Question 4 (15 Points): Consider a light, single -engine, propeller-driven airplane The airplane weight is 1470 kg and the wing reference area is 16m^2 Assume a steady level flight at sea level.where the ambient atmospheric density is 1.225kg/m^3 The drag coefficient of the airplane C_(D) is a function of the lift coefficient C_(L) ; this function for the given airplane is C_(D)=0.025+0.054C_(L^2)^2 For two flight velocities of V=40m/s calculate C_(L) and C_(D) coefficients.

To calculate the lift coefficient ($C_L$) and drag coefficient ($C_D$) for the given airplane at a velocity of $V = 40 \, \text{m/s}$, we need to use the relationship between lift force, drag force, and weight of the airplane. <br /><br />First, let's calculate the lift force ($L$) required for steady level flight. The lift force must balance the weight of the airplane:<br /><br />\[ L = W \]<br /><br />where $W$ is the weight of the airplane. Given that the weight of the airplane is $1470 \, \text{kg}$ and assuming the acceleration due to gravity is $9.81 \, \text{m/s}^2$, we have:<br /><br />\[ W = 1470 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 14418.6 \, \text{N} \]<br /><br />Next, we use the lift equation to find the lift coefficient. The lift force is also given by:<br /><br />\[ L = \frac{1}{2} \rho V^2 S C_L \]<br /><br />where $\rho$ is the ambient atmospheric density, $V$ is the velocity, and $S$ is the wing reference area. Rearranging this equation to solve for $C_L$, we get:<br /><br />\[ C_L = \frac{2L}{\rho V^2 S} \]<br /><br />Substituting the given values, we have:<br /><br />\[ C_L = \frac{2 \times 14418.6 \, \text{N}}{1.225 \, \text{kg/m}^3 \times (40 \, \text{m/s})^2 \times 16 \, \text{m}^2} \]<br /><br />\[ C_L = \frac{28837.2}{1.225 \times 1600 \times 16} \]<br /><br />\[ C_L = \frac{28837.2}{31360} \]<br /><br />\[ C_L \approx 0.919 \]<br /><br />Now that we have the lift coefficient, we can calculate the drag coefficient using the given relationship:<br /><br />\[ C_D = 0.025 + 0.054 C_L^2 \]<br /><br />Substituting the value of $C_L$ we found:<br /><br />\[ C_D = 0.025 + 0.054 \times (0.919)^2 \]<br /><br />\[ C_D = 0.025 + 0.054 \times 0.842 \]<br /><br />\[ C_D = 0.025 + 0.045588 \]<br /><br />\[ C_D \approx 0.0706 \]<br /><br />Therefore, for a velocity of $V = 40 \, \text{m/s}$, the lift coefficient ($C_L$) is approximately $0.919$ and the drag coefficient ($C_D$) is approximately $0.0706$.