Introduction to Fluid
Mechanics.
Chapter 1.
Phys 404
Dr Nazir Mustapha
1
Al Imam University
College of Sciences
Dr Nazir Mustapha – 323-223-C-1
Sunday
1
2
3
4
Phys 404
(291)
Phys 404
(291)
Phys 404
(291)
Phys 404
(291)
Phys 404
(291)
Phys 404
(291)
Monday
Tuesday
Wednesday
Phys 404
(291)
Phys 404
(291)
Thursday
Phys 404
(291)
Phys 404
(291)
5
6
Exams
- Midterm 1: week: 2
- Midterm 2: week: 4
- Final:
week: 6 – or - 7
Grading:
- Midterm1: 20
- Midterm 2: 20
- Home works, participation &
quizzes: 20
-Final Exam: 40
-Please note that a quiz will be given at the end of
each chapter.
3
PHYISCS 404 - Chapter 1-2
Physics 404 –Fluid Mechanics
Dr. Nazir Mustapha
Office: SR 104, Third Floor
Phone: 2582167
e-mail: 94@imamm.org
nazirmustapha63@gmail.com
OFFICE HOURS
Sun:10:00 – 11:20am,
LECTURE HOURS
Sun & Wed: 1,2,3 &4 and Thurs: 1, 2
SCHEDULE
Lectures are on: Sunday, Wednesday and Thursday.
4
Textbooks:
1. Yunus A. Çengel and John M. Cimbala.
Fluid
Mechanics:
Fundamentals
and
Applications.Second Edition, McGraw Hill, 2010.
2.Fox, McDonald & Pritchard. Introduction to
Fluid Mechanics. 5th Edition, Wiley, 2004
5
Mechanical properties of Matter and
fluids.
Contents:
• Definition of a fluid
• Density,
• Pressure,
• Fluid flow,
• Pascal Law
• Boyle’s Law
• Bernoulli's equation and its application,
6
Introduction
• Fluid mechanics is a study of the behavior of
fluids, either at rest (fluid statics) or in motion
(fluid dynamics).
• The analysis is based on the fundamental laws
of mechanics, which relate continuity of mass
and energy with force and momentum.
• An understanding of the properties and
behavior of fluids at rest and in motion is of
great importance in Physics.
7
1.1 Definition of Fluid
 A fluid is a substance, which deforms
continuously, or flows, when subjected to
shearing force.
 In fact if a shear stress is acting on a fluid it
will flow and if a fluid is at rest there is no
shear stress acting on it.
Fluid Flow
Shear stress – Yes
Fluid Rest
Shear stress – No
8
Definition of a Fluid
A fluid is a substance that flows under the action of shearing
forces ‫ قوى القص‬. If a fluid is at rest, we know that the forces
on it are in balance.
A gas is a fluid that is easily compressed. It fills any vessel in
which it is contained.
A liquid is a fluid which is hard to compress. A given mass of
liquid will occupy a fixed volume, irrespective of the size of
the container.
A free surface is formed as a boundary between a liquid and a
gas above it.
9
9
Definition of a Fluid
• “a fluid, such as water or air, deforms
continuously when acted on by shearing
stresses of any magnitude.”
- Munson, Young, Okiishi
Water
Oil
Air
Why isn’t steel a fluid?
10
Fluid Deformation between
Parallel Plates
F
U
b
Side view
Force F causes the top plate to have velocity U.
What other parameters control how much force is
required to get a desired velocity?
Distance between plates (b)
Area of plates (A)
Viscosity!
11
Shear stress in moving fluid
• If fluid is in motion, shear stress are developed if the particles of the
fluid move relative to each other. Adjacent particles have different
velocities, causing the shape of the fluid to become distorted
• On the other hand, the velocity of the fluid is the same at every
point, no shear stress will be produced, the fluid particles are at rest
relative to each other.
Shear force
Moving plate
Fluid particles
New particle position
Fixed surface
12
Shearing Forces
13
Fluid Mechanics
•Liquids and gases have the ability to flow.
•They are called fluids.
•There are a variety of “LAWS” that fluids obey.
•Need some definitions.
14
Differences between liquid and gases.
Liquid
Gases
Difficult to compress and often
regarded as incompressible.
Easily to compress – changes of volume
is large, cannot normally be neglected
and are related to temperature.
Occupies a fixed volume and will
take the shape of the container.
No fixed volume, it changes volume to
expand to fill the containing vessels.
A free surface is formed if the
volume of container is greater
than the liquid.
Completely fill the vessel so that no free
surface is formed.
15
Pressure in a Fluid
•The pressure is just the weight of all the fluid above
you.
•Atmospheric pressure is just the weight of all the air above
an area on the surface of the earth.
•In a swimming pool the pressure on your body surface is
just the weight of the water above you (plus the air pressure
above the water).
16
Pressure in a Fluid
•So, the only thing that counts in fluid pressure is the
gravitational force acting on the mass ABOVE you.
•The deeper you go, the more weight above you and the
more pressure.
•Go to a mountaintop and the air pressure is lower.
17
Pressure in a Fluid
Pressure
acts
perpendicular to the
surface and increases
at greater depth.
18
Pressure in a Fluid
19
Buoyancy
Net upward
force is called
the buoyant
force!!!
Easier to lift a
rock
in
water!!
20
Displacement of Water
The amount of
water displaced is
equal
to
the
volume of the
rock.
21
Archimedes’ Principle
• An immersed body is buoyed up by a force
equal to the weight of the fluid it displaces.
• If the buoyant force on an object is greater
than the force of gravity acting on the
object, the object will float.
• The apparent weight of an object in a liquid
is gravitational force (weight) minus the
buoyant force.
22
Flotation
• A floating object displaces a weight of fluid
equal to its own weight.
23
Flotation
24
Gases
• The primary difference between a liquid and a
gas is the distance between the molecules.
• In a gas, the molecules are so widely separated,
that there is little interaction between the
individual molecules.
• IDEAL GAS.
• Independent of what the molecules are.
25
Boyle’s Law
26 26
Boyle’s Law
• Pressure depends on density of the gas.
• Pressure is just the force per unit area
exerted by the molecules as they collide
with the walls of the container.
• Double the density, double the number of
collisions with the wall and this doubles the
pressure.
27
Boyle’s Law
Density
is
mass
divided by volume.
Halve the volume and
you double the density
and thus the pressure.
28
Boyle’s Law
• At a given temperature for a given quantity
of gas, the product of the pressure and the
volume is a constant
P1V1  P2 V2
29
Elements that exist as gases at 250C and 1 atmosphere
The Gas Laws
Figure 1.7: The Pressure-Volume
Relationship:Boyle’s Law
The Gas Laws
The Pressure-Volume Relationship:
Boyle’s Law
• Mathematically:
1
V  constant 
P
PV  constant
P1  V1  P2  V2
• A sample of gas contained in a flask with a volume of
1.53 L and kept at a pressure of 5.6×103 Pa. If the
pressure is changed to 1.5×104 Pa at constant
temperature, what will be the new volume? (in-class
example).
Boyle’s Law
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
A sample of chlorine gas occupies a volume of 946 mL at
a pressure of 726 mmHg. What is the pressure of the gas
(in mmHg) if the volume is reduced at constant
temperature to 154 mL?
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
Atmospheric Pressure
• Just the weight of the air above you.
• Unlike water, the density of the air
decreases with altitude since air is
compressible and liquids are only very
slightly compressible.
• Air pressure at sea level is about 105
newtons/meter2.
35
Barometers
36
Buoyancy in a Gas
• An object surrounded by air is buoyed up by
a force equal to the weight of the air
displace.
• Exactly the same concept as buoyancy in
water. Just substitute air for water in the
statement.
• If the buoyant force is greater than the
weight of the object, it will rise in the air.
37
Buoyancy in a Gas
Since air gets less
dense with altitude,
the buoyant force
decreases
with
altitude. So helium
balloons don’t rise
forever!!!
38
Bernoulli’s Principle
39
Bernoulli’s Principle.
• Flow is faster when the pipe is narrower.
• Put your thumb over the end of a garden hose.
• Energy conservation requires that the pressure be
lower in a gas that is moving faster.
• Has to do with the work necessary to compress a
gas (PV is energy, more later).
40
Bernoulli’s Principle
• When the speed of a fluid increases,
internal pressure in the fluid decreases.
41
Bernoulli’s Principle
42
Bernoulli’s Principle
Why the streamlines are compressed is
quite complicated and relates to the air
boundary layer, friction and turbulence.
43
Bernoulli’s Principle
44
Bernoulli’s Equation
45
Bernoulli’s Equation
A fluid flowing through a constricted
pipe with streamline flow. The fluid
in the section of length Δx1 moves to
the section of length Δx2 . The
volumes of fluid in the two sections
are equal.
The speed of water
spraying from the end of a
hose increases as the size of
the opening is decreased
with thumb.
46
Bernoulli’s Equation
• The sum of the pressure, kinetic energy per
unit volume, and gravitational potential
energy per unit volume has the same value at
all points along a streamline.
• This result is summarized in Bernoulli’s
equation:
• P + 1/2 ρ v2 + ρ gy = constant
4747
The Venturi tube
The horizontal constricted pipe illustrated in figure known as a Venturi tube, can be used
to measure the flow speed of an incompressible fluid. Let us determine the flow speed
at point 2 if the pressure difference P1 ̶ P2 is known.
Solution: Because the pipe is horizontal,
y1 = y2, and applying Bernoulli’s equation:
Figure 15: Venturi Tube
(a) Pressure P1 is greater Than the pressure
P2 since v1 < v2. This device can be used
to measure the speed of fluid flow.
4848
Bernoulli’s Equation
For steady flow, the speed, pressure, and elevation of an
incompressible and nonviscous fluid are related by an
equation discovered by Daniel Bernoulli (1700–1782).
Figure 16
4949
Bernoulli’s Equation
In the steady flow of a nonviscous,
incompressible fluid of density ρ,
the pressure P, the fluid speed v,
and the elevation y at any two
points (1 and 2) are related by:
Figure 17:
5050
Bernoulli’s Equation
A fluid flowing through a
constricted
pipe
with
streamline flow. The fluid in
the section of length Δx1 moves
to the section of length Δx2 .
The volumes of fluid in the two
sections are equal.
The speed of water spraying from the
end of a hose increases as the size of
the opening is decreased with thumb.
5151
Bernoulli’s Equation
A fluid can also change its
height. By looking at the work
done as it moves, we find:
This is Bernoulli’s equation.
One thing it tells us is that as
the speed goes up, the
pressure goes down.
52
53
Fluids at rest
54
Stress on an object submerged in a static fluid
• The force exerted by a
static fluid on an object is
always perpendicular to the
surfaces of the object.
• The force exerted by the
fluid on the walls of the
container is perpendicular
to the walls at all points.
55
Density
The density of a fluid is defined as its mass per unit
volume. It is denoted by the Greek symbol,  .
= m
kgm-3
V
kg
m3
 water= 998 kgm-3
air =1.2kgm-3
If the density is constant (most liquids), the flow is
incompressible.
If the density varies significantly (e.g. some gas flows), the
flow is compressible.
(Although gases are easy to compress, the flow may be treated as incompressible
if there are no large pressure fluctuations)
56
Pressure
Pressure is the force per unit area, where the force is
perpendicular to the area.
Nm-2
(Pa)
p=
F
A
N
m2
pa= 105 Nm-2
1psi =6895Pa
This is the Absolute pressure, the pressure compared to
a vacuum.
The pressure measured in your tyres is the gauge pressure,
p-pa. 1 Pa = 1 N/m2, in the SI system.
57
Figure 1.2: Pressure
Atmosphere Pressure and the
Barometer
F
P
A
•Standard atmospheric
pressure = 760 mm of Hg
•1 atm = 760 mmHg = 760 torr = 101.325 kPa.
Force
Pressure = Area
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
Barometer
10 miles
4 miles
Sea level
0.2 atm
0.5 atm
1 atm
Absolute Pressure
61
Example: (a) What are the total force and the absolute pressure
on the bottom of a swimming pool 22.0 m by 8.5 m whose
uniform depth is 2.0 m? (b) What will be the pressure against
the side of the pool near the bottom?
(a)The absolute pressure is given by P=P0+ρgh, and the total
force is the absolute pressure times the area of the bottom of
the pool.

P  P0   gh  1.013  10 N m  1.00  10 kg m
5
2
3
3
9.80 m s   2.0 m 
2
 1.21 105 N m 2


F  PA  1.21 105 N m 2  22.0 m  8.5 m   2.3  107 N
62
(b) The pressure against the side of the pool,
near the bottom, will be the same as the
pressure at the bottom,
P  1.21 10 N m
5
2
63
Uniform Pressure
Uniform Pressure:
If the pressure is the same at all points on a
surface.
64
Properties of Fluids
Density: SI Units: Kg/m3
65
Properties of fluids
66
Properties of fluids
67
Blood as a fraction of body weight
• The body of a man whose weight is 690 N contains
about 5.2 × 10 ̶ 3 m3 of blood.
• (a) Find the blood’s weight and?
• (b) Express it as percentage of the body weight?
( in class example).
68
A simple device for measuring the
pressure exerted by a fluid
• The device consists of an
evacuated cylinder that
encloses a light piston
connected to a spring.
• As the device is submerged
in a fluid, the fluid presses
on the top of the piston and
compresses the spring until
the inward force exerted by
the fluid is balanced by the
outward force exerted by the
spring.
69
Force and Pressure
• dF =P dA
• Where P is the pressure
at the location of the
area dA.
70
Example1: The water bed
• The mattress of a water bed is 2m long by 2m wide and 30 cm
deep.
• (a) Find the volume of the mattress?
• V = (2m)(2m)(0.3m) = 1.2 m3
• (b) Find the weight of the water in the mattress?
• M =  V= (1000 kg/m3)(1.2m3) = 1.2 x 103 kg
• and its weight is: Mg = (1.2 x 103 kg)( 9.80m/s2) = 1.18x 104N
• (c) Find the pressure exerted by the water on the floor when
the bed rests in its normal position. Assume that the entire
lower surface of the bed makes contact with the floor?
71
The water Bed
• When the bed is in its normal position, the area in contact with
the floor is: 2 x 2 = 4 m2 ; thus we find that:
• (d) What if the water bed is replaced by a 50 kg ordinary bed
that is supported by four legs? Each leg has a circular cross
section of radius 2 cm. what pressure does this bed exert on
the floor? HW
72
73
Variation of Pressure with depth
Pressure in a fluid acts equally in all directions
Pressure in a static liquid increases linearly with depth
p= g  h
pressure
increase
increase in
depth (m)
The pressure at a given depth in a continuous, static body of
liquid is constant.
p1
p2
p3
p1 = p2 = p3
74
Problem:
In a movie, Tarzan evades his captors by hiding underwater for
many minutes while breathing through a long, thin reed.
Assuming the maximum pressure difference his lungs can
manage and still breathe is calculate the deepest he could have
been.
Solution: The pressure difference on the lungs is the pressure
change from the depth of water
P   g h  h 
P
g

 133 N m 2 
 85 mm-Hg  

1
mm-Hg


1.00 10
3
kg m
3
 9.80 m s 
2
 1.154 m  1.2 m
75
A parcel of fluid (darker region) in a larger volume of fluid is
singled out. The net force exerted on the parcel of fluid must be
zero because it is in equilibrium.
76
Exercises: A Pain in your Ear
•
Estimate the force exerted on your
eardrum due to the water above
when you are swimming at the
bottom of a pool is 5 m deep.
• g = 9.8 m/s2 (SI)
• ρwater = 1000 kg / m3
•
Solution: We estimate the surface
area of the eardrum to be
approximately:
•
•
1cm2 = 1x10-4 m2.
The air inside the middle ear is
normally at atmospheric pressure P0.
•
The difference between the total
pressure at the bottom of the pool and
atmospheric pressure:
Pbot – P0 = ρ
gh
= (1000 kg / m3)(9.8 m/s2 )(5m)
= 4.9 × 104 Pa
The force on the eardrum:
F =( Pbot – P0 ) A = 5N
•
77
Example: A pain in the Ear.
b) We estimate the surface area of the eardrum to be approximately:
1 cm2 = 1 × 10 ̶ 4 m2, This means that the force on it is:
F = ( Pbot – P0) A =( 4.9 × 10 4 pa) (1×10 ̶ 4 cm2) = 5 N
Because a force on the eardrum of this magnitude
is extremely uncomfortable, swimmers “pop their
ears” while under water, an action that pushes air
from the lungs into the middle ear. Using this
technique equalizes the pressure on the two sides
of the eardrum and relieves the discomfort.
78
Ideal Gas Law
• Gases are highly compressible in comparison to
liquids, with changes in gas density directly related to
changes in pressure and temperature through the
equation:
– P =  RT
• where
– P is the absolute pressure,
–
is the gas density
– R is the gas constant
– Ru is the universal gas constant,
– T is the absolute temperature
• SI unit of the temperature is the Kelvin scale.
79
Density Form of Ideal Gas Law
P=  RT
Ideal gas Equation
T(K) = T(0C) + 273.15
The gas constant R is different for each gas.
Many gases can be treated as ideal gases :
•Air – nitrogen, oxygen, hydrogen, helium, argon,
neon, krypton and carbon dioxide.
•Dense gases such as water vapor in steam power
plants and refrigerant vapor in refrigerators should not
be treated as ideal gases.
81
Density, Specific Gravity, and Mass of Air
in a room
Determine the density, specific gravity, and mass of the air in a
room whose dimensions are 4 m × 5 m × 6 m at 100 kPa and 25
0C.
Solution: The density, specific gravity, and mass of the air in a
room are to be determined.
Assumptions At specific conditions, air can be treated as an
ideal gas.
Properties The gas constant of air is R = 0.287 kPa. m3/ kg. K.
Analysis The density of the air is determined from the ideal-gas
relation:
P=  RT
82
Density, Specific Gravity, and
Mass of Air in a room
P
100kPa


RT (0.287kPa.m3 / kg.K )( 25  273.15) K
  1.17kg / m
3
83
Density, Specific Gravity, and Mass of Air
in a room
Then the specific gravity of the air becomes:
 air
1.17 kg / m
SG 

 0.00117
 water 1000kg / m3
3
84
Density, Specific Gravity, and
Mass of Air in a room
Finally, the volume and the mass of the air in the room
are:
Volume: V = (4 m) (5 m)(6 m) = 120 m3
mass:
m = ρ V = (1.17 kg/m3)(120 m3) = 140 kg.
Discussion: Note that that we converted the temperature
to the unit K from 0C before using it in the ideal-gas
relation.
85
General Characteristics of Gases
•
•
•
•
Highly compressible.
Occupy the full volume of their containers.
When gas is subjected to pressure, its volume decreases.
Gases always form homogeneous mixtures with other
gases.
• Gases only occupy about 0.1 % of the volume of their
containers.
Four Physical Quantities for
Gases
Phys.
Qty.
pressure
Other common
Symbol SI unit
units
Pascal atm, mm Hg, torr,
P
(Pa)
psi
volume
V
m3
dm3, L, mL, cm3
temp.
T
K
°C, °F
moles
n
mol
Diagram of a hydraulic press
Because the increase in pressure is the same on the
two sides, a small force F1 at the left produces a
much greater force F2 at the right.
A vehicle undergoing repair is
supported by a hydraulic lift in a
garage.
88
Pascal Law
French Scientist Blaise Pascal
( 1623 – 1662)
Pascal ‘s Law: a change in the pressure applied to a
fluid is transmitted undiminished to every point of
the fluid and to the walls of the container.
89
hydraulic press
• In the hydraulic press illustrated before:
90
Hydrostatic Application: Transmission of Fluid Pressure
• Mechanical advantage can be
gained with equality of
pressures
• A small force applied at the
small piston is used to
develop a large force at the
large piston.
• This is the principle between
hydraulic jacks, lifts, presses,
and hydraulic controls
A2
F2 
F1
A1
A1
F1 
F2
A2
91
Example: the car lift
92
Measuring pressure (1)
Manometers
p1
p2=pa
x
liquid
density 
h
p1 = px
(negligible pressure
change in a gas)
px = py
(since they are at
the same height)
z
pz= p2 = pa
y
py - pz = gh
p1 - pa = gh
So a manometer measures gauge pressure.
93
Figure 1.3: Pressure
• Closed Systems => manometers
Manometer, how it works?
)‫مقياس ضغط ( الغاز أو السائل‬
• Manometer is an instrument used to measure the
pressure of a gas or vapor. There are several types of
manometers. The simplest kind consists of a Ushaped tube with both ends open. The tube contains a
liquid, often mercury, which fills the bottom of the U
and rises in each of the arms. The person using this
type connects one of the arms to the gas whose
pressure is to be measured. The other arm remains
open to the atmosphere. In this way, the liquid is
exposed to the pressure of the gas in one arm and
atmospheric pressure in the other.
95
Measuring Pressure (2)
Barometers
A barometer is used to measure
the pressure of the atmosphere.
The simplest type of barometer
consists of a column of fluid.
p2 - p1 = gh
pa = gh
vacuum
p1 = 0
h
p2 = pa
examples
water: h = pa/g =105/(103*9.8) ~10m
mercury: h = pa/g =105/(13.4*103*9.8)
~800mm
96
Measurement of Pressure: Barometers
Evangelista Torricelli (1608-1647)
The first mercury barometer was
constructed in 1643-1644 by
Torricelli.
He showed that the height of
mercury in a column was 1/14
that of a water barometer, due to
the fact that mercury is 14 times
more dense that water.
He also noticed that level of mercury
varied from day to day due to
weather changes, and that at the
top of the column there is a
vacuum.
97
Measuring Pressure
• Two devices for measuring
pressure:
• a) an open-tube manometer
and
• b) a mercury barometer.
98
99
Archimedes’ Principle
• An immersed body is buoyed up by a force
equal to the weight of the fluid it displaces.
• If the buoyant force on an object is greater
than the force of gravity acting on the object,
the object will float.
• The apparent weight of an object in a liquid is
gravitational force (weight) minus the buoyant
force.
100
Buoyancy
• Net upward force is
called the buoyant
force!!!
• Easier to lift a rock in
water!!
101
Displacement of Water
• The amount of water
displaced is equal to the
volume of the rock.
102
Flotation
• A
floating
object
displaces a weight of
fluid equal to its own
weight.
103
Flotation
104
Buoyant Forces and Archimedes’s
Principle
• The upward force exerted by water on any
immersed object is called the Buoyant force.
• We can determine the magnitude of a
buoyant force by applying some logic and
Newton’s second Law.
• Archimedes’s Principle States that the
magnitude of the Buoyant Force always
equals the weight of the Fluid displaced by
the object.
105
Buoyant Forces and Archimedes’s
Principle
• The buoyant force acting on the steel cube is
the same as the buoyant force acting on a
cube of liquid of the same dimensions.
Buoyant Forces and Archimedes’s
Principle
The external forces acting on
the cube of liquid are:
1- The force of gravity Fg and
2-The buoyant force FB.
Under equilibrium conditions:
FB = Fg
h: height of the cube
Buoyant Force:
• FB = mg
• V is the volume of the cube.
• Or FB = Vρ g
•
• FB = mg = Vρ g
• ΔP = FB/A
• FB = (ΔP)A = (ρ g h)A
• FB
= ρ gV
mg is the weight of the fluid
in the cube.
• The mass of the fluid in the
cube is m= ρ V.
• The pressure difference ΔP
between the bottom and
top faces of the cube is
equal to the buoyant force
per unit area of those faces.
Example of a buoyant force.
•Hot-air balloons. Because
hot air is less dense than
cold air.Then:
•A net upward force acts
on the balloons.
Buoyant Forces and Archimedes’s
Principle
• (a) A totally submerged
object that is less dense
than the fluid in which
it
is
submerged
experiences
a
net
upward force.
• (b) A totally submerged
object that is denser
than the fluid sinks.
Buoyant Forces and Archimedes’s
Principle
Quiz:
• A glass of Water contains a
single floating ice cube. When
the ice melts, does the water
level go up, go down, or remain
the same?
• When a person in a rowboat in a
small pond throws an anchor
overboard, does the water level
of the pond go up, go down, or
remain the same?
Exercises: A Pain in your Ear
•
Estimate the force exerted on your
eardrum due to the water above
when you are swimming at the
bottom of a pool is 5 m deep.
• g = 9.8 m/s2 (SI)
• ρwater = 1000 kg / m3
•
Solution: We estimate the surface
area of the eardrum to be
approximately:
•
•
1cm2 = 1×10 ̶ 4 m2.
The air inside the middle ear is
normally at atmospheric pressure P0.
•
The difference between the total
pressure at the bottom of the pool and
atmospheric pressure:
Pbot – P0 = ρ
gh
= (1000 kg / m3)(9.8 m/s2 )(5m)
= 4.9 x 104 Pa
The force on the eardrum:
F =( Pbot – P0 ) A = 5N
•
Newtonian and Non-Newtonian Fluid
Fluid
obey
Newton’s law
of viscosity
refer
Newton’s’ law of viscosity is given by;
du

dy
(1.1)

= shear stress

= viscosity of fluid
du/dy = shear rate, rate of strain or velocity gradient
Newtonian fluids
Example:
Air
Water
Oil
Gasoline
Alcohol
Kerosene
Benzene
Glycerine
• The viscosity  is a function only of the condition of the fluid, particularly its
temperature.
• The magnitude of the velocity gradient (du/dy) has no effect on the magnitude of .
113
Velocity gradient
114
So:
115
116
Viscosity
117
118
Newtonian and Non-Newtonian
Fluid
Non- Newtonian
Newton’s law
Fluid
Do not obey
of viscosity
fluids
• The viscosity of the non-Newtonian fluid is dependent on the
velocity gradient as well as the condition of the fluid.
Newtonian Fluids
 a linear relationship between shear stress and the velocity gradient (rate
of shear),
 the slope is constant
 the viscosity is constant
non-Newtonian fluids
 slope of the curves for non-Newtonian fluids varies
119
Figure 1.1
Shear stress vs. velocity gradient
Bingham plastic : resist a small shear stress but flow easily under large shear
stresses, e.g. sewage sludge, toothpaste, and jellies.
Pseudo plastic : most non-Newtonian fluids fall under this group. Viscosity
decreases with increasing velocity gradient, e.g. colloidal
substances like clay, milk, and cement.
Dilatants
: viscosity decreases with increasing velocity gradient, e.g.
quicksand.
98
Viscosity
 Viscosity is the resistance that a fluid offers to
flow when subject to a shear stress.
Isaac Newton: “the resistance
which arises from the lack of
slipperiness of the parts of a fluid,
other things being equal, is
proportional to the velocity with
which the parts of the liquid are
separated from each other”
121
SI. Units
Primary Units:
Quantity
SI Unit
Length
Metre, m
Mass
Kilogram, kg
Time
Seconds, s
Temperature
Kelvin, K
Current
Ampere, A
Luminosity
Candela
In fluid mechanics we are generally only interested in the top four units from this table.
122
Derived Units
Quantity
SI Unit
velocity
m/s
-
acceleration
m/s2
-
force
Newton (N)
N = kg.m/s2
energy (or work)
Joule (J)
J = N.m = kg.m2/s2
power
Watt (W)
W = N.m/s = kg.m2/s3
pressure (or stress)
Pascal (P)
P = N/m2 = kg/m/s2
density
kg/m3
-
specific weight
N/m3 = kg/m2/s2
N/m3 = kg/m2/s2
relative density
a ratio (no units)
dimensionless
viscosity
N.s/m2
N.s/m2 = kg/m/s
surface tension
N/m
N/m = kg/s2
123
Fluid Properties (Continue)
Specific weight:
Specific weight of a fluid, 
• Definition: weight of the fluid per unit volume
• Arising from the existence of a gravitational force
• The relationship  and g can be found using the following:
Since
therefore
 = m/V
 = g
(1.3)
Units: N/m3
Typical values:
Water = 9814 N/m3;
Air = 12.07 N/m3
124
Fluid Properties (Continue)
Specific gravity
The specific gravity (or relative density) can be defined in two ways:
Definition 1: A ratio of the density of a substance to the density
of water at standard temperature (4C) and
atmospheric pressure, or
Definition 2: A ratio of the specific weight of a substance to the
specific weight of water at standard temperature
(4C) and atmospheric pressure.
SG 
s
w @ 4C

s
 w @ 4C
(1.4)
Unit: dimensionless.
125
Fluid Properties (Continue)
Viscosity:
• Viscosity, , is the property of a fluid, due to cohesion and
interaction between molecules, which offers resistance to shear
deformation.
• Different fluids deform at different rates under the same shear
stress. The ease with which a fluid pours is an indication of its
viscosity. Fluid with a high viscosity such as syrup deforms
more slowly than fluid with a low viscosity such as water. The
viscosity is also known as dynamic viscosity.
Units: N.s/m2 or kg/m/s
Typical values:
Water = 1.14 ×10 ̶ 3 kg/m/s;
Air = 1.78 ×10 ̶ 5 kg/m/s
126
Properties of fluids
Kinematic viscosity, 
Definition: is the ratio of the viscosity to the density;
  /
• will be found to be important in cases in which significant
viscous and gravitational forces exist.
Units: m2/s
Typical values:
Water = 1.14x10-6 m2/s;
Air = 1.46x10-5 m2/s;
In general,
viscosity of liquids with temperature, whereas
viscosity of gases with
in temperature.
127
Bulk Modulus
 All fluids are compressible under the application of an external
force and when the force is removed they expand back to their
original volume.
 The compressibility of a fluid is expressed by its bulk modulus of
elasticity, K, which describes the variation of volume with change
of pressure, i.e.
K
change in pressure
volumetric strain
 Thus, if the pressure intensity of a volume of fluid, , is increased
by Δp and the volume is changed by Δ, then
p
K
 / 
p
K

Typical values:Water = 2.05x109 N/m2;
Oil = 1.62x109 N/m2
128
Vapor Pressure
 A liquid in a closed container is subjected to a partial
vapor pressure in the space above the liquid due to the
escaping molecules from the surface;
 It reaches a stage of equilibrium when this pressure
reaches saturated vapor pressure.
 Since this depends upon molecular activity, which is a
function of temperature, the vapor pressure of a fluid
also depends on its temperature and increases with it.
 If the pressure above a liquid reaches the vapor pressure
of the liquid, boiling occurs; for example if the pressure
is reduced sufficiently boiling may occur at room
temperature.
129
Example 1.2
•A reservoir of oil has a mass of 825 kg. The reservoir has a
volume of 0.917 m3. Compute the density, specific weight,
and specific gravity of the oil.
•Solution:
 oil 
 oil
mass
m
825
 
 900kg / m 3
volume  0.917
weight
mg


 g  900x9.81  8829 N / m 3
volume

SG oil 
 oil
 w @ 4 C
900

 0.9
1000
130
End of Chapter 1.
131
Chapter 2: Forces in Static Fluids
This section will study the forces acting on or generated by
fluids at rest.
Objectives:
• Introduce the concept of pressure.
• Prove it has a unique value at any particular elevation.
• Show how it varies with depth according to the
hydrostatic equation and
• Show how pressure can be expressed in terms of head of
fluid.
This understanding of pressure will then be used to
demonstrate methods of pressure measurement that will be
useful later with fluid in motion and also to analyse the
forces on submerges surface/structures.
Chapter 2: Forces in Static
Fluids
Objectives:
•Define and apply the concepts of density
and fluid pressure to solve physical
problems.
•Define and apply concepts of absolute,
gauge, and atmospheric pressures.
•State Pascal’s law and apply for input and
output pressures.
•State and apply Archimedes’ Principle to
solve physical problems.
133
Mass Density
mass
m
Density 
; 
volume
V
Wood
Lead: 11,300 kg/m3
2 kg, 4000 cm3
Wood:
500 kg/m3
4000 cm3
Lead
45.2 kg
Same volume
Lead
177 cm3
2 kg
Same mass
134
Example 1: The density of steel is 7800
kg/m3. What is the volume of a 4-kg block of
steel?
m
 ;
V
m
4 kg
V 
 7800 kg/m3
4 kg
V = 5.13 × 10 ̶ 4 m3
What is the mass if the volume is 0.046 m3?
m  V  (7800 kg/m3 )(0.046 m3 );
m = 359 kg
135
Relative Density
The relative density r of a material is the ratio of
its density to the density of water (1000 kg/m3).
r 
x
1000 kg/m3
Examples:
Steel (7800 kg/m3)
r = 7.80
Brass (8700 kg/m3)
r = 8.70
Wood (500 kg/m3)
r = 0.500
136
Pressure
Pressure is the ratio of a force F to the area A
over which it is applied:
Force
Pressure 
;
Area
A = 2 cm2
1.5 kg
F
P
A
F (1.5 kg)(9.8 m/s 2 )
P 
A
2 x 10-4 m 2
P = 73,500 N/m2
137
Pressure
Pressure is the ratio of a force F to the area A
over which it is applied:
Force
Pressure 
;
Area
A = 2 cm2
1.5 kg
F
P
A
F (1.5 kg)(9.8 m/s 2 )
P 
A
2 x 10-4 m 2
P = 73,500 N/m2
138
Fluid Pressure
A liquid or gas cannot sustain a shearing stress - it is
only restrained by a boundary. Thus, it will exert a
force against and perpendicular to that boundary.
• The force F exerted by a
fluid on the walls of its
container always acts
perpendicular to the walls.
Water flow
shows  F
139
Fluid Pressure
Fluid exerts forces in many directions. Try to submerse
a rubber ball in water to see that an upward force acts
on the float.
• Fluids exert pressure in
all directions.
F
140
Pressure vs. Depth in Fluid
Pressure = force/area
mg
P
;
A
P
m  V ; V  Ah
Vg
A

 Ahg
A
• Pressure at any point in a
fluid is directly proportional
to the density of the fluid
and to the depth in the fluid.
h
Area mg
Fluid Pressure:
P = gh
141
Independence of Shape and Area.
Water seeks its own level,
indicating that fluid pressure
is independent of area and
shape of its container.
• At any depth h below the surface of the water
in any column, the pressure P is the same.
The shape and area are not factors.
142
Properties of Fluid Pressure
• The forces exerted by a fluid on the walls of its
container are always perpendicular.
• The fluid pressure is directly proportional to the
depth of the fluid and to its density.
• At any particular depth, the fluid pressure is the
same in all directions.
• Fluid pressure is independent of the shape or
area of its container.
143
Example 2. A diver is located 20 m below the surface of a
lake (ρ = 1000 kg/m3). What is the pressure due to the water?
The difference in pressure
from the top of the lake to
the diver is:
P = gh
 = 1000 kg/m3
h
h = 20 m; g = 9.8 m/s2
P  (1000 kg/m3 )(9.8 m/s2 )(20 m)
P = 196000 Pa = 196 kPa
144
Atmospheric Pressure
One way to measure atmospheric
pressure is to fill a test tube with
mercury, then invert it into a
bowl of mercury.
Density of Hg = 13,600 kg/m3
Patm = gh
P=0
atm atm
h
Mercury
h = 0.760 m
Patm = (13,600 kg/m3)(9.8 m/s2)(0.760 m)
Patm = 101,300 Pa = 1.013 × 105 Pa
145
Absolute Pressure
Absolute Pressure: The sum of the
pressure due to a fluid and the
pressure due to atmosphere.
Gauge Pressure: The difference
between the absolute pressure and
the pressure due to the atmosphere:
1 atm = 101.3 kPa
P = 196 kPa
h
Absolute Pressure = Gauge Pressure + 1 atm
P = 196 kPa
1 atm = 101.3 kPa
Pabs = 196 kPa + 101.3 kPa
Pabs = 297 kPa
146
Pascal’s Law
Pascal’s Law: An external pressure applied
to an enclosed fluid is transmitted uniformly
throughout the volume of the liquid.
Fin Ain Fout Aout
Pressure in = Pressure out
Fin Fout

Ain Aout
147
Example 3. The smaller and larger pistons of a
hydraulic press have diameters of 4 cm and 12 cm.
What input force is required to lift a 4000 N weight
with the output piston?
Fin Fout
Fout Ain

; Fin 
Ain Aout
Aout
D
R ;
2
Fin A Fout Aout
in
t
Area   R 2
(4000 N)( )(2 cm) 2
Fin 
 (6 cm)2
Rin= 2 cm; R = 6 cm
F = 444 N
148
Archimedes’ Principle
• An object that is completely or partially submerged in
a fluid experiences an upward buoyant force equal to
the weight of the fluid displaced.
2 kg
2 kg
The buoyant force is due
to the displaced fluid.
The block material
doesn’t matter.
149
Calculating Buoyant Force
The buoyant force FB is due to
the difference of pressure P
between the top and bottom
surfaces of the submerged block.
Area
FB
P 
 P2  P1 ; FB  A( P2  P1 )
A
FB  A( P2  P1 )  A(  f gh2   f gh1 )
FB  (  f g ) A(h2  h1 ); V f  A(h2  h1 )
Vf is volume of fluid displaced.
FB
mg
h1
h2
Buoyant Force:
FB = f gVf
150
Example 4: A 2-kg brass block is attached to a string and
submerged underwater. Find the buoyant force FB and the
tension in the rope T.
All forces are balanced:
FB + T = mg
b 
FB = wgVw
mb
m
2 kg
; Vb  b 
Vb
b 8700 kg/m3
Vb = Vw = 2.30 x 10-4 m3
Fb = (1000 kg/m3)(9.8 m/s2)(2.3 x 10-4 m3)
FB = 2.25 N
T
FB = gV
Force
diagram
mg
151
Example 4 (Cont.): A 2-kg brass block is attached
to a string and submerged underwater. Now find
the tension in the rope. (T=?)
FB = 2.25 N
FB + T = mg
T = mg - FB
T = (2 kg)(9.8 m/s2) - 2.25 N
T = 19.6 N - 2.25 N
T = 17.3 N
This force is sometimes referred to as
the apparent weight.
T
FB = gV
Force
diagram
mg
152
Floating objects:
When an object floats, partially submerged, the buoyant
force exactly balances the weight of the object.
FB
FB = f gVf
mx g = xVx g
f gVf = xVx g
mg
Floating Objects:
If Vf is volume of displaced
water Vwd, the relative density
of an object x is given by:
f Vf = xVx
Relative Density:
 x Vwd
r 

 w Vx 153
Example 5: A student floats in a salt lake with one-third of his
body above the surface. If the density of his body is 970 kg/m3,
what is the density of the lake water?
Assume the student’s volume is 3 m3.
Vs = 3 m3; Vwd = 2 m3; s = 970 kg/m3
w Vwd = sVs
 s Vwd 2 m3
3 s


; w 
3
 w Vs 3 m
2
3 s 3(970 kg/m3 )
w 

2
2
1/3
2/3
w = 1460 kg/m3
154
Problem Solving Strategy
1. Draw a figure. Identify givens and what is to be
found. Use consistent units for P, V, A, and .
2. Use absolute pressure Pabs unless problem
involves a difference of pressure P.
3. The difference in pressure P is determined by the
density and depth of the fluid:
m
F
P2  P1   gh;  = ; P =
V
A
155
Problem Strategy (Cont.)
4. Archimedes’ Principle: A submerged or floating
object experiences an buoyant force equal to the
weight of the displaced fluid:
FB  m f g   f gV f
5. Remember: m, ρ and V refer to the displaced
fluid. The buoyant force has nothing to do with
the mass or density of the object in the fluid. (If
the object is completely submerged, then its
volume is equal to that of the fluid displaced.)
156
Problem Strategy (Cont.)
6. For a floating object, FB is
equal to the weight of that
object; i.e., the weight of the
object is equal to the weight of
the displaced fluid:
mx g  m f g
or
FB
mg
 xVx   f V f
157
Summary
mass
m
Density 
; 
volume
V
Force
Pressure 
;
Area
F
P
A
r 
x
1000 kg/m3
Fluid Pressure:
P = gh
Pascal: 1 Pa = 1 N/m2
158
Summary (Cont.)
Pascal’s
Law:
Fin Fout

Ain Aout
Archimedes’
Principle:
Buoyant Force:
FB = f gVf
159
2.1 Fluids Statics
The general rules of statics ( as applied in solid
mechanics) apply to fluids at rest . From earlier
we know that:
• a static fluid can have no shearing force acting
on it, and that
• any force between the fluid and the boundary
must be acting at right angles to the boundary.
Force normal to the boundary
Pressure Force normal to the boundary
•This statement is also true for curved surfaces, in this
case the force acting at any point is normal to the
surface at that point.
•This statement is also true for any imaginary plane in
a static fluid
•We use this fact in our analysis by considering
elements of fluid bounded by imaginary
Planes.
We also know that:
•For an element of fluid at rest, the element
will be in equilibrium – the sum of the
components of forces in any direction will
be zero.
• The sum of the moments of forces on the
element about any point must also be zero.
2.2. Pressure
• If the force exerted on each unit area of a boundary is
the same, the pressure is said to be uniform.
Example
Pressure and manometers
exercise 1.1
Pressure and manometers
exercise 1.2
Pressure and manometers
exercise 1.3
Exercise 1.3
Exercise 1.4
Exercise 1.5
End of Chapter 2
172