ME438 Aerodynamics is offered by Dr. Bilal Siddiqui to senior mechanical engineeing undergraduates at DHA Suffa University. This lecture set deals with thin airfoil theory.

1. Aerodynamics
ME-438
Spring’16
ME@DSU
Dr. Bilal A. Siddiqui

3. Philosophy of the Vortex Panel Method
• Replace the airfoil surface with a vortex sheet of strength 𝛾(𝑠)
• Find the distribution of 𝛾 along s such that the induced velocity field
from the vortex sheet when added to the uniform velocity 𝑉∞ will
make the vortex sheet (airfoil surface) a streamline of the flow.
• The circulation around the airfoil will be given by Γ = ∫ 𝛾𝑑𝑠
• The lift per unit span can then be calculates by the K-J theorem
𝐿′ = 𝜌∞ 𝑉∞Γ
Developed by Ludwig Prandtl during 1912–1922

4. Kutta Condition
• For a cusped trailing edge, the edge angle is almost zero, therefore we can
have nonzero speed at the edge. However, since only one pressure can
exist at ‘a’, V1=V2
• We can summarize the statement of the Kutta condition as follows:
1. For a given airfoil at a given angle of attack, the value of Γ around the airfoil is
such that the flow leaves the trailing edge smoothly.
2. If the trailing-edge angle is finite, then the trailing edge is a stagnation point.
3. If the trailing edge is cusped, then the velocities leaving the top and bottom
surfaces at the trailing edge are finite and equal in magnitude and direction.
• Since, we have 𝛾 = 𝑢1 − 𝑢2, this means the Kutta condition is
𝜸 𝑻𝑬 = 𝟎

5. Thin Airfoil Theory
• Most airfoils have negligible thickness compared to the chord length
(less that 20%)
• The camber is also rather small mostly (less than 5% of the chord)
• This allows for some assumptions to be valid for many airfoils
• More importantly, it allows exact analytic close formed solution for
the calculation of lift, drag and pitching moment of
• “thin” airfoils (
𝑡
𝑐
≤ 12%)
• at low angles of attack (𝛼 ≤ 10°)
• and low speeds (incompressible; 𝑀∞ ≤ 0.3)
• in inviscid flow (irrotational)

6. Thin Airfoil Theory - 2
• 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.

7. Thin Airfoil Theory - 3
• For the vortex sheet placed on the camber line, 𝛾′ =
𝛾′ 𝑠
• Normal velocity induced by the sheet is 𝑤(𝑠).
• For the camber line to be a stream line 𝑉∞ 𝑛
+
𝑤′
𝑠 = 0
• For another sheet placed on the chord line, 𝛾 = 𝛾(𝑥)
• Normal velocity induced by the sheet is 𝑤(𝑥). For
small camber 𝑤′
𝑠 ≅ 𝑤(x)
• For the camber line to be a stream line 𝑉∞ 𝑛
+
𝑤 𝑥 = 0
• Kutta condition: 𝑤 𝑐 = 0

8. Thin Airfoil Theory-4
• From geometry
𝑉∞ 𝑛
= 𝑉∞ sin 𝛼 + tan−1 −
𝑑𝑧
𝑑𝑥
• For small angles sin 𝜃 ≅ 𝜃
• Both camber and angle of attack
are small, so
𝑉∞ 𝑛
≈ 𝑉∞ 𝛼 −
𝑑𝑧
𝑑𝑥

9. Thin Airfoil Theory-5
• Induced normal velocity at point x due to the vortex filament at 𝜉
𝑑𝑤 = −
𝛾 𝜉 𝑑𝜉
2𝜋 𝑥 − 𝜉
Total induced normal velocity at x is
𝑤 = −
1
2𝜋 0
𝑐
𝛾 𝜉
𝑥 − 𝜉
𝑑𝜉

10. Thin Airfoil Theory - 6
• Therefore, the ‘no penetration’ (abstinence?) boundary condition is
𝑽∞ 𝜶 −
𝒅𝒛
𝒅𝒙
=
𝟏
𝟐𝝅 𝟎
𝒄
𝜸 𝝃
𝒙 − 𝝃
𝒅𝝃
• This is the fundamental equation of the thin airfoil theory.
• For a given airfoil, both 𝛼 and
𝑑𝑧
𝑑𝑥
are known.
• We need to find 𝛾 𝑥 which
• makes the camber line 𝑧(𝑥) a streamline of the flow
• and satisfied the Kutta condition 𝛾 𝑐 = 0

11. Specializing Thin Airfoil Theory for Symmetric Airfoils
• For a symmetric airfoil, the fundamental equation simplifies to
𝑽∞ 𝜶 =
𝟏
𝟐𝝅 𝟎
𝒄
𝜸 𝝃
𝒙 − 𝝃
𝒅𝝃
• Let us transform the variable 𝜉 to another variable 𝜃
𝜉 =
𝑐
2
1 − cos 𝜃 → 𝑑𝜉 =
𝑐
2
sin 𝜃𝑑𝜃
For any particular value of 𝜉 = 𝑥, there is a corresponding particular 𝜃 𝑥
𝑥 =
𝑐
2
1 − cos 𝜃 𝑥 [𝜃0 = 0 and 𝜃𝑐 = 𝜋]
• The fundamental equation can then be written equivalently as
𝑽∞ 𝜶 =
𝟏
𝟐𝝅 𝟎
𝝅
𝜸 𝜽 𝒔𝒊𝒏 𝜽
𝒄𝒐𝒔𝜽 − 𝒄𝒐𝒔 𝜽 𝒙
𝒅𝜽

12. Thin Airfoil Theory for Symmetric Airfoils-2
• The solution to this integral equation can be shown to be
𝜸 𝜽 =
𝟐𝜶𝑽∞ 𝟏 + 𝐜𝐨𝐬 𝜽
𝐬𝐢𝐧 𝜽
• It can be easily shown to satisfy the fundamental equation as well as
the Kutta condition 𝛾 𝑐 = 𝛾 𝜋 = 0.
• Total circulation is found by Γ = ∫0
𝑐
𝛾 𝜉 𝑑𝜉 =
𝑐
2
∫0
𝜋
𝛾 𝜃 sin 𝜃 𝑑𝜃
• This can be evaluation as
Γ = 𝑉∞ 𝛼
0
𝜋
1 + cos 𝜃 𝑑𝜃 = 𝜋𝛼𝑐𝑉∞

13. Thin Airfoil Theory for Symmetric Airfoils-3
• The lift can now be calculated by the K-J theorem
𝐿′ = 𝜌∞ 𝑉∞Γ = (𝜌∞ 𝑉∞
2 𝑐)𝜋𝛼
Thus,
𝒄𝒍 = 𝟐𝝅𝜶
𝒄𝒍 𝜶
=
𝒅𝒄𝒍
𝒅𝜶
= 𝟐𝝅
These is the fundamental result of thin airfoil theory. The lift curve
slope is 0.11 per degree angle of attack! It fits well with experiment.

14. Thin Airfoil Theory for Symmetric Airfoils-4
• For calculating moment about leading edge, the incremental lift 𝑑𝐿 =
𝜌∞ 𝑉∞ 𝑑Γ due to circulation 𝛾 𝜉 𝑑𝜉 caused by a small portion 𝑑𝜉 of
the vortex sheet is multiplied by its moments arm (𝑥 − 𝜉).
• The total moment is the integration of these small moments
𝑀′ 𝐿𝐸 = −
0
𝑐
𝜉𝑑𝐿 = −𝜌∞ 𝑉∞
0
𝑐
𝜉𝛾 𝜉 𝑑𝜉 = −
1
4
𝜌∞ 𝑉∞
2 𝑐2 𝜋𝛼
Therefore the moment coefficient is 𝑐 𝑚 𝐿𝐸
= −
𝜋𝛼
2
= −
𝑐 𝑙
2

15. Thin Airfoil Theory for Symmetric Airfoils-5
• Shifting this moment to the quarter chord point
𝑀𝑐/4
′
= 𝑀′ 𝐿𝐸 + 𝐿
𝑐
4
=
1
2
𝜌∞ 𝑉∞
2
𝑐 −
𝜋𝛼𝑐
2
+
𝜋𝛼
2
𝑐 = 0
• 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.
• This means the quarter chord point on a symmetric airfoil is both the
center of pressure and the aerodynamic center!
• This is only true for thin, symmetric airfoils at low speeds!
• This too holds up very well in experiments.

16. Experimental Validation of Thin Airfoil Theory
For Symmetric Airfoils

17. An example
Consider a thin flat plate at 5 deg. angle of attack. Calculate
• lift coefficient
• moment coefficient about the leading edge
• moment coefficient about the quarter-chord point
• moment coefficient about the trailing edge.