Airfoils

The panel method

Laboratory Session #1 – Aerodynamics

Oct 2014

Alvaro Dosantos Moreno 100283186

1) Value of the elements of the matrix [Aij]First the values for beta, theta and r must be found.

r [i,j]=
0.5

0.5

0.354

0.5

0.877 0.1885 0.1885 0.877
0.39

0.6474 0.39

0.39theta [j]=
π

π /4 - π/9

Once these values have been obtained A can be solved:
A=
0 -0.1346 0.1763
0.0750

0 -0.2387

-0.0649 0.1144

0Also the RHS of the equation:
R=

The first column stands for alpha=0º and the second for alpha=10º
0

-0.1736

-0.7071 -0.5736
0.3420

0.52) Show that the matrix is singular before applying Kutta’s condition

DeterminantOfA = -5.7246e-04≈0

3) Apply Kutta’s condition and write thesystem of equations (in matrix
form)

A1 =
0 -0.1346 0.1763
0.075

0

-0.2387

1

0

1

As before, in the matrix R1, the first column is foralpha=0º and the second for alpha=10º
[A] * [ɣ] = [R] would be the system used to solve for ɣ

4) Solve γi for α = 0° and α =10°, and compute the lift inboth cases.

gamma0 =
-2.2539
2.9514
2.2539

To calculate the 𝑐𝑙 the total circulation is needed:
𝛤 = ∑𝑛𝑖 𝛾𝑖 ∗ 𝑙𝑖

where l is the length of eachpanel

𝛤0º =
0.6168

𝑐𝑙 = 2 ∗ 𝛤 ∗ ρ ∗ U∞ as as assuming ρ and 𝑈∞ are equal to 1:

CL1 =
1.2336

5) Compare the result with the prediction of theThin Airfoil Theory.
Which result do you think is more accurate and why?

Applying Thin Airfoil Theory:

1
dzc 
 2
dx 
 0.1819
1

0