Transcript E - OoCities

Physics Beyond 2000
Chapter 7
Properties of Matter
States of matter
• Solid state
• Liquid state
• Gas state (will be studied in Chapter 8.)
Points of view
• Macroscopic: Discuss the relation among
physics quantities.
• Microscopic: All matters consist of particles.
The motions of these particles are studied.
Statistics are used to study the properties.
Solids
• Extension and compression (deformation) of
solid objects
– elasticity .
• Hooke’s law for springs
– The deformation e of a spring is proportional
to the force F acting on it, provided the
deformation is small.
F = k.e where k is the force constant of
the spring
Hooke’s law for springs
Force F
Natural length
F = k.e
extension
Extension
e1
F1
0
compression
e2
Deformation e
F2
Compression
http://www.phy.ntnu.edu.tw/~hwang/springForce/springForce.html
http://webphysics.davidson.edu/Applets/animator4/demo_hook.html
Hooke’s law for springs
Force F
The slope of the graph
represents the stiffness
of the spring.
A spring with large slope
is stiff.
A spring with small slope
is soft.
F = k.e
extension
0
compression
Deformation e
Energy stored in a spring
• It is elastic potential energy.
• It is equal to the work done W by the
external force F to extend (or compress) the
spring by a deformation e.
e
1 2
U e   Fdx  ke
2
0
Energy stored in a spring
• It is also given by the area under the F-e
graph.
Force
F
1
1 2
U e  F .e  ke
2
2
extension
0
e
Example 1
• Find the extension of a spring from energy
changes.
• Find the extension of the spring by using
Hooke’s law.
Example 2
• A spring-loaded rifle
Example 2
• A spring-loaded rifle
More than springs
• All solid objects follow Hooke’s law
provided the deformation is not too large.
• The extension depends on
–
–
–
–
the nature of the material
the stretching force
the cross-sectional area of the sample
the original length
stress 
• Stress  is the force F on unit crosssectional area A.
F

A
A
F
Unit: Pa
Stress is a measure of the cause of a deformation.
Note that A is the cross-sectional area of the wire
before any stress is applied.
strain 
• Strain  is the extension e per unit length.
• If  is the natural length of the wire,
e
then  

F

e
Strain expresses the effect of the strain on the wire.
Example 3
• Find the stress and the strain of a wire.
F

A
stress
e


strain
Young modulus E
• Young modulus E is the ratio of the tensile
stress σ applied to a body to the tensile
strain ε produced.
 F .
E 
 e. A
Unit: Pa
Young modulus E
The value of E is dependent on the material.
 F .
E 
 e. A
Young modulus E and
force constant k
 F .
E 
 e. A

and
F = k.e
E. A
k

So k depends on E (the material), A (the
thickness) and  (the length).
Example 4
• Find the Young modulus.
Experiment to find
Young modulus
reference
wire
sample
wire
level
meter
vernier
scale
weight
Suspend two long thin
wire as shown.
The reference wire can
compensate for the
temperature effect.
The vernier scale is to
measure the extension
of the sample wire.
Experiment to find
Young modulus
reference
wire
sample
wire
vernier
scale
weight
Adjust the weight so
that vernier scale to
read zero.
Measure the diameter
of the sample wire and
calculate its crosssection area A.
The bubble in in the
middle.
Level meter
zero
Diameter of
the wire
Experiment to find
Young modulus
reference
wire
sample
wire
vernier
scale
weight
Measure the length
of the sample wire.


Experiment to find
Young modulus
reference
wire
sample
wire
vernier
scale
Add weight W to the
sample wire and measure
its extension e .
The force on the wire is
F = W = mg.
F = W = mg
weight
where m is the added mass.
More weights
More weights
The bubble moves
to the left
This end
is higher.
This end
is lower.
It is because the sample wire, which
is on the right, extends.
Turn this
screw
(vernier
scale) to
raise up
the end of
the level
meter
suspended by
the sample
wire.
This end of
the level meter
is suspended
by the sample
wire.
The bubble in in the
middle again.
The reading
on the screw
shows the
extension
of the sample
wire.
Experiment to find
Young modulus
reference
wire
sample
wire
Plot the graph of stress σ
against strainε.
σ
σ
elastic
limit
vernier
scale
F
weight
0
ε
Experiment to find
Young modulus
What is the slope of this
graph? Young modulus
reference
wire
sample
wire
σ
elastic
limit
vernier
scale
F
weight
0
ε
Experiment to find
Young modulus
The linear portion of the graph gives Hooke’s law.
The stress applied to any solid is proportional to
the strain it produces for small strain.
σ
elastic
limit
0
ε
The stress-strain curve
stress σ
C
B
L
D
A
O
permanent
strain
strain ε
A: proportional limit
L: elastic limit
B: yield point
C: breaking stress
D: breaking point
The stress-strain curve
stress σ
A: proportional limit
Between OA, the stress
C
B
L
D
A
O
permanent
strain
strain ε
is proportional to the strain.
Point A is the limit of this
proportionality.
The stress-strain curve
stress σ
C
B
L
D
A
O
permanent
strain
strain ε
L: elastic limit
Between AL, the
strain can be back to
zero when the stress is
removed.i.e. the wire
is still elastic.
Usually the elastic
limit coincides with
the proportional limit.
The stress-strain curve
stress σ
C
B
L
D
A
O
permanent
strain
strain ε
B: yield point
Between LB, the wire
has a permanent
deformation when
the stress is removed.
i.e. the wire is plastic.
At point B, there is
a sudden increase of
strain a small increase
in stress.
The stress-strain curve
stress σ
C
B
L
D
A
O
permanent
strain
strain ε
C: breaking stress
This is the maximum
stress.
Beyond this point,
the wire extends
and narrows quickly,
causing a constriction
of the cross-sectional
area.
The stress-strain curve
stress σ
C
B
L
D
A
O
permanent
strain
strain ε
D: breaking point
The wire breaks
at this point.
This is the maximum
strain of the wire.
Example 5
• Refer to table 7.1 on p.112.
Energy stored in
the extended wire
The area under the stress-strain graph =
stress
1
Fe
1
2
 
2
A
1
where
2 Fe is the elastic potential
energy and
A  is the volume of the
wire.
σ
ε
strain
Properties of materials
•
•
•
•
Stiffness
Strength
Ductility
Toughness
Stiffness
• It indicates how the material opposes to
deformation.
• Young modulus is a measure of the stiffness
of a material.
• A material is stiff if its Young modulus is
large.
• A material is soft if its Young modulus is
small.
Strength
• It indicates how large the stress the material
can stand before breaking.
• The breaking stress is a measure of the
strength of the material.
• A material is strong if it needs a large stress
to break it.
• A material is weak if a small stress can
break it.
Ductility
• It indicates how the material can become a
wire or a thin sheet.
• A ductile material enters
its plastic stage
ε
with a small stress.
Toughness
• A tough material is one which does not
crack readily.
• The opposite is a brittle material.
• A brittle material breaks over a very short
time without plastic deformation.
Graphical representation
strongest
stress σ
stiffest
most flexible
weakest
strain ε
Graphical representation
stress σ
tough
brittle
ductile
strain ε
Graphical representation for
various materials
stress σ
metal
glass
rubber
strain ε
Elastic deformation
and plastic deformation
• In elastic deformation,
the object will be back
to its original shape
when the stress is
removed.
• In plastic deformation,
there is a permanent
strain when the stress
is removed.
Plastic deformation
stress σ
elastic
limit
loading
unloading
0
permanent
strain
strain ε
Fatigue
• Metal fatigue is a cumulative effect causing a
metal to fracture after repeated applications of
stress, none of which exceeds the breaking stress.
Creep
• Creep is a gradual elongation of a metal
under a constant stress which is well below
its yield point.
Plastic deformation of glass
• Glass does not have any plastic deformation.
• When the applied stress is too large, the
glass has brittle fracture.
Plastic deformation of rubber
• Deformation of rubber would produce
internal energy.
• The area in the loop represents the internal
energy produced per unit volume.
stress σ
Hysteresis loop
loading
unloading
strain ε
Model of a solid
• Microscopic point of view
• A solid is made up of a large number of identical
hard spheres (molecules).
• The molecules are attracted to each other by a
large force.
• The molecules are packed closely in an orderly
way.
• There are also repulsion to stop the molecules
penetrating into each other.
Structure of solid
• Crystalline solid: The molecules have
regular arrangement. e.g. metal.
• Amorphous solid: The molecules are
packed disorderly together. e.g. glass.
Elastic and plastic deformation
of metal
• Metal has a structure of layers.
• Layers can slide over each other under an
external force.
layer
layer
Elastic and plastic deformation
of metal
• When the force is small, the layer displaces
slightly.
Force
Elastic and plastic deformation
of metal
• When the force is removed, the layer moves
back to its initial position.
• The metal is elastic.
Elastic and plastic deformation
of metal
• When the force is large, the layer moves a
large displacement.
Force
Elastic and plastic deformation
of metal
• When the force is removed, the layer settles
down at a new position.
• The metal has a plastic deformation.
New structure
Initial structure
Intermolecular forces
• The forces are basically electrostatic in
nature.
• The attractive force results from the
electrons of one molecule and the protons of
an adjacent molecule.
• The attractive force increases as their
separation decreases.
Intermolecular forces
• The forces are basically electrostatic in
nature.
• When the molecules are too close, their
outer electrons repel each other. This
repulsive force prevents the molecules from
penetrating each other.
Intermolecular forces
• The forces are basically electrostatic in
nature.
• Normally the molecules in a solid have a
balance of the attractive and repulsive
forces.
• At the equilibrium position, the net
intermolecular force on the molecule is zero.
Intermolecular separation r
• It is the separation between the centres of
two adjacent molecules.
ro
ro is the equilibrium
distance.
r = ro
The force on each molecule
is zero.
Intermolecular separation r
• It is the separation between the centres of
two adjacent molecules.
r
r > ro
The force on the molecule
is attractive.
ro
Intermolecular separation r
• It is the separation between the centres of
two adjacent molecules.
r
ro
r < ro
The force on the molecule
is repulsive.
Intermolecular forces
Intermolecular
force
ro is the equilibrium separation
repulsive
0
attractive
ro
r
The dark line is the
resultant curve.
Intermolecular separation
• Suppose that a solid consists of N molecules
with average separation r.
• The volume of the solid is V.
• What is the relation among these quantities?
V
N .r  V  r  3
N
3
Intermolecular separation
• Example 6.
• Mass = density × volume
• The separation of molecules in solid and
liquid is of order 10-10 m.
Intermolecular potential energy
ro is the equilibrium separation
Intermolecular
force
0
r
ro
The potential energy
is zero for large separation.
Potential energy
-ε
The potential energy is a minimum
at the equilibrium separation.
Intermolecular potential energy
ro is the equilibrium separation
Intermolecular
force
0
r
ro
Potential energy
-ε
The potential energy is a minimum
at the equilibrium separation.
When they move towards
each other from far away, the
potential energy decreases
because there is attractive force.
The work done by external
force is negative.
Intermolecular potential energy
ro is the equilibrium separation
0
When they are further towards
each other after the equilibrium
Intermolecular position, the potential energy increases
because there is repulsive force.
force
The work done by external force is
ro
positive.
r
Potential energy
-ε
The potential energy is a minimum
at the equilibrium separation.
Force and Potential Energy
• U = potential energy
• F = external force

U    Fdr
r
and
dU
F 
dx
Variation of molecules
• If the displacement of two neighbouring
molecules is small, the portion of forceseparation is a straight line with negative
slope.
F
Intermolecular
force
repulsive
r
r
0
attractive
ro
ro
Variation of molecules
• The intermolecular force is
F = -k. Δr
where k is the force constant
between molecules
and Δr is the displacement
from the equilibrium position.
So the molecule is in
simple harmonic motion.
F
r
ro
Variation of molecules
So the molecule is in
simple harmonic motion.
with k
ω2 =
m
where m is the mass
of each molecule.
F
r
ro
Variation of molecules
• However this is only a
highly simplified model.
• Each molecule is under
more than one force
from neighbouring
molecules.
The three phases of matter
• Solid, liquid and gas states.
• In solid and liquid states, the average
separation between molecules is close to ro.
0
Intermolecular
force
ro
Potential energy
-ε
r
The three phases of matter
• Solid, liquid and gas states.
• In gas state, the average separation between
molecules is much longer than ro.
0
Intermolecular
force
ro
Potential energy
-ε
r
Elastic interaction of molecules
• All the interactions between molecules in
any state are elastic. i.e. no energy loss on
collision between molecules.
Solids
When energy is supplied to a solid, the molecules
vibrate with greater amplitude until melting occurs.
0
Intermolecular
force
ro
Potential energy
-ε
r
Solids
On melting, the energy is used to break
the lattice structure.
0
Intermolecular
force
ro
Potential energy
-ε
r
Liquids
• Molecules of liquid move underneath the
surface of liquid.
• When energy is supplied to a liquid, the
molecules gain kinetic energy and move
faster. The temperature increases.
Liquids
• At the temperature of vaporization (boiling
point), energy supplied is used to do work
against the intermolecular attraction.
• The molecules gain potential energy. The
state changes.
• The temperature does not change.
Gases
Molecules are moving at very high speed in
random direction.
0
Intermolecular
force
ro
Potential energy
-ε
r
Gases
The average separation between molecules
is much longer than ro
0
Intermolecular
force
ro
Potential energy
-ε
r
Gases
The intermolecular force is so small that it is
insignificant.
0
Intermolecular
force
ro
Potential energy
-ε
r
Example 7
• There are 6.02  1023 molecules for one
mole of substance.
• The is the Avogadro’s number.
Example 8
• The separation between molecules depend
on the volume.
Thermal expansion
In a solid, molecules are vibrating about their
equilibrium position.
Potential energy
0
-ε
ro
r
Thermal expansion
Suppose a molecule is vibrating between
positions A and B about the equilibrium
position.
Potential energy
A
0
-ε
B
ro
r
Thermal expansion
Note that the maximum displacement from the equilibrium
position is not the same on each side because the energy curve
is not symmetrical about the equilibrium position.
Potential energy
B
A
0
-ε
r
ro
A’
B’
C’
Thermal expansion
The potential energy of the molecule varies
along the curve A’C’B’ while the molecule is
oscillating along AB.
Potential energy
A
B
0
A’
-ε
r
ro
B’
C’
Thermal expansion
The centre of oscillation M is mid-way
from the positions A and B. So point M
is slightly away from the equilibrium position.
Potential energy
M
A
r
ro
0
A’
-ε
B
B’
C’
Thermal expansion
When a solid is heated up, it gains more potential energy
and the points A’ and B’ move up the energy curve. The
amplitude of oscillation is also larger.
Potential energy
M
A
r
ro
0
A’
-ε
B
B’
C’
Thermal expansion
The molecule is vibrating with larger
amplitude between new positions AB.
Potential energy
M
A
r
ro
0
A’
-ε
B
B’
C’
Thermal expansion
The centre of oscillation M , which is the mid-point
of AB, is further away from the equilibrium position.
Potential energy
M
A
r
ro
0
A’
-ε
B
B’
C’
Thermal expansion
As a result, the average separation between molecules
increases by heating. The solid expands on heating.
Potential energy
M
A
r
ro
0
A’
-ε
B
B’
C’
Absolute zero temperature
At absolute zero, the molecule does not vibrate. The
separation between molecules is ro. The potential
energy of the molecule is a minimum.
Potential energy
0
-ε
ro
C’
r
Young Modulus in
microscopic point of view
• Consider a wire made up of layers of closely
packed molecules.
• When there is not any stress, the separation
between two neighbouring layer is ro.
• ro is also the diameter of each molecule.
wire
ro
Young Modulus in
microscopic point of vies
• The cross-sectional area of the wire is
A  N .r
2
o
area of one molecule
o
r
= 2
ro
ro
where N is the number of
molecules in each layer
A
ro
Young Modulus in
microscopic point of vies
• When there is not an external force F, the
separation between two neighbouring layer
increases by r.
F
ro+ r
Young Modulus in
microscopic point of vies
• The strain is
r

ro
F
ro+ r
Young Modulus in
microscopic point of vies
• Since the restoring force between two molecules
in the neighbouring layer is directly proportional
to N and r, we have F = N.k.r where k is the
force constant between two molecules.
F
ro+ r
Young Modulus in
microscopic point of vies
A  Nr
2
o
and F = N.k.r
F
F k
 stress    
A ro
ro+ r
Young Modulus in
microscopic point of vies
Thus, the Young modulus is
F
 k
E 
 ro
ro+ r
Example 9
• Find the force constant k between the
molecules.
Density
• Definition: It is the mass of a substance per
unit volume.
m

V
where m is the mass
and V is the volume
Unit: kg m-3
Measure the density of liquid
• Use hydrometer
upthrust
weight
Pressure
• Definition: The pressure on a point is the
force per unit area on a very small area
around the point.
F
P
A
or
F
P  lim
A 0 A
Unit: N m-2 or Pa.
Pressure in liquid
• Pressure at a point inside a liquid acts
equally in all directions.
• The pressure increases with depth.
Find the pressure inside a liquid
•  = density of the liquid
• h = depth of the point X
surface of
liquid
h
X
Find the pressure inside a liquid
• Consider a small horizontal area A around
point X.
surface of
liquid
h
X
A
Find the pressure inside a liquid
• The force from the liquid on this area is the weight
W of the liquid cylinder above this area
surface of
liquid
W
X
h
A
Find the pressure inside a liquid
• W=?
W = hAg
surface of
liquid
W
X
h
A
Find the pressure inside a liquid
W
W = hAg and P =
A
surface of
liquid
W
X
h
A
A
P
 hg
W
Find the pressure inside a liquid
As there is also atmospheric pressure Po on the liquid
surface, the total pressure at X is
Po
P  Po  hg
surface of
liquid
h
P
X
A
Example 10
• The hydraulic pressure.
Force on a block in liquid
Consider a cylinder of area A and height L in
a liquid of density .
P1
A
h1
h2
L
P2
Force on a block in liquid
The pressure on its top area is P1 = h1g + Po
The pressure on its bottom area is P2 = h2g + Po
P1
A
h1
h2
L
P2
Force on a block in liquid
The pressure difference P = P2 – P1 = Lg
with upward direction.
P1
A
h1
h2
L
P2
Force on a block in liquid
So there is an upward net force F = P.A
= Vg where V is the volume of the
cylinder.
A
h1
h2
L
F
Force on a block in liquid
This is the upthrust on the cylinder.
Upthrust = Vg
h1
V
F
h2
Force on a block in liquid
Upthrust = Vg
Note that it is also equal to the weight of the
liquid with volume V.
h1
V
F
h2
Force on a block in liquid
The conclusion: If a solid is immersed in a
liquid, the upthrust on the solid is equal to
the weight of liquid that the solid displaces.
h1
V
F
h2
Force on a block in liquid
The conclusion is correct for a solid in
liquid and gas (fluid).
h1
V
F
h2
Archimedes’ Principle
• When an object is wholly or partially
immersed in a fluid, the upthrust on the
object is equal to the weight of the fluid
displaced.
upthrust
upthrust
Measuring upthrust
W
beaker
B
spring-balance The reading of the
spring-balance is W,
object
which is the weight
of the object.
liquid
compression
balance
The reading of the
compression
balance is B, which
is the weight of liquid
and beaker.
Measuring upthrust
beaker
spring-balance Carefully immerse
half the volume of
object
the object in liquid.
liquid
compression
balance
What would happen to
the reading of the
spring-balance and that
of the compression
balance?
Measuring upthrust
beaker
spring-balance The reading of the springbalance decreases.
object
Why?
liquid
The difference in the readings
of the spring-balance gives
the upthrust on the object.
compression
balance
Measuring upthrust
beaker
spring-balance The reading of the
compression balance
object
increases.
Why?
liquid
compression
balance
The difference in the readings
of the compression balance
gives the upthrust on the
object.
Measuring upthrust
spring-balance
object
beaker
liquid
compression
balance
Carefully immerse
the whole object in liquid.
What would happen to
the reading of the
spring-balance and that
of the compression balance?
Measuring upthrust
spring-balance
object
beaker
liquid
compression
balance
Carefully place
the object on the
bottom of the beaker.
What would happen to
the reading of the
spring-balance and that
of the compression
balance?
Law of floatation
• A floating object displaces its own weight
of the fluid in which it floats.
weight of the object
= upthrust
= weight of fluid displaced
weight
upthrust
float or sink?
•
•
•
•
‘ = density of the object
 = density of the fluid
If ‘ > , then the object sinks in the fluid.
If ‘ < , then the object floats in the fluid.
density is larger than 
density is smaller than 
Manometer
• A manometer can measure the pressure difference
of fluid.
• Note that the pressure on the same level in the
liquid must be the same.
connect
to the fluid
X
Y
Same level
liquid of
density 
Manometer
Po = atmospheric
pressure
Po+P
= fluid pressure
X
Y
h = difference
in height
liquid of
density 
Manometer
The pressures at points
A and B are equal.
Po = atmospheric
pressure
Po+P
= fluid pressure
A
B
h = difference
in height
liquid of
density 
Manometer
The pressure at A = Po+P
The pressure at B = Po + hg
Po = atmospheric
pressure
Po+P
= fluid pressure
A
B
h = difference
in height
liquid of
density 
Manometer
The pressure difference of the fluid P = hg
Po = atmospheric
pressure
Po+P
= fluid pressure
A
B
h = difference
in height
liquid of
density 
Liquid in a pipe
• Consider a pipe of non-uniform crosssectional area with movable piston at each
end.
• The fluid is in static equilibrium.
Same level
hx = h Y
Y
X
static fluid
Liquid in a pipe
• The manometers show that the
pressures at points X and Y are equal.
Same level
hx = h Y
Y
X
static fluid
Liquid in a pipe
• The pressures at points M and N on the
pistons are also equal.
Same level
hx = h Y
M
N
static fluid
Liquid in a pipe
• There must be equal external pressures on
the pistons to keep it in equilibrium.
PM = PN
Same level
hx = h Y
PM
M
N
static fluid
PN
Liquid in a pipe
• As F = P.A , the external forces are different
on the two ends.
FM > FN
Same level
hx = h Y
FM
M
N
static fluid
FN
Liquid in a pipe
• Note that the net force on the liquid is still zero to
keep it in equilibrium.
• There are forces towards the left from the inclined
surface.
Same level
hx = h Y
FM
M
N
static fluid
FN
Fluid Dynamics
• Fluid includes liquid and gas which can
flow.
• In this section, we are going to study the
force and motion of a fluid.
• Beurnoulli’s equation is the conclusion of
this section.
Turbulent flow
• Turbulent flow: the fluid flows in irregular
paths.
• We will not study this kind of flow.
Streamlined flow
• Streamlined flow (laminar flow) : the fluid
moves in layers without fluctuation or
turbulence so that successive particles
passing the same point with the same
velocity.
Streamlined flow
• We draw streamlines to represent the
motion of the fluid particles.
Equation of continuity
• Suppose that the fluid is incompressible.
That is its volume does not change. Though
the shape (cross-sectional area A) may
change.
Equation of continuity
• At the left end, after time t, the volume
passing is A1.v1. t
• At the right end, after the same time t, the
volume passing is A2.v2. t
Equation of continuity
• As the volumes are equal for an
incompressible fluid,
A1.v1. t = A2.v2. t

A1.v1 = A2.v2
Equation of continuity
• Example 12
Pressure difference
and work done
• Suppose that an
incompressible fluid
flows from position 1 to
position 2 in a tube.
• Position 2 is higher than
x1
position 1.
A1
• There is a pressure
difference P at the two P+P
ends.
Position 1
h1
x2
A2
p
Position 2
h2
Pressure difference
and work done
• Work done by the
external forces is
(P+P).A1.x1 - P.A2.x2
x2
A2
P
Position 2
x1
h2
A1
P+P
Position 1
h1
Pressure difference
and work done
• Work done by the
external forces is
(P+P).A1.x1 - P.A2.x2
• A1x1=A2x2=V
= volume of fluid that
moves
x2
A2
P
Position 2
x1
h2
A1
P+P
Position 1
h1
Pressure difference
and work done
• Work done by the
external forces is
(P+P).A1.x1 - P.A2.x2
= P .V
m
• With V =

m
x2
A2
p
Position 2
x1
A1
Work done = P

where m is the mass of the P+P
fluid and ρis the density
of the fluid
Position 1
h1
h2
Bernoulli’s principle
• In time t, the fluid
moves x1 at position
1 and x2 at position 2.
• x1 = v1.t and
x2 = v2.t
x2
A2
v2
P2
Position 2
x1
h2
A1
P1
v1
Position 1
h1
Bernoulli’s principle
• In time t, the fluid
moves x1 at position
1 and x2 at position 2.
• x1 = v1.t and
x2 = v2.t
• Work done by
external pressure =
m
(P1-P2)

x2
A2
v2
P2
Position 2
x1
h2
A1
P1
v1
Position 1
h1
Bernoulli’s principle
A2
• Work done by
external pressure =
m
(P1-P2)

• Increase in kinetic
energy =
1 2 1 2
mv2  mv1
2
2
x2
v2
P2
Position 2
x1
h2
A1
P1
v1
Position 1
h1
Bernoulli’s principle
A2
• Increase in kinetic
energy =
v2
P2
1 2 1 2
mv2  mv1
2
2
• Increase in
gravitatioanl
potential energy =
mgh2 – mgh1
x2
Position 2
x1
h2
A1
P1
v1
Position 1
h1
Bernoulli’s principle
m
1 2 1 2
( P1  P2 ).  ( mv2  mv1 )  (mgh2  mgh1 )

2
2
The left hand side is the work
done by external pressure. It
is also the energy supplied
x1
to the fluid.
A1
The right hand side is
P1
the increase in energy
v1
of the fluid.
Position 1
h1
x2
A2
v2
P2
Position 2
h2
Bernoulli’s principle
x2
A2
1 2
1 2
P1   .g.h1   .v1  P2   .g.h2   .v2
2
2
v2
P2
Position 2
x1
h2
A1
P1
v1
Position 1
h1
Bernoulli’s principle
x2
A2
1 2
P  gh  v  constant
2
v2
P2
Position 2
x1
h2
A1
P1
v1
Position 1
h1
Bernoulli’s principle
1 2
1 2
P1  gh1  v1  P2  gh2  v2
2
2
If h1 = h2 , then
1 2
1 2
P1  v1  P2  v2
2
2
Position 1
Position 2
Bernoulli’s principle
If h1 = h2 , then
or
1 2
1 2
P1  v1  P2  v2
2
2
1 2
P  v  constant
2
Position 1
Position 2
Bernoulli’s principle
If h1 = h2 , then
1 2
P  v  constant
2
So for horizontal flow, where the speed is high,
the pressure is low.
Low speed, high pressure
Position 1
High speed, low pressure
Position 2
Bernoulli’s principle
So for horizontal flow, where the speed is high,
the pressure is low.
Low speed, high pressure
High speed, low pressure
h1
h2
Position 1
Position 2
Bernoulli’s principle
So for horizontal flow, where the speed is high,
the pressure is low.
Low speed, high pressure
P1 = h1..g
Position 1
High speed, low pressure
h1
P2 = h2..g
h2
Position 2
Simple demonstration of
Bernoulli’s principle
•Held two paper strips
vertically with a small
gap between them.
•Blow air gently into
the gap.
•Explain what you
observe.
air
Simple demonstration of
Bernoulli’s principle
In the gap, the speed
of airflow is high.
So the pressure is low
in the gap.
The high pressure
outside presses
high pressure
the strips together.
high pressure
air
Examples of Bernoulli’s effect
• Airfoil: the airplane is flying to the left.
Examples of Bernoulli’s effect
• Airfoil: There is a pressure difference
between the top and the bottom of the wing.
A net lifting force is produced.
Example 13
• Airfoil and Bernoulli’s effect
• To find the lifting force on an airplane.
Examples of Bernoulli’s effect
• Spinning ball: moving to the left and
rotating clockwise.
Examples of Bernoulli’s effect
• Spinning ball: the pressure difference
produces a deflection force and the ball
moves along a curve.
Examples of Bernoulli’s effect
• Spinning ball: the pressure difference
produces a deflection force and the ball
moves along a curve.
spinning ball
not spinning
Examples of Bernoulli’s effect
• Spinning ball: the pressure difference
produces a deflection force and the ball
moves along a curve.
spinning ball
not spinning
Ball floating in air
air
air
Ball floating in air
force due to spinning
thrust from the
air blower
air
air
weight of the ball
What is the direction of spinning of the ball?
Ball floating in air
force due to pressure
difference
thrust from the
air blower
air
air
weight of the ball
It is spinning in clockwise direction.
Ball floating in air
Force due to
pressure
difference
Air blown out through a funnel
What would happen to the light ball?
Air blown out through a funnel
Force due to
pressure difference
Force due to
pressure difference
weight
It is sucked to the top of the funnel.
Yacht sailing
• A yacht can sail
against the wind.
• Note that the sail is
curved.
Yacht sailing
• A yacht can sail
against the wind.
• The pressure
difference produces
a net force F.
• A component of F
pushes the yacht
forward.
Yacht sailing
• The yacht must
follow a zig-zag
path in order to sail
against the wind.
wind
path
Jets
• When a stream of fluid is ejected rapidly
out of a jet, air close to the stream would be
dragged along and moves at higher speed.
• This results in a low pressure near the
stream.
low pressure
air
fluid
air
Jets: Bunsen burner
• The pressure
near the jet is
low.
• Air outside is
pulled into the
bunsen burner
through the
air hole.
air
gas
Jets: Paint sprayer
Jets: Filter pump
Jets: Carburretor
Roofs, window and door
• The pressure
difference
makes the door
close.
door
fast wind,
low pressure
Example 14
• Strong wind on top of the roof.
tile
Example 14
• Strong wind on top of the roof.
fast wind on top of the tiles (outside the house)
no wind under the tiles
(inside the house)
tile
Example 14
low pressure
fast wind on top of the tiles (outside the house)
no wind under the tiles
(inside the house)
high pressure
tile
A hole in a water tank
• The speed of
water on the
surface is almost
zero.
• The speed of
water at the hole
is v.
Po
h
v
Po
A hole in a water tank
• The water
pressure on the
surface is Po.
• The water
pressure at the
hole is Po.
Po
h
v
Po
A hole in a water tank
• The height of
water on the
surface is h
• The height of
water at the hole
is 0.
Po
h
v
Po
A hole in a water tank
• Apply Bernoulli’s
equation,
1 2
Po  0  gh  P0  v  0
2
 v  2 gh
Po
h
v
Po
This is the same speed of an object falling through
a distance h freely.
Example 15
• A hole in a tank
Pitot tube
• Pitot tube is used to measure the speed of
fluid.
static
tube
h
h1
v
total
tube
h
2
Pitot tube
• The pressure below the static tube is h1g.
• The pressure at the mouth of the total tube is h2g.
static
tube
h
h1
v
total
tube
h
2
Pitot tube
• The fluid speed below the static tube is v.
• The fluid speed at the mouth of the total tube is 0.
static
tube
h
h1
v
total
tube
h
2
Pitot tube
• Apply Bernoulli’s equation,
1 2
h1 g  v  h2 g  0
2
static
tube
h
h1
v
total
tube
h
2
Pitot tube
v
2 g (h2  h1 )

static
tube
total
tube
h
h1
v
 2 gh
h
2