# Project2A - Cal State LA

```ME 408 Project#2
Lift Coefficient of Joukowski Airfoil (Due Wednesday 2/7/07)
1. Joukowski Transformation:
An airfoil shape can be created by mapping a circle in the (x1, y1) plane to a new (x2, y2) plane with
the transformation:
1
z2 z1 
z1
z1
x1  1  y1
x2  1  y2
z2
1 is the imaginary
are complex variables, and
number.
Also, the x-coordinate of the center of the circle must be negative (x0 &lt; 0), its y-coordinate must be
positive (y0 &gt; 0), and the circle must pass through the point (1, 0) on the x1-axis. Hence the radius of
the circle is:
a
 1  x0 2  y02
Hence the coordinates of the circle for each angular location (0 &lt;  &lt; 2) are:
a cos    x0
x1
y1
a sin    y0
(a) Show that the coordinates of the resulting airfoil shape are:
x2

2
2
x1  x 1  y1  1

y2
x1 2  y1 2

2
2
y1  x1  y 1  1

x1 2  y1 2
(b) Select a center location (x0, y0) for the circle, and calculate the radius and the coordinates of the
circle for each . Calculate the resulting Joukowski airfoil coordinates, and plot both the circle
and the airfoil shape. Use the coordinates on the x2-axis to determine the chord length c.
Notice that the point (1, 0) in the (x1, y1) plane is mapped to the sharp trailing edge of the
airfoil.
Joukowski Transformation
1.5
1.0
0.5
0.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
-0.5
-1.0
0.5
1.0
1.5
2.0
2.5
2. Kutta Condition and Lift Coefficient at =0
(a) Recall that at zero angle of attack (=0) and with no circulation (=0), the point at  = 0 on the
circle is a stagnation point, and the streamline from this stagnation point is simply the horizontal line
y1 = y0. Map this streamline onto the (x2, y2) plane of the airfoil. Notice that the streamline
along the lower surface of the airfoil has to make a 180-degree sharp turn to the stagnation point on
the upper surface.
(b) Since it is physically impossible for a streamline to change direction at the sharp trailing edge, the
flow will add to itself a clockwise circulation to move the stagnation point on the circle clockwise to
the point (1, 0), that is, through an angle:
 y0 
 sin  1 
 a
This requirement is known as Kutta condition, which ensures the stagnation point on the airfoil to
locate at the sharp trailing edge. Show that the circulation strength required is:

4  a U  sin  
4  U  y0
where U is the free stream velocity. Since the stream function is 0 at the stagnation point on a circle,
the streamline shape from the trailing stagnation point in polar coordinate is simply:



U r 1 
2
a 
r 2 
 sin  

 a
 ln  
2  r 
Trace this streamline from r = a to r = 3 by computing the corresponding (x1, y1) coordinates.
Map the stagnation streamline onto the (x2, y2) plane, and verify that it indeed starts from the
sharp trailing edge.
(c) As we conclude from the study of potential flow past a circle, this circulation will generate a lift on
the circle and therefore also on the airfoil:
L  U  
where is the fluid density.
Compute the lift coefficient of the airfoil defined as: CL
2L
U 2c
3. Lift Coefficient at Non-Zero Angle of Attack (&gt; 0)
For &gt; 0, the stagnation point on the circle simply has to shift (clockwise to satisfy Kutta
condition. Hence the clockwise circulation required is:
 4aU sin    
(a) The lift is defined as the net force normal to the free stream. Show that the lift coefficient is:
CL
a
8   sin     
c
(b) Compute and plot the lift coefficient from (the zero-lift angle of attack) to  100.
```