Page: Velocity of sound or pressure wave in a fluid
0

In solid, liquid or gas, the disturbance is transmitted from one point to the other. The velocity with which the disturbance is transmitted depends upon the distance between molecules of the medium.

In the case of solids, the disturbance is transmitted instantaneously as the molecules are closely packed. In the case of liquid and gases, the molecules are relatively apart, has the disturbance is transmitted from one molecule to the other or next molecule.

But in case of fluids, there is some distance between two adjacent molecules, hence each molecule travel a distance before transmitting the disturbance.

The distance between the molecules is related to the density, which in turn depends upon the pressure in case of fluids.

Hence the velocity disturbance depends on the charge in pressure and density of the fluid.

Velocity of sound in terms of bulk modulus.

$K=\dfrac {\text{Increase in Pressure}}{\dfrac{\text{Decrease in Volume}}{\text{Original Volume}}}$

$K=\dfrac {dP}{-\left(\dfrac{dV}V\right)}$

Now as we know

$\rho\times \text{Volume}=\text{constant}$

$\rho\times V=\text{constant}$

Differentiating the above equation ($\rho$ and V)

$\rho dV+Vd\rho=0$

$\rho dV=-Vd\rho$

$-\dfrac {dV}V=\dfrac {d\rho}\rho$

Substituting the value $\left(-\dfrac {dV}V\right)$, we get

$K=\dfrac {dP}{\dfrac {d\rho}\rho}$

$K=\rho\dfrac {dP}{d\rho}$

$\dfrac K\rho=\dfrac {dP}{d\rho}$

Therefore, the Velocity of the sound wave is

$c=\sqrt{\dfrac K\rho}=\sqrt{\dfrac {dP}{d\rho}}$

Velocity of sound for isothermal process

$\dfrac P\rho=\text{constant}$

or, $P\rho^{-1}=\text{constant}$

Differentiating the above equation, we get

$P(-1)\rho^{-2}d\rho+\rho^{-1}dP=0$

Dividing by '$\rho^{-1}$', we get

$-P\rho^{-1}d\rho+dP=0$

or, $\dfrac {-P}\rho d\rho+dP=0$

$\therefore, dP=\dfrac P\rho d\rho$

$\dfrac {dP}{d\rho}=\dfrac P\rho=RT$

$\therefore c=\sqrt{\dfrac P\rho}=\sqrt {RT}$

Velocity of sound for adiabatic process

$\dfrac P{\rho^K}=\text{constant}$

or, $P\rho^{-K}=\text{constant}$

Differentiating the above equation, we get

$P(-K)\rho^{-K-1}d\rho+\rho^{-K}dP=0$

Dividing by '$\rho^{-K}$', we get

$-PK\rho^{-1}d\rho+dP=0$

$\therefore, dP=\dfrac {PK}\rho d\rho$

$\dfrac {dP}{d\rho}=\dfrac P\rho K=RTK=KRT$

$\therefore c=\sqrt {KRT}$

page fluid mechanics 2 fm2 • 36 views