Performance Constraint Analysis Questions and Answers

Answer: A

Correct answer: 110.3 m\(^2\)

Wing area is the design weight divided by the chosen wing loading:

\[ S = \frac{W}{W/S} = \frac{640}{5.8} = 110.3\ \mathrm{m}^2 \]

The distractor \( 3712 \) multiplies instead of dividing, which is the error to guard against - a quick sanity check is that a wing loading in kilonewtons per square metre divided into a weight in kilonewtons must leave square metres. This area, with the installed thrust from the other coordinate, is the principal output of the sizing exercise.

Answer: B

Correct answer: 198.4 kN

The installed thrust follows from the vertical coordinate of the design point:

\[ T = W \times \frac{T}{W} = 640 \times 0.31 = 198.4\ \mathrm{kN} \]

The distractor \( 99.2 \) halves this, as if the figure applied to one engine of a twin; the diagram's \( T/W \) is always the total installed thrust, and the per-engine rating comes from dividing by the number of engines afterwards. This total is what determines which engines from the available catalogue can be considered.

Answer: C

Correct answer: 4.86 kN/m\(^2\)

Convert the mass to weight before dividing by the area:

\[ W = 52000 \times 9.81 = 510.1\ \mathrm{kN} \]

\[ \frac{W}{S} = \frac{510.1}{105} = 4.86\ \mathrm{kN/m}^2 \]

The distractor \( 495 \) divides the mass in kilograms by the area without converting to weight, giving kg/m\(^2\) - a perfectly valid unit but not the one asked for. Mixing the two conventions is the commonest error in this calculation; the factor between them is \( g \).

Answer: D

Correct answer: 56.0 m/s

The stalling speed follows from equating lift to weight at maximum lift coefficient:

\[ V_s = \sqrt{\frac{2(W/S)}{\rho C_{L,max}}} = \sqrt{\frac{2 \times 5000}{1.225 \times 2.6}} = \sqrt{3140} = 56.0\ \mathrm{m/s} \]

Note the wing loading must be in newtons per square metre, not kilonewtons. The distractor \( 79.2 \) omits the \( C_{L,max} \), and \( 28.0 \) halves the result. This relation is what turns the landing requirement into a vertical line on the diagram: fix the approach speed and \( C_{L,max} \), and the wing loading is bounded.

Answer: A

Correct answer: 68.9 m/s

The approach speed is the required multiple of the stalling speed:

\[ V_{app} = 1.23 \times V_s = 1.23 \times 56.0 = 68.9\ \mathrm{m/s} \]

The factor of \( 1.23 \) on the reference stalling speed is the modern certification minimum, replacing the older \( 1.3 \) on \( V_{S0} \). It exists to guarantee manoeuvre and gust margin on the approach, and because landing distance scales roughly with the square of this speed, it is what ultimately fixes the maximum wing loading a runway length will permit.

Answer: B

Correct answer: 0.107

In a steady climb the thrust must overcome the drag and supply the gradient:

\[ \frac{T}{W} = \frac{1}{L/D} + \gamma = \frac{1}{12} + 0.024 = 0.0833 + 0.024 = 0.107 \]

This is the thrust the remaining engine must produce, expressed as a fraction of the whole aircraft weight - which is why a twin must install roughly twice this, about \( 0.214 \), to have it available after a failure. That doubling is what makes second segment climb such a demanding constraint for twins, and why the required gradient is lower for them than for three- and four-engine aircraft.

Answer: C

Correct answer: 0.059

In steady level flight thrust equals drag and lift equals weight, so the ratio of thrust to weight is the reciprocal of the lift-to-drag ratio:

\[ \frac{T}{W} = \frac{1}{L/D} = \frac{1}{17} = 0.059 \]

This is the thrust required at the cruise condition. Referring it to sea level static, where the diagram is plotted, requires dividing by the thrust lapse - typically around \( 0.25 \) to \( 0.30 \) at cruise altitude, which raises the plotted value to roughly \( 0.20 \).

Answer: D

Correct answer: 1.41

Stalling speed varies as the square root of wing loading:

\[ V_s \propto \sqrt{\frac{W}{S}} \quad \Rightarrow \quad \frac{V_{s,2}}{V_{s,1}} = \sqrt{2} = 1.41 \]

So doubling the wing loading raises the stalling speed by about 41 per cent, not by 100 per cent. Because landing distance scales roughly with the square of approach speed, doubling the wing loading roughly doubles the landing run - which is exactly the trade the vertical landing constraint enforces.

Answer: A

Correct answer: 138.5 m\(^2\)

The landing constraint caps the wing loading, and a capped wing loading sets a floor on wing area:

\[ S_{min} = \frac{W}{(W/S)_{max}} = \frac{720}{5.2} = 138.5\ \mathrm{m}^2 \]

Anything smaller would put the wing loading above the limit and the approach speed above what the runway allows. Because the wing may not be made smaller than this, the landing requirement frequently determines the wing area of a transport outright, with the other constraints then determining only the thrust.

Answer: B

Correct answer: 84 kN

Find the total installed thrust first, then divide among the engines:

\[ T_{total} = 900 \times 0.28 = 252\ \mathrm{kN} \]

\[ T_{engine} = \frac{252}{3} = 84\ \mathrm{kN} \]

The distractor \( 252 \) is the total rather than the per-engine figure. The per-engine number is what matters for engine selection, since it must be matched against actual catalogue ratings - and the availability of a suitable engine often forces the design point to move slightly from where the diagram alone would put it.