Dissipative Hydrodynamics

Abstract

In the previous chapter, the resonance frequency, ω_o, of a Helmholtz resonator was calculated. When driven at that frequency, the predicted pressure amplitude inside the resonator’s volume (compliance) became infinite. This was because the theory used to model that inertance and compliance network in Figs. 8.11 and 8.15, and in Eq. (8.50), did not include any dissipation. By introducing DeltaEC, we were able to calculate the amount of power dissipated in the neck (inertance) and volume (compliance) of a 500 ml boiling flask. In this chapter, those losses will be calculated from hydrodynamic “first principles.”

In the previous chapter, the resonance frequency, ω_o, of a Helmholtz resonator was calculated. When driven at that frequency, the predicted pressure amplitude inside the resonator’s volume (compliance) became infinite. This was because the theory used to model that inertance and compliance network in Figs. 8.11 and 8.15, and in Eq. (8.50), did not include any dissipation. By introducing DeltaEC, we were able to calculate the amount of power dissipated in the neck (inertance) and volume (compliance) of a 500 ml boiling flask. In this chapter, those losses will be calculated from hydrodynamic “first principles.”

Just as we used lumped acoustical elements in Chap. 8 to begin our exploration of the linearized, nondissipative forms of the hydrodynamic equations, we will again use lumped elements to introduce thermal and viscous dissipative effects, restricting our attention (temporarily) to thermoviscous effects at solid-fluid boundaries. Later, in Chap. 14, those dissipative mechanisms will be generalized to include dissipative effects (attenuation), with the addition of molecular relaxation effects (see Sect. 4.4), which will be incorporated by the introduction of a “bulk viscosity,” to account for dissipation of sound waves propagating in spaces that are far from any boundaries.

1 The Loss of Time Reversal Invariance

There are certain phenomena that can be “played backward” in a movie or video that do not look different than when played forward. If we took a video of a wavelike pulse propagating on a string or a slinky, like the Gaussian pulses in Figs. 3.2, 3.3, and 3.4, it would be very difficult to tell whether the video was being played forward or backward. This indifference to the “arrow-of-time” is called time reversal invariance.

On the other hand, there are certain things that do not look right played backward. An egg dropping to the floor and cracking does not make sense played backward—a yellow slimy mess that self-organizes into an unbroken shell and then levitates. The murder at the start of Memento [1] shows the bullet case reentering the gun and a Polaroid™ photo fading out and then reentering its camera.

When dissipation is included in our hydrodynamic equations, time reversal invariance must be abandoned. This can be demonstrated mathematically by focusing on a traveling wave moving to the right, along the x axis, as described in Eq. (1.9), which represented complex exponential form.

$$ p\left(x,t\right)={p}_m+\mathit{\Re e}\left[\hat{\mathbf{p}}{e}^{j\left(\omega\;t- kx\right)}\right] $$

(9.1)

The complex amplitude of the wave, $\hat{\mathbf{p}}$, can introduce an additional phase factor to compensate for some arbitrary choice of the time origin, t = 0. If we reverse the sign of the time, the phase factor changes from j(ωt − kx) to −j(ωt + kx), which is the equivalent of making the wave propagate in the negative x direction. When we change the direction of time, the sign of the velocity also changes (Figs. 9.1 and 9.2).

Fig. 9.1

Sketches of pulses in a nondissipative medium showing the (transverse) pulse amplitude propagating reversibly in time along one dimension. We could assume that the pulse started at t = 0, as the single lump shown on the left, and then became two pulses of the same width and shape having amplitudes equal to half the original pulse height, traveling in opposite directions. We could also assume that the arrows showing the direction of motion for the two pulses at the right could be reversed. Then the pulses are moving toward each other; they superimpose to form the larger pulse on the left and then pass through each other to become the two pulses shown at the right with their arrows (as drawn) again showing the direction of their subsequent motion after they had passed through each other

Fig. 9.2

Unlike Fig. 9.1, the time evolution of the pulse at the left can only proceed in the direction indicated by the arrow. The progression of the pulse is determined by a diffusion equation like Eq. (9.4). If the pulse at the left represents an initial temperature distribution in a fluid, over time the temperature would diffuse as shown due to the thermal conductivity of the medium. If the pulse represented the injection of dye into a stagnant fluid, the diffusion of the dye would be represented by the subsequent pulse shapes. The most diffuse pulse, shown at the right, would never spontaneously self-focus into the more concentrated temperature or dye distribution shown at the left

In the continuity equation (7.32), the reversal of time changes the sign of the time derivative, ∂ρ/∂t. The term that includes the divergence, $\nabla \bullet \left(\rho \overrightarrow{v}\right)$, also has its sign reversed because the velocity changes sign. Both terms on both sides of the continuity equation are thus “negated” by the time reversal so the equation is unaffected; just multiply the whole equation by (−1) and we are back where we started.

This is not the case for the Navier-Stokes equation (7.34), even in its linearized form that discards the convective term, $\left(\overrightarrow{v}\bullet \nabla \right)\overrightarrow{v}$, and ignores gravity.

$$ \rho \frac{\partial \overrightarrow{v}\left(\overrightarrow{x},t\right)}{\partial t}=-\overrightarrow{\nabla}p\left(\overrightarrow{x},t\right)+\mu\;{\nabla}^2\overrightarrow{v}\left(\overrightarrow{x},t\right) $$

(9.2)

Since both the velocity and the time change sign, the derivative, $\partial \overrightarrow{v}/\partial t$, does not change sign. On the right-hand side of Eq. (9.2), the pressure, $p\left(\overrightarrow{x},t\right)$, is a thermodynamic variable and therefore a Galilean invariant; its value is a property of the medium that is not a function of the reversal of time or of uniform motion of the coordinate system.

We’d be in fine shape with both the time derivative and the gradient of the pressure, since the sign of neither changes under time reversal (i.e., the Euler equation is time reversal invariant). But the Laplacian operator, ∇², that is multiplied the shear viscosity, μ, operates on the velocity, which does change sign when time is reversed. The Navier-Stokes equation, as written in Eq. (9.2), is therefore not time reversal invariant. Neither is the linearized entropy equation, as written in Eq. (7.43), even if the viscous entropy generation term is neglected and we assume that the thermal conductivity, κ, is independent of position, where s is the specific entropy (per unit mass) [J/K − kg] = [m²/s² − K].

$$ \rho T\frac{\partial s\left(\overrightarrow{x},t\right)}{\partial t}=\kappa\;{\nabla}^2T\left(\overrightarrow{x},t\right) $$

(9.3)

Here again, both $s\left(\overrightarrow{x},t\right)$ and $T\left(\overrightarrow{x},t\right)$ are Galilean invariant thermodynamic properties of the fluid that do not change under time reversal. The sign of ∂s/∂t does change since the sign of the time is reversed. Equation (9.3) can be transformed into the Fourier Diffusion Equation after substituting the relation between changes in heat per unit mass, dq, and changes in entropy per unit mass, ds. We have dq = T ds, from Eq. (7.5), and the relation between the addition of heat and the change in temperature at constant pressure, dq = ρc_p dT, for a polytropic substance, from Eq. (7.14).

$$ \frac{\partial T\left(\overrightarrow{x},t\right)}{\partial t}=\frac{\kappa }{\rho {c}_P}{\nabla}^2T\left(\overrightarrow{x},t\right) $$

(9.4)

Equations that have the form of Eq. (9.2) and Eq. (9.4) are diffusion equations. They produce dissipation and they violate time reversal invariance. If those irreversible effects are present, then energy will be dissipated. We can clearly see a difference if the video is played forward or backward in the presence of dissipation.

2 Ohm’s Law and Electrical Resistivity

We will start our investigation of dissipative hydrodynamics with thermal conduction. To do so, we will make an analogy to dissipation in direct current (dc) electrical circuit theory, since most students who go on to careers in engineering and science learn Ohm’s law in high school.

Our previous representation of ideal compliances and inertances were analogous to ideal capacitors and inductors; energy could be stored as compressive (elastic) potential energy in a compliance or as kinetic energy in an inertance, but it would not be dissipated, as it could in an electrical circuit where the current flows through an electrical resistance, R_dc. The relationship between the current flowing through the resistance, I, and the voltage difference, ΔV, across the resistance is known as Ohm’s law: I = ΔV/R_dc. The electrical resistance is a (mathematically) real quantity (usually considered to be a constant that is independent of current¹), so that the current and the voltage are in-phase; hence, ϕ = 0.

The time-averaged electrical power, 〈Π_el〉_t, dissipated in the resistor is a positive-definite function of either the current or the voltage.

$$ {\left\langle {\Pi}_{el}\right\rangle}_t=\frac{I\kern0.5em \Delta V}{2}\cos \phi =\frac{I^2{R}_{dc}}{2}=\frac{{\left(\Delta V\right)}^2}{2{R}_{dc}} $$

(9.5)

As we have done with our acoustical variables, ΔV and I are designated by their peak values, not the root-mean-squared values that would be displayed by an ac voltmeter. The phase angle between I (t) and ΔV (t) is ϕ, but since I (t) is in-phase with ΔV (t), cos ϕ = 1. It does not matter whether the current is flowing to the right or to the left; power will be dissipated in a resistor and entropy will be created by an irreversible process. This electrical power dissipation in a resistor is usually called Joule heating, after James Prescott Joule (1818–1889).

Considering a material, shown schematically in Fig. 9.3, having a constant electrical conductivity, σ, with units of [(Ω − m)⁻¹],² the electrical resistance measured across its length, R_dc, will depend directly upon its length, L, and inversely upon its cross-sectional area, A. For that resistor, the current passing through the resistor, I, and the voltage difference, ΔV, across the resistor are related by the direct current (dc) version of Ohm’s law.

$$ I=\frac{\varDelta V}{R_{dc}}=\frac{\sigma A}{L}\left({V}_{in}-{V}_{out}\right) $$

(9.6)

If two identical pieces of the same material, like that shown in Fig. 9.3, were put side-by-side (in parallel), their areas would add, and the resistance would be cut in half. If ΔV = V_in − V_out remains constant, the current would double. If two pieces were placed end-to-end (in series), their lengths would add without changing their areas, and the resistance would be doubled. Again, if ΔV remained constant, the current would be reduced by half, relative to the single sample.

Fig. 9.3

Electrical current, I, flows through a conductor with electrical conductivity, σ, from a higher electrical potential, V_in, to a lower electrical potential, V_out. The electrical resistance, R_dc, of the resistor of length, L, and cross-sectional area, A, is R_dc = L/σA

3 Thermal Conductivity and Newton’s Law of Cooling

Newton’s Law of Cooling has exactly the same form as Ohm’s law, as written in Eq. (9.6), where we recognize I as the rate of electrical charge flow, with units of amperes [A] or coulombs/second [C/s]. By analogy, the rate of heat flow, $\dot{Q}$[J/s], is linearly related to the temperature difference, ΔT, across the length, L, of the sample. It is also proportional to the thermal conductivity of the material, κ, having units of [W/K-m] or [W/°C-m].

$$ \dot{Q}=\frac{\varDelta T}{R_{th}}=\frac{\kappa A}{L}\left({T}_{in}-{T}_{out}\right)=\kappa A\frac{\varDelta T}{L} $$

(9.7)

The SI units of heat flow are watts [W] or joules/second [J/s]. As expressed in the right-hand term of Eq. (9.7), the heat flow (thermal power flow) is proportional to the temperature gradient, ΔT/L (Fig. 9.4).

Fig. 9.4

Heat power, $\dot{Q}$, flows through a thermal conductor made from a material with constant thermal conductivity, κ, from a higher temperature, T_in, to a lower temperature, T_out. The thermal resistance, R_th, of the sample of length, L, and cross-sectional area, A, is R_th = L/κA

As shown in Fig. 9.5, analysis of the heat flow, $\dot{Q}$, through a sample like that in Fig. 9.4, but with length, dx, converts Newton’s Law of Cooling in Eq. (9.7) into the Fourier Diffusion Equation (9.4). The heat flux, ${\dot{q}}_{in}$, with units of heat (energy) per unit area per unit time [W/m²], enters the slab at x, and ${\dot{q}}_{out}$ exits at x + dx.

Fig. 9.5

A different heat flux, ${\dot{q}}_{in}$, flows into a differential “slab” of thickness, dx, and area, A, than flows out, ${\dot{q}}_{out}$, at x + dx. As a result, the heat that remains within the differential element, ${\dot{Q}}_{net}$, causes the temperature of that element to change in accordance with the definition of heat capacity, C_P = (∂Q/∂T)_P, where we assume the sample is held at constant pressure

The thermal power, ${\dot{Q}}_{net}$, that is deposited in the slab results in an increase in the temperature of the slab with time. If we assume that the differential element of length, dx, has cross-sectional area, A, and the material has a specific heat per unit mass, c_p, that is independent of temperature, then the heat capacity of the differential element is the mass of the element, ρA dx, times the specific heat (per unit mass). By Eq. (9.8), this net input of thermal power must result in a change in the temperature of the slab with time.

$$ {\dot{Q}}_{net}=\rho {c}_pA\; dx\frac{dT}{dt}=\dot{Q}(x)-\dot{Q}\left(x+ dx\right)=A\left[\dot{q}(x)-\dot{q}\left(x+ dx\right)\right] $$

(9.8)

Since the heat flows in response to a temperature gradient, as expressed by Newton’s Law of Cooling (9.7), and Eq. (9.8) produces the net heat transport in terms of the (one-dimensional) temperature gradients, ∂T/∂x, evaluated at x and x + dx.

$$ \rho {c}_pA\; dx\frac{dT}{dt}=A\left[-\kappa {\left(\frac{\partial T}{\partial x}\right)}_x+\kappa {\left(\frac{\partial T}{\partial x}\right)}_{x+ dx}\right] $$

(9.9)

At this point, we are only one Taylor series expansion away from Fourier’s Diffusion Equation.

$$ {\displaystyle \begin{array}{l}{\left(\frac{\partial T}{\partial x}\right)}_{x+ dx}\cong {\left(\frac{\partial T}{\partial x}\right)}_x+\frac{\partial }{\partial x}{\left(\frac{\partial T}{\partial x}\right)}_x dx+\dots \kern0.5em \\ {}\Rightarrow \kern0.5em {\left(\frac{\partial T}{\partial x}\right)}_{x+ dx}-{\left(\frac{\partial T}{\partial x}\right)}_x\cong {\left(\frac{\partial^2T}{\partial {x}^2}\right)}_x dx\end{array}} $$

(9.10)

Since dx is a small quantity (compared to what?), we can neglect the terms containing higher powers (dx)ⁿ of dx and combine Eq. (9.9) and Eq. (9.10), while bringing the thermal conductivity, κ, outside the derivative, assuming that it is spatially uniform.

$$ \frac{\partial T}{\partial t}=\frac{\kappa }{\rho {c}_p}\frac{\partial^2T}{\partial {x}^2}=\frac{\kappa }{\rho {c}_p}{\nabla}^2T=\alpha {\nabla}^2T $$

(9.11)

The earlier expression is identical to the Fourier Heat Diffusion Equation that we derived from our linearization of the entropy equation (7.43) under the same assumption of a constant temperature-independent thermal conductivity. The new constant introduced in the right-hand term of (9.11), α = κ/ρc_p, is the thermal diffusivity.³ Energy and temperature units cancel in this ratio, so α has units of length-squared divided by time [m²/s] [2].

The thermal diffusivity is useful because it is a measure of the ability of a material to conduct thermal energy relative to its ability to store thermal energy. Materials with a large α will respond quickly to changes in their thermal environment, while materials with small α will respond more sluggishly, taking longer to reach a new equilibrium condition if the temperature of the surroundings is changed. The thermal diffusivity of most metallic solids and gases (near room temperature) is α ≅ 10⁻⁵ m²/s, while for insulating solids and many liquids, α ≅ 10⁻⁸ to 10⁻⁷ m²/s. Of course, if the fluid is in motion, the convective heat transport (e.g., “wind chill”) can dominate conductive heat transport [3].

3.1 The Thermal Boundary Layer

As acousticians, we are interested in the acoustical solutions to Eq. (9.11) that involve time-harmonic temperature deviations, T(t) = T_s cos (ωt), typically at a single frequency, ω, from some mean temperature, T_m. As claimed before, harmonic analysis is the acoustician’s most powerful mathematical tool. We will begin by applying Eq. (9.11) to a semi-infinite wall defining a plane surface at x = 0 that has an oscillating temperature, T(0, t) = T_m + T_s cos (ωt). By letting T_s be a scalar, we are defining the phase of the temperature response with respect to the oscillating temperature of the boundary. The wall is in contact with a fluid that has a thermal diffusivity, α = κ/ρc_p, as diagrammed in Fig. 9.6.

Fig. 9.6

A semi-infinite solid with an oscillatory surface temperature, T_solid(0, t) = T_m + T_s cos (ωt), is in contact at the plane x = 0 with a fluid at the same mean temperature, T_m. Since there is only fluid at x ≥ 0, we will assume a right-going wavelike space and time dependence for the oscillating component of the fluid’s temperature, ${T}_{fluid}\left(x,t\right)={T}_m+\mathit{\Re e}\left[\hat{\mathbf{T}}{e}^{j\left(\omega\;t-\mathbf{k}x\right)}\right]$

We will assume that the space-time behavior of the fluid at x ≥ 0 is that of a wave traveling to the right, ${T}_{fluid}\left(x,t\right)={T}_m+\mathit{\Re e}\left[\hat{\mathbf{T}}{e}^{j\left(\omega\;t-\mathbf{k}x\right)}\right]$, and substitute this expression into the one-dimensional version of Eq. (9.11) to convert the differential equation into an algebraic equation.

$$ j\omega \hat{\mathbf{T}}=-\alpha {\mathbf{k}}^2\hat{\mathbf{T}} $$

(9.12)

Cancelling the $\hat{\mathbf{T}}$s and dividing both sides by –α, Eq. (9.12) becomes k² =  − jω/α = ω/jα. Taking the square root⁴ of k² (see Sect. 1.5.2), we find that k is a complex number, having real and imaginary parts of equal magnitude.

$$ \mathbf{k}=\sqrt{\frac{\omega }{j\alpha}}=\frac{1-j}{\sqrt{2}}\sqrt{\frac{\omega }{\alpha }}=\left(1-j\right)\sqrt{\frac{\omega }{2\alpha }}\equiv \frac{1-j}{\delta_{\kappa }} $$

(9.13)

It is convenient to define a real scalar physical length, δ_κ, based on the reciprocal wavenumber that is the (exponential) thickness of the oscillatory thermal boundary layer. The scale length of thermal diffusion, δ_κ, is very important in acoustics (see “delta_kappa” in Fig. 8.19) and is called the thermal penetration depth.

$$ {\delta}_{\kappa}\equiv \sqrt{\frac{2\kappa }{\rho {c}_p\omega }}=\sqrt{\frac{2\alpha }{\omega }}=\sqrt{\frac{\alpha }{\pi f}} $$

(9.14)

The complex wave number, k, has equal real and imaginary parts. In electromagnetism, this similar behavior occurs when an electromagnetic wave impinges on an electrically conducting (usually metallic) solid or the surface of an ionic solution (e.g., seawater). In that case, the equivalent to Eq. (9.14) is known as the “skin depth” (see Sect. 9.4.2).

This behavior is different from that which occurs at the interface between two optically transparent media when the angle of incidence exceeds the critical angle⁵ or at the interface between two acoustical media with different specific acoustical impedances (see Sect. 11.2.1). For the case of total internal reflection, the wavenumber within the excluded medium is entirely imaginary and the disturbance is known as an evanescent wave.

The complex wavenumber, k, in Eq. (9.13) still has the units of reciprocal length [m⁻¹]. Substitution of k from (9.13) back into the assumed solution, ${T}_{fluid}\left(x,t\right)={T}_m+\mathit{\Re e}\left[\hat{\mathbf{T}}{e}^{j\left(\omega\;t-\mathbf{k}x\right)}\right]$, shown in Fig. 9.6, and application of the boundary condition at x = 0, provides an explicit expression for the spatial distribution of temperature oscillations within the fluid.

$$ {T}_1\left(x,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{T}}}_s\;{e}^{-x/{\delta}_{\kappa }}{e}^{j\left(\omega\;t-x/{\delta}_{\kappa}\right)}\right]\kern1em \mathrm{for}\kern1em x\ge 0 $$

(9.15)

Expanding Eq. (9.15) in terms of real trigonometric functions, Euler’s formula in Eq. (1.53) will facilitate plotting of the real and imaginary parts of the spatial dependence of T₁ (x), shown in Fig. 9.7.

$$ {T}_1(x)={T}_s{e}^{-x/{\delta}_{\kappa }}\left[\cos \left(\frac{x}{\delta_{\kappa }}\right)-j\sin \left(\frac{x}{\delta_{\kappa }}\right)\right]\kern1em \mathrm{for}\kern1em x\ge 0 $$

(9.16)

It is clear from Fig. 9.7 that Eq. (9.15) and Eq. (9.16) satisfy the boundary condition requiring that the temperatures of the solid and the fluid, at their plane-of-contact, x = 0, are exactly equal, ℜe[T₁(0, t)] = T_s, and in-phase, ℑm[T₁(0, t)] = 0, at all times. If the adjacent temperatures were not equal, then the discontinuity would require Eq. (9.11) to produce infinite heat flows. The temperature disturbance in the fluid is localized quite near the wall. At distances greater than x = 4δ_κ, the effects of the wall’s oscillating temperature correspond to temperature oscillations in the fluid that are less than 2% of those at the interface between the wall and the fluid.

Fig. 9.7

The real (solid) and imaginary (dashed) components of the oscillatory portion of the temperature near a wall that has an oscillating temperature at its surface. The y axis is normalized so that T_s = 1. The x axis is scaled by the thermal penetration depth, δ_κ. The real part of Eq. (9.16) is in-phase with the temperature oscillations at the surface of the solid and is equal to the magnitude of those oscillations, thus satisfying the boundary condition at x = 0: T_fluid(0, t) = T_solid(0, t) at all times. At a distance of 0.7854δ_κ from the wall, the amplitudes of the real and imaginary parts have equal magnitude

The situation encountered more commonly in acoustical systems is that the temperature of the fluid is oscillating while the wall temperature remains constant, typically due to the solid’s higher thermal conductivity and heat capacity per unit volume. For example, in an ideal gas, the amplitude of adiabatic temperature oscillations, T_1,adiab, far from any solid boundaries, were given by Eq. (7.25): ${T}_{1,\kern0.5em adiab}={T}_m\left[\left(\gamma -1\right)/\gamma \right]\left(\left|\hat{\mathbf{p}}\right|/{p}_m\right)$. Typically, a solid boundary will keep an ideal gas’s temperature constant on the solid-gas interface plane, which is defined as x = 0 in Fig. 9.6. In that case, the boundary condition is that T₁ (0, t) = 0.

$$ {T}_1\left(x,t\right)={T}_{1,\kern0.5em adiab}\mathit{\Re e}\left[{e}^{j\omega\;t}\left(1-{e}^{-\left(1+j\right)x/{\delta}_{\kappa }}\right)\right]\kern1em \mathrm{for}\kern1em x\ge 0 $$

(9.17)

The real and imaginary parts of that solution, provided in Eq. (9.18), are plotted in Fig. 9.8.⁶

$$ {T}_1\left(x,t\right)={T}_{1,\kern0.5em adiab}\mathit{\Re e}\left[{e}^{j\omega\;t}\left\{1-{e}^{-x/{\delta}_{\kappa }}\cos \left(\frac{x}{\delta_{\kappa }}\right)+j\left[{e}^{-x/{\delta}_{\kappa }}\sin \left(\frac{x}{\delta_{\kappa }}\right)\right]\right\}\right]\kern0.5em \mathrm{for}\kern0.5em x\ge 0 $$

(9.18)

Fig. 9.8

The real (solid line) and imaginary (dashed line) parts of the oscillatory portion of the temperature near an isothermal surface are plotted for the situation where the fluid far from the wall has a normalized oscillating temperature magnitude, |T₁(x/δ_κ ≫ 1)| = 1, that might be caused by adiabatic expansion and compression of an ideal gas given by Eq. (7.25). The x axis is again scaled by the thermal penetration depth, δ_κ. The isothermal surface at x = 0 requires that both the real and imaginary parts of Eq. (9.18) vanish at the interface: T_s(0, t) = 0. The thermal influence of the wall extends only a distance of about four thermal penetration depths into the fluid

3.2 Adiabatic Compression Within a Bounded Volume

At various places in Chaps. 7 and 8, I have claimed that the acoustic compressions and expansions of an ideal gas take place adiabatically. Having produced an acoustical solution the Fourier Heat Diffusion Equation (9.11), and having defined the thermal penetration depth, δ_κ, in Eq. (9.14), we are now equipped to determine the circumstances that are necessary so that acoustic compressions and expansions occur nearly isothermally in the vicinity of solid surfaces.

Before addressing this question of adiabatic vs. isothermal in a more formal context, it might be useful to examine the effects of thermal conduction in an acoustical compliance, such as the air-filled 500 ml volume of the Helmholtz resonator, shown in the photograph of Fig. 8.16. As shown in Fig. 8.27, the resonance frequency of the empty resonator was (0b) = 241.7 Hz. Figure 9.9 shows the DeltaEC Thermophysical Property output for that case. The value of the thermal penetration depth is visible at the lower right-hand corner: delta_kappa = δ_κ = 168.62 μm.

Fig. 9.9

Screenshot of the thermophysical properties of dry air at T_m = 295.65 = 22.5 °C for the 500 ml boiling flask analyzed in Sect. 8.5.2 and modeled in DeltaEC in Fig. 8.27

The radius of the 500 ml spherical volume, R = (3 V/4π)^1/3 = 4.92 cm. At the interface between the glass and the air, the gas must remain isothermal. The exaggerated boundary layer’s thickness is shown schematically by the dashed spherical surface in Fig. 9.10.

Fig. 9.10

Schematic representation of the spherical volume of radius, R, that forms the acoustical compliance of the Helmholtz resonator depicted in Fig. 8.16. Within a spherical shell of thickness, δ_κ, adjacent to the glass, the pressure oscillations of the gas are nearly isothermal. Throughout the remaining volume, the compressions and expansions of the gas are adiabatic

At a distance of 2δ_κ from the glass, the magnitude of the temperature oscillations of the gas (9.18) is 86.5% of their value far from that boundary, based on Eq. (9.18) as determined by the magnitude of the pressure oscillations within the volume, p_cav, according to Eq. (7.25). We can “split the difference” and approximate the effect of the wall by considering the gas a distance, δ_κ, or less from the glass as being compressed and expanded isothermally and the gas farther than δ_κ from the wall undergoing adiabatic compressions and expansions. The “isothermal region” of the 500 ml sphere has the volume of a spherical shell, V_isothermal = 4πδ_κ R², with the shell radius, R = 4.92 cm, and a thickness, δ_κ = 168.6 μm, for the conditions specified in Fig. 9.9. The ratio of the “isothermal” volume to the “adiabatic” volume can be calculated.

$$ \frac{V_{isothermal}}{V_{adiabatic}}\cong \frac{4\pi {R}^2{\delta}_{\kappa }}{\left(4\pi /3\right){R}^3}=\frac{3{\delta}_{\kappa }}{R}\kern1em \mathrm{if}\kern1em R>>{\delta}_{\kappa } $$

(9.19)

For our case, δ_κ /R = 3.43 × 10⁻³, so the volumetric ratio in Eq. (9.19) is 1.0%. This shows that use of γ in the expression for the compliance of the volume, C = V/γp_m, in Eq. (8.26), was justified. Since γ_air ≅ 7/5 = 1.40, the isothermal compliance (γ_iso ≡ 1) is 40% larger than its adiabatic value, so we would expect the resonance frequency predicted by Eq. (8.51) to be lower by about 0.2%, due to the increased compressibility of the gas within the isothermal boundary layer.

3.3 Energy Loss in the Thermal Boundary Layer*

Heat is transferred from the compressed (hence, hotter) gas to the solid (isothermal) substrate during one-half of the acoustic cycle and is transferred from the solid substrate back to the expanded (hence, cooler) gas during the other half of the acoustic cycle. The entropy lost during these two heat transfers is non-zero even though the amount of heat transferred is nearly identical. The reason is that the change in entropy is related to the ratio of the heat transfer, dQ, to the absolute temperature at which that heat transfer takes place, T. Since the heat transferred from the substrate to the gas occurs at a lower average temperature, approximately T_m − (T₁/2), than the heat transfer from the gas to the substrate, taking place at approximately T_m + (T₁/2), where T₁ is the adiabatic temperature change far from the substrate’s surface, derived previously in Sects. 1.1.3 and 7.1.3, there will be a net increase in the total entropy, (ΔS)_net > 0, during the complete cycle.

Although an exact calculation of this acoustic energy loss per unit surface area per unit time, ${\dot{e}}_{th}$, occurring in the thermal boundary layer requires an integral over a complete cycle, that derivation is somewhat more complicated than the analogous calculation of the energy loss due to viscous shear stresses in the fluid, ${\dot{e}}_{vis}$, provided in Sect. 9.4.3. Instead of the exact calculation, we can use Newton’s Law of Cooling to estimate the heat transferred per unit surface area during each half-cycle assuming a square-wave pressure change, ±p₁, rather than a sinusoidal variation of pressure.

The heat transferred from the gas to the substrate, dQ_+½, per unit area, A, during one-half of an acoustic period, (2f)⁻¹, can be approximated, assuming T₁ is constant during the transfer and ∂T/∂x ≅ T₁/δ_κ. The heat transfer from the substrate to the gas is dQ_-½.

$$ \frac{dQ_{\pm \frac{1}{2}}}{A}\cong \frac{\pm \kappa }{2}\frac{T_1}{f{\delta}_{\kappa }}\propto \frac{\pm \kappa }{2f{\delta}_{\kappa }}\left(\frac{\left|\hat{\mathbf{p}}\right|}{p_m}\right) $$

(9.20)

The entropy change, dS_±½, for each half-cycle depends upon the temperature at which each heat transfer, dQ_±½, takes place.

$$ \frac{dS_{\pm \frac{1}{2}}}{A}\cong \frac{dQ_{\pm \frac{1}{2}}}{A}\left(\frac{1}{T_m\pm \left({T}_1/2\right)}\right)=\frac{\pm \kappa }{2{fT}_m{\delta}_{\kappa }}\left(\frac{T_1}{1\pm \left({T}_1/2{T}_m\right)}\right) $$

(9.21)

The net increase in entropy, (ΔS)_net = dS_+½ − dS_-½, can be approximated by a binomial expansion of the factor in parentheses in Eq. (9.21), since T₁/2T_m ≪ 1.

$$ \frac{{\left(\varDelta S\right)}_{net}}{A}\propto \frac{\kappa }{f{\delta}_{\kappa }}\frac{T_1}{2{T}_m}\left[\left(\frac{1}{1-\left({T}_1/2{T}_m\right)}\right)-\left(\frac{1}{1+\left({T}_1/2{T}_m\right)}\right)\right]\propto {\left(\frac{T_1}{T_m}\right)}^2\propto {\left(\frac{\left|\hat{\mathbf{p}}\right|}{p_m}\right)}^2 $$

(9.22)

The conversion of this net entropy increase into a net energy dissipation per cycle per unit area simply requires multiplication by T_m. It should be clear that the thermal boundary layer’s acoustic energy dissipation per unit area, per unit time, ${\dot{e}}_{th}$, is both positive-definite and proportional to ${\left|\hat{\mathbf{p}}\right|}^2$. An exact calculation [4], assuming the proper time averaging over a sinusoidal modulation of the acoustic pressure, is analogous to the calculation for ${\dot{e}}_{vis}$ that will result in Eq. (9.37).

$$ {\dot{e}}_{th}=-\frac{\left(\gamma -1\right)}{4\gamma}\frac{{\left|\hat{\mathbf{p}}\right|}^2}{p_m}{\delta}_{\kappa}\omega $$

(9.23)

As expected, ${\dot{e}}_{th}$ vanishes for an isothermal process where γ = 1.

3.4 Adiabatic vs. Isothermal Propagation in an Ideal Gas

Having introduced the thermal penetration depth as the real (exponential) length that characterizes the thickness of oscillatory thermal boundary layer, we can use δ_κ to show that sound waves in ideal gases are very nearly adiabatic. To do this, let us start with the assumption that sound propagation is adiabatic and calculate the frequency above which that assumption breaks down.

If we refer to the sinusoidal waves shown in Fig. 3.5, the crests (displacement maxima) correspond to a temperature, T₁, that is higher than T_m by the amount specified in Eq. (7.25): T₁ = T_m[(γ − 1)/γ](p₁/p_m). Similarly, the temperature of the gas at the troughs (pressure minima) is lower than T_m by –T₁. The crests and troughs are separated in space by one-half wavelength, λ/2. Can heat diffuse between the warmer crests and the cooler troughs during a half-cycle? The diffusion distance is given by the thermal penetration depth. At 345 Hz, λ/2 = 0.50 m in air. At the same frequency, δ_κ = 141 μm. During one half-cycle, T/2 = 1.45 ms, heat cannot diffuse far enough to equalize the temperatures of the peaks and troughs of the wave.

The above argument can be made more general by setting the wavenumber, k = ω/c, for the adiabatic sound wave equal to the wavenumber for thermal diffusion, ${\delta}_{\kappa}^{-1}$, and solving for the frequency, ω_crit, at which those two are equal. This is also equivalent to setting the adiabatic sound speed, c_s = (∂p/∂ρ)_s = ω/k, equal to the “speed of heat,” c_th = ωδ_κ.

$$ {\omega}_{crit}=\frac{\rho_m{c}_p{c}^2}{2\kappa } $$

(9.24)

For air at 300 K and standard pressure, the critical frequency, ω_crit = 2.72 × 10⁹ rad/s, or f_crit = ω_crit /2π = 432 MHz. At that frequency, the wavelength of sound is less than one micron, so our assumption of adiabatic sound propagation is quite good for all frequencies of interest in gases.⁷

At f_crit for air, the half-wavelength of sound corresponds to only 11 times the average distance between molecules. At these frequencies, our hydrodynamical approach, which assumes that fluids can be represented as a continuum, is starting to break down and corpuscular effects start to become important.

Of course, at lower pressures, this effect occurs at lower frequencies, so our analysis has to transition from hydrodynamical to ballistic, where the individual particle collisions dominate the propagation.⁸ It has recently been shown that ballistic propagation should be considered in the study of sound propagation high in the Earth’s atmosphere and in the atmospheres other planets in our solar system [5].

4 Viscosity

The irreversibility in the Navier-Stokes equation (7.34) arises from the term proportional to the shear viscosity, $\mu\;{\nabla}^2\overrightarrow{v}$. The fundamental difference between a solid and a fluid (liquid or gas) is that shear deformations of a solid are restored elastically (see Sect. 4.2.3), while fluids cannot sustain a state of static shear indefinitely. Figure 9.11 depicts two parallel plates separated by a distance, d, that are in relative motion at a velocity, v_x, and contain a viscous fluid in between.

Fig. 9.11

Two-dimensional representation of a fluid that is being sheared by the relative motion of two parallel plates. The lower plate is shown as stationary and the upper plate is moving in the x direction at a uniform speed, v_x. The outlined arrow (⇒) represents the constant force, F_x, that is required to overcome the fluid’s frictional drag. That force is applied in the x direction, on the plate with area, A_y, normal to the y axis. Due to the non-slip boundary condition at the interface between the viscous fluid and the plates, the velocity of the fluid is the same as that of the plates over the planes where the substrates and fluid are in contact

A “no-slip” boundary condition is applied to the viscous fluid at the interface between the fluid and the solid surface. The force, F_x, per unit area, A_y, of the plate, is given by the equivalent of Ohm’s law for the relevant component of the shear stress, τ_xy, as a function of the velocity gradient.

$$ {\tau}_{xy}=\frac{F_x}{A_y}=\mu \frac{\partial {v}_x}{\partial y} $$

(9.25)

We can take Eq. (9.25) to be our definition of the shear viscosity, μ, which has the units of [Pa-s].

This differs slightly from Ohm’s law (9.6) and Newton’s Law of Cooling (9.7) because the electrical potential difference (voltage) and the temperature difference are both real scalars. Velocity is a vector, so the shear stress, $\overleftrightarrow{\boldsymbol{\uptau}}$, is a tensor. If the shear viscosity, μ, is not a function of the shear rate, then fluids that obey Eq. (9.25) are known as Newtonian fluids.⁹

The shear viscosity, μ, is also sometimes called the dynamic viscosity or absolute viscosity to distinguish it from the kinematic viscosity, ν ≡ μ/ρ. We will see in the next sub-section that the kinematic viscosity plays the same role for viscous flow as the thermal diffusivity, α, plays for heat transfer. Both ν and α are diffusion constants and have the SI units of [m²/s].

4.1 Poiseuille Flow in a Pipe of Circular Cross-Section

When one plate moves parallel to a stationary plate, as shown in Fig. 9.11, the velocity of the fluid varies linearly across the gap (as symbolized by the arrows of different length) and the y component of the gradient of the velocity, ∇_yv_x = ∂v_x/∂y, is a constant. We can calculate the velocity profile for steady-state flow (i.e., ∂v_x /∂t = 0) in a tube of circular cross-section, with radius, a, by solving the Navier-Stokes equation (7.34). Due to the cylindrical symmetry of the problem, the x component of the Navier-Stokes equation for a Newtonian fluid can be expressed in cylindrical coordinates [6]:¹⁰

$$ {\displaystyle \begin{array}{c}\rho \left(\frac{\partial {v}_x}{\partial t}+{v}_r\frac{\partial {v}_x}{\partial r}+\frac{v_{\theta }}{r}\frac{\partial {v}_x}{\partial \theta }+{v}_x\frac{\partial {v}_x}{\partial x}\right)=\\ {}-\frac{\partial p}{\partial x}+\mu \left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial {v}_x}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2{v}_x}{\partial {\theta}^2}+\frac{\partial^2{v}_x}{\partial {x}^2}\right]+\rho {g}_x\end{array}} $$

(9.26)

Although Eq. (9.26) looks rather intimidating,¹¹ in most applications many of the terms vanish. In the case of steady-state flow in a pipe, only two terms from Eq. (9.26) survive in Eq. (9.27).

$$ \frac{\mu }{r}\frac{\partial }{\partial r}\left(r\frac{\partial {v}_x}{\partial r}\right)=\frac{\partial p}{\partial x}=\frac{\varDelta p}{L} $$

(9.27)

The last term on the right-hand side of Eq. (9.27) assumes that the flow is in response to a pressure difference, Δp, in a pipe of length, L. The equation can be integrated twice to produce an expression for the velocity as a function of radius, v_x(r).

$$ {v}_x(r)=\frac{\varDelta p}{4\mu L}{r}^2+{C}_1\ln \left[r\right]+{C}_2 $$

(9.28)

C₁ and C₂ are constants of integration. Since v_x (0) must remain finite, C₁ = 0. (For flow in the annular space between two pipes, C₁ becomes useful for matching the boundary condition at the inner radius.¹²) The constant C₂ is determined from the non-slip boundary condition that requires v_x (a) = 0.

$$ {v}_x(r)=\frac{\varDelta p}{4\mu L}\left({a}^2-{r}^2\right) $$

(9.29)

The pipe’s “discharge” (i.e., mass flow, $\dot{m}$, or volume velocity, $U=\dot{m}/\rho$) can be calculated by integrating the parabolic velocity profile of Eq. (9.29), shown in Fig. 9.12, over the pipe’s cross-sectional area.

$$ \dot{m}=\rho U=2\pi \rho {\int}_0^a{v}_x(r)r\kern0.5em dr=\frac{\pi \rho {a}^4}{8\mu}\frac{\varDelta p}{L} $$

(9.30)

This result, known as Poiseuille’s formula, is valid as long as the parabolic profile has been established¹³ and the flow velocity in the pipe is slow enough that the flow remains laminar (like Fig. 9.12) and does not become turbulent.¹⁴

Fig. 9.12

Schematic representation of the velocity vectors for steady flow through a pipe of cylindrical cross-section with radius, a. The velocity profile, given in Eq. (9.29), is parabolic. This flow behavior is known as “Poiseuille flow.” (Jean Léonard Marie Poiseuille (1797–1869) was a French physicist and physiologist who first published this result, in 1840, during his investigation of blood flow in narrow capillaries. The CGS unit of viscosity, the poise, was named in his honor.) When looking at this two-dimensional representation, it is important to keep its cylindrical symmetry in mind. Imagine the entire sketch rotated about the dashed centerline

4.2 The Viscous Boundary Layer

Just as we used a surface with a time-dependent temperature in contact with a stagnant fluid to derive an expression for the thermal penetration depth, δ_κ, in Eq. (9.14), we can assume transverse oscillations of a solid surface, as shown in Fig. 9.13, to derive an expression for the oscillatory viscous boundary layer.

Fig. 9.13

A semi-infinite solid surface oscillates in the y direction with a transverse oscillatory velocity, v_s cos (ωt). The surface is in contact with a fluid along the x = 0 plane. At that interface, the fluid moves with the same velocity as the solid due to the non-slip boundary condition at x = 0. Since there is only fluid at x ≥ 0, we will assume a right-going wavelike space and time dependence for the oscillating component of the fluid velocity, v_y(x, t), as we did for temperature oscillations in Fig. 9.6

We will assume that the space-time behavior of the fluid for x ≥ 0 is that of a wave traveling to the right, ${v}_y\left(x,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{v}}}_{\mathbf{y}}\kern0.5em {e}^{j\left(\omega\;t-k\;x\right)}\right]$. We assume that the fluid is otherwise at rest so there is no mean flow: ${\overrightarrow{v}}_m=0$. Since the transverse motion of the wall does not compress any fluid, the pressure is constant throughout the fluid, allowing us to ignore the $\overrightarrow{\nabla}p$ term in Eq. (9.2).

$$ \rho \frac{\partial {v}_y\left(x,t\right)}{\partial t}=\mu \kern0.1em {\nabla}^2{v}_y\left(x,t\right) $$

(9.31)

We can substitute the wavelike expression into the two remaining terms in this one-dimensional version of the linearized Navier-Stokes equation to again obtain the required relationship between ω and k:

$$ j\omega {\hat{\mathbf{v}}}_{\mathbf{y}}=-\frac{\mu }{\rho }{\mathbf{k}}^{\mathbf{2}}{\hat{\mathbf{v}}}_{\mathbf{y}}=-\nu {\mathbf{k}}^{\mathbf{2}}{\hat{\mathbf{v}}}_{\mathbf{y}} $$

(9.32)

We have again utilized ν, called the kinematic viscosity, which has the units of [m²/s], just like the thermal diffusivity, α, which was introduced in Eq. (9.11). As before, we have solved Eq. (9.31) for the complex wavenumber, k, and introduce a viscous penetration depth, δ_ν, to characterize the shear wave within the fluid.

$$ \mathbf{k}=\sqrt{\frac{- j\omega}{\nu }}=\frac{1-j}{\sqrt{2}}\sqrt{\frac{\omega }{\nu }}=\left(1-j\right)\sqrt{\frac{\omega }{2\nu }}\equiv \frac{1-j}{\delta_{\nu }}\kern1em \Rightarrow \kern1em {\delta}_{\nu }=\sqrt{\frac{2\mu }{\rho \kern0.1em \omega }}=\sqrt{\frac{2\nu }{\kern0.1em \omega }} $$

(9.33)

From this point on, the results for v_y (x,t) in the viscous fluid are identical to those already presented for T₁ (x,t).¹⁵ Figure 9.7 would describe the velocity field in front of the transversely oscillating boundary of Fig. 9.13 if the x axis were scaled by δ_ν instead of by δ_κ. Similarly, if the boundary were stationary and the fluid far from the boundary were moving with an oscillatory velocity amplitude, v_y, then Fig. 9.8 would describe the motion of the fluid if the x axis were scaled by δ_ν instead of δ_κ.¹⁶

The solutions would be identical for Maxwell’s equations governing the electric field, $\overrightarrow{E}$, due to an electromagnetic wave impinging on an electrically conducting medium.

$$ \frac{\partial \overrightarrow{E}}{\partial t}=\frac{1}{\sigma \mu}{\nabla}^2\overrightarrow{E} $$

(9.34)

In that case, δ = (2/σμω)^1/2 is known as the electromagnetic “skin depth.” Here μ is the magnetic permeability (not viscosity), and σ is the same electrical conductivity as introduced earlier in Ohm’s law (9.6).

4.3 Viscous Drag in the Neck of a Helmholtz Resonator

We now return once again to the example of our 500 ml flask in Fig. 8.16 and calculate the effects of viscosity on the flow through the neck of our Helmholtz resonator. From Fig. 9.9, we see that the viscous penetration depth for that example is δ_ν = 141.9 μm. Since the radius of the neck, a = D_neck/2 = 12.5 mm, the ratio, δ_ν/a = 0.0114 << 1, so the effect of the non-slip boundary condition at the stationary neck surface does not extend very far into the air that fills the neck. In that limit, we describe the motion of the fluid in the neck as plug flow that is diagrammed schematically in Fig. 9.14.

Fig. 9.14

Schematic representation of oscillatory “plug flow” in the cylindrical neck of a Helmholtz resonator with radius, a ≫ δ_ν. In the region far from the walls, the velocity is independent of the radial distance from the central axis of the neck. The velocity decays exponentially over a characteristic exponential distance, δ_ν, to zero at the walls where it must vanish to satisfy the non-slip boundary condition for a viscous fluid. When looking at this drawing, it is important to keep the cylindrical symmetry in mind. Imagine the entire sketch rotated about the dashed centerline

The relevant component of the shear stress on the surface of the neck, τ_xy, is determined by Eq. (9.25). In this case, the fluid near the neck’s axis is moving with speed, v_x, in the axial direction, and the fluid in contact with the neck must be stationary to satisfy the non-slip boundary condition. Since |k| ∝ |1/δ_ν|, it is easy to evaluate the velocity gradient, ∂v_x/∂y ∝ v_x/δ_ν, where v_x is the peak fluid velocity in the neck far from surface of the neck and y is the radial direction normal to the surface of the neck.

$$ {\tau}_{xy}=\frac{F_x}{A_y}=\mu \frac{\partial {v}_x\left(y,t\right)}{\partial y}=\mu \frac{\left|{\hat{\mathbf{v}}}_{\mathbf{x}}\right|}{\delta_{\nu }}\propto \mu \frac{v_x}{\delta_{\nu }} $$

(9.35)

In this application, A_y is the surface area on the inside of the neck. The wall and the fluid are in relative motion since the wall is stationery and the fluid is oscillating. Since the physical neck length, L = 49.2 mm, and the diameter, D_neck = 25.0 mm, A_y = πD_neckL = 3.864 x 10⁻³ m². The time-averaged power dissipated, 〈Π_vis〉_t, per unit area by the viscous shear stress, ${\dot{e}}_{vis}={\left\langle {\Pi}_{vis}\right\rangle}_t/{A}_y$, is one-half the time average of the product of the peak stress, τ_xy, times the peak velocity, v_x, where T = f⁻¹ is the period.

$$ {\dot{e}}_{vis}=\frac{{\left\langle {\Pi}_{vis}\right\rangle}_t}{A_y}=\frac{1}{T}{\int}_0^T\left[{\tau}_{xy}\cdotp {v}_x(t)\right]\; dt\propto \frac{\mu }{T}{\int}_0^T\frac{v_x^2(t)}{\delta_{\nu }}\; dt $$

(9.36)

The oscillatory flow velocity, v_x (r, t), through the neck of the resonator is uniform (except in the thin viscous boundary layer). The sinusoidal time dependence of the velocity can be represented as v_x(t) = v_x cos (ω t).

$$ {\dot{e}}_{vis}=\frac{{\left\langle {\Pi}_{vis}\right\rangle}_t}{A_y}\propto \frac{\mu }{2}\frac{v_x^2}{\delta_{\nu }}=\frac{\rho_m}{4}{\left|\frac{\hat{\mathbf{U}}}{A_y}\right|}^2{\delta}_{\nu}\omega $$

(9.37)

The far right expression for ${\dot{e}}_{vis}$ is obtained by using the definition of the viscous penetration depth in Eq. (9.33) to substitute for the shear viscosity, $\mu =\left({\delta}_{\nu}^2{\rho}_m\omega \right)/2$, and by letting <v_x> = |<U₁>|/A. As with any dissipative mechanism, like Joule heating in Eq. (9.5), ${\dot{e}}_{vis}$ is positive-definite and independent of the sign (direction) of v_x or U.

Although the result for ${\dot{e}}_{vis}$ in Eq. (9.37) is correct, the use of proportionalities instead of equalities was motivated by the fact that the actual expression for the viscous stress tensor is not the one provided in Eq. (9.35) but a complex phasor, ${\hat{\boldsymbol{\uptau}}}_{\mathbf{xy}}$, that lags $\hat{\mathbf{v}}$ by 45° in time.¹⁷

$$ \left|{\hat{\boldsymbol{\uptau}}}_{\mathbf{xy}}\right|=\frac{\sqrt{2}\mu {v}_x}{\delta_{\nu }}{e}^{-y/{\delta}_{\nu }} $$

(9.38)

By examining the DeltaEC model of the 500 ml flask shown in Fig. 8.27, we can calculate the average value of <v_x> = |<U₁>|/A based on input (2a), the cross-sectional area of the neck, A = 4.909 × 10⁻⁴ m². The volume velocity entering the neck is given in Segment #0, where (0f) gives |U₁| = 4.0777 × 10⁻⁴ m³/s at the open end of the neck. The volume velocity leaving the neck and entering the compliance is given in Segment #1, where result (1C) gives |U₁| = 3.9806 × 10⁻⁴ m³/s for an average volume velocity |<U₁>| = 4.0292 × 10⁻⁴ m³/s, so <v_x> = |<U₁>|/A = 0.8208 m/s. Using the viscosity from Fig. 9.9, μ = 1.835 × 10⁻⁵ Pa−s, and δ_ν = 141.9 μm, Eq. (9.37) yields an average viscous power dissipation per unit area of ${\dot{e}}_{vis}={\left\langle {\Pi}_{vis}\right\rangle}_t/A$ = 42.0 mW/m².

The surface area of the neck A_y = π D_neck L = 3.86 × 10⁻³ m², so the time-averaged power dissipated in the neck due to viscous drag in the oscillatory boundary layer <Π_vis>_t = ${\dot{e}}_{vis}\left(\pi {D}_{neck}L\right)$ = 167.4 μW. This is quite close to the value in the DeltaEC model that provides for the dissipation in the neck (1E)–(1F) = 174 μW. The small discrepancy (about 3.7%) arises from the fact that we took an average of the flow velocity that was not weighted (i.e., integrated) to accommodate the v_x² dependence of ${\dot{e}}_{vis}$ in Eq. (9.37). DeltaEC is correct. Using the same results from the DeltaEC model of Fig. 8.27, to calculate the average of the squared velocity, $\left\langle {v}_x^2\right\rangle =$0.6818 m²/s², so <Π_vis>_t = ${\dot{e}}_{vis}\left(\pi {D}_{neck}L\right)$ = 170.2 μW. This result is even closer to the correct DeltaEC result. We have avoided evaluation of any integrals, but in the process have developed a fundamental understanding of dissipation in the viscous boundary layer.

Although we have not derived the corresponding power dissipation per unit area due to thermal conduction, ${\dot{e}}_{th}$, the strategy was presented in Sect. 9.3.3 and is illustrated in the THERMAL.EXE animation.¹⁸

$$ \dot{e}={\dot{e}}_{vis}+{\dot{e}}_{th}=-\frac{\rho_m}{4}{\left|\frac{\hat{\mathbf{U}}}{A}\right|}^2{\delta}_{\nu}\omega -\frac{\left(\gamma -1\right)}{4\gamma}\frac{{\left|\hat{\mathbf{p}}\right|}^2}{p_m}{\delta}_{\kappa}\omega $$

(9.39)

The second term in Eq. (9.39) can be applied to the 500 ml Helmholtz resonator of Fig. 8.16 to calculate the power dissipated by the irreversible thermal conduction between the (isothermal) walls of the compliance and the gas undergoing adiabatic expansions and compressions far from the walls.

Using Fig. 8.27, the DeltaEC result (2A) = (3A) gives |p_cav| = 74.355 Pa. Figure 9.9 provides the air’s thermal conductivity, κ = 2.59 × 10⁻² W/K-m, and the thermal penetration depth, δ_κ = 168.6 μm, resulting in ${\dot{e}}_{th}$ = 9.98 mW/m². The surface area of the spherical compliance is input (2a): A_sphere = 2.9974 × 10⁻² m². The product gives the time-averaged power dissipated in on the surface of the spherical volume due to thermal relaxation loss in the oscillatory boundary layer ${\left\langle {\Pi}_{th}\right\rangle}_t={\dot{e}}_{th}{A}_{sphere}$ = 29.91 μW. This is exactly the value that the DeltaEC model provides as (1F)–(2F) = 29.92 μW.

4.4 Quality Factors for a Helmholtz Resonator

Having expressions for thermoviscous boundary layer dissipation, we can calculate the quality factor, Q, of the Helmholtz resonator shown in Fig. 8.16. Q is a dimensionless measure of the sharpness of the resonance (see Appendix B). Without dissipation, Q = ∞, as seen in Eq. (8.52). Based on the DeltaEC model in Fig. 8.28 and Eq. (C.1), the Q = |p_cav/p| = result (3A)/input (0d) = 74.355.

Figure 8.32 displays the incremental plot file, ∗.ip, for the same resonator providing that amplitude and phase of p_cav (f). We can approximate the derivative of the phase with respect to frequency at resonance by fitting the three phases (the one closest to resonance and the two above and below it) vs. frequency for an approximate value of ${\left(\partial \theta /\partial f\right)}_{f_o}\cong$ 34.6°/Hz. Substitution into Eq. (B.4) gives Q ≅ 73.6. Running DeltaEC over a finer frequency step size around f_o should produce a slightly greater value for the slope of the phase vs. frequency.

It will be instructive to calculate the Q for both the viscous and thermal losses individually since the final results will be simple and rather intuitively satisfying. To begin, recall the definition of Q based on the ratio of the energy stored in the system, E_stored, to the energy dissipated in one cycle, E_{dissipated/cycle} = 〈Π_dis〉_tT = 〈Π_dis〉_t/f. From Eq. (B.2),

$$ Q=2\pi \frac{E_{stored}}{E_{dissipated/ cycle}}=\frac{\omega {E}_{stored}}{{\left\langle {\Pi}_{dissipated}\right\rangle}_t} $$

(9.40)

For our Helmholtz resonator, the quality factors for each dissipative process will be calculated individually. Since the energy dissipation is additive, the total quality factor, Q_tot, will be the reciprocal of the sum of the reciprocals of the quality factors for the viscous dissipation, Q_vis, for thermal conduction, Q_th, and for radiation of sound, Q_rad (actually, an “accounting loss” as discussed in Sect. 3.7 for a string).

$$ \frac{1}{Q_{tot}}=\frac{1}{Q_{vis}}+\frac{1}{Q_{th}}+\frac{1}{Q_{rad}} $$

(9.41)

Like any single-degree-of-freedom simple harmonic oscillator, the stored energy oscillates between its kinetic and potential forms with the sum being constant at steady state. As in Eq. (2.18), the maximum kinetic energy, (KE)_max, can be used to represent the value of the stored energy. For the Helmholtz resonator, the kinetic energy is determined by the velocity of the fluid in the resonator’s neck, with cross-sectional area, A = πa². Because the viscous penetration depth is much less that the radius of the neck, δ_ν ≪ a, we will assume “plug flow,” as shown schematically in Fig. 9.14, and approximate the moving mass of the gas in the neck of our Helmholtz resonator to be all of the mass, m, of the gas within the neck, as was done in Sect. 8.4.4.

$$ m={\rho}_m{AL}_{neck}={\rho}_m\left(\pi {a}^2\right){L}_{neck} $$

(9.42)

Just as the thin viscous boundary layer was neglected in the calculation of the moving mass, the gas velocity, v₁ = |U₁|/A_neck, will be assumed to be independent of the radial distance from the neck’s center line (i.e., plug flow).

$$ {E}_{stored}={(KE)}_{\mathrm{max}}=\frac{1}{2}{mv}_1^2=\frac{\rho_m{A}_{neck}{L}_{neck}}{2}{\left|\frac{\hat{\mathbf{U}}}{\pi {a}^2}\right|}^2 $$

(9.43)

Since our results will be compared to the DeltaEC model of the 500 ml flask in Fig. 8.27, we will neglect the kinetic energy of the gas beyond the ends of the neck that produced the effective length correction necessary to match the theoretically calculated and experimentally determined resonance frequencies.

To calculate the energy dissipated by viscous shear on the surface of the neck of the resonator having surface area, A_neck = 2πaL_neck, and cross-sectional area, πa², the power dissipation per unit area, ${\dot{e}}_{vis}$, given in Eq. (9.37), will be used.

$$ {E}_{dis/ cycle}=\frac{{\dot{e}}_{vis}{A}_{neck}}{f_o}=\frac{\rho_m}{4{f}_o}{\left|\frac{\hat{\mathbf{U}}}{\pi {a}^2}\right|}^2{\delta}_{\nu }{\omega}_o{A}_{neck}=\frac{\pi }{2}{\left|\frac{\hat{\mathbf{U}}}{\pi {a}^2}\right|}^2{\rho}_m{\delta}_{\nu }2\pi {aL}_{neck} $$

(9.44)

When we take the ratio of Eq. (9.43) to Eq. (9.44) to calculate Q, using Eq. (9.40), the amplitude of the oscillation squared, proportional to |U₁/A_neck|², will cancel. This must be the case, since we are considering a linear system and linearity demands that the quality factor must be amplitude independent.

$$ {Q}_{vis}=2\pi \frac{E_{stored}}{E_{dis/ cycle}}=2\pi \frac{\left({\rho}_m\pi {a}^2{L}_{neck}\right)/2}{\left(\pi /2\right){\delta}_{\nu }2\pi {aL}_{neck}}=\frac{2\pi {a}^2{L}_{neck}}{\delta_{\nu }2\pi {aL}_{neck}}=\frac{a}{\delta_{\nu }} $$

(9.45)

This is a wonderfully simple and intuitively satisfying result: Q_vis = a/δ_ν. The quality factor due to viscous dissipation at the surface of the neck of a Helmholtz resonator is simply the dimensionless ratio of the neck’s radius to the viscous penetration depth in the gas.

A numerical calculation for Q_vis in a cylindrical duct involves the integration of the flow field using a J_o Bessel function with a complex argument, (j−1)r/δ_ν, to represent the velocity, resulting in an approximation that is valid to within 0.3% for a/δ_ν > 3 [7].

$$ {Q}_{vis}\cong \frac{a}{\delta_{\nu }}+\frac{\delta_{\nu }}{5a}-\frac{3}{4}\kern1em \mathrm{for}\kern1em \frac{a}{\delta_{\nu }}\ge 3 $$

(9.46)

The same approach can be used to calculate the quality factor for thermal dissipation on the surface of the Helmholtz resonator’s volume (i.e., compliance). Since the amplitude of the pressure oscillations within the volume is proportional to p₁, as shown in Eq. (9.39), it will be convenient to convert |U₁/πa²| in Eq. (9.43) to a pressure amplitude, since the quality factor for thermal losses, Q_th, must also be amplitude independent in the linear acoustics limit. From Eq. (8.25), the adiabatic compliance of a volume, V, can be used to relate pressure amplitude,$\left|\hat{\mathbf{p}}\right|$, to the volume velocity amplitude,$\left|\hat{\mathbf{U}}\right|$.

$$ \hat{\mathbf{p}}=\frac{1}{j\omega}\frac{\gamma {p}_m}{V}\hat{\mathbf{U}}\kern1em \Rightarrow \kern1em {E}_{stored}={(KE)}_{\mathrm{max}}=\frac{\rho_m{L}_{neck}}{2{A}_{neck}}\frac{\omega_o^2{V}^2}{\gamma^2}\frac{{\left|\hat{\mathbf{p}}\right|}^2}{p_m^2} $$

(9.47)

Substitution of ${\dot{e}}_{th}$ from Eq. (9.39), times the surface area of the spherical compliance, S_Vol = 4πR², into Eq. (9.40) produces the expression for the thermal contribution to the quality factor, Q_th.

$$ {Q}_{th}=\frac{2\pi \frac{\rho_m{L}_{neck}}{2{A}_{neck}}\frac{\omega_o^2{V}^2}{\gamma^2}\frac{{\left|\hat{\mathbf{p}}\right|}^2}{p_m^2}}{\frac{\left(\gamma -1\right)}{4\gamma {f}_o}\frac{{\left|\hat{\mathbf{p}}\right|}^2}{p_m}{S}_{Vol}{\delta}_{\kappa }{\omega}_o}=\left[{\omega}_o^2\frac{\rho_m}{\gamma {p}_m}\frac{L_{neck}V}{A_{neck}}\right]\frac{2\left(4\pi /3\right){R}^3}{\left(\gamma -1\right){\delta}_{\kappa }4\pi {R}^2} $$

(9.48)

The Helmholtz resonance frequency, ω_o, is given in Eq. (8.51), and c² = γp_m /ρ_m, so the term in Eq. (9.48) contained within the square brackets is just unity, proving again that “substitution is the most powerful technique in mathematics” (see Sect. 1.1).

$$ {Q}_{th}=\frac{2}{3\left(\gamma -1\right)}\frac{R}{\delta_{\kappa }} $$

(9.49)

As expected, the thermal quality factor becomes infinite (i.e., lossless) if the expansions and compressions of the gas within the entire volume of the Helmholtz resonator are isothermal, γ_iso = 1. We also see again that the quality factor is proportional to the ratio of a length characterizing the size of the spherical volume, R, and the thermal penetration depth, δ_κ.

The results for Q_vis in Eq. (9.45) and Q_th in Eq. (9.49) can be used to estimate the quality factor for the 500 ml boiling flask example that was used to create the DeltaEC file in Fig. 8.27, using the penetration depths at f_o = 241.73 Hz, provided in Fig. 9.9, when combined as shown in Eq. (9.41). This results in a Q_vis = 88, and a Q_th = 482, for Q_tot = 74.4. The DeltaEC model also gives Q = 74.4.

Although the time-averaged radiated power, 〈Π_rad〉_t, will not be derived until Chap. 12 (see Eq. 12.18), it was mentioned in Footnote 24 in Chap. 8.

$$ {\left\langle {\Pi}_{rad}\right\rangle}_t=\pi \frac{\rho_m{f}^2}{2c}{\left|\hat{\mathbf{U}}\right|}^2 $$

(9.50)

The dependence upon |U₁|² means that it is easy to calculate a contribution to the quality factor from the radiation “loss” using Eq. (9.43) to express the stored energy.

$$ {Q}_{rad}=2\pi \frac{\frac{\rho_m\pi {a}^2{L}_{neck}}{2}{\left|\frac{\hat{\mathbf{U}}}{\pi {a}^2}\right|}^2}{\pi \frac{\rho_m{f}_o}{2c}{\left|\hat{\mathbf{U}}\right|}^2}=2\frac{\left(\lambda {L}_{neck}\right)}{\pi {a}^2} $$

(9.51)

The expression on the far right in Eq. (9.51) uses c = λf to express the quality factor as proportional to the ratio of two areas. For the 500 ml flask DeltaEC model in Fig. 8.27, where λ = 1.426 m at resonance, Q_rad = 286. This is more than three times greater than Q_vis, reinforcing the earlier result from the DeltaEC model that showed that viscous loss in the neck was the dominant loss mechanism in that example.

5 Kinetic Theory of Thermal and Viscous Transport

Thus far, in this chapter, we have taken a phenomenological approach to describe the dissipation caused by thermal conduction and viscous shear. Those mechanisms have been characterized by diffusion equations that introduced two phenomenological constants: the thermal conductivity, κ, and the shear viscosity, μ. Those constants are properties of the fluid that might depend upon pressure and/or temperature. By introducing a simple microscopic model for these diffusive processes, we can gain additional insight into the relationship between these constants and the microscopic properties of ideal gases.

The same microscopic model that was used in Chap. 7 to derive the Ideal Gas Law and to introduce the mean squared particle velocity and the Equipartition Theorem in Eq. (7.2) can be resurrected to calculate κ and μ for an ideal gas by adding the concept of a mean free path, $\overline{\mathrm{\ell}}$, characterizing an average distance that an atom or molecule in a gas will travel before colliding with another of its own kind.¹⁹

For the following calculations, we assume that the gas is “dilute” [8]. This means that we will assume that each particle spends a relatively large fraction of its time at distances far from other particles so that the time between collisions is much greater than the time involved in a collision. We also assume that the probability of a simultaneous collision between more than two particles can be neglected. Finally, we assume de Broglie wavelength, λ_B = h/mv, for the particles with momentum, mv, is much shorter than the separation between particles so quantum-mechanical effects can be ignored: ${\lambda}_B\ll \overline{\mathrm{\ell}}$.

5.1 Mean Free Path

If we assume that our “point particles” have a mass, m, and an effective hard sphere diameter, D, then two identical particles in the gas will collide if their centers are separated by a distance that is less than or equal to D. We focus our attention on a single particle that is moving with a mean velocity, ${\overline{v}}^{\hbox{'}}$, determined by the Equipartition Theorem, as applied in Eq. (7.26), ${\left\langle {v}^2\right\rangle}^{\frac{1}{2}}\equiv {\overline{v}}^{\hbox{'}}=\sqrt{3{k}_BT/m}$. This mean thermal velocity is designated ${\overline{v}}^{\hbox{'}}$ to remind us that we have assumed that all other particles are not moving. Assuming (temporarily) that all other particles are stationary, then the moving particle sweeps out a cylindrical volume, ${V}_{Swept}=\pi {D}^2{\overline{v}}^{\hbox{'}}t$, in a time, t.

The number of (stationary) particles within that swept volume depends upon the (number) density of the gas, n = ρ/m, where m is the mass of an individual particle. During the time interval, t, there will be W = nV_Swept stationary gas particles within that cylinder. The collision rate, $\dot{n}=\left(W/t\right)=\pi {D}^2{\overline{v}}^{\hbox{'}}n$. The mean free path, $\overline{\mathrm{\ell}}$, is the distance the particle travels between collisions that take place, on average, every $\overline{\tau}={\dot{n}}^{-1}$ seconds.

Of course, all the other particles are not stationary but are moving with the same mean (thermal) velocity, ${\overline{v}}^{\hbox{'}}$, so the actual velocity that we need in our expression for the mean free path is the mean relative particle velocity, ${v}_{rel}\equiv \overline{v}$. If two particles are traveling in exactly the same direction, their mean relative velocity would be zero. If they are moving directly toward each other, their mean relative velocity would be $2{\overline{v}}^{\hbox{'}}$. Integration over all possible directions could determine the average relative velocity, but it is easier to utilized the “reduced mass,” μ = m₁m₂/(m₁ + m₂), to determine mean relative velocity. Placing the reduced mass into the Equipartition Theorem produces the mean relative velocity in the same way the reduced mass was used to determine the antisymmetric frequency of two otherwise free particles that were joined by a spring, as discussed in Sect. 4.3.1, then applied to the Tonpilz transducer.

Since the particles are assumed to be identical, μ = m/2, so the value of the relative velocities of the particles,${\overline{v}}_{rel}$, is given by the Equipartition Theorem.

$$ {\left\langle {v}_{rel}^2\right\rangle}^{\frac{1}{2}}\equiv \overline{v}={\left(3{k}_BT/\mu \right)}^{\frac{1}{2}}={\left(6{k}_BT/m\right)}^{\frac{1}{2}}={\overline{v}}^{\hbox{'}}\sqrt{2} $$

(9.52)

That result can be used to calculate the mean free path as introduced when we initially assumed that the other particles were stationary.

$$ \overline{\mathrm{\ell}}=\frac{\overline{v}}{\pi {D}^2{\overline{v}}^{\hbox{'}}n}=\frac{1}{\pi \sqrt{2}{D}^2n} $$

(9.53)

It is useful to recognize that the mean free path is independent of $\overline{v}$ or ${\overline{v}}^{\hbox{'}}$ and therefore independent of temperature. At higher temperatures, the particles are moving faster, but V_Swept is correspondingly larger.

Near room temperature, for air at atmospheric pressure, the number density can be calculated from the Ideal Gas Law or from the molar volume and Avogadro’s number: n = ρ/m = p_m/k_BT ≅ 2.5 × 10²⁵ particles/m³ ≅ N_A/(22.4 × 10⁻³ m³). For air, a typical molecular diameter, D_air ≅ 2 × 10⁻¹⁰ m = 2 Å, so πD² ≅ 1.3 × 10⁻¹⁹ m², making $\overline{\mathrm{\ell}}$ ≅ 2.2 × 10⁻⁷ m = 0.22 microns. Using the same values, the time between collisions is $\overline{\mathrm{\ell}}/\overline{v}\cong$ 6 × 10⁻¹⁰ seconds, or the collision rate, $\dot{n}$, is 2 × 10⁹/second = 2 GHz. Each molecule of air at atmospheric pressure and room temperature experiences about two billion collisions each second.

5.2 Thermal Conductivity of an Ideal Gas

With our understanding of the mean free path, we are now able to determine a value of the thermal conductivity of an ideal gas from the microscopic model. If we assume a linear temperature gradient in a gas, then the thermal energy that is transported by a gas particle will be based on the temperature that particle had at the position where it suffered its last collision. Figure 9.15 provides a suitable geometry for such a calculation by assuming that the temperature gradient, ∂T/∂z ≠ 0, exists in the z direction.

Fig. 9.15

Coordinate system for the calculation of the thermal conductivity, κ, of an ideal gas. The heat flux, represented by the arrows, is determined by number of particles crossing the line per unit time and the temperature of the particles that was determined by the last collision they suffered before crossing the line

The number of particles crossing a unit area per unit time (i.e., the particle flux) in the z direction is determined by the particle density, n, and the mean particle velocity, ${\overline{v}}^{\hbox{'}}$. If there are n particles per unit volume, roughly one-third of them have velocities in the z direction and half of those, or n/6 particles per unit volume, have mean velocities in the (−z) direction.

Particles which cross the dashed line in Fig. 9.15 from below have, on average, experienced the last collision at a distance, $\overline{\mathrm{\ell}}$, below that plane. But the mean energy per particle, $\overline{\varepsilon}$, is a function of T and, since T = T(z), the mean energy is also a function of z, $\overline{\varepsilon}(z)$. The particles crossing from below carry with them a mean energy, $\overline{\varepsilon}\left(z-\overline{\mathrm{\ell}}\right)$, and the ones above experienced the last collision at a distance, $\overline{\mathrm{\ell}}$, above that plane, $\overline{\varepsilon}\left(z+\overline{\mathrm{\ell}}\right)$. Each “free particle,” which has three degrees of freedom, carries an average kinetic energy of $\overline{\varepsilon}=3{k}_BT/2$, corresponding to a heat capacity of 3k_B/2 per particle (see Sect. 7.2.1).

$$ {\left(\frac{\partial Q}{\partial t}\right)}_{above}=\frac{1}{6}n{\overline{v}}^{\hbox{'}}\overline{\varepsilon}\left(z+\overline{\mathrm{\ell}}\right)=\left(\frac{n{\overline{v}}^{\hbox{'}}}{6}\right)\left[\frac{3{k}_B}{2}\left(T+\overline{\mathrm{\ell}}\frac{\partial T}{\partial z}\right)\right] $$

(9.54)

Similarly, the heat flux going in the opposite direction from below, ${\dot{q}}_{below}$, is determined by the last collision that took place at a slightly lower temperature.

$$ {\left(\frac{\partial Q}{\partial t}\right)}_{below}=\frac{1}{6}n{\overline{v}}^{\hbox{'}}\overline{\varepsilon}\left(z-\overline{\mathrm{\ell}}\right)=\left(\frac{n{\overline{v}}^{\hbox{'}}}{6}\right)\left[\frac{3{k}_B}{2}\left(T-\overline{\mathrm{\ell}}\frac{\partial T}{\partial z}\right)\right] $$

(9.55)

The net energy flux, ${\dot{q}}_{net}$, from hot to cold, is given by the difference of the fluxes calculated in Eqs. (9.54) and (9.55).

$$ {\left(\frac{\partial Q}{\partial t}\right)}_{net}={\left(\frac{\partial Q}{\partial t}\right)}_{below}-{\left(\frac{\partial Q}{\partial t}\right)}_{above}=-\left(\frac{n{\overline{v}}^{\hbox{'}}\overline{\mathrm{\ell}}{k}_B}{2}\right)\frac{\partial T}{\partial z} $$

(9.56)

As indicated in Fig. 9.15, the thermal conductivity of a gas, κ_gas, can be expressed in terms of the result in Eq. (9.56).

$$ \frac{\partial Q}{\partial t}=-\kappa \frac{\partial T}{\partial z}\kern1em \Rightarrow \kern1em {\kappa}_{gas}=\frac{1}{2}n{\overline{v}}^{\hbox{'}}\overline{\mathrm{\ell}}{k}_B $$

(9.57)

A more rigorous kinetic theory for thermal conductivity of inert gases at low pressures was developed by Chapman in England and Enskog in Sweden. Their approach calculates the numerical pre-factor in Eq. (9.57) to be 0.37, rather than 0.5, since the particles that are not going straight up or down along the z direction experienced their last collision at a distance less than$\overline{\mathrm{\ell}}$ [9]. Their calculations also describe the thermal conductivity of polyatomic molecules quite accurately.

Using the Equipartition Theorem to let ${\left\langle {v}^2\right\rangle}^{\frac{1}{2}}\equiv {\overline{v}}^{\hbox{'}}=\sqrt{3{k}_BT/m}$ provides a microscopic expression for the thermal conductivity of an ideal gas, κ_gas.

$$ {\kappa}_{gas}=0.37\frac{nk_B{\overline{v}}^{\hbox{'}}}{\pi \sqrt{2}{D}^2n}=0.37\frac{k_B}{2\pi {D}^2}\sqrt{\frac{3{k}_BT}{m}}\cong \frac{0.1}{D^2}\frac{k_B^{3/2}}{m^{\frac{1}{2}}}\sqrt{T} $$

(9.58)

Our application of a microscopic model, based on the collision of hard spheres, demonstrates that the thermal conductivity of an ideal gas is independent of the pressure or density of the gas (at least as long as the dimensions of the system are much larger than the mean free path) for most gases at pressures below 1 MPa and is inversely proportional to the square root of the particle’s mass and inversely proportional to the particle’s cross-sectional area.

The model also predicts that the thermal conductivity will be proportional to the square root of the absolute temperature, $\sqrt{T}$. For real gases, the observed temperature dependence of the thermal conductivity of ideal gases is closer to T^0.7 [10], due to the fact that the gas particles are not “hard spheres” but interact through a molecular force field like the Lennard-Jones potential shown in Fig. 2.39.

5.3 Viscosity of an Ideal Gas

A microscopic determination of the viscosity can be obtained in exactly the same way using the momentum transported by a particle of mass, m, across a plane. We can use an approach similar to that expressed in the geometry of Fig. 9.15. As before, we assume that the density of the gas is sufficiently low that the mean free path is much greater than the particle diameter but much less than the typical dimensions of the system (e.g., the spacing between the moving plates in Fig. 9.11 or the diameter of the tube in Figs. 9.12 and 9.14).

The mean shear velocity, v_x, is again determined by the last collision the particle suffered prior to crossing the plane from above or from below, and the momentum in the x direction transported by each particle is mv_x. The particle density can be used to calculate the momentum transported per unit area, per unit time, P.

$$ {P}_{above}=\frac{n\overline{v}m}{6}\left[{v}_x+\frac{\partial {v}_x}{\partial z}\overline{\mathrm{\ell}}\right]\kern1em \mathrm{and}\kern1em {P}_{below}=\frac{n\overline{v}m}{6}\left[{v}_x-\frac{\partial {v}_x}{\partial z}\overline{\mathrm{\ell}}\right] $$

(9.59)

The net momentum change per unit area is the difference of the two expressions in Eq. (9.59).

$$ {P}_{above}-{P}_{below}={\tau}_{xz}=\frac{n{\overline{v}}^{\hbox{'}}m\overline{\mathrm{\ell}}}{3}\left(\frac{\partial {v}_x}{\partial z}\right) $$

(9.60)

Comparison of Eq. (9.60) to the phenomenological expression for one component of shear stress, τ_xy, in Eq. (9.25), produces a microscopic expression for the shear viscosity of an ideal gas, μ_gas.

$$ {\mu}_{gas}=\frac{n{\overline{v}}^{\hbox{'}}m\overline{\mathrm{\ell}}}{3}=\frac{\rho {\overline{v}}^{\hbox{'}}\overline{\mathrm{\ell}}}{3}=\frac{1}{\pi {D}^2}{\left(\frac{mk_BT}{6}\right)}^{\frac{1}{2}}=\frac{0.13}{D^2}{k}_B^{\frac{1}{2}}{m}^{\frac{1}{2}}{T}^{\frac{1}{2}} $$

(9.61)

Just like the thermal conductivity of an ideal gas, this estimate of the shear viscosity is independent of the pressure or density and proportional to the square root of the absolute temperature and to the square root of the particle mass. Like the ideal gas thermal conductivity, the actual temperature dependence of the viscosity for real gases is also closer to T^0.7 [10].

5.4 Prandtl Number of an Ideal Gas and Binary Gas Mixtures*

The relative importance of thermal conductivity and viscosity can be expressed in a dimensionless ratio that is known as the Prandtl number, Pr = μc_P/κ, where c_P is the specific heat (per unit mass) at constant pressure. This ratio is particularly important for convective heat transfer where the viscosity determines the energy dissipation from imposed flow and the thermal conductivity determines the heat transport. Liquid metals, like mercury or sodium-potassium eutectic (NaK), have a very small Prandtl number because of their low viscosity (hence, the popular designation as “quicksilver,” in the case of mercury) and high thermal conductivity, due to the efficient heat transport provided by the conduction electrons. On the other hand, viscous fluids with low thermal conductivity, like molasses, have large Prandtl numbers.

For monatomic ideal gases, the isobaric heat capacity per particle (see Sect. 7.2.1) is (5k_B/2) so the isobaric specific heat per particle is (5k_B/2m). In combination with the ideal gas viscosity in Eq. (9.61) and thermal conductivity in Eq. (9.58), it is easy to demonstrate that the Prandtl number for ideal gases should be a constant.

$$ {\Pr}_{gas}\equiv \frac{\mu_{gas}{c}_P}{\kappa_{gas}}=\frac{\delta_{\nu}^2}{\delta_{\kappa}^2}=\frac{0.13}{0.37}\frac{5{k}_B}{2m}\frac{D^2}{D^2}\frac{k_B^{\frac{1}{2}}{m}^{\frac{1}{2}}{T}^{\frac{1}{2}}}{k_B^{3/2}{T}^{\frac{1}{2}}/{m}^{\frac{1}{2}}}\cong 0.9 $$

(9.62)

Measured values for the Prandtl number of monoatomic gases are not 0.9 but closer to $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$. For other polyatomic gases, the Prandtl numbers are also constant and range from about 0.7 to 0.9 [11]. This is due to the fact that it is those “hard spheres” that are responsible for both the heat and the momentum transport.

A similar result, known as the Wiedemann-Franz Law [12], shows that the ratio of the thermal conductivity, κ, and the electrical conductivity, σ, of metallic solids is a constant, independent of the particular metal, that depends only upon temperature.

$$ L\equiv \frac{\kappa }{\sigma T}=\frac{\pi^2}{3}{\left(\frac{k_B}{e}\right)}^2 $$

(9.63)

L is the Lorenz number and depends only upon fundamental constants (the charge on the electron, e, and Boltzmann’s constant, k_B). Again, this is due to the fact that it is the electrons in metals that provide both heat transport and electrical conductivity.

For ideal gases, a more rigorous calculation was made by Eucken [13] that related the thermal conductivity of gases to their viscosity.

$$ {\kappa}_{gas}=\left({c}_P+\frac{5\Re }{4M}\right){\mu}_{gas}\kern1em \Rightarrow \kern1em {\Pr}_{gas}=\frac{c_P}{c_P+1.25\left(\Re /M\right)} $$

(9.64)

Eucken’s formula gives the correct result for monatomic (noble) gases using the universal gas constant, ℜ ≡ 8.314462 J/mol-K, and is a very good approximation for most polyatomic gases.

Although the Prandtl number for ideal gases is a constant, it is possible to reduce the Prandtl number for gas mixtures. In a mixture of a light and heavy gas, the light species can dominate the heat transfer while the heavier species dominates the momentum transfer. Swift has demonstrated this by a simple calculation for inert gas mixtures based on the results from the microscopic model, using the subscripts “1” and “2” to designate two different gases.

$$ {\displaystyle \begin{array}{c}{k}_BT\kern0.5em \propto \kern1em \left(\frac{1}{2}\right){m}_1{\overline{v}}_1^2\propto \kern1em \left(\frac{1}{2}\right){m}_2{\overline{v}}_2^2\\ {}{\mu}_{mix}\propto \kern1em \left({n}_1{\overline{v}}_1{m}_1+{n}_2{\overline{v}}_2{m}_2\right)\overline{\mathrm{\ell}}\kern1em \propto \kern1em {n}_1\sqrt{m_1}+{n}_2\sqrt{m_2}\\ {}{\kappa}_{mix}\propto \kern1em \left({n}_1{\overline{v}}_1+{n}_2{\overline{v}}_2\right){c}_P\overline{\mathrm{\ell}}\kern1em \propto \kern1em {n}_1/\sqrt{m_1}+{n}_2/\sqrt{m_2}\\ {}{c}_P\propto \frac{\Re }{n_1{m}_1+{n}_2{m}_2}\end{array}} $$

(9.65)

These proportions can be used to form the ratio of a mixture’s Prandtl number, Pr_mix, to the pure gas Prandtl number, Pr_gas, in terms of the molar concentration of species “1,” x₁ = n₁/(n₁ + n₂), and to the square root of their mass ratio, $\beta =\sqrt{m_1/{m}_2}$.

$$ \frac{\Pr_{mix}}{\Pr_{gas}}=\frac{n_1\sqrt{m_1}+{n}_2\sqrt{m_2}}{\left(\frac{n_1}{\sqrt{m_1}}+\frac{n_2}{\sqrt{m_2}}\right)\left({n}_1{m}_1+{n}_2{m}_2\right)}=\frac{1-\left(1-\beta \right){x}_1}{\left[1+\left(\frac{1}{\beta }-1\right){x}_1\right]\left[1-\left(1-{\beta}^2\right){x}_1\right]} $$

(9.66)

Figure 9.16 plots the gas mixture Prandtl number as a function of the mole fraction of helium, x₁, for helium-argon mixtures with β_He/Ar = 0.316 and helium-xenon mixtures with β_He/Xe = 0.175. For pure gases (i.e., x₁ = 0% or x₁ = 100%), Pr_mix = 2/3, but a mixture of a light and a heavy gas can produce lower Prandtl numbers. The results in Fig. 9.16 demonstrate that the Prandtl number of a helium-argon mixture can be reduced to a minimum value of Pr_He/Ar = 0.37 and a helium-xenon mixture can produce a minimum Prandtl number as low as Pr_He/Xe = 0.20. The lowest inert gas Prandtl number would be achieved with a mixture of ³He and radon, Pr_He/Rn = 0.14, although such a mixture would be highly radioactive. A more detailed kinetic theory calculation produces nearly identical results [14].

Fig. 9.16

The Prandtl number of inert gas mixtures as a function of the mole fraction of helium. The dashed line represents helium-argon mixtures and the solid line represents helium-xenon mixtures

The use of gas mixtures that reduce Prandtl number have been shown to improve the performance of thermoacoustic engines and refrigerators [15], although it has also been shown that such high-amplitude sound waves can also introduce concentration gradients in mixtures [16].

6 Not a Total Loss

Just as we had examined the lossless equations of hydrodynamics in Chap. 8 by applying them to simple “lumped elements,” this chapter has introduced the diffusion equations that govern dissipation in fluidic acoustical systems (thermal conduction also contributes to dissipation in solids) by applying them to a Helmholtz resonator that had been modeled by DeltaEC in the previous chapter. In doing so, two new length scales were introduced: the thermal and viscous penetration depths, δ_κ and δ_ν. Those length scales play the same critical role that wavelength and wavenumber played in the nondissipative equations. Comparison of the thermal penetration depth to the wavelength provided quantitative justification for the assumption we will exploit in the following chapter where sound waves are treated as being adiabatic oscillatory excursions away from equilibrium.

As in Chap. 7, the microscopic models and the phenomenological models of dissipative processes in ideal gases provided complementary insights into the behavior of the phenomenological constants, κ and μ, with respect to changes in density and temperature, as well as supplying an intuitive picture of the processes by which heat and momentum are transported. The simple kinetic theory, based on a “hard sphere” collision assumption, introduced an additional important length scale: the mean free path, $\overline{\mathrm{\ell}}$. When $\overline{\mathrm{\ell}}$ ≪ λ, our continuum model of diffusive processes provides an appropriate description.

With the thermodynamic, hydrodynamic, and microscopic analyses introduced in this chapter and in Chap. 7, the fundamentals necessary to understand wave propagation in fluids, and particularly in ideal gases, will be put to use in the remainder of this textbook.

Talk Like an Acoustician

Time reversal invariance	Ballistic propagation
Shear viscosity	Newtonian fluids
Dynamic (or absolute) viscosity	Diffusion constant
Fourier Diffusion Equation	Non-slip boundary condition
Ohm’s law	Shear (or dynamic or absolute) viscosity
Joule heating	Kinematic viscosity
Newton’s Law of Cooling	Poiseuille’s formula
Thermal diffusivity	Oscillatory plug flow
Thermometric conductivity	Viscous penetration depth
Thermal penetration depth	Entrance length
Skin depth	Thermal velocity
Instantaneous value	Mean free path
Acoustic approximation	Prandtl number

Exercises

• 1. Moon free path. Atmospheric pressure on the moon is about 10⁻¹³ atm. = 10⁻⁸ Pa. Assume that the moon’s atmosphere has the same chemical composition as Earth’s and calculate the mean free path for molecules in our moon’s atmosphere. Is the concept of pressure meaningful for such a long mean free path? Why or why not?

• 2. Vacuum insulation. As long as the characteristic dimensions of a gas-filled space are much greater than the mean free path of the gas particles, the thermal conductivity is independent of gas pressure. To reduce thermal conduction through a gas, the vacuum space gap, g, must be less than the mean free path, $\overline{\mathrm{\ell}}$, of the gas molecules trapped in the vacuum insulation space.

  The inverse relationship between number density, n, and mean free path, $\overline{\mathrm{\ell}}$, illustrated in Eq. (9.53), is no longer valid when $\overline{\mathrm{\ell}}$ becomes larger than g. Beyond that point, the thermal resistance of the insulation space will increase linearly in proportion to increase of the mean free path because there are fewer gas particles to transport heat and those particles are more likely to collide with the walls than they are to collide with each other.

  How low must the gas pressure in the insulation space be so that the thermal resistance of the vacuum space is ten times smaller than the thermal resistance when g ≪$\overline{\mathrm{\ell}}$if g = 10 mm, assuming that the vacuum space contains some air?

• 3. Relaxation frequency for a capacitive microphone. The volume inside a condenser microphone that is behind the diaphragm (see Fig. 6.14) must be isolated from the acoustical pressure variations which drive the motion of the microphone’s diaphragm. Because the microphone diaphragm must be able to withstand slow changes in pressure encountered during transportation (e.g., shipping by air), a capillary tube is provided so that changes in ambient gas pressure, p_m, can be relieved, but acoustical pressure variations at the frequencies of interest are not allowed to influence the gas pressure within the microphone’s back volume.

  • (a) Poiseuille resistance. Use Eq. (9.30) to write an expression for the acoustic flow resistance, R_ac = Δp/U, appropriate to steady gas flow in the Poiseuille regime.

  • (b) Relaxation time. Using the result from part (a), write an expression for the exponential relaxation time, τ = R_acC, where the compliance, C, given in Eq. (8.26), is determined by the microphone’s back volume, V_back, assuming that the compressions and expansions of the gas within that volume are adiabatic. Discuss whether or not the adiabatic assumption is valid.

  • (c) Viscous penetration depth. For audio applications, the low-frequency cut-off is usually taken to be 20 Hz, corresponding to the nominal lower limit of human hearing [17]. Determine the viscous penetration depth at that frequency in air at 20 °C with p_m = 100 kPa.

  • (d) 0.010″ diameter capillary. If the back volume of the microphone is V_back = 674 mm³ and it is connected to a capillary tube that has an inside diameter of 250 microns, and a length of 10.0 mm, determine the exponential relaxation time, τ, for pressure equilibration and corresponding cut-off frequency, f_-3dB = (2πτ)⁻¹.

• 4. Viscous damping of an oscillating spar buoy. The spar buoy shown in Fig. 2.33 has a diameter of 25 cm. The bottom 3.0 m of the buoy is submerged. Its vertical oscillations have a period of 3.6 seconds. The buoy’s effective moving mass, m_o = 154 kg.

  • (a) Viscous penetration depth. If the water has a density, ${\rho}_{H_2O}$ = 1026 kg/m³, and a shear viscosity, ${\mu}_{H_2O}$ = 1.07 × 10⁻³ Pa-s, how large is the viscous penetration depth at the buoy’s natural frequency of vertical oscillations?

  • (b) Viscously entrained mass. What is the additional mass of water trapped in the viscous boundary layer if we assume that mass is the mass of water within one viscous penetration depth? (The circular bottom of the buoy can be ignored since it is not applying any shear stresses on the water.)

  • (c) Viscous damping. Determine the mechanical resistance, R_m, by calculating the viscous drag of the water on the buoy. Use your value of R_m to determine the exponential relaxation time, τ = 2m_o/ R_m, if the only source of damping is the viscous drag provided by the surrounding water.

• 5. Greenspan viscometer. Shown in Fig. 9.17 is a schematic cross-section of a cylindrically symmetric (about the “Duct” axis) double-Helmholtz resonator that has been used by the National Institute of Standards and Technology (formerly, the Bureau of Standards) to make an acoustical determination of the viscosity of gases [18]. Both volumes, V, are identical and A is the total surface area of each volume. “S” and “D” are PZT stacks that act as the excitation and detection transducers that are covered by a “Diaphragm” which can be considered perfectly rigid.

  For this problem, assume that the viscometer is filled with neon gas at a mean pressure, p_m = 1.0 MPa. Both compliances have the same volume, V = 29 cm³, and the same surface area, A = 55 cm². The duct length, L_d = 3.1 cm, and the radius of the duct, r_d = 2.3 mm. This viscometer produced measurements of viscosity that differed from published results obtained by other methods by less than ±0.5% and sound speeds that differed by less than ±0.2% [18].

  • (a) Helmholtz resonance frequency. Assuming that the neon gas is lossless, calculate the resonance frequency of this dual Helmholtz resonator using a sound speed in neon of c_Ne (20 °C) = 449 m/s.

  • (b) Gas displacement in the duct. If the amplitude of the acoustic pressure in either volume is 10.0 Pa, what is the peak-to-peak displacement of the gas in the duct?

  • (c) Viscous penetration depth. Based on the resonance frequency, calculate the viscous penetration depth if the neon has a density, ρ_Ne (20 °C) = 8.28 kg/m³, and a viscosity, μ_Ne (20 °C) = 3.13 × 10⁻⁵ Pa-s.

  • (d) Quality factor. Calculate the quality factor, Q, for this Helmholtz resonance.

  • (e) Viscosity. In the limit that δ_ν ≪ r_d, express the gas viscosity, μ, in terms of the Helmholtz resonance frequency, f_o, the gas density, ρ, and the resonance quality factor, Q.

  • (f) Thermal relaxation. Investigate how negligible the thermal boundary layer losses are in the compliances.

Fig. 9.17

Cross-sectional schematic representation of a Greenspan viscometer [19]