(a) For \(0\leq \frac{x}{c}\leq 0.4\)\[\left ( \frac{dz}{dx} \right )_{1}=0.2-0.5\left ( \frac{x}{c} \right )\]

For \(0.4\leq \frac{x}{c}\leq 1\)\[\left ( \frac{dz}{dx} \right )_{2}=0.0888-0.2222\left ( \frac{x}{c} \right )\]

Since , \(x=\frac{c}{2}\left ( 1-cos\theta \right )\) , then

\[\left ( \frac{dz}{dx} \right )_{1}=-0.05+0.25cos\theta\;\textrm{for}\;0\leq \theta\leq 1.3694\]

\[\left ( \frac{dz}{dx} \right )_{2}=-0.0223+0.1111cos\theta\;\textrm{for}\;1.3694\leq \theta\leq \pi\]

\[\alpha _{L=0}=-\frac{1}{\pi}\int_{0}^{\pi}\frac{dz}{dx}\left ( cos\theta-1 \right )d\theta\]

\(=\frac{1}{\pi}\int_{0}^{1.3694}\left ( -0.05+0.25cos\theta \right )\left ( cos\theta-1 \right )d\theta\)-\(\frac{1}{\pi}\int_{1.3694}^{\pi}\left ( -0. 0223+0.1111cos\theta\right )\left ( cos\theta-1 \right )d\theta\)

\(=-\frac{1}{\pi}\int_{0}^{1.3694}\left ( 0.05-0.3cos\theta+0.25cos2\theta \right )d\theta\)\(\;-\)\(\frac{1}{\pi}\int_{1.3694}^{\pi}\left ( 0.0223-0.13334cos\theta+0.1111cos2\theta \right )d\theta\)

\(=-\frac{1}{\pi}\left [ 0.050-0.3sin\theta+0.25\left ( \frac{\theta}{2}+\frac{1}{4} sin2\theta\right ) \right ]_{0}^{1.3694}\)\(\;-\)\(\frac{1}{\pi}\left [ 0.0223\theta-0.1334sin\theta+0.111\left ( \frac{\theta}{2} +\frac{1}{4}sin2\theta\right ) \right ]_{1.3694}^{\pi}\)

\(=-\frac{1}{\pi}\left [ 0.06847-0.2939+0.1712+0.0245 \right ]-\)\(\frac{1}{\pi}\left [ 0.0701+0.1745 \right ]+\frac{1}{\pi}\left [ 0.0305-0.1307+0.0761+0.0109 \right ]\)

\(=-\frac{0.2281}{\pi}=-0.0726\;\textrm{radian}=-4.16^{\circ}\)

(b) \(c_{l}=2\pi\left ( \alpha-\alpha_{L=0} \right )\;\textrm{where}\;\alpha\;\textrm{is in radian}\)

\[=\frac{2\pi}{57.3}\left ( 3-\left ( -4.16 \right ) \right )=0.782\]

We need to find here the location of aerodynamic center. Slope of the lift curve is \[a_{0}=\frac{0.65-(-0.39)}{4-(-6)}=0.104\;\textrm{per degree}\]

Slope of the moment coefficient curve \[m_{0}=\frac{-0.037-(-0.045)}{4-(-6)}=8\times10^{-4}\;\textrm{per degree}\]

Location of aerodynamic center is given as \[\bar{x}_{ac}=-\frac{m_{0}}{a_{0}}+0.25
\\\Rightarrow \bar{x}_{ac}= -\frac{8\times10^{-4}}{0.104}+0.25=0.242\]

Kelvin’s circulation theorem tells  that the time rate of change of circulation around a closed curve consisting of the same fluid elements is zero,that is \(\frac{D\tau}{Dt}=0\).

Circulation is defined as \[\tau=\oint_{c}\overrightarrow{V}\cdot \overrightarrow{ds}\]

\[  \frac{D\tau}{Dt}=\oint_{c}\frac{D\overrightarrow{V}}{Dt}\cdot \overrightarrow{ds}+\oint_{c}\overrightarrow{V}\cdot \overrightarrow{ds} \]

\[\frac{D\overrightarrow{ds}}{Dt}=\overrightarrow{dV}\]

\[\oint_{c}\overrightarrow{v}\cdot \overrightarrow{dV}=\oint_{c}d\left ( \frac{V^{2}}{2} \right )=0\]

\[\frac{D\overrightarrow{V}}{dt}=-\frac{1}{\rho}\nabla p\]

\[\oint_{c}\frac{D\overrightarrow{V}}{Dt}\cdot \overrightarrow{ds}=-\oint_{c}\frac{1}{\rho}\nabla p\cdot \overrightarrow{ds}=-\oint_{c}\frac{dp}{\rho}\]

when \(\rho\)=constant or\( \rho\)=\(\rho(p)\),then

\[-\oint_{c}\frac{dp}{\rho }=0\]

Therefore \[\oint_{c}\frac{D\overrightarrow{V}}{Dt}\cdot \overrightarrow{ds}=0\]

or \[\frac{D\tau}{Dt}=0\]

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 shape 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.