Next: Speed of sound in Up: Speed of Sound Previous: Motivation Index

Introduction

**Figure:** `A very slow moving piston in a still gas.`
$\begin{figure}\centerline{\includegraphics {cont/sound/illustrations/piston}} \end{figure}$

The people had recognized for several hundred years that sound is a variation of pressure. The ears sense the variations by frequency and magnitude which are transferred to the brain which translates to voice. Thus, it raises the question: what is the speed of the small disturbance travel in a ``quiet'' medium. This velocity is referred to as the speed of sound.

To answer this question consider a piston moving from the left to the right at a relatively small velocity (see Figure 3.1). The information that the piston is moving passes thorough a single ``pressure pulse.'' It is assumed that if the velocity of the piston is infinitesimally small, the pulse will be infinitesimally small. Thus, the pressure and density can be assumed to be continuous.

**Figure:** `Stationary sound wave and gas moves relative to the pulse.`
$\begin{figure}\centerline{\includegraphics {cont/sound/illustrations/stationaryWave}} \end{figure}$

In the control volume it is convenient to look at a control volume which is attached to a pressure pulse. Applying the mass balance yields

$\displaystyle \rho c = (\rho + d\rho)(c-dU)$

(3.1)

or when the higher term $dUd\rho$ is neglected yields

$\displaystyle \rho d U = c d\rho \Longrightarrow dU = {c d \rho \over \rho}$

(3.2)

From the energy equation (Bernoulli's equation), assuming isentropic flow and neglecting the gravity results

$\displaystyle { ( c - dU)^2 - c^{2} \over 2} + {dP \over \rho} = 0$

(3.3)

neglecting second term (

) yield

$\displaystyle -c dU + {dP \over \rho} = 0$

(3.4)

Substituting the expression for

from equation (3.2) into equation (3.4) yields

$\displaystyle c^{2} \left( { d\rho \over \rho } \right) = {dP \over \rho} \Longrightarrow c^2 = {dP \over d\rho}$

(3.5)

An expression is needed to represent the right hand side of equation (3.5). For an ideal gas,

is a function of two independent variables. Here, it is considered that $P= P(\rho, s)$ where

is the entropy. The full differential of the pressure can be expressed as follows:

$\displaystyle dP = \left. {\Pxy{P}{\rho} } \right\vert _{s} d\rho + \left. {\Pxy{P}{s} } \right\vert _{\rho} ds$

(3.6)

In the derivations for the speed of sound it was assumed that the flow is isentropic, therefore it can be written

$\displaystyle \Dxy{P}{\rho} = \left. {\Pxy{P}{\rho} } \right\vert _{s}$

(3.7)

Note that the equation (3.5) can be obtained by utilizing the momentum equation instead of the energy equation.
$\begin{examl} Demonstrate that equation \eqref{sound:eq:cvCombined2} can be derived from the momentum equation. \end{examl}$
Solution

The momentum equation written for the control volume shown in Figure (3.2) is

$\displaystyle \overbrace{(P + dP) - P}^{\sum F} = \overbrace{(\rho +d\rho)(c - dU)^{2} - \rho c^2}^ {\int_{cs} U(\rho U dA)}$

(3.8)

Neglecting all the relative small terms results in

$\displaystyle dP = (\rho + d\rho) \left( c^{2} - \cancelto{\sim 0}{2cdU} + \cancelto{\sim 0}{\rule{20pt}{0pt} dU^{2}}\qquad\;\right) - \rho c^{2}$

(3.9)

$\displaystyle dP = c ^{2} d \rho$

(3.10)

This yields the same equation as (3.5).

Next: Speed of sound in Up: Speed of Sound Previous: Motivation Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21