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Unit Seven:Solids and Fluids

John Elberfeld

JElberfeld@itt-tech.edu

518 872 2082

GE253 Physics

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Schedule

• Unit 1 – Measurements and Problem Solving• Unit 2 – Kinematics• Unit 3 – Motion in Two Dimensions• Unit 4 – Force and Motion• Unit 5 – Work and Energy• Unit 6 – Linear Momentum and Collisions• Unit 7 – Solids and Fluids• Unit 8 – Temperature and Kinetic Theory• Unit 9 – Sound• Unit 10 – Reflection and Refraction of Light• Unit 11 – Final

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Chapter 7 Objectives• Distinguish between stress and strain, and use

elastic moduli to compute dimensional changes.• Explain the pressure-depth relationship and state

Pascal's principle and describe how it is used in practical applications.

• Relate the buoyant force and Archimedes' principle and tell whether an object will float in a fluid, on the basis of relative densities.

• Identify the simplifications used in describing ideal fluid flow and use the continuity equation and Bernoulli's equation to explain common effects of ideal fluid flow.

• Describe the source of surface tension and its effect and discuss fluid viscosity.

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Reading Assignment

• Read and study College Physics, by Wilson and Buffa, Chapter 7, pages 219 to 250

• Be prepared for a quiz on this material

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Written Assignments

• Do the homework on the handout.• You must show all your work, and carry

through the units in all calculations• Use the proper number of significant

figures and, when reasonable, scientific notation

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Introduction

• Until now, we have beendealing with “ideal” objects

• Now we will look at objects that can stretch, bend, break, flow, and compress when forces are applied

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Introduction

• In this week, we will study real, macroscopic objects, such as wires, blocks, and containers of fluids.

• We will keep the concept of force and see what happens to macroscopic objects when forces act on them and see what forces macroscopic objects exert.

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Stress and Strain

• The stress of taking physics may cause a strain on your brain

• But in physics, these terms have different and precise meanings.

• We will learn to use these terms as we study the effects of forces on solids and fluids – in this unit, not to accelerate objects, but to deform them.

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Stress and Strain

• Stress (force/area) causes strain (% deformation of some type)

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Stress and Strain

• In physics, stress is related to the force applied to an object

• Strain is related to how the object is deformed as a result.

• Both also depend on the size of the object.

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Stress

• The diagram shows a wire or rod of length Lo and area A.

• When you apply a force F on the wire, it stretches by a distance .

• When you apply a force F in the opposite direction on a rod, the rod is compressed by a distance .

• The greater the force F, the greater is the amount of stretching or compressing.

L0A F

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Stress

• It is clear that a thick wire is harder to stretch than a thin one.

• For this reason, you can define the concept of stress as:

• Stress = F / A• The units of stress are N/m2.

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Strain

• The longer the wire or rod, the greater is the amount of stretching. For this reason, you can define strain as:

• Strain = | ΔL | / L• The vertical lines stand for absolute value,

which means the same strain results from positive ΔL due to stretching or negative Δ L due to compressing.

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Strain

• A strain of 0.05 on a wire means that the wire has been stretched to a length 5% greater than its original length.

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Young’s Modulus

• NOTE: This graph shows Young’s Modulus: how much a wire will stretch if you apply different forces on it.

Young’sModulusRegion

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Graph

• Engineers need to know the force need to stretch material a certain distance

• What force cause the bridge to sag into the waves?

• Graphs show strain on the X axis and stress on the Y axis

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Young’s Modulus

• Young’s Modulus works in the straight line part of the graph only

• The ratio of stress to strain in a material is known as Young’s modulus:

• Y = (F/A) / ( ΔL / L)• Where, F is force; A is cross sectional

area perpendicular to the applied force; and L is length of the object being stressed.

• NOTE: Young’s Modulus is an ELASTIC Modulus because materials bounce back to their original shape in that part of the graph.

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Young’s Modulus • The bigger the modulus, the stronger the

material, and the more force it takes to cause a specific deformation.

• Because material bounce back to their original shapes, moduli (plural) are elastic

• The units of Young’s modulus are the same as those of stress, N/m2.

• Young’s modulus for steel is 20 x 1010 N/m2 and for bone is 1.5 x 1010 N/m2

for example.

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Practice

• The femur (upper leg bone) is the longest and strongest bone in the body.

• Let us assume that a typical femur is circular and has a radius of 2.0 cm.

• How much force is required to extend the bone by 0.010%?

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Calculations

• Y = (F/A) / ( ΔL / L)• From the chart: Ybone = 1.5 x 1010N/m2

• R = 2cm(1m/100cm) = .02m• A = R2 = (.02m)2 = 1.26 x 10-3m2

• ΔL / L = .01% = .01/100 = 1 x 10-4

• Y ( ΔL / L) = (F/A) • F = A Y ( ΔL / L) • F = 1.26 x 10-3m2 1.5 x 1010N/m2 1 x 10-4

• F = 1.89 x 103 N (about 425 pounds)

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Practice

• A mass of 16 kg is suspended from a steel wire of 0.10-cm-diameter.

• By what percentage does the length of the wire increase?

• Young’s modulus for steel is 20 x 1010 N/m2.

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Calculation

• Y = (F/A) / ( ΔL / L)• F = W = mg =16kg 9.8m/s2 = 157 N• R = .1cm(1m/100cm)/2 = .0005m• A = R2 = (.0005m)2 = 7.85 x 10-7m2

• ( ΔL / L) = (F/A) / Y = (F/ R2 ) / Y • (ΔL/L) = (157N/ 7.85 x 10-7m2) / 20x1010N/m2

• (ΔL/L) = .001 = 0.1 %• NOTE: The change in length is inversely

proportional to the SQUARE of the radius!

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Shear Modulus

• To distort a rectangular solid, apply a force on one of its faces in a direction parallel to the face.

• Simultaneously, you must also apply a force on the opposite face in the opposite direction.

• The diagram on the screen shows how you can distort an object by applying force.

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Shear

• The diagram on the screen shows that the amount of distortion can be measured by the new angle Ø , which the faces make.

• Shearing stress is defined as F/A, where F is the tangential force and A is the area of the surface that the force acts on.

• Shearing strain is the angle Ø . • Similar to linear distortions, shearing

strain is directly proportional to shearing stress.

• The constant of proportionality is called the shear modulus (S):

• S = (F/A) / Φ

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Shear

• Shear modulus has the same units as Young’s modulus, N/m2.

• For many substances, shear modulus is approximately one-third of the Young’s modulus.

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Bulk Modulus

• You can also distort a rectangular solid by applying forces perpendicular to its surfaces.

• This stress causes the solid to become smaller and is known as volume stress or pressure.

• Volume stress is defined as F/A, where F is the force perpendicular to the surfaces, and A is the area of the surface.

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Bulk Modulus

• Volume strain is the change in volume divided by the original volume.

• Similar to shearing stress and stress, strain is directly proportional to volume stress.

• The constant of proportionality is called bulk modulus (B) and is defined as:

• B = (F/A) / ( ΔV / V)

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Bulk Modulus

• Solids are usually surrounded by air that exerts a compressing force on their surface.

• Pressure = Force/Area• Therefore, when you apply a force F, you

increase this force, or more precisely, the volume stress.

• Bulk modulus is usually written as: • B = (F/A) / ( ΔV / V) • B = Δp / ( ΔV / V) • Δp is the increase in pressure above

normal air pressure

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Table

1.0 x 10 9

4.5 x 109

26 x 109

2.2 x 109

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Compare Bulk Modulus

• NOTE: • B = Δp / ( ΔV / V)• For a gas, it takes the smallest pressure to

create a change in volume, so gasses have the smallest bulk modulus

• Solids require a big pressure to have a change in volume, so they have the biggest bulk modulus

• Bigger modulus implies a stronger material

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Practice

• By how much must you change the pressure on a liter of water to compress it by 0.10%?

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Calculation

• B = Δp / ( ΔV / V) • B = 2.2 x 109N/m2 • ( ΔV / V) = .1% (1/100%) = .001• Δp = B ( ΔV / V) • Δp = 2.2 x 109 N/m2 .001 = 2.2 x 106 N/m2

• Because there is always air pressure, this is the INCREASE in pressure to cause the change in volume

33 Pressure in a Fluid

• The molecules in a solid are tightly bound.• In a liquid or a gas, however, the

molecules are in motion and free to move.• The diagram shows a cubical container

filled with gas. • The molecules of the gas are

represented by the red dots. • The blue arrows show the

direction they are moving in.

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Pressure

• When a gas molecule with momentum p hits the wall and bounces back, it exerts an impulse equal to 2p on the wall.

• To calculate the force exerted by a single molecule, you need to know how long the collision lasted.

• The effect of a single molecule is very small, but the effect of a room full, like in a tornado, can be huge.

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Measuring Pressure

• The diagram shows a device that measures the pressure of a gas.

• Because gas molecules can move freely, the pressure of a gas will be the same no matter where you place the gauge.

• Furthermore, any changes in pressure applied to the container will be transmitted throughout the gas volume by the moving molecules.

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Pascal’s Principle

• Because liquid molecules are also free to move, the same principles also apply to liquids and thus all fluids.

• Pascal’s principle states this effect:• Pressure applied to an enclosed fluid is

transmitted undiminished to every point in the fluid and to the walls of the container.

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Atmospheric Pressure

• When a gas is in a container, it exerts uniform pressure throughout the container.

• In the case of atmosphere, however, the air stays at the Earth’s surface because of the force of gravity.

• Air pressure is greatest at sea-level and decreases as the altitude increases.

• The pressure of our atmosphere at sea-level is:

• This amount of pressure is defined as 1 atmosphere (atm).

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Density

• A liquid is much more dense than a gas. • As a result, the effect of gravity on a liquid

is much greater than that on a gas in the same sized container on the earth.

• In ordinary sized containers, a gas exerts uniform pressure throughout the container it is in, but a liquid does not.

• To refresh your memory, density is the mass of an object or system of particles divided by the volume it occupies.

• ρ = m / V

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Depth and Pressure in a Liquid

• The diagram explains the relationship between pressure and depth in a liquid.

• It shows a container of water with an imaginary rectangular column of water.

• The column of water has a surface area A, and a weight mg.

• Hence, a person holding the column of water experiences a force mg, and a pressure mg/A, which is the pressure of the water at the bottom of the container.

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Demonstration

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Pressure

• You can derive the equation for pressure using the relationship between mass and density:

• ρ = m/V => m = ρ V• V = A h => m = ρ Ah• p = F/A = mg/A = ρ Ah g / A• p = ρ g h• The total pressure of the liquid at the bottom

of the container is:

• p = p0 + ρ g h where p0 = normal air pressure

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Density

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Practice

• What is the total pressure exerted on the back of a scuba diver in a lake at a depth of 8.00 m?

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Total Pressure

• Pressing down on the divers back is all the water above him, plus all the AIR above the water.

• pwater = ρ g h = 1000kg/m3 x 9.8m/s2 x 8m

• pwater = 7.84 x 104 N/m2 = 7.84 x 104 Pa

• pAir = 1 Atmosphere = 101kPa

• Ptotal = 7.84 x 104 Pa + 101kPa

• Ptotal = 1.79 x 105 Pa

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Barometers

• A barometer is a device that measures atmospheric pressure.

• To make a barometer at home, turn a glass upside down under water.

• When you lift it straight up, it will remain filled with water.

• Air pressure holdsthe water up

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Barametric Pressure

• At the surface of the Earth, the mercury in a barometer rises to a height of 760 mm.

• This means atmospheric pressure is:

•   • The density of mercury is taken from the density

table shown previously. • Here we have introduced another unit of pressure

called the torr, named after a famous scientist, and representing the pressure corresponding to 1 mm of mercury.

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Pascal’s Principle

• The pressure of a gas in a container is the same at every point in the gas and on the walls of the container.

• Pressure in a liquid varies with depth • The diagram on the screen shows a

piston that exerts a force F over an area A on a body of water.

• This creates a pressure p that can be experienced throughout the fluid.

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Pressure

• Total pressure is the sum of the pressure from the weight of the fluid AND the added pressure from the piston

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Hydraulic Lift

• Two pistons are connected to a hydraulic lift.• When you press the input piston, the output

piston rises. • The pressure on the input piston is the same as

that on the output piston. • Keep in mind that the force is given by the

pressure times the area. F = P A• The output piston has a much larger area than

the input piston. • Therefore, the output piston exerts a much larger

force than the input piston, and you can easily lift an automobile. F = P A

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Lift a Car

• p1 = p2

• F1/A1 = F0/A0

• F0 = F1 (A0 /A1 )

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Practice

• The input and lift (output) pistons of a garage lift have diameters of 10 cm and 30 cm, respectively. The lift raises a car with a weight of 1.4 x 104 N.

• (a) What is the force on the input piston?• (b) What is the pressure to the input

piston?

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Calculations

• p1 = p2

• F1/A1 = F0/A0

• R1 = 10cm (1m/100cm)/2 = .05m• R2 = 30cm (1m/100cm)/2 = .15m• A1 = R1

2 = (.05m)2 = 7.85x10-3m2

• A2 = R22 = (.15m)2 = 7.07x10-2m2

• F1 = F0(A1/A0)• F1 = 1.4 x 104 N (7.85x10-3m2/ 7.07x10-2m2)• F1 =1,550N• p1 = F1/A1 =1,554N / 7.85x10-3m2= 2.0x105N/m2

• p2 = F2/A2 = 1.4 x 104 N / 7.07x10-2m2

• p2 = 2.0x105N/m2 = checks!

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Thought Experiment

• If the output piston of a hydraulic lift has a very large area, a two-year-old can lift the Empire State Building.

• Does this make sense?

• Is energy conserved?

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Results

• Yes, it does make sense. • Simple machines, such as levers,

transform small input forces into large output forces.

• Energy is the ability to do work. • It is the product of force and distance. • Machines transform small forces applied

over large distances to large forces applied over small distances.

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Practice

• A 60-kg athlete does a single-hand handstand.

• If the area of the hand in contact with the floor is 100 cm2, what pressure is exerted on the floor?

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Calculation

• p = F/A• A = 100 cm2 (1m/100cm)2 = 1.00x10-2m2

• p = mg/A = 60kg 9.8m/s2/ 1.00x10-2m2

• p = 5.88x104N/m2

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Practice

• An oak barrel with a lid of area 0.20 m2 is filled with water. A long, thin tube of cross-sectional area 5.0 X 10-5 m2 is inserted into a hole at the center of the lid, and water is poured into the tube.

• When the water reaches 12 m high, the barrel bursts.

• What was the weight of the water in the tube?

• What was the pressure of the water on the lid of the barrel?

• What was the net force on the lid due to the water pressure?

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Calculations

• W = m g = ρ V g = ρ Ah g • A = 5.0 X 10-5 m2 , h = 12 m • W=1000kg/m3 5.0 X 10-5 m2 12 m 9.8m/s2 • W = 5.88 N• p = F/A = 5.88 N / 5.0 X 10-5 m2 • p = 1.18x105N/m2

• OR• p =ρgh= 1000kg/m3 9.8m/s2 12m = 1.18x105N/m2

• Checks!

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Calculations

• p = F / A• F = p A• F = 1.18x105N/m2 0.20 m2 • F = 2.36 x 104N (More than 5000 pounds!)

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Archimedes

• Have you ever yelled, 'Eureka!' when you figured out how to solve a challenging problem?

• Well the story goes that when Archimedes figured out the principle named after him, he was so excited that he not only yelled, 'Eureka,' but also went running through town from the public bath … naked!

• You may not get that excited over Archimedes principle, but you will learn some useful and interesting things about fluid buoyancy and fluid flow in this lesson.

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Archimedes' Principle

• The upward force experienced by an object when it is immersed in a liquid or a gas is called buoyant force.

• Buoyant force is equal in magnitude to the weight of the volume of fluid displaced.

• This rule is called Archimedes' principle.

• BF = mg = ρ V g

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Archimedes’ Principle

• The diagrams on the screen illustrate Archimedes’ principle.

• Part (a) shows buoyant force as the force that pushes up when you hold a piece of wood under water.

• Part (b) shows buoyant force as the force that reduces the weight of an object weighing 10 N to 8 N.

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Practice

• A spherical helium balloon has a radius of 30 cm. What is the buoyant force acting on it in air?

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Calculations

• Find the volume if R = 30 cm = .3 m• V = (4/3) R3 = .113 m3 • BF = mg = ρ V g = 1.29kg/m3 .113 m3 9.8m/s2

• BF = 1.43 N (about 1/3 of a pound)• If the balloon and the air in the balloon has

weight less then 1.43 N, it will float: if more, it will sink

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Practice

• The diagram shows a partially submerged iceberg at rest.

• It is in equilibrium because its buoyant force is equal to the weight of the iceberg.

• Suppose a uniform solid cube of material 10 cm on each side has a mass of 700 g. Will the cube float? If it will float, how much of the cube will be submerged?

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Calculations

• Suppose the cube were completely SUBMERGED!• LSide = 10 cm = .1m• V = LWH = (.1m)(.1m)(.1m) = 10-3 m3

• BF = ρ V g = 1x103Kg/m3 10-3 m3 9.8m/s2

• BF = 9.8N• Weight of the object = mg • W = 700 g (1kg/1000g) 9.8m/s2 = 6.86 N• The buoyant force is greater than the weight!• The object will be pushed up to the surface and

will rise above the surface until the weight of the fluid displaced just balances the weight of the object

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Floating Object

• V = LWH = (.1m)(.1m)(.1m) = 10-3 m3

• W = 700 g (1kg/1000g) 9.8m/s2 = 6.86 N• BF = 6.86 N = ρ V g = 1x103Kg/m3 V 9.8m/s2

• V = 6.86 N / (1x103Kg/m3 9.8m/s2 )• V = 7 x 10-4 m3

• 7 x 10-4 m3 is the volume of the cube below the surface

• The remainder:• 10-3 m3 - 7 x 10-4 m3 = 3 x 10-4 m3 is above the

surface• It is 70% submerged

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Ideal Fluid Flow

• To mathematically describe fluid flow like we have projectile motion, for example can be very difficult in real cases.

• Consider the upper portion of the rising smoke – it looks almost random!

• In order to obtain a basic description of fluid flow, we will make some simplifications and consider ideal fluid flow, so the flow looks more like the lower part of the rising smoke stream.

• The conditions for an ideal fluid are that its flow is:

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Ideal Fluid Flow

1. Steady: The flow is not turbulent; particles in the fluid do not collide, and their paths do not cross. The flow looks smooth.

2. Irrotational: There is no rotational motion in small volumes of the fluid, so there are no whirlpools. A paddle wheel completely       embedded in the stream does not rotate.

3. Nonviscous: Viscosity is negligible. This means the fluid flows easily, without significant friction or resistance to the flow.

4. Incompressible: The density is constant throughout the fluid.

• When a fluid's flow has these four characteristics, we say it is an ideal fluid. Unless specified otherwise, we will consider these four conditions to be true in the fluids we consider.

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Equation of Continuity

• When a liquid flows through a tube, the amount of liquid entering the tube must equal the amount of liquid coming out of it.

• If the cross-sectional area of the tube varies across its length, then the speed of the liquid must vary too.

• The diagram on the screen shows what happens to the speed of water when the nozzle has a smaller cross-sectional area than the tube.

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Equation of continuity

• If the liquid is incompressible, the density of the liquid remains the same throughout the tube. If the liquid is incompressible, the density of the liquid remains the same throughout the tube.

• Hence, as the above equation simplifies to: • Av = constant• Area x velocity = constant• Volume/time = constant• Equation of continuity is an equation that

describes the fact that the amount of fluid entering a tube is equal to the amount of fluid leaving the tube.

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Equation of Continuity

• High cholesterol in the blood can cause fatty deposits called plaques to form on the walls of blood vessels.

• Suppose plaque reduces the effective radius of an artery by 25%.

• Because the area is smaller, what must happen to the velocity?

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Bernoulli’s Equation

• Bernoulli's equation states that when a liquid is moving fast, its pressure is reduced.

• The diagram below shows a tube of varying cross-sectional areas.

• The pressure indicators show that the pressure is lower in the middle region, where the smaller cross sectional area results in the flow rate being greater.

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Bernoulli's Equation

• To observe Bernoulli's equation, conduct the following experiment.

• Hold two corners of a sheet of paper with both hands.

• Rotate your hands so that the paper takes the shape of the wing of an airplane.

• Blow over the top of the paper. • You will observe that the sheet of paper rises. • It lifts up because air flows on one side of the

paper, but is stationary on the other. • This is the principle behind the lifting of air

planes: the orientation and shape of the wing of an airplane causes the air flowing over it to flow faster than the air under it.

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Wing Design

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Surface Tension• When there is a boundary between a liquid and a

gas, some remarkable phenomena occur. • The figures on the screen show two water/air

phenomena and one soap solution/air boundary phenomena.

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Surface Tension

• 1) An insect is able to walk on water. 2) Water in air forms circular droplets.3) Soap solutions form bubbles.

• In the phenomenon where bubbles are made, there are two surfaces: the inside and the outside of the bubble.

• Although the bubble is thin compared to a piece of paper, it is very thick compared to the size of a molecule.

• The surface in an air-liquid boundary is only two or three molecules thick.

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Surface Tension

• The forces of attraction between the molecules of a liquid are lower than those of a solid.

• All the same, they do exist because otherwise the molecules of a liquid would escape into the atmosphere.

• In a liquid, a molecule, which is at a distance from the surface, is surrounded by other molecules.

• A molecule on the surface, however, is not entirely surrounded by molecules.

• The molecules on the surface are held together by a horizontal force between them, which is called surface tension.

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Viscosity

• You must have noticed that in January, molasses flows slower than water or kerosene.

• This happens because of viscosity. • Viscosity is a fluid’s internal resistance to

flow.• It can be measured with a device called a

viscosimeter. • There are various types of viscosimeters.

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Flow Rate

• When fluid is flowing in a pipe, as a result of viscosity, the speed of the fluid near the surface of the pipe is less than that at the center.

• The flow rate is the average volume of a fluid that flows beyond a given point during a time interval

• The diagram on the screen shows that is not easy to calculate the average flow rate.

• Flow rate has the unit of m3/s.

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Summary

• What are the elastic modulie?

• What is the relationship between pressure and depth in a fluid?

• What is Archimedes’ principle?

• What are the essential principles of fluid flow?

• What causes surface tension and what is viscosity?