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web notes: lect6.ppt
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Ideal fluid
Real fluid
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What is the speed with which liquid flows from a hole at
the bottom of a tank?
(1)
Surface of liquid
v1 ~ 0 m.s-1
p1 = patm
y1
h = (y1 - y2)
y2
Draw flow tubes
v2 = ? m.s-1
(2) Just outside hole p = p
2
atm
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Assume liquid behaves as an ideal fluid and that Bernoulli's
equation can be applied
p1 + ½  v12 +  g y1 = p2 + ½  v22 +  g y2
A small hole is at level (2) and the water level at (1) drops
slowly  v1 = 0
p1 = patm
p2 = patm
 g y1 = ½ v22 +  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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What is the speed of flow in section 1 of the system?
(1)
(2)
F
v1 = ?
h
m
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Assume liquid behaves as an ideal fluid and that Bernoulli's equation
can be applied for the flow along a streamline
p1 + ½  v 1 2 +  g y 1 = p2 + ½  v 2 2 +  g y 2
y1 = y2
p1 – p2 = ½ F (v22 - v12)
p1 - p2 =  m g h
A1 v1 = A2 v2

v2 = v1 (A1 / A2)
m g h = ½ F { v12 (A1 / A2)2- v12 } = ½ F v12 {(A1 / A2)2 - 1}
v1 
2 g h m
 A 2
F  1 
 A2 

 1

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pC
yC
Q: What do we know?
Continuous flow
Pressure in top section > 0
otherwise there will be
a vacuum
pC  0
How does a
siphon work?
Focus on falling water
not rising water
patm - pC  0
patm   g yC
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C
yC
A
yA
B
Assume that the
liquid behaves as an
ideal fluid, the
equation of continuity
and Bernoulli's
equation can be
used.
yB
D
yD = 0
pA = patm = pD
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Consider points C and D and apply Bernoulli's principle.
pC + ½  vC2 +  g yC = pD + ½  vD2 +  g yD
From equation of continuity vC = vD
pC = pD +  g (yD - yC) = patm +
 g (yD - yC)
The pressure at point C can not be negative
pC  0 and yD = 0
pC = patm -  g yC  0
yC  patm / ( g)
For a water siphon
patm ~ 105 Pa
g ~ 10 m.s-1
 ~ 103 kg.m-3
yC  105 / {(10)(103)} m
yC  10 m
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How fast does the liquid come out?
Consider a point A on the surface of the liquid in the
container and the outlet point D.
Apply Bernoulli's principle
pA + ½  vA2 +  g yA = pD + ½  vD2 +  g yD
vD2 = 2 (pA – pD) /  + vA2 + 2 g (yA - yD)
pA – pD = 0
yD = 0
assume vA2 << vD2
vD = (2 g yA )
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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?
Why does a golf ball have dimples?
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FORCES ACTING ON OBJECT MOVING THROUGH FLUID
Resultant FR
Lift FL
drag FD
Motion of object through fluid
Fluid moving around stationary object
Forward thrust
by engine
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Uniform motion of an object through an ideal fluid ( = 0)
The pattern is symmetrical
 FR = 0
C
B
A
D
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Drag force
frictional drag (viscosity)
pressure drag (eddies – lower pressure)
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Drag force due
to pressure difference
motion of air
motion of object
low pressure region
rotational KE of eddies 
heating effect  increase
in internal energy 
temperature increases
NO
CURVE
high pressure region
Drag force is
opposite to the
direction of motion
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motion of air
motion of object
Drag force due
to pressure difference
flow speed (high) vair + v
 reduced pressure
v
vair (vball)
MAGNUS EFFECT
flow speed (low) vair - v
 increased pressure
v
high pressure region
low pressure region
Boundary layer – air
sticks to ball
(viscosity) – air
dragged around with
ball
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Professional golf drive
Initial speed v0 ~ 70 m.s-1
Angle ~ 6°
Spin  ~ 3500 rpm
Range ~ 100 m (no Magnus effect)
Range ~ 300 m (Magnus effect)
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The trajectory of a
golf ball is not
parabolic
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.
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lift
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Direction plane is moving w.r.t. the air
Direction air is moving w.r.t. plane
low
pressure

lift
low pressure drag
attack angle
momentum transfer
high
pressure
downwash
huge vortices
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Semester 1, 2004 Exam question
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) The velocity profile across the artery (diagram may be helpful).
(v) The pressure drop along the segment of the artery.
(vi) The significance of the value of the Reynolds number calculated in
part (iii).
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Solution
radius R = 4.0010-3 m
volume flow rate Q = 1.0010-6 m3.s-1
viscosity of blood  = 2.08410-3 Pa.s
density of blood  = 1.06010-3 kg.m-3
(i) v = ? m.s-1
(ii) p = ? Pa
(iii) Re = ?
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(i)
Equation of continuity: Q = A v
A =  R2 =  (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  R4 / (8  L)
L = 0.100 m
P = 8  L Q / ( R4)
P = (8)(2.08410-3)(0.1)(1.0010-6) / {()(4.0010-3)4} Pa
P = 2.07 Pa
(iii) Reynolds Number
Re =  v L / 
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
(v) Viscosity  internal friction  energy dissipated as thermal energy 
pressure drop along artery
(vi) Re very small  laminar flow (Re < 2000)
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
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