB  w  ma
x
w
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Drag
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Buoyant
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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 
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ms g  lVs g

 2.4 Pa sec
6rvt
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Surface Tension
The surface of a fluid acts like 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 
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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 onesixth 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
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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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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.
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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.
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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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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.
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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-PHY 1401
Ch09e - Fluids-Revised: 7/12/2010
65
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-PHY 1401
2. the same.
4. more than twice.
Ch09e - Fluids-Revised: 7/12/2010
66
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-PHY 1401
Ch09e - Fluids-Revised: 7/12/2010
67
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-PHY 1401
Ch09e - Fluids-Revised: 7/12/2010
68