Question 12.10: Oblique Shock Calculations Supersonic air at Ma1 = 2.0 and 7...
Oblique Shock Calculations
Supersonic air at Ma_1 = 2.0 and 75.0 kPa impinges on a two-dimensional wedge of half-angle 𝛿 = 10° (Fig. 12–44). Calculate the two possible oblique shock angles, \beta_{weak} and \beta_{strong}, that could be formed by this wedge. For each case, calculate the pressure and Mach number downstream of the oblique shock, compare, and discuss.
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_{weak} = 39.3° and \beta_{strong} = 83.7°. From these values, we use the first part of Eq. 12–44 to
calculate upstream normal Mach number Ma_{1, n},
Ma _{1, n}= Ma _1 \sin \beta \quad \text { and } \quad Ma _{2, n}= Ma _2 \sin (\beta – \theta) (12.44)
\tan \theta=\frac{2 \cot \beta\left( Ma _1^2 \sin ^2 \beta – 1\right)}{ Ma _1^2(k + \cos 2 \beta) + 2} (12.46)
Weak shock: Ma_{1, n} = Ma_1 \sin \beta → Ma_{1, n} = 2.0 \sin 39.3° = 1.267
and
Strong shock: Ma_{1, n} = Ma_1 \sin \beta → Ma_{1, n} = 2.0 \sin 83.7° = 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}. For the weak shock, Ma_{2, n} = 0.8032, and for the strong shock, 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 kPaand
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}= 3 3 3 kP aFinally, 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.
