# An airplane is cruising at \(50\,m/s\) at standard sea level in straight and level flight. The airplane has a wing area of \(15\,m^2\) and a wing span of \(10\,m\), and has the maximum gross weight of \(1000\,kg\). The airfoil of the wing has lift slope of \(0.102\) per degree and angle of attack at zero lift of \(-2.5^{\circ}\). Considering, \(\tau = 0.1\), find the angle of attack for the wing. Also, find the induced drag for the airplane, if span efficiency factor is \(0.6\).

An airplane is cruising at \(50\,m/s\) at standard sea level in straight and level flight. The airplane has a wing area of \(15\,m^2\) and a wing span of \(10\,m\), and has the maximum gross weight of \(1000\,kg\). The airfoil of the wing has lift slope of \(0.102\) per degree and angle of attack at zero lift of \(-2.5^{\circ}\). Considering, \(\tau = 0.1\), find the angle of attack for the wing. Also, find the induced drag for the airplane, if span efficiency factor is \(0.6\).

Induced drag is created due to downwash. Downwash is produced by the wing-tip vortices during flight. Aspect ratio of the wing is, \[AR=\frac{b^{2}}{S}=\frac{\left ( 10 \right )^{2}}{15}=6.67\]

At, standard sea level density of air is \(1.225\,kg/m^{3}\)

Velocity of airplane is \(50\,m/s\)

Dynamic pressure, \[q_{\infty} = \frac{1}{2}\rho _{\infty}V_{\infty}^{2}=\frac{1}{2}\times1.225\times\left ( 50 \right )^{2}=1531.25\,N/m^{2}\]

Lift slop of the airfoil of the wing is \(0.102 \,per\,degree = 5.8442\,per\,radian\)

Lift produced by the wing is, \[L=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}SC_{L}=\rho _{\infty}SC_{L}\]\[\Rightarrow C_{L}=\frac{L}{\rho _{\infty}S}\]

L = Weight of the airplane = \(1000\times9.8 = 9800\,N\)

Therefore, \[C_{L} = \frac{9800}{1531.25\times15}=0.427\]

Lift slope of the wing will be, \[a = \frac{a_{0}}{1 + \frac{a_{0}}{\pi AR}\left ( 1+\tau \right ) }=\frac{5.8442}{1+\frac{5.8442}{\pi\left ( 6.67 \right )}\left ( 1+0.1 \right )}=4.472\,per\,radian = 0.0781\,per\,degree\]

Also, Coefficient of lift is given as \[C_{L}=a\left ( \alpha -\alpha _{L=0} \right )\Rightarrow 0.427=0.0781\left ( \alpha -\left ( -2.5^{\circ} \right ) \right )\Rightarrow \alpha = 2.97\,degree\]

Induced drag is given as \[D_{i}=q_{\infty}SC_{D,i}\]\[ C_{D,i}=\frac{C_{L}^{2}}{\pi e AR}=\frac{\left ( 0.427 \right )^{2}}{\pi \left ( 0.6 \right )\left ( 6.67 \right )}=0.0145\]

Therefore, \(D_{i} = \left ( 1531.25 \right )\left ( 15 \right )\left ( 0.0145 \right ) = 333.05\,N\)