Aircraft Performance in Level Flight Questions and Answers

Answer: D

The service ceiling is the altitude at which the maximum rate of climb is 100 ft/min.

The maximum rate of climb falls as altitude increases because excess power shrinks. The altitude where it reaches zero is the absolute ceiling, the theoretical limit of sustained level flight. Because that limit is approached only asymptotically, a practical ceiling is defined where the airplane can still climb at 100 ft/min (about 0.5 m/s).

"Beyond which the airplane cannot sustain level flight" describes the absolute ceiling. 100 m/min is about 330 ft/min, the wrong standard. The altitude for maximum power required is not a ceiling definition at all.

Answer: A

The possible set is \(\phi = 0, \theta = 2^{o}, \psi = 0, \alpha = 2^{o}, \beta = 0\).

In straight, level, constant-velocity flight the wings are level (\(\phi = 0\)), the flight path is horizontal so the pitch angle equals the angle of attack (\(\gamma = \theta - \alpha = 0\)), and a conventional airplane flies with no sideslip (\(\beta = 0\)). The heading \(\psi\) can be any constant value. The first set meets every condition.

A 5-degree bank with zero sideslip tilts the lift vector and turns the airplane, and \(\theta = 0\) with \(\alpha = 2^{o}\) would mean a descent, so the second set is neither straight nor level. The third and fourth sets have a 5-degree sideslip with the wings level, which leaves an unbalanced side force, and the fourth also has \(\theta - \alpha = 3^{o}\), a climb. The official key accepts only the first set.

Answer: A

The maximum cruise speed is 87 m/s.

In level flight thrust equals drag and lift equals weight. Writing \(q\) for the dynamic pressure, \(C_{L} = W/(qS)\), so

\[ \begin{aligned} T &= qSC_{D_{0}} + \frac{KW^{2}}{qS} \\ 1000 &= 0.375q + \frac{333333}{q} \end{aligned} \]

using \(S = 15\), \(C_{D_{0}} = 0.025\), \(K = 0.05\) and \(W = 10^{4}\) N. This is a quadratic in \(q\):

\[ \begin{aligned} 0.375q^{2} - 1000q + 333333 &= 0 \\ q &= \frac{1000 \pm 707.1}{0.75} \\ q_{max} &= 2276\ \text{Pa} \end{aligned} \]

\[ \begin{aligned} V_{max} &= \sqrt{2q_{max}/\rho} \\ &= \sqrt{2 \times 2276 / 0.60} = 87\ \text{m/s} \end{aligned} \]

The smaller root, \(q = 390.5\) Pa, gives 36 m/s, the minimum speed at which 1 kN of thrust can hold level flight, which is the trap in the options. 30 m/s and 101 m/s do not satisfy the thrust-drag balance.