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.”
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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 solidfluid 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 “arrowoftime” 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 selforganizes 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.
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).
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 NavierStokes equation (7.34), even in its linearized form that discards the convective term, \( \left(\overrightarrow{v}\bullet \nabla \right)\overrightarrow{v} \), and ignores gravity.
Since both the velocity and the time change sign, the derivative, \( \partial \overrightarrow{v}/\partial t \), does not change sign. On the righthand 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 NavierStokes 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] = [m^{2}/s^{2} − K].
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).
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 current^{Footnote 1}), so that the current and the voltage are inphase; hence, ϕ = 0.
The timeaveraged electrical power, 〈Π_{el}〉_{t}, dissipated in the resistor is a positivedefinite function of either the current or the voltage.
As we have done with our acoustical variables, ΔV and I are designated by their peak values, not the rootmeansquared 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 inphase 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}],^{Footnote 2} the electrical resistance measured across its length, R_{dc}, will depend directly upon its length, L, and inversely upon its crosssectional 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.
If two identical pieces of the same material, like that shown in Fig. 9.3, were put sidebyside (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 endtoend (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.
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/Km] or [W/°Cm].
The SI units of heat flow are watts [W] or joules/second [J/s]. As expressed in the righthand term of Eq. (9.7), the heat flow (thermal power flow) is proportional to the temperature gradient, ΔT/L (Fig. 9.4).
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/m^{2}], enters the slab at x, and \( {\dot{q}}_{out} \) exits at x + dx.
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 crosssectional 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.
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 (onedimensional) temperature gradients, ∂T/∂x, evaluated at x and x + dx.
At this point, we are only one Taylor series expansion away from Fourier’s Diffusion Equation.
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.
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 temperatureindependent thermal conductivity. The new constant introduced in the righthand term of (9.11), α = κ/ρc_{p}, is the thermal diffusivity.^{Footnote 3} Energy and temperature units cancel in this ratio, so α has units of lengthsquared divided by time [m^{2}/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} m^{2}/s, while for insulating solids and many liquids, α ≅ 10^{−8} to 10^{−7} m^{2}/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 timeharmonic 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 semiinfinite 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.
We will assume that the spacetime 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 onedimensional version of Eq. (9.11) to convert the differential equation into an algebraic equation.
Cancelling the \( \hat{\mathbf{T}} \)s and dividing both sides by –α, Eq. (9.12) becomes k^{2} = − jω/α = ω/jα. Taking the square root^{Footnote 4} of k^{2} (see Sect. 1.5.2), we find that k is a complex number, having real and imaginary parts of equal magnitude.
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.
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 angle^{Footnote 5} 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.
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 T_{1} (x), shown in Fig. 9.7.
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 planeofcontact, x = 0, are exactly equal, ℜe[T_{1}(0, t)] = T_{s}, and inphase, ℑm[T_{1}(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.
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 solidgas interface plane, which is defined as x = 0 in Fig. 9.6. In that case, the boundary condition is that T_{1} (0, t) = 0.
The real and imaginary parts of that solution, provided in Eq. (9.18), are plotted in Fig. 9.8.^{Footnote 6}
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 airfilled 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 righthand corner: delta_kappa = δ_{κ} = 168.62 μm.
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.
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} R^{2}, 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.
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 onehalf 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 nonzero 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} − (T_{1}/2), than the heat transfer from the gas to the substrate, taking place at approximately T_{m} + (T_{1}/2), where T_{1} 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 halfcycle assuming a squarewave pressure change, ±p_{1}, rather than a sinusoidal variation of pressure.
The heat transferred from the gas to the substrate, dQ_{+½}, per unit area, A, during onehalf of an acoustic period, (2f)^{−1}, can be approximated, assuming T_{1} is constant during the transfer and ∂T/∂x ≅ T_{1}/δ_{κ}. The heat transfer from the substrate to the gas is dQ_{½}.
The entropy change, dS_{±½}, for each halfcycle depends upon the temperature at which each heat transfer, dQ_{±½}, takes place.
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 T_{1}/2T_{m} ≪ 1.
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 positivedefinite 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).
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, T_{1}, that is higher than T_{m} by the amount specified in Eq. (7.25): T_{1} = T_{m}[(γ − 1)/γ](p_{1}/p_{m}). Similarly, the temperature of the gas at the troughs (pressure minima) is lower than T_{m} by –T_{1}. The crests and troughs are separated in space by onehalf wavelength, λ/2. Can heat diffuse between the warmer crests and the cooler troughs during a halfcycle? 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 halfcycle, 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}.
For air at 300 K and standard pressure, the critical frequency, ω_{crit} = 2.72 × 10^{9} 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.^{Footnote 7}
At f_{crit} for air, the halfwavelength 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.^{Footnote 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 NavierStokes 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.
A “noslip” 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.
We can take Eq. (9.25) to be our definition of the shear viscosity, μ, which has the units of [Pas].
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.^{Footnote 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 subsection 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 [m^{2}/s].
4.1 Poiseuille Flow in a Pipe of Circular CrossSection
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, ∇_{y}v_{x} = ∂v_{x}/∂y, is a constant. We can calculate the velocity profile for steadystate flow (i.e., ∂v_{x} /∂t = 0) in a tube of circular crosssection, with radius, a, by solving the NavierStokes equation (7.34). Due to the cylindrical symmetry of the problem, the x component of the NavierStokes equation for a Newtonian fluid can be expressed in cylindrical coordinates [6]:^{Footnote 10}
Although Eq. (9.26) looks rather intimidating,^{Footnote 11} in most applications many of the terms vanish. In the case of steadystate flow in a pipe, only two terms from Eq. (9.26) survive in Eq. (9.27).
The last term on the righthand 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).
C_{1} and C_{2} are constants of integration. Since v_{x} (0) must remain finite, C_{1} = 0. (For flow in the annular space between two pipes, C_{1} becomes useful for matching the boundary condition at the inner radius.^{Footnote 12}) The constant C_{2} is determined from the nonslip boundary condition that requires v_{x} (a) = 0.
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 crosssectional area.
This result, known as Poiseuille’s formula, is valid as long as the parabolic profile has been established^{Footnote 13} and the flow velocity in the pipe is slow enough that the flow remains laminar (like Fig. 9.12) and does not become turbulent.^{Footnote 14}
4.2 The Viscous Boundary Layer
Just as we used a surface with a timedependent 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.
We will assume that the spacetime behavior of the fluid for x ≥ 0 is that of a wave traveling to the right, \( {v}_y\left(x,t\right)=\mathit{\Re e}\left[{\hat{\mathbf{v}}}_{\mathbf{y}}\kern0.5em {e}^{j\left(\omega\;tk\;x\right)}\right] \). 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).
We can substitute the wavelike expression into the two remaining terms in this onedimensional version of the linearized NavierStokes equation to again obtain the required relationship between ω and k:
We have again utilized ν, called the kinematic viscosity, which has the units of [m^{2}/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.
From this point on, the results for v_{y} (x,t) in the viscous fluid are identical to those already presented for T_{1} (x,t).^{Footnote 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}.^{Footnote 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.
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 nonslip 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.
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 nonslip 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.
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_{neck}L = 3.864 x 10^{−3} m^{2}. The timeaveraged 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 onehalf 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.
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).
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}> = <U_{1}>/A. As with any dissipative mechanism, like Joule heating in Eq. (9.5), \( {\dot{e}}_{vis} \) is positivedefinite 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.^{Footnote 17}
By examining the DeltaEC model of the 500 ml flask shown in Fig. 8.27, we can calculate the average value of <v_{x}> = <U_{1}>/A based on input (2a), the crosssectional area of the neck, A = 4.909 × 10^{−4} m^{2}. The volume velocity entering the neck is given in Segment #0, where (0f) gives U_{1} = 4.0777 × 10^{−4} m^{3}/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 U_{1} = 3.9806 × 10^{−4} m^{3}/s for an average volume velocity <U_{1}> = 4.0292 × 10^{−4} m^{3}/s, so <v_{x}> = <U_{1}>/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/m^{2}.
The surface area of the neck A_{y} = π D_{neck} L = 3.86 × 10^{−3} m^{2}, so the timeaveraged 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_{x}^{2} 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 m^{2}/s^{2}, 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.^{Footnote 18}
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/Km, and the thermal penetration depth, δ_{κ} = 168.6 μm, resulting in \( {\dot{e}}_{th} \) = 9.98 mW/m^{2}. The surface area of the spherical compliance is input (2a): A_{sphere} = 2.9974 × 10^{−2} m^{2}. The product gives the timeaveraged 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}〉_{t}T = 〈Π_{dis}〉_{t}/f. From Eq. (B.2),
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).
Like any singledegreeoffreedom 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 crosssectional area, A = πa^{2}. 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.
Just as the thin viscous boundary layer was neglected in the calculation of the moving mass, the gas velocity, v_{1} = U_{1}/A_{neck}, will be assumed to be independent of the radial distance from the neck’s center line (i.e., plug flow).
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 crosssectional area, πa^{2}, the power dissipation per unit area, \( {\dot{e}}_{vis} \), given in Eq. (9.37), will be used.
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 U_{1}/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.
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].
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 p_{1}, as shown in Eq. (9.39), it will be convenient to convert U_{1}/πa^{2} 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,\( \left\hat{\mathbf{U}}\right \).
Substitution of \( {\dot{e}}_{th} \) from Eq. (9.39), times the surface area of the spherical compliance, S_{Vol} = 4πR^{2}, into Eq. (9.40) produces the expression for the thermal contribution to the quality factor, Q_{th}.
The Helmholtz resonance frequency, ω_{o}, is given in Eq. (8.51), and c^{2} = 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).
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 timeaveraged 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.
The dependence upon U_{1}^{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.
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.^{Footnote 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 quantummechanical 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 = m_{1}m_{2}/(m_{1} + m_{2}), 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.
That result can be used to calculate the mean free path as introduced when we initially assumed that the other particles were stationary.
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_{B}T ≅ 2.5 × 10^{25} particles/m^{3} ≅ N_{A}/(22.4 × 10^{−3} m^{3}). For air, a typical molecular diameter, D_{air} ≅ 2 × 10^{−10} m = 2 Å, so πD^{2} ≅ 1.3 × 10^{−19} m^{2}, 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 × 10^{9}/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.
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 onethird 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).
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.
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).
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).
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 prefactor 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}.
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 crosssectional 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 LennardJones 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.
The net momentum change per unit area is the difference of the two expressions in Eq. (9.59).
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}.
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 sodiumpotassium 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.
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 WiedemannFranz 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 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.
Eucken’s formula gives the correct result for monatomic (noble) gases using the universal gas constant, ℜ ≡ 8.314462 J/molK, 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.
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,” x_{1} = n_{1}/(n_{1} + n_{2}), and to the square root of their mass ratio, \( \beta =\sqrt{m_1/{m}_2} \).
Figure 9.16 plots the gas mixture Prandtl number as a function of the mole fraction of helium, x_{1}, for heliumargon mixtures with β_{He/Ar} = 0.316 and heliumxenon mixtures with β_{He/Xe} = 0.175. For pure gases (i.e., x_{1} = 0% or x_{1} = 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 heliumargon mixture can be reduced to a minimum value of Pr_{He/Ar} = 0.37 and a heliumxenon 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 ^{3}He 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].
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 highamplitude 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  Nonslip 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 gasfilled 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_{ac}C, 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 lowfrequency cutoff 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 mm^{3} 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 cutoff frequency, f_{3dB} = (2πτ)^{−1}.

(a)

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/m^{3}, and a shear viscosity, \( {\mu}_{H_2O} \) = 1.07 × 10^{−3} Pas, 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.

(a)

5.
Greenspan viscometer. Shown in Fig. 9.17 is a schematic crosssection of a cylindrically symmetric (about the “Duct” axis) doubleHelmholtz 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 cm^{3}, and the same surface area, A = 55 cm^{2}. 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 peaktopeak 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/m^{3}, and a viscosity, μ_{Ne} (20 °C) = 3.13 × 10^{−5} Pas.

(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.

(a)
Notes
 1.
William Hewlett, an electrical engineering student at Stanford University, used a currentdependent resistance (a flashlight bulb) to stabilize the amplitude of an audio oscillator circuit as his master’s thesis. That oscillator subsequently became the HP200. It was the first product developed commercially by the original Silicon Valley startup: HewlettPackard. The first five of their “production” models were purchased by Walt Disney to produce sound effects in the animated feature film, Fantasia.
 2.
Before the Frenchdominated Le Système International d’Unités was adopted, the unit of conductance was the mho (ohm spelled backward). They renamed the unit of electrical conductance the siemens [S]. These are the same folks, who, in their collective wisdom, renamed the unit of frequency from the obscure cyclespersecond (cps) to the hertz [Hz].
 3.
This material parameter is also sometimes called the thermometric conductivity and abbreviated as χ. For instance see, L. D. Landau and E. M. Lifshitz, Fluid Mechanics, 2nd ed. (ButterworthHeinemann, 1987); ISBN 0 7506 2767 0. See §50.
 4.
To take the square root of –j = j^{−1}, it is sometimes useful to draw a diagram on the complex plane by expressing j^{−1} as –j, drawn vertically downward at −90° from the positive real axis. The square root operation divides that angle by two, resulting in a unit vector rotated only −45° away from the real axis having a projection of (2)^{–½} along the real axis and –(2)^{–½} along the imaginary axis demonstrating that one root of √–j is (1–j)/√2. Regarding −90° as +270°, and again dividing by two, yields the other root, (j–1)/√2.
 5.
Evolution has found these waves useful for the control of light reaching the optic nerve of insects with compound eyes. Did you ever notice that it is hard to swat a fly under a wide range of lighting conditions? How can you control the light levels when your eye has a thousand lenses? One iris for each lens is clearly out of the question, since each lens has a diameter of only about 30 μm. The mechanism employed by insects to control light levels makes each lens the entrance to an optical waveguide (called an ommatidium by the entomologists), much like an optical fiber used in telecommunications, that channels light from the lens down to the optic nerve (see Fig. 11.6). In bright light, a chemical change in the fluid surrounding the waveguides causes a precipitate to form that scatters the wave decaying (exponentially) beyond the waveguide into the fluid. Scattering of this diffusion wave reduces the light that makes it down the waveguide to the optic nerve. For further information on the physics of insect vision and some relevant graphics, see R. P. Feynman, Lectures on Physics, Vol. I (AddisonWesley 1963), §364.
 6.
It is difficult for most people to visualize the spatial and temporal dependence of the temperature based only on plots of the real and imaginary parts of the solution as a function of position, such as those provided in Figs. 9.7 and 9.8. An animation of the temperature variation for an ideal gas near a solid (isothermal) wall is available at the Los Alamos National Laboratory Thermoacoustics Home Page: http://www.lanl.gov/thermoacoustics/Book/index.html. In the second paragraph on the page at that site, you have the option to download a zipped animation file. You can “unzip” the file and then run the DOSexecutable animation THERMAL.EXE. The animation starts with the pressure and velocity in a standing wave and then zooms into the solid boundary to animate the temperature in the fluid as a function of space and time. The animation goes further to calculate the work done by an imaginary piston moving with the fluid to demonstrate that power is dissipated during the transition from adiabatic compressions far from the wall to isothermal compressions at the fluidsolid boundary.
 7.
In liquids, ω_{crit} is even higher. In water, f_{crit} = 1210 GHz, although this is less significant because the difference between the adiabatic and isothermal sound speeds are so small due to the smaller thermal expansion coefficient. At 4 °C, the density of water reaches its maximum value, so the thermal expansion coefficient vanishes and the isothermal and adiabatic sound speeds are equal. The existence of life on this planet probably owes much to the fact that ice is less dense than water.
 8.
This transition from hydrodynamic to ballistic propagation is known as the “Knudsen limit.” In that regime, the real and imaginary components of the wavenumber become equal. Based on M. Greenspan, “Propagation of Sound in Five Monatomic Gases,” J. Acoust. Soc. Am. 28(4), 644–648 (1956), the hydrodynamic results should not be used for frequencies above ω ≅ c^{2}/10α. An excellent review article by Greenspan appears in Physical Acoustics, Vol. II A, edited by W. P. Mason and R. N. Thurston (Academic Press, 1964), pp. 1–45.
 9.
The entire field of rheology is dedicated to the study of nonNewtonian fluids and plastic flows. Nondrip paints are thixotropic fluids; their viscosities decrease with increasing strain rate. You want a paint that has a low viscosity when it is being applied to reduce drag on the paint brush (and wrist of the painter), but a high viscosity once it is applied to a surface to keep it from dripping. Viscoelastic fluids like Silly Putty™ are also nonNewtonian fluids. A detailed discussion of viscosity and of nonNewtonian flow is provided by R. B. Bird, W. E. Steward, and E. N. Lightfoot, Transport Phenomena (J. Wiley & Sons, 1960); ISBN 047107392X, in Chapter 1.
 10.
 11.
There are two more of these for the (radial) r component and the (azimuthal) θ component of Eq. (9.26) that are just as ugly!
 12.
See Ref. [6], for solutions in tubes of other crosssections, in §17, pp. 53–54 for (Problem 1) annular, (Problem 2) elliptical, and (Problem 3) triangular ducts.
 13.
When flow enters a smaller tube from a larger reservoir, it must travel some distance before the flow becomes “organized” into the parabolic profile shown in Figure 9.12. This distance is known as the “entrance length,” L, and is usually expressed in terms of the tube’s inner diameter, D, and the Reynolds number, Re (see the next footnote): L/D ≅ 0.06 Re.
 14.
The criterion for the transition to turbulence in smoothwalled round pipes is usually given in terms of the nondimensional Reynolds number, Re = ρ < v > D/μ, where <v> is the flow velocity averaged over the pipe’s area, πD^{2}/4. The transition from laminar (Fig. 9.12) to turbulent flow is usually taken to occur at a Reynolds number greater than 2200 ± 100 in circular pipes with a “smooth” surface finish. Further details regarding the transition to turbulence are given in most fluid dynamics textbooks, such as Refs. [2, 3, 6], and presented as a “Moody chart,” where the nondimensional drag is plotted vs. Reynolds number for various values of surface roughness.
 15.
Mathematicians would call the Fourier Heat Diffusion equation (9.11) and the NavierStokes equation (9.2) “isomorphic.” The fact that T is the variable in Eq. (9.11) and v_{y} is the variable in Eq. (9.31) only bores most mathematicians. The forms of the solutions to both equations must be identical.
 16.
For an animated visualization of the fluid in the oscillatory viscous boundary layer, we can again turn to the Los Alamos Thermoacoustics Home Page and run OSCWALL.EXE to see exactly the situation (rotated by 90°) that is diagrammed in Fig. 9.13. The reverse case of a stationary wall and fluid moving uniformly far from the wall is animated in VISCOUS.EXE. In that animation, a vibrating piston sets up the fundamental λ/2 = L standing wave in a tube and then focuses in on a portion of the resonator’s wall at the velocity antinode near the center of the tube.
 17.
A more rigorous general derivation of the viscous boundary layer dissipation and of the acoustic power is provided by Swift in his textbook, Thermoacoustics [10]. See §4.4.2 and Chap. 5.
 18.
A derivation of Eq. (9.39) is provided in G. W. Swift, Thermoacoustics: A unifying perspective for some engines and refrigerators, 2nd ed. (Acoust. Soc. Am., 2017); ISBN 9783319669328, Chapter 5.1.
 19.
A more detailed and systematic discussion of these concepts is given by E. H. Kennard in his classic textbook, Kinetic Theory of Gases with an Introduction to Statistical Mechanics (McGrawHill, 1938).
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Garrett, S.L. (2020). Dissipative Hydrodynamics. In: Understanding Acoustics. Graduate Texts in Physics. Springer, Cham. https://doi.org/10.1007/9783030447878_9
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