We need to find here the angle of attack.Lift per unit span is \[{L}’=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}sc_{l}\]

\[q_{\infty}=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}=\frac{1}{2}\times1.23\times50^{2}=1538\;N/m^{2}\]

\[c_{l}=\frac{{L}’}{q_{\infty}s}=\frac{1353}{1538\times2}=0.44\]

For NACA \(2412\) airfoil coefficient of lift of \(0.44\) is at \(2\) degrees angle of attack.

Dynamic pressure \[q_{\infty}=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}=\frac{1}{2}\times0.002377\times50^{2}=2.97\;lb/ft^{2}\].Fo r NACA 2412 airfoil coefficient of lift \(c_{l}=0.64,c_{m,c/4}=-0.036\).

Lift per unit span \[{L}’=q_{\infty}sc_{l}=2.97\times2\times1\times0.64=3.80\;lb/span\]

Moment about the quarter chord \[{M}’=q_{\infty}scc_{m,c/4}=2.97\times2\times1\times2\times\left ( -0.36 \right )
\\=-0.428\;ft/lb\] per unit span.

clear;clc
% Step 1: Select the parameters that define the airfoil of interest.
% (1) Select the a == angle of attack alpha
a = -3; % in degrees
a = a*pi/180; % Conversion to radians
% (2) Select the parameter related to thichkness of the airfoil:
e = .1;
% (3) Select the shift of y-axis related to camber of the airfoil:
f = .1;
% (4) Select the trailing edge angle parameter:
te = .06; % 0 < te < 1 (0 ==> cusped trailing edge)
n = 2 - te; % Number related to trailing edge angle.
tea = (n^2-1)/3; % This is a Karman-Trefftz extension.
% Step 2: Compute the coordinates of points on circle in zp-plane:
R = 1 + e;
theta = 0:pi/200:2*pi;
yp = R * sin(theta);
xp = R * cos(theta);
% Step 3: Transform coordinates of circle from zp-plane to z-plane:
z = (xp - e) + 1i.*(yp + f);
% Step 4: Transform circle from z-plane to airfoil in w-plane
% (the w-plane is the "physical" plane of the airfoil):
rot = exp(1i*a); % Application of angle of attack.
w = rot .* (z + tea*1./z); % Joukowski transformation.
% Step 5: Plot of circle in z-plane on top of airfoil in w-plane
plot(xp,yp), hold on
plot(real(w),imag(w),'r'),axis image, hold off

Mapping of a circle to airfoil

mu = 1; gam = -2; V=1;
x = -4:.02:4;
y = -3:.02:3;
for m = 1:length(x)
for n = 1:length(y)
xx(m,n) = x(m); yy(m,n) = y(n);
psis(m,n) = V * y(n) - mu * y(n)/(x(m)^2+(y(n)+.01)^2) - (gam/4/pi)*log(x(m)^2+(y(n)+.01)^2);
end
end
contour(xx,yy,psis,[-3:.3:3],'k'), axis image

Streamline flow over a cylinder

\[\frac{V_{r}}{V_{\infty}}=\left ( 1-\frac{R^{2}}{r^{2}} \right )cos\theta\]

\[\frac{V_{\theta}}{V_{\infty}}=-\left ( 1+\frac{R^{2}}{r^{2}} \right )sin\theta\]

At \(\left ( r,\theta \right ) \) ,\(V_{r}\) and \(V_{\infty}\) is proportional to \(V_{\infty}\). Therefore \(\overrightarrow{V}\) is same  what \(V_{\infty} \)  may be. The same of the streamlines will remain same it will not change.

For a doublet,the  stream function is \[\psi=\frac{-k}{2\pi}\frac{\sin\theta}{r}\]

\[V_{r}=\frac{\partial \phi}{\partial r}=\frac{1}{r}\frac{\partial \psi}{\partial \theta}\]

Here \[\psi=\frac{-k}{2\pi}\frac{\sin\theta}{r}
\\\Rightarrow \frac{\partial \psi}{\partial \theta}=\frac{-k}{2\pi}\frac{\cos\theta}{r}
\\\frac{\partial \phi}{\partial r}=\frac{1}{r}\left ( \frac{-k}{2\pi}\frac{\cos\theta}{r} \right )=\frac{-k}{2\pi}\frac{cos\theta}{r^{2}}\]

On integrating with respect to ‘r’

\[\phi=\left ( \frac{-k}{2\pi}cos\theta \right )\left ( -\frac{1}{r} \right )
\\\Rightarrow \phi=\frac{k}{2\pi}\frac{\cos\theta}{r}\]

For a uniform flow velocity potential \(\phi=V_{\infty}x\) and stream function \(\psi=V_{\infty}y\).Laplace equation is given as \[\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0\;;\;\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}=0\]

Velocity potential\[\phi=V_{\infty}x
\\\frac{\partial \phi}{\partial x}=V_{\infty}\;,\;\frac{\partial^2 \phi}{\partial x^2}=0
\\\frac{\partial \phi}{\partial y}=0\;,\;\frac{\partial^2 \phi }{\partial y^2}=0\]

Therefore Laplace equation \[\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0+0=0\]

is satisfied.

Stream function \[\psi=V_{\infty}y\]

\[\frac{\partial\psi }{\partial x}=0\;,\;\frac{\partial^2 \psi}{\partial x^2}=0
\\\frac{\partial \psi}{\partial y}=V_{\infty}\;,\;\frac{\partial^2 \psi}{\partial y^2}=0\]

Therefore Laplace equation \[\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}=0+0=0\]

is satisfied.

For a source flow ;

\[\overrightarrow{V}=V_{r}\overrightarrow{e}_{r}=\frac{\Lambda }{2\pi r}\overrightarrow{e_{r}}\]

In polar coordinates :

\[\nabla \cdot \overrightarrow{V}=\frac{1}{r}\frac{\partial }{\partial r}\left ( rV_{r} \right )+\frac{1}{r}\frac{\partial V_{\theta}}{\partial \theta}
\\\Rightarrow\nabla \cdot \overrightarrow{V}=\frac{1}{r}\frac{\partial }{\partial r}\left [ r\;\frac{\Lambda }{2\pi r} \right ]+\frac{1}{r}\frac{\partial(0)}{\partial \theta}
\\\Rightarrow\nabla \cdot \overrightarrow{V}=0+0=0\]

This shows that the flow is incompressible.For the flow to be irrotational

\[\nabla\times \overrightarrow{V}=0\]

\[\nabla \times \overrightarrow{V}=\frac{1}{r}\begin{vmatrix}
\overrightarrow{e}_{r} &r\overrightarrow{e}_{\theta} & \overrightarrow{e}_{z}\\
\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta} & \frac{\partial }{\partial z}\\
V_{r}& rV_{\theta} & V_{z}
\end{vmatrix}=\frac{1}{r}\begin{vmatrix}
\overrightarrow{e}_{r} &r\overrightarrow{e}_{\theta} & \overrightarrow{e}_{z}\\
\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta} & \frac{\partial }{\partial z}\\
\frac{\Lambda }{2\pi r}&0 & 0
\end{vmatrix}\]

\[\Rightarrow\nabla\times \overrightarrow{V}=-r\overrightarrow{e_{\theta}}\left ( \frac{\partial (0)}{\partial r} -\frac{\partial }{\partial z}\left ( \frac{ \Lambda }{2\pi r} \right )\right )+\overrightarrow{e}_{z}\left ( \frac{\partial (0)}{\partial r} -\frac{\partial }{\partial \theta}\left ( \frac{ \Lambda }{2\pi r} \right ) \right )\]

\[\Rightarrow\nabla\times \overrightarrow{V}=-r\overrightarrow{e_{\theta}}\left (0-0\right )+\overrightarrow{e}_{z}\left ( 0-0 \right )=0\]

This shows that it is irrotational everywhere.

Let a uniform flow with velocity =\( V_{\infty}\).

\(\overrightarrow{V}=V_{\infty}\hat{i},V_{\infty}=u=\mathrm{constant}\)

\[\nabla \cdot \overrightarrow{V}=\frac{\partial u}{\partial x}+\frac{\partial v }{\partial y}+\frac{\partial w}{\partial z}=0+0+0=0\]

Therefore the flow is incompressible.

\[\nabla \times\overrightarrow{V} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
\frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\
u& v& w
\end{vmatrix}\]

\[\Rightarrow\overrightarrow{i}\left ( 0-0 \right )-\overrightarrow{j}\left ( 0-0 \right )+\overrightarrow{k}\left ( 0-0 \right )
=0\]

Therefore

 \[\nabla \times\overrightarrow{V}=0\]

The flow is irrotational.

We need to calculate here the pressure coefficient at a point on the surface of the wing of the airplane. Pressure coefficient is given as \[C_{p}=1-\left ( \frac{V}{V_{\infty}} \right )^{2}\]

Therefore \[C_{p}=1-\left ( \frac{130}{98.8} \right )^{2}
\\=-0.73\]