A uniform supersonic flow of air with Mach number $M_1 = 2.5$ and static pressure $p_1 = 50\text{ kPa}$ passes over a wedge generating an attached straight oblique shock at an angle $\beta = 35.0^\circ$. What is the normal component of the upstream Mach number ($M_{n1}$) to two decimal places?

Answer Key:

1.43

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Governing Concept:** For an oblique shock wave with wave angle $\beta$, the kinematics across the shock are governed strictly by the component of velocity normal to the wave front. 2. **Stage 2: Mathematical Derivation:** * $M_{n1} = M_1 \sin\beta$. * Given $M_1 = 2.5$ and $\beta = 35.0^\circ$. * $\sin(35.0^\circ) \approx 0.573576$. * $M_{n1} = 2.5 \times 0.573576 = 1.43394 \approx 1.43$. 3. **Stage 3: Downstream Shock Behavior:** Since $M_{n1} > 1.0$, a physical compression shock occurs, reducing $M_{n2} < 1$. 4. **Stage 4: Common Trap:** Using deflection angle $\theta$ instead of shock wave angle $\beta$ in $M_1 \sin\theta$. 5. **Stage 5: Official Reference:** John D. Anderson Jr., *Modern Compressible Flow*, Chapter 4 (Oblique Shock Waves).

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