Compressible Flow Fundamentals Questions and Answers

Answer: C

The speed of sound is \( a = \sqrt{\gamma R T} \). The product is \( 1.4 \times 287 \times 288 = 115\,718 \), whose square root is \( 340.2 \) m/s. The distractor 402 is what results from forgetting the temperature, and 116 from forgetting to take the root at all.

Answer: D

Applying \( a = \sqrt{\gamma R T} \) gives \( \sqrt{1.4 \times 287 \times 216.7} = \sqrt{87\,070} = 295.1 \) m/s. Equivalently, since the speed of sound goes as the square root of temperature, \( 340.2 \times \sqrt{216.7/288} = 340.2 \times 0.867 = 295 \) m/s. This is why a jet cruising at a fixed Mach number flies more slowly in true airspeed as it climbs into colder air.

Answer: A

Mach number is the flow speed divided by the local speed of sound, so \( 250 / 340 = 0.735 \). The reciprocal, 1.36, is the standard slip. At this Mach number the aircraft is well into the range where compressibility must be accounted for, since the threshold is around 0.3.

Answer: B

The relation is \( T_0 / T = 1 + 0.2 M^2 \) for air. Here \( 1 + 0.2 \times 4 = 1.8 \), so \( T_0 = 300 \times 1.8 = 540 \) K. The large rise shows why kinetic heating becomes a structural problem at supersonic speed even before hypersonic conditions are reached.

Answer: C

The isentropic relation is \( p_0 / p = (1 + 0.2 M^2)^{3.5} \). At \( M = 1 \) that is \( 1.2^{3.5} = 1.893 \). Its reciprocal, \( 0.528 \), is the critical pressure ratio, the value the back pressure must fall below before a converging nozzle will choke.

Answer: D

The Mach angle satisfies \( \sin\mu = 1/M \), so \( \mu = \arcsin(0.5) = 30 \) degrees. Taking the arcsine of the Mach number instead of its reciprocal is impossible here, since 2 lies outside the domain, which is a useful check on the formula.

Answer: A

From \( \sin\mu = 1/1.4 = 0.714 \), the angle is \( 45.6 \) degrees. Note the trend: as the Mach number falls towards one the cone opens out towards 90 degrees, and as it rises the cone becomes steadily more slender, which is why hypersonic bodies sit inside very thin shock layers.

Answer: B

At the critical condition the Mach number is one, so \( T^*/T_0 = 1 / (1 + 0.2) = 2/(\gamma + 1) = 2/2.4 = 0.833 \). The value 0.528 is the corresponding pressure ratio and 0.634 the density ratio; all three are worth remembering, since they recur throughout nozzle work.

Answer: C

Raising the critical temperature ratio to the power \( \gamma/(\gamma-1) = 3.5 \) gives \( 0.833^{3.5} = 0.528 \). A converging nozzle chokes as soon as the back pressure falls below this fraction of the stagnation pressure, after which the mass flow is fixed.

Answer: D

The throat is at Mach one, so \( T^* = 0.833 \times T_0 = 0.833 \times 300 = 250 \) K. Note that the reservoir temperature is the stagnation temperature, since the gas there is at rest, and it is conserved through an adiabatic nozzle even if losses occur.