Compressible Flow Fundamentals Questions and Answers

Answer: A

True airspeed is the Mach number times the local speed of sound, and that speed follows the square root of temperature. Climbing from 288 K to about 217 K takes it from roughly 340 to 295 m/s, so the true airspeed falls by about thirteen per cent even though the Mach number has not changed.

Answer: B

In supersonic flow the gas outruns its own pressure signals, so nothing can propagate upstream; that is the defining feature of the regime and the reason choked nozzles are insensitive to back pressure. The other three all follow from it directly.

Answer: C

Sonic conditions can be reached only where the area is stationary, and in a purely converging duct the minimum area is the exit plane. The nozzle therefore chokes with Mach one at the exit and goes no further, whatever the back pressure. A divergent section is required to continue accelerating the flow.

Answer: D

Newton treated the compressions and rarefactions as isothermal, giving the square root of pressure over density and a result roughly fifteen per cent below measurement. Laplace pointed out that the oscillations are far too rapid for heat to be exchanged, so the process is adiabatic, and inserting the specific heat ratio closed the gap.

Answer: A

A Mach wave is the limiting case of a vanishingly weak disturbance, so the property changes across it are infinitesimal and the entropy rise is of second order and negligible. A shock of finite strength turns the flow abruptly and raises entropy measurably, destroying stagnation pressure in the process.

Answer: B

The aircraft trails a shock system that persists for as long as it flies supersonically, and the ground observer hears the characteristic double bang as that pressure signature sweeps over. It is emphatically not confined to the moment of crossing Mach one, a persistent popular misconception.

Answer: C

The diffuser's second throat slows the supersonic stream through a series of weak shocks rather than one strong normal shock, which greatly reduces the stagnation pressure loss and so the power the tunnel needs. Sizing it correctly is one of the harder parts of supersonic tunnel design, since it must also allow the tunnel to start.

Answer: D

The relation gives \( A/A^* = (1/M)[(2/(\gamma+1))(1 + 0.2M^2)]^3 \) for air. The bracket is \( 0.833 \times 1.8 = 1.5 \), and \( 1.5^3 = 3.375 \); dividing by \( M = 2 \) gives \( 1.6875 \). So a nozzle exit must be about 1.69 times the throat area to reach Mach 2.

Answer: A

Applying \( T_0/T = 1 + 0.2M^2 \) gives \( 1 + 0.2 \times 0.64 = 1.128 \). Even at this high subsonic Mach number the temperature rise on bringing the flow to rest is under thirteen per cent, which is why kinetic heating is not a structural concern for subsonic transports.

Answer: B

Raising the temperature ratio to the power 3.5 gives \( 1.128^{3.5} = 1.524 \). Comparing this with the temperature ratio of 1.128 shows how much more sharply pressure responds to Mach number, which is why pitot-static systems need compressibility corrections well below Mach one.