Transcript AE 301 Aerodynamics I - Prescott Campus, Arizona

Source Panels
• Let’s now go back and see how we can setup a numeric
solution for a distributed source.
• We will use what we learned about stagnation points
and offset the start and finish of the line source by half
the radius of curvature at the two ends.
q(t )
x
q
c
xs
t
xf
xs  12  LE
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x f  1 12 TE
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Source Panels [2]
• The stream function for this case is given by:
x f q (t )
 y 
 x, y   V y  
atan
dt
xs 2
 x t 
• We could solve this equation directly, however, lets
consider changing this relation by doing an integration
by parts.
• Remember that the general rule of integration by parts
is:
b
b
b
 F (t )G(t )dt  F (t )G(t ) a   F (t )G(t )dt
a
a
• Thus, let’s introduce a new function, m(x) defined by:
dm ( x)
q( x) 
 m( x)
dx
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Source Panels [3]
• Thus, our stream function is:
x f m(t )
 y 
 x, y   V y  
atan
dt
xs
2
 x t 
• Which by integration by parts becomes:
xf
x f m(t )
m(t )
y
 y 
 x, y   V y 
atan
dt
  x
2
2
s
2
2 x  t   y
 x  t  xs
• However, to find m(x) we integrate q(x) over the line
source up to that point:
xs
m( xs )   q( x)dx  0
xs
xf
m( x f )   q( x)dx  0
xs
• We know the expression for m(xf) is zero because the
integral show is the total mass flux – which we want to
be zero!
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Source Panels [4]
• Thus, we finally get the equation:
x f m(t )
y


 x, y  V y  
dt
2
2
xs 2 
x  t  y
• But, what is this new function m(x)?
• Well, if you look at the form of the integrand, you will
realize this looks like the equation a doublet.
• Thus, m(x) is a doublet distribution! No wonder it
automatically satisfies mass conservation.
• We could attempt a numerical solution of our original
equation, but solving the above equation will not be any
harder, and our body will automatically be closed.
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Doublet Panels
• For our numerical approach, let’s approximate m(x) a
smooth function, with a stair-step type function shown
below.
m
c
x
xf
xs
• If we break m(x) up into into n steps or “panels”, then
we need to define n+1 points along the axis at each
jump – with t1 = xs and tn+1 = xf .
• On each step, the value of the function will be mj where
j=1…n.
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Doublet Panels [2]
• With this approximation, we can now rewrite out
equation as the summation of integrals over each panel:
n
 x, y   V y  
j 1
mj
y
t j 1
2  x  t 
2
tj
y
2
dt
• The doublet line strength came out of the integral since
it is constant on the panel.
• However, the remaining integral is very familiar:
t j 1

x, y 
t j 1
tx
y


dt

at
an
t j x  t 2  y 2
 j 1
 y  tj
j
  j 1   j
tj
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t j 1
Doublet Panels [3]
• So that our equation is simply:
n m
 x, y   V y   j  j 1   j 
j 1 2
• This equation applies to any point, (x,y), in space.
• From our earlier solutions for a doublet in uniform flow
we also know that  = 0 on the surface. Thus:
• However, to generate a solution for the mj‘s, we will
apply this equation to the surface where y=Y(x).
• In particular, we will apply this equation and n surface
points, (xi,Yi) as shown on the following page.
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Doublet Panels [4]
i= 1
2
4
3
5
6
n-1
n
• At these points, it must be true that:
n m
 ( xi , Yi )  0  VYi   j i , j 1  i , j 
j 1 2
• or
n
mj
i, j 1  i, j 
Yi  
j 1 2V
mj 
 t j 1  xi 
 t j  xi  
 atan
  atan
 




j 1 2V 
 yi 
 yi  
n
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Doublet Panels [5]
• So once again we have a matrix formulation, this time
looking like:
Am  Y 
• where the elements of the A matrix are:
 t j 1  xi 
 t j  xi 
  atan

Ai , j  atan
 yi 
 yi 
• and I am using a simplified doublet strength:
mj
mj 
2V
• The main complication from our earlier solution is
keeping track of two x-coordinates: xi and tj.
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Doublet Panels [6]
• Once we have solved for the doublet strengths, the
velocities can be found by:
m j (t  x)
u x, y 
 1 
2
2
V
(
t

x
)

y
j 1
t j 1
n
mj y
vx, y 
 
2
2
V
j 1 (t  x)  y
n
tj
t j 1
tj
• And, as always:
2
 u   v 
c p  1      
 V   V 
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Doublet Panels versus Point Sources
• As part of your homework, you will program this method
into Matlab.
• What you will find is that there is very little difference
between this approach and our earlier point sources.
• In fact there are some very close similarities between
the two methods.
• The advantage of the panel approach is that our
sources/doublets do not have to be on y=0 axis – they
could be on the body surface itself.
• Later on we will do exactly that – but that adds a
greater difficulty of having the panels at an angle to the
horizontal.
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Complex Variable Method/Joukowski Airfoils
• The final portion of Moran’s Chapter 3 deals with an
analytic approach which conformally maps the solution
on a cylinder to that on an ellipse or airfoil shape.
• While interesting and intellectually stimulating (for
Yurick) the limitation of the method is that it only works
exactly for one family of airfoils: the Joukowski Airfoils.
• These airfoils work alright invisicidly, but do not have
either low drag or high Clmax characteristics.
• For all other airfoil types, a numerical approach must
again be used to approximate the conformal mapping.
• If you have to do a numerical approach anyway, I prefer
the panel method approach.
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