Question 17.12: Prandtl–Meyer Expansion Wave Calculations Supersonic air at ...

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 $\text { (i.e., } \theta \cong \delta=10^{\circ} \text { ) }$. With $Ma _{1}$ = 2.0, we solve Eq. 17–49 for the upstream Prandtl–Meyer function,

$\nu( Ma )=\sqrt{\frac{k+1}{k-1}} \tan ^{-1}\left[\sqrt{\frac{k-1}{k+1}\left( Ma ^{2}-1\right)}\right]-\tan ^{-1}\left(\sqrt{ Ma ^{2}-1}\right)$

$\begin{aligned}\nu( Ma ) &=\sqrt{\frac{k+1}{k-1}} \tan ^{-1}\left[\sqrt{\frac{k-1}{k+1}\left( Ma ^{2}-1\right)}\right]-\tan ^{-1}\left(\sqrt{ Ma ^{2}-1}\right) \\&=\sqrt{\frac{1.4+1}{1.4-1}} \tan ^{-1}\left[\sqrt{\frac{1.4-1}{1.4+1}\left(2.0^{2}-1\right)}\right]-\tan ^{-1}\left(\sqrt{2.0^{2}-1}\right)=26.38^{\circ}\end{aligned}$

Next, we use Eq. 17–48 to calculate the downstream Prandtl–Meyer function,

Turning angle across an expansion fan:  $\theta=\nu\left( Ma _{2}\right)-\nu\left( Ma _{1}\right)$

$\theta=\nu\left( Ma _{2}\right)-\nu\left( Ma _{1}\right) \rightarrow \nu\left( Ma _{2}\right)=\theta+\nu\left( Ma _{1}\right)=10^{\circ}+26.38^{\circ}=36.38^{\circ}$

Ma2 is found by solving Eq. 17–49, which is implicit—an equation solver is helpful. We get $Ma _{2}=2.385$. 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,

$\nu( Ma )=\sqrt{\frac{k+1}{k-1}} \tan ^{-1}\left[\sqrt{\frac{k-1}{k+1}\left( Ma ^{2}-1\right)}\right]-\tan ^{-1}\left(\sqrt{ Ma ^{2}-1}\right)$

$P_{2}=\frac{P_{2} / P_{0}}{P_{1} / P_{0}} P_{1}=\frac{\left[1+\left(\frac{k-1}{2}\right) Ma _{2}^{2}\right]^{-k /(k-1)}}{\left[1+\left(\frac{k-1}{2}\right) Ma _{1}^{2}\right]^{-k /(k-1)}}(230 kPa )=126 k P a$

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.