Transcript Chapter 9: Fluids

Chapter 9: Fluids
•Introduction to Fluids
•Pressure
•Pascal’s Principle
•Gravity and Fluid Pressure
•Measurement of Pressure
•Archimedes’ Principle
•Continuity Equation
•Bernoulli’s Equation
•Viscosity and Viscous Drag
•Surface Tension
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§9.1 Fluids
A liquid will flow to take the shape of the container that holds
it. A gas will completely fill its container.
Fluids are easily deformable by external forces.
A liquid is incompressible. Its volume is fixed and is
impossible to change.
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§9.2 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
4
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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§9.3 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.
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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
F2
500 N
5000 N
50,000 N
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The work done pressing the smaller piston (#1) equals the
work done by the larger piston (#2).
F1d1  F2 d2
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Example: In the previous example, for the case A2/A1 = 10,
it was found that F2/F1 = 10. If the larger piston needs to
rise by 1 m, how far must the smaller piston be depressed?
Using the result on the previous slide,
F2
d1 
d 2  10 m
F1
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§9.4 Gravity’s Effect on Fluid
Pressure
FBD for the fluid cylinder
P1A
A cylinder
of fluid
w
P2A
12
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 9.15): 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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§9.5 Measuring Pressure
A manometer
is a U-shaped
tube that is
partially filled
with liquid.
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
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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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Example (text problem 9.22): An IV is connected to a
patient’s vein. The blood pressure in the vein has a gauge
pressure of 12 mm of mercury. At least how far above the
vein must the IV bag be placed in order for fluid to flow into
the vein? Assume that the density of the IV fluid is the same
as blood.
Blood:
Pgauge   Hg gh1
h1  12 mm
IV:
The pressure is equivalent
to raising a column of
mercury 12 mm tall.
Pgauge   bloodgh2
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Example continued:
At a minimum, the gauge pressures must be equal. When
h2 is large enough, fluid will flow from high pressure to low
pressure.
Pgauge   Hg gh1   bloodgh2
 Hg gh1
h2 
 bloodg
  Hg 
 13,600 kg/m 3 
h1  
12 mm 
 
3 
 1060 kg/m 
  blood 
 154 mm
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§9.6 Archimedes’ Principle
F1
An FBD for an object
floating submerged in
a fluid.
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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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 9.30): 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?
F  F
B
FB
w0
FBD
for the
barge
FB  w
mw g   wVw g  mb g
w
 wVw  mb
 w  Ad   mb
mb
3.0 105 kg
 1. 5 m

d
3
 w A 1000 kg/m 20.0 m *10.0 m 


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Example (text problem 9.34): 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.)
FB
FBD
for the
metal
w
F  F
B
 w  ma
The buoyant force is the
weight of the fluid displaced
by the object
FB   waterVg
 ρ waterV

ρ waterVg
FB
g 
 g  g
 1
Solve for a: a 
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


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§9.7 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
V2 =
constant
v1v2
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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 cross-sectional
area A1 is the same as the mass that flows through crosssectional area A2.
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m
 Av
t
is the mass flow rate (units kg/s)
V
 Av
t
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 0.2 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
 0.2 cm 
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§9.8 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
Potential
energy
per unit
volume
Kinetic
energy
per unit
volume
Points 1 and 2
must be on the
same streamline
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Example (text problem 9.50): A nozzle is connected to a
horizontal hose. The nozzle shoots out water moving at 25.0
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 = 25m/s and v1 is unknown. Use the continuity equation.
  d2 2 
  
2
 A2 
 d2 
  2 
v1   v2  
v    v2
2  2
 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
2
2
1
2
 1000 kg/m 3 25.0 m/s 
2
 3.1 105 Pa




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§9.9 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.
(a) What must the pressure of the fluid in the syringe be
in order to inject the solution at a rate of 0.150 mL/sec?
Solve Poiseuille’s Law for the pressure difference:


3
8L V 8 2.00 10 2 Pa sec 3.00 cm 
cm
P  4

 0.15
4
s

1
r t
 0.3 10 cm


 2830 Pa
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Example continued:
This pressure difference is between the fluid in the
syringe and the fluid in the vein; it is the given gauge
pressure.
P  Ps  Pv
Ps  Pv  P
 2130 Pa  2830 Pa  4960 Pa
43
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 = 0.496 N.
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§9.10 Viscous Drag
The viscous drag force on a sphere is given by Stokes’ law.
FD  6rv
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Example (text problem 9.64): 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
FBD for
sphere
FB FD
x
w
Apply Newton’s Second Law
to the sphere
F  F
D
 FB  w  ma
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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 
ms g   lVs g

 2.4 Pa sec
6rvt
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§9.11 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.
48
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
F  PA  
 r
 2
6
r  9 10 N

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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.
F
1
6F
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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