Problem 3.10

Air at a temperature of ^oC flows in a supersonic wind tunnel over a very narrow wedge. A shadowgraph photograph of the flow reveals weak waves emanating from the front of the wedge at an angle of ^o to the undisturbed flow. Find the Mach number and velocity in the flow approaching the wedge.

Given:
\( T= \) 45^oC = 318 K,
Mach angle \( = \mu = \) 35^o.

To calculate: \( M=?, V=? \)

The problem description is schematically shown in Fig. 1.

Fig. 1: Schematic of the problem description

The Mach angle is given by the equation,

\[ \sin\left(\mu\right)=\frac{1}{M} \]

\[ M=\frac{1}{\sin\left(\mu\right)}=\frac{1}{\sin\left(35\right)} \]

\[ \boxed{M=1.74345}\ . \]

The speed of sound ahead of the wedge can be calculated as,

\[ a=\sqrt{\gamma\,R\,T}=\sqrt{1.4\times287\times318} \]

The velocity can be calculated as,

\[ V=M\,a=1.74345\times\sqrt{1.4\times287\times318} \]

\[ \boxed{V = 623.2\,\text{m/s}}\ . \]

Problem 3.10:

Solution: