Transcript AE 301 Aerodynamics I

Rankine Ovals
• The previous flow example has limited application since
the body isn’t closed but extends downstream to x.
• A simple way of closing the body is to add a sink
downstream of the source and of equal strength.
• For symmetry purposes, we will also move the origin to
be halfway between the source and sink. Thus:
+Q
y
-Q
V
x
-xo
xo
• Since the center of the source and sink are no longer on
the origin, the definitions of r and  in terms of x and y
also change.
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Rankine Ovals [2]
• For any point (x,y) in space, the following relations
(x,y)
apply:
r2
1
r1
2
(-xo,0)
r2 
x  xo 2  y 2
y
 2  atan
x  x0
(xo,0)
r1 
x  xo 2  y 2
1  atan
y
x  x0
• And the stream function solution is:
Q
Q
  V y   2  1
2
2
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Rankine Ovals [3]
• The stream (and potential) function solution to this case
looks lie:
• This body type is known as a Rankine Oval.
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Rankine Ovals [4]
• First lets find the length of this body.
• Note that for this flow, there are two stagnations points
– one at both the leading edge and trailing edges.
• To determine the location of these points on the y axis,
once again find where u=0.

 
Q
Q 
u
 V y 
 2  1 
y y 
2
2 


x  xo 
x  xo 
Q
Q
 V 

2
2
2 x  xo   y
2 x  xo 2  y 2
• At the stagnation point (xs,0)
Q
1
Q
1
u  0  V 

2 xs  xo  2 xs  xo 
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Rankine Ovals [5]
• With reduction, this expression becomes:
Qxo
xs2  x02 
2V
• Or
xs   x02 
Qx o
 a
2V
• The streamline which forms the body shape is found
from the value of psi at the right stagnation point:
Q
Q
 a,0  V (0)  (0)  (0)  0
2
2
• Since psi is zero on the surface, the following
relationship exists:
Q
Q
  VY   2  1  0
2
2
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Rankine Ovals [6]
• Or, the difference in the two angles is:
 2V  Y
 
1   2  Y 
 Q  qa
• Where a nondimensional source strength, q, is defined
by:
Q
q
2aV
• Taking the tangent of the first equation, substituting for
the tangent of the angles, and then solving for x, the
following equation is found for the surface:
2 x0Y
2
2
2
x  x0  Y 
tan(Y / aq)
• Which is really x(Y) rather than Y(x) – but is the best we
can do.
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Rankine Ovals [6]
• Just for completeness, the u and v components of this
flow velocity are:

x  xo  
Q  x  xo 
u  V 


2
2
2
2
2  x  xo   y
x  xo   y 

Qy 
1
1
v


2
2
2  x  xo   y
x  xo 2  y 2 
• And, once again, the pressure coefficients can be found
2
2
from:
 u   v 
c p  1      
 V   V 
• But I really doubt it this can be simplified to a short
simple equation.
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Doublets and Cylinders
• While Rankine ovals are not a common aerodynamic
shape, they do lead to one that is – cylinders.
• First, consider a source/sink pair and consider the limit
as they closer and closer (xo  0) while their strengths
increase such that Q 2xo = = constant, we get:
(x,y)
 


  Limx0 0 
2 
1 
2x0 
 4x0
 2  1 


Limx0 0 

2
2
x
0 

r2
2
(-xo,0)
r1
1
(xo,0)
• This limit can be evaluated by looking at the details of
the triangle formed…
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Doublets and Cylinders [2]
d=1-2
• By right triangle rules:
a  2 x0 sin  2
a
 2 x0 sin  2 

1   2  d  asin   asin
r1
 r1 


r1

a
2
2xo
• In the limit, d is very small, r1= r2 =r, and 1= 2 = .
• Thus:
 sin 
 y
 

2 r
2 r 2
• Or similarly for the potential function:

AE 401 Advanced Aerodynamics
 cos 
 x

2 r
2 r 2
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Doublets and Cylinders [3]
• The streamlines defined by this Doublet look like:
• One nice feature of a doublet is that it has a zero net
mass flux – what flows out flows back in!
• The other nice feature is when a doubled is combined
with a uniform flow.
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Doublets and Cylinders [4]
• When a uniform flow is superimposed on a doublet, the
flow looks like that around a circular cylinder.
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Doublets and Cylinders [5]
• The stream and potential functions for the flow just
shown are usually written as:
 a2 
  V y1  2 
 r 
 a2 
  V x1  2 
 r 
• As part of your homework, you will show that the
velocity on the surface of the cylinder is given by:
V  u  2V sin 
• And thus the surface pressure coefficient is:
2
V 
c p  1     1  4 sin 2 
 V 
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