Transcript BE105_27_lift

Lecture 27: Lift
Many biological devices (Biofoils) are used to create Lift.
How do these work?
First, some definitions…
wing section,c
(chord)
wing
length, R
chord section analysis….
(normal to U)
lift
wing
area, S
total
(normal to wing)
force
angle of
attack = a
drag (parallel to U)
wing
velocity = U
Two ways to derive lift:
1) mass deflection
total
force
a
U
Surface area, S
Force  dtd (mU ) ~ SU  U
Force  12 Ctotal SU 2
Ctotal  f (Re,a )
air deflected
downward by wing
Pressure always acts normal to the surface of an object.
Therefore, this mass deflection force acts roughly perpendicular to surface of biofoil.
1) Mass
deflection
lift
a
U
Force  12 Ctotal SU 2
2
1
Lift  2 Clift SU
Drag  12 Cdrag SU 2
Clift  Ctotal cosa
Cdrag  Ctotal sin a  Cviscous
total
force
Lift and drag are defined as
components perpendicular
and parallel to direction
of motion.
drag
Surface area, S
air deflected
downward by wing
RoboFly
dimensionless scaling parameters
amplitude · length2
Reynolds number =
frequency · viscosity
reduced frequency =
forward velocity
length · angular velocity
total
force
CL
Fs
a
q
CD
3.5
90o
80
total force orientation
q (degs)
3.0
total force coefficient
CT
100
2.5
2.0
1.5
1.0
60
40
20
0
-20
0.5
-40
0.0
-60
-9 0
9 18 27 36 45 54 63 72 81 90
angle of attack
a (degs)
-9 0
9 18 27 36 45 54 63 72 81 90
angle of attack
a (degs)
4
3
CT
CT
CT cos a
a
2
CT = 3.5 sin a
1
CT sin a
0
0
15
30
45
60
75
90
angle of attack (a)
3
4
CD = CT sin a
CL = CT cos a
3
2
CL
CD
2
1
1
viscous
drag {
0
0
0
15
30
45
60
75
angle of attack (a)
90
0
15
30
45
60
75
angle of attack (a)
90
lift
a
total
force
drag
Surface area, S
U
Lift  Clift SU
1
2
2
Drag  12 Cdrag SU 2
Clift  k sin a cosa
Cdrag  k sin a sin a  Cviscous
k ~
Polar plot of lift and drag:
3
highest
lift:drag ratio
2
force coefficients
3.0
CD
2.5
2.0
1.5
CL
lift coefficient
3.5
a=45
a=22.5
1
1.0
0.5
a=-9
0
0.0
-0.5
a=90
a=-9
-9 0 9 18 27 36 45 54 63 72 81 90
angle of attack (degs)
-1
0
1
2
drag coefficient
3
4
2. Circulation
fluid travels faster
over to of biofoil
Flow is tangential
at trailing edge
U
Flow separates
at leading edge
Law of continuity
applies to streamline
Difference in velocity
across surface is equivalent
to net circular flow
around biofoil = Circulation, G
mathematically:
G   U  dS
G     dA
U
Kutta-Joukowski Theorem:
Lift   UG
(lift per unit span)
1
combine with 2
previous
definition:
CL SU 2 / R  UG
G
CL 
Uc
R=biofoil length
c= biofoil width
Consider 2D biofoil starting from rest:
G=0
G=0
G
starting
vortex
bound
vortex
-G
Required by
Kelvin’s Law
Consider 3D biofoil starting from rest:
Helmholtz’ Law requires that a vortex filament cannot end abruptly:
bound
vortex
Downward flow
through center
of vortex ring
starting
vortex
tip
vortex
Circulation, G, is constant
along vortex ring
How is structure of vortex ring related to lift on biofoil?
forward
velocity, U
R
Area = A
Circulation, G
Ring momentum = mass flux through ring= GA
Force = d/dt (GA)
= G d/dt(A)
= G R U
Force/R = GU = Kutta-Joukwski
Therefore, elongation of vortex ring is manifestation of force on biofoil.
Three important descriptors of fluid motion:
1. velocity, u(x,y)
ux
uy
2. vorticity, (x,y)
Duy

Dx
x
y
u(x,y)
3. circulation, G
Dux
Dy
GS
Fslap = m U / t
Fstroke =  G A /t
Momentum
of vortex ring
 G A
where m is bolus of
accelerated water, moving at
velocity, u
impulse (F x t) = mass x velocity
A
G = circulation