Expansion Waves Questions and Answers

Answer: C

The trailing characteristic sits at the downstream Mach angle, \( \arcsin(1/3) = 19.47 \) degrees, measured from the downstream flow direction. Because this is smaller than the leading angle of 30 degrees, the fan opens out, which is the geometric signature of an expansion.

Answer: D

Since the stagnation temperature is constant, the ratio is the inverse ratio of the two stagnation-to-static factors: \( (1 + 0.2 \times 4)/(1 + 0.2 \times 9) = 1.8/2.8 = 0.643 \). The gas has cooled by more than a third simply by being accelerated.

Answer: A

Stagnation pressure is conserved through an isentropic expansion, so the ratio is \( (1.8/2.8)^{3.5} = 0.643^{3.5} \). Taking logarithms, \( 3.5 \times \ln 0.643 = -1.546 \), so the ratio is \( 0.213 \). Pressure falls far more steeply than temperature, as the exponent of 3.5 implies.

Answer: B

Correct answer: 41.8 degrees

\[ \mu = \arcsin\left(\frac{1}{M}\right) = \arcsin\left(\frac{1}{1.5}\right) = \arcsin(0.667) = 41.8^{\circ}. \] The angle falls as the Mach number rises - 90 degrees at \( M=1 \), where the wave is normal to the flow, and approaching zero at high Mach number, where the Mach cone closes down about the flight path. The 48.2 offered is its complement.

Answer: C

The Prandtl-Meyer function is zero at Mach one, so the downstream value is simply the turning angle itself, \( 26.38 \) degrees, which corresponds to Mach 2. This is the cleanest illustration of what the function means: it measures turning accumulated from sonic conditions.

Answer: D

The maximum function value is \( 130.45 \) degrees and the flow has already used \( \nu(3) = 49.76 \) degrees of it, so \( 130.45 - 49.76 = 80.7 \) degrees remain. Note how much turning is still nominally available even at Mach 3, though the pressures involved are by then extremely low.

Answer: A

Adding to the upstream value, \( \nu_2 = 11.90 + 37.86 = 49.76 \) degrees, which is \( \nu(3) \), so the flow reaches Mach 3. Working the same problem in two stages, first to Mach 2 and then on to Mach 3, gives the same answer, which is a useful check on the additivity.

Answer: B

An expansion fan is isentropic, so the stagnation pressure is unchanged and the ratio is exactly one. The value 0.213 is the static pressure ratio, and 0.721 belongs to a normal shock at Mach 2; mixing them up is the standard trap in a question of this shape.

Answer: C

Starting from Mach one, where the function is zero, the turning needed is just \( \nu(1.5) = 11.90 \) degrees. The distractor 41.8 degrees is the Mach angle at that speed, a quantity often confused with the Prandtl-Meyer function because both are angles attached to the same Mach number.

Answer: D

The static pressure ratio for this expansion is 0.213, so \( 100 \times 0.213 = 21.3 \) kPa. Nearly four fifths of the pressure has been given up in exchange for one unit of Mach number, and none of it has been lost to entropy: the stagnation pressure is untouched.