AerodynamicsME-438

Spring’16ME@DSU

Dr. Bilal A. Siddiqui

Thin Airfoil Theory – Recall 1

• For thin airfoils, we can basically replace the airfoil with a single vortex sheet. For this case, Prandtl found closed form analytic solutions.• Looking at the airfoil from far, one can neglect

the thickness and consider the airfoil as just the camber line. The airfoil camber is z(x).• If we neglect the camber also, we can

basically place all the vortices on the chord line for the same effect.

Thin Airfoil Theory- Recall 2• From geometry

• For small angles • Both camber and angle of attack

are small, so

Thin Airfoil Theory - Recall 3• Induced normal velocity at point x due to the vortex filament at

Total induced normal velocity at x is

Thin Airfoil Theory – Recall 4• Therefore, the ‘no penetration’ (abstinence?) boundary condition is

• This is the fundamental equation of the thin airfoil theory.• For a given airfoil, both and are known.• We need to find which • makes the camber line a streamline of the flow• and satisfied the Kutta condition

Thin Airfoil Theory – General case

• Let us again transform the variable to another variable

For any particular value of , there is a corresponding particular [ and ]

• The fundamental equation can then be written equivalently as

Thin Airfoil Theory for Cambered Airfoils-2• The solution to this integral equation can be shown to be

• This distribution has the 1st term similar to the symmetric airfoil distribution and the 2nd term is the contribution due to camber. As expected!

• It can be shown that,

• Thus, the first term depends on the angle of attack as well as the camber, but the second term only depends on the camber.

• It can be easily shown to satisfy the fundamental equation as well as the Kutta condition .

Thin Airfoil Theory for Cambered Airfoils-3• Total circulation is found by • This can be evaluation as

• Using trigonometric identities,

• we can show

Thin Airfoil Theory for Cambered Airfoils-4• The lift can now be calculated by the K-J theorem

Thus,

This means the lift curve slope of a cambered airfoil is equivalent to the lift curve slope of a symmetric airfoil. They only differ in the zero AoA lift.

Kia karain hai…Lift lift hota hai

Thin Airfoil Theory for Cambered Airfoils-5• The angle of zero lift is denoted by αL=0 and

is a negative value.

• It is easy to see that

The more highly cambered an airfoil, the more negative its zero lift AoA

Thin Airfoil Theory for Cambered Airfoils-4• For calculating moment about leading edge, the incremental lift due

to circulation caused by a small portion of the vortex sheet is multiplied by its moments arm .• The total moment is the integration of these small moments

Moment coefficient then is Thus the moment too is that of symmetric airfoil, plusa constant term due to camber.

Thin Airfoil Theory for Symmetric Airfoils-5• Shifting this moment to the quarter chord point

• Now recall that:• Center of pressure is point on the chord about which there is zero moment• Aerodynamic center is the point on the chord about which the moment about the

chord does not change with the angle of attack.• A1 and A2 do not depend on AoA

• This means the quarter chord point on a cambered airfoil is the aerodynamic center, but not the center of pressure! • Center of pressure can be found by using the relation

This is clearly dependent on angle of attack.

Experimental Validation of Thin Airfoil Theory For Cambered Airfoils• Consider the NACA 23012 airfoil. Its camber line is given as

• Calculate (a) the angle of attack at zero lift, (b) the lift coefficient when α = 40, (c) the moment coefficient about the quarter chord, and (d) the location of the center of pressure in terms of xcp/c, when α = 40. Compare the results with experimental data.

Experimental Validation