Techno Tips | Machine Technology | Robotic Technology: Derivation of thin airfoil theory

Friday, 16 December 2011

Derivation of thin airfoil theory

From top to bottom:
• Laminar flow airfoil for a RC park flyer
• Laminar flow airfoil for a RC pylon racer
• Laminar flow airfoil for a manned propeller aircraft
• Laminar flow at a jet airliner airfoil
• Stable airfoil used for flying wings
• Aft loaded airfoil allowing for a large main spar and late stall
• Transonic supercritical airfoil
• Supersonic leading edge airfoil

Colours:
Black = laminar flow,
red = turbulent flow,
grey = subsonic stream,
blue = supersonic flow volume

The airfoil is modeled as a thin lifting mean-line (camber line). The mean-line, y(x), is considered to produce a distribution of vorticity

γ(s)

along the line, s. By the Kutta condition, the vorticity is zero at the trailing edge. Since the airfoil is thin, x (chord position) can be used instead of s, and all angles can be approximated as small.

From the Biot-Savart law, this vorticity produces a flow field

w (x)

where

$w(x) = \frac{1} {(2 \pi)} \int_{0}^{c} \frac {\gamma (x')}{(x-x')} dx'$

where

x

is the location where induced velocity is produced,

x'

is the location of the vortex element producing the velocity and

c

is the chord length of the airfoil.

Since there is no flow normal to the curved surface of the airfoil,

w (x)

balances that from the component of main flow

V

, which is locally normal to the plate—the main flow is locally inclined to the plate by an angle

α - d y / d x

. That is:

$V . (\alpha - dy/dx) = w(x) = \frac{1} {(2 \pi)} \int_{0}^{c} \frac {\gamma (x')}{(x-x')} dx'$

This integral equation can by solved for

γ(x)

, after replacing x by

$\ x = c(1 - cos (\theta ))/2$ ,

as a Fourier series in

A n s i n (n θ)

with a modified lead term

A 0 (1 + c o s (θ)) / s i n (θ)

That is $\frac{\gamma(\theta)} {(2V)} = A_0 \frac {(1+cos(\theta))} {sin(\theta)} + \sum A_n . sin (n \theta))$

(These terms are known as the Glauert integral).

The coefficients are given by $A_0 = \alpha - \frac {1}{\pi} \int_{0}^{\pi} (dy/dx) . d\theta$

and $A_n = \frac {2}{\pi} \int_{0}^{\pi} cos (n \theta) (dy/dx) . d\theta$

By the Kutta–Joukowski theorem, the total lift force F is proportional to

$\rho V \int_{0}^{c} \gamma (x). dx$

and its moment M about the leading edge to $\rho V \int_{0}^{c} x.\gamma (x) . dx$

The calculated Lift coefficient depends only on the first two terms of the Fourier series, as

$\ C_L = 2 \pi (A_0 + A_1/2)$

The moment M about the leading edge depends only on

A 0, A 1

and

A 2

, as

$\ C_M = - 0.5 \pi (A_0+A_1-A_2/2)$

The moment about the 1/4 chord point will thus be,

$\ C_M(1/4c) = - \pi /4 (A_1 - A_2)$ .

From this it follows that the center of pressure is aft of the 'quarter-chord' point 0.25 c, by

$\ \Delta x /c = \pi /4 ((A_1-A_2)/C_L)$

The aerodynamic center, AC, is at the quarter-chord point. The AC is where the pitching moment M' does not vary with angle of attack, i.e.,

$\frac { \partial (C_{M'}) }{ \partial (C_L)} = 0$

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