Question 12.10: Oblique Shock Calculations Supersonic air at Ma_1 = 2.0 and ...

SOLUTION We are to calculate the shock angle, Mach number, and pressure downstream of the weak and strong oblique shock formed by a two-dimensional wedge.

Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin.

Properties The fluid is air with k = 1.4.

Analysis Because of assumption 2, we approximate the oblique shock deflection angle as equal to the wedge half-angle, i.e., 𝜃 ≅ 𝛿 = 10°. With Ma _{1} = 2.0 and 𝜃 = 10°, we solve Eq. 12–46 for the two possible values of oblique shock angle \beta: \beta_{\text {weak }} = 39.3° and \beta_{\text {strong }} = 83.7°. From these values, we use the first part of Eq. 12–44 to calculate upstream normal Mach number Ma _{1, n,}

The 𝜃-𝛽-Ma relationship: \tan \theta=\frac{2 \cot \beta\left( Ma _{1}^{2} \sin ^{2} \beta-1\right)}{ Ma _{1}^{2}(k+\cos 2 \beta)+2} (Eq. 12-46)

Ma _{1, n}= Ma _{1} \sin \beta \quad \text { and } \quad Ma _{2, n}= Ma _{2} \sin (\beta-\theta) (Eq. 12-44)

Weak shock: Ma _{1, n}= Ma _{1} \sin \beta \rightarrow Ma _{1, n}=2.0 \sin 39.3^{\circ}=1.267

and

Strong shock: Ma _{1, n}= Ma _{1} \sin \beta \rightarrow Ma _{1, n}=2.0 \sin 83.7^{\circ}=1.988

We substitute these values of Ma _{1, n} into the second equation of Fig. 12–36 to calculate the downstream normal Mach number Ma _{2, n} = 0.5794. We also calculate the downstream pressure for each case, using the third equation of Fig. 12–36, which gives

Weak shock: \frac{P_{2}}{P_{1}}=\frac{2 k Ma _{1, n}^{2}-k+1}{k+1} \rightarrow P_{2}=(75.0 kPa ) \frac{2(1.4)(1.267)^{2}-1.4+1}{1.4+1}=128 kPa

and

Strong shock: \frac{P_{2}}{P_{1}}=\frac{2 k Ma _{1, n}^{2}-k+1}{k+1} \rightarrow P_{2}=(75.0 kPa ) \frac{2(1.4)(1.988)^{2}-1.4+1}{1.4+1}=333 kPa

Finally, we use the second part of Eq. 12–44 to calculate the downstream Mach number,

Weak shock: Ma _{2}=\frac{ Ma _{2, n}}{\sin (\beta-\theta)}=\frac{0.8032}{\sin \left(39.3^{\circ}-10^{\circ}\right)}= 1 . 6 4

and

Strong shock: Ma _{2}=\frac{ Ma _{2, n}}{\sin (\beta-\theta)}=\frac{0.5794}{\sin \left(83.7^{\circ}-10^{\circ}\right)}= 0 . 6 0 4

The changes in Mach number and pressure across the strong shock are much greater than the changes across the weak shock, as expected.

Discussion Since Eq. 12–46 is implicit in 𝛽, we solve it by an iterative approach or with an equation solver. For both the weak and strong oblique shock cases, Ma _{1, n} is supersonic and Ma _{2, n} is subsonic. However, Ma _{2} is supersonic
across the weak oblique shock, but subsonic across the strong oblique shock. We could also use the normal shock tables in place of the equations, but with loss of precision.