A wing of \( 120 \) m\( ^2 \) has an aspect ratio of \( 9 \). Its span is nearest to
Answer: A
Correct answer: 32.9 m
From \( A = b^2/S \),
\[ b = \sqrt{AS} = \sqrt{9 \times 120} = \sqrt{1080} \approx 32.9 \text{ m} \]The 13.3 m answer divides the area by the aspect ratio, which gives the mean chord rather than the span, and 1080 m forgets the square root altogether.
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A wing has a root chord of \( 6 \) m and a taper ratio of \( 0.3 \). Its mean aerodynamic chord is nearest to
Answer: B
Using \( \text{MAC} = \tfrac{2}{3}c_r(1+\lambda+\lambda^2)/(1+\lambda) \), the bracket is \( 1.39/1.3 = 1.069 \), and \( 4 \times 1.069 = 4.28 \) m. The tip chord itself would be \( 0.3 \times 6 = 1.8 \) m, offered as the distractor.
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An aircraft weighing \( 883 \) kN cruises at \( 250 \) m/s where the density is \( 0.365 \) kg/m\( ^3 \), on \( 120 \) m\( ^2 \) of wing. Its cruise lift coefficient is nearest to
Answer: C
The dynamic pressure is \( 0.5 \times 0.365 \times 250^2 = 11\,406 \) Pa, so \( C_L = 883\,000/(11\,406 \times 120) = 0.645 \). A cruise value near 0.5 to 0.6 is typical, and the aerofoil's design lift coefficient should match it.
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A wing is swept \( 30 \) degrees and the aircraft cruises at \( M = 0.82 \). The section sees a normal Mach number of
Answer: D
Correct answer: 0.71
Only the velocity component normal to the leading edge governs the section’s compressible behaviour, so
\[ M_n = M\cos\Lambda = 0.82 \times \cos 30^\circ = 0.82 \times 0.866 \approx 0.71 \]The section therefore behaves as though it were flying at Mach 0.71, comfortably below its critical value, while the aircraft cruises at 0.82. This is the whole purpose of sweep.
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An aircraft of \( 90 \) t is to have a wing loading of \( 6000 \) N/m\( ^2 \). The wing area required is nearest to
Answer: A
Correct answer: 147 m^2
The weight must first be expressed as a force:
\[ W = 90\,000 \times 9.81 = 882\,900 \text{ N} \]\[ S = \frac{W}{W/S} = \frac{882\,900}{6000} \approx 147 \text{ m}^2 \]The 15 m squared answer comes from dividing the mass in tonnes by the wing loading without converting to newtons - a factor of nearly ten, and the commonest slip in this calculation.
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Deploying flaps raises the maximum lift coefficient from \( 1.4 \) to \( 2.8 \). The stalling speed falls by a factor of
Answer: B
Stalling speed varies inversely with the square root of maximum lift coefficient, so doubling the coefficient divides the speed by \( \sqrt{2} = 1.414 \). A thirty per cent reduction in approach speed is the whole justification for a high-lift system.
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A wing of \( 147 \) m\( ^2 \) and aspect ratio \( 9.5 \) has a span of nearest to
Answer: C
Correct answer: 37.4 m
Again from \( A = b^2/S \),
\[ b = \sqrt{9.5 \times 147} = \sqrt{1396.5} \approx 37.4 \text{ m} \]A span of 37 metres on a 147 square metre wing is typical of a narrow-body transport, and it sits inside the 36 metre gate box only just - which is exactly the constraint that stops designers raising the aspect ratio further and drives the use of winglets and folding tips instead.
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A tapered wing has a root chord of \( 5.5 \) m and a tip chord of \( 1.65 \) m. Its taper ratio is
Answer: D
Correct answer: 0.30
Taper ratio is the tip chord divided by the root chord:
\[ \lambda = \frac{c_t}{c_r} = \frac{1.65}{5.5} = 0.30 \]Values around 0.25 to 0.35 are usual on transports, being close to the elliptical loading that minimises induced drag while remaining simple to build. Inverting the ratio gives 3.33, which is a common error: taper ratio is always defined tip over root and so is always less than one for a tapered wing.
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Wing fuel volume scales roughly with thickness ratio and chord squared. Reducing thickness from \( 12 \) to \( 10 \) per cent at constant chord changes the volume by
Answer: A
The volume falls in proportion to thickness, so \( 10/12 = 0.833 \), a reduction of about seventeen per cent. On a design already tight for tank space this can be the constraint that stops the wing being thinned further.
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An aircraft stalls at \( 60 \) m/s clean. With flaps raising \( C_{L\max} \) by \( 60 \) per cent, the stalling speed becomes nearest to
Answer: B
Correct answer: 47 m/s
Stalling speed varies inversely as the square root of the maximum lift coefficient, so
\[ V_{s2} = \frac{V_{s1}}{\sqrt{1.6}} = \frac{60}{1.265} \approx 47 \text{ m/s} \]A 60 per cent increase in lift coefficient buys only a 21 per cent reduction in speed, because of the square root. That diminishing return is why high-lift systems become so elaborate: each further increment of speed reduction costs disproportionately more device.
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