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 provide a full solution to this question, I need to see “Figure 1” mentioned in your prompt. Since no figure was provided, I will assume a generic double-ramp oblique shock problem involving a 2.4 Mach number inlet flow, a shock from a compression ramp, and a reflected shock off a surface like an engine cowling. If your figure has different geometry or angles, please upload or describe it.
Assumptions (for a typical double-ramp flow):
- Incoming flow: Mach M1=2.4M_1 = 2.4
- First ramp deflection angle: θ=10∘\theta = 10^\circ
- Perfect gas: γ=1.4\gamma = 1.4
Step 1: Oblique Shock Relations (First Shock)
Using oblique shock tables (or θ-β-M chart) for:
- M1=2.4M_1 = 2.4
- θ=10∘\theta = 10^\circ
We find the shock wave angle β1≈33.3∘\beta_1 \approx 33.3^\circ, and the Mach number after the first shock:
- M2≈2.05M_2 \approx 2.05
Step 2: Reflected Oblique Shock (Second Shock)
This shock turns the flow back to the freestream direction (i.e., undoes the first deflection).
- Using M2=2.05M_2 = 2.05 and a deflection angle θ=−10∘\theta = -10^\circ,
- From oblique shock charts again, β2≈29.1∘\beta_2 \approx 29.1^\circ, and the Mach number after the reflected shock:
- M3≈1.7M_3 \approx 1.7
Step 3: Total Pressure and Entropy
Across shock waves:
- Total pressure decreases.
- Entropy increases.
Using normal shock relations:
First Shock:
- p02/p01≈0.85p_{0_2}/p_{0_1} \approx 0.85
Second Shock:
- p03/p02≈0.83p_{0_3}/p_{0_2} \approx 0.83
Total pressure ratio across both shocks:
- p03/p01=0.85×0.83≈0.7055p_{0_3}/p_{0_1} = 0.85 \times 0.83 \approx 0.7055
Step 4: Entropy Change
Entropy change Δs=Rln(p01/p03)\Delta s = R \ln(p_{0_1}/p_{0_3})
Assuming air: R=287 J/kg\cdotpKR = 287 \, \text{J/kg·K}
- Δs=287ln(1/0.7055)≈287ln(1.417)≈287×0.348≈99.9 J/kg\cdotpK\Delta s = 287 \ln(1/0.7055) \approx 287 \ln(1.417) \approx 287 \times 0.348 \approx 99.9 \, \text{J/kg·K}
Step 5: Angle of Incidence vs. Reflection
In oblique shock reflection:
- The angle of incidence is NOT necessarily equal to the angle of reflection due to the asymmetry in post-shock Mach numbers and deflection angle changes.
- This phenomenon is different from specular reflection in optics.
Summary of Results:
| Parameter | Value |
|---|---|
| Mach after 1st shock M2M_2 | ≈ 2.05 |
| Mach after reflection M3M_3 | ≈ 1.7 |
| Total pressure ratio p03/p01p_{0_3}/p_{0_1} | ≈ 0.7055 |
| Entropy increase Δs\Delta s | ≈ 99.9 J/kg·K |
| Angle of incidence = reflection? |