A strong normal shock wave, with a pressure ratio of 29 across it, is traveling into still air ($T_1 = 300$ K, $\gamma = 1.4$). The Mach number of the shock wave is _________ (round off to 2 decimal places).
Official Answer Key:
5.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2017 Q.35)**
**1. Intuitive Physical Setup & Given Parameters**:
- Moving normal shock traveling into stationary air.
- Upstream quiescent state: $T_1 = 300\text{ K}$, $\gamma = 1.4$.
- Static pressure ratio across the shock: $\frac{P_2}{P_1} = 29.0$.
**2. Step-by-Step Mathematical Derivation**:
1. **Normal Shock Pressure Ratio Relation**:
For any normal shock with upstream relative Mach number $M_s$:
\frac{P_2}{P_1} = 1 + \frac{2\gamma}{\gamma + 1}(M_s^2 - 1)
2. **Substituting Given Properties ($\gamma = 1.4$)**:
\frac{2\gamma}{\gamma + 1} = \frac{2(1.4)}{2.4} = \frac{2.8}{2.4} = \frac{7}{6}
29.0 = 1 + \frac{7}{6}(M_s^2 - 1)
3. **Solving for Shock Mach Number $M_s$**:
28.0 = \frac{7}{6}(M_s^2 - 1) \implies M_s^2 - 1 = 28.0 \times \frac{6}{7} = 4.0 \times 6 = 24.0
M_s^2 = 24.0 + 1.0 = 25.0
M_s = \sqrt{25.0} = 5.00
The shock wave propagates at exactly Mach 5.00 into the still gas.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Evaluation**: **5.00** (Acceptable Range: [4.95, 5.05]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **5.0**
- **Exam Rule**: $M_s = \sqrt{1 + \frac{\gamma+1}{2\gamma}\left(\frac{P_2}{P_1}-1\right)}$. For $\frac{P_2}{P_1} = 29$, $M_s = \sqrt{1 + \frac{6}{7}(28)} = \sqrt{25} = 5$.