Fluid Dynamics - Applications
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Transcript of Fluid Dynamics - Applications
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1http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm
web notes: Fluidslect5.pdf
flow4.pdf flight.pdf
Lecture 5
Dr Julia Bryant
Fluid dynamics - Applications
of Bernoulli’s principleFluid statics• What is a fluid?Density
Pressure • Fluid pressure and depthPascal’s principle
• BuoyancyArchimedes’ principle
Fluid dynamics • Reynolds number
• Equation of continuity • Bernoulli’s principle
• Viscosity and turbulent flow
• Poiseuille’s equation
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Bernoulli’s Equationfor any point along a flow tube or streamline
p + 1 v 2 + g y = constant2
Between any two points along a flow tube or streamline p1 + 1 " v 1
2 + " g y 1 = p2 + 1 " v 22 + " g y 2
2 2 Dimensions
p [Pa] = [N.m-2] = [N.m.m-3] = [J.m-3]
1 " v 2 [kg.m-3.m2.s-2] = [kg.m-1.s-2] = [N.m.m-3] = [J.m-3]2
" g h [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.
1 " v 2
KE of bulk motion of fluid2 " g h GPE for location of fluid
p pressure energy density arising from internal forces within
moving fluid (similar to energy stored in a spring)
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Solving Bernoulli’s Equation
1. Establish points 1 and 2 along the flow.
2. Define the coordinate system - where is
y=0?
3. List your known and unknown variables.4. Solve for the unknown variables, possibly
using the continuity equation.
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(1) Surface of liquid
(2) Just outside hole
v 2 = ? m.s-1
What is the speed with which liquid flows from a hole at
the bottom of a tank? Where does it hit the ground?
R?
Fluid is flowing from thesurface ==> Bernoulli’s applies
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(1) Surface of liquid
(2) Just outside hole
v 2 = ? m.s-1
y 2 = 0
v 1 ~ 0 m.s-1 (large tank)
y = h
p1 = patm
p2 = patm
Firstly, what is the speed with which liquid flows from a hole at
the bottom of a tank?
y 1
y 2
h = (y 1 - y2 )
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Assume liquid behaves as an ideal fluid, Bernoulli's equation
can be applied
p1 + 1 " v
1
2 + " g y 1 = p
2 + 1 " v
2
2 + " g y 22 2
A small hole is at level (2) and the water level at (1) drops
slowly (if tank is large) # v 1 = 0
p1 = patm p2 = patm
" g y 1 = 1 "v 22 + " g y 2
2
v 22 = 2 g (y 1 – y 2) = 2 g h h = (y 1 - y2 )
v 2 = $ (2 g h) Torricelli formula (1608 – 1647)
This is the same velocity as a particle falling freely through a
height h
y 1
y 2
h
y = 0
p 1
p 2
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Recall d=1 a t2
2 t= # (2d/a)
So, time taken to fall d= y 1 - h ist= # {2(y
1
- h)/g}
Distance,R = v2 t
= $ (2 g h) . # {2(y 1 - h)/g} =2 # {h(y
1- h)}
v 2 = $ (2 g h)
Secondly, where does it hit the ground?
y 1
y 2
h
R?
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0
5
10
15
20
25
0 2 4 6 8 10 12 14 16 18 20
h
R
R =2 # {h(y 1 - h)}
Secondly, where does it hit the ground?
h=4
R?
DEMO
h=8.5
h=14
y 1=20
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R =2 # {h(y 1 - h)}
Secondly, where does it hit the ground?
Y 1=16
R?
DEMO
4.5
10
0
5
10
15
20
0 2 4 6 8 10 12 14 16
h
R
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Venturi Meter
A Venturi meter is used to measure the speed of the flow
in a pipe.
What is the speed v1 of flow in section 1 of the system?
DEMO
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Venturi Meter
Bernoulli’s equation only applies along the same
streamline. Therefore we must choose points 1 and 2
along the pipe, but not in the vertical tubes.
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Venturi Meter
Q: What do we know? y 1 = y 2 no height difference
We will need to use:
• Bernoulli$s equation ( Assume liquid behaves as an ideal fluid)
p1 + 1 " v 12 + " g y 1 = p2 + 1 " v 2
2 + " g y 22 2
• Continuity equation A1 v 1 = A2 v 2
• hydrostatic equation for the vertical pipes p 1 - p 2 = "m g h
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From Bernoulli:
p 1 + 1 " v 12 + " g y 1 = p 2 + 1 " v 2
2 + " g y 2 2 2
y 1
= y 2
p 1 – p 2 = 1 " (v 22 - v 1
2) From the continuity equation v 2 = v 1 (A1 / A2) 2
p 1 p 2
d
v 1 = 2 gh
{(A1 / A2)2 - 1}
So, p 1 – p 2 = 1 " ((A1 / A2)2 v 1
2 - v 12)
2
p 1 – p 2 = 1 " v 12 ((A1 / A2)
2 - 1) A1 > A2 so p 1 – p 2>0 p 1 > p 2
2 From hydrostatics p 1 = p 0 + " g (d+h)
p 2 = p 0 + " g d p 1 - p 2 = " g h
"g h = 1 " v 12 {(A1 / A2)
2 - 1} 2
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DEMO
or
Bernoulli’s Bar
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How to impress your mother in a pub!From Bernoulli:
p 1 + 1 " v 12 + " g y 1 = p 2 + 1 " v 2
2 + " g y 2
2 2y 1 = y 2 = 0
p 1 – p 2 = 1 " (v 22 - v 1
2)
2
Continuity equation A2 v 2 = A1 v 1% A2 v 1 p 2
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How does a
siphon work?
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C
B
A
D
y A
y B
y C
Assume that the
liquid behaves as an
ideal fluid, theequation of continuity
and Bernoulli's
equation can be
used.
y D = 0
p A = patm = pD
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C
B
A
D
y A
y B
y C
y D = 0
What do we know?
p A = patm = pD
v A =0 approximately
What do we need to find?
vD = ?
yC = ?
Focus on falling water
not rising water
patm - pC % 0 patm % " g y C
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pC + 1 " v C2 + " g y C = pD + 1 " v D
2 + " g y D2 2
From equation of continuity v C = v D pC = pD + " g (y D - y C) = patm + " g (y D - y C)
The pressure at point C can not be negative
pC % 0 and y D = 0
pC = patm - " g y C % 0y C & patm / (" g )
For a water siphon
patm ~ 105 Pa
g ~ 10 m.s-1
" ~ 103
kg.m-3
y C & 105 / {(10)(103)} m
y C 10 m
Consider points C and D and apply Bernoulli's principle.
C
B
A
D
y A
y B
y C
y D = 0
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C
B
A
D
y A
y B
y C
How fast does
the liquidcome out?
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p A + 1 " v A2 + " g y A = pD + 1 " v D
2 + " g y D2 2
v D2 = 2 ( p A – pD) / " + v A
2 + 2 g (y A - y D)
p A – pD = 0 y D = 0 assume v A2
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FLUID FLOW
MOTION OF OBJECTS IN FLUIDS
How can a plane fly?
Why does a cricket ball swing or a baseball curve?
web notes: flow4.pdf flight.pdf
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Lift F L
drag F D
Resultant F R
Motion of object through fluid
Fluid moving around stationary object
FORCES ACTING ON OBJECT MOVING THROUGH FLUID
Forward thrust
by engine
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C
D
B A
Uniform motion of an object through an ideal fluid
(' = 0)
• Consider a cylinder.• The fluid will slide freely over the
surface.
• Pressure near A and B are equal and
greater than undisturbed flow(streamlines further apart).
• Pressure near C and D are equal andlower then undisturbed flow
# Resultant force = zero
• No drag or lift.• The pattern is symmetrical
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When real fluids flow they have a certain
internal friction called viscosity. It exists in
both liquids and gases and is essentially a
frictional force between different layers of
fluid as they move past one another.
In liquids the viscosity is due to the cohesive
forces between the molecules whilst in
gases the viscosity is due to collisions
between the molecules.
“VISCOSITY IS DIFFERENT TO DENSITY”
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Drag force
In a viscous fluid, a thin layer of fluid sticks to thesurface of an object and the resulting friction leads to a
drag force on the object.
The flow is no longer complete around the object, and
the flow lines break away from the surface resulting ineddies behind the object. The pressure in the eddies is
lowered and the pressure difference gives pressure drag
force.
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low pressure region
high pressure region
rotational KE of eddies # heating effect # increasein internal energy #
temperature increases
Drag force due to pressure difference
Drag force is
opposite to thedirection of motion
motion of air
motion of object
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high pressure region
Drag force due
to pressure difference
v
v
flow speed (high) v air + v
# reduced pressure
flow speed (low) v air - v # increased pressure
v air (v ball)
Boundary layer – airsticks to ball(viscosity) – airdragged around withball
MAGNUS EFFECT
motion of air
motion of object
What happens if the object is spinning?
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Golf ball with backspin (rotating CW) with air stream going from
left to right. Note that the air stream is deflected downward with a
downward force. The reaction force on the ball is upward. This
gives the longer hang time and hence distance carried.
The trajectory of a
golf ball is notparabolic
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Professional golf drive
Initial speed v 0 ~ 70 m.s-1
Angle ~ 6°
Spin ( ~ 3500 rpm
Range ~ 100 m (no Magnus effect)
Range ~ 300 m (Magnus effect)
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Stagnation line
Higher v, lower pressure
Lift increases
with angle ofattack, untilstall.
0°
5°
10°
How can a plane fly?
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Direction plane is moving w.r.t. the air
Direction air is moving w.r.t. plane
low pressure drag
!
attack angle
lift
downwash
huge vortices
momentum transfer
low
pressure
high
pressure
DEMO
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Why do some liquidssplash more?
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Why do cars need different oils in hot
and cold countries?
Why do engines run more freely as it
heats up?
Have you noticed that skin lotions are
easier to pour in summer than winter?
Why is honey sticky?
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When real fluids flow they have a certain
internal friction called viscosity. It exists in
both liquids and gases and is essentially africtional force between different layers of
fluid as they move past one another.
In liquids the viscosity is due to the cohesive
forces between the molecules whilst in
gases the viscosity is due to collisions
between the molecules.
“VISCOSITY IS DIFFERENT TO DENSITY”
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stationary wall
L A viscous fluid
plate
X
Z
A useful model
Assumptions
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stationary wall
L
Viscous fluid will flow
plate moves with speed v
X
Z
A useful model
Q: Direction?
Q: Highest speeds?Q: Lowest speeds?
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stationary wall
low speed
high speed
plate moves with speed v
X
Z
v x = 0
v x = v
A useful model:
Newtonian fluids
water, most gases
linearvelocity
gradient
d
L v x
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stationary wall
plate exerts force F
velocitygradient
is proportional toshearstress
(F/A) = ' (v / L)
A useful model:
Newtonian fluidsover area A