There are four mechanisms that cause sound energy to be absorbed and sound waves to be attenuated as they propagate in a single-component, homogeneous fluid:

• Viscous (shear) effects in bulk fluids

• Thermal conduction in bulk fluids

• Molecular absorption in bulk fluids

• Thermoviscous boundary layer losses

The decrease in the amplitude of acoustical disturbances or in the amplitude of vibrational motion (due to dissipative mechanisms) has been a topic of interest throughout this textbook. In this chapter, we will capitalize on our investment in such analyses to develop an understanding of the attenuation of sound waves in fluids that are not influenced by proximity to solid surfaces. Such dissipation mechanisms are particularly important at very high frequencies and short distances or very low frequencies over geological distances.

The parallel addition of a mechanical resistance element to the stiffness and mass of a simple harmonic oscillator led to an exponential decay in the amplitude of vibration with time in Sect. 2.4. The (mechanically) series combination of a stiffness element and a mechanical resistance in the Maxwell model of Sect. 4.4.1, and in the Standard Linear Model of viscoelasticity in Sect. 4.4.2 introduced the concept of a relaxation time, τR, that had significant effects on the elastic (in-phase) and dissipative (quadrature) responses as a function of the nondimensional frequency, ωτR. Those response curves were “universal” in the sense that causality linked the elastic and dissipative responses through the Kramers-Kronig relations, as presented in Sect. 4.4.4.

That relaxation time perspective, along with its associated mathematical consequences, will be essential to the development of expressions for attenuation of sound in media that can be characterized by one or more relaxation times related to those internal degrees of freedom that make the equation of state a function of frequency. Examples of these relaxation time effects include the rate of collisions between different molecular species in a gas (e.g., nitrogen and water vapor in air), the pressure dependence of ionic association-dissociation of dissolved salts in seawater (e.g., MgSO4 and H3BO3), and evaporation-condensation effects when a fluid is oscillating about equilibrium with its vapor (e.g., fog droplets in air or gas bubbles in liquids).

The viscous drag on a fluid oscillating within the neck of a Helmholtz resonator, combined with the thermal relaxation of adiabatic temperature changes at the (isothermal) surface of that resonator’s compliance, led to energy dissipation in lumped-element fluidic oscillators in Sect. 9.4.4, producing damping that limited the quality factor of those resonances in exactly the same way as mechanical resistance limited the quality factor of a driven simple harmonic oscillator, which was first introduced as a consequence of similitude (i.e., dimensional analysis) in Sect. 1.7.1.

The thermoviscous boundary layer dissipation, summarized in Eq. (9.38), was used to calculate the attenuation of plane waves traveling in a waveguide in Sect. 13.5.5. As will be demonstrated explicitly in this chapter, thermoviscous boundary layer losses provide the dominant dissipation mechanism at low frequencies (i.e., lumped element systems and waveguides below cut-off) for most laboratory-sized objects (including the laboratory itself when treated as a three-dimensional enclosure). For fluid systems that are not dominated by dissipation on solid surfaces in close proximity to the fluids they contain, the dissipation due to losses within the fluid itself (i.e., bulk losses) can be calculated directly from the hydrodynamic equations of Sect. 7.3.

To reintroduce the concepts complex wavenumber or complex frequency that typically characterize the attenuation of sound over space or time, a simple solution of the Navier-Stokes equation will first be derived. That approach will not provide the correct results for attenuation of sound, even in the absence of relaxation effects, because it does not properly take into account the relationship between shear deformation and hydrostatic compression in fluids that are necessary to produce plane waves (see Fig. 14.1). That relationship was used to relate the modulus of unilateral compression for isotropic solids (aka the dilatational modulus) to other isotropic moduli in Sect. 4.2.2 and in Fig. 4.3. The complete solution for bulk losses due to a fluid’s shear viscosity, μ; thermal conductivity, κ; and the relaxation of internal degrees of freedom (i.e., “bulk viscosity), ζ, will follow that nearly correct introductory treatment and will be based upon arguments related to entropy production, like the analysis in Sect. 9.3.3.

## 1 An Almost Correct Expression for Viscous Attenuation

Because we started with a complete hydrodynamic description of homogeneous, isotropic, single-component fluids using the Navier-Stokes equation, we are now well-prepared to investigate the dissipation mechanisms that attenuate the amplitude of sound waves propagating far from the influence of any solid boundaries. A one-dimensional linearized version of the Navier-Stokes equation (9.2) is reproduced below:

$${\rho}_m\frac{\partial {v}_1\left(x,t\right)}{\partial t}+\frac{\partial {p}_1\left(x,t\right)}{\partial x}-\mu \frac{\partial^2{v}_1\left(x,t\right)}{\partial {x}^2}=0$$
(14.1)

The linearized, one-dimensional continuity equation (10.1) is not affected by the inclusion of viscosity in the Navier-Stokes equation.

$$\frac{\partial {\rho}_1\left(x,t\right)}{\partial t}+{\rho}_m\frac{\partial {v}_1\left(x,t\right)}{\partial x}=0$$
(14.2)

Since the thermal conductivity of the fluid is ignored, κ = 0, the linearized version of the adiabatic equation of state can still be invoked to eliminate ρ1(x, t) in favor of p1(x, t), allowing expression of the continuity equation in terms of the same variables used in Eq. (14.1).

$$\begin{array}{c}{\rho}_1\left(x,t\right)={\left(\frac{\partial \rho }{\partial p}\right)}_s{p}_1\left(x,t\right)=\frac{p_1\left(x,t\right)}{c^2}\\ {}\Rightarrow \kern1em \frac{1}{\rho_m{c}^2}\frac{\partial {p}_1\left(x,t\right)}{\partial t}+\frac{\partial {v}_1\left(x,t\right)}{\partial x}=0\end{array}}$$
(14.3)

As done so many times before, the dispersion relation, ω (k), will be calculated by assuming a right-going traveling wave to convert the homogeneous partial differential equations (14.1) and (14.3) to coupled algebraic equations.

$$\begin{array}{c}-{jkp}_1+\left(j{\omega \rho}_m+{k}^2\mu \right){v}_1=0\\ {} j\omega \frac{p_1}{\rho_m{c}^2}-{jkv}_1=0\end{array}}$$
(14.4)

There are now both real and imaginary terms in the coupled algebraic equations, unlike their nondissipative equivalents, such as Eqs. (10.16) and (10.17). The existence of a nontrivial solution to Eq. (14.4) requires that the determinant of the coefficients vanish.

$$\left|\begin{array}{cc}\frac{j\omega}{\rho_m{c}^2}& - jk\\ {}- jk& \left(j{\omega \rho}_m+{k}^2\mu \right)\end{array}\right|=0$$
(14.5)

The evaluation of this determinant leads to the secular equation that will specify the complex wavenumber, k, in terms of the angular frequency,Footnote 1 ω.

$${\left(\frac{\omega }{c}\right)}^2={k}^2\left(1+\frac{j\omega \mu}{\rho_m{c}^2}\right)$$
(14.6)

If ωμ/ρmc2 < < 1, the binomial expansion can be used twice to approximate the spatial attenuation coefficient, α.

$$\mathbf{k}\cong \frac{\omega }{c}-j\frac{\omega^2\mu }{2{\rho}_m{c}^3}=k-j{\alpha}_{almost}$$
(14.7)

To remind ourselves that this result is not completely correct, this spatial attenuation coefficient has been designated αalmost  = ω2(μ/2ρmc3). The form of this result, specifically the fact that the attenuation is proportional to ω2/ρm, suggests that experimental results could be plotted as a function of the square of frequency divided by mean pressure, as shown in Fig. 14.4.

It is also useful to notice that μ/ρmc2 in Eq. (14.6) has the units of time, so the “small parameter” in those binominal expansions of Eq. (14.6) is of the form $$j{\omega \tau}_{\overline{\mathrm{\ell}}}$$. Equally important is the recognition that such a relaxation time, $${\tau}_{\overline{\mathrm{\ell}}}$$, is on the order of the collision time in a gas, based on the mean free path, $$\overline{\mathrm{\ell}}$$, derived from simple kinetic theory of gases in Sect. 9.5.1 .

$${\tau}_{\overline{\mathrm{\ell}}}=\frac{\overline{\mathrm{\ell}}}{c}\cong \frac{3{\mu}_{gas}}{\rho_m{c}^2}=\frac{3}{\gamma}\frac{\mu_{gas}}{p_m}$$
(14.8)

This is different from the relaxation times, τR, which can characterize the time dependence of the equation of state or the response of a viscoelastic medium described in Sect. 4.4, where ωτR ⪒ 1. At frequencies above $${\omega}_{\overline{\mathrm{\ell}}}\cong {\left({\tau}_{\overline{\mathrm{\ell}}}\right)}^{-1}$$, the assumptions that underlie the hydrodynamic approach are no longer valid (see Chap. 7, Problem 1). For air near room temperature and at atmospheric pressure, $${\tau}_{\overline{\mathrm{\ell}}}$$ ≅ 400 ps, so f = ω /2π ≅ 400 MHz. This is identical with the result obtained in Eq. (9.24) for the critical frequency, ωcrit, at which sound propagation in air transitions from adiabatic at low frequencies to isothermal at high frequencies. The regime where $${\omega \tau}_{\overline{\mathrm{\ell}}}$$ > 1 becomes questionable within the context of a (phenomenological) hydrodynamic theory .

Unlike the complex wavenumbers of exponentially decaying thermal and viscous waves near boundaries, examined in Sects. 9.3.1 and 9.4.2, where ℜe[k] = ℑm[k], most attenuation mechanisms in bulk fluids far from boundaries have ℜe[k] ≫ ℑm[k] or |k| ≫ α. If k is substituted into the expression for the pressure associated with a single-frequency one-dimensional plane wave traveling in the +x direction, p1 (x, t), it is easy to see that α leads to an exponential decay in the amplitude with propagation distance for the sound wave.

$${p}_1\left(x,t\right)=\Re \mathrm{e}\left[\hat{\mathbf{p}}{e}^{j\left(\omega\;t-\mathbf{k}x\right)}\right]=\Re \mathrm{e}\left[\hat{\mathbf{p}}{e}^{j\;\left[\omega\;t-\left(k- j\alpha \right)x\right]}\right]={e}^{-\alpha x}\Re \mathrm{e}\left[\hat{\mathbf{p}}{e}^{j\left(\omega\;t- kx\right)}\right]$$
(14.9)

Since space and time can be transformed by the sound speed, a real temporal attenuation coefficient can be defined, β  = α c, to describe the rate at which the amplitude of the plane wave decays in time.

$${p}_1\left(x,t\right)={e}^{-\beta t}\mathit{\Re e}\left[\hat{\mathbf{p}}{e}^{j\left(\omega\;t- kx\right)}\right]\kern1em \mathrm{where}\kern1em \beta =\alpha c$$
(14.10)

Although the result for αalmost is not exactly correct, it does exhibit a feature of the correct result for the spatial attenuation coefficient that includes thermoviscous dissipation that is given by αclassical in Eq. (14.31) where internal relaxation effects are discussed in Sect. 14.5.Footnote 2

Unlike the spatial attenuation coefficient for dissipation of plane waves in a waveguide, given in Eq. (13.78), which is proportional to$$\sqrt{\omega }$$, Eq. (14.7) shows that αalmost is proportional to the square of the frequency, as is αclassical.

## 2 Bulk Thermoviscous Attenuation in Fluids

Although the previous results for αalmost are incomplete, it both has provided an introduction to the complex wavenumber, k, that determines the attenuation distance and has introduced a relaxation time, $${\tau}_{\overline{\mathrm{\ell}}}$$, that sets an upper limit to the frequencies above which the continuum model of a fluid is not appropriate. One reason that previous result for the viscous attenuation is not complete is that we have ignored the fact that the fluid deformation corresponding to the passage of a plane wave requires the superposition of two shear deformations and a hydrostatic compression. This superposition of shear strain and hydrostatic strain is illustrated schematically in Fig. 14.1. Of course, the result for the attenuation coefficient, αalmost, in Sect. 14.1, also does not yet include the thermal conductivity, κ, of the fluid.

To incorporate all of the dissipative effects in a fluid, it is necessary to start from the complete expression for entropy production in a single-component homogeneous fluid. The mechanical energy dissipation, Emech, is the maximum amount of work that can be done in going from a given non-equilibrium state of energy, Eo, back to equilibrium, E(S), which occurs when the transition is reversible (i.e., without a change in entropy) . $${\dot{E}}_{mech}$$ is the rate at which the mechanical energy is dissipated by the periodic transitions from the non-equilibrium state to the equilibrium state as orchestrated by the wave motion.

$${\dot{\Pi}}_{mech}=-\dot{E}(S)=-\left(\frac{\partial E}{\partial S}\right)\dot{S}={T}_m\dot{S}$$
(14.11)

The right-most expression in Eq. (14.11) uses the fact that the derivative of the energy with respect to the entropy is the equilibrium value of the mean absolute temperature, Tm. The entropy equation (7.43) can be written so that the shear stresses and the hydrostatic stresses can be expressed symmetrically in Cartesian components.Footnote 3

$$\rho T\left(\frac{\partial s}{\partial t}+\overrightarrow{v}\cdotp \overrightarrow{\nabla}s\right)=\nabla \cdotp \left(\kappa\;\overrightarrow{\nabla}T\right)+\frac{1}{2}\mu {\left(\frac{\partial {v}_i}{\partial {x}_k}+\frac{\partial {v}_k}{\partial {x}_i}-\frac{2}{3}\frac{\partial {v}_i}{\partial {x}_i}\right)}^2+\zeta {\left(\nabla \cdotp \overrightarrow{v}\right)}^2$$
(14.12)

The two cross derivatives, (∂vi/∂xk) and (∂vk/∂xi), represent the two shear deformations with the hydrostatic component removed: $$\raisebox{1ex}{2}\!\left/ \!\raisebox{-1ex}{3}\right.\;\left(\partial {v}_i/\partial {x}_i\right)$$ . The square of the hydrostatic deformation is represented by $${\left(\nabla \cdotp \overrightarrow{v}\right)}^2$$. The hydrostatic deformation is multiplied by a new positive scalar coefficient, ζ, that must have the same units as the shear viscosity [Pa-s].

Having the form of a conservation equation (see Sect. 10.5), the right-hand side of Eq. (14.12) represents the rate of entropy production, $$\dot{S}$$, caused by thermal conduction, viscous shear, and some possible entropy production mechanism (unspecified at this point but eventually related to the time dependence of the equation of state) associated with the hydrostatic deformation. Using Eq. (14.12), the dissipated mechanical power, Πmech, can be evaluated by integrating over a volume element that includes the plane wave disturbance, dV.

$${\Pi}_{mech}=-\frac{\kappa }{T_m}\int {\left(\nabla T\right)}^2 dV-\frac{\mu }{2}\int {\left(\frac{\partial {v}_i}{\partial {x}_k}+\frac{\partial {v}_k}{\partial {x}_i}-\frac{2}{3}\frac{\partial {v}_i}{\partial {x}_i}\right)}^2 dV-\zeta \int {\left(\nabla \cdotp \overrightarrow{v}\right)}^2 dV$$
(14.13)

Since we are still attempting a solution in the linear limit, the lowest-order contribution to the power dissipation must be second order in the wave’s displacement from equilibrium; in this case, $${T}_1^2\kern0.5em \mathrm{and}\kern0.5em {\left|{\overrightarrow{v}}_1\right|}^2$$, hence it is positive definite (see Sect. 10.5). For that reason, the absolute temperature, T, can be taken outside the integral and represented by Tm, since allowing for acoustical variation of that temperature term would add a correction to the thermal conduction loss that is third order in displacements from equilibrium.

For a plane wave propagating in the x direction, it is convenient to express vx = v1 sin (ω t − kx), setting vy = vz = 0. Substitution into the last two terms of Eq. (14.13) produces the (nonthermal) mechanical dissipation.

$$-\left(\frac{4}{3}\mu +\zeta \right)\int {\left(\frac{\partial {v}_1}{\partial x}\right)}^2 dV=-{k}^2\left(\frac{4}{3}\mu +\zeta \right){v}_1^2\int {\cos}^2\left(\omega t- kx\right)\; dV$$
(14.14)

Since we are only interested in the time-averaged power dissipation, the contributions from the nonthermal terms in Eq. (14.13) is$$-\left({k}^2/2\right)\left[\left(4\mu /3\right)+\zeta \right]{v}_1^2\kern0.1em {V}_o$$, where Vo is the volume of the fluid under consideration through which the plane wave is propagating.

It is worth comparing the appearance of the factor, 4/3, that multiplies the shear viscosity, μ, with the corresponding expression for the modulus of unilateral compression, D (aka the dilatational modulus), introduced in Sect. 4.2.2, to the shear modulus, G, and bulk modulus, B, in Table 4.1: D = (4G/3) + B. Again, this is a direct consequence of the fact that the distortion produced by a plane wave can be decomposed into two shears (related to G) and a hydrostatic compression (related to B).

The result in Eq. (14.14), without ζ, was first produced by Stokes who expressed the result as the temporal attenuation coefficient . The lack of agreement between his theoretical predictions and experimental measurements provided the starting point for the modern attempts to account for attenuation in terms of molecular relaxation . The spatial attenuation coefficient, due to viscous dissipation, was first introduced by Stefan in 1866Footnote 4 . The first calculation to include both the effects of thermal conductivity and shear viscosity on the absorption of sound was published by Kirchhoff in 1868 .

To evaluate the contribution of thermal conduction to the mechanical dissipation in Eq. (14.13), the temperature change needs to be related to the pressure change to evaluate the one-dimensional temperature gradient, (∂T/∂x). For an ideal gas, this relation should be familiar, having been derived in Eqs. (1.21) and (7.25).

$${\left(\frac{\partial T}{\partial p}\right)}_s=\left(\frac{\gamma -1}{\gamma}\right)\frac{T_m}{p_m}$$
(14.15)

Since we seek an attenuation coefficient that would be applicable to all fluids, a more general expression for (∂T/∂p)s needs to be calculated to evaluate (∂T/∂x).

$${\left(\frac{\partial T}{\partial x}\right)}_s={\left(\frac{\partial T}{\partial p}\right)}_s{\left(\frac{\partial p}{\partial v}\right)}_s{\left(\frac{\partial v}{\partial x}\right)}_s=-{\rho}_mc{\left(\frac{\partial T}{\partial p}\right)}_s{kv}_1\cos \left(\omega\;t- kx\right)$$
(14.16)

The derivative of pressure with respect to velocity for a nearly adiabatic plane wave is a direct consequence of the Euler equation: p1 = ρmcv1.

The derivative of temperature with respect to density can be evaluated using the enthalpy function, H(S, p) = U + pV, that sums the internal energy, U, introduced in Sect. 7.1.2 to calculate heat capacities, with the mechanical work, W = pV. From Eqs. (7.8, 7.9, and 7.10), the internal energy, U(S,V), can be transformed into the enthalpy, H(S, p), using the product rule for differentiation. In thermodynamics, this operation is known as a Legendre transformation .

$$\begin{array}{c} dU= TdS- pdV= TdS-d(pV)+ Vdp\\ {}\Rightarrow d\left(U+ pV\right)\equiv dH= TdS+ Vdp\end{array}}$$
(14.17)

The change in enthalpy, dH(S, p), can be expanded in a Taylor series, retaining only the linear terms.

$$dH={\left(\frac{\partial H}{\partial S}\right)}_p dS+{\left(\frac{\partial H}{\partial p}\right)}_S dp$$
(14.18)

Comparison of Eqs. (14.17) and (14.18) can be used to evaluate those derivatives.

$${\left(\frac{\partial H}{\partial S}\right)}_p=T\kern1em \mathrm{and}\kern1em {\left(\frac{\partial H}{\partial p}\right)}_S=V$$
(14.19)

Since the order of differentiation is irrelevant, the mixed partial derivatives must be equal.

$$\frac{\partial^2H}{\partial p\partial S}=\frac{\partial^2H}{\partial S\partial p}\kern1em \Rightarrow \kern1em {\left(\frac{\partial T}{\partial p}\right)}_S={\left(\frac{\partial V}{\partial S}\right)}_p$$
(14.20)

This result is one of several thermodynamic identities known as the Maxwell relations .

$${\left(\frac{\partial V}{\partial S}\right)}_p={\left(\frac{\partial V}{\partial T}\right)}_p{\left(\frac{\partial T}{\partial S}\right)}_p$$
(14.21)

The result in Eq. (14.21) can be expressed in terms of tabulated material properties  using the definition of the (extensive) heat capacity at constant pressure, Cp, or the (intensive) specific heat (per unit mass) at constant pressure, cp, from Eq. (7.14), and the definition of the isobaric (constant pressure) volume coefficient of thermal expansion, βp.

$${C}_p=T{\left(\frac{\partial S}{\partial T}\right)}_p\kern1em \mathrm{or}\kern1em {c}_p=\kern0.5em T{\left(\frac{\partial s}{\partial T}\right)}_p\kern1em \mathrm{and}\kern1em {\beta}_p=\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_p=-\frac{1}{\rho_m}{\left(\frac{\partial \rho }{\partial T}\right)}_p$$
(14.22)

These results can be combined to produce an expression for the temperature gradient required to evaluate the thermal conduction integral in Eq. (14.13) using Eq. (14.16).

$$\frac{\partial T}{\partial x}=c\frac{\beta_p{T}_m}{c_p}\frac{\partial {v}_1}{\partial x}=-c\frac{\beta_p{T}_m}{c_p}{v}_1k\cos \left(\omega\;t- kx\right)$$
(14.23)

As before, our interest will be in the time-averaged value for evaluation of the thermal conduction term in the integral expression for Πmech in Eq. (14.12).

$${\left\langle -\frac{\kappa }{T_m}\int {\left(\nabla T\right)}^2 dV\right\rangle}_t=\frac{-\kappa {c}^2{\beta}_p^2{v}_1^2{k}^2}{2{c}_p^2}{V}_o$$
(14.24)

This result can be evaluated in terms of the difference in the specific heats that was shown by thermodynamic arguments to be CPCV = ℜ, in Eq. (7.14), for an ideal gas, or cpcv = ℜ/M, where ℜ is the universal gas constant and M is the mean molecular or atomic mass of the ideal gas or ideal gas mixture. By the same thermodynamic arguments, the general result for the specific heat difference can be expressed in terms of the fluid parameters in Eq. (14.22) .

$${c}_p-{c}_v=T{\beta}_p^2{\left(\frac{\partial p}{\partial \rho}\right)}_T=T{\beta}_p^2\left(\frac{c_v}{c_p}\right){\left(\frac{\partial p}{\partial \rho}\right)}_s=T{\beta}_p^2{c}^2\left(\frac{c_v}{c_p}\right)$$
(14.25)

The relationship between the square of the isothermal sound speed, (∂p/∂ρ)T, and the square of the adiabatic sound speed, (∂p/∂ρ)s = c2, should be familiar since (cv/cp) ≡ γ−1.

Substitution of Eq. (14.25) into Eq. (14.24) provides a compact expression for the time-averaged power dissipation that is valid for ideal gases as well as for all other homogeneous fluids.

$${\left\langle -\frac{\kappa }{T_m}\int {\left(\nabla T\right)}^2 dV\right\rangle}_t=-\left(\frac{1}{2}\right)\kappa {k}^2{v}_1^2{V}_o\left(\frac{1}{c_v}-\frac{1}{c_p}\right)$$
(14.26)

Combining Eq. (14.26) with Eq. (14.14) provides an expression for the time-averaged mechanical power dissipation due to all of the irreversible dissipation mechanisms.

$${\left\langle {\Pi}_{mech}\right\rangle}_t=-\left(\frac{1}{2}\right){k}^2{v}_1^2{V}_o\left[\left(\frac{4}{3}\mu +\zeta \right)+\kappa \left(\frac{1}{c_v}-\frac{1}{c_p}\right)\right]$$
(14.27)

The total energy, E, of the plane wave occupying the volume, Vo, can be expressed in terms of the maximum kinetic energy density, (KE)max.

$$E={\left(\mathrm{KE}\right)}_{\mathrm{max}}{V}_o=\left(\frac{1}{2}\right){\rho}_m{v}_1^2{V}_o$$
(14.28)

Since the decay rate of the energy is twice that of the amplitude decay rate, the spatial attenuation constant that reflects thermoviscous losses (and whatever ζ represents!), αT-V, can be written in terms of the time-averaged power dissipation, 〈Πmecht, in Eq. (14.27), and the average total energy, E, in Eq. (14.28).

$${\alpha}_{T-V}=\frac{\left|{\left\langle {\Pi}_{mech}\right\rangle}_t\right|}{2 cE}=\frac{\omega^2}{2{\rho}_m{c}^3}\left[\left(\frac{4}{3}\mu +\zeta \right)+\kappa \left(\frac{1}{c_v}-\frac{1}{c_p}\right)\right]$$
(14.29)

This final result for αT-V is valid for all fluids as long as the decrease in the sound wave’s amplitude over the distance of a single wavelength is relatively small, αT −V λ ≪ 1, since the stored energy was calculated for an undamped sound wave.

We also see that this result is similar to the “almost correct” result, αalmost, calculated from the Navier-Stokes equation in Sect. 14.1. The dependence on frequency, ω; mass density, ρm; and sound speed, c, is identical, but the shear viscosity, μ, is no longer the only transport property of the medium that contributes to attenuation of the sound wave; in Eq. (14.29), μ has been replaced by the term within the square bracket.

As demonstrated earlier in Eq. (14.8), the expression for attenuation given in Eq. (14.29) will always be valid for sound in gases, since the kinematic viscosity, νgas = μgas/ρm, is on the order of the product of the mean free path, $$\overline{\mathrm{\ell}}$$, times the mean thermal velocity of the gas molecules or the sound speed.

$$\frac{\nu_{gas}\omega }{c^2}\cong \overline{\mathrm{\ell}}\frac{\omega }{c}\cong \frac{\overline{\mathrm{\ell}}}{\lambda}\ll 1$$
(14.30)

## 3 Classical Thermoviscous Attenuation

Before the role of molecular relaxation was appreciated and the associated dissipative coefficient, ζ, was introduced, attenuation of sound due to thermoviscous losses was calculated by Kirchhoff . That result is often called the classical absorption coefficient, αclassical.

$${\alpha}_{classical}=\frac{\omega^2}{2{\rho}_m{c}^3}\left[\frac{4}{3}\mu +\kappa \left(\frac{1}{c_v}-\frac{1}{c_p}\right)\right]=\frac{\omega^2}{2{c}^3}\left[\frac{4}{3}\frac{\mu }{\rho_m}+\frac{\kappa }{\rho_m{c}_p}\left(\frac{c_p}{c_v}-1\right)\right]$$
(14.31)

For an ideal gas, the classical attenuation coefficient can be expressed more transparently in terms of the kinematic viscosity, ν = μ/ρm; the polytropic coefficient, γ = cp/cv; and the dimensionless ratio of the thermal and viscous diffusion constants, known as the Prandtl number, Pr ≡ (μ/cpκ) = (δν/δκ)2, that was introduced in Sect. 9.5.4.

$${\alpha}_{classical}=\frac{\omega^2\nu }{2{c}^3}\left[\frac{4}{3}+\frac{\left(\gamma -1\right)}{\Pr}\right]\kern1em \Rightarrow \kern1em \frac{\alpha_{classical}}{f^2}=\frac{2{\pi}^2\nu }{c^3}\left[\frac{4}{3}+\frac{\left(\gamma -1\right)}{\Pr}\right]$$
(14.32)

Most single-component gases and many gas mixtures have Pr ≅ 2/3. For air at atmospheric pressure and 20 °C, ν = 1.51 ×10−5 m2/s, Pr = 0.709, γ = 1.402, and c = 343.2 m/s. Under those conditions, αclassical / f 2 = 1.40 × 10−11 s2/m. The accepted value of α/f2 in the high-frequency limit is 1.84 × 10−11 s2/m. This discrepancy is due to the absence of ζ in Eq. (14.32) .

Since the difference between the specific heats at constant pressure and constant volume is small for liquids, the viscous contribution to the classical attenuation constant is dominant. For pure water at 280 K, $${\nu}_{H_2O}$$ = 1.44 × 10−6 m2/s and$${\Pr}_{H_2O}$$ = 10.4,  with $${c}_{H_2O}$$ = 1500 m/s, making αclassical/f 2 = 1.1 × 10−14 s2/m for freshwater. The accepted value of α/f2 in the high-frequency limit is 2.5 × 10−14 s2/m, again due to the absence of ζ in Eq. (14.32) .

## 4 The Time-Dependent Equation of State

The distortion of a fluid element caused by passage of a plane wave was decomposed into shear deformations, which changed the shape of the element, and a hydrostatic deformation, which changed the volume of the element, as diagrammed schematically in Fig. 14.1. Each of those deformations introduced irreversibility that increased entropy as expressed in Eq. (14.12), leading to energy dissipation as expressed in Eq. (14.13). Based on the discussion in Sect. 9.4, the shear viscosity, μ, was introduced to relate the shear deformations to the dissipative shear stresses.

Another coefficient, ζ, was introduced to relate entropy production to the divergence of the fluid’s velocity field, $$\nabla \cdotp \overrightarrow{v}$$. The continuity equation requires that when $$\nabla \cdotp \overrightarrow{v}\ne 0$$, the density of the fluid must also be changing: (∂ρ/∂t) ≠ 0. Why should a change in the fluid’s density be related to irreversible entropy production?

When the phenomenological model was introduced, it was assumed that only five variables were required to completely specify the state of a homogeneous, isotropic, single-component fluid: one mechanical variable (p or ρ) and one thermal variable (s or T), along with the three components of velocity (e.g., vx, vy, and vz). For a static fluid, |v| = 0, only two variables were required, resulting in the laws of equilibrium thermodynamics (i.e., energy conservation and entropy increase) rather than the laws of hydrodynamics (that also incorporate thermodynamics). The evolution of those variables was determined by the imposition of five conservation equations (i.e., mass, entropy, and vector momentum). That assertion included an implicit assumption that an equation of state existed and it could be used to relate the thermodynamic variables (mechanical and thermal) to each other instantaneously.

For some fluids, the assumption of an instantaneous response of the density to changes in the pressure is not valid. (Noble gases are one notable exception, since they do not have any rotational degrees of freedom.) The microscopic models of gases that were based on the kinetic theory introduced the concept of collision times between the constituent particles (atoms and/or molecules) and the Equipartition Theorem in Eq. (7.2) that stated that through these collisions, an equilibrium could be established that distributed the total thermal energy of the system equitably (on average) among all of the available degrees of freedom. What has been neglected (to this point) was the fact that the collisions take a non-zero time to establish this equilibrium; if the conditions of the fluid element are changing during this time, the system might never reach equilibrium.

How did we get away with this “five-variable fraud” for so long? One answer is hidden in the transition from adiabatic sound speed in an ideal gas to the isothermal sound speed. Equation (9.24) defined a critical frequency, ωcrit, at which the speed of thermal diffusion was equal to the speed of sound propagation. At that frequency, the wavelength of sound corresponded to a distance, which was about 20 times the average spacing between particles, known as the mean free path between collisions, $$\overline{\mathrm{\ell}}$$. Since that collision frequency was so much higher than our frequencies of interest, the equilibration between translational and rotational degrees of freedom in gases of polyatomic molecules occurred so quickly that the equation of state appeared to act instantaneously . Even though a vibrating object couples to the translational degrees of freedom in a gas, the translational and rotational degrees of freedom came into equilibrium in much less time than the period of the vibrating object’s oscillations. That “fraud” was obscured by our use of γ = 7/5 in the expression for sound speed which treated the air as instantly sharing energy between the internal translational and rotational degrees of freedom.

When there are other components in a gas or liquid, they may have relaxation times that are sufficiently close to the acoustic periods of interest that their “equilibration” to the acoustically induced changes cannot be considered to occur instantaneously. In air, for example, if there is water vapor present, it will equilibrate with the O2 and N2 over times that are comparable to the periods of sound waves of interest for human perception (i.e., 20 Hz ≲ f ≲ 20 kHz). Figure 14.2 shows the relaxation frequencies as a function of the mole fraction, h, of H2O and also relative humidity as a percentage .

Those equilibration times are dependent upon the gas mixture’s temperature, pressure, and mixture concentration (i.e., mole fraction of water vapor, h, or relative humidity, RH) . It is apparent that the relaxation frequencies in Fig. 14.2 are in the audio range for ordinary values of temperature and humidity .

$$\begin{array}{c}{f}_{rO}=\frac{p_m}{p_{ref}}\left\{24+\left[\frac{\left(4.04x{10}^4h\right)\left(0.02+h\right)}{0.391+h}\right]\right\}\\ {}{f}_{rN}=\frac{p_m}{p_{ref}}{\left(\frac{T}{T_{ref}}\right)}^{-\frac{1}{2}}\left(9+280h\kern0.5em \exp \left\{-4.170\left[{\left(\frac{T}{T_{ref}}\right)}^{-1/3}-1\right]\right\}\right)\end{array}}$$
(14.33)

The relaxation frequencies for water vapor and nitrogen, frN, and for water vapor and oxygen, frO, assume a standard atmospheric composition with 78.1% nitrogen, 20.9% oxygen, and 314 ppm carbon dioxide at a reference pressure, pref = 101,325 Pa, and reference temperature, Tref = 293.15 K = 20.0 °C. In Eq. (14.33), the molar concentration of water vapor, h, is expressed in percent. For ordinary atmospheric conditions near sea level, 0.2% ≲ h ≲ 2.0%.

To relate RH to h (both in %), it is first necessary to calculate the saturated vapor pressure of water in air, psat, relative to ambient pressure, pref = 101.325 kPa, using the triple-point isotherm temperature, T01 = 273.16 K = +0.01 °C.

$$\frac{p_{sat}}{p_{ref}}={10}^C;\kern1em C=-6.8346{\left(\frac{T_{01}}{T}\right)}^{1.261}+4.6151$$
(14.34)

The molar concentration of water vapor, h, in percent, can then be expressed in terms of the relative humidity, RH, also in percent .

$$h= RH\left(\frac{p_{sat}}{p_m}\right)$$
(14.35)

A similar effect is observed in seawater where boric acid, B(OH)3, and magnesium sulfate, MgSO4, have relaxation frequencies of 1.18 kHz and 145 kHz, respectively, at 20 °C . In the case of these salts, the relaxation time represents the pressure-dependent association-dissociation reaction between the dissolved salts and their ions.

## 5 Attenuation due to Internal Relaxation Times

“If a system is in stable equilibrium, then any spontaneous change of its parameters must bring about processes which tend to restore the system to equilibrium.” H. L. Le ChâtelierFootnote 5

A new positive scalar coefficient, ζ, was introduced in the entropy conservation Eq. (14.12) to scale the irreversibility of hydrostatic fluid deformations. It has the same units as the shear viscosity [Pa-s]Footnote 6 and is usually of about the same magnitude. If the medium does not possess any additional internal degrees of freedom that have to be brought into equilibrium, then its value can be identically zero. That constant is zero for the noble gases (He, Ne, Ar, Kr, Xe, and Rn) that are intrinsically monatomic with atoms that are spherically symmetrical, thus lacking rotational degrees of freedom (see Sect. 7.2). On the other hand, as suggested in Fig. 14.2, if there are processes with relaxation times that are near the frequencies of interest, the value of ζ can be orders of magnitude greater than μ near those frequencies.

With acoustical compressions or expansions, as in any rapid change of state, the fluid cannot remain in thermodynamic equilibrium. Following Le Châtlier’s Principle,5 the system will attempt to return to a new equilibrium state that is consistent with the new parameter values that moved it away from its previous state of equilibrium. In some cases, this equilibration takes place very quickly so that the medium behaves as though it were in equilibrium at all times. In other cases, the equilibration is slow, and the medium never catches up. In either case, the processes that attempt to reestablish equilibrium are irreversible and therefore create entropy and dissipate energy.

If ξ represents some physical parameter of the fluid and ξo represents the value of ξ at equilibrium, then if the fluid is not in equilibrium, ξ will vary with time. If the fluid is not too far from equilibrium, so the difference, ξ − ξo, is small (i.e., |ξ − ξo|/ξo ≪ 1), and then the rate of change of that parameter, $$\dot{\xi}$$, can be expanded in a Taylor series retaining only the first term and recognizing that any zero-order contribution to $$\dot{\xi}$$ must vanish since ξ = ξo at equilibrium.

$$\dot{\xi}=-\frac{\left(\xi -{\xi}_o\right)}{\tau_R}$$
(14.36)

This suggests an exponential relaxation of the system toward its new equilibrium state. Le Châtelier’s Principle requires that the rate must be negative and that the relaxation time, τR, must be positive.

For acoustically induced sinusoidal variations in the parameter, ξ, at frequency, ω, the sound speed will depend upon the relative values of the period of the sound, T = 2π/ω, and the relaxation time, τR. If the period of the disturbance is long compared to the exponential equilibration time, τR, so that ωτR = 2πτR/T ≪ 1, then the fluid will remain nearly in equilibrium at all times during the acoustic disturbance. In that limit, the sound speed will be the equilibrium sound speed, co. In the opposite limit, ωτR = 2πτR/T ≫ 1, the medium’s sound speed, c > co, will be determined by the fluid’s elastic response if the internal degrees of freedom cannot be excited by the disturbance. Said another way, the internal degrees of freedom are “frozen out” in that limit; they simply do not have enough time to participate before the state of the system has changed.

One way to think about this effect is to consider the sound speed in a gas of diatomic molecules that possess three translational degrees of freedom and two rotational degrees of freedom. The specific heats of monatomic and polyatomic gases were discussed in Sect. 7.2, and the relationship between the sound speed in such gases and the specific heat ratio, γ = cp/cv, is provided in Eq. (10.22). If the rotational degrees of freedom are not excited, then the gas behaves as though it were monatomic, so γ = 5/3. If the rotational and translational degrees of freedom are always in equilibrium, then γ = 7/5 < 5/3, so $${c}_{\infty }=\sqrt{25/21}\kern0.5em {c}_o$$.

The mathematical “machinery” needed to represent the attenuation and dispersion of sound waves in a homogeneous medium with an internal degree of freedom, or a “relaxing sixth variable,” has already been developed to describe viscoelastic solids in Sect. 4.4.2. Figure 4.25 could just as well describe the propagation speed (solid line) as a function of the nondimensional frequency, ωτR, with co being the limiting sound speed for ωτR = 2πτR/T ≪ 1 and c being the sound speed for ωτR = 2πτR/T ≫ 1. In addition, the Kramers-Kronig relations of Sect. 4.4.4 would still apply; the variation in sound speed with frequency requires a frequency-dependent attenuation, shown in Fig. 4.25 as the dashed line, and vice versa.

The transformation of the results derived for the stiffness and damping of a viscoelastic medium simply requires that the sound speed is proportional to the square root of the elastic modulus (i.e., stiffness) as expressed in Eq. (10.21). That substitution allows Eq. (4.67) to produce the propagation speed as a function of the nondimensional frequency, ωτR.

$${c}^2={c}_o^2+\left({c}_{\infty}^2-{c}_o^2\right)\frac{{\left({\omega \tau}_R\right)}^2}{1+{\left({\omega \tau}_R\right)}^2}$$
(14.37)

The same approach applied to Eq. (4.70) provides the attenuation per wavelength, αλ, as function of the dimensionless frequency, ωτR.

$$\left(\alpha \lambda \right)=2\pi \frac{\left({c}_{\infty}^2-{c}_o^2\right)\left({\omega \tau}_R\right)}{c_o^2\left[1+{\left({\omega \tau}_R\right)}^2\right]+\left({c}_{\infty}^2-{c}_o^2\right){\left({\omega \tau}_R\right)}^2}$$
(14.38)

Following Eq. (4.71) or Eq. (4.89), the maximum value of attenuation per wavelength in Eq. (14.38) will occur at a unique value of the nondimensional frequency, (ωτR)max.

$${\left({\omega \tau}_R\right)}_{\mathrm{max}}=\frac{c_o}{c_{\infty }}\kern1em \mathrm{and}\kern1em {\left(\alpha \lambda \right)}_{\mathrm{max}}=\pi \frac{\left({c}_{\infty}^2-{c}_o^2\right)}{c_{\infty }{c}_o}$$
(14.39)

The consequence of the Kramers-Kronig relations for such single relaxation time phenomena, as emphasized in Sect. 4.4.2, is that the attenuation is entirely determined by the dispersion, and vice versa.

Using these results, it is possible to write simple universal expressions for attenuation due to excitation of internal degrees of freedom in terms of the relaxation frequency, fR = (2πτR)−1.

$$\frac{\alpha \lambda}{{\left(\alpha \lambda \right)}_{\mathrm{max}}}=\frac{2}{\frac{f_R}{f}+\frac{f}{f_R}}\kern1em \Rightarrow \kern1em \alpha (f)=\left[\frac{2{\left(\alpha \lambda \right)}_{\mathrm{max}}}{cf_R}\right]\frac{f^2}{1+{\left(\frac{f}{f_R}\right)}^2}$$
(14.40)

The variation in the attenuation per wavelength, αλ, and the propagation speed, c, as a function of nondimensional frequency, ωτR, is plotted in Fig. 14.3 and should be compared to the plot for a viscoelastic solid in Fig. 4.25, which exhibits identical behavior. For relaxation frequencies, fR, that are much higher than the frequency of interest, f, the attenuation constant’s quadratic frequency dependence is recovered, as was derived in Eq. (14.7) for αalmost and in Eq. (14.29) for αT-V.

### 5.1 Relaxation Attenuation in Gases and Gas Mixtures

The first example of the effects of the relaxation of an internal degree of freedom on sound speed and attenuation in gas is taken from the measurements of Shields in fluorine . Halogen vapors (e.g., chlorine, fluorine, bromine, iodine) are unique in that they consist of homonuclear diatomic molecules that have appreciable vibrational energy, EV = (n + 1/2) ℏωV, at room temperature (see Sect. 7.2.2). Because the diatomic molecules behave as simple (quantum mechanical) harmonic oscillators, that internal degree of freedom can be characterized by a single relaxation time corresponding to the radian period of their harmonic oscillations.

Figure 14.4 shows the measured values of attenuation per wavelength, αλ, and sound speed, c, as a function of frequency (also normalized by pressure), in units of kHz/atm., for fluorine gas at 102 ± 2 °C, after correction for boundary layer losses at the surface of the tube containing the gas .

The vibrational relaxation time for diatomic fluorine at 102 °C is τR = 10.7 μs, corresponding to a relaxation frequency, fR = (2πτR)−1 = 14.9 kHz. Based on the fit to the sound speed, the limiting speeds are co = 332 m/s and c = 339 m/s. The peak in the attenuation per wavelength, (αλ)max = 0.13, occurs at (ωτR)max = 0.98, based on Eq. (14.39), in excellent agreement with the data in Fig. 14.4.

The relaxation attenuation in humid air is more complicated since the two relaxation frequencies for equilibration of the water vapor with the nitrogen and with the oxygen are different, as expressed in Eq. (14.33) and plotted in Fig. 14.2. Since energy loss is cumulative, it is possible to express the attenuation constant, αtot, as the sum of the attenuation caused by the classical value, αclassical, and the contributions from the two relaxation processes.

$${\alpha}_{tot}={\alpha}_{classical}+{\alpha}_{O_2}+{\alpha}_{N_2}$$
(14.41)

The pressure, frequency, and temperature dependence for the total attenuation coefficient is provided in combination with the relaxation times of Eq. (14.33) and plotted as a function of the frequency/pressure ratio for various values of the relative humidity in Fig. 14.5 .

$$\begin{array}{l}\frac{\alpha_{Air}}{f^2}=1.84\times {10}^{-11}{\left(\frac{p_m}{p_{ref}}\right)}^{-1}{\left(\frac{T}{T_{ref}}\right)}^{\frac{1}{2}}+{\left(\frac{T}{T_{ref}}\right)}^{-5/2}\\ {}\times \left\{0.01275\kern0.5em {e}^{-2,239/T}\left[\frac{f_{rO}}{f_{rO}^2+{f}^2}\right]+0.1068{e}^{-3,352/T}\left[\frac{f_{rN}}{f_{rN}^2+{f}^2}\right]\right\}\end{array}}$$
(14.42)

The difference between the classical attenuation constant and the total is clearly very large. At 2 kHz and 1 atm., αclassical = 0.02 dB/m, but with 10% relative humidity, the attenuation is αtot = 0.80 dB/m.

Values for the attenuation in dB/km are tabulated in a standard for different values of temperature from −25 °C to +50 °C and 10% ≤ RH ≤ 100% for pure tones with frequencies from 50 Hz to 10 kHz in $$\raisebox{1ex}{1}\!\left/ \!\raisebox{-1ex}{3}\right.$$ -octave increments . A small subset of that data are presented in Table 14.1.

### 5.2 Relaxation Attenuation in Fresh and Salt Water

As earlier noted in Sect. 14.3, the measured attenuation of sound in water is greater than αclassical /f2 based on the shear viscosity by more than a factor of two. In the calculation of αclassical for water, the thermal conductivity was neglected. It can be shown that neglect of the thermal conductivity is not the cause of this discrepancy. Measurements of attenuation at 4 °C, where water has its density maximum and the thermal expansion coefficient vanishes, mean that cp = cv, so according to Eq. (14.25), there are no temperature changes associated with the acoustical pressure changes .

The excess attenuation has been ascribed to a structural relaxation process wherein a molecular rearrangement is caused by the acoustically produced pressure changes. During acoustic compression, the water molecules are brought closer together and are rearranged by being repacked more closely. This repacking takes a non-zero amount of time and leads to relaxational attenuation that makes ζ ≠ 0 . The relaxation time as a function of water temperature for this process, τR, is on the order of picoseconds and is plotted as a function of temperature in Fig. 14.6.

At 4 °C, τR ≅ 3.5 ps, corresponding to a relaxation frequency, fR = (2πτR)−1 = 45 GHz, well above experimentally accessible frequencies. For that reason, there are no “relaxation bumps” in the attenuation vs. frequency, as seen in Fig. 14.7. Nonetheless, this structural relaxation makes ζ > μ, accounting for the excess attenuation in pure water.

The attenuation of sound in seawater is similar to that in humid air where the relaxation frequencies are within a frequency range of interest. In seawater, there are two pressure-dependent ionic association-dissociation reactions due to the dissolved boric acid, B(OH)3, and the dissolved magnesium sulfate, MgSO4. Their contributions to the attenuation have the generic form introduced in Eq. (14.40).

$$\begin{array}{c}{\mathrm{Mg}\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}\leftrightarrow {\mathrm{Mg}}^{+3}+{\mathrm{SO}}_4^{-2}+{\mathrm{H}}_2\mathrm{O};\kern0.5em {\alpha}_{{\mathrm{Mg}\mathrm{SO}}_4}\cong \frac{4.6\times {10}^{-3}{\left(\mathrm{kHz}\right)}^2}{4100+{\left(\mathrm{kHz}\right)}^2}\\ {}\mathrm{B}{\left(\mathrm{OH}\right)}_3+{\left(\mathrm{OH}\right)}^{-1}\leftrightarrow \mathrm{B}{\left(\mathrm{OH}\right)}_4^{-1};\kern0.5em {\alpha}_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}\cong \frac{1.2\times {10}^{-5}{\left(\mathrm{kHz}\right)}^2}{1+{\left(\mathrm{kHz}\right)}^2}\end{array}}$$
(14.43)

The approximate attenuation values, $${\alpha}_{{\mathrm{MgSO}}_4}\kern0.5em \mathrm{and}\kern0.5em {\alpha}_{B{\left(\mathrm{OH}\right)}_3}$$, in Eq. (14.43) are in units of [m−1] when the frequency is expressed in kHz.

Those reactions that have relaxation frequencies that depend upon absolute temperature, T, and salinity, S, that is expressed in parts per thousand, ‰, and those relaxation frequencies, $${f}_{{\mathrm{rBH}}_3{\mathrm{O}}_3}\kern0.5em \mathrm{and}\kern0.5em {f}_{{\mathrm{rMgSO}}_4}$$, are in hertz [24, 25].

$$\begin{array}{c}{f}_{{\mathrm{rBH}}_3{\mathrm{O}}_3}=2,800\sqrt{\mathrm{S}/35}\kern0.5em \times {10}^{\left[4-\left(1,245/T\right)\right]}\\ {}{f}_{{\mathrm{rMfSO}}_4}=\frac{8,170\times {10}^{\left[8-\left(1,990/T\right)\right]}}{1+0.008\left(\mathrm{S}-35\right)}\end{array}}$$
(14.44)

For salinity, S = 35‰, and T = 293 K, $${f}_{\mathrm{rB}{\left(\mathrm{OH}\right)}_3}$$ = 1.58 kHz and$${f}_{{\mathrm{rMgSO}}_4}$$ = 132 kHz.

The attenuation coefficient has the expected form, based on Eq. (14.40), and is plotted as a function of frequency in Fig. 14.7.

$$\frac{\alpha }{f^2}=\frac{A_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}{f}_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}}{f^2+{f}_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}^2}+\frac{P_{{\mathrm{MgSO}}_4}{A}_{{\mathrm{MgSO}}_4}{f}_{{\mathrm{MgSO}}_4}}{f^2+{f}_{{\mathrm{MgSO}}_4}^2}+{A}_o{P}_o$$
(14.45)

Approximate expressions for the coefficients representing the relaxation strengths, A, and pressure correction factors, P, are provided by Fisher and Simmons  with more accurate values provided by François and Garrison [24, 25].

## 6 Transmission Loss

The fact that the bulk attenuation of sound in fluids is quadratic in the frequency has important consequences for ultrasonics (f  > 20 kHz) and for long-range sound propagation. At the extremely low frequencies, infrasound in the Earth’s atmosphere can propagate around the entire globe, and the sound of breaking waves generated by a storm on the Pacific coast of the United States has been detected by a low-frequency microphone at the Bureau of Standards in Washington, DC. An International Monitoring System with 60 infrasound monitoring stations has been deployed globally to detect violations of the Comprehensive Nuclear-Test-Ban Treaty .

### 6.1 Short and Very Short Wavelengths

As discussed in Sect. 12.8.1, the Rayleigh resolution criterion implies that the smallest feature that can be resolved in an ultrasonic image will be limited by the wavelength of the sound used to produce the image. Since many ultrasonic imaging systems are used in biomedical applications, we can assume a speed of sound in biological tissue that is approximately equal to the speed of sound in water, $${c}_{{\mathrm{H}}_2\mathrm{O}}$$ = 1500 m/s . To resolve an object that is about a millimeter would then require sound at a frequency, f = c/λ ≅ 1.5 MHz. At that frequency, the attenuation of sound in liver tissue is over 2 dB/cm, so a roundtrip transmission loss to go to a depth of 10 cm is 40 dB.

Because the speed of sound in liquids is typically 200,000 times slower than the speed of light, it is possible to achieve optical wavelength resolution of about 5000 Å = 0.5 μm using sound at a frequency of 3 GHz. In addition, acoustical microscopy produces image “contrast” due to variation in the acoustic absorption of the specimens and scattering that arises from the acoustic impedance mismatch between the specimen and the surrounding material due density and compressibility differences (see Sects. 12.6.1 and 12.6.2). Such sources of contrast will reveal completely different information about a specimen than can be deduced due to changes in optical index of refraction or optical reflectance. In addition, sound can penetrate an optically opaque object, and staining is not required for contrast enhancement.Footnote 7

As mentioned, acoustic microscopy is limited by the fact that attenuation is such a strong function of frequency. At 1.0 GHz, the attenuation of sound in water is 200 dB/mm . Despite the high attenuation loss, acoustic microscopes can image red blood cells acoustically at 1.1 GHz with a resolution equivalent to an oil-immersion optical microscope at a magnification of 1000 . Subcellular details as small as 0.1–0.2 μm (e.g., nuclei, nucleoli, mitochondria, and actin cables) have been resolved due to the extraordinary contrast that can differentiate various cytoplasmic organelles .

The greatest resolution that has been achieved using acoustical microscopy has been accomplished in superfluid helium at temperatures near absolute zero, which has a sound speed, c1 ≅ 240 m/s, a speed that is even lower than the speed of sound in air. Since the dynamics of liquid helium at temperatures below Tλ = 2.17 K are determined by quantum mechanics, the attenuation mechanisms are different than those for classical fluids, which also gives it a much smaller attenuation. At low temperatures, T < 0.5 K, the phonon mean free path is controlled by scattering from “rotons,” which are quantized collective excitation of the superfluid . Sound wavelengths in liquid helium shorter than 2000 Å = 0.2 μm, in a non-imaging experiment, at frequencies of 1.0 GHz, had been studied before 1970 by Imai and Rudnick .

### 6.2 Very Long Wavelengths

At the opposite extreme, at much lower frequencies, the absorption can be quite small. Although the worldwide network of infrasound monitoring sites, using electronic pressure sensors and sophisticated signal processing, that is being used to assure compliance with the Comprehensive Nuclear-Test-Ban Treaty has already been mentioned , the most famous measurement of long-distance infrasound propagation was made using barometers.

On 27 August 1883, the island of Krakatoa, in Indonesia east of Java, was destroyed by an immense volcanic explosion. The resulting pressure wave was recorded for days afterward at more than 50 weather stations worldwide. Several of those stations recorded as many as seven passages of the wave as it circled the globe:

“The barograph in Glasgow recorded seven passages: at 11 hours, 25 hours, 48 hours, 59 hours, 84 hours, 94 hours, and 121 hours (5 days) after the eruption.” 

About 4 h after the explosion, the pressure pulse appeared on a barograph in Calcutta. In 6 h, the pulse reached Tokyo; in 10 h, Vienna; and in 15 h, New York. The period of the pulse was between 100 and 200 s corresponding to a fundamental frequency of about 7 mHz. Its propagation velocity was between 300 and 325 m/s . Although that is close to the speed of sound in air near room temperature, the wave was similar to a shallow water gravity wave in which the height of the atmosphere rose and fell with the passage of the wave .

## 7 Quantum Mechanical Manifestations in Classical Mechanics

“The major role of microscopic theory is to derive phenomenological theory.” G. E. UhlenbeckFootnote 8

Although acoustics is justifiably identified as a field of classical phenomenology, there are many acoustical effects that have their origin in the microscopic theory of atoms and therefore manifest macroscopic behaviors that can only be explained in terms of quantum mechanics. The effects of these “hidden variables” have been manifest throughout this textbook starting with the damping of simple harmonic oscillators that connects the “system” to the environment, thus producing Brownian motion , which was related to the more general theory coupling fluctuations and dissipation  in Chap. 2.Footnote 9

This theme recurred in Chap. 7 when the quantization of energy levels for molecular vibration and rotation influenced the specific heat of gases and in Chap. 9 where a simple kinetic theory of gases was used to determine the pressure and temperature variation of viscosity and thermal conductivity. Now we see in this chapter how structural relaxations in water , like those in Fig. 5.23 for the four crystalline structures of plutonium , scattering of phonons and rotons in superfluids , or molecular vibrations in F2  and collision times in gas mixtures , or chemical reactions [24, 25], manifest themselves in the attenuation of sound.

These internal relaxation effects have been incorporated into our phenomenological theory through the introduction of an additional dissipative process that has been quantified by the introduction of a frequency-dependent parameter, ζ, that shares the same units with the coefficient of shear viscosity, μ. That coincidence has led to this new parameter being called the coefficient of “bulk viscosity” (or sometimes “second viscosity”), even though it is independent of the shear deformation of the fluid and is not the source of momentum transport.

### Talk Like an Acoustician

 Viscoelasticity Mean free path Viscous drag Kinetic theory Thermal relaxation Einstein summation convention Thermoviscous boundary layer Bulk viscosity Spatial attenuation coefficient Second viscosity Temporal attenuation coefficient Le Châtelier’s principle Shear strain Nondimensional frequency Hydrostatic strain Vibrational relaxation time Collision time Structural relaxation Enthalpy function Association-dissociation reactions Legendre transformation Kramers-Kronig relations Maxwell relations Relaxation time

### Exercises

1. 1.

Bulk attenuation and reverberation time. An expression was provided in Eq. (13.29) to incorporate the attenuation in air contained within an enclosure into the expression for reverberation time. A “useful correlation” was provided in Eq. (13.30) that was applicable for 1500 Hz ≤ f ≤ 10,000 Hz and for relative humidity in the range 20% ≤ RH ≤ 70%.

1. (a)

Average frequency dependence. Table 14.1 (left) provides the attenuation in dB/km from the ANSI/ASA standard for the frequencies within the range specified at 20 °C for RH = 50% . Plot the log10 of the spatial attenuation, α, in m−1 vs. the log10 of frequency, f, in kHz, to determine the power law dependence on frequency (see Sect. 1.9.3). Is your result proportional to f1.7 to within the statistical uncertainty of your least-squares fit? Keep in mind that attenuation expressed in [dB/m] must be multiplied by 0.1151 ≅ [10log10(e2)]−1 to convert to m−1 (sometimes including the dimensionless “Nepers” to report results in Nepers/m).

2. (b)

Humidity dependence. Table 14.1 (right) also includes the attenuation in dB/km at 4.0 kHz and 20 °C for 20% ≤ RH ≤ 70%. The “useful correlation” claims that the correction to frequency dependence for variations in relative humidity should be linear in (50%/RH). How close is that presumed humidity dependence to values in the table for variation in relative humidity at 4.0 kHz?

2. 2.

The mother of all PA systems. Shown in Fig. 14.8 is a loudspeaker that can produce 30,000 watts of acoustic power by modulating a pressurized air stream (like a siren) using a cylindrical “valve” mounted on a voice coil, like that used for an electrodynamic loudspeaker. That sound source, located at the apex of the horn, is called a “modulated airstream loudspeaker.” 

Such a public address system was developed to tell illiterate enemy combatants, in their native language, to put down their weapons and surrender from a distance that is greater than the distance that could be traversed by artillery shells. Although the bandwidth of telephone speech for very good intelligibility is generally 300 Hz to 3.4 kHz, for this problem, we will focus on the propagation of a 1 kHz pure tone. Since this system was deployed in desert terrain, assume that RH = 10%, Tm = +50 °C = 122 °F, and pm = 100 kPa.

1. (a)

Sound level at 100 m. Assuming hemispherical spreading and no sound absorption, what is the root-mean-square acoustic pressure amplitude if the source produces 30 kW of acoustic power and any nonlinear effects that might cause harmonic distortion (see Sect. 15.2.3) can be neglected? Also neglect any refractive effects due to sound speed gradients caused by temperature or wind as discussed in Sect. 11.3. Report your results as both in r.m.s. pressure amplitude and in dB re: 20 μParms.

2. (b)

Greater distances. Repeat part a, but determine the sound pressure at 1.0 km and 3.0 km, again neglecting attenuation.

3. (c)

Include attenuation. Determine the spatial attenuation coefficient under these conditions at 1.0 kHz and use it to reduce the sound pressure at 0.1 km, 1.0 km, and 3.0 km below that obtained due only to hemispherical spreading.

4. 3.

Pump wave attenuation for a parametric array. The generation of highly directional sound beams from the nonlinear acoustical interaction of two colinear high-frequency sound beams will be discussed in Sect. 15.3.3. Calculate the exponential attenuation length,  = α−1, of a typical 40 kHz beam in dry air, RH = 0%, and moist air with RH = 60% using the graph in Fig. 14.5.

5. 4.

Siren. The siren shown in Fig. 14.9 consumed 2500 ft3/min of air at a pressure of 5 psi above ambient to produce 50 horsepower of acoustic power at 500 Hz, and did so with 72% efficiency .

1. (a)

Hydraulic power. How much time-averaged power, 〈Πhyt = (Δp)|U|, is available, in watts and in horsepower, from the specified volume flow rate, U, and the available pressure drop, Δp?

2. (b)

Hemispherical spreading. Assuming the mean temperature during the measurement was Tm = 20 °C and pm = 100 kPa, what would be the root-mean-square pressure a distance of 1000 ft. from the siren if it produced a sound power of 50 horsepower? Report your results both in pressure and dB re: 20 μParms.

3. (c)

Include attenuation. If RH = 50%, how much additional loss, in dB, would be produced by absorption at the same distance?

6. 5.

The SOFAR channel. One method for locating pilots who crash over the ocean is to drop an explosive sound source that is set to detonate at a depth that is equal to the axis of the deep sound channel that is created by a sound speed profile like the one shown in Fig. 11.8. In that figure, the axis of the sound channel is 1112 m below the ocean’s surface. Sound that is trapped in that channel will spread cylindrically, rather than spherically, beyond a transition distance that will be assumed to be much shorter than the distance of interest, so the sound amplitude will decrease in proportion to $$\sqrt{R}$$, where R is the distance between the source and the receiver.

1. (a)

Cylindrical spreading. What would be the loss, in dB, due to cylindrical spreading over 5000 km relative to the level 1 km from the source?

2. (b)

Include attenuation. Using the results for seawater in Fig. 14.7, what would be the spatial attenuation, in dB, that would have to multiply the cylindrical spreading loss over 5.0 km calculated in part (a) for sound with a frequency of 100 Hz? Repeat for the loss due to attenuation in seawater after 5000 km.

7. 6.

High-frequency relaxational attenuation constant for air and water.

1. (a)

Limiting frequency dependence. Using Eq. (14.40), show that for frequencies well above the highest relaxational frequency, fR, the attenuation of sound is independent of frequency.

2. (b)

Relaxational attenuation constant (bulk viscosity) of air. In Eq. (14.42), the measured high-frequency limit of the spatial attenuation constant in air is αair /f2 = 1.84 × 10−11 s2/m. Calculate αclassical/f2 for air at atmospheric pressure and 20 °C, and use both high-frequency results (i.e., with and without relaxation effects) to determine the value for ζair in the high-frequency limit.

3. (c)

Relaxational attenuation constant (bulk viscosity) of pure water. The measured high-frequency limit of the spatial attenuation constant in pure water at 4 °C is $$\underset{f\gg {f}_R}{\lim }{\alpha}_{{\mathrm{H}}_2\mathrm{O}}/{f}^2=$$ 2.5 × 10−14 s2/m. Based on αclassical / f2 for pure water at atmospheric pressure and 4 °C, determine the value for $${\zeta}_{H_2O}$$ in the high-frequency limit.