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4.23. Wave equation for an infinite lossy tube

Assumptions

The circular tube of radius a shown in Fig. 4.40 has z as the axial ordinate and w as the radial ordinate. In the following discussion, it is assumed that the radial pressure distribution is uniform and that the pressure variations are purely axial. This has been shown to be valid provided that a(meters) ≤ 10⁴/f ^3/2 [11]. Also, it is assumed that the radial velocity is zero, but the axial velocity is allowed to vary radially because of laminar flow resulting from viscous losses. Thermal conduction through the tube wall is also taken into consideration, where the wall is at ambient temperature T ₀. However, boundary slip is allowed for, whereby the axial particle velocity adjacent to the tube wall can be nonzero and the air temperature there can be nonambient. This is particularly relevant in the case of very narrow tubes, where the viscous and thermal losses are less than would be predicted if we were to assume “no slip”. By “no slip”, we mean if the axial velocity at the wall were zero and the temperature there were ambient. Furthermore, the degree of slip is proportional to the gradient of the radial distribution of the velocity or temperature at the tube wall. In this section, we shall introduce the concept of the viscous boundary layer, which is a region adjacent to the wall in which the axial velocity is less than it would be in a loss-free tube. Outside the boundary layer, the tube is considered to be loss-free such that the axial velocity is unaffected by the wall.

The momentum conservation equation

In accordance with the conservation of momentum law, we can write the linearized Navier–Stokes equation [12].

(ρ_{0} \frac{\partial}{\partial t} - μ \nabla^{2}) u (w) = - \frac{\partial p}{\partial z},

$(ρ_{0} \frac{\partial}{\partial t} - μ \nabla^{2}) u (w) = - \frac{\partial p}{\partial z},$

(4.175)

where

\nabla^{2} = \partial^{2} / (\partial w^{2}) + w^{- 1} \partial / (\partial w)

$\nabla^{2} = \partial^{2} / (\partial w^{2}) + w^{- 1} \partial / (\partial w)$

and u is the axial velocity, p is the axial pressure, ρ ₀ = 1.18 kg/m³ and μ = 18.6 × 10⁻⁶ N·s/m² are the density and viscosity of air, respectively, and z is the axial ordinate. Replacing the time derivative with jω gives

(\nabla^{2} + k_{V}^{2}) \tilde{u} (w) = - \frac{k_{V}^{2}}{j ω ρ_{0}} \frac{\partial \tilde{p}}{\partial z},

$(\nabla^{2} + k_{V}^{2}) \tilde{u} (w) = - \frac{k_{V}^{2}}{j ω ρ_{0}} \frac{\partial \tilde{p}}{\partial z},$

(4.176)

where k _V is the viscous (or shear) wave number given by

k_{V} = \sqrt{- j ω ρ_{0} / μ} .

$k_{V} = \sqrt{- j ω ρ_{0} / μ} .$

(4.177)

Thermal conduction (entropy) and the gas law

Fourier's law for thermal conduction gives

κ \nabla^{2} \tilde{τ} (w) = j ω T_{0} (ρ_{0} C_{V} \tilde{p} / P_{0} - C_{P} \tilde{δ}),

$κ \nabla^{2} \tilde{τ} (w) = j ω T_{0} (ρ_{0} C_{V} \tilde{p} / P_{0} - C_{P} \tilde{δ}),$

(4.178)

where

\tilde{p}

$\tilde{p}$

\tilde{δ}

$\tilde{δ}$

, and

\tilde{τ}

$\tilde{τ}$

are the small pressure, density, and temperature fluctuations respectively. Also, κ = 25.4 × 10⁻³ N⋅s⁻¹⋅K⁻¹ is the thermal conductivity, T ₀ = 295°K is the ambient temperature, C _V is the specific heat capacity under constant volume, and C _P is the specific heat capacity under constant pressure. For an ideal gas we can write the following linearized equation of state [12]

\frac{\tilde{p}}{P_{0}} = \frac{\tilde{δ}}{ρ_{0}} + \frac{\tilde{τ} (w)}{T_{0}} .

$\frac{\tilde{p}}{P_{0}} = \frac{\tilde{δ}}{ρ_{0}} + \frac{\tilde{τ} (w)}{T_{0}} .$

(4.179)

Eliminating

\tilde{δ}

$\tilde{δ}$

from Eqs. (4.178) and (4.179) gives

κ \nabla^{2} \tilde{τ} (w) = j ω ρ_{0} T_{0} ((C_{V} - C_{P}) \frac{\tilde{p}}{P_{0}} + C_{P} \frac{\tilde{τ} (w)}{T_{0}}) .

$κ \nabla^{2} \tilde{τ} (w) = j ω ρ_{0} T_{0} ((C_{V} - C_{P}) \frac{\tilde{p}}{P_{0}} + C_{P} \frac{\tilde{τ} (w)}{T_{0}}) .$

(4.180)

We also note that C _P − C _V

P ₀/(ρ ₀ T ₀) [11] so that

(\nabla^{2} + P_{r} k_{V}^{2}) \tilde{τ} (w) = \frac{P_{r} k_{V}^{2}}{ρ_{0} C_{P}} \tilde{p},

$(\nabla^{2} + P_{r} k_{V}^{2}) \tilde{τ} (w) = \frac{P_{r} k_{V}^{2}}{ρ_{0} C_{P}} \tilde{p},$

(4.181)

where P _r is the (dimensionless) Prandtl number given by

P_{r} = μ C_{P} / κ

$P_{r} = μ C_{P} / κ$

(4.182)

which is the ratio of the viscous diffusion rate to the thermal diffusion rate.

Solution of the velocity and temperature radial equations

Eqs. (4.176) and (4.181) for the radial velocity and temperature distributions, respectively, are subject to the following slip boundary conditions:

\tilde{u} (a) = {- a B_{u} \frac{\partial \tilde{u} (w)}{\partial w} |}_{w = a'}

$\tilde{u} (a) = {- a B_{u} \frac{\partial \tilde{u} (w)}{\partial w} |}_{w = a'}$

(4.183)

\tilde{τ} (a) = {- a B_{e} \frac{\partial \tilde{τ} (w)}{\partial w} |}_{w = a'}

$\tilde{τ} (a) = {- a B_{e} \frac{\partial \tilde{τ} (w)}{\partial w} |}_{w = a'}$

(4.184)

where the boundary slip factors B _u and B _e are given by

B_{u} = (2 α_{u}^{- 1} - 1) K_{n},

$B_{u} = (2 α_{u}^{- 1} - 1) K_{n},$

(4.185)

B_{e} = \frac{2 γ}{P_{r} (1 + γ)} (\frac{2}{α_{e}} - 1) K_{n},

$B_{e} = \frac{2 γ}{P_{r} (1 + γ)} (\frac{2}{α_{e}} - 1) K_{n},$

(4.186)

which are zero in the case of no slip, where

\tilde{u} (a) = \tilde{τ} (a) = 0

$\tilde{u} (a) = \tilde{τ} (a) = 0$

We note that γ = C _P/C _v is the specific heat ratio, α _u and α _e are the accommodation coefficients, both of which are assumed to have a value of 0.9, and K _n is the (dimensionless) Knudsen number given by

K_{n} = λ_{m} / a,

$K_{n} = λ_{m} / a,$

(4.187)

where λ _m = 60 nm is the molecular mean free path length between collisions [10]. We let

\tilde{u} (w) = - \frac{1}{j ω ρ_{0}} \frac{\partial \tilde{p}}{\partial z} (1 - F (k_{V}, w, B_{u})),

$\tilde{u} (w) = - \frac{1}{j ω ρ_{0}} \frac{\partial \tilde{p}}{\partial z} (1 - F (k_{V}, w, B_{u})),$

(4.188)

\tilde{τ} (w) = \frac{\tilde{p}}{ρ_{0} C_{P}} ((1 - F (k_{T}, w, B_{e})),

$\tilde{τ} (w) = \frac{\tilde{p}}{ρ_{0} C_{P}} ((1 - F (k_{T}, w, B_{e})),$

(4.189)

which after substituting in Eqs. (4.176) and (4.181), respectively, leads to a new pair of equations:

(\nabla^{2} + k_{V}^{2}) F (k_{V}, w, B_{u}) = 0,

$(\nabla^{2} + k_{V}^{2}) F (k_{V}, w, B_{u}) = 0,$

(4.190)

(\nabla^{2} + k_{T}^{2}) F (k_{T}, w, B_{e}) = 0,

$(\nabla^{2} + k_{T}^{2}) F (k_{T}, w, B_{e}) = 0,$

(4.191)

where k _T is the thermal (or entropy) wave number given by

k_{T} = \sqrt{P_{r}} k_{V} .

$k_{T} = \sqrt{P_{r}} k_{V} .$

(4.192)

Eqs. (4.190) and (4.191) are subject to the boundary conditions

{F (k_{V}, a, B_{u}) + a B_{u} \frac{\partial}{\partial w} F (k_{V}, w, B_{u}) |}_{w = a} = 1,

${F (k_{V}, a, B_{u}) + a B_{u} \frac{\partial}{\partial w} F (k_{V}, w, B_{u}) |}_{w = a} = 1,$

(4.193)

{F (k_{T}, a, B_{e}) + a B_{e} \frac{\partial}{\partial w} F (k_{T}, w, B_{e}) |}_{w = a} = 1

${F (k_{T}, a, B_{e}) + a B_{e} \frac{\partial}{\partial w} F (k_{T}, w, B_{e}) |}_{w = a} = 1$

(4.194)

Solutions to Eqs. (4.190) and (4.191) are given by

F (k_{V}, w, B_{u}) = A J_{0} (k_{V} w),

$F (k_{V}, w, B_{u}) = A J_{0} (k_{V} w),$

(4.195)

F (k_{T}, w, B_{e}) = B J_{0} (k_{T} w) .

$F (k_{T}, w, B_{e}) = B J_{0} (k_{T} w) .$

(4.196)

The unknown coefficients can be found by substituting Eqs. (4.195) and (4.196) in the boundary conditions of Eqs. (4.193) and (4.194), respectively, to give

A = {(J_{0} (k_{V} a) - B_{u} k_{V} a J_{1} (k_{V} a))}^{- 1}

$A = {(J_{0} (k_{V} a) - B_{u} k_{V} a J_{1} (k_{V} a))}^{- 1}$

and

B = {(J_{0} (k_{T} a) - B_{e} k_{T} a J_{1} (k_{T} a))}^{- 1} .

$B = {(J_{0} (k_{T} a) - B_{e} k_{T} a J_{1} (k_{T} a))}^{- 1} .$

The average values across the tube cross section are defined by

\begin{matrix} 〈 F (k_{V}, a, B_{u}) 〉 = \frac{1}{π a^{2}} \int_{0}^{2 π} \int_{0}^{a} F (k_{V}, w, B_{u}) w d w d ϕ \\ = \frac{Q (k_{V} a)}{1 - 0.5 B_{u} k_{V}^{2} a^{2} Q (k_{V} a)}, \end{matrix}

$\begin{matrix} 〈 F (k_{V}, a, B_{u}) 〉 = \frac{1}{π a^{2}} \int_{0}^{2 π} \int_{0}^{a} F (k_{V}, w, B_{u}) w d w d ϕ \\ = \frac{Q (k_{V} a)}{1 - 0.5 B_{u} k_{V}^{2} a^{2} Q (k_{V} a)}, \end{matrix}$

(4.197)

where

Q (x) = \frac{2 J_{1} (x)}{x J_{0} (x)} = {(\sum_{n = 0}^{\infty} \frac{x^{2}}{x^{2} - β_{n}^{2}})}^{- 1},

$Q (x) = \frac{2 J_{1} (x)}{x J_{0} (x)} = {(\sum_{n = 0}^{\infty} \frac{x^{2}}{x^{2} - β_{n}^{2}})}^{- 1},$

(4.198)

where β _n are the zeros of J ₁(x), such that J ₁(β _n) = 0, and for large n, we may use β _n|_n _→ _∞ ≈ (n + ¼)π. Unlike the conventional individual expansions for J ₀(x) and J ₁(x) given by Eq. (A2.71) of Appendix II, this expansion does not blow up for larger values of x when the expansion limit is truncated.

Similarly,

〈 F (k_{T}, a, B_{e}) 〉 = \frac{Q (k_{T} a)}{1 - 0.5 B_{e} k_{T}^{2} a^{2} Q (k_{T} a)} .

$〈 F (k_{T}, a, B_{e}) 〉 = \frac{Q (k_{T} a)}{1 - 0.5 B_{e} k_{T}^{2} a^{2} Q (k_{T} a)} .$

(4.199)

In Fig. 4.41, the axial velocity along the radius of a narrow tube is plotted at a frequency of 100 Hz using Eq. (4.188), where the radius of the tube is 1 μm. Also, the axial velocity along the radius of a wide tube is plotted in Fig. 4.42 at a frequency of 10 kHz, where the radius is 1 mm.

Figure 4.41 Variation of normalized velocity along radius of ultra-narrow tube of radius 1 μm at a frequency of 100 Hz. The effective boundary layer thickness is 224 μm.

Figure 4.42 Variation of normalized velocity along radius of medium-sized tube of radius 1 mm at a frequency of 10 kHz. The effective boundary layer thickness is 22.4 μm.

The effective boundary layer thickness δ _visc can be calculated using the formula [12].

δ_{v i s c} = \sqrt{\frac{2 μ}{ρ_{0} ω}}

$δ_{v i s c} = \sqrt{\frac{2 μ}{ρ_{0} ω}}$

(4.200)

In Fig. 4.41, the effective boundary layer thickness of 224 μm is much greater than the radius of the tube, so the normalized velocity never reaches the theoretical maximum value of unity even at the center (w = 0). At the wall, the boundary slip condition is clearly visible as the velocity does not reach zero. By contrast, the effective boundary layer thickness in Fig. 4.42 of 22.4 μm is only 2.24% of the radius, which explains why the normalized velocity is unity over most of the radius and only falls rapidly close to the wall (w = a), albeit after a small peak. At the edge of the boundary layer, it is about 86% of the velocity at the center. At the wall, the velocity is virtually zero, so there is no appreciable slip.

Mass conservation and Helmholtz wave equation

Finally, we use the following mass conservation equation or equation of continuity [12].

j ω 〈 \tilde{δ} 〉 + ρ_{0} \frac{\partial}{\partial z} 〈 \tilde{u} 〉 = 0 .

$j ω 〈 \tilde{δ} 〉 + ρ_{0} \frac{\partial}{\partial z} 〈 \tilde{u} 〉 = 0 .$

(4.201)

For the average velocity, we can write from Eq. (4.188)

〈 \tilde{u} 〉 = - \frac{1}{j ω ρ_{0}} \frac{\partial \tilde{p}}{\partial z} (1 - 〈 F (k_{V}, a, B_{u}) 〉) .

$〈 \tilde{u} 〉 = - \frac{1}{j ω ρ_{0}} \frac{\partial \tilde{p}}{\partial z} (1 - 〈 F (k_{V}, a, B_{u}) 〉) .$

(4.202)

Differentiating Eq. (4.202) with respect to z and inserting it in Eq. (4.201) yields

〈 \tilde{δ} 〉 = - \frac{1}{ω^{2}} (1 - 〈 F (k_{V}, a, B_{u}) 〉) \frac{\partial^{2} \tilde{p}}{\partial z^{2}} .

$〈 \tilde{δ} 〉 = - \frac{1}{ω^{2}} (1 - 〈 F (k_{V}, a, B_{u}) 〉) \frac{\partial^{2} \tilde{p}}{\partial z^{2}} .$

(4.203)

Also, from the gas law of Eq. (4.178)

\frac{\tilde{p}}{P_{0}} = \frac{〈 \tilde{δ} 〉}{ρ_{0}} + \frac{〈 \tilde{τ} 〉}{T_{0}},

$\frac{\tilde{p}}{P_{0}} = \frac{〈 \tilde{δ} 〉}{ρ_{0}} + \frac{〈 \tilde{τ} 〉}{T_{0}},$

(4.204)

where the average temperature is derived from Eq. (4.189) as follows

〈 \tilde{τ} 〉 = \frac{\tilde{p}}{ρ_{0} C_{P}} (1 - 〈 F (k_{T}, a, B_{e}) 〉) .

$〈 \tilde{τ} 〉 = \frac{\tilde{p}}{ρ_{0} C_{P}} (1 - 〈 F (k_{T}, a, B_{e}) 〉) .$

(4.205)

Substituting Eq. (4.205) in Eq. (4.204) while noting that

C_{P} - C_{v} = P_{0} / (ρ_{0} T_{0}) and γ = C_{P} / C_{v}

$C_{P} - C_{v} = P_{0} / (ρ_{0} T_{0}) and γ = C_{P} / C_{v}$

so that

C_{P} = \frac{γ P_{0}}{(γ - 1) ρ_{0} T_{0}}

$C_{P} = \frac{γ P_{0}}{(γ - 1) ρ_{0} T_{0}}$

(4.206)

gives

〈 \tilde{δ} 〉 = \frac{ρ_{0}}{γ P_{0}} (1 + (γ - 1) 〈 F (k_{T}, a, B_{e}) 〉) \tilde{p} .

$〈 \tilde{δ} 〉 = \frac{ρ_{0}}{γ P_{0}} (1 + (γ - 1) 〈 F (k_{T}, a, B_{e}) 〉) \tilde{p} .$

(4.207)

Equating Eqs. (4.203) and (4.207) then leads to the following Helmholtz wave equation:

\frac{\partial^{2} \tilde{p}}{\partial z^{2}} + k^{2} \tilde{p} = 0,

$\frac{\partial^{2} \tilde{p}}{\partial z^{2}} + k^{2} \tilde{p} = 0,$

(4.208)

where

k^{2} = \frac{(1 + (γ - 1)) 〈 F (k_{T}, a, B_{e}) 〉) ω^{2} ρ_{0}}{(1 - 〈 F (k_{V}, a, B_{u}) 〉) γ P_{0}}

$k^{2} = \frac{(1 + (γ - 1)) 〈 F (k_{T}, a, B_{e}) 〉) ω^{2} ρ_{0}}{(1 - 〈 F (k_{V}, a, B_{u}) 〉) γ P_{0}}$

(4.209)

or, using Eqs. (4.197) and (4.199),

k^{2} = \frac{(1 + (γ - 1) \frac{Q (k_{T} a)}{1 - 0.5 B_{e} k_{T}^{2} a^{2} Q (k_{T} a)}) ω^{2} ρ_{0}}{(1 - \frac{Q (k_{V} a)}{1 - 0.5 B_{u} k_{V}^{2} a^{2} Q (k_{V} a)}) γ P_{0}} .

$k^{2} = \frac{(1 + (γ - 1) \frac{Q (k_{T} a)}{1 - 0.5 B_{e} k_{T}^{2} a^{2} Q (k_{T} a)}) ω^{2} ρ_{0}}{(1 - \frac{Q (k_{V} a)}{1 - 0.5 B_{u} k_{V}^{2} a^{2} Q (k_{V} a)}) γ P_{0}} .$

(4.210)

Dynamic density

To simplify the expressions for the wave number k and characteristic impedance Z ₀, we can use the following shorthand known as the dynamic density where 〈u〉 is given by Eq. (4.202) so that

\begin{matrix} ρ = - \frac{1}{j ω 〈 \tilde{u} 〉} \frac{\partial \tilde{p}}{\partial z} = \frac{ρ_{0}}{1 - 〈 F (k_{V}, a, B_{u}) 〉} \\ = ρ_{0} {(1 - \frac{Q (k_{V} a)}{1 - 0.5 B_{u} k_{V}^{2} a^{2} Q (k_{V} a)})}^{- 1} . \end{matrix}

$\begin{matrix} ρ = - \frac{1}{j ω 〈 \tilde{u} 〉} \frac{\partial \tilde{p}}{\partial z} = \frac{ρ_{0}}{1 - 〈 F (k_{V}, a, B_{u}) 〉} \\ = ρ_{0} {(1 - \frac{Q (k_{V} a)}{1 - 0.5 B_{u} k_{V}^{2} a^{2} Q (k_{V} a)})}^{- 1} . \end{matrix}$

(4.211)

Dynamic compressibility

Also, the dynamic compressibility is defined by

C = \frac{〈 \tilde{δ} 〉}{ρ_{0} \tilde{p}} .

$C = \frac{〈 \tilde{δ} 〉}{ρ_{0} \tilde{p}} .$

(4.212)

From the ideal gas law of Eq. (4.207), we obtain

\frac{〈 \tilde{δ} 〉}{ρ_{0} \tilde{p}} = \frac{1}{γ P_{0}} (1 + (γ - 1) 〈 F (k_{T}, a, B_{e}) 〉),

$\frac{〈 \tilde{δ} 〉}{ρ_{0} \tilde{p}} = \frac{1}{γ P_{0}} (1 + (γ - 1) 〈 F (k_{T}, a, B_{e}) 〉),$

(4.213)

which is inserted in Eq. (4.212) to give

C = \frac{1}{γ P_{0}} (1 + \frac{(γ - 1) Q (k_{T} a)}{1 - 0.5 B_{e} k_{T}^{2} a^{2} Q (k_{T} a)}) .

$C = \frac{1}{γ P_{0}} (1 + \frac{(γ - 1) Q (k_{T} a)}{1 - 0.5 B_{e} k_{T}^{2} a^{2} Q (k_{T} a)}) .$

(4.214)

Wave number and characteristic impedance

Using the expressions for the dynamic density ρ and dynamic compressibility C from Eqs. (4.211) and (4.214), respectively, the wave number of Eq. (4.210) simply becomes

k = ω \sqrt{ρ C}

$k = ω \sqrt{ρ C}$

(4.215)

By comparing this with the wave number for a loss-free plane wave (from Eqs. 2.19 to 2.45)

k = ω \sqrt{\frac{ρ_{0}}{γ P_{0}}}

$k = ω \sqrt{\frac{ρ_{0}}{γ P_{0}}}$

(4.216)

we see that when there are no viscous or thermal losses ρ = ρ ₀ and C = 1/(γP ₀). Hence, the compressibility is the inverse bulk modulus of the medium. Similarly, from Eq. (2.124), we see that the characteristic specific impedance of an infinite tube is

Z_{s} = \sqrt{ρ / C} .

$Z_{s} = \sqrt{ρ / C} .$

(4.217)

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Table of Contents for
Acoustics

4.23. Wave equation for an infinite lossy tube

Assumptions

Categories

The momentum conservation equation

Thermal conduction (entropy) and the gas law

Solution of the velocity and temperature radial equations

Mass conservation and Helmholtz wave equation

Dynamic density

Dynamic compressibility

Wave number and characteristic impedance

Table of Contents for Acoustics

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4.23. Wave equation for an infinite lossy tube

Assumptions

Categories

The momentum conservation equation

Thermal conduction (entropy) and the gas law

Solution of the velocity and temperature radial equations

Mass conservation and Helmholtz wave equation

Dynamic density

Dynamic compressibility

Wave number and characteristic impedance

Table of Contents for
Acoustics