Transcript AE 301 Aerodynamics I

Biot-Savart Law
• It will be very important in our analysis of wings to be
able to predict the induced velocity of a vortex filament
of finite length.
• In 2-D flow, we saw that the velocity due to a vortex by
y
itself was given by:
u
z

V ( r )  u 
2r
x


r
• But, this is for a vortex filament which extends to plus or
minus infinity.
• The relation for predicting the induced velocity of a finite
length filament is known as the Biot-Savart Law.
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Biot-Savart Law [2]
• The equation for this law is an integral:

V ( x, y , z ) 
4
dl  r
 r3
V

z
y
r ( x, y, z)
dl
x
• Note the integration is from one end of the vortex to the
other and includes the possibility of a curved or closed
vortex line.
• For a straight line vortex, the problem is a little simpler
since the cross product yields:
e

dl  r  r sin dle
• Where e is a unit azimuthal vector.
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
r
dl
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Biot-Savart Law [3]
• The incremental induced velocity for a short segment of
the filament is then:
e sin dl
dV ( x, y, z ) 
4
r2
• However, if we introduce the normal distance to the
point as d, then:
l
d  r sin 
l  d cot 
d
dl 
d
2
sin 
dl

r
• Therefore:
e
 sin    d  e sin 
dV 
sin  
d
  2  
4
 d   sin   4 d
2
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d
Biot-Savart Law [4]
• When this integrated over the entire length of the
filament, the result is:
e  2
V 
sin d

4d 1
e
cos 1  cos  2 
V 
4d
d
1
r2
r1
• Note that if the filament were indeed infinite, then:
1  0
2  
e

1  (1) 
V 
e
4d
2d
• Which is our result in 2-D flow (with a change in sign
due to the change in axis).
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2
Biot-Savart Law [5]
• Of course, Hemholtz’s theorems say that we can’t have a
single vortex filament like we have shown.
• Instead, we could have a horseshoe vortex with both a
bound vortex and two trailing vortices.
• The total contribution from this horseshoe vortex to a
point which lies in the same plane is:
l2
2

cos 1  cos  2 
w
d
l
1
4d
z
y
 


1

1  cos  1  
4l1 
2

x
  
 

 cos  2    1
4l2  
2 
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Lifting Line Model
•
The simplest analytic model of a wing which yields
useful results is called the Lifting Line Model.
•
In this model, we consider a side, slender wing (i.e high
Aspect Ratio) and use only a single bound vortex with a
trailing vortex sheet.
z
b/2
y
x
 t ( y)
-b/2
•
trailing vortex
sheet
The local force per unit span is proportional to the
strength of the local bound vortex as dictated by the
Kutta-Joukowski theorem.
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Lifting Line Model [2]
•
•
•
However, unlike 2-D flow, the velocity should be the
local, not freestream velocity.
dF ( y)
f ( y) 
 V   ( y)
dy
The difference is due to the fact that there is a local
downwash on the bound vortex being produced by the
trailing vortices.
V
wi
i
V
If we assume the induced angle of attack is small, then
we get two force contributions at each station:
l ( y)  V( y)
d ( y)  wi ( y)  iV( y)
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Lifting Line Model [3]
•
The total wing lift and drag will be found be integrating
these section lift and drags.
•
Because the wing is long and slender, it is consistent to
assume that the local lift coefficient is also given by our
2-D result – including the downwash term:
cl  2 (   L0  i )  m0e
•
In this equation we are introducing both the 2-D lift
curve slope, mo , and the effective angle of attack:
 e     L0  i
•
To get this induced angle of attack, we need to sum up
the induced downwash of all the trailing vortices.
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Lifting Line Model [3]
•
Since we have no sweep, the induced velocity on the
bound vortex at point y due to the trailing vortex at
point t is just:
(t )
z
y
 (t )dt
(t )dt
 (t )
dwi ( y) 

4 t  y  4  y  t 
dw ( y)
x
•
i
Where we have made use of the earlier observation
that:
d(t )
 t (t )  
 (t )
dy
•
Thus, the total downwash at this point, is given by:
1
i 
V
1
b / 2 dwi ( y)  4V
b/2
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(t )
b / 2  y  t  dt
b/2
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Lifting Line Model [4]
•
In practice, we see that this integral is singular at y=t,
thus we should treat it as a Cauchy Principal Value
integral:
1 b / 2 (t )
i 
dt
P

4V b / 2  y  t 
•
Putting all this together gives the result:
•
Or
l ( y)
2( y)
cl  1

 m0 (   L 0   i )
2
V c
2 V c

1
( y)  cV m0    L 0 
4V

1
2
•
(t ) 
Pb / 2  y  t  dt
b/2
Which, at least technically, could be solved.
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Elliptic Lift Distribution
•
Before we jump in and attempt solutions, let’s consider
one special case which was discovered.
•
If the circulation (and thus lift) were described by an
elliptic function in the form:
( y )
s
2 1/ 2
  2y  
( y)  s 1    
  b  
-b/2
•
•
Then:
2 1/ 2 

d    2 y    s
( y ) 
s 1      
dy    b    2


  2y 
1   
  b 
b/2
2



1 / 2
  8y 
 2 
 b 
Which when substituted into our induced alpha term is:
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y
Elliptic Lift Distribution [2]
 s
i 
Vb 2
•
•
•
t
b/2
P 1  4t
b / 2
2
/b
 y  t 
2 1/ 2
dt
This integral can be evaluated using trigonometric
variable substitutions similar to our 2-D analysis:
b
b
b
y  cos  0
t  cos 
dt   sin 
2
2
2
With these, the integral becomes:
 s 
cos
P
i 
d

2Vb 0 cos 0  cos 
Appendix A on Cauchy Integrals gives the result:

P
0
 sin n o
cos n
d 
cos  cos 0 
sin  o
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Elliptic Lift Distribution [3]
•
•
Thus, an elliptical loading has a downwash of:
s

 i ,e 
wi ,e   i ,eV  s
2Vb
2b
Both of which are constants and not functions of y.
•
Thus, the elliptical spanload yields a uniform, constant
downwash from wing tip to wing tip.
•
The lift coefficient for an elliptic distribution is found by:
  2y 
L  V  ( y)dy  V s  1   
b / 2
b / 2
  b 
b/2
•
or
b/2
b  2
b
L  V s  sin  o d o  V s
2 0
4
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2 1/ 2

 dy

b
L  V s
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Elliptic Lift Distribution [4]
•
Another way to look at this is the circulation required to
give a require lift:
4L
s 
V b
•
For the drag, the integration is:
D  i ,eV 
b/ 2
b / 2
•
( y)dy   i ,e L
2L2
D
V2b 2
In terms of coefficients:
sb sAR
L
CL  1


2
2V S
2Vb
2 V S
D
4L2
CL2
CD  1
 2 4 2 
2
 V Sb  AR
2 V S
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Elliptic Lift Distribution [5]
•
The last thing we want to discuss about the elliptic
distribution is the lift curve slope.
•
We have the local lift coefficient given by:

s 
CL 

  mo    L 0 
cl  mo (   L 0   i )  mo    L 0 

2Vb 
AR 


• Thus, we can also calculate the wing lift coefficient by:
•
•
1 b/2
CL 

CL   cl cdy  mo    L 0 


b
/
2
S
AR 

Solving for the lift coefficient gives:
mo    L 0 
CL 
 m   L 0 
1  mo AR
mo
m
1  mo AR
Thus we have a lower lift curve slope on wings then for
the airfoils they are made of.
AE 401 Advanced Aerodynamics
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