Normal and Oblique Shock Waves Questions and Answers

Answer: C

As the deflection tends to zero the shock weakens to a Mach wave, whose angle is the Mach angle. That is the lower limit. The upper limit is ninety degrees, at which the oblique shock has become a normal shock and the deflection has fallen back to zero from the other direction.

Answer: D

There is no attached oblique shock solution beyond the maximum deflection, so the shock detaches and stands off ahead of the body as a curved bow wave. Immediately in front of the nose it is effectively normal, giving a subsonic pocket there, and it weakens to a Mach wave far out to the sides.

Answer: A

A blunt nose demands an infinite deflection at the stagnation streamline, far beyond any attached solution, so the shock must stand off. This is not merely tolerated but sought after in re-entry design, because the detached shock keeps the hottest gas away from the surface and eases the heating problem.

Answer: B

The limiting density ratio is the specific heat ratio plus one over the specific heat ratio minus one, which is 2.4 over 0.4, that is six. Density cannot be compressed without limit however strong the shock, whereas the pressure and temperature ratios do grow without bound. Real gas effects raise the limit somewhat at hypersonic speeds.

Answer: C

The limiting value is the square root of the specific heat ratio minus one over twice the specific heat ratio, that is the square root of 0.4 over 2.8, which is 0.378. So however strong the shock, the downstream flow never becomes arbitrarily slow in Mach number terms, even though its temperature has risen enormously.

Answer: D

An oblique shock is adiabatic just as a normal shock is, so the stagnation temperature comes through it unchanged. Stagnation pressure is the quantity that suffers, and it falls less for an oblique shock than for a normal shock at the same upstream Mach number, which is the whole basis of external-compression intake design.

Answer: A

Entropy rise grows very rapidly with shock strength, so several weak shocks cost far less stagnation pressure than one strong one doing the same total compression. The Concorde and the SR-71 intakes both work this way, and the recovered stagnation pressure directly sets the thrust the engine can produce.

Answer: B

The wall imposes flow tangency, so the deflection produced by the incident shock must be cancelled. A reflected shock of the opposite deflection does exactly that. The reflection is not mirror-symmetric in angle, because the reflected shock acts on a slower stream than the incident one did.

Answer: C

The reflected shock has to turn a stream that the incident shock has already slowed, and if the deflection needed exceeds what that slower stream can achieve, no regular reflection exists. A Mach reflection then forms instead, with a normal shock stem standing on the wall and a triple point above it.

Answer: D

Expanding the entropy change for a weak shock shows the first and second order terms vanish, leaving a cubic dependence on the strength. This is why a vanishingly weak wave may be treated as isentropic, and why splitting a compression into several weak shocks is so much cheaper than doing it with one strong one.