Whenever there is any disturbance in a compressible fluid, the disturbance is propagated in all the directions with a velocity of sound. The nature of propagation depends on the Mach number.

• Mach angle:- It is defined as half of the angle of Mach cone. It is denoted by 'a'.

  $\therefore \sin a=\dfrac 1M$

• Zone of action:- When the Mach number is greater than 1 (M>1), the effect of disturbance is full only in the region inside the Mach cone. This region is known as the zone of action.

• Zone of silence:- When the Mach number is greater than '1.0' (M>1), the effect of disturbance is not full in the region outside the Mach cone. This region which is outside the Mach cone is known as zone of silence.

Numericals

Q1. Find the velocity of bullet fired in standard air if the Mach angle is $30^\circ C$. Take $R=287.14J/kg-K$ and K=1.4 for air. Assume temperature as $15^\circ C$.

Solution: Given:

Mach angle, $a=30^\circ C$

$R=287.14J/kg-K$

K=1.4

$t=15^\circ C$

$\therefore T=15+273=288K$

Therefore, the velocity of sound is given by,

$c=\sqrt{KRT}$

$c=\sqrt{1.4\times 287.14\times 288}$

c=340.25m/s

Using the relation,

$\sin s=\dfrac cV$

$\therefore, \sin 30^\circ=\dfrac {340.25}V$

$\Rightarrow V=\dfrac {340.25}{\sin 30^\circ}$

$V=680.50 \ m/s$

Q2. A projectile travels in air of pressure $10.1043N/cm^2$ at $10^\circ C$ at a speed of 1500 km per hour. Find the Mach number and Mach angle. Take $K=1.4, R=287 J/kg-K$

Solution: Given:

Pressure, $P=10.1043N/cm^2=10.1043\times10^4N/m^2$

$R=287.14J/kg-K$

K=1.4

$t=10^\circ C$

$\therefore T=10+273=283K$

Spped, $V=1500km/hr$

$V=\dfrac {1500\times1000}{60\times60}=416.67m/s$

For adiabatic process, the velocity of sound is given by

$c=\sqrt{KRT}$

$c=\sqrt{1.4\times 287.14\times 283}$

c=337.20m/s

Mach Number,

$M=\dfrac Vc$

$M=\dfrac {416.67}{337.20}$

$M=1.235$

Using the relation,

$\sin a=\dfrac cV=\dfrac 1M$

$\therefore \sin a=\dfrac 1{1.235}$

$\therefore \sin a=0.8097$

$a=\sin^{-1}(0.8097)$

$\therefore a=54.06^\circ$ is the mach angle.