# 11.3.1: Introduction

People had recognized for several hundred years that sound is a variation of pressure. This velocity is referred to as the speed of sound and is discussed first.

Fig. 11.2 Stationary sound wave and gas moves relative to the pulse.

To answer this question consider a piston moving from the left to the right at a relatively small velocity (see Figure 11.1). 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. In the control volume it is convenient to look at a control volume which is attached to a pressure pulse (see Figure 11.2). Applying the mass balance yields

$\rho\, c = (\rho + d\rho)\,(c-dU) \label{gd:sd:eq:cvMass1}$

or when the higher term $$dU\,d\rho$$ is neglected yields

$\rho d U = c \, d\rho \Longrightarrow dU = \dfrac{c d \rho }{ \rho} \label{gd:sd:eq:cvMass2}$ From the energy equation (Bernoulli's equation), assuming isentropic flow and neglecting the gravity results

$\dfrac{ \left( c - dU\right)^2 - c^{2} }{ 2} + \dfrac{dP }{ \rho} = 0 \label{gd:sd:eq:cvEnergy1}$ neglecting second term ($$dU^2$$) yield

$-c dU + \dfrac{dP }{ \rho} = 0 \label{gd:sd:eq:cvCombined}$

Substituting the expression for $$dU$$ from equation (2) into equation (4) yields

Sound Speed

$\label{gd:sd:eq:cvCombined2} c^{2} \left( { d\rho \over \rho } \right) = {dP \over \rho} \Longrightarrow c^2 = \dfrac{dP }{ d\rho}$

An expression is needed to represent the right hand side of equation (5). For an ideal gas, $$P$$ is a function of two independent variables. Here, it is considered that $$P= P(\rho, s)$$ where $$s$$ is the entropy. The full differential of the pressure can be expressed as follows:

$dP = \left. \dfrac{\partial P}{\partial \rho} \right|_{s} d\rho + \left. {\dfrac{\partial P}{\partial s} } \right|_{\rho} ds \label{gd:sd:eq:insontropic1}$

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

$\dfrac{\partial P}{\partial \rho} = \left. \dfrac{\partial P}{\partial \rho} \right|_{s} \label{gd:sd:eq:insontropic2}$ Note that the equation (5) can be obtained by utilizing the momentum equation instead of the energy equation.

Example 11.1

Demonstrate that equation (5) can be derived from the momentum equation.

Solution 11.1

The momentum equation written for the control volume shown in Figure 11.2 is

$\overbrace{(P + dP) - P}^{\sum F} = \overbrace{(\rho +d\rho)(c - dU)^{2} - \rho \, c^2}^{\int_{cs} U\,(\rho\, U\, dA)} \label{gd:sd:eq:cvMomentumEx}$

Neglecting all the relative small terms results in

$dP = (\rho + d\rho) \left( c^{2} - \cancelto{\sim 0}{2\,c\,dU} + \cancelto{\sim 0}{dU^{2}} \right) - \rho c^{2} \label{gd:sd:eq:mem}$ And finally it becomes

$dP = c ^{2} \, d \rho \label{gd:sd:eq:memSimple}$ This yields the same equation as (5).