Normal and Oblique Shock Waves Questions and Answers

Answer: C

No normal shock can leave the flow supersonic, whatever the upstream Mach number or the gas. The supersonic-to-supersonic branch would require entropy to fall. A downstream value above one is therefore an arithmetic error, and it is worth using this as a routine check on any shock calculation.

Answer: D

Concentrating the whole compression into one strong normal shock maximises the entropy rise and so the loss, which is the opposite of what is wanted. The other three all spread the compression over weaker waves, and since the loss grows as the cube of shock strength, that is a substantial saving.

Answer: A

The shock is adiabatic, so the total energy and with it the stagnation temperature survive, but it is irreversible, so entropy rises and stagnation pressure is destroyed. This asymmetry is the single most examined point about shocks and is worth stating to oneself before attempting any such question.

Answer: B

The shock stands further off as the Mach number falls, not closer; at hypersonic speeds it wraps in tight against the body and produces the characteristically thin shock layer. The other three statements correctly describe a bow shock.

Answer: C

The maximum deflection at Mach 2 is close to 23 degrees, so a 30 degree half-angle cannot be accommodated by any attached solution, weak or strong. A curved bow shock forms ahead of the leading edge instead, with a subsonic pocket behind its central portion.

Answer: D

The tangential component is not reduced at all, which is precisely why the other three statements hold. If both components fell in the same proportion the flow direction would be unchanged and an oblique shock would produce no turning whatever.

Answer: A

Every shock of finite strength raises entropy, and the rise grows as the cube of the strength. Only in the vanishing limit of a Mach wave is the rise negligible enough to treat the process as isentropic. Adiabatic does not imply isentropic, which is the trap in the second option.

Answer: B

The isentropic relations assume constant entropy, which a shock plainly violates, so applying them through it is simply invalid. They remain perfectly good on either side of the shock; the Rankine-Hugoniot relations are what carry the flow across it.

Answer: C

Correct answer: The stagnation temperature falls across an oblique shock

Stagnation temperature is constant across any shock, oblique or normal, because a shock is adiabatic and does no work - the energy equation gives it directly. What falls is the stagnation *pressure*, since the process is irreversible. The other three do follow from the decomposition: the normal shock relations apply to the normal component, that component determines the loss and is smaller than the full Mach number, and the tangential component passes through unchanged so the downstream flow may still be supersonic.

Answer: D

The loss can be made small by splitting the compression among many weak waves, and real intakes push hard in that direction, but it cannot be made zero. Every shock of finite strength raises entropy, and only a wave of vanishing strength, which performs no compression, is isentropic.