Question: Problem on Mach angle and Mach number

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**Given :**

**Pressure** = 1 bar

**Temperature** t = 10˚ C

T = 10 + 273 = 283 k

**Speed** V = 1800 km/hr

= $\frac{1800 \times 1000}{60 \times 60} = 500m/s$

**K** = 1.4

**R** = 287 J/kg K.

Now, For adiabatic process, the velocity of sound is,

$C = \sqrt{KRT} = \sqrt{1.4 \times 287 \times 283}$

**C** = 337.20 m/s.

$\therefore$ Mach Number = $\frac{v}{c} = \frac{500}{337.20} = 1.48$

and, Mach angle is, Sin $\alpha$ $\frac{c}{v} = \frac{1}{m}$

$\therefore \alpha = sin^{-1} (\frac{1}{1.48})$

**Mach angle** is $\alpha = 42.40˚$

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