Physics 201 Chapter 13 Lecture 1 - UW-Madison … Physics 201, UW-Madison 1 Physics 201 Chapter 13...

11/30/2009 Physics 201, UW-Madison 1

Physics 201Chapter 13Lecture 1

Fluid Statics Pascal’s Principle Archimedes Principle (Buoyancy)

Fluid Dynamics Continuity Equation Bernoulli Equation


Fluids

Atmospheric PressureEven when there is no breeze air molecules are continuously

bombarding everything around - results in pressure

normal atmospheric pressure = 1.01 x 105 Pa (14.7 lb/in2)

Pressure (P)P = Force/Area [N/m2]

1 N/m2 = 1 Pascal (Pa)

Density = Mass/Volumeρ = M / V units = kg/m3

Pressure variation with depthP = ρ g h

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Densities of substances

3


Density & Pressure are related by the Bulk Modulus

LIQUID: incompressible (density almost constant)

GAS: compressible (density depends a lot on pressure)

Compressiblity

B =Δp

(−ΔV /V )


Variation of pressure with depth

m = ρV; V = Ah⇒ m = ρAh

P =FA=mgA

; i.e., P =ρAh( )gA

⇒ P = hρg

True for all shapes of containers


Pressure difference of 3 m of water compares to the change of descending 3000m in air

6


Pascal’s Principle A change in pressure in an enclosed fluid is

transmitted undiminished to all the fluid and to its container.

This principle is used in hydraulic system P1 = P2

(F1 / A1) = (F2 / A2)

F2


Pascal’s Principle

This principle is used in hydraulic system P1 = P2

(F1 / A1) = (F2 / A2) Can be used to achieve a mechanical advantage F2 = F1 (A2 / A1)

» Work done is the same: height by which the surface A2 rises is smaller than the change in the height of surface with area A1.

F1

A1

F2A2


Using Fluids to Measure Pressure

1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H20

• Use Barometer to measure Absolute Pressure

Barometer Top of tube evacuated (p=0) Bottom of tube submerged into pool of mercury

open to atmosphere (p=p0) Pressure dependence on depth:

• Use Manometer to measure Gauge Pressurep0

Δh

Manometerp1

Measure pressure of volume (p1) relative to the atmospheric pressure (≡ gauge pressure )

The height difference (Δh) measures the gauge pressure:

Δh =(p1 − p0 )

ρg

h =p0

ρg


Measurement of Pressure

Manometer If both sides of an U-tube are open to atmosphere

the levels of the fluid are the same on both sides If one side is connected to a “pressurized side” the

level difference between the two sides can be used to measure pressure.


Measuring the tire pressure:Is this a manometer or a barometer?


Measuring Blood Pressure Blood pressure is quite high, 120/80 mm of Hg Use higher density fluid in a manometer: Mercury


Atmosphere - pressure vs height

13

P = P0e−ρ0P0

gh

--> whiteboard

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Pressure in a fluidPressure in a fluid

• Impulse to wall:

Fx!t = !px = !(Mvx)

Fx = !(Mvx)/ !t

• Force is perpendicular to surface

• Force proportional to area of surface

• pressure (p)

p = Force/area [N/m2]

1 N/m2 = 1 Pascal (Pa)

v

v

Fx

wall

molecule

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PressurePressurey

Atmospheric PressureAtmospheric Pressure

Even when there is no breeze air molecules are continuously

bombarding everything around - results in pressure

normal atmospheric pressure = 1.01 x 105 Pa (14.7 lb/in2)

average vertical force = fy

=

!py

!t=

! mvy

"

#$$

%

&''

!t


ArchimedesObject immersed in a fluid is subject to a “buoyant force”.

Force on sides cancel

Force on top Ft = ρghT A

Force on bottom Fb = ρghB A

ΔF = ρg A Δh

FB = (mg)disp


The pressure is a function of the depth only (for a givendensity of the fluid and of g)


Float

Weight of object = ρ0gV

Buoyant force is the weight of the displaced fluid

Weight of fluid = ρfgV

Displace just enough fluid such that forces = 0!


Archimedes PrincipleBuoyant Force (B)

weight of fluid displaced (P=F/A, P=ρgh)» B = ρfluid g Vdisplaced

» W = ρobject g Vobject

» object sinks if ρobject > ρfluid

» object floats if ρobject < ρfluid

» Eureka!

If object floats….» B=W» Therefore ρfluid g Vdisplaced = ρobject g Vobject

» Therefore Vdisplaced/Vobject = ρobject / ρfluid


Float Buoyant force is the weight of the displaced fluid Weight of object = ρIceVtotal gWeight of fluid = ρSeaWatergVsubmersed

Displace just enough fluid such that forces = 0!

General solution: Vdisplaced/Vobject = ρobject / ρfluid


The weight of a glass filled to the brim with water is Wb. A cube of ice is placed in it, causing some water to spill. After the spilled water is cleaned up, the weight of the glass with ice cube is Wa. How do the weights compare: 1. Wb > Wa. 2. Wb < Wa.3. Wb = Wa.

Archimedes’ Principle: The buoyant force on an object equals the weight of the fluid it displaces.

Weight of water displaced = Buoyant force = Weight of ice

Archimedes Principle


QuestionSuppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up causing the water to spill. 2. Go down.3. Stay the same.

Archimedes’ Principle: The buoyant force on an object equals the weight of the fluid it displaces.

Weight of water displaced = Buoyant force = Weight of ice

When ice melts it will turn into water of same volume


Buoyancy

Two cups hold water at the same level. One of the two cups has plastic balls (projecting above the water surface) floating in it. Which cup weighs more?

Archimedes principle tells us that the cups weigh the same. Each plastic ball displaces an amount of water that is exactly

equal to its own weight.

Cup I Cup II

1) Cup I2) Cup II3) Both the same


Two identical glasses are filled to the same level with water. Solid steel balls are at the bottom in one of the glasses. Which of the two glasses weighs more? 1. The glass without steel balls 2. The glass with steel balls 3. Both glasses weigh the same

The steel balls sink. The buoyant force equal to the weight of the displaced water is not sufficient to counter the weight of the steel balls. Therefore, the glass with steel balls weighs more.

Sunken Balls


Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greater depth. The force needed to hold brick B in place is:

1. larger

2. the same as

3. smaller

than the force required to hold brick A in place.

The buoyant force on each brick is equal to the weight of the water it displaces and does not depend on depth.

Buoyant force and depth


Fluid Flow

• Volume flow rate: ΔV/Δt = A Δd/Δt = Av (m3/s)

• Continuity: A1 v1 = A2 v2

i.e., flow rate the same everywhere

e.g., flow of river

Fluid flow without friction


ProblemTwo hoses, one of 20-mm diameter, the other of 15-mm diameter are connected one behind the other to a faucet. At the open end of the hose, the flow of water measures 10 liters per minute. Through which pipe does the water flow faster? 1. The 20-mm hose 2. The 15-mm hose 3. Water flows at the same speed in both cases4. The answer depends on which of the two hoses comes first in the flow

When a tube narrows, the same volume occupies a greater length. For the same volume to pass through points 1 and 2 in a given time, the velocity must be greater at point 2. The process is reversible.


Faucet

A stream of water gets narrower as it falls from a faucet (try it & see).

The velocity of the liquid increases as the water falls due to gravity. If the volume flow rate is conserved, them the cross-sectional area must decrease in order to compensate

A1

A2

V1

V2

The density of the water is the same no matter where it is in space and time, so as it falls down and accelerates because of gravity,the water is in a sense stretched, so it thins out at the end.

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Types of Fluid Flow

Laminar flow Steady flow Each particle of the fluid follows a smooth path The paths of the different particles never cross each

other The path taken by the particles is called a streamline

Turbulent flow An irregular flow characterized by small whirlpool like

regions Turbulent flow occurs when the particles go above some

critical speed

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Viscosity

Characterizes the degree of internal friction in the fluid This internal friction, viscous force, is associated with the

resistance that two adjacent layers of fluid have to moving relative to each other

It causes part of the kinetic energy of a fluid to be converted to internal energy

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Ideal Fluid Flow

There are four simplifying assumptions made to the complex flow of fluids to make the analysis easier1. The fluid is nonviscous – internal friction is

neglected2. The flow is steady – the velocity of each point

remains constant3. The fluid is incompressible – the density remains

constant4. The flow is irrotational – the fluid has no angular

momentum about any point

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Streamlines

The path the particle takes in steady flow is a streamline

The velocity of the particle is tangent to the streamline

A set of streamlines is called a tube of flow


Streamlines


Continuity equation

Volume Flow rate

Mass flow rate

Δm1 = ρ1ΔV1 = ρ1Av1Δt

IM1 =Δm1

Δt= ρ1A1v1

IV =ΔVΔt

= Av

Δm1

Δt=Δm2

Δt

IM 2 − IM1 =dm2

dt−dm1

dt=dm12

dtContinuity equation

In steady state

General case: mass may be accumulated or decreased in the volume between A1 and A2

ρ2A2v2 = ρ1A1v1


Continuity equationVolume Flow rate

Mass flow rate IM1 =Δm1

Δt= ρ1A1v1

IV =ΔVΔt

= Av

In steady state

Case of incompressible fluid: density constant

ρ2A2v2 = ρ1A1v1

A2v2 = A1v1


Bernoulli’s Equation As a fluid moves through a region

where its speed and/or elevation above the Earth’s surface changes, the pressure in the fluid varies with these changes

Consider the two shaded segments The volumes of both segments are

equal The net work done on the segment

is W =(P1 – P2) V Part of the work goes into changing

the kinetic energy and some to changing the gravitational potential energy


Bernoulli’s Equation

The change in kinetic energy: ΔK = 1/2 mv2

2 - 1/2 mv12

The masses are the same since the volumes are the same

The change in gravitational potential energy: ΔU = mgy2 – mgy1

The work also equals the change in energy Combining: W = (P1 – P2)V =1/2 mv2

2 - 1/2 mv12 + mgy2 – mgy1

Rearranging and expressing in terms of density: P1 + 1/2 ρv1

2 + mgy1 = P2 + 1/2 ρv22 + mgy2

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Bernoulli’s Equation

Pressure drops in a rapidly moving fluid whether or not the fluid is confined to a tube

For incompressible, frictionless fluid:

P +12ρv2 + ρgh = constant

12ρv2 = 1

2mv2

1V

=KEV

ρgh = mghV

=PEV

Bernoulli equation states conservation of energyFor Static Fluids:P1 + ρgh1 = P2 + ρgh2

Bernoulli's Principle (constant depth):P1 +12ρv1

2 = P2 +12ρv2

2

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What is the pressure of an incompressible fluid in the constricted region?

Continuity equation gives velocity in the constricted region (increases with A1/A2)):

Bernoulli equation

says that pressure drops as 40

P1 +12ρv1

2 = const

A2v2 = A1v1

P ∝1v2

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Applications of Bernoulli’s Principle

Wings and sails Higher velocity on one side of sail versus the

other results in a pressure difference that can even allow the boat to sail into the wind

Entrainment Reduced pressure in high velocity fluid pulls in

particles from static or lower velocity fluid» Bunsen burner, Aspirator, …

Velocity measurement

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Problem(a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6 m/s parallel to its front surface and 3.5 m/s along its back surface. Take the density of air to be 1.29 kg/m3. (b) Discuss whether this force is great enough to be effective for propelling a sail boat.

Force, F = (P1 − P2 )A =12ρ(v2

2 − v12 )A = 15.3 N

The force is small. However, when the sails are large, the force can be high enough to propel a sail boat. For larger boats, one can add more than one sail to increase the surface area.One can even sail into the wind, where (P1 − P2 ) is small.

P1 +12ρv1

2 = P2 +12ρv2

2Bernoulli eq.for constant height

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Applications of Fluid Dynamics

Streamline flow around a moving airplane wing

Lift is the upward force on the wing from the air

Drag is the resistance The lift depends on the

speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal

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Lift – General

In general, an object moving through a fluid experiences lift as a result of any effect that causes the fluid to change its direction as it flows past the object

Some factors that influence lift are: The shape of the object The object’s orientation with respect to the fluid flow Any spinning of the object The texture of the object’s surface

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Golf Ball

The ball is given a rapid backspin

The dimples increase friction Bernoulli says: Higher relative

velocity will reduce the pressure. Increases lift

It travels farther than if it was not spinning

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Problem(a) What is the pressure drop due to Bernoulli effect as water goes into a 3 cm diameter nozzle from a 9 cm diameter fire hose while carrying a flow of 40 L/s? (b) To what maximum height above the nozzle can this water rise neglecting air resistance.

v1 =F1A1

=40 ×10−3m3 /sπ (0.045)2 = 6.29 m/s

v2 =F2A2

=40 ×10−3m3 /sπ (0.015)2 = 56.6 m/s

P1 − P2 =12ρ(v2

2 − v12 ) = 1.58 ×106 N/m2

h =v2

2g=

(56.6)2

2 × 9.8m=163 m

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Torricelli’s Theorem

P1, v1, h1

P2=P1 , v2 , h2

Bernoulli's equation at constant pressure (P1 = P2 )

P1 +12ρv1

2 + ρgh1 = P2 +12ρv2

2 + ρgh2

12ρv1

2 + ρgh1 =12ρv2

2 + ρgh2

v22 = v1

2 + 2g(h1 − h2 )h = h1 − h2 Same as kinematics equation for any object falling with negligible friction.

h1

h2

Density is constant

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Viscosity

V0

V=0

F

η =

FAvL

L

•Friction in fluids

shearing stress

“strain”

Newton’s lawLaminar flow -- no turbulence

Pressure, τ = η dvdz

η is the coefficient of viscosity'A' is the moving surface

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Real fluid flow

At constant velocity net force is zero.F = P1 − P2( )πr2 and the area on which the force is acting is A = 2πrL

τ = F A =ΔPr2L

η =τ

−ΔvΔror dv =

− P1 − P2( )r2Lη

dr

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More Viscosity

for a given situation P1,P2 ,η, and L are constant

let b =ΔP2ηL

and dv = −brdr and integrating

v = −br2

2+ C

v = 0 at the boundary r = R and, substituting,

v =P1 − P2( )4ηL

R2 − r2( )

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Flow and Viscosity

ΔV = vt( )2πrΔr and again let B = π tb

and, then dV = BR2rdr − Br3dr

V = BR2 rdr0

R

∫ − B r3dr0

R

∫ = BR4

2− B

R4

4= B

R4

4Finally,

Vt=π P1 − P2( )R4

8ηLPoiseuille’s Eqn.

•no turbulence•no sized particles•constant η

Physics 201 Chapter 13 Lecture 1 - UW-Madison … Physics 201, UW-Madison 1 Physics 201 Chapter 13...

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