106 phys - ch13
Chapter 13
The Mechanics of Nonviscous Fluids
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Dr. Abdallah M. Azzeer
Three (common) phases of matter:
1. Solid: Maintains shape & size (approx.), even under large forces.
2. Liquid: No fixed shape. Takes shape of container.
3. Gas: Neither fixed shape, nor fixed volume. Expands to fill container.
Lump 2. & 3. into category of FLUIDS.
Fluids: Have the ability to flow.
Fluids
GAS
LIQUID
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106 phys - ch13
Significance
Fluids omnipresent
¾ Weather & climate
¾ Vehicles: automobiles, trains, ships, and planes, etc.
¾ Environment
¾ Physiology and medicine
¾ Sports & recreation
¾ Many other examples!
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Dr. Abdallah M. Azzeer
Weather & Climate
Tornadoes
Thunderstorm
Global Climate
Hurricanes
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106 phys - ch13
Vehicles
Surface ships
Aircraft
Submarines
High-speed rail
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Environment
Air pollution
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River hydraulics
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106 phys - ch13
Physiology and Medicine
Blood pump
Ventricular assist device
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Dr. Abdallah M. Azzeer
Sports & Recreation
Water sports
Cycling
Auto racing
Surfing
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Offshore racing
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106 phys - ch13
Density and Pressure
In fluids, we are interested in properties that can vary
from point to point. Thus, it is more useful to speak of
density and pressure than of mass and force.
m
(uniform density)
V
•Density is a scalar, the SI unit is kg/m3.
ρ=
9
Dr. Abdallah M. Azzeer
TABLE
Some Densities
Density (kg/m3)
Material or Object
Interstellar space
10-20
Best laboratory vacuum
10-17
Air: 20°C and 1 atm pressure
1.21
20°C and 50 atm
Styrofoam
60.5
Ice
Water: 20°C and 1 atm
20°C and 50 atm
0.917 x 103
0.998 x 103
1.000 x 103
Seawater: 20°C and 1 atm
1.024 x 103
Whole blood
Iron
1.060 x 103
7.9 x 103
Mercury (the metal)
Earth: average
core
crust
Sun: average
core
White dwarf star (core)
Uranium nucleus
Neutron star (core)
Black hole (1 solar mass)
13.6 x 103
5.5 x 103
9.5 x 103
2.8 x 103
1.4 x 103
1.6 x 105
1010
3 x 1017
1018
1019
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1 x 102
10
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106 phys - ch13
Pressure
p=
F
(pressure of uniform force on flat area)
A
¾ F is the magnitude of the normal force on area A.
¾ The SI unit of pressure is N/m2 , called the pascal (Pa).
¾ The tire pressure of cars are in kilopascals.
TABLE
1 atm = 1.01 3x 105 Pa = 76 cm Hg = 760 mm Hg
2
= 760 torr = 14.7 lb/in
• The torr (named for Evangelista
Torricelli,
who
invented
the
mercury barometer) is equal to 1
mm of Hg.
Some Pressures
Pressure (Pa)
Center of the Sun
2 x 1016
Center of Earth
Highest sustained laboratory
pressure
4 x 1011
1.5 x 1010
Deepest ocean trench (bottom)
1.1 x 108
Spike heels on a dance floor
1 x 106
Automobile tirea
2 x 105
Atmosphere at sea level
1.0 x 105
Normal blood pressureab
1.6 x 104
Best laboratory vacuum
10-12
Pressure in excess of atmospheric pressure.
The systolic pressure, corresponding to 120 torr
on the physician's pressure gauge.
a
b
Dr. Abdallah M. Azzeer
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Example ;
A living room has floor dimensions of 3.5 m and 4.2 m and a height of 2.4 m.
(a) What does the air in the room weigh when the air pressure is 1.0 atm? (b) What is the
magnitude of the atmosphere's force on the floor of the room?
SOLUTION:
mg = ( ρ V ) g
( Use ρ of air from Table )
3
= ( 1.21 kg / m ) ( 3.5 m x 4.2 m x 2.4 m ) ( 9.8 m / s 2 ) = 418 N ≈ 420 N
This is the weight of about 110 cans of soda.
(b)
⎛ 1 .01 x 10 5 N / m 2
F = p A = ( 1 .0 atm ) ⎜
⎜
1 .0 atm
⎝
) ⎞⎟
( 3 .5 m ) ( 4 .2 m ) = 1 .5 x 10 6 N
⎟
⎠
This enormous force is equal to the weight of the column of air that
covers the floor and extends all the way to the top of the
atmosphere.
(( Need
Need to
to understand
understand the
the Kinetic
Kinetic Theory
Theory of
of gases)
gases)
Dr. Abdallah M. Azzeer
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106 phys - ch13
Fluids at Rest
F2=F1+mg
F1=P1A and F2=P2A
P2A=P1A+ρ gAh
¾ If we choose P1=P0 and h=y1-y2 then we get
P =P0 + ρ g h
(pressure at depth h)
¾ The pressure at a point in a fluid in static
equilibrium depends on the depth of that point but
not on any horizontal dimension of the fluid or its
container.
¾ P is the absolute pressure, P0 is the atmospheric
pressure and ( P-P0 ) is the gauge pressure.
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Dr. Abdallah M. Azzeer
Example ;
The U-tube in Figure contains two liquids in static equilibrium: Water
of density ρw (= 998 kg/m3) is in the right arm, and oil of unknown
density ρx is in the left. Measurement gives l = 135 mm and d = 12.3
mm. What is the density of the oil?
SOLUTION:
We equate the pressure in the two arms at the level of the interface :
Pint = p0 + ρ w g l
( right arm )
Pint = p0 + ρ x g ( l + d ) ( left arm )
1
135 mm
ρ x = ρw
= ( 998 kg / m 3 )
l+d
135 mm + 12.3 mm
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= 915 kg / m 3
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106 phys - ch13
The Mercury Barometer
p0 = ρ g h
For normal atmospheric pressure, h is 76 cm Hg.
Dr. Abdallah M. Azzeer
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The Open Tube Manometer
• The gauge pressure is the difference between the
absolute pressure and the atmospheric pressure.
pg = p A − p0 = ρ g h
• The gauge pressure is directly proportional to h. It
can be positive or negative depending on whether the
absolute pressure is greater or less than the
atmospheric pressure.
• We can suck fluids up a straw because at that time
the absolute pressure in the lungs is less than the
atmospheric pressure.
Dr. Abdallah M. Azzeer
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106 phys - ch13
Example ;
Estimate the force exerted on your eardrum due to the water above when you are
swimming at the bottom of the pool with a depth 5.0 m.
We first need to find out the pressure difference that is being exerted on the eardrum. Then
estimate the area of the eardrum to find out the force exerted on the eardrum.
Since the outward pressure in the middle of the eardrum is the same as normal air pressure
P − P0 = ρ W gh = 1000 × 9 . 8 × 5 . 0 = 4 . 9 × 10 4 Pa
Estimating the surface area of the eardrum at 1.0cm2=1.0x10-4 m2, we obtain
F = (P − P0 )A ≈ 4 . 9 × 10 4 × 1 . 0 × 10 − 4 ≈ 4 . 9 N
Dr. Abdallah M. Azzeer
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Fluid flow
¾ Main difference between a moving fluid and a static fluid is that a moving fluid exerts
a force parallel to a surface.
¾ Steady flow – velocity of fluid “particles” constant as time passes (laminar).
¾ Unsteady flow – the velocity at a point changes as time passes.
™Turbulent flow - extreme kind of unsteady flow. The velocity changes erratically
from moment to moment both in time and direction.
Fluid flow can be compressible or incompressible.
• Most liquids virtually incompressible.
• Most gases are compressible.
(certain situations can be treated as incompressible)
Fluid Flow can be viscous or nonviscous.
Vortex shedding behind cylinder
• Viscous (has large viscosity).
• Water is less viscous (low viscosity) and flows more readily.
Ideal Fluid is incompressible and nonviscous
Dr. Abdallah M. Azzeer
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106 phys - ch13
Fluid flow can be rotational or irrotational
• Rotational flow occurs when fluid has a rotational as well as a translational
motion.
Streamlines
• Streamlines – used in steady fluid flow to represent the trajectories of
the fluid particles
Dr. Abdallah M. Azzeer
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Fluids in Motion
The Equation of Continuity
ΔV = A 1v 1 Δt = A 2v 2 Δt
A 1v 1 = A 2v 2 (equation of continuity)
Q = Av = constant
(volume flow rate, equation of continuity )
¾ The quantity Q is the volume flow rate (volume per unit time). The SI unit is
m3/s.
¾ If the density of the fluid is uniform, the mass flow rate Qm (mass per unit
time) is constant.
Q m = ρ Q = ρ Av = constant
(mass flow rate)
• The SI unit of Q is kg/s
Dr. Abdallah M. Azzeer
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106 phys - ch13
Example
The cross-sectional area A0 of the aorta (the major blood vessel emerging from the heart) of
a normal resting person is 3 cm2, and the speed v0 of the blood through it is 30 cm/s. A
typical capillary (diameter ≈ 6 μm) has a cross-sectional area A of 3 x 10-7 cm2 and a flow
speed v of 0.05 cm/s. How many capillaries does such a person have?
A0 v 0 = nAv
n=
A0 v 0
( 3 cm 2 ) ( 30 cm / s )
=
Av
( 3 x 10 − 7 cm 2 )( 0 .05 cm / s )
= 6 x 109
or
6 billion
The combined cross-sectional area of the capillaries is about 600 times the
cross-sectional area of the aorta.
Dr. Abdallah M. Azzeer
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Example;
Figure shows how the stream of water emerging from a faucet “necks down” as it falls. The
indicated cross-sectional areas are A0 = 1.2 cm2 and A = 0.35 cm2. The two levels are
separated by a vertical distance h = 45 mm. What is the volume flow rate from the tap?
A0v0 = Av
v0 =
=
Q v 2 = v0 2 + 2 g h
2 g h A2
A0 2 − A2
( 2 ) ( 9.8 m / s 2 ) ( 0.045 m ) ( 0.35 cm 2 )2
( 1.2 cm 2 )2 − ( 0.35 cm 2 )2
g
= 0.286 m / s = 28.6 cm / s
The volume flow rate is :
Qv = A0v0 = ( 1.2 cm 2 ) ( 28.6 cm / s ) = 34 cm 3 / s
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106 phys - ch13
Ideal Fluids
¾ Laminar Flow
™No turbulence
¾ Non-viscous
™No friction between fluid layers
¾ Incompressible
™Density is same everywhere
Dr. Abdallah M. Azzeer
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Bernoulli’s Equation
p1 + 1 ρ v1 2 + ρ g y1 = p 2 + 1 ρ v 2 2 + ρ g y 2
2
2
p + 1 ρ v 2 + ρ g y = a constant (Bernoulli ' s equation)
2
• If y1 = y2, then
p1 + 1 ρ v12 = p2 + 1 ρ v 2 2
2
2
• If the speed of a fluid element
increases as it travels along a
horizontal streamline, the pressure of
the fluid must decrease, and
conversely.
• Large speed means small pressure.
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106 phys - ch13
Proof of Bernoulli’s Equation
• Work done on system = (K.E. + P. E.) gained
p 1 = F1Δ x 1 = p 1A 1Δ x 1 = p 1ΔV
p 2 = p 2 A 2 Δ x 2 = p 2 ΔV
• Work done on system at
• Work done by system at
•
ΔV = A 1Δ x 1 = A 2 Δ x 2
( p1 − p2 ) ΔV
ρ (v2 − v12 ) ΔV
• Net work done on system =
1
2
• Net K.E. gained =
2
ρ g ( y2 − y1 )
• Net P.E. gained =
• Therefore,
p1 + 1 ρ v1 2 + ρ g y1 = p 2 + 1 ρ v 2 2 + ρ g y 2
2
2
Dr. Abdallah M. Azzeer
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Example
Ethanol of density ρ = 791 kg/m3 flows smoothly through a horizontal pipe that tapers in
cross-sectional area from A1 = 1.20 x 10-3 m2 to A2 = A1/ 2. The pressure difference between
the wide and narrow sections of pipe is 4120 Pa. What is the volume flow rate QV of the
ethanol?
Qv = v1 A1 = v 2 A2
p1 + 1 ρ v12 + ρ g y = p 2 + 1 ρ v 22 + ρ g y
2
2
Qv
Q
2 Qv
and v 2 = v =
A1
A2
A1
p1 + 1 ρ v12 = p2 + 1 ρ v 22
v1 =
2
2
3Qv2
2( p1 − p2 )
= v 22 − v12 =
ρ
A12
Qv = A 1
2 ( p1 − p2 )
3ρ
The lower speed v1 means that p1 is greater, we have :
Qv = 1.20 x 10 − 3 m 2
( 2 ) ( 4120 Pa )
( 3 ) ( 791kg / m 3 )
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106 phys - ch13
These views of a baseball are from above, looking down toward the ground,
with the ball moving to the right. (a) Without spin, the ball does not curve to
either side. (b) A spinning ball curves in the direction of the deflection force.
(c) The spin in part b causes the ball to curves as shown here.
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Dr. Abdallah M. Azzeer
Torricelli’s equation
A tank is open to atmosphere at the top. What is the velocity of the liquid leaving the
pipe at the bottom (assume ideal fluid).
Apply Bernoullis equation
2
P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22
Since at (1) and (2) P1 = P2 = Patm
ρgh1 + ½ρv12 = ρgh2 + ½ρv22 dividing by density and
since h2 – h1 = h
h
1
v12 = v22 + 2gh
If rate of fall of surface very slow can set v2 = 0.
v1 =
2 gh
Torricelli’s equation
Dr. Abdallah M. Azzeer
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106 phys - ch13
Venturi Tube - application of Bernoullis equation to
measure fluid flow
• Fluid flows through different cross sectional
areas in different portions of the tube.
• As tube narrows (2), velocity of fluid increases
thus dropping pressure
• Velocity of the fluid can be found by
measuring the pressure
h1
h2
v1
v2
Apply Bernoulli’s equation to points at the same height in the flow stream just below the
columns.
P1 +
1
1
ρ v 12 = P 2 + ρ v 22
2
2
A 1v1 = A 2 v 2
⎞
⎛ 2
1
2⎜ A1
− 1⎟
P1 − P 2 = ρ v 1
⎟
⎜ A2
2
⎠
⎝ 2
From the continuity equation:
Hence,
So: v 2 =
A1
v
A2 1
Thus measuring P1-P2 and a knowledge of the areas determines v1
v2 can then be found from the continuity equation
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Dr. Abdallah M. Azzeer
Water circulates throughout a house in a hot-water heating system. If the water is pumped
at a speed of 0.5 m/s through a 4.0 cm diameter pipe in the basement under a pressure of
3.0 atm, what will be the flow speed and pressure in a 2.6 cm diameter pipe on the second
5.0 m above? Assume the pipes do not divide into branches.
Using the equation of continuity, flow speed on the second floor is
2
A1 v 1
π r1 2 v 1
0 .0 2 0 ⎞
⎛
=
v2 =
= 0 .5 × ⎜
⎟ = 1 .2 m / s
A2
π r22
⎝ 0 .0 1 3 ⎠
Using Bernoulli’s equation, the pressure in the pipe on the second floor is
(
)
1
ρ v12 − v 22 + ρ g ( y1 − y 2 )
2
1
= 3.0 × 10 5 + 1 × 10 3 0.5 2 − 1.2 2 + 1 × 10 3 × 9.8 × ( − 5 )
2
5
= 2.5 × 10 N / m 2
P2 = P1 +
(
Dr. Abdallah M. Azzeer
‫ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ‬.‫د‬
)
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106 phys - ch13
Example Vascular/Heart Flutter
If artery area A1 is reduced by factor of 5 (A2 = 1/5 A1) what is the effect on the
arterial blood pressure? (Blood density 1060 kgm-3 , person lying down, velocity in a
healthy artery is 0.3ms-1).
P1
From the continuity equation:
⎛A
v 2 = ⎜⎜ 1
⎝ A2
⎞
⎟⎟ v1 = 5v1 = 5(0.3ms −1 ) = 1.5ms −1
⎠
v1
P2
P1
v2
v1
Applying Bernoulli’s equation (y1 = y2 since horizontal)
P1 + ½ρv12 = P2 + ½ρv22
Pressure difference = P2 - P1 = ½ρ(v12 - v22)
= ½(1060kgm-3)(0.32 – 1.52) = -1145 Pa
Dr. Abdallah M. Azzeer
31
Applications of Bernoullis equation
Aortic Aneurysm:
Opposite effect to heart flutter:
The blood will travel more slowly through the
aneurysm thus increasing pressure.
P1 + ½ρv12 = P2 + ½ρv22
Air wing designed so that air velocity above
wing faster than below.
Hence above wing has lower pressure thus get
lift.
Angle of attack
Dr. Abdallah M. Azzeer
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106 phys - ch13
The Flight of Animals and Aeroplanes
• A full discussion requires a combination
of
elaborate
mathematics
and
experimental data
• Bernoulli’s equation can derive some
qualitative results about flying.
• Note air around plane disturbed only
briefly. Not steady state. To person on
plane airflow around wing approx.
steady state. Hence take coordinate
frame at rest relative to plane.
Dr. Abdallah M. Azzeer
33
Applying Bernoulli above & below the wing
Since va > vb then the pressure Pa < Pb.
Note the ρgh terms can be ignored because the wing is thin.
From Bernoullis equation can write:
Pb − Pa =
(
1
ρ v a2 − v 2b
2
)
Multiplying by Area can get the
lift force FL
FL = A(P b − Pa ) = A
(
1
ρ v a2 − v 2b
2
)
Note va and vb are proportional to the air velocity v and this equation can be
rewritten in terms of a proportionality factor CL called the lift coefficient
(which is dimensionless)
FL = AC L
1
ρv 2
2
Dr. Abdallah M. Azzeer
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