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Asked on 11th November 2019 in Aerodynamics.
(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
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Asked on 10th November 2019 in Aerodynamics.
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
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Asked on 9th November 2019 in Aerodynamics.
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
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Asked on 8th November 2019 in Aerodynamics.
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.
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Asked on 8th November 2019 in Aerodynamics.
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.
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Asked on 6th November 2019 in Aerodynamics.
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
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Asked on 6th November 2019 in Aerodynamics.
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
- 1631 views
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Asked on 4th November 2019 in Aerodynamics.
\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.
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Asked on 4th November 2019 in Aerodynamics.
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}
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Asked on 3rd November 2019 in Aerodynamics.
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.
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