Consider a ramp as shown in Figure 1. If the Mach number of the incoming flow is 2.4, compute the Mach number after the first oblique shock, and the Mach number after the reflected oblique shock. Compute the total and stagnation pressure ratios across the two shock system. Compute also the net change in entropy across the two shock system. Is the angle of incidence equal to the angle of reflection as the shock reflects off the engine cowling? Comment on what you observe.
The Correct Answer and Explanation is:
To answer this question, we need to analyze a double oblique shock system, which typically occurs in supersonic flow over a compression corner followed by shock reflection off a surface, such as an engine cowling. Let’s go step by step. Since the figure is not provided, we’ll assume a standard case often found in textbooks: a compression ramp followed by a reflected oblique shock, with a turning angle θ = 15° for the ramp. The freestream Mach number is M₁ = 2.4.
Given:
- Freestream Mach number, M₁ = 2.4
- Deflection angle (θ) = 15°
- γ (ratio of specific heats) = 1.4 (air)
Step 1: First oblique shock
Using θ-β-M relation (from oblique shock tables or solver), for:
- M₁ = 2.4
- θ = 15°
We find: - Shock angle β₁ ≈ 39.7°
- M₂ (after first shock) ≈ 1.87
Step 2: Reflected shock (off engine cowling)
The flow is now turning back to horizontal — i.e., an expansion of −15°, so we expect:
- A reflected shock to realign flow with the wall
- New deflection angle = −15° (opposite to original)
Using M₂ = 1.87 and θ = −15°, the reflected shock yields:
- Shock angle β₂ ≈ 52.3°
- M₃ (after second shock) ≈ 1.45
Step 3: Pressure and stagnation pressure ratios
Across first shock:
- p₂/p₁ ≈ 2.83
- p₀₂/p₀₁ ≈ 0.841 (total pressure loss)
Across second shock:
- p₃/p₂ ≈ 1.77
- p₀₃/p₀₂ ≈ 0.857
So total pressure ratio:
p₀₃/p₀₁ = 0.841 × 0.857 ≈ 0.72
Step 4: Entropy change
Entropy change per unit mass:
$$
\Delta s = c_p \ln\left(\frac{T_3}{T_1}\right) – R \ln\left(\frac{p_3}{p_1}\right)
$$
Or using:
$$
\Delta s = R \ln\left(\frac{p_{01}}{p_{03}}\right)
$$
$$
\Delta s ≈ R \ln\left(\frac{1}{0.72}\right) ≈ R \cdot 0.328
$$
Thus, entropy increases, consistent with shock behavior.
Angle of incidence vs. reflection
No, the angle of reflection is not equal to the angle of incidence unless symmetry is enforced. Here, the geometry and flow turnings mean the reflection angle adjusts to satisfy flow turning requirements — usually resulting in a different β angle.
Conclusion:
- M₂ ≈ 1.87, M₃ ≈ 1.45
- p₀₃/p₀₁ ≈ 0.72, showing total pressure loss
- Entropy increases across the shock system
- Reflection angle ≠ incidence angle due to flow geometry constraint