physics 102 part ii thermal physics moza m. al-rabban professor of physics [email protected] fluids

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Physics 102 Part II Thermal Physics Moza M. Al-Rabban Professor of Physics [email protected] Fluids

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Physics 102 Part II

Thermal Physics

Physics 102 Part II

Thermal Physics

Moza M. Al-Rabban

Professor of Physics

[email protected]

FluidsFluids

Text Book

• Physics, by James S. Walker

• Chapters 15, 16, and 17

OutlineOutline15-1 Density15-2 Pressure15-3 Static Equilibrium in Fluids: Pressure and

Depth15-4 Archimedes' Principle and Buoyancy15-5 Applications of Archimedes' Principle15-6 Fluid Flow and Continuity 15-7 Bernoulli's Equation15-8 Applications of Bernoulli's Equation*15-9 Viscosity and Surface Tension

States Of Matter

• Solids

• Liquids

• gases

Solids:Solids:

Solids tends to hold their shapes.

Although the atoms vibrate around fixed equilibrium positions, they do not have enough energy to break the bound with their neighbors.

FluidsLiquids and Gases

• Liquids and Gases, do not hold their shapes.

• A liquid flows and takes the shape of its container and a gas expands to fill its container.

• The atoms or molecules in Fluid do not have fixed positions, so a fluid does not have a definite shape.

DensityDensity

• Definition of Density,

3/:

/

mkgunitSI

VM

The denser a material, the more mass it has in any given volume.

PressurePressure

• Definition of Pressure, P

2/:

/

mNunitsSI

AFP

Example 1:

Find the pressure exerted on the skin of a balloon if you press with a force of 2.1 N using (a) your finger or (b) a needle. Assume the area of your fingertip is , and the area of the needle tip is (c) Find the minimum force necessary to pop the balloon with the needle, given that the balloon pops with a pressure of

24100.1 m27105.2 m

25 /100.3 mN

Solve

(a) Calculate the pressure exerted by the finger:

24

24/101.2

100.1

1.2mN

m

N

A

FP

2627

/104.8105.2

1.2mN

m

N

A

FP

(b) Calculate the pressure exerted by the needle:

(c)

NmmNPAF 075.0105.2/100.3 2725

Atmospheric Pressure and Atmospheric Pressure and Gauge PressureGauge Pressure

• Atmospheric Pressure, Pat

ata

aat

a

at

PPbar

kPP

mNpascalP

mNunitSI

mNP

1101

101

/1)(1

/:

/1001.1

5

2

2

25

Gauge pressure, Pg

• The pressure in a fluid acts equally in all directions, and acts at right angles to any surface.

• In many cases we are interested in difference between a given pressure and atmospheric pressure.

atg PPP

Gauge Pressure Does the flat tire on your automobile have zero air pressure? If it is completely flat, it still has the atmospheric pressure air in it. To be sure, it has zero useful pressure in it, and your tire gauge would read zero Newton per square meter. Most gauges read the excess

of pressure over atmospheric pressure and this excess is called "gauge pressure ."

When a system is at atmospheric pressure like the left image above, the gauge pressure is said to be zero. In this image, the system has been opened so that it is at equilibrium with the atmosphere. In the right image, the system has been closed and the plunger

pushed down until the pressure reads about 15 lb/in2 .

atabsulutegauge PPP

Static Equilibrium in Fluids:Pressure and Depth

ghPP

ghPA

ghAAP

A

FP

ghAAPWFF

ghAVgMgW

hAVVM

APF

at

atatbottom

bottom

attopbottom

attop

,

This expression holds for any liquid with constant density and a pressure Pat at its upper surface.

The relation can be applied to any two points in a fluid.

ghPP at

Dependence of Pressure on depth

ghPP 12

Example3:

A cubical box 20.00 cm on a side is completely immersed in a fluid. At the top of the box the pressure is 105.0 kPa; at the bottom the pressure is 106.8 kPa. What is the

density of the fluid?

Example3:

A cubical box 20.00 cm on a side is completely immersed in a fluid. At the top of the box the pressure is 105.0 kPa; at the bottom the pressure is 106.8 kPa. What is the density of the fluid?

Solve:

3

2

55

12

12

/9202000.0/81.9

10050.110068.1mkg

msm

PaPa

gh

PP

ghPP

A Simple Barometer

An interesting application of the variation of pressure with depth is the barometer, which can be used to measure atmospheric pressure.

The pressure in the tube at depth h below the vacuum is 0 + gh = gh

At the level of fluid in the bowel we know that the pressure is one atmosphere, therefore, Pat = gh

A fluid that is often used in such a barometer is mercury (Hg), with a density of

34 /103595.1 mkg

The corresponding height for a column of mercury is

mmHgPatmosphere

mmsmmkg

Pa

g

Ph

at

at

7601

760/81.9/103595.1

10013.1234

5

Fluid Seeks Its Own LevelFluid Seeks Its Own Level

(a)When the levels are equal, the pressure is the same at the base of each arm of the U tube. As a result, the fluid in the horizontal section of the U is in equilibrium.

(b) With unequal heights, the pressure are different. In this case, the pressure is greater at the base of the right arm, hence fluid will flow toward the left and the level will equalize.

Example 4: Oil and Water Don’t Mix

A U-shaped tube is filled mostly with water, but a small amount of vegetable oil has been added to one side, as shown in the sketch. The density of the water is

, and the density of the oil is

If the depth of the oil is 5.00 cm, what is the difference in level h between the top of the oil on one side of the U and the top of the water on the other side?

33 /1000.1 mkg 32 /1020.9 mkg

Solution

1 .Find the pressure at point A, where the depth of the water is h1

1ghPP wateratA

2 .Find the pressure at point B, where the depth of the water is h2 = 5.00 cm

2ghPP oilatB

3 .Set PA equal PB :21 ghPghP oilatwaterat

Example 4: Oil and Water Don’t Mix

A U-shaped tube is filled mostly with water, but a small amount of vegetable oil has been added to one side, as shown in the sketch. The density of the water is

, and the density of the oil is

If the depth of the oil is 5.00 cm, what is the difference in level h between the top of the oil on one side of the U and the top of the water on the other side?

4. Solve for the depth of the water,h1

cmcmcmhhh

cmmkg

mkgcm

hhwater

oil

40.060.400.5

60.4/1000.1

/1020.900.5

12

33

32

21

Pascal’s PrincipalPascal’s Principal

11 / AFP

11

212112

22

/ FA

AFAAFF

APF

An external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid.

An external pressure applied to an enclosed fluid is transmitted unchanged to every point within thefluid.

A hydraulic lift

2

112

2211

A

Add

dAdA

End of Lecture 13