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7/29/2019 Science lect6.ppt
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FLUID FLOWIDEAL 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
http://daye.en.alibaba.com/product/50015092/50084257/Sprayers/Pressure_Sprayer_IT1.htmlhttp://daye.en.alibaba.com/product/50015092/50084257/Sprayers/Pressure_Sprayer_IT1.html7/29/2019 Science lect6.ppt
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Daniel Bernoulli (1700 1782)
Floating ball
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A1A2
v1v2
A1
v1
Low speedLow KEHigh pressure
high speedhigh KElow pressure
Low speedLow KE
High pressure
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vsmall vsmallvlarge
plarge plarge
psmall
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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 topof the roof faster flow above house reduced pressure above roofthan 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 acrossthe 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?
http://images.google.com/imgres?imgurl=http://www.endurancesportscar.com/photo/r%26sc2.jpg&imgrefurl=http://allcars.persianblog.com/&h=311&w=498&sz=14&tbnid=DJen0LI1B-sJ:&tbnh=79&tbnw=126&prev=/images%3Fq%3Dracing%2Bcars%26hl%3Den%26lr%3D&oi=imagesr&start=2http://images.google.com/imgres?imgurl=http://www.gel-communications.co.uk/animalfun/assets/images/rabbit.jpg&imgrefurl=http://www.gel-communications.co.uk/animalfun/html/rabbit.html&h=384&w=512&sz=44&tbnid=ynKTnoGke70J:&tbnh=96&tbnw=128&prev=/images%3Fq%3Drabbit%26hl%3Den%26lr%3D&oi=imagesr&start=1http://images.google.com/imgres?imgurl=http://www.endurancesportscar.com/photo/r%26sc2.jpg&imgrefurl=http://allcars.persianblog.com/&h=311&w=498&sz=14&tbnid=DJen0LI1B-sJ:&tbnh=79&tbnw=126&prev=/images%3Fq%3Dracing%2Bcars%26hl%3Den%26lr%3D&oi=imagesr&start=27/29/2019 Science lect6.ppt
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velocity increasedpressure decreased
low pressurehigh
pressure
(patm)
VENTURI EFFECT
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high speedlow pressure
force
force
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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artery
External forces causesartery to collapse
Flow speeds up atconstrictionPressure is lowerInternal force acting on
artery wall is reduced
Arteriosclerosis and vascular flutter
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y1
y2
x1
x2 p2
A2
A1
v1
v2
p1
X
Y
time 1
time 2
m
m
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Bernoullis Equation
for any point along a flow tube or streamline
p+ v2 + g y = constantDimensionsp [Pa] = [N.m-2] = [N.m.m-3] = [J.m-3]
v2 [kg.m-3.m2.s-2] = [kg.m-1.s-2] = [N.m.m-3] = [J.m-3]
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.
v2 KE of bulk motion of fluid
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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y1
y2
x1
x2 p2
A2
A1
v1
v2
p1
X
Y
time 1
time 2
m
m
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Mass element mmoves from (1) to (2)
m = A1x1 = A2x2 = V whereV= A1x1 = A2x2
Equation of continuity A V= constant
A1v1 = A2v2 A1 > A2 v1 < v2
Since v1
< v2
the mass element has been accelerated by the net force
F1F2 = p1A1p2A2
Conservation of energy
A pressurized fluid must contain energy by the virtue that work mustbe done to establish the pressure.
A fluid that undergoes a pressure change undergoes an energychange.
Derivation of Bernoulli's equation
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K= m v22 - m v12 = Vv22 - V v12
U = m g y2m g y1 = V g y2 = V g y1
Wnet = F1x1F2x2 = p1A1x1p2A2x2
Wnet = p1Vp2V = K + U
p1Vp2V = Vv22 - V v12 + V g y2 - V g y1
Rearranging
p1 + v12 + g y1 = p2 + v22 + g y2
Applies only to an ideal fluid (zero viscosity)
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Ideal fluid
Real fluid
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(1) Point on surface of liquid
(2) Point just outside hole
v2 = ? m.s-1
y1
y2
Flow of a liquid from a hole at the bottom of a tank
A li id b h id l fl id d th t B lli'
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Assume liquid behaves as an ideal fluid and that Bernoulli'sequation can be applied
p1 + v12 + g y1 = p2 + v22 + g y2
A small hole is at level (2) and the water level at (1) dropsslowly v1 = 0
p1 = patm p2 = patm
g y1 = v22 + g y2
v22 = 2 g(y1y2) = 2 g h h =(y1 - y2)
v2 = (2 g h) Torricelli formula (1608 1647)
This is the same velocity as a particle falling freely through a
height h
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(1)
(2)
F
m
h
v1 =?
How do you measure the speed of flow for a fluid?
A li id b h id l fl id d h B lli' i
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Assume liquid behaves as an ideal fluid and that Bernoulli's equationcan be applied for the flow along a streamline
p1 + v12 + g y1 = p2 + v2
2 + g y2
y1 = y2
p1 p2 = F (v22 - v1
2)
p1 - p2 = mg h
A1v1 = A2v2 v2 = v1 (A1 /A2)
mg h = F { v12 (A1 /A2)
2- v12 } = Fv1
2 {(A1 /A2)2 - 1}
m
1 2
1
F
2
2
1
g hv
A
A
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C
B
A
D
yA
yB
yC
How does a siphon
work?
How fast does theliquid
come out?
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Assume that the liquid behaves as an ideal fluid and thatboth the equation of continuity and Bernoulli's equation canbe used.
Heights: yD = 0 yB yA yCPressures: 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'sprinciple to these points
From equation of continuity vC = vD
The pressure at point C can not be negative
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pA + vA2 + g yA = pD + vD2 + g yD
vD2 = 2 (pApD) / + vA2 + 2 g (yA - yD)
pApD = 0 yD= 0 assume vA2
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pC + vC2 + g yC = pD + vD2 + g yD
vC = vDpC = 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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A large artery in a dog has an inner radius of 4.0010-3 m. Blood flowsthrough the artery at the rate of 1.0010-6 m3.s-1. The blood has aviscosity of 2.08410-3 Pa.s and a density of 1.06103 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
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Solution
radius R= 4.0010-3 m
volume flow rate Q= 1.0010-6 m3.s-1
viscosity of blood = 2.08410-3 Pa.s
density of blood = 1.06010-3 kg.m-3
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(i) Equation of continuity: Q= A v
A = R2 = (4.0010-3)2 = 5.0310-5 m2
v= Q/A = 1.0010-6 / 5.0310-5 m.s-1 = 1.99 10-2 m.s-1
(ii) Poiseuilles EquationQ= 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} PaP= 2.07 Pa
(iii) Reynolds NumberRe = v L / where L = 2 R (diameter of artery)Re = (1.060 10
3)(1.99 10-2)(2)(4.00 10-3) / (2.084 10-3)Re = 81
use diameter not length
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Flow of a viscous newtonain fluid through a pipeVelocity Profile
Adhesive forces between fluid and surface fluidstationary at surface
Parabolic velocity
profile
Cohesive forces
between molecules layers of fluid slide pasteach other generating
frictional forces
energy dissipated (likerubbing hands together)
(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)