Question 4 (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 a 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. <br /><br />Given that the angle of attack is $\alpha = 2^{\circ}$, we can convert it to radians using the conversion factor $\frac{\pi}{180}$. <br /><br />So, $\alpha = 2^{\circ} \times \frac{\pi}{180} = \frac{\pi}{90}$ radians.<br /><br />Now, we can substitute the value of $\alpha$ into the lift coefficient equation:<br /><br />$C_{L} = 2\pi \alpha$<br /><br />$C_{L} = 2\pi \times \frac{\pi}{90}$<br /><br />$C_{L} = \frac{2\pi^2}{90}$<br /><br />$C_{L} = \frac{\pi^2}{45}$<br /><br />Therefore, the lift coefficient $(C_{L})$ is $\frac{\pi^2}{45}$.<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 />Given that the chord length is 1.5 m, the wing span is 2 m, and the airspeed is 30 m/s, we can calculate the wing area $S$ as:<br /><br />$S = \frac{1}{2} \times \text{chord length} \times \text{wing span}$<br /><br />$S = \frac{1}{2} \times 1.5 \, \text{m} \times 2 \, \text{m}$<br /><br />$S = 1.5 \, \text{m}^2$<br /><br />Now, we can substitute the values of $\rho_{\infty}$, $v$, $S$, and $C_{L}$ into the lift equation:<br /><br />$L = \frac{1}{2} \times 1.225 \, \text{kg/m}^3 \times (30 \, \text{m/s})^2 \times 1.5 \, \text{m}^2 \times \frac{\pi^2}{45}$<br /><br />$L = \frac{1}{2} \times 1.225 \, \text{kg/m}^3 \times 900 \, \text{m}^2/\text{s}^2 \times 1.5 \, \text{m}^2 \times \frac{\pi^2}{45}$<br /><br />$L = 1.225 \, \text{kg/m}^3 \times 450 \, \text{m}^2/\text{s}^2 \times \frac{\pi^2}{90}$<br /><br />$L = 1.225 \, \text{kg/m}^3 \times 5 \, \text{m}^2/\text{s}^2 \times \frac{\pi^2}{90}$<br /><br />$L = 6.125 \, \text{kg} \times \frac{\pi^2}{90}$<br /><br />$L = \frac{6.125\pi^2}{90} \, \text{N}$<br /><br />$L = \frac{\pi^2}{14.6} \, \text{N}$<br /><br />Therefore, the lift force is $\frac{\pi^2}{14.6} \, \text{N}$.