Physics 121,Lecture 20,Pg 1
Physics 121,Lecture 20
Today’s Agenda
? Announcements
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,8,# 7,22,28,33,35,44,45,50,54,61,and 65.
? Today’s topics
?Fluid and solids
?Pressure measurement
?Archimedes’ principle
?Fluids in motion
Physics 121,Lecture 20,Pg 2
Fluids, Chapter 9
? At ordinary temperature,matter exists in one of three states
?Solid - has a shape and forms a surface
?Liquid - has no shape but forms a surface
?Gas - has no shape and forms no surface
? What do we mean by,fluids”?
?Fluids are,substances that flow”….,substances that
take the shape of the container”
?Atoms and molecules are free to move.
?No long range correlation between positions.
Physics 121,Lecture 20,Pg 3
Fluids
? What parameters do we use to describe fluids?
?Density
?Pressure
??
? ? mV
units,
kg/m3 = 10-3 g/cm3
??
p ? FA
units,
1 N/m2 = 1 Pa (Pascal)
1 bar = 105 Pa
1 mbar = 102 Pa
1 torr = 133.3 Pa
nF ?pA? A
n
? Any force exerted by a fluid is perpendicular to a surface of
contact,and is proportional to the area of that surface.
?Force (a vector) in a fluid can be expressed in terms of
pressure (a scalar) as:
Physics 121,Lecture 20,Pg 4
Fluids
? Bulk Modulus
)V/V(
pB
??
??
?LIQUID,incompressible (density almost constant)
?GAS,compressible (density depends a lot on pressure)
??? ??? ??? ??? ??? ???? ???????
Bulk modulus (Pa=N/m2)
H2O SteelGas (STP) Pb
Physics 121,Lecture 20,Pg 5
? When the pressure is much less
than the bulk modulus of the fluid,
we treat the density as constant
independent of pressure:
incompressible fluid
? For an incompressible fluid,the
density is the same everywhere,
but the pressure is NOT!
Pressure vs,Depth
Incompressible Fluids (liquids)
? Consider an imaginary fluid volume (a cube,face area A)
?The sum of all the forces on this volume must be ZERO as
it is in equilibrium,F2 - F1 - mg = 0
y
1
y
2
A
p
1
p
2
F
1
F
2
mg
0
p
ApApFF 1212 ???
Ag)yy(mg 12 ??? )yy(gpp 1212 ????
Physics 121,Lecture 20,Pg 6
? If the pressures were different,fluid would flow in the tube!
? However,if fluid did flow,then the system was NOT in equilibrium
since no equilibrium system will spontaneously leave equilibrium.
Pressure vs,Depth
? For a fluid in an open container
pressure same at a given depth
independent of the container p ( y )
y
? Fluid level is the same
everywhere in a connected
container,assuming no surface
forces
? Why is this so? Why does the
pressure below the surface
depend only on depth if it is in
equilibrium?
? Imagine a tube that would connect two regions at the same depth.
Physics 121,Lecture 20,Pg 7
Lecture 20,ACT 1
Pressure
? What happens with two fluids Consider a U
tube containing liquids of density ?1 and ?2 as
shown:
?Compare the densities of the liquids:
A) ?1 < ?2 B) ?1 = ?2 C) ?1 > ?2
?1
?2
dI
Physics 121,Lecture 20,Pg 8
Pascal’s Principle
? So far we have discovered (using Newton’s Laws):
?Pressure depends on depth,?p = ?g?y
? Pascal’s Principle addresses how a change in pressure is
transmitted through a fluid.
Any change in the pressure applied to an enclosed fluid is
transmitted to every portion of the fluid and to the walls
of the containing vessel.
? Pascal’s Principle explains the working of hydraulic lifts
?i.e,the application of a small force at one place can result
in the creation of a large force in another.
?Does this,hydraulic lever” violate conservation of energy?
?Certainly hope not.,Let’s calculate.
Physics 121,Lecture 20,Pg 9
Pascal’s Principle
? Consider the system shown:
?A downward force F1 is applied
to the piston of area A1.
?This force is transmitted through
the liquid to create an upward
force F2.
?Pascal’s Principle says that
increased pressure from F1
(F1/A1) is transmitted throughout
the liquid.
F F
1
2
d
2
d
1
A A 2
1
2
2
1
1
A
F
A
F ?
1
212
A
AFF ?
? F2 > F1, Have we violated conservation of energy
Physics 121,Lecture 20,Pg 10
Pascal’s Principle
? Consider F1 moving through a
distance d1.
?How large is the volume of the
liquid displaced?
F F
1
2
d
2
d
1
A A
21
111 dV A??
12 VV ??? 2112 A
Add ?
? Therefore the work done by F1 equals the work done by F2
We have NOT obtained,something for nothing”.
1211121222 WA
Ad
A
AFdFW ???
?This volume determines the
displacement of the large piston.
Physics 121,Lecture 20,Pg 11
A1 A
10
A2 A
10
A1 A20
M
M
MdC
dB
dA
A) dA = (1/2)dB B) dA = dB C) dA = 2dB
? If A10 = 2′A20,compare dA and dC.
A) dA = (1/2)dC B) dA = dC C) dA = 2dC
Lecture 20,ACT 2
Hydraulics
? Consider the systems shown to the right.
?In each case,a block of mass M is
placed on the piston of the large
cylinder,resulting in a difference dI in
the liquid levels.
? If A2 = 2′A1,compare dA and dB.
Physics 121,Lecture 20,Pg 12
? At what depth is the water pressure two atmospheres?
It is one atmosphere at the surface,What is the
pressure at the bottom of the deepest oceanic trench
(about 104 meters)?
P2 = P1 + ?gd
2.02?105 Pa = 1.01?105 Pa
+ 103 kg/m3*9.8m/s2*d
d = 10.3 m
P2 = 1.01?105 Pa + 103 kg/m3*9.8m/s2*104 m
= 9.81?107 Pa = 971 Atm
Solution:
d is the depth.
The pressure increases one
atmosphere for every 10m.
This assumes that water is
incompressible.
For d = 104 m:
Example Problems (1)
Physics 121,Lecture 20,Pg 13
? Two masses rest on two pistons in a U-shaped tube of
water,as shown,If M1 = 3 kg and M2 = 4.5 kg,what is
the height difference,h? The area of piston 1 is A1 =
200 cm2,and the area of piston 2 is A2 = 100 cm2.
Example Problems (2)
Solution:
M1
M2h
The pressure difference due to the two
masses must be balanced by the water
pressure due to h.
Note that g does not appear in the
answer.
M1g/A1 + ?gh = M2g/A2
h = (M2/A2-M1/A1)/?
= 0.3 m
Physics 121,Lecture 20,Pg 14
Using Fluids to Measure Pressure
v a c u u m
p = 0
h
a tm o s p h e r e
p = p
0
Barometer gph 0??
? Use Barometer to measure Absolute Pressure
?Top of tube evacuated (p=0)
?Bottom of tube submerged into pool of mercury open
to atmosphere (p=p0)
?Pressure dependence on depth:
Physics 121,Lecture 20,Pg 15
Using Fluids to Measure Pressure
1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H20
p0
?h
Manometer
p1
g
)pp(h 01
?
???
? Use Manometer to measure Gauge Pressure
?Measure pressure of volume (p1) relative to
atmospheric pressure (??gauge pressure)
?The height difference (?h) measures the gauge
pressure
Physics 121,Lecture 20,Pg 16
Archimedes’ Principle
? Suppose we weigh an object in air (1) and in water (2).
?How do these weights compare?
W2?W1
W1 < W2 W1 = W2 W1 > W2
?Why?
? Since the pressure at the bottom of
the object is greater than that at the
top of the object,the water exerts a
net upward force,the buoyant force,
on the object.
Physics 121,Lecture 20,Pg 17
? The buoyant force is equal to the difference in the
pressures times the area.
W2?W1
)Ay-g ( y)( 12????? AppF 12B
l i q u i dl i q u i dl i q u i dl i q u i dB WgMgVF ???? ?
Archimedes:
The buoyant force is equal to the
weight of the liquid displaced.
? The buoyant force determines
whether an object will sink or float.
How does this work?
y
1
y
2
A
p
1
p
2
F
1
F
2
Archimedes’ Principle
Physics 121,Lecture 20,Pg 18
Sink or Float?
? The buoyant force is equal to the weight
of the liquid that is displaced.
? If the buoyant force is larger than the
weight of the object,it will float; otherwise
it will sink.
F mgB
y
? We can calculate how much of a floating object will be
submerged in the liquid:
?Object is in equilibrium mgFB ?
o b j e c to b j e c tl i q u i dl i q u i d VgVg ????? ??
l i q u i d
o b je c t
o b je c t
l i q u i d
V
V
?
??
Physics 121,Lecture 20,Pg 19
The Tip of the Iceberg
? What fraction of an iceberg is submerged?
F mgB
y
l i q u i d
o b je c t
o b je c t
l i q u i d
V
V
?
??
%90k g / m 1 0 2 4 k g / m 9 1 7VV 3
3
w a t e r
i c e
i c e
w a t e r ???
?
?
Physics 121,Lecture 20,Pg 20
Lecture 20,ACT 3
Buoyancy
? A lead weight is fastened to a large
styrofoam block and the combination
floats on water with the water level with
the top of the styrofoam block as shown.
?If you turn the styrofoam+Pb upside
down,what happens?
styrofoam
Pb
A) It sinks C)B) sty ro fo amPb
sty ro fo am
Pb D) styrofoamPb
Physics 121,Lecture 20,Pg 21
ACT 3-A
More Fun With Buoyancy
? Two cups are filled to the same level with
water,One of the two cups has plastic
balls floating in it,
?Which cup weighs more?
?Archimedes principle tells us that the cups weigh the
same.
?Each plastic ball displaces an amount of water that is
exactly equal to its own weight.
Cup I Cup II
(A) Cup I (B) Cup II (C) the same (D) can’t tell
Physics 121,Lecture 20,Pg 22
ACT 3-B
Even More Fun With Buoyancy
? A plastic ball floats in a cup of water with
half of its volume submerged,Next some
oil (?oil < ?ball < ?water) is slowly added to
the container until it just covers the ball,
?Relative to the water level,the ball will,water
(A) move up (B) move down (C) stay in same place
?For oil to cover the ball,the ball must have,displaced”
some oil.
?Therefore,the buoyant force on the ball increases.
?Therefore,the ball moves up (relative to the water).
?Note that initially we assumed the buoyant force of the air
on the ball was negligible (it is!); the buoyant force of the
oil is not.
Physics 121,Lecture 20,Pg 23
Fluids in Motion
? Up to now we have described fluids in terms of
their static properties:
?density ?
?pressure p
? To describe fluid motion,we need something that
can describe flow:
?velocity v
? There are different kinds of fluid flow of varying complexity
? non-steady / steady
? compressible / incompressible
? rotational / irrotational
? viscous / ideal
Physics 121,Lecture 20,Pg 24
? Simplest situation,consider
ideal fluid moving with steady
flow - velocity at each point in
the flow is constant in time
? In this case,fluid moves on
streamlines
A
1
A
2
v
1
v
2
streamline
Ideal Fluids
? Fluid dynamics is very complicated in general (turbulence,
vortices,etc.)
? Consider the simplest case first,the Ideal Fluid
?no,viscosity” - no flow resistance (no internal friction)
?incompressible - density constant in space and time
Physics 121,Lecture 20,Pg 25
? Flow obeys continuity equation
? volume flow rate Q = A·v is constant along flow
tube.
? follows from mass conservation if flow is
incompressible.
A
1
A
2
v
1
v
2
streamline
A1v1 = A2v2
Ideal Fluids
? streamlines do not meet or cross
? velocity vector is tangent to
streamline
? volume of fluid follows a tube of
flow bounded by streamlines
Physics 121,Lecture 20,Pg 26
Steady Flow of Ideal Fluids
(actually laminar flow of real fluid)
Physics 121,Lecture 20,Pg 27
1) Assuming the water moving in the pipe is an ideal
fluid,relative to its speed in the 1” diameter pipe,
how fast is the water going in the 1/2” pipe?
Lecture 20 Act 4
Continuity
? A housing contractor saves
some money by reducing the
size of a pipe from 1”
diameter to 1/2” diameter at
some point in your house,
v1 v1/2
a) 2 v1 b) 4 v1 c) 1/2 v1 c) 1/4 v1
Physics 121,Lecture 20,Pg 28
? Recall the standard work-energy relation
?Apply the principle to a section of flowing fluid with
volume dV and mass dm = ??dV (here W is work done
on fluid)
V)pp( xApxApW 21 222111pr e s s ur e d dd?? ??
)yy(gV )yy(gmW 12 12gr a v i t y ??? ??? d?d
)vv(VvmvmKW 21222121212221 ?????? d?dd
222212121211 gyvpgyvp ???? ?????? Bernoulli Equation
y
1
y
2
v
1
v
2
p
1
p
2
dV
p r e ssu r eg r a vi t y WWW ??
KW ??
Conservation of Energy for
Ideal Fluid
Physics 121,Lecture 20,Pg 29
Lecture 20 Act 5
Bernoulli’s Principle
? A housing contractor saves
some money by reducing the
size of a pipe from 1”
diameter to 1/2” diameter at
some point in your house,
2) What is the pressure in the 1/2” pipe relative to the
1” pipe?
a) smaller b) same c) larger
v1 v1/2
Physics 121,Lecture 20,Pg 30
Some applications
? Lift for airplane wing
? Enhance sport performance
? More complex phenomena,ex,turbulence
Physics 121,Lecture 20,Pg 31
More applications
? Vortices,ex,Hurricanes
? And much more …
Physics 121,Lecture 20,Pg 32
Recap for today:
? Todays’ topics
?Fluid and solids
?Pressure measurement
?Archimedes’ principle
?Fluids in motion
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,8,# 7,22,28,33,35,44,45,50,54,61,and 65.
Physics 121,Lecture 20
Today’s Agenda
? Announcements
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,8,# 7,22,28,33,35,44,45,50,54,61,and 65.
? Today’s topics
?Fluid and solids
?Pressure measurement
?Archimedes’ principle
?Fluids in motion
Physics 121,Lecture 20,Pg 2
Fluids, Chapter 9
? At ordinary temperature,matter exists in one of three states
?Solid - has a shape and forms a surface
?Liquid - has no shape but forms a surface
?Gas - has no shape and forms no surface
? What do we mean by,fluids”?
?Fluids are,substances that flow”….,substances that
take the shape of the container”
?Atoms and molecules are free to move.
?No long range correlation between positions.
Physics 121,Lecture 20,Pg 3
Fluids
? What parameters do we use to describe fluids?
?Density
?Pressure
??
? ? mV
units,
kg/m3 = 10-3 g/cm3
??
p ? FA
units,
1 N/m2 = 1 Pa (Pascal)
1 bar = 105 Pa
1 mbar = 102 Pa
1 torr = 133.3 Pa
nF ?pA? A
n
? Any force exerted by a fluid is perpendicular to a surface of
contact,and is proportional to the area of that surface.
?Force (a vector) in a fluid can be expressed in terms of
pressure (a scalar) as:
Physics 121,Lecture 20,Pg 4
Fluids
? Bulk Modulus
)V/V(
pB
??
??
?LIQUID,incompressible (density almost constant)
?GAS,compressible (density depends a lot on pressure)
??? ??? ??? ??? ??? ???? ???????
Bulk modulus (Pa=N/m2)
H2O SteelGas (STP) Pb
Physics 121,Lecture 20,Pg 5
? When the pressure is much less
than the bulk modulus of the fluid,
we treat the density as constant
independent of pressure:
incompressible fluid
? For an incompressible fluid,the
density is the same everywhere,
but the pressure is NOT!
Pressure vs,Depth
Incompressible Fluids (liquids)
? Consider an imaginary fluid volume (a cube,face area A)
?The sum of all the forces on this volume must be ZERO as
it is in equilibrium,F2 - F1 - mg = 0
y
1
y
2
A
p
1
p
2
F
1
F
2
mg
0
p
ApApFF 1212 ???
Ag)yy(mg 12 ??? )yy(gpp 1212 ????
Physics 121,Lecture 20,Pg 6
? If the pressures were different,fluid would flow in the tube!
? However,if fluid did flow,then the system was NOT in equilibrium
since no equilibrium system will spontaneously leave equilibrium.
Pressure vs,Depth
? For a fluid in an open container
pressure same at a given depth
independent of the container p ( y )
y
? Fluid level is the same
everywhere in a connected
container,assuming no surface
forces
? Why is this so? Why does the
pressure below the surface
depend only on depth if it is in
equilibrium?
? Imagine a tube that would connect two regions at the same depth.
Physics 121,Lecture 20,Pg 7
Lecture 20,ACT 1
Pressure
? What happens with two fluids Consider a U
tube containing liquids of density ?1 and ?2 as
shown:
?Compare the densities of the liquids:
A) ?1 < ?2 B) ?1 = ?2 C) ?1 > ?2
?1
?2
dI
Physics 121,Lecture 20,Pg 8
Pascal’s Principle
? So far we have discovered (using Newton’s Laws):
?Pressure depends on depth,?p = ?g?y
? Pascal’s Principle addresses how a change in pressure is
transmitted through a fluid.
Any change in the pressure applied to an enclosed fluid is
transmitted to every portion of the fluid and to the walls
of the containing vessel.
? Pascal’s Principle explains the working of hydraulic lifts
?i.e,the application of a small force at one place can result
in the creation of a large force in another.
?Does this,hydraulic lever” violate conservation of energy?
?Certainly hope not.,Let’s calculate.
Physics 121,Lecture 20,Pg 9
Pascal’s Principle
? Consider the system shown:
?A downward force F1 is applied
to the piston of area A1.
?This force is transmitted through
the liquid to create an upward
force F2.
?Pascal’s Principle says that
increased pressure from F1
(F1/A1) is transmitted throughout
the liquid.
F F
1
2
d
2
d
1
A A 2
1
2
2
1
1
A
F
A
F ?
1
212
A
AFF ?
? F2 > F1, Have we violated conservation of energy
Physics 121,Lecture 20,Pg 10
Pascal’s Principle
? Consider F1 moving through a
distance d1.
?How large is the volume of the
liquid displaced?
F F
1
2
d
2
d
1
A A
21
111 dV A??
12 VV ??? 2112 A
Add ?
? Therefore the work done by F1 equals the work done by F2
We have NOT obtained,something for nothing”.
1211121222 WA
Ad
A
AFdFW ???
?This volume determines the
displacement of the large piston.
Physics 121,Lecture 20,Pg 11
A1 A
10
A2 A
10
A1 A20
M
M
MdC
dB
dA
A) dA = (1/2)dB B) dA = dB C) dA = 2dB
? If A10 = 2′A20,compare dA and dC.
A) dA = (1/2)dC B) dA = dC C) dA = 2dC
Lecture 20,ACT 2
Hydraulics
? Consider the systems shown to the right.
?In each case,a block of mass M is
placed on the piston of the large
cylinder,resulting in a difference dI in
the liquid levels.
? If A2 = 2′A1,compare dA and dB.
Physics 121,Lecture 20,Pg 12
? At what depth is the water pressure two atmospheres?
It is one atmosphere at the surface,What is the
pressure at the bottom of the deepest oceanic trench
(about 104 meters)?
P2 = P1 + ?gd
2.02?105 Pa = 1.01?105 Pa
+ 103 kg/m3*9.8m/s2*d
d = 10.3 m
P2 = 1.01?105 Pa + 103 kg/m3*9.8m/s2*104 m
= 9.81?107 Pa = 971 Atm
Solution:
d is the depth.
The pressure increases one
atmosphere for every 10m.
This assumes that water is
incompressible.
For d = 104 m:
Example Problems (1)
Physics 121,Lecture 20,Pg 13
? Two masses rest on two pistons in a U-shaped tube of
water,as shown,If M1 = 3 kg and M2 = 4.5 kg,what is
the height difference,h? The area of piston 1 is A1 =
200 cm2,and the area of piston 2 is A2 = 100 cm2.
Example Problems (2)
Solution:
M1
M2h
The pressure difference due to the two
masses must be balanced by the water
pressure due to h.
Note that g does not appear in the
answer.
M1g/A1 + ?gh = M2g/A2
h = (M2/A2-M1/A1)/?
= 0.3 m
Physics 121,Lecture 20,Pg 14
Using Fluids to Measure Pressure
v a c u u m
p = 0
h
a tm o s p h e r e
p = p
0
Barometer gph 0??
? Use Barometer to measure Absolute Pressure
?Top of tube evacuated (p=0)
?Bottom of tube submerged into pool of mercury open
to atmosphere (p=p0)
?Pressure dependence on depth:
Physics 121,Lecture 20,Pg 15
Using Fluids to Measure Pressure
1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H20
p0
?h
Manometer
p1
g
)pp(h 01
?
???
? Use Manometer to measure Gauge Pressure
?Measure pressure of volume (p1) relative to
atmospheric pressure (??gauge pressure)
?The height difference (?h) measures the gauge
pressure
Physics 121,Lecture 20,Pg 16
Archimedes’ Principle
? Suppose we weigh an object in air (1) and in water (2).
?How do these weights compare?
W2?W1
W1 < W2 W1 = W2 W1 > W2
?Why?
? Since the pressure at the bottom of
the object is greater than that at the
top of the object,the water exerts a
net upward force,the buoyant force,
on the object.
Physics 121,Lecture 20,Pg 17
? The buoyant force is equal to the difference in the
pressures times the area.
W2?W1
)Ay-g ( y)( 12????? AppF 12B
l i q u i dl i q u i dl i q u i dl i q u i dB WgMgVF ???? ?
Archimedes:
The buoyant force is equal to the
weight of the liquid displaced.
? The buoyant force determines
whether an object will sink or float.
How does this work?
y
1
y
2
A
p
1
p
2
F
1
F
2
Archimedes’ Principle
Physics 121,Lecture 20,Pg 18
Sink or Float?
? The buoyant force is equal to the weight
of the liquid that is displaced.
? If the buoyant force is larger than the
weight of the object,it will float; otherwise
it will sink.
F mgB
y
? We can calculate how much of a floating object will be
submerged in the liquid:
?Object is in equilibrium mgFB ?
o b j e c to b j e c tl i q u i dl i q u i d VgVg ????? ??
l i q u i d
o b je c t
o b je c t
l i q u i d
V
V
?
??
Physics 121,Lecture 20,Pg 19
The Tip of the Iceberg
? What fraction of an iceberg is submerged?
F mgB
y
l i q u i d
o b je c t
o b je c t
l i q u i d
V
V
?
??
%90k g / m 1 0 2 4 k g / m 9 1 7VV 3
3
w a t e r
i c e
i c e
w a t e r ???
?
?
Physics 121,Lecture 20,Pg 20
Lecture 20,ACT 3
Buoyancy
? A lead weight is fastened to a large
styrofoam block and the combination
floats on water with the water level with
the top of the styrofoam block as shown.
?If you turn the styrofoam+Pb upside
down,what happens?
styrofoam
Pb
A) It sinks C)B) sty ro fo amPb
sty ro fo am
Pb D) styrofoamPb
Physics 121,Lecture 20,Pg 21
ACT 3-A
More Fun With Buoyancy
? Two cups are filled to the same level with
water,One of the two cups has plastic
balls floating in it,
?Which cup weighs more?
?Archimedes principle tells us that the cups weigh the
same.
?Each plastic ball displaces an amount of water that is
exactly equal to its own weight.
Cup I Cup II
(A) Cup I (B) Cup II (C) the same (D) can’t tell
Physics 121,Lecture 20,Pg 22
ACT 3-B
Even More Fun With Buoyancy
? A plastic ball floats in a cup of water with
half of its volume submerged,Next some
oil (?oil < ?ball < ?water) is slowly added to
the container until it just covers the ball,
?Relative to the water level,the ball will,water
(A) move up (B) move down (C) stay in same place
?For oil to cover the ball,the ball must have,displaced”
some oil.
?Therefore,the buoyant force on the ball increases.
?Therefore,the ball moves up (relative to the water).
?Note that initially we assumed the buoyant force of the air
on the ball was negligible (it is!); the buoyant force of the
oil is not.
Physics 121,Lecture 20,Pg 23
Fluids in Motion
? Up to now we have described fluids in terms of
their static properties:
?density ?
?pressure p
? To describe fluid motion,we need something that
can describe flow:
?velocity v
? There are different kinds of fluid flow of varying complexity
? non-steady / steady
? compressible / incompressible
? rotational / irrotational
? viscous / ideal
Physics 121,Lecture 20,Pg 24
? Simplest situation,consider
ideal fluid moving with steady
flow - velocity at each point in
the flow is constant in time
? In this case,fluid moves on
streamlines
A
1
A
2
v
1
v
2
streamline
Ideal Fluids
? Fluid dynamics is very complicated in general (turbulence,
vortices,etc.)
? Consider the simplest case first,the Ideal Fluid
?no,viscosity” - no flow resistance (no internal friction)
?incompressible - density constant in space and time
Physics 121,Lecture 20,Pg 25
? Flow obeys continuity equation
? volume flow rate Q = A·v is constant along flow
tube.
? follows from mass conservation if flow is
incompressible.
A
1
A
2
v
1
v
2
streamline
A1v1 = A2v2
Ideal Fluids
? streamlines do not meet or cross
? velocity vector is tangent to
streamline
? volume of fluid follows a tube of
flow bounded by streamlines
Physics 121,Lecture 20,Pg 26
Steady Flow of Ideal Fluids
(actually laminar flow of real fluid)
Physics 121,Lecture 20,Pg 27
1) Assuming the water moving in the pipe is an ideal
fluid,relative to its speed in the 1” diameter pipe,
how fast is the water going in the 1/2” pipe?
Lecture 20 Act 4
Continuity
? A housing contractor saves
some money by reducing the
size of a pipe from 1”
diameter to 1/2” diameter at
some point in your house,
v1 v1/2
a) 2 v1 b) 4 v1 c) 1/2 v1 c) 1/4 v1
Physics 121,Lecture 20,Pg 28
? Recall the standard work-energy relation
?Apply the principle to a section of flowing fluid with
volume dV and mass dm = ??dV (here W is work done
on fluid)
V)pp( xApxApW 21 222111pr e s s ur e d dd?? ??
)yy(gV )yy(gmW 12 12gr a v i t y ??? ??? d?d
)vv(VvmvmKW 21222121212221 ?????? d?dd
222212121211 gyvpgyvp ???? ?????? Bernoulli Equation
y
1
y
2
v
1
v
2
p
1
p
2
dV
p r e ssu r eg r a vi t y WWW ??
KW ??
Conservation of Energy for
Ideal Fluid
Physics 121,Lecture 20,Pg 29
Lecture 20 Act 5
Bernoulli’s Principle
? A housing contractor saves
some money by reducing the
size of a pipe from 1”
diameter to 1/2” diameter at
some point in your house,
2) What is the pressure in the 1/2” pipe relative to the
1” pipe?
a) smaller b) same c) larger
v1 v1/2
Physics 121,Lecture 20,Pg 30
Some applications
? Lift for airplane wing
? Enhance sport performance
? More complex phenomena,ex,turbulence
Physics 121,Lecture 20,Pg 31
More applications
? Vortices,ex,Hurricanes
? And much more …
Physics 121,Lecture 20,Pg 32
Recap for today:
? Todays’ topics
?Fluid and solids
?Pressure measurement
?Archimedes’ principle
?Fluids in motion
? Homework 8,due Friday Nov,11 @ 6:00 PM.
?Chap,8,# 7,22,28,33,35,44,45,50,54,61,and 65.
