Question 12.11: Prandtl–Meyer Expansion Wave Calculations Supersonic air at ...

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.

12.45
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We are to calculate the Mach number and pressure downstream of a sudden expansion along a wall.
Assumptions   1  The flow is steady. 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,

ν( 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)             (12.49)

=\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}

Next, we use Eq. 12–48 to calculate the downstream Prandtl–Meyer function,

\theta=ν\left( Ma _2\right)-ν\left( Ma _1\right) \rightarrow ν\left( Ma _2\right)=\theta+ν\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 ~ kP 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.

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