Lecture 27: Lift
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Transcript of Lecture 27: Lift
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Lecture 27: Lift
Many biological devices (Biofoils) are used to create Lift.How do these work?
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First, some definitions…
wing section,c(chord)
totalforce(normal to wing)
drag
lift(normal to U)
(parallel to U)
chord section analysis….
wingvelocity = U
angle ofattack =
winglength, R wing
area, S
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Two ways to derive lift:1) mass deflection
totalforce
U
air deflecteddownward by wing
USUmUForce dtd ~)(
Surface area, S
)(Re,
221
fC
SUCForce
total
total
Pressure always acts normal to the surface of an object.Therefore, this mass deflection force acts roughly perpendicular to surface of biofoil.
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1) Massdeflection
totalforce
drag
lift
U
air deflecteddownward by wing
Surface area, S
221 SUCForce total
Lift and drag are defined ascomponents perpendicularand parallel to direction of motion.
viscoustotaldrag
totallift
drag
lift
CCC
CC
SUCDrag
SUCLift
sin
cos
221
221
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RoboFly
amplitude · length2
frequency · viscosityReynolds number =
reduced frequency =forward velocity
length · angular velocity
dimensionless scaling parameters
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totalforce
-9 0 9 18 27 36 45 54 63 72 81 90
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
tota
l for
ce c
oeff
icie
ntC
T
angle of attack (degs)
-9 0 9 18 27 36 45 54 63 72 81 90
-60
-40
-20
0
20
40
60
80
10090o
tota
l for
ce o
rien
tati
on
degs
angle of attack (degs)
CL
CD
Fs
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0 15 30 45 60 75 90
0
1
2
3
4
CT
angle of attack ()
CT = 3.5 sin
0 15 30 45 60 75 90
0
1
2
3
4
CD
angle of attack ()
CD = CT sin
0 15 30 45 60 75 90
0
1
2
3
angle of attack ()
CL = CT cos
CL
{viscous
drag
CT sin
CT
CT cos
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totalforce
drag
lift
U
Surface area, S
~
sinsin
cossin
221
221
k
CkC
kC
SUCDrag
SUCLift
viscousdrag
lift
drag
lift
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-9 0 9 18 27 36 45 54 63 72 81 90-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
angle of attack (degs)
forc
e co
effi
cien
ts
CL
CD
Polar plot of lift and drag:
0 1 2 3 4-1
0
1
2
3
drag coefficient
lift
coe
ffic
ient
=-9
=-9
=22.5
=45
=90
highestlift:drag ratio
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U
Flow is tangentialat trailing edge
Flow separatesat leading edge
2. Circulation
Law of continuity applies to streamline
fluid travels fasterover to of biofoil
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U
Difference in velocityacross surface is equivalent
to net circular flow around biofoil = Circulation,
dSUmathematically:
dA
Kutta-Joukowski Theorem:
UtLif (lift per unit span) Uc
C
URSUC
L
L
/221
combine withpreviousdefinition: R=biofoil length
c= biofoil width
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Consider 2D biofoil starting from rest:
=0
=0
startingvortex
boundvortex
Required byKelvin’s Law
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Consider 3D biofoil starting from rest:
startingvortex
tipvortex
boundvortex
Downward flowthrough centerof vortex ring
Circulation, , is constantalong vortex ring
Helmholtz’ Law requires that a vortex filament cannot end abruptly:
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How is structure of vortex ring related to lift on biofoil?
Circulation, Area = A
forward velocity, U
Ring momentum =mass flux through
ring=A
Force = d/dt (A)
= d/dt(A)
= R U
Force/R = U = Kutta-Joukwski
R
Therefore, elongation of vortex ring is manifestation of force on biofoil.
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Three important descriptors of fluid motion:
2. vorticity, (x,y)
1. velocity, u(x,y)
u(x,y)
ux
uy
x
y
uxyx
uy
3. circulation,
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Fslap = m U / t
where m is bolus of accelerated water, moving atvelocity, u
impulse (F x t) = mass x velocity
Fstroke = A /t
Momentum of vortex ring A
= circulation
A
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