Lift and moment coefficients of a flat plate aerofoil.
A Joukowski mapping can be used of known flow around a circular cylinder to predict flow over a flat plate aerofoil at angle of attack.
$$z_1=x_1+iy_1\text" "z_2=x_2+iy_2$$
$$z_2=z_1+k^2/z_1$$
where $ k=a $, the radius of circle
The equation for the surface of the circle can be set in polar coordinates as, $z_1=ae^{iθ}$ hence the equation for the surface of the plate will be found from the transformation,
$$\table z_2, =, ae^{iθ}+a^2/{ae^{iθ}}; , =, ae^{iθ}+ae^{-iθ}; , =, a(\cos(θ)+i\sin(θ)+\cos(θ)-i\sin(θ)); z_2, =, 2a\cos(θ); x_2, =, 2a\cos(θ)\text" "y_2=0$$
To obtain a correct Kutta condition on the plate the circulation required will be
$$Γ=4πaV_∞\sin(α)$$
This means rotation is added to the circle sufficient to move the rear stagnation point to the location that maps to the trailing edge of the plate.
Velocity $ V_1 $ at any point on the surface of the circle in $z_1$ will be the sum of the standard circle flow, rotated by an angle $ α $ , and the contribution from the vortex on the origin.
$$\table V_1, =, 2V_∞\sin(θ-α)+Γ/{2πa}; , =, 2V_∞\sin(θ-α)+{4πaV_∞\sin(α)}/{2πa}; , =, 2V_∞\sin(θ-α)+2V_∞\sin(α); V_1, =, 2V_∞(\sin(θ)\cos(α)-\cos(θ)\sin(α)+\sin(α))$$
The velocity on the surface of the plate will be based on circle flow velocity and the absolute value of the derivative of the transformation function.
$$V_2=V_1/{|{dz_2}/{dz_1}|}$$
The derivative can be evaluated as
$${dz_2}/{dz_1}=1-k^2/z_1^2$$
and on the surface or the circle this will be
$$\table {dz_2}/{dz_1}, =,1-a^2/{a^2e^{2iθ}}; , =,1-e^{-2iθ}; , =,1-(\cos(2θ)-i\sin(2θ)); , =,\cos^2(θ)+\sin^2(θ)-\cos^2(θ)+\sin^2(θ)+2i\cos(θ)\sin(θ); , =,2\sin(θ)(\sin(θ)+i\cos(θ)); \text"|"{dz_2}/{dz_1}\text"|", =,2\sin(θ)$$
Velocity on the plate can then be calculated as,
$$\table V_2, =, {2V_∞(\sin(θ)\cos(α)-\sin(α)\cos(θ)+\sin(α))}/{2\sin(θ)}; , =, V_∞(\cos(α)+\sin(α)(1-\cot(θ))); , =, V_∞(\cos(α)+\sin(α)\tan(θ/2))$$
If only small angles of attack are considered so that no flow separation occurs,
$ α$ is small , $ \cos(α) ≈ 1$ and $\sin(α)=α$ then
$$V_2=V_∞(1+α\tan(θ/2))$$
Surface pressure coefficient will be found by applying the incompressible version of Bernoulli equation,
$$\table Cp, =, 1-V_2^2/{V_∞^2}; , =, 1-(1+α\tan(θ/2))^2; Cp, =, -2α\tan(θ/2)-α^2tan^2(θ/2)$$
The pressure distribution over the aerofoil can be observed by looking at upper surface ( $ θ$ ) and the lower surface ( $−θ$ ) pressure coefficients. Cp At the trailing edge ( $θ=0$ ) is zero. On the upper surface the coefficient becomes more and more negative until the leading edge ( $θ=π$ ) where it produces infinite suction. On the lower surface going forward tan (${−θ}/2$ ) is positive and $\tan(θ/2) > \tan^2(θ/2)$ so initially the coefficient becomes more positive. This continues up to the point when $V_2=0 $ , the stagnation point near the leading edge of the plate, where $ α\tan({−θ}/2)=−1 $ and $Cp=1$ . In front of this $ α\tan(θ/2) > 1$ and $α\tan(θ/2) < α^2\tan^2(θ/2)$ so the coefficient reduces to zero and then becomes more and more negative, eventually equaling the infinite suction pressure at the leading edge.
The resultant of upper and lower surface pressures pushing on the surface of the plate will result in a normal force. The component of this normal force perpendicular to the stream will be the lift. For small angles of attack $\cos(α)≈1 $, thus the normal force is the same as the lift. Lift can then be calculated as the sum of pressure forces acting normal to the surface.
$$\text"Lift"=∫_0^c(P_{lower}-P_{upper}).(1.dx)$$
where
for 2D sections the force acts on a 1 unit depth of wing.
The lift coefficient will then be,
$$\table C_L, =, {\text"Lift"}/{1/2ρV_∞^2S}; , =, ∫_0^1{{P_{lower}-P_{upper}}/{1/2ρV_∞^2}}.{dx}/{c.1}; C_L, =, ∫_0^1(Cp_{lower}-Cp_{upper}).{dx}/c$$
Substituting for the previous expression for pressure coefficient,
$$\table C_L, =,∫_0^1(-2α\tan(-θ/2)-α^2\tan^2(-θ/2)+2α\tan(θ/2)+α^2\tan^2(θ/2)).{dx}/c; , =,∫_0^1(2α\tan(θ/2)-α^2\tan^2(θ/2)+2α\tan(θ/2)+α^2\tan^2(θ/2)).{dx}/c; C_L, =,∫_0^1(4α\tan(θ/2)).{dx}/c;$$
From the mapping geometry,
$$dx=dx_2\text" and "dx_2={dx_2}/{dθ}dθ$$
$$dx=-2a\sin(θ)dθ$$
In addition c≈4a, $θ=π$ when $x/c=0$ and $θ=0$ when $x/c=1$ so
$$\table C_L, =, ∫_π^0 4α\tan(θ/2){-2a\sin(θ)}/{4a}.dθ; , =, ∫_0^{π} 2α\tan(θ/2)\sin(θ).dθ; C_L, =, 2πα$$
As the lowest pressure region is toward the front a pitching moment will be caused about the centre of the plate by the distribution of lift over the chord of the plate. There will be a balance point toward the leading edge of the plate and this can be calculated by summing the moments caused by individual lift elements along the chord the moment about the leading edge, $x_2=-2a, x=0$
$$M_0=∫-(P_{lower}-P_{upper}).x.dx.1$$
$$C_{M0}={M_0}/{1/2ρV_∞^2Sc}=∫_0^1-(Cp_{lower}-Cp_{upper})x/c.{dx}/c$$
By definition, moment is positive nose up, hence the negative sign for leading edge moment coefficient. In this case the moment arm can also be found from the geometry of the transformation, $x=x_2 + 2a.$
$$x=2a\cos(θ)+2a$$
so the substitution can be made for pressure coefficient, moment arm and integration variable.
$$\table C_{M0}, =, ∫_0^1(2α\tan(-{θ/2})+α^2\tan^2(-{θ/2})-2α\tan(θ/2)-α^2\tan^2(θ/2)){2a+2a\cos(θ)}/c.{dx}/c; , =, ∫_π^0(-4α\tan(θ/2)){2a(1+\cos(θ))}/{4a}{-2a\sin(θ)}/{4a}.dθ; , =, ∫_0^π(-α\tan(θ/2)(1+\cos(θ)))\sin(θ).dθ; C_{M0}, =,-π/2α$$
This can be used to predict pitching moment about the standard reference point at ¼ chord location.
$$\table C_{M1\/4c}, =,C_{M0}+C_L{0.25c}/c=C_{M0}+{C_L}/4; , =,-π/2α+{2πα}/4; C_{M1\/4c}, =,0$$
Hence the centre of pressure will be at the ¼ chord location on the flat plate. As $C_{M1\/4c}$ is independent of α the aerodynamic centre will be at ¼c.