Oblique Shock Calculator — θ-β-M Relations

🌊 Oblique Shock Wave Schematic

📝 Configuration

🌬️ Gas Selection

Gas Preset:

💥 Shock Conditions

Upstream Mach Number ($M_1$):

Deflection Angle ($\theta$) [deg]:

Specific Heat Ratio ($\gamma$):

📐 Upstream State

Upstream Pressure ($p_1$) [Pa]:

Upstream Temperature ($T_1$) [K]:

Gas Constant ($R$) [J/kg·K]:

Key Equation (θ-β-M):

$$\tan\theta = \frac{2\cot\beta\;\left(M_1^2 \sin^2\!\beta - 1\right)}{M_1^2\left(\gamma + \cos 2\beta\right) + 2}$$

Normal component:
$M_{1n} = M_1 \sin\beta$

Downstream Mach:
$M_2 = \dfrac{M_{2n}}{\sin(\beta - \theta)}$

📊 Results & Visualization

Configure inputs and click Calculate to view results.

ℹ️ About Oblique Shock Waves

When a supersonic flow encounters a wedge or compression corner, an oblique shock wave forms at an angle β to the incoming flow. The flow is deflected by angle θ through the shock.

Key characteristics:
- For each M₁ and θ, two solutions exist: a weak shock (smaller β, usually supersonic downstream) and a strong shock (larger β, subsonic downstream).
- A maximum deflection angle θ_max exists for each M₁. Beyond it, the shock detaches.
- The weak shock is the solution most commonly observed in nature and engineering applications.
- At θ = 0, the weak solution degenerates to a Mach wave (β = μ = sin⁻¹(1/M₁)).

📘 Calculation Methodology

Mathematical Model & Theory

The oblique shock relations are derived by decomposing the upstream velocity into components normal and tangential to the shock wave. The normal component undergoes a normal shock, while the tangential component is unchanged.

$$\tan\theta = \frac{2\cot\beta\left(M_1^2\sin^2\!\beta - 1\right)}{M_1^2\left(\gamma + \cos 2\beta\right) + 2}$$ $$M_{1n} = M_1\sin\beta, \quad M_2 = \frac{M_{2n}}{\sin(\beta - \theta)}$$ $$\frac{p_2}{p_1} = 1 + \frac{2\gamma}{\gamma+1}\left(M_{1n}^2 - 1\right)$$ $$\frac{\rho_2}{\rho_1} = \frac{(\gamma+1)M_{1n}^2}{(\gamma-1)M_{1n}^2 + 2}$$

The θ-β-M equation is cubic in sin²β and is solved numerically via bisection. Two physical roots exist for θ < θ_max.

Worked Engineering Example

Problem:
Air (γ = 1.4) at M₁ = 2.0 encounters a 10° wedge (θ = 10°). Find the weak shock angle and downstream conditions.

Step-by-step (weak solution):
1. Solve θ-β-M for β:
$\beta_{\text{weak}} \approx 39.31°$

2. Normal Mach: $M_{1n} = 2.0 \times \sin(39.31°) = 1.2671$

3. Normal shock on M₁ₙ = 1.2671:
$p_2/p_1 = 1.7070$, $\rho_2/\rho_1 = 1.4557$
$T_2/T_1 = 1.1726$, $M_{2n} = 0.8031$

4. Downstream Mach:
$M_2 = \dfrac{0.8031}{\sin(39.31° - 10°)} = 1.6405$

Result: Weak shock at β = 39.31°, M₂ = 1.64 (still supersonic).

Assumptions & References

Assumptions: Steady, two-dimensional flow. Calorically perfect gas (γ = const). Adiabatic. Thin shock. Uniform upstream conditions. Infinite wedge (no three-dimensional effects).

References:

Anderson, J. D. Modern Compressible Flow: With Historical Perspective, McGraw-Hill, 4th ed., 2021 — Ch. 4 (Oblique Shock and Expansion Waves).
Shapiro, A. H. The Dynamics and Thermodynamics of Compressible Fluid Flow, Vol. 1, Wiley, 1953.
Gupta, S. C. Applied Computational Fluid Dynamics — §1.19 (Shock wave relations).
NACA Report 1135 — Equations, Tables, and Charts for Compressible Flow, 1953.

ThermoFluidCalc Report

💥 Oblique Shock Calculator — θ-β-M Relations