Supersonic airflow at $M_1 = 3.0$ passes through an oblique shock wave inclined at an angle $\beta = 30^\circ$ to the freestream. What is the downstream normal Mach number $M_{n2}$? (Take $\gamma = 1.4$, round off to two decimal places).

Answer Key:

0.70

### Solution: Upstream normal Mach number: $M_{n1} = M_1 \sin\beta = 3.0 \times \sin(30^\circ) = 3.0 \times 0.5 = 1.5$. Normal shock relation for $M_{n2}$: $M_{n2}^2 = \frac{2 + (\gamma - 1)M_{n1}^2}{2\gamma M_{n1}^2 - (\gamma - 1)} = \frac{2 + (0.4)(2.25)}{2(1.4)(2.25) - 0.4} = \frac{2 + 0.90}{6.30 - 0.40} = \frac{2.90}{5.90} \approx 0.4915$. Downstream normal Mach $M_{n2} = \sqrt{0.4915} \approx 0.701 \approx 0.70$. Correct: 0.70 (range 0.68 to 0.72).

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