Question 2 (10+10=20 Points): An airfoil with a chord length of 1.5 m is in a flow with V_(infty )=50m/s and rho =1.2kg/m^3 The airfoil is cambered and has a zero-lift angle of attack of alpha _(L=0)=-1^circ (a) Calculate the lift coefficient at alpha =5^circ . (b) Determine the total lift force per unit span acting on the airfoil.

(a) To calculate the lift coefficient at $\alpha = 5^{\circ}$, we can use the lift coefficient equation for a cambered airfoil:<br /><br />$C_L = \frac{2 \pi \rho V_{\infty} b}{c} \left( \frac{\sin(\alpha + \alpha_{L=0}) - \sin(\alpha_{L=0})}{\cos(\alpha + \alpha_{L=0})} \right)$<br /><br />where:<br />$C_L$ is the lift coefficient,<br />$\rho$ is the air density,<br />$V_{\infty}$ is the free stream velocity,<br />$b$ is the half-chord length,<br />$c$ is the chord length,<br />$\alpha$ is the angle of attack, and<br />$\alpha_{L=0}$ is the zero-lift angle of attack.<br /><br />Given:<br />$\rho = 1.2 \, \text{kg/m}^3$<br />$V_{\infty} = 50 \, \text{m/s}$<br />$b = \frac{c}{2} = \frac{1.5}{2} = 0.75 \, \text{m}$<br />$\alpha_{L=0} = -1^{\circ}$<br />$\alpha = 5^{\circ}$<br /><br />Substituting the given values into the equation:<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \left( \frac{\sin(5^{\circ} - 1^{\circ}) - \sin(-1^{\circ})}{\cos(5^{\circ} - 1^{\circ})} \right)$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \left( \frac{\sin(4^{\circ}) + \sin(1^{\circ})}{\cos(4^{\circ})} \right)$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \left( \frac{0.0698 + 0.0175}{0.9976} \right)$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \left( \frac{0.0873}{0.9976} \right)$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times 50 \times 0.75}{1.5} \times 0.0875$<br /><br />$C_L = \frac{2 \pi \times 1.2 \times