# Physics 102 Part II Thermal Physics

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```Physics 102
Part II
Thermal Physics
Fluids
Moza M. Al-Rabban
Professor of Physics
[email protected]
Text Book
• Physics, by James S. Walker
• Chapters 15, 16, and 17
Outline
15-1 Density
15-2 Pressure
15-3 Static Equilibrium in Fluids: Pressure and Depth
15-4 Archimedes' Principle and Buoyancy
15-5 Applications of Archimedes' Principle
15-6 Fluid Flow and Continuity
15-7 Bernoulli's Equation
15-8 Applications of Bernoulli's Equation
*15-9 Viscosity and Surface Tension
States Of Matter
• Solids
• Liquids
• gases
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.
Fluids
Liquids 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.
Density
• Definition of Density, 
  M /V
SI unit : kg / m
3
The denser a material, the more mass it has in
any given volume.
Pressure
• Definition of Pressure, P
P F /A
SI units: N / m
2
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 1.0 104 m 2 , and
the area of the needle tip is 2.5 107 m2 (c) Find the minimum force necessary to pop
the balloon with the needle, given that the balloon pops with a pressure of 3.0 105 N / m2
Solve
(a) Calculate the pressure exerted by the finger:
P
F
2.1 N
4
2


2
.
1

10
N
/
m
A 1.0 10  4 m 2
(b) Calculate the pressure exerted by the needle:
P
F
2.1 N
6
2


8
.
4

10
N
/
m
A 2.5 10 7 m 2
(c)



F  PA  3.0 105 N / m2 2.5 107 m2  0.075N
Atmospheric Pressure
and Gauge Pressure
• Atmospheric Pressure, Pat
Pat  1.01 105 N / m 2
SI unit : N / m 2
1Pa ( pascal)  1N / m 2
Pat  101kPa
1bar  105 Pa  1Pat
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.
Pg  P  Pat
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 &quot;gauge pressure&quot;.
Pgauge  Pabsulute  Pat
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.
Static Equilibrium in Fluids:
Pressure and Depth
Ftop  Pat A
M  V , V  hA
W  Mg  Vg   hAg
Fbottom  Ftop  W  Pat A   hAg
Pbottom
Fbottom Pat A   hAg


 Pat  gh
A
A
P  Pat  gh
This expression holds for any liquid with constant density  and a pressure Pat at its
upper surface.
The relation P  P
at
two points in a fluid.
 gh
Dependence of Pressure on depth
P2  P1  gh
can be applied to any
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:
P2  P1  gh

P2  P1
gh
1.068105 Pa  1.050105 Pa
3


920
kg
/
m
9.81m / s 2 0.2000m 


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
  1.3595104 kg / m3
The corresponding height for a column of mercury is
Pat
1.013105 Pa
h

 760m m
4
3
2
g 1.359510 kg / m 9.81m / s


1 atm osphere Pat  760m m Hg

Fluid Seeks Its Own Level
When the levels are equal, the pressure is the same at the base of each (a)
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
1.00103 kg / m3
, and the density of the oil is 9.20102 kg / m3
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?
Solution
1. Find the pressure at point A, where
the depth of the water is h1
PA  Pat  water gh1
2. Find the pressure at point B, where
the depth of the water is h2 = 5.00 cm
PB  Pat  oil gh2
3. Set PA equal PB:
Pat  water gh1  Pat  oil gh2
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
  oil 

h1  h2 
  water 
 9.20102 kg / m3 
  4.60 cm
 5.00 cm
3
3 
 1.0010 kg / m 
h  h2  h1  5.00 cm  4.60 cm  0.40 cm
Pascal’s Principal
An external pressure applied to an enclosed fluid is
transmitted unchanged to every point within the fluid.
P  F1 / A1
F2  P A2
 A2 
 F2  F1 / A1 A2  F1    F1
 A1 
A1d1  A2 d 2
 A1 
d 2  d1  
 A2 
A hydraulic lift
End of Lecture 13
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