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Propagation of pressure waves in a compressible fluid

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.

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