Question 3 (10+10=20 Point): A symmetric airfoil operates at a small angle of attack (alpha =2^circ ) in subsonic flow. The lift coefficient is given by C_(L)=2pi alpha (with alpha in radians). (a) Calculate the lift coefficient (C_(L)) (b) If the chord length is 1.5 m, wing span is 2 m and the airspeed is 30m/s,rho _(infty )=1.225kg/m^3 determine the lift force.

(a) To calculate the lift coefficient $(C_{L})$, we need to convert the angle of attack from degrees to radians. We know that $1^{\circ} = \frac{\pi}{180}$ radians. Therefore, $\alpha = 2^{\circ} = 2 \times \frac{\pi}{180} = \frac{\pi}{90}$ radians.<br /><br />Now, we can substitute the value of $\alpha$ into the lift coefficient formula:<br /><br />$C_{L} = 2\pi \alpha = 2\pi \times \frac{\pi}{90} = \frac{2\pi^2}{90} \approx 0.21$<br /><br />(b) To determine the lift force, we can use the lift equation:<br /><br />$L = \frac{1}{2} \rho_{\infty} v^2 S C_{L}$<br /><br />where $L$ is the lift force, $\rho_{\infty}$ is the air density, $v$ is the airspeed, $S$ is the wing area, and $C_{L}$ is the lift coefficient.<br /><br />First, let's calculate the wing area $S$:<br /><br />$S = \frac{1}{2} b c = \frac{1}{2} \times 2 \times 1.5 = 1.5 \, m^2$<br /><br />Now, we can substitute the values into the lift equation:<br /><br />$L = \frac{1}{2} \times 1.225 \times (30)^2 \times 1.5 \times 0.21$<br /><br />$L \approx 34.29 \, N$<br /><br />Therefore, the lift force is approximately $34.29 \, N$.