Subsonic and Supersonic Flow over Wings Questions and Answers

Answer: C

To a simple approximation the free-stream Mach number may be raised by the factor \( 1/\cos\Lambda \) before the section sees the same normal component, so \( 0.72 / \cos 30^\circ = 0.72/0.866 = 0.83 \). Real wings fall a little short of this estimate, because the simple sweep theory ignores root and tip effects.

Answer: D

Volume wave drag scales with the square of the thickness ratio, so halving the thickness divides the drag by \( 2^2 = 4 \). This quadratic payoff is why supersonic wings are made as thin as the structure will possibly allow, typically three to five per cent.

Answer: A

From \( \sin\mu = 1/1.6 = 0.625 \), the half-angle is \( 38.7 \) degrees. A wing leading edge would therefore have to be swept more than \( 90 - 38.7 = 51.3 \) degrees to stay subsonic at this speed, less sweep than the Mach 2 case demands.

Answer: B

The component is \( 0.80 \times \cos 25^\circ = 0.80 \times 0.906 = 0.725 \). Modest sweep buys a correspondingly modest benefit; getting the section down near Mach 0.7 from a cruise of 0.85 requires something closer to 35 degrees.

Answer: C

The result is \( c_l = 4\alpha/\sqrt{M^2-1} \), with \( \alpha = 0.0698 \) radians and \( \sqrt{3} = 1.732 \), giving \( 4 \times 0.0698/1.732 = 0.161 \). The subsonic two pi alpha value of 0.439 does not apply above Mach one and is offered as the trap.

Answer: D

The area of a triangular planform is \( \tfrac{1}{2} b c_r = 0.5 \times 8 \times 12 = 48 \) m\( ^2 \), so the aspect ratio is \( 64/48 = 1.33 \). Values well below two are typical of slender deltas and are the reason such aircraft approach at very high angles of attack.

Answer: A

The Mach angle is \( \arcsin(1/3) = 19.47 \) degrees, so the sweep must exceed \( 90 - 19.47 = 70.5 \) degrees. The requirement tightens rapidly with speed, and beyond about Mach 3 keeping a subsonic edge demands so much sweep that designers usually accept a sharp supersonic edge instead.

Answer: B

A supersonic pocket has appeared, since the critical value has been passed, but the shock closing it is not yet strong enough to produce the sharp drag rise. This band between the two Mach numbers is precisely where a supercritical section is designed to operate.

Answer: C

Wave drag due to lift varies with the square of the lift coefficient, and by linearised theory the lift coefficient is proportional to incidence, so doubling the angle multiplies that drag component by four. The volume-dependent part is unaffected, since it depends on thickness alone.

Answer: D

Computing \( 0.90 \times \cos 40^\circ = 0.90 \times 0.766 = 0.689 \). Notice that greater sweep has allowed a higher cruise Mach number while presenting the section with almost exactly the same normal component as a 35-degree wing at Mach 0.85.