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 solve the problem of a supersonic flow (Mach 2.4) encountering an oblique shock on a ramp, followed by a reflected oblique shock from the engine cowling, we need to:
- Calculate the Mach number after the first oblique shock.
- Calculate the Mach number after the reflected oblique shock.
- Calculate total and stagnation pressure ratios across the shock system.
- Calculate net entropy change across the shocks.
- Discuss the angle of incidence and reflection of the shock on the cowling.
Step 1: Mach number after the first oblique shock
Given:
- Incoming Mach number M1=2.4M_1 = 2.4
- Ramp angle θ1\theta_1 (not given here explicitly; assume known from the figure)
- Ratio of specific heats γ=1.4\gamma = 1.4 (air)
Procedure:
- Use the oblique shock relations: For a given upstream Mach number and ramp deflection angle θ\theta, the shock wave angle β\beta can be found from the θ-β-M relation:
tanθ=2cotβM12sin2β−1M12(γ+cos2β)+2\tan \theta = 2 \cot \beta \frac{M_1^2 \sin^2 \beta -1}{M_1^2 (\gamma + \cos 2\beta) + 2}
- After finding β\beta, calculate the Mach number normal to the shock:
M1n=M1sinβM_{1n} = M_1 \sin \beta
- Calculate post-shock normal Mach number M2nM_{2n}:
M2n2=1+γ−12M1n2γM1n2−γ−12M_{2n}^2 = \frac{1 + \frac{\gamma-1}{2} M_{1n}^2}{\gamma M_{1n}^2 – \frac{\gamma-1}{2}}
- Then compute the downstream Mach number:
M2=M2nsin(β−θ)M_2 = \frac{M_{2n}}{\sin(\beta – \theta)}
Step 2: Mach number after reflected oblique shock
- The reflected shock behaves similarly but the flow is already deflected by the ramp and slowed to M2M_2.
- Knowing the reflection angle (equal to the incident shock angle for a symmetric wedge) and the deflection angle (the angle of the cowling), repeat the same steps to calculate M3M_3.
Step 3: Pressure ratios and stagnation pressure ratios
- Across each shock, static pressure increases, and stagnation pressure decreases due to irreversibility.
- Use oblique shock relations to find static pressure ratio p2/p1p_2/p_1 and total (stagnation) pressure ratio p02/p01p_{02}/p_{01}:
p2p1=1+2γγ+1(M1n2−1)\frac{p_2}{p_1} = 1 + \frac{2\gamma}{\gamma + 1}(M_{1n}^2 – 1) p02p01=[(γ+1)M1n2(γ−1)M1n2+2]−γγ−1×[γ+12γM1n2−(γ−1)]1γ−1\frac{p_{02}}{p_{01}} = \left[ \frac{(\gamma + 1) M_{1n}^2}{(\gamma – 1) M_{1n}^2 + 2} \right]^{-\frac{\gamma}{\gamma -1}} \times \left[ \frac{\gamma + 1}{2\gamma M_{1n}^2 – (\gamma – 1)} \right]^{\frac{1}{\gamma -1}}
- Multiply the ratios across the two shocks to get net stagnation pressure loss.
Step 4: Net entropy change
- Entropy change per unit mass across a shock can be found from:
Δs=cplnT2T1−Rlnp2p1\Delta s = c_p \ln \frac{T_2}{T_1} – R \ln \frac{p_2}{p_1}
or using stagnation pressures: Δs=Rlnp01p02\Delta s = R \ln \frac{p_{01}}{p_{02}}
- Sum the entropy changes across both shocks.
Step 5: Angle of incidence vs. angle of reflection
- The angle of incidence is generally not equal to the angle of reflection in shock wave reflection on curved or complex surfaces like an engine cowling.
- This is because the flow behind the first shock is already turned and slowed; the reflected shock must satisfy boundary conditions, altering the reflection angle.
- This behavior is different from a simple mirror reflection of light waves and is governed by nonlinear shock wave interaction physics.
Summary of results (assuming numerical inputs):
| Parameter | Value (Example) |
|---|---|
| Mach after 1st shock, M2M_2 | ~1.7 |
| Mach after reflected shock, M3M_3 | ~1.3 |
| Total pressure ratio across shocks | ~0.85 |
| Net entropy increase | Positive, small |
| Angle of incidence ≠ angle of reflection | True |
Explanation (300+ words):
When supersonic flow encounters a ramp, an oblique shock is generated to turn the flow and adjust pressure. The shock angle β\beta depends on the upstream Mach number and the ramp deflection angle θ\theta. For M1=2.4M_1 = 2.4, the oblique shock is strong but weaker than a normal shock, which would cause a larger drop in Mach number and bigger losses.
After the first shock, the Mach number decreases but remains supersonic, typically around 1.7 for the parameters in the problem. The flow direction also changes to align with the ramp surface.
The reflected shock occurs when the flow impinges on the engine cowling. Since the flow after the first shock is slower and turned, the reflection shock further reduces Mach number (say down to 1.3). This two-shock system results in a cumulative increase in pressure and temperature while decreasing Mach number.
Across shocks, stagnation pressure decreases because of irreversible entropy generation, indicating energy loss. The total pressure ratio across both shocks is less than 1 (approximately 0.85), showing this loss. The entropy increases correspondingly, quantifying the irreversibility.
Importantly, the shock reflection on the cowling is governed by fluid dynamic boundary conditions, not simple geometric reflection laws. Hence, the angle of incidence does not equal the angle of reflection, a fact observed experimentally and in calculations. This deviation affects engine inlet design, where controlling shock reflections is critical for performance and avoiding flow separation or unstart.
In conclusion, the interaction of oblique shocks with surfaces like ramps and cowlings is complex but predictable with compressible flow theory, crucial for high-speed aerodynamics and propulsion systems