# Chapter 7 Properties of Matter

## Presentation on theme: "Chapter 7 Properties of Matter"— Presentation transcript:

Chapter 7 Properties of Matter
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 F = k.e where k is the force constant of the spring
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 Deformation e extension compression F = k.e e1 e2 F1 F2 Natural length Extension Compression

Hooke’s law for springs
Force F Deformation e extension compression F = k.e 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.

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.

Energy stored in a spring
It is also given by the area under the F-e graph. extension F e Force

Example 1 Find the extension of a spring from energy changes.
Find the extension of the spring by using Hooke’s law.

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.
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, then e F Strain expresses the effect of the strain on the wire.

Example 3 Find the stress and the strain of a wire. strain stress

Young modulus E Young modulus E is the ratio of the tensile stress σ applied to a body to the tensile strain ε produced. Unit: Pa

Young modulus E The value of E is dependent on the material.

Young modulus E and force constant k
F = k.e So k depends on E (the material), A (the thickness) 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 to measure the extension of the sample wire. sample wire reference wire level meter vernier scale weight

Experiment to find Young modulus
Adjust the weight so that vernier scale to read zero. Measure the diameter of the sample wire and calculate its cross-section area A. sample wire reference wire vernier scale weight

The bubble in in the middle. Level meter

zero

Diameter of the wire

Experiment to find Young modulus
reference wire sample weight vernier scale Measure the length of the sample wire.

Experiment to find Young modulus
Add weight W to the sample wire and measure its extension e . The force on the wire is F = W = mg. sample wire reference wire vernier scale F = W = mg where m is the added mass. weight

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.

This end of the level meter is suspended by the sample wire. Turn this screw (vernier scale) to raise up the end of the level meter 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
Plot the graph of stress σ against strainε. sample wire σ ε elastic limit reference wire vernier scale F weight

Experiment to find Young modulus
What is the slope of this graph? Young modulus sample wire σ ε elastic limit reference wire vernier scale F weight

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 ε elastic limit

The stress-strain curve
O A L B C D A: proportional limit L: elastic limit B: yield point C: breaking stress D: breaking point permanent strain

The stress-strain curve
O A L B C D A: proportional limit Between OA, the stress is proportional to the strain. Point A is the limit of this proportionality. permanent strain

The stress-strain curve
O A L B C D 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. permanent strain

The stress-strain curve
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. stress σ strain ε O A L B C D permanent strain

The stress-strain curve
O A L B C D 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. permanent strain

The stress-strain curve
O A L B C D D: breaking point The wire breaks at this point. This is the maximum strain of the wire. permanent strain

Example 5 Refer to table 7.1 on p.112.

Energy stored in the extended wire
The area under the stress-strain graph = stress where 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 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.

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. 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 is the equilibrium distance. r = ro The force on each molecule is zero. ro

Intermolecular separation r
It is the separation between the centres of two adjacent molecules. ro r r > ro The force on the molecule is attractive.

Intermolecular separation r
It is the separation between the centres of two adjacent molecules. ro r r < ro The force on the molecule is repulsive.

Intermolecular forces
ro is the equilibrium separation repulsive r ro attractive 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?

Intermolecular separation
Example 6. Mass = density × volume The separation of molecules in solid and liquid is of order m.

Intermolecular potential energy
ro is the equilibrium separation Intermolecular force r ro The potential energy is zero for large separation. Potential energy The potential energy is a minimum at the equilibrium separation.

Intermolecular potential energy
ro is the equilibrium separation Intermolecular force r ro 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. Potential energy The potential energy is a minimum at the equilibrium separation.

Intermolecular potential energy
ro is the equilibrium separation When they are further towards each other after the equilibrium position, the potential energy increases because there is repulsive force. The work done by external force is positive. Intermolecular force ro r Potential energy The potential energy is a minimum at the equilibrium separation.

Force and Potential Energy
U = potential energy F = external force and

Variation of molecules
If the displacement of two neighbouring molecules is small, the portion of force-separation is a straight line with negative slope. F attractive Intermolecular force repulsive r ro 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 ro So the molecule is in simple harmonic motion.

Variation of molecules
So the molecule is in simple harmonic motion. with ω2 = 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. Intermolecular force r ro Potential energy

The three phases of matter
Solid, liquid and gas states. In gas state, the average separation between molecules is much longer than ro. Intermolecular force r ro Potential energy

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. Intermolecular force r ro Potential energy

Solids On melting, the energy is used to break the lattice structure.
Intermolecular force r ro Potential energy

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. -ε
Intermolecular force r ro Potential energy

Gases The average separation between molecules is much longer than ro
Intermolecular force r ro Potential energy

Gases The intermolecular force is so small that it is insignificant.
r ro Potential energy

Example 7 There are 6.02  1023 molecules for one mole of substance.

Example 8 The separation between molecules depend on the volume.

Thermal expansion In a solid, molecules are vibrating about their
equilibrium position. Potential energy r ro

Thermal expansion Suppose a molecule is vibrating between
positions A and B about the equilibrium position. Potential energy r ro A B

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 r ro A B 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 r ro A B A’ 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 r ro A B A’ B’ C’ M

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 B r ro A’ B’ C’

Thermal expansion The molecule is vibrating with larger
amplitude between new positions AB. Potential energy M A B r ro A’ 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 B r ro A’ B’ C’

Thermal expansion As a result, the average separation between molecules increases by heating. The solid expands on heating. Potential energy M A B r ro A’ 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 r ro 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 where N is the number of molecules in each layer area of one molecule = ro ro A

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 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
and F = N.k.r F ro+ r

Young Modulus in microscopic point of vies
Thus, the Young modulus is F ro+ r

Example 9 Find the force constant k between the molecules.

Density Definition: It is the mass of a substance per unit volume.
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. or 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 X h

Find the pressure inside a liquid
Consider a small horizontal area A around point X. X h surface of liquid 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 of liquid A W

Find the pressure inside a liquid
W = ? W = hAg X h surface of liquid A W

Find the pressure inside a liquid
W = hAg and P = X h surface of liquid A 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 X h A P surface of liquid

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 . L h1 h2 P1 P2 A

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 L h1 h2 P1 P2 A

Force on a block in liquid
The pressure difference P = P2 – P1 = Lg with upward direction. L h1 h2 P1 P2 A

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. L h1 h2 A F

Force on a block in liquid
This is the upthrust on the cylinder. Upthrust = Vg h1 h2 F V

Force on a block in liquid
Upthrust = Vg Note that it is also equal to the weight of the liquid with volume V. h1 h2 F V

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 h2 F V

Force on a block in liquid
The 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 upthrust W spring-balance The reading of the
spring-balance is W, which is the weight of the object. object beaker liquid The reading of the compression balance is B, which is the weight of liquid and beaker. compression balance B

Measuring upthrust spring-balance Carefully immerse half the volume of
the object in liquid. object beaker liquid What would happen to the reading of the spring-balance and that of the compression balance? compression balance

Measuring upthrust spring-balance The reading of the spring-
balance decreases. Why? object beaker liquid The difference in the readings of the spring-balance gives the upthrust on the object. compression balance

Measuring upthrust spring-balance The reading of the
compression balance increases. Why? object beaker liquid The difference in the readings of the compression balance gives the upthrust on the object. compression balance

Measuring upthrust spring-balance Carefully immerse object
the whole object in liquid. object beaker liquid What would happen to the reading of the spring-balance and that of the compression balance? compression balance

Measuring upthrust spring-balance Carefully place the object on the
bottom of the beaker. object beaker liquid What would happen to the reading of the spring-balance and that of the compression balance? 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 smaller than  density is larger than 

Manometer liquid of density  connect to the fluid Same level X Y
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 of density  connect to the fluid Same level X Y

Manometer Po = atmospheric pressure Po+P = fluid pressure
X Y Po+P = fluid pressure h = difference in height liquid of density 

Manometer The pressures at points A and B are equal. Po = atmospheric
Po+P = fluid pressure h = difference in height A B liquid of density 

Manometer The pressure at A = Po+P The pressure at B = Po + hg
liquid of density  Po = atmospheric pressure Po+P = fluid pressure h = difference in height B A

Manometer The pressure difference of the fluid P = hg
liquid of density  Po = atmospheric pressure Po+P = fluid pressure h = difference in height B A

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 hx = hY Y X static fluid

Liquid in a pipe The manometers show that the pressures at points X and Y are equal. Same level hx = hY Y X static fluid

Liquid in a pipe The pressures at points M and N on the pistons are also equal. Same level hx = hY 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 = hY M N PM PN static fluid

Liquid in a pipe As F = P.A , the external forces are different on the two ends. FM > FN Same level hx = hY M N FM FN static fluid

Liquid in a pipe Same level hx = hY M N FM FN static fluid
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 = hY M N FM FN static fluid

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
x2 A2 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. p Position 2 x1 h2 A1 P+P Position 1 h1

Pressure difference and work done
x2 A2 Work done by the external forces is (P+P).A1.x1 - P.A2.x2 P Position 2 x1 h2 A1 P+P Position 1 h1

Pressure difference and work done
x2 A2 Work done by the external forces is (P+P).A1.x1 - P.A2.x2 A1x1=A2x2=V = volume of fluid that moves P Position 2 x1 h2 A1 P+P Position 1 h1

Pressure difference and work done
x2 A2 Work done by the external forces is (P+P).A1.x1 - P.A2.x2 = P .V With V = Work done = P where m is the mass of the fluid and ρis the density of the fluid p Position 2 x1 h2 A1 P+P Position 1 h1

Bernoulli’s principle
x2 A2 v2 In time t, the fluid moves x1 at position 1 and x2 at position 2. x1 = v1.t and x2 = v2.t P2 Position 2 x1 h2 A1 P1 v1 Position 1 h1

Bernoulli’s principle
x2 A2 v2 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) P2 Position 2 x1 h2 A1 P1 v1 Position 1 h1

Bernoulli’s principle
x2 A2 v2 Work done by external pressure = (P1-P2) Increase in kinetic energy = P2 Position 2 x1 h2 A1 P1 v1 Position 1 h1

Bernoulli’s principle
x2 A2 v2 Increase in kinetic energy = Increase in gravitatioanl potential energy = mgh2 – mgh1 P2 Position 2 x1 h2 A1 P1 v1 Position 1 h1

Bernoulli’s principle
x2 A2 v2 P2 The left hand side is the work done by external pressure. It is also the energy supplied to the fluid. The right hand side is the increase in energy of the fluid. Position 2 x1 h2 A1 P1 v1 Position 1 h1

Bernoulli’s principle
x1 x2 Position 1 Position 2 A1 A2 v1 v2 h2 h1

Bernoulli’s principle
x1 x2 Position 1 Position 2 A1 A2 v1 v2 h2 h1 constant

Bernoulli’s principle
If h1 = h2 , then Position 2 Position 1

Bernoulli’s principle
If h1 = h2 , then or constant Position 2 Position 1

Bernoulli’s principle
If h1 = h2 , then constant So for horizontal flow, where the speed is high, the pressure is low. Low speed, high pressure High speed, low pressure Position 2 Position 1

Bernoulli’s principle
So for horizontal flow, where the speed is high, the pressure is low. High speed, low pressure Low speed, high pressure h1 h2 Position 2 Position 1

Bernoulli’s principle
So for horizontal flow, where the speed is high, the pressure is low. High speed, low pressure Low speed, high pressure h1 P1 = h1..g P2 = h2..g h2 Position 2 Position 1

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
air high pressure In the gap, the speed of airflow is high. So the pressure is low in the gap. The high pressure outside presses the 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 force due to spinning thrust from the air blower
weight of the ball air What is the direction of spinning of the ball?

Ball floating in air thrust from the force due to pressure air blower
difference air weight of the ball air 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

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.
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 Po The speed of water on the surface is almost zero. The speed of water at the hole is v. h v Po

A hole in a water tank Po The water pressure on the surface is Po.
The water pressure at the hole is Po. h v Po

A hole in a water tank Po The height of water on the surface is h
The height of water at the hole is 0. h v Po

A hole in a water tank Po Apply Bernoulli’s equation, 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. total
v static tube total h1 h2

Pitot tube total static tube h h2 h1 v
The pressure below the static tube is h1g. The pressure at the mouth of the total tube is h2g. h v static tube total h1 h2

Pitot tube total static tube h h2 h1 v
The fluid speed below the static tube is v. The fluid speed at the mouth of the total tube is 0. h v static tube total h1 h2

Pitot tube Apply Bernoulli’s equation, h v static tube total h1 h2

Pitot tube h v static tube total h1 h2