A diamond or double-wedge aerofoil in supersonic flow experiences wave drag that varies, for small angles, as
Answer: A
Linearised supersonic theory splits the wave drag into a thickness contribution and a lift contribution, each entering as a square. This is why supersonic sections are made as thin as structure allows, and why cruising at low incidence matters so much for range.
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The lift coefficient of a flat plate in supersonic flow, by linearised theory, is
Answer: B
The Ackeret result replaces the two pi alpha of subsonic thin-airfoil theory. The slope now falls as the Mach number rises, which is the opposite of the subsonic Prandtl-Glauert behaviour, and it is one reason supersonic aircraft need large wings or high incidence to manoeuvre.
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On a diamond aerofoil at zero incidence, the pressure on the rear faces is
Answer: C
The flow is compressed by shocks at the leading edge and expanded round the shoulder, so it arrives at the rear faces at lower pressure than it had at the front. The imbalance is a net rearward force, and it exists with no viscosity at all.
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The aerodynamic centre of a thin aerofoil in supersonic flow lies close to
Answer: D
Supersonic loading is far more uniform along the chord than subsonic loading, so the resultant acts near the middle rather than at the quarter chord. The rearward shift of the aerodynamic centre as an aircraft accelerates through Mach one is a major trim and stability problem, and Concorde pumped fuel between tanks to deal with it.
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A flat plate is set at \( 5 \) degrees incidence in a stream at \( M = 2 \). By linearised theory its lift coefficient is nearest to
Answer: A
The Ackeret result gives \( c_l = 4\alpha/\sqrt{M^2-1} \). Here \( \alpha = 0.0873 \) radians and \( \sqrt{3} = 1.732 \), so \( c_l = 4 \times 0.0873 / 1.732 = 0.202 \). The distractor 0.548 is the subsonic two pi alpha value, which does not apply above Mach one.
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The same flat plate at \( 5 \) degrees is now flown at \( M = 3 \). Its lift coefficient becomes
Answer: B
With \( \sqrt{M^2-1} = \sqrt{8} = 2.828 \), the coefficient is \( 4 \times 0.0873 / 2.828 = 0.123 \). Lift falls as the Mach number rises in supersonic flow, the exact reverse of the subsonic Prandtl-Glauert trend, and it is why supersonic manoeuvre is so expensive in incidence.
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Assertion (A): The aerodynamic centre moves rearward as an aircraft accelerates through the transonic range.
Reason (R): Supersonic chordwise loading is far more uniform than subsonic loading.
Select the correct answer from the options below.
Answer: C
Both are true and the second explains the first. Subsonic loading is concentrated near the leading edge, placing the resultant at the quarter chord, whereas supersonic loading is spread along the chord and places it near mid-chord. The shift produces a large nose-down trim change, which Concorde managed by moving fuel.
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Assertion (A): The lift curve slope of a supersonic aerofoil falls as the Mach number rises.
Reason (R): The linearised supersonic result divides by the square root of the Mach number squared minus one.
Select the correct answer from the options below.
Answer: D
Both statements are true and the second is the reason for the first. That radical grows with Mach number, so the slope shrinks. It is worth contrasting with the subsonic case, where the corresponding factor sits in the denominator the other way round and the slope rises instead.
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Consider the following statements about wave drag on a supersonic aerofoil:
1. It is present even in a purely inviscid calculation.
2. It has a contribution from thickness and a contribution from lift.
3. It vanishes if the aerofoil has a finite span.
Which of the statements given above is or are correct?
Answer: A
The first two are correct and are the basis of supersonic section design. The third is nonsense: a finite span adds induced drag on top of the wave drag rather than removing it, and the two mechanisms are quite independent of each other.
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Consider the following statements about the design of a rocket nozzle:
1. The area ratio fixes the exit Mach number for a given gas.
2. Perfect expansion occurs at only one ambient pressure.
3. A first stage is typically underexpanded at sea level.
Which of the statements given above is or are correct?
Answer: B
The first two are correct. The third is the wrong way round: at sea level the ambient pressure is at its highest, so a nozzle sized for altitude is overexpanded near the ground and becomes underexpanded as the vehicle climbs and the ambient pressure falls away.
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