FLUID FLOW
IDEAL FLUID
BERNOULLI'S PRINCIPLE
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?
web notes: lect6.ppt
flow3.pdf
Daniel Bernoulli (1700 – 1782)
Floating ball
A1
A1
A2
v1
Low speed
Low KE
High pressure
v2
high speed
high KE
low pressure
v1
Low speed
Low KE
High pressure
p large
p large
p small
v small
v large
v small
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.
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?
VENTURI EFFECT
high
pressure
(patm)
low pressure
velocity increased
pressure decreased
force
high speed
low pressure
force
What happens when two ships or trucks pass alongside each other?
Have you noticed this effect in driving across the Sydney Harbour Bridge?
artery
Flow speeds up at
constriction
Pressure is lower
Internal force acting on
artery wall is reduced
External forces causes
artery to collapse
Arteriosclerosis and vascular flutter
x2
Y
p2
m
v2
X
time 2

p1
x1
y2
A1
m
y1
v1
time 1
A2
Bernoulli’s Equation
for any point along a flow tube or streamline
p + ½  v2 +  g y = constant
Dimensions
p [Pa] = [N.m-2] = [N.m.m-3] = [J.m-3]
½  v2
[kg.m-3.m2.s-2] = [kg.m-1.s-2] = [N.m.m-3] = [J.m-3]
gh
[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v2
KE of bulk motion of fluid
gh
GPE for location of fluid
p
pressure energy density arising from internal forces within
moving fluid (similar to energy stored in a spring)
x2
Y
p2
m
v2
X
time 2

p1
x1
y2
A1
m
y1
v1
time 1
A2
Mass element m moves from (1) to (2)
Derivation of Bernoulli's equation
m =  A1 x1 =  A2 x2 =  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
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.
K = ½ m v22 - ½ m v12 = ½  V v22 - ½  V v12
U = m g y2 – m g y1 =  V g y2 =  V g y1
Wnet = F1 x1 – F2 x2 = p1 A1 x1 – p2 A2 x2
Wnet = p1 V – p2 V = K + U
p1 V – p2 V =
½  V v22 - ½  V v12 +  V g y2 -  V g y1
Rearranging
p1 + ½  v12 +  g y1 = p2 + ½  v22 +  g y2
Applies only to an ideal fluid (zero viscosity)
Ideal fluid
Real fluid
Flow of a liquid from a hole at the bottom of a tank
(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  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
How do you measure the speed of flow for a fluid?
(1)
(2)
F
v1 =
?
h
m
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

C
yC
A
yA
B
yB
D
How does a siphon
work?
How fast does the
liquid
come out?
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
yB
yA
yC
Pressures: 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
Now consider the points C and D and apply Bernoulli's
principle to these points
From equation of continuity vC = vD
The pressure at point C can not be negative
pA + ½  vA2 +  g yA = pD + ½  vD2 +  g yD
vD2 = 2 (pA – pD) /  + vA2 + 2 g (yA - yD)
pA – pD = 0
vD = (2 g yA )
yD = 0
assume vA2 << vD2
pC + ½  vC2 +  g yC = pD + ½  vD2 +  g yD
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
yC  105 / {(10)(103)} m
yC  10 m
 ~ 103 kg.m-3
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).
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  = 2.08410-3 Pa.s
density of blood  = 1.06010-3 kg.m-3
(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
(ii) Poiseuille’s Equation
Q = P  R4 / (8  L)
m.s-1
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
(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