Physics 101: Chapter 11 Fluids

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# Physics 101: Chapter 11 Fluids - PowerPoint PPT Presentation

Physics 101: Chapter 11 Fluids. Textbook Sections 11.1-11.6 Density Pressure Pascal’s Principle. Textbook Sections 11.6-11.10 Archimedes Principle &amp; Buoyancy Fluids in motion: Continuity &amp; Bernoulli’s equation Some problems, homework hints. Physics 101: Density. Density = Mass/Volume

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Physics 101: Chapter 11Fluids
• Textbook Sections 11.1-11.6
• Density
• Pressure
• Pascal’s Principle
• Textbook Sections 11.6-11.10
• Archimedes Principle & Buoyancy
• Fluids in motion: Continuity & Bernoulli’s equation
• Some problems, homework hints
Physics 101: Density
• Density = Mass/Volume
•  = M/V
• units = kg/m3
• Densities of some common things (kg/m3)
• Water 1000
• ice 917 (floats on water)
• blood 1060 (sinks in water)
• Copper 8890
• Mercury 13,600
• Aluminum 2700
• Wood 550
• air 1.29
• Helium 0.18
Atmospheric Pressure
• normal atmospheric pressure = 1.01 x 105 Pa (14.7 lb/in2)

PD

PU

PU

PU

In Denver

In Urbana

Chapter 11,Preflight

You buy a bag of potato chips in Urbana Illinois and forget them (un-opened) under the seat of your car. You drive out to Denver Colorado to visit a friend for Thanksgiving, and when you get there you discover the lost bag of chips. The odd thing you notice right away is that the bag seems to have inflated like a balloon (i.e. it seems much more round and bouncy than when you bought it). How can you explain this ??

Due to the change in height, the air is much thinner in Denver. Thus, there is less pressure on the outside of the bag of chips in Denver, so the bag seems to inflate because the air pressure on the inside is greater than on the outside.

p1=0

h

p2=patm

• p2 = p1 + gh
• patm = gh
• Measure h, determine patm
• example--Mercury
•  = 13,600 kg/m3
• patm = 1.05 x 105 Pa
•  h = 0.757 m = 757 mm = 29.80” (for 1 atm)

CORRECT

p1=0

h

p2=patm

Chapter 11,Preflights

Suppose you have a barometer with mercury and a barometer with water. How does the height hwater compare with the height hmercury?

1. hwater is much larger than hmercury

2. hwater is a little larger than hmercury

3. hwater is a little smaller than hmercury

4. hwater is much smaller than hmercury

water is much less dense than mercury, so the same amount of pressure will move the water farther up the column.

CORRECT

Chapter 11, Preflights

Is it possible to stand on the roof of a five story (50 foot) tall house and drink, using a straw, from a glass on the ground?

1. No

2. Yes

Even if a person could completely remove all of the air from the straw, the height to which the outside air pressure moves the water up the straw would not be high enough for the person to drink the water.

p=0

h

pa

Chapter 11, Preflight

Evacuate the straw by sucking

How high will water rise?

no more than h = Pa/g = 10.3m = 33.8 feet

no matter how hard you suck!

Summary
• Density
• Pressure
• P2 = P1 + gh
• Pascal’s Principle

Note: Everything we do assumes fluid is non-viscous and incompressible.

• Textbook Sections 11.6-11.10
• Archimedes Principle & Buoyancy
• Fluids in motion: Continuity & Bernoulli’s equation
• Some problems, homework hints
Archimedes Principle (summary)
• Buoyant Force (B)
• weight of fluid displaced
• B = fluid x Vdispl g
• W = Mg = object Vobject g
• object sinks if object > fluid
• object floats if object < fluid
• Eureka!
• If object floats….
• B=W
• Therefore fluid gVdispl. = object gVobject
• Therefore Vdispl./Vobject = object / fluid

CORRECT

Chapter 11, Preflight

Suppose 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 out of the glass.

2. Go down.

3. Stay the same.

B = W gVdisplaced

W = ice gVice  W gV

Must be same!

CORRECT

Tub of water + ship

Tub of water

Overflowed water

Chapter 11, Preflight

Which weighs more:

1. A large bathtub filled to the brim with water.

2. A large bathtub filled to the brim with water with a battle-ship floating in it.

3. They will weigh the same.

Weight of ship = Buoyant force =

Weight of displaced water

r

A1 P1

v1

A2 P2

v2

Chapter 11: Fluid Flow (summary)
• Mass flow rate: Av (kg/s)
• Volume flow rate: Av (m3/s)
• Continuity: A1 v1 = A2 v2
• i.e., mass flow rate the same everywhere
• e.g., flow of river
• For fluid flow without friction:
• Bernoulli: P1 + 1/2v12 + gh1 = P2 + 1/2v22 + gh2
Chapter 11, Preflight

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

Explain this phenomenon using the equation of continuity

hmm, gravity stretches it out

A1

V1

The water's velocity is increasing as it flows down, so in order to compensate for the increase in velocity, the area must be decreased because the density*area*speed must be conserved

A2

V2

First off the mass of the water is conserved, and water is a incompressible fluid. The equation of continuity states that A1V1= A2V2. We know that the velocity or V2 will be increasing due to gravity so in order for the equation to work the area must decrease or get narrower.

CORRECT

Chapter 11, Preflight

A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which bent is downward such that the bottom of this pipe even with the other hole, like in the picture below:

Though which drain is the water spraying out with the highest speed?

1. The hole

2. The pipe

3. Same

The speed of the water depends on the pressure at the opening through which the water flows, which depends on the depth of the opening from the surface of the water. Since both the hole and the opening of the downspout are at [the same] depth below the surface, water from the two will be discharged with the same speeds.