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11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle (Buoyancy) Fluid Dynamics Continuity Equation Bernoulli Equation

Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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Page 1: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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

Page 2: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 3: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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 )

Page 4: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 5: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 6: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 7: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 8: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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.

Page 9: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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

Page 10: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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

Page 11: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 12: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 13: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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!

Page 14: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 15: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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!

Page 16: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 17: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

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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

Page 19: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 20: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 21: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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

Page 22: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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.

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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.

Page 24: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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Streamlines

Page 25: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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Continuity equation

Volume Flow rate

Mass flow rate

Δm1 = ρ1ΔV1 = ρ1Av1Δt

IM1 =Δm1

Δt= ρ1Av1

IV =ΔVΔt

= Av

Δm1

Δt=Δm2

Δt

IM 2 − IM1 =dm2

dt−dm1

dt=dm12

dt Continuity equation

In steady state

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

Page 26: Physics 201 Chapter 13 Lecture 1 - Department of Physics · 11/30/2009 Physics 201, UW-Madison 1 Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal’s Principle Archimedes Principle

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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