Using thin airfoil theory, calculate \( (a) α_{ L=0}\) (b) \(cl\) when \(α = 3^{\circ} \).
The NACA 4412 airfoil has a mean camber line given by\[\frac{z}{c} = \left\{ {\begin{array}{*{20}{c}}
{0.25\left[ {0.8\frac{x}{c} – {{\left( {\frac{x}{c}} \right)}^2}} \right]\,\textrm{for}\;0 \le \frac{x}{c} \le 0.4}\\
{0.111\left[ {0.2 + 0.8\frac{x}{c} – {{\left( {\frac{x}{c}} \right)}^2}} \right]\textrm{for}\;0.4 \le \frac{x}{c} \le 1}
\end{array}} \right\}\]
Using thin airfoil theory, calculate \( (a) α_{ L=0}\) (b) \(cl\) when \(α = 3^{\circ} \).
(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) \(c_{l}=2\pi\left ( \alpha-\alpha_{L=0} \right )\;\textrm{where}\;\alpha\;\textrm{is in radian}\)
\[=\frac{2\pi}{57.3}\left ( 3-\left ( -4.16 \right ) \right )=0.782\]
