Prandtl–Meyer Expansion Wave Calculations
Supersonic air at Ma _{1} = 2.0 and 230 kPa flows parallel to a flat wall that suddenly expands by 𝛿 = 10° (Fig. 12–45). Ignoring any effects caused by the boundary layer along the wall, calculate downstream Mach number Ma _{2} and pressure P _{2}.
SOLUTION We are to calculate the Mach number and pressure downstream of a sudden expansion along a wall.
Assumptions 1 The flow is steady. 2 The boundary layer on the wall is very thin.
Properties The fluid is air with k = 1.4.
Analysis Because of assumption 2, we approximate the total deflection angle as equal to the wall expansion angle, i.e., 𝜃 ≅ 𝛿 = 10°. With Ma _{1} = 2.0, we solve Eq. 12–49 for the upstream Prandtl–Meyer function,
v( Ma )=\sqrt{\frac{k+1}{k-1}} \tan ^{-1}\left(\sqrt{\frac{k-1}{k+1}\left( Ma ^{2}-1\right)}\right)-\tan ^{-1}\left(\sqrt{ Ma ^{2}-1}\right) (Eq. 12-49)
\begin{aligned}& v( Ma )=\sqrt{\frac{k+1}{k-1}} \tan ^{-1}\left(\sqrt{\frac{k-1}{k+1}\left( Ma ^{2}-1\right)}\right)-\tan ^{-1}\left(\sqrt{ Ma ^{2}-1}\right) \\=& \sqrt{\frac{1.4+1}{1.4-1}} \tan ^{-1}\left(\sqrt{\frac{1.4-1}{1.4+1}\left(2.0^{2}-1\right)}\right)-\tan ^{-1}\left(\sqrt{2.0^{2}-1}\right)=26.38^{\circ}\end{aligned}
Next, we use Eq. 12–48 to calculate the downstream Prandtl–Meyer function,
Turning angle across an expansion fan: \theta=v\left( Ma _{2}\right)-v\left( Ma _{1}\right) (Eq. 12-48)
\theta=v\left( Ma _{2}\right)-v\left( Ma _{1}\right) \rightarrow v\left( Ma _{2}\right)=\theta+v\left( Ma _{1}\right)=10^{\circ}+26.38^{\circ}=36.38^{\circ}
Ma _{2} is found by solving Eq. 12–49, which is implicit—an equation solver is helpful. We get Ma _{2} = 2.38. There are also compressible flow calculators on the Internet that solve these implicit equations, along with both normal and oblique shock equations; e.g., see www.aoe.vt.edu/~devenpor/aoe3114/calc.html.
We use the isentropic relations to calculate the downstream pressure,
P_{2}=\frac{P_{2} / P_{0}}{P_{1} / P_{0}} P_{1}=\frac{\left[1+\left(\frac{k-1}{2}\right) Ma _{2}^{2}\right]^{-k /(k-1)}}{\left[1+\left(\frac{k-1}{2}\right) Ma _{1}^{2}\right]^{-k /(k-1)}}(230 kPa )=126 k P a
Since this is an expansion, Mach number increases and pressure decreases, as expected.
Discussion We could also solve for downstream temperature, density, etc., using the appropriate isentropic relations.