Consider a centered Prandtl-Meyer expansion fan at a \(\theta = 4^{\circ}\) corner in a Mach \(1.78\) air flow, as shown in the figure below. The angle \(\psi\) (see figure) made by the ending wave of the fan with respect to the incoming stream is _______ degrees (rounded off to 1 decimal place).
An excerpt from the table of Prandtl-Meyer function for air is provided below.
M
\(\nu\) [deg]
1.72
18.40
1.74
18.98
1.76
19.56
1.78
20.15
1.80
20.73
1.82
21.30
1.84
21.88
1.86
22.45
1.88
23.02
1.90
23.59
1.92
24.15
1.94
24.71
1.96
25.27
1.98
25.83
2.00
26.38
Show Hint
Use \(\nu(M_2) = \nu(M_1) + \theta\) to find the downstream Mach number, then note the ending wave sits at the Mach angle \(\mu_2\) measured from the downstream (turned) flow direction, so \(\psi = \mu_2 - \theta\) measured from the incoming stream.
Step 1: Recall the Prandtl-Meyer relation for an expansion corner.
The Prandtl-Meyer function \(\nu(M)\) is the angle through which a sonic flow must turn isentropically to reach Mach number \(M\). Across a corner that turns the flow by \(\theta\), the function simply adds:
\[ \nu(M_2) = \nu(M_1) + \theta \]
Step 2: Read \(\nu(M_1)\) from the table and find \(\nu(M_2)\).
At \(M_1 = 1.78\), the table gives \(\nu_1 = 20.15^{\circ}\).
\[ \nu_2 = 20.15 + 4 = 24.15^{\circ} \]
The table shows \(\nu = 24.15^{\circ}\) exactly at \(M_2 = 1.92\), so \(M_2 = 1.92\).
Step 3: Find the Mach angles at the start and end of the fan.
The Mach angle is \(\mu = \arcsin(1/M)\).
\[ \mu_1 = \arcsin\left(\frac{1}{1.78}\right) = \arcsin(0.5618) = 34.2^{\circ} \]
\[ \mu_2 = \arcsin\left(\frac{1}{1.92}\right) = \arcsin(0.5208) = 31.4^{\circ} \]
Step 4: Relate the ending wave angle to the incoming stream.
The fan is bounded by a leading Mach wave, inclined at \(\mu_1\) to the incoming (upstream) flow direction, and an ending Mach wave, inclined at \(\mu_2\) to the outgoing (downstream) flow direction. The downstream flow direction itself has already been turned by \(\theta\) away from the incoming stream direction (the flow follows the wall through the corner). So, measured from the original incoming stream direction, the ending wave sits at
\[ \psi = \mu_2 - \theta = 31.4 - 4 \]
Final Answer:
\[ \boxed{\psi \approx 27.4^{\circ}} \]
