Range and Endurance Questions and Answers

Answer: B

The relation is \(\frac{C_{D}}{C_{L}^{2}} = \frac{4}{3\pi eAR}\).

With the parabolic drag polar \(C_{D} = C_{D0} + KC_{L}^{2}\), \(K = 1/(\pi eAR)\), minimise \(C_{D}/C_{L}^{3/2}\) by setting its derivative with respect to \(C_{L}\) to zero:

\[ \begin{aligned} C_{L}^{3/2}(2KC_{L}) &= \tfrac{3}{2}C_{L}^{1/2}(C_{D0} + KC_{L}^{2}) \\ C_{D0} &= \tfrac{1}{3}KC_{L}^{2} \\ C_{D} &= \tfrac{4}{3}KC_{L}^{2} \end{aligned} \]

Dividing by \(C_{L}^{2}\) gives \(4/(3\pi eAR)\). This is the condition for maximum endurance of a propeller airplane, where induced drag is three times zero-lift drag. \(C_{D}/C_{L}^{2} = 1/(\pi eAR)\) would mean zero parasite drag, and the other two forms do not follow from the polar.

Answer: C

The pilot should increase thrust and reduce angle of attack.

For a jet, fuel flow is roughly proportional to thrust, so maximum endurance is flown at minimum drag, the speed of maximum L/D. Flying faster than that speed raises the drag, so steady flight at the higher speed needs more thrust.

Lift must still equal weight: \(W = \frac{1}{2}\rho V^{2}SC_{L}\). At the same altitude, a higher V needs a smaller \(C_{L}\), that is, a smaller angle of attack.

More thrust alone, with the same angle of attack, would give more lift than weight as the speed rose, so the aircraft would climb. Increasing the angle of attack makes that worse. Reducing the angle of attack alone loses lift without the thrust to hold the higher speed, so the aircraft would descend.