(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.