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It can be defined as the square root of the ratio of the inertia force of flowing liquid to the elastic force.
Mach Number=M
$\therefore M=\sqrt{\dfrac{\text{Inertia Force}}{\text{Elastic Force}}}$
$M=\sqrt{\dfrac{\rho AV^2}{KA}}$
$M=\sqrt{\dfrac{V^2}{\dfrac K\rho}}$
$M=\dfrac{V}{\sqrt{\dfrac K\rho}}$
Hence, $\left[M=\dfrac Vc\right]$
Significance of Mach number:-
The flow is defined according to the mach number, it is an important non-dimensional parameter
1. Sub-sonic flow:- When the mach number is less than '1.0' (M<1), i.e., velocity of flow is less than the velocity of sound wave (V<c). Such floor is known as Sub-sonic flow.</p>
2. Sonic flow:- When the mach number is equal to 1.0, i.e., the velocity of flow is equal to the velocity of sound (V = c). Such flow is known as Sonic flow.
3. Supersonic flow:- When the Mach number is greater than 1.0, i.e., the velocity of flow is greater than velocity of sound (V>C), such flow is known as Supersonic flow.
Numericals
Q1. Find the Sonic velocity for the following fluids:-
I. Crude oil sp.gr. 0.8 and bulk modulus $153036N/cm^2$
II. Mercury having a bulk modulus of $2648700N/cm^2$
Solution:
I. For crude oil,
sp.gr=0.8
$K=153036N/cm^2=153036\times10^4N/m^2$
$\rho=0.8\times1000=800kg/m^3$
$\therefore c=\sqrt{\dfrac K\rho}$
$c=\sqrt{\dfrac {153036\times10^4}{800}}=1383.09m/s$
II. For mercury,
sp.gr=13.6
$K=2648700N/cm^2=2648700\times10^4N/m^2$
$\rho=13.6\times1000kg/m^3$
$\therefore c=\sqrt{\dfrac K\rho}$
$c=\sqrt{\dfrac {2648700\times10^4}{13.6\times1000}}=1395.55m/s$
Q2. Calculate the Mach number at a point on a jet-propelled aircraft, which is flying at 1100 km per hour at sea level where temperature is $20^\circ C$. Take K = 1.4, $R=287J/kg-K$
Solution: Given:
Speed, V=1100km/hr
$V=\dfrac {1100\times1000}{60\times60}$
V=305.55m/s
Temperature, $t=20^\circ C$
$T=273+20=293K$
$K=1.4, R=287J/kg-K$
$\therefore c=\sqrt{KRT}$
$c=\sqrt{1.4\times287\times 293}$
$c=343.11 \ m/s$
Mach Number, $M=\dfrac Vc=\dfrac {305.55}{343.11}$
$M=0.89$