Download FLUID FLOW IDEAL FLUID BERNOULLI`S PRINCIPLE

VIS U AL P HYS ICS
S ch ool of P h ysi cs
U n i v er si t y of S yd n ey Au st r a l i a
FLUID FLOW
IDEAL FLUID
BERNOULLI'S PRINCIPLE
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How can a plane fly?
How does a perfume spray work?
What is the venturi effect?
Why does a cricket ball swing or a baseball curve?
IDEAL FLUID
Fluid motion is usually very complicated. However, by making a set
of assumptions about the fluid, one can still develop useful models of
fluid behaviour. An ideal fluid is
• Incompressible – the density is constant
• Irrotational – the flow is smooth (streamline or laminar), no
turbulence
• Nonviscous – fluid has no internal friction ( η= 0)
• Steady flow – the velocity of the fluid at each point is constant
in time.
BERNOULLI'S EQUATION (conservation of ENERGY)
An interesting effect is that, for a fluid (e.g. air) flowing through a pipe
with a constriction in it, the fluid pressure is lowest at the constriction.
In terms of the equation of continuity the fluid pressure falls as the
flow speed increases.
The reason is easy to understand. The fluid has different speeds
and hence different kinetic energies at different parts of the tube.
The changes in energy must result from work being done on the fluid
and the only forces in the tube that might do work on the fluid are the
driving forces associated with changes in pressure from place to
place.
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• Conservation of energy
• A pressurized fluid must contain energy by the virtue that work
must be done to establish the pressure.
• A fluid that undergoes a pressure change undergoes an energy
change.
A1
A1
A2
v2
v1
Low speed
Low KE
High pressure
v1
high speed
high KE
low pressure
Low speed
Low KE
High pressure
Derivation of Bernoulli's equation
∆x2
Y
p2
m
X
v2
A2
time 2
ρ
p1
∆x1
y2
A1
m
y1
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v1
time 1
2
Mass element m moves from (1) to (2)
m = ρ A1 ∆x1 = ρ A2 ∆x2 = ρ ∆V where ∆V = A1 ∆x1 = A2 ∆x2
Equation of continuity A V = constant
A1 v1 = A2 v2 A1 > A2 ⇒ v1 < v2
Since v1 < v2 the mass element has been accelerated by the net
force
F1 – F2 = p1 A1 – p2 A2
Mass element has an increase in KE
∆K = ½ m v22 - ½ m v12 = ½ ρ ∆V v22 - ½ ρ ∆V v12
Mass element has an increase in GPE
∆U = m g y2 – m g y1 = ρ ∆V g y2 = ρ ∆V g y1
The increase in KE and GPE comes from the net work done on the
mass element by the forces F1 and F2 (the sample of mass m, in
moving from a region of higher pressure to a region of lower
pressure, has positive work done on it by he surrounding fluid)
Wnet = F1 ∆x1 – F2 ∆x2 = p1 A1 ∆x1 – p2 A2 ∆x2
Wnet = p1 ∆V – p2 ∆V = ∆K + ∆U
p1 ∆V – p2 ∆V =
½ ρ ∆V v22 - ½ ρ ∆V v12 + ρ ∆V g y2 - ρ ∆V g y1
Rearranging
!
p1 + ½ ρ v12 + ρ g y1 = p2 + ½ ρ v22 + ρ g y2
This is Bernoulli's equation.
We can also write, for any point along a flow tube or streamline
p + ½ ρ v 2 + ρ g y = constant
This is Bernoulli's theorem.
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!
Bernoulli's theorem only applies to an ideal fluid. Can not use
Bernoulli's Principle when the viscosity is significant. Can be applied
to gases provided there are only small changes in pressure.
Dimensions
p
[Pa] = [N.m-2] = [N.m.m-3] = [J.m-3]
½ρ v 2
[kg.m-3.m2.s-2] = [kg.m.s-2.m.m-3] = [N.m.m-3] = [J.m-3]
ρgh
[kg.m-3 m.s-2. m] = [kg.m.s-2.m.m-3] = [N.m.m-3] = [J.m-3]
Each term has the dimensions of energy / volume or energy density.
½ ρ v2
KE of bulk motion of fluid
ρgh
GPE for location of fluid
p
pressure energy density arising from internal forces within
moving fluid (similar to energy stored in a spring)
Consequences of Bernoulli's Theorem
• A portion of a fluid made to move rapidly sustains a lower
pressure than does the portion of fluid moving slowly. Using
Galilean principle of relatively it is easier to consider the
objects to be stationary and the fluid moving
What happens when two ships or trucks pass alongside each other?
?
Have you noticed this effect in driving across the Sydney Harbour
Bridge?
.
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force
high speed
low pressure
force
p large
p large
p small
v small
!
v large
v small
• Venturi Effect
A fast jet of air emerging from a small nozzle will have a lower
pressure than the surrounding atmospheric pressure. The Venturi
effect is used in many types of drainage systems. When you are in
the dentist's chair, the dentist uses a device based on the Venturi
effect to remove saliva out of your mouth. The device is connected
to a tap. Water flows fast past the constriction, causing the pressure
to drop inside the long tube. When the other end of this tube is
immersed in a pool of saliva, the higher pressure outside forces the
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saliva up the tube, and away. In a cotton picker, air flows instead of
water to pick up cotton!
high velocity flow
high
pressure
(patm)
low pressure
velocity increased
pressure decreased
The chimney effect Just the venturi effect being used to pick
material up. In most automobiles, petrol is pushed into the
carburettor in this way.
Atomizer This same effect makes atomizers, perfume sprayers,
insect sprayers, nebulizers and spray guns work. It is most
important that the free surface of the liquid should be open to the
atmosphere, else the high pressure outside the container and the low
pressure inside will result in the container being crushed (the
enclosure must be vented). Fly sprays always have a small air hole.
Remember the fluid is caused to move from a container by a
pressure difference – the fluid is pushed not sucked
?
?
How can you drink from a straw?
• Model for a knee in fluid flow (river going around a bend)
Where does the flow speed up and slow down?
How does the pressure change as the liquid goes round the corner?
It can be observed that the streamlines hug the sharp corner. But far
enough before the corner, and far enough after it, they are parallel
and equally spaced. Consider the liquid flowing between lines 1 and
2. Its cross-sectional area decreases near the corner, so the liquid
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speeds up there. The fluid between lines 3 and 4 has its area
increased near the corner, so it slows down.
Remembering that pressure can also be thought of as force per unit
area, we see that the fluid must be exerting an extra large force on
the tube at the outside corner. Bernoulli's principle – slower speed
higher pressure, the higher pressure is on the outside corner of the
bend.
slow flow
(streamlines
further apart)
high pressure
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1
Same speed
and pressure
across river
faster flow
(streamlines
closer together)
low pressure
• Arteriosclerosis and vascular flutter
Arteriosclerosis arises when plaque builds up on the inner walls of
arteries, restricting blood flow. Any such obstruction results in a
Venturi pressure drop.
In an advanced stage of the disease, pressure differences across
artery causes it to collapse. Backed up blood will produce extra
pressure which will force it open, only to collapse again. This gives
vascular flutter which can be heard with a stethoscope.
A similar effect occurs in snoring.
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artery
Flow speeds up at
constriction
Pressure is lower
Internal force acting on
artery wall is reduced
arm
External forces causes
artery to collapse
head
arm
lung
lung
heart
trunk
leg
?
leg
• Why does smoke go up a chimney?
Partly hot air rises (air less dense)
Partly Bernoulli's principle (wind blows across top of chimney
reducing pressure at the top)
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?
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• In a serve storm how does a house loose its roof?
Air flow is disturbed by the house. The "streamlines" crowd around
the top of the roof ⇒ faster flow above house ⇒ reduced pressure
above roof than inside the house ⇒ room lifted off because of
pressure difference.
• How can a table tennis ball float in a moving air stream?
The flow of moving air around the ball can be asymmetrical so that
there is a higher pressure on one side of the ball than the other. If the
ball begins to leave the jet stream of air, the higher pressure outside
the jet of air pushed the ball back into the flow.
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• Why do rabbits not suffocate in the burrows?
Air must circulate. The burrows must have two entrances. Air flows
across the two holes is usually slightly different ⇒ slight pressure
difference ⇒ forces flow of air through burrow.
One hole is usually higher than the other and the a small mound is
built around the holes to increase the pressure difference.
• Why do racing cars wear skirts?
Racing cars are designed so that the air beneath the car moves
faster than above it ⇒ large downward force on the car ⇒
increased normal force ⇒ increased frictional force ⇒ better grip
and propulsion of the car.
How can we describe the flow of a liquid from a tank (many
applications)
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(1)
Point on surface of liquid
y1
v2 = ? m.s-1
y2
(2) Point just outside hole
Assume liquid behaves as an ideal fluid and that Bernoulli's equation
can be applied
p1 + ½ ρ v12 + ρ g y1 = p2 + ½ ρ v22 + ρ g y2
A small hole is at level (2) and the water level at (1) drops slowly ⇒
v1 = 0
p1 = patm
p2 = patm
∴ ρ g y1 = ½ ρv22 + ρ g y2
v22 = 2 g (y1 – y2) = 2 g h
v2 = √(2 g h)
h = (y1 - y2)
Torricelli formula (1608 – 1647)
This is the same velocity as a particle falling freely through a height h
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?
How do you measure the speed of flow for a fluid?
One method is to use a manometer type venturi meter.
(1)
(2)
ρF
v1 = ?
h
ρm
Assume liquid behaves as an ideal fluid and that Bernoulli's equation
can be applied for the flow along a streamline
p1 + ½ ρ v12 + ρ g y1 = p2 + ½ ρ v22 + ρ g y2
y1 = y 2
p1 – p2 = ½ ρF (v22 - v12)
p1 - p2 = ρm g h
A1 v1 = A2 v2
⇒
v2 = v1 (A1 / A2)
ρm g h = ½ ρF {v12 (A1 / A2)2- v12} = ½ ρF v12 {(A1 / A2)2 - 1}
v1 =
2 g h ρm
 A 2
1

 A2 
ρ F 
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
− 1

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?
How does a siphon work?
How fast does the liquid come out?
C
yC
A
yA
B
yB
D
Assume that the liquid behaves as an ideal fluid and that both the
equation of continuity and Bernoulli's equation can be used.
Heights: yD = 0
Pressures:
yB
yA
yC
pA = patm = pD
Consider a point A on the surface of the liquid in the container and
the outlet point D.
Apply Bernoulli's principle to these points
pA + ½ ρ vA2 + ρ g yA = pD + ½ ρ vD2 + ρ g yD
vD2 = 2 (pA – pD) / ρ + vA2 + 2 g (yA - yD)
pA – pD = 0
vD =
yD = 0
assume vA2 << vD2
√(2 g yA )
Now consider the points C and D and apply Bernoulli's principle to
these points
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pC + ½ ρ vC2 + ρ g yC = pD + ½ ρ vD2 + ρ g yD
From equation of continuity vC = vD
pC = pD + ρ g (yD - yC) = patm +
ρ g (yD - yC)
The pressure at point C can not be negative
patm +
ρ g (yD - yC) ≥ 0
pC ≥ 0
and yD = 0
pC = patm - ρ g yC ≥ 0
yC ≤ patm / (ρ g)
For a water siphon
patm ~ 105 Pa
g ~ 10 m.s-1
ρ ~ 103 kg.m-3
yC ≤ 105 / {(10)(103)} m
yC ≤ 10 m
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For streamline steady flow what is the difference between a
nonviscous and viscous fluid flowing through a pipe?
Ideal fluid
Real fluid
For the viscous fluid there is a gradual decrease in pressure along
the pipe.
?
Viscosity ⇒ frictional forces ⇒ energy lost by heating fluid ⇒
increase in internal energy of fluid ⇒ increase in temperature of
fluid.
Where does the energy come from?
Equation of continuity still applies ⇒ energy can not come from KE
of fluid ⇒ energy must be provided from the energy of the fluid ⇒
pressure drop along tube.
Home activity
Shake a can of soft drink vigorously when outside. Then carefully
open the can. Becareful not to wet yourself or someone else.
What happens and why?
Where did the energy come from and go?
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Get a table tennis ball. Blow a steam of air from your mouth vertically
and balance the ball in the air stream.
Hold a piece of paper and blow across it – the paper will rise.
Hold a piece of paper in each hand – let the papers hang vertically –
blow a stream of air between the pages – they should come together.
Floating ball
Daniel Bernoulli (1700 – 1782)
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A large artery in a dog has an inner radius of 4.00×10-3 m. Blood
flows through the artery at the rate of 1.00×10-6 m3.s-1. The blood has
a viscosity of 2.084×10-3 Pa.s and a density of 1.06×103 kg.m-3.
Calculate:
(i)
The average blood velocity in the artery.
(ii) The pressure drop in a 0.100 m segment of the artery.
(iii) The Reynolds number for the blood flow.
Briefly discuss each of the following:
(iv)
(v)
(vi)
The velocity profile across the artery (diagram may be
helpful).
The pressure drop along the segment of the artery.
The significance of the value of the Reynolds number
calculated in part (iii).
Semester 1, 2004 Exam question
Solution
radius R = 4.00×10-3 m
volume flow rate Q = 1.00×10-6 m3.s-1
viscosity of blood η = 2.084×10-3 Pa.s
density of blood ρ = 1.060×10-3 kg.m-3
(i) Equation of continuity: Q = A v
A = π R2 = π (4.00×10-3)2 = 5.03×10-5 m2
v = Q / A = 1.00×10-6 / 5.03×10-5 m.s-1 = 1.99×10-2
m.s-1
(ii) Poiseuille’s Equation
Q = ∆P π R4 / (8 η L)
L = 0.100 m
∆P = 8 η L Q / ( π R 4 )
∆P = (8)(2.084×10-3)(0.1)(1.00×10-6) / {(π)(4.00×10-3)4} Pa
∆P = 2.07 Pa
(iii) Reynolds Number
Re = ρ v L / η
where L = 2 R (diameter of artery)
Re = (1.060×103)(1.99×10-2)(2)(4.00×10-3 ) / (2.084×10-3)
Re = 81
use diameter not length
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(iv) Parabolic velocity profile: velocity of blood zero at sides of artery
Flow of a viscous newtonain fluid through a pipe
Velocity Profile
Cohesive forces
between molecules ⇒
layers of fluid slide past
each other generating
frictional forces ⇒
energy dissipated (like
rubbing hands together)
Parabolic velocity
profile
Adhesive forces between fluid and surface ⇒ fluid
stationary at surface
(v) Viscosity ⇒ internal friction ⇒ energy dissipated as thermal energy ⇒
pressure drop along artery
(vi) Re very small ⇒ laminar flow (Re < 2000)
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?
A pitot tube can be used to determine the speed of an
aeroplane relative to the surrounding air. The device consists
of a U shaped tube containing a fluid with a density ρ. One
end of the tube A opens to the air at the side of the plane,
the other end B is open to the air in the direction the plane is
flying so that the air stagnant at this location (zero speed).
Show that the speed of the plane can be expressed in terms
of the difference in the height h of the fluid in the manometer
tube as
v = 2gh
ρ
ρ air
where ρair is the density of the air.
v
B
A
h
Solution
Apply Bernoulli’s Principle to the points at the front (B) and side (A) of
the plane. For air the elevations yA and yB are essentially the same and
the speed vB = 0 m.s-1.
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p A + 1 2 ρ air vA 2 + ρ air gy A = pB + 1 2 vB 2 + ρ air gyB
∆p = p B − p A =
1
2
ρ air v A2
For the liquid of density ρ in the manometer
∆p = ρ gh
Equating the two expressions for ∆p ⇒
∆p = ρ gh =
1
2
v = v A = 2 gh
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ρ air v A2
ρ
ρ air
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