physics beyond 2000 chapter 7 properties of matter
TRANSCRIPT
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 so
lid 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
e1
e2
F1
F2
Natural length
Extension
Compression
Force F
Deformation e
extension
compression0
F = k.e
http://www.phy.ntnu.edu.tw/~hwang/springForce/springForce.html
http://webphysics.davidson.edu/Applets/animator4/demo_hook.html
Hooke’s law for springs
The slope of the graph represents the stiffness of the spring.A spring with large slope is stiff.A spring with small slopeis soft.
Force F
Deformation e
extension
compression0
F = k.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.
2
0 2
1keFdxU
e
e
Energy stored in a spring
• It is also given by the area under the F-e graph.
2
2
1.
2
1keeFU e
extension
F
e0
Force
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 cross-sectional area A.
A
F
FA
Unit: Pa
Stress is a measure of the cause of a deformation.Note that A is the cross-sectional area of the wirebefore any stress is applied.
strain
• Strain is the extension e per unit length.
• If is the natural length of the wire,
then
e
e
F
Strain expresses the effect of the strain on the wire.
Example 3
• Find the stress and the strain of a wire.
A
F
e
stress 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.
Ae
FE
.
.
Unit: Pa
Young modulus E
Ae
FE
.
.
The value of E is dependent on the material.
Young modulus E and force constant k
Ae
FE
.
.
and F = k.e
AE
k.
So k depends on E (the material), A (thethickness) and (the length).
Example 4
• Find the Young modulus.
Experiment to find Young modulus
Suspend two long thin wire as shown.The reference wire can compensate for the temperature effect.The vernier scale is tomeasure the extensionof the sample wire.
referencewire
sample wire
weight
levelmeter
vernierscale
Experiment to find Young modulus
Adjust the weight so that vernier scale to read zero.Measure the diameter of the sample wire andcalculate its cross-section area A.
referencewire
sample wire
weight
vernierscale
The bubble in in themiddle.
Level meter
zero
Diameter ofthe wire
Experiment to find Young modulus
Measure the length of the sample wire.
referencewire
sample wire
weight
vernierscale
Experiment to find Young modulus
referencewire
sample wire
weight
vernierscale
Add weight W to the sample wire and measure its extension e .The force on the wire isF = W = mg.
F = W = mg where m is the added mass.
More weights
More weights
The bubble movesto the left
It is because the sample wire, whichis on the right, extends.
This endis higher.
This endis lower.
Turn thisscrew (vernierscale) toraise upthe end ofthe levelmetersuspended bythe samplewire.
This end ofthe level meteris suspendedby the sample wire.
The bubble in in themiddle again.
The readingon the screwshows theextensionof the samplewire.
Experiment to find Young modulus
referencewire
sample wire
weight
vernierscale
F
Plot the graph of stress σagainst strainε.
σσ
ε
elastic limit
0
Experiment to find Young modulus
referencewire
sample wire
weight
vernierscale
F
What is the slope of thisgraph?
σ
ε
elastic limit
0
Young modulus
Experiment to find Young modulus
σ
ε
elastic limit
0
The linear portion of the graph gives Hooke’s law.The stress applied to any solid is proportional tothe strain it produces for small strain.
The stress-strain curve
permanentstrain
stress σ
strain εO
AL
B
C
D
A: proportional limitL: elastic limitB: yield pointC: breaking stressD: breaking point
The stress-strain curve
permanentstrain
stress σ
strain εO
AL
B
C
D
A: proportional limit Between OA, the stressis proportional to the strain.Point A is the limit of thisproportionality.
The stress-strain curve
permanentstrain
stress σ
strain εO
AL
B
C
D
L: elastic limitBetween AL, the strain can be back to zero when the stress isremoved.i.e. the wireis still elastic.Usually the elastic limit coincides withthe proportional limit.
The stress-strain curve
permanentstrain
stress σ
strain εO
AL
B
C
D
B: yield pointBetween LB, the wirehas a permanent deformation whenthe stress is removed.i.e. the wire is plastic.At point B, there isa sudden increase ofstrain a small increasein stress.
The stress-strain curve
permanentstrain
stress σ
strain εO
AL
B
C
D
C: breaking stressThis is the maximumstress.Beyond this point,the wire extendsand narrows quickly,causing a constrictionof the cross-sectionalarea.
The stress-strain curve
permanentstrain
stress σ
strain εO
AL
B
C
D
D: breaking pointThe wire breaksat this point.This is the maximumstrain of the wire.
Example 5
• Refer to table 7.1 on p.112.
Energy stored in the extended wire
stress
strain
σ
ε
The area under the stress-strain graph = A
Fe21
2
1
where Fe is the elastic potential energy and A is the volume of the wire.
2
1
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 with a small stress.
ε
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
stress σ
strain ε
stiffest
weakest
strongest
most flexible
Graphical representation
stress σ
strain ε
brittle
tough
ductile
Graphical representation for various materials
stress σ
strain ε
glass
metal
rubber
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
0 strain εpermanentstrain
loading
unloading
elastic limit
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
strain ε
loading
unloading
Hysteresis loop
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.
layerlayer
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 equilibriumdistance.r = ro
The force on each moleculeis zero.
Intermolecular separation r
• It is the separation between the centres of two adjacent molecules.
r > ro
The force on the moleculeis attractive.
ro
r
Intermolecular separation r
• It is the separation between the centres of two adjacent molecules.
r < ro
The force on the moleculeis repulsive.
ro
r
Intermolecular forcesIntermolecularforce
0
repulsive
attractive
rro
ro is the equilibrium separation
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?
33.N
VrVrN
Intermolecular separation
• Example 6.
• Mass = density × volume
• The separation of molecules in solid and liquid is of order 10-10 m.
Intermolecular potential energy
Intermolecularforce
0rro
ro is the equilibrium separation
Potential energy
-ε
The potential energyis zero for large separation.
The potential energy is a minimumat the equilibrium separation.
Intermolecular potential energy
Intermolecularforce
0rro
ro is the equilibrium separation
Potential energy
-ε
When they move towardseach other from far away, the potential energy decreases because there is attractive force.The work done by external force is negative.
The potential energy is a minimumat the equilibrium separation.
Intermolecular potential energy
Intermolecularforce
0r
ro
ro is the equilibrium separation
Potential energy
-ε
When they are further towards each other after the equilibrium position, the potential energy increasesbecause there is repulsive force.The work done by external force is positive.
The potential energy is a minimumat the equilibrium separation.
Force and Potential Energy
• U = potential energy
• F = external force
r
FdrU anddx
dUF
Variation of molecules• If the displacement of two neighbouring mo
lecules is small, the portion of force-separation is a straight line with negative slope.
attractive
Intermolecularforce
0
repulsiver
ro
F
r
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.
F
r
roSo the molecule is insimple harmonic motion.
Variation of molecules
F
r
ro
So the molecule is insimple harmonic motion.with ω2 =
where m is the massof each molecule.
m
k
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.
Intermolecularforce
0rro
Potential energy
-ε
The three phases of matter
• Solid, liquid and gas states.
• In gas state, the average separation between molecules is much longer than ro.
Intermolecularforce
0rro
Potential energy
-ε
Elastic interaction of molecules
• All the interactions between molecules in any state are elastic. i.e. no energy loss on collision between molecules.
Solids
Intermolecularforce
0rro
Potential energy
-ε
When energy is supplied to a solid, the moleculesvibrate with greater amplitude until melting occurs.
Solids
Intermolecularforce
0rro
Potential energy
-ε
On melting, the energy is used to break the lattice structure.
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
Intermolecularforce
0rro
Potential energy
-ε
Molecules are moving at very high speed inrandom direction.
Gases
Intermolecularforce
0rro
Potential energy
-ε
The average separation between moleculesis much longer than ro
Gases
Intermolecularforce
0rro
Potential energy
-ε
The intermolecular force is so small that it isinsignificant.
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
Potential energy
0rro
-ε
In a solid, molecules are vibrating about their equilibrium position.
Thermal expansionSuppose a molecule is vibrating betweenpositions A and B about the equilibrium position.
Potential energy
0rro
-ε
A B
Thermal expansionNote 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
0rro
-ε
A B
A’ B’
C’
Thermal expansionThe potential energy of the molecule variesalong the curve A’C’B’ while the molecule is oscillating along AB.
Potential energy
0rro
-ε
A B
A’ B’
C’
Thermal expansionThe centre of oscillation M is mid-wayfrom the positions A and B. So point Mis slightly away from the equilibrium position.
Potential energy
0rro
-ε
A B
A’ B’
C’
M
Thermal expansionWhen 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
0rro
-ε
A B
A’ B’
C’
M
Thermal expansionThe molecule is vibrating with largeramplitude between new positions AB.
Potential energy
0rro
-ε
A B
A’ B’
C’
M
Thermal expansionThe centre of oscillation M , which is the mid-pointof AB, is further away from the equilibrium position.
Potential energy
0rro
-ε
A B
A’ B’
C’
M
Thermal expansionAs a result, the average separation between moleculesincreases by heating. The solid expands on heating.
Potential energy
0rro
-ε
A B
A’ B’
C’
M
Absolute zero temperatureAt absolute zero, the molecule does not vibrate. Theseparation between molecules is ro. The potential energy of the molecule is a minimum.
Potential energy
0rro
-ε C’
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.
ro
wire
Young Modulus in microscopic point of vies
• The cross-sectional area of the wire is
ro
2. orNA where N is the number ofmolecules in each layer
Aarea of one molecule= 2
or
ro
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.
ro+ r
F
Young Modulus in microscopic point of vies
• The strain is
ro+ r
F
or
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.
ro+ r
F
Young Modulus in microscopic point of vies
ro+ r
F
2oNrA and F = N.k.r
or
k
A
Fstress
Young Modulus in microscopic point of vies
ro+ r
F
Thus, the Young modulus is
or
kE
Example 9
• Find the force constant k between the molecules.
Density
• Definition: It is the mass of a substance per unit volume.
V
m
where m is the mass and V is the volumeUnit: 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.
A
FP or
A
FP
A
0
lim
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 Xsurface ofliquid
X
h
Find the pressure inside a liquid
• Consider a small horizontal area A around point X.
X
h
surface ofliquid
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
X
h
surface ofliquid
A
W
Find the pressure inside a liquid• W = ?
X
h
surface ofliquid
A
W
W = hAg
Find the pressure inside a liquid
X
h
surface ofliquid
A
W
W = hAg and P =
g hA
WP
A
W
Find the pressure inside a liquid
surface ofliquid
As there is also atmospheric pressure Po on the liquidsurface, the total pressure at X is
Po
X
h
A
P
ghPP o
Example 10
• The hydraulic pressure.
Force on a block in liquid
L
h1
h2
P1
P2
A
Consider a cylinder of area A and height L ina liquid of density .
Force on a block in liquid
L
h1
h2
P1
P2
A
The pressure on its top area is P1 = h1g + Po
The pressure on its bottom area is P2 = h2g + Po
Force on a block in liquid
L
h1
h2
P1
P2
A
The pressure difference P = P2 – P1 = Lg with upward direction.
Force on a block in liquidSo there is an upward net force F = P.A = Vg where V is the volume of the cylinder.
L
h1
h2
A
F
Force on a block in liquidThis is the upthrust on the cylinder.Upthrust = Vg
h1
h2
F
V
Force on a block in liquidUpthrust = Vg Note that it is also equal to the weight of theliquid with volume V.
h1
h2
F
V
Force on a block in liquidThe 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
h2
F
V
Force on a block in liquidThe conclusion is correct for a solid in liquid and gas (fluid).
h1
h2
F
V
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 upthrustspring-balance
object
liquid
compressionbalance
The reading of thespring-balance is W,which is the weightof the object.
The reading of the compressionbalance is B, which is the weight of liquidand beaker.
beaker
W
B
Measuring upthrustspring-balance
object
liquid
Carefully immersehalf the volume ofthe object in liquid.
What would happen tothe reading of thespring-balance and thatof the compression balance?
beaker
compressionbalance
Measuring upthrustspring-balance
object
liquid
The reading of the spring-balance decreases.Why?
The difference in the readingsof the spring-balance givesthe upthrust on the object.
beaker
compressionbalance
Measuring upthrustspring-balance
object
liquid
The reading of the compression balance increases.Why?
The difference in the readingsof the compression balance gives the upthrust on the object.
beaker
compressionbalance
Measuring upthrustspring-balance
object
liquid
Carefully immersethe whole object in liquid.
What would happen tothe reading of thespring-balance and thatof the compression balance?
beaker
compressionbalance
Measuring upthrustspring-balance
object
liquid
Carefully placethe object on thebottom of the beaker.
What would happen tothe reading of thespring-balance and thatof the compression balance?
beaker
compressionbalance
Law of floatation
• A floating object displaces its own weight of the fluid in which it floats.
weight upthrust
weight of the object= upthrust= weight of fluid displaced
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.
liquid ofdensity
connectto the fluid X Y Same level
Manometer
liquid ofdensity
Po = atmospheric pressure
Po+P= fluid pressure
h = difference in height
X Y
Manometer
liquid ofdensity
Po = atmospheric pressure
Po+P= fluid pressure h = difference
in heightBA
The pressures at pointsA and B are equal.
Manometer
liquid ofdensity
Po = atmospheric pressure
Po+P= fluid pressure h = difference
in heightBA
The pressure at A = Po+PThe pressure at B = Po + hg
Manometer
liquid ofdensity
Po = atmospheric pressure
Po+P= fluid pressure h = difference
in heightBA
The pressure difference of the fluid P = hg
Liquid in a pipe• Consider a pipe of non-uniform cross-sectional
area with movable piston at each end.• The fluid is in static equilibrium.
Same level
X Y
hx = hY
static fluid
Liquid in a pipe
• The manometers show that the pressures at points X and Y are equal.
Same level
X Y
hx = hY
static fluid
Liquid in a pipe
• The pressures at points M and N on the pistons are also equal.
Same level
M N
hx = hY
static fluid
Liquid in a pipe
• There must be equal external pressures on the pistons to keep it in equilibrium.
PM = PN
Same level
M N
hx = hY
static fluid
PM PN
Liquid in a pipe
• As F = P.A , the external forces are different on the two ends.
FM > FN
Same level
M N
hx = hY
static fluid
FM 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
M N
hx = hY
static fluid
FM 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 position 1.
• There is a pressure difference P at the two ends.
h2
P+P
p
x1
x2
Position 1
Position 2
A1
A2
h1
Pressure difference and work done
• Work done by the external forces is
(P+P).A1.x1 - P.A2.x2
P+P
P
x1
x2
Position 1
Position 2
A1
A2
h2
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
P+P
P
x1
x2
Position 1
Position 2
A1
A2
h2
h1
Pressure difference and work done
• Work done by the external forces is
(P+P).A1.x1 - P.A2.x2 = P .V• With V =
Work done = Pwhere m is the mass of the fluid
and ρis the density of the fluid
P+P
p
x1
x2
Position 1
Position 2
A1
A2
m
m h2
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.tP1
P2
x1
x2
Position 1
Position 2
A1
A2
v1
v2
h2
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 =
(P1-P2)P1
P2
x1
x2
Position 1
Position 2
A1
A2
v1
v2
m
h2
h1
Bernoulli’s principle
• Work done by external pressure =
(P1-P2)
• Increase in kinetic energy =
P1
P2
x1
x2
Position 1
Position 2
A1
A2
v1
v2
m
21
22 2
1
2
1mvmv
h2
h1
Bernoulli’s principle
• Increase in kinetic energy =
• Increase in gravitatioanl potential energy =
mgh2 – mgh1P1
P2
x1
x2
Position 1
Position 2
A1
A2
v1
v2
h2
h1
21
22 2
1
2
1mvmv
Bernoulli’s principle
P1
P2
x1
x2
Position 1
Position 2
A1
A2
v1
v2
h2
h1
)()2
1
2
1().( 12
21
2221 mghmghmvmv
mPP
The left hand side is the work done by external pressure. It is also the energy suppliedto the fluid.The right hand side isthe increase in energy of the fluid.
Bernoulli’s principle
P1
P2
x1
x2
Position 1
Position 2
A1
A2
v1
v2
h2
h1
2222
2111 .
2
1...
2
1.. vhgPvhgP
Bernoulli’s principle
P1
P2
x1
x2
Position 1
Position 2
A1
A2
v1
v2
h2
h1
2
2
1vghP constant
Bernoulli’s principle
2222
2111 2
1
2
1vghPvghP
If h1 = h2 , then
222
211 2
1
2
1vPvP
Position 1Position 2
Bernoulli’s principle
If h1 = h2 , then
222
211 2
1
2
1vPvP
Position 1Position 2
or
2
2
1vP constant
Bernoulli’s principle
If h1 = h2 , then
Position 1Position 2
2
2
1vP constant
So for horizontal flow, where the speed is high,the pressure is low.
Low speed, high pressureHigh speed, low pressure
Bernoulli’s principle
Position 1Position 2
So for horizontal flow, where the speed is high,the pressure is low.
Low speed, high pressure High speed, low pressure
h1
h2
Bernoulli’s principle
Position 1Position 2
So for horizontal flow, where the speed is high,the pressure is low.
P1 = h1..gh1
h2P2 = h2..g
Low speed, high pressure High speed, low pressure
Simple demonstration of Bernoulli’s principle
•Held two paper strips vertically with a smallgap between them.•Blow air gently into the gap.•Explain what you observe.
air
Simple demonstration of Bernoulli’s principle
air
high pressure
high pressure
In the gap, the speedof airflow is high.So the pressure is lowin the gap.The high pressureoutside pressesthe strips together.
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
air
air
weight of the ball
thrust from theair blower
force due to spinning
What is the direction of spinning of the ball?
Ball floating in air
air
air
weight of the ball
thrust from theair blower
force due to pressuredifference
It is spinning in clockwise direction.
Ball floating in air
Force due topressuredifference
Air blown out through a funnel
What would happen to the light ball?
Air blown out through a funnel
It is sucked to the top of the funnel.
weight
Force due topressure difference
Force due topressure difference
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.
air
air
fluid
low pressure
Jets: Bunsen burner
• The pressure near the jet is low.
• Air outside is pulled into the bunsen burner through the air hole. gas
air
Jets: Paint sprayer
Jets: Filter pump
Jets: Carburretor
Roofs, window and door
• The pressure difference makes the door close.
fast wind,low pressure
door
Example 14
• Strong wind on top of the roof.
tile
Example 14
• Strong wind on top of the roof.
tile
fast wind on top of the tiles (outside the house)
no wind under the tiles (inside the house)
Example 14
tile
fast wind on top of the tiles (outside the house)
no wind under the tiles (inside the house)
high pressure
low pressure
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.
v
h
Po
Po
A hole in a water tank
• The water pressure on the surface is Po.
• The water pressure at the hole is Po. v
h
Po
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. v
h
Po
Po
A hole in a water tank
• Apply Bernoulli’s equation,
v
h
Po
Po
02
10 2
0 vPghPo
ghv 2
This is the same speed of an object falling througha distance h freely.
Example 15
• A hole in a tank
Pitot tube
• Pitot tube is used to measure the speed of fluid.
h
v
statictube
total tube
h1h2
Pitot tube
• The pressure below the static tube is h1g.
• The pressure at the mouth of the total tube is h2g.
h
v
statictube
total tube
h1h2
Pitot tube
• The fluid speed below the static tube is v.• The fluid speed at the mouth of the total tube is 0.
h
v
statictube
total tube
h1h2
Pitot tube
• Apply Bernoulli’s equation,
h
v
statictube
total tube
h1h2
02
12
21 ghvgh
Pitot tube
h
v
statictube
total tube
h1h2
ghhhg
v 2)(2 12