Normal and Oblique Shock Waves Questions and Answers

Answer: A

The shock polar maps the locus of possible downstream velocity vectors for all admissible shock angles at a given upstream Mach number. It shows at a glance the two solutions for each deflection, the maximum deflection where they merge, and the point at which the downstream flow becomes sonic.

Answer: B

The Hugoniot relates the states either side of a shock, and because the process is dissipative it reaches a hotter, higher-entropy state than an isentropic compression to the same pressure would. The two curves are tangent to second order at the initial state, which is why weak shocks are nearly isentropic.

Answer: C

The sharp adverse pressure gradient imposed by the shock is felt by the already slow fluid near the wall, which thickens and frequently separates. The interaction feeds back on the shock itself, smearing it into a lambda foot. It is a major source of buffet and drag on transonic wings.

Answer: D

For back pressures in this range the flow accelerates supersonically past the throat and then must be brought back to match the higher exit pressure. A normal shock forms in the divergent section, and it sits further downstream the lower the back pressure, until at the design condition it leaves the exit altogether.

Answer: A

Near the wall the boundary layer is subsonic, so it transmits pressure information upstream and the compression begins ahead of the main front. The result is an oblique leading branch meeting the near-normal shock above it, giving the characteristic lambda shape visible in schlieren photographs.

Answer: B

Substituting into \( M_2^2 = (1 + 0.2 M_1^2)/(1.4 M_1^2 - 0.2) \), the numerator is \( 1 + 1.8 = 2.8 \) and the denominator \( 12.6 - 0.2 = 12.4 \), so \( M_2^2 = 0.2258 \) and \( M_2 = 0.475 \). It is approaching, but has not reached, the limiting value of 0.378.

Answer: C

Using \( 1 + 1.1667(M_1^2 - 1) \) with \( M_1^2 - 1 = 8 \) gives \( 1 + 9.333 = 10.33 \). Comparing with 4.5 at Mach 2, the pressure ratio has more than doubled for a fifty per cent rise in Mach number, which shows how steeply shock strength climbs.

Answer: D

The relation \( (\gamma+1)M_1^2/((\gamma-1)M_1^2+2) \) gives \( 21.6/(3.6+2) = 21.6/5.6 = 3.857 \). It is climbing towards the ceiling of six but slowly, which illustrates how much harder density is to compress than pressure is to raise.

Answer: A

Correct answer: 2.83

\[ M_{n1} = M_1\sin\beta = 4\sin 45^{\circ} = 4 \times 0.707 = 2.83. \] Only that normal component is processed by the shock; the tangential component is unchanged. It is why an oblique shock at a shallow angle is so much weaker than a normal one in the same stream, and why the wave angle is the designer’s lever for controlling the loss in a supersonic intake.

Answer: B

Each shock passes through the other and continues, refracted in direction, and because the two streams either side have passed through different shock sequences they emerge with different entropies and velocities. They share a pressure and a direction but not a speed, so a slip line separates them.