c- 2: unit c- 3: list of subjects

AE301 Aerodynamics I

UNIT C: 2-D Airfoils

ROAD MAP . . .

C-1: Aerodynamics of Airfoils 1

C-2: Aerodynamics of Airfoils 2

C-3: Panel Methods

C-4: Thin Airfoil Theory

AE301 Aerodynamics I

Unit C-3: List of Subjects

Problem Solutions? How?

Source Panel Method

Vortex Panel Method

This is a tutorial (self-study) material for the COURSE PROJECT

• Not covered in class as a standard lecture of AE301

• Online lecture (YouTube) Available

PANEL METHODS / THIN AIRFOIL THEORY: PURE THEORETICAL SCHEMES

Panel methods and thin airfoil theory are built upon potential flow analysis. In other words, these are

pure theoretical schemes. Similar to the unit B-3, as we deal with the pure theory, one must be very

carefully recognize (and keep track of) the followings:

(1) What are the assumptions made to simplify the equation?

• Steady-state?

• Inviscid?

• No body forces?

• Incompressible?

• Irrotational?

(2) Because of the assumptions made, how your theoretical solution different (or apart) from the actual

(or real) flow field phenomena?

Unit CUnit C--33Page Page 11 of 12of 12

Problem Solutions? How?Problem Solutions? How?

SOURCE PANEL METHOD

A source sheet can be defined as infinite number of line sources (side by side) where the strength of

each line source is infinitesimally small.

Source strength per unit length can be defined as: ( )s = .

Consider an arbitrary point ( , )P x y . . . a small section of the source sheet of strength ds induces an

infinitesimally small velocity potential d :

ln2

dsd r

=

Thus, the complete velocity potential at point P induced by the entire source sheet is:

( , ) ln2

b

a

dsx y r

=


Source Panel Method (1)Source Panel Method (1)

Let us approximate the source sheet by a series of straight panels, moreover, let the source strength per

unit length be constant over a given panel.

For the total number of n panels, the source strengths per unit length for each panel can be represented

by: 1 2 3, , , , ,j n

Then, solve: , 1, 2,3 ,j j n = such that the body surface becomes a streamline of the flow: means that

the boundary condition is imposed numerically by j ’s such that the normal component of the flow

velocity is zero at the midpoint of each panel (called, the “control point”).

Again, consider a point P (x,y) in the flow field, and let pjr be the distance from any point on the jth

panel to P.

The velocity potential induced at P due to the jth panel is:

ln2

j

j pj j

j

r ds

=



The potential at P due to all panels is:

( )1 1

ln2

n nj

j pj j

j j j

P r ds

= =

= =

where, the distance 2 2( ) ( )pj j jr x x y y= − + −

Next, let us put P at the control point of the jth panel ( , )i iP x y :

( )1

, ln2

nj

i i ij j

j j

x y r ds

=

= 2 2( ) ( )ij i j i jr x x y y= − + −

This is the contribution of all panels to the potential at the control point of the ith panel.

The free stream component normal to the panel is:

,ˆ cosn i iV V n V = =

The normal component of velocity induced at ( , )i ix y by a panel is:

( , )n i i

i

V x yn

=



Combining this with the potential ( , )i ix y , which is contribution of all the panels to the potential at the

control point of the ith panel:

( )1

( 1)

ln2 2

nji

n ij j

j ijj

V r dsn

=

= +

Applying the boundary condition, , 0n nV V + = , yields:

( )1

( 1)

ln cos 02

ni

ij j i

j ijj

r ds Vn

=

+ + =

Let: , ,(ln )i j i j j

ij

I r dsn

=

=>

( ),

1 1

cos 02 2

nji

i j i

j j

I V

=

+ + =

This is a linear algebraic equation with n unknowns ( 1 2 3, , , , ,j n ) => these unknowns can

be solved (n equations with n unknowns)

Once ’s are all solved ( , 1,2,3 ,i i n = ) . . .

The component of free stream velocity tangent to the surface is:

, sins iV V =

The tangential velocity sV at the control point of the ith panel induced by all the panels is:

( )1

ln2

ni

s ij j

j j

V r dss s

=

= =

The total surface velocity at the ith control point iV is the sum of the contribution from the free stream

and from the source panels:

( ),

1

sin ln2

ni

i s s i ij j

j j

V V V V r dss

=

= + = +

Finally, the pressure coefficient at the ith panel can be given as: 2

, 1 ( )p i iC V V= −



Source panel code has two fundamental problems:

• Source panel does not simulate circulation. Therefore, the source panel code’s application is

inherently limited to a nonlifting flow around a body only.

• Source panel cannot enforce the Kutta condition at the trailing edge. Therefore, the stagnation

point is formed at an arbitrary location as the angle of attack is increased, and this causes the

nonphysical solutions.


Class Example Problem CClass Example Problem C--33--11

Related Subjects . . . Related Subjects . . . ““Source Panel MethodSource Panel Method””

There is a “theoretical” airfoil, called a “Joukowski” airfoil. The shape of this airfoil

can be obtained from a circle by a mathematical transformation between two domains.

Since this airfoil is a theoretically derived shape, the surface pressure coefficient

distribution (Cp) can also be obtained theoretically. Using source panel method,

calculate the pressure coefficient distribution (Cp) of symmetic Joukowski airfoil at

AOA = 0. Compare the theoretical distribution of Cp (Joukowski airfoil: known

values) against Cp computed by the source panel method.

Symmetric Joukowski Airfoil

VORTEX SHEET

A vortex filament can be defined as a straight line perpendicular to the page, going through point O

(extend to infinity both sides).

A vortex sheet can be defined as infinite number of vortex filaments side by side, with the strength of

each filament is infinitesimally small.

Consider an arbitrary point P (x, z). The small section of vortex sheet of strength ds induces a small

velocity potential and velocity dV :

2

dsd

= − and

2

dsdV

r

= −

The velocity potential at P due to entire vortex sheet from a to b is, therefore:

1( , )

2

b

a

x z ds

= − where, the circulation is:

b

a

ds =


Vortex Panel Method (1)Vortex Panel Method (1)

Consider a dashed path enclosing a vortex sheet of strength ds :

2 1 1 2 1 2 1 2( ) ( ) ( )C

v dn u ds v dn u ds u u ds v v dn = − = − − − + = − + − V ds

Let the top and bottom of the dashed line approach the vortex sheet:

0dn→ , and 1 2u u = −

Important conclusion: the local jump in tangential velocity across the vortex sheet is equal to the local

sheet strength. This is the fundamental concept of thin airfoil theory.

The vortex panel method is built upon the idea that, if we replace the 2-D airfoil shape by a vortex sheet,

it is possible to obtain the lift using the Kutta-Joukowski theorem.

Thin airfoil theory and vortex panel method are built upon the same fundamental models of vortex sheet.

• If airfoil is thin => thin airfoil theory

• If airfoil is not thin => vortex panel method



1 2u u = −ds =

VORTEX PANEL METHOD

Very similar approach (as discussed in source panel method).

A series of vortex sheets approximates the body of arbitrary shape (vortex panels):

Total number of n panels with the vortex strengths per unit length of: 1 2 3, , , , ,j n

At the control points, the normal component of the velocity is zero:

, 0n nV V + =

1

cos 02

nj ij

i j

j ij

V dsn

=

− =

or

1

cos 02

nj

i ij

j

V I

=

− =

Note that the Kutta condition can be applied precisely at the trailing edge and is given by (TE) 0 = ,

means at trailing edge: 1i i −= −

This Kutta condition forces the flow field to have a proper stagnation point at (or near) the trailing edge

of the flow field.



Once the unknown vortex strength, 1 2 3, , , , ,j n , are obtained, the total circulation for the given

flow field can be calculated as:

1

n

j j

j

s=

=

Therefore, the lift per unit span is (from the Kutta-Joukowski theorem):

1

'n

j j

j

L V s

=

=



VARIETY OF SOURCE / VORTEX PANEL METHOD SCHEMES

• 1st order vortex panel method: the strength of vortex panel is constant for a given panel.

• 2nd order vortex panel method: the strength of vortex panel varies linearly over a given panel.

• Combination of source + vortex panels: use source panel to simulate the airfoil thickness, use vortex

panel to introduce circulation (lift).



There are a few important restrictions of using Vortex Panel Java Applet

• Airfoil coordinate data must be arranged that:

Trailing Edge (1,0) => Lower Surface Data => Leading Edge (0,0) => Trailing Edge

• Too much panels clustered near the trailing edge will make it difficult to properly apply Kutta

condition: you may want to manually remove some panels.

• Total number of panels should be in the range of 30 to 100. Using too many panels will slow down

computations and will not necessarily improve the accuracy.


Class Example Problem CClass Example Problem C--33--22

Related Subjects . . . Related Subjects . . . ““Vortex Panel MethodVortex Panel Method””

Using Virginia Tech vortex panel method (Java Applet), determine the lift coefficient

of NACA 0009 airfoil over a range of angles of attack from 0 to 16 degrees. Plot a lift

curve (cl v.s. ) and compare it against NACA airfoil data.

http://www.engapplets.vt.edu/

Lift Coefficient v.s. AOA

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 2 4 6 8 10 12 14 16

AOA (degrees)

Se

cti

on

Lif

t C

oe

ffic

ien

t

Vortex Panel Method NACA 0009 Data

AOA

(degrees) Vortex Panel Method NACA 0009 Data

0 0 0

1 0.118183

2 0.23633 0.2

3 0.354404

4 0.472371 0.4

5 0.590194

6 0.707837 0.6

7 0.825265

8 0.942441 0.85

9 1.05933

10 1.175896 1.08

11 1.292105

12 1.407919 1.26

13 1.523305

14 1.638227 1.3

15 1.752649

16 1.866538 1.14

Lift Coefficient

c- 2: unit c- 3: list of subjects

Documents