# 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 $$0.6$$.

techAir Asked on 12th August 2021 in

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

Worldtech Answered on 23rd August 2021.