# Sound Transmission in a Duct with Sudden Area Expansion, Extended Inlet, and Lined Walls in Overlapping Region

Authors: Ahmet Demir
Journal: Advances in Acoustics and Vibration
 (2016)

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## Abstract

The transmission of sound in a duct with sudden area expansion and extended inlet is investigated in the case where the walls of the duct lie in the finite overlapping region lined with acoustically absorbent materials. By using the series expansion in the overlap region and using the Fourier transform technique elsewhere we obtain a Wiener-Hopf equation whose solution involves a set of infinitely many unknown expansion coefficients satisfying a system of linear algebraic equations. Numerical solution of this system is obtained for various values of the problem parameters, whereby the effects of these parameters on the sound transmission are studied.


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## Body

## 1. Introduction

One can reduce the unwanted noise propagating along a duct by using a reactive or a dissipative silencer. In reactive silencers sudden area changes in cross-sectional area help to reduce the energy in the transmitted wave via internal reflections. Having a sudden area expansion together with a sudden area contraction simple expansion chambers works in accordance with this principle and is widely investigated in literature [1–4]. In further investigations it has been shown that the extension of inlet and outlet tubes into the expansion chamber increased the acoustic attenuation performance [5–7].On the other hand, it has been proved that the treatment of the duct walls with an acoustically absorbent lining is another effective method in reducing unwanted noise [8]. Application of locally reacting linings or expansion chambers in ducts are efficient methods for noise reduction. These two methods were combined in [9] to discover transmission properties of a combination silencer consist of an expansion chamber whose walls are treated by acoustic liners and have been analysed by the author previously.In this paper, the transmission of sound in an extended tube resonator whose walls are in overlapping region, where extended inlet and expanding duct walls overlap, are treated by locally reacting lining is investigated. So the main objective of this paper is to reveal the influence of the partial lining on the transmitted field and to present an alternative method of formulation. The method previously employed in [10, 11] consists of expanding the field in the overlap region into a series of complete set of orthogonal eigenfunctions and using the Fourier transform technique elsewhere. The problem is then reduced directly into a Wiener-Hopf equation whose solution involves a set of infinitely many unknown expansion coefficients satisfying an infinite systems of linear algebraic equations. Numerical solution to these systems is obtained for various values of the parameters of the problem such as the radii of the semi-infinite waveguides, the overlap length, and the impedance loading whereby the effects of these parameters on the transmitted field are presented graphically.The time dependence is assumed to beexp⁡-iωt with ω being the angular frequency and suppressed throughout.


## 2. Materials and Methods

Consider two opposite semi-infinite circular cylindrical waveguides of different radii with common longitudinal axis, sayz, in a cylindrical polar coordinate system ρ,ϕ,z. They occupy the regions ρ=a and  z<l and ρ=b>a and  z>0, respectively, where l represents the overlap length. These two waveguides are connected with a vertical wall at a<ρ<b and   z=0. The parts of the surfaces r=a+0 and ρ=b-0 lying in the overlap region 0<z<l of the waveguides and the vertical wall are assumed to be treated by acoustically absorbing linings which are characterized by constant but different surface admittances, say η1, η2, and η3, respectively, while the remaining parts are perfectly rigid (see Figure 1). The waveguides are immersed in an inviscid and compressible stationary fluid of density ρ~0 and sound speed c. A plane sound wave is incident from the positive z-direction, through the waveguide of radius ρ=a. From the symmetry of the geometry of the problem and the incident field the acoustic field everywhere will be independent of the ϕ coordinate. We shall therefore introduce a scalar potential uρ,z which defines the acoustic pressure and velocity by p=iωρ~0u and v=gradu, respectively.Figure 1
Geometry of the problem.Let the incident field be given by(1)ui=exp⁡ikz,where k=ω/c denotes the wave number. For the sake of analytical convenience we will assume that the surrounding medium is slightly lossy and k has a small positive imaginary part. The lossless case can be obtained by letting Imk→0 at the end of the analysis.The total fielduT(ρ,z) can be written as(2)uTρ,z=u1ρ,z+uiρ,z,ρ∈0,a,z∈-∞,∞,u21ρ,zHz-Hz-l+u22ρ,zHz-l,ρ∈a,b,z∈0,∞.u1ρ,z and u2jρ,z(j=1,2) denote the scattered fields which satisfy the Helmholtz equation(3)1ρ∂∂ρρ∂∂ρ+∂2∂z2+k2u1ρ,zu2jρ,z=0,j=1,2and are to be determined with the help of the following boundary and continuity relations:(4a)∂∂ρu1a,z=0,z<l,(4b)∂∂zu21ρ,0=0,a<ρ<b,(4c)ikη1+∂∂ρu21a,z=0,0<z<l,(4d)ikη2-∂∂ρu21b,z=0,0<z<l,(4e)ikη3+∂∂zu21ρ,0=0,a<ρ<b,(4f)∂∂ρu22b,z=0,z>l,(4g)u21ρ,l-u22ρ,l=0,a<ρ<b,(4h)∂∂zu21ρ,l-∂∂zu22ρ,l=0,a<ρ<b,(4i)u1a,z+uia,z-u22a,z=0,z>l,(4j)∂∂ρu1a,z+∂∂ρuia,z-∂∂ρu22a,z=0,z>l.In addition to these boundary and continuity relations one has to take into account the following radiation and edge conditions to ensure the uniqueness of the mixed boundary value problem stated by (3) and (4a)–(4j):(5)u~eikrr,r=ρ2+z2⟶∞,(6a)uTρ,z=O1,z⟶l,(6b)∂∂ρuTρ,z=Oz-l-1/2,z⟶l.


### 2.1. The Wiener-Hopf Equations

Consider the Fourier transform of the Helmholtz equation satisfied by the scattered fieldu1ρ,z in the region ρ<a for z∈(-∞,∞); namely,(7)1ρ∂∂ρρ∂∂ρ+K2αFρ,α=0,where Fρ,α is the Fourier transform of the field u1ρ,z defined to be(8a)Fρ,α=∫-∞∞u1ρ,zeiαzdz=eiαlF+ρ,α+F-ρ,αwith(8b)F-ρ,α=∫-∞lu1ρ,zeiαz-ldz,(8c)F+ρ,α=∫l∞u1ρ,zeiαz-ldz.Owing to the analytical properties of Fourier integrals, F+(ρ,α) and F-(ρ,α) are regular functions in the upper half-plane Imα>Im(-k) and in the lower half-plane Imα<Imk, respectively. The solution of (7) reads(9)Fρ,α=-AαJ0KρKαJ1Ka,where A(α) is a spectral coefficient to be determined and K(α) is the square-root function(10)Kα=k2-α2which is defined in the complex α-plane cut as shown in Figure 2 such that K(0)=k. Consider now the Fourier transform of (4a); namely,(11)F˙-a,α=0.The differentiation of (9) with respect to ρ and putting ρ=a gives(12)eiαlF˙+a,α=Aα.Substituting (12) into (9) yields(13)F+ρ,α=-F˙+a,αJ0KρKαJ1Ka-F-ρ,α.In the region a<ρ<b the field u22ρ,z satisfies the Helmholtz equation for z∈l,∞ as denoted in (3). The Fourier transform of this equation for the region in question is(14)1ρ∂∂ρρ∂∂ρ+K2αG+ρ,α=fρ-iαgρ,where(15a)fρ=∂∂zu22ρ,l,(15b)gρ=u22ρ,l.In (14), G+ρ,α is a regular function in the upper half of the complex α-plane which is defined as(16)G+ρ,α=∫l∞u22ρ,zeiαz-ldz.Figure 2
Complexα-plane.Particular solutions to (14) can be found easily by using Green’s function which satisfies the Helmholtz equation(17)1ρ∂∂ρρ∂∂ρ+K2αGρ,α=0,ρ≠t,ρ,t∈a,bwith the following conditions:(18a)Gt+0,t,α=Gt-0,t,α,(18b)∂∂ρGt+0,t,α-∂∂ρGt-0,t,α=1t,(18c)∂∂ρGb,t,α=0,(18d)∂∂ρGa,t,α=0.The solution is(19)Gρ,t,α=1K2αMαQρ,t,αwith(20a)Qρ,t,α=π2J0KρKY1Ka-KJ1KaY0KρJ0KtKY1Kb-KJ1KbY0Kt,a≤ρ≤t,J0KρKY1Kb-KJ1KbY0KρJ0KtKY1Ka-KJ1KaY0Kt,t≤ρ≤b,(20b)Mα=J1KaY1Kb-J1KbY1Ka.Together with the boundary condition (4f) the solution can be written as(21)G+ρ,α=1K2αMαBαJ0KρKY1Kb-Y0KρKJ1Kb+∫abft-iαgtQt,ρ,αtdt.Here, B(α) is a spectral coefficient to be determined.The continuity relation in (4j) requires(22)G˙+a,α=F˙+a,α.B(α) can be solved uniquely from (22) as(23)Bα=-F˙+a,α.The substitution of (23) into (21) gives(24)G+ρ,α=1K2αMα-F˙+a,αJ0KρKY1Kb-Y0KρKJ1Kb+∫abft-iαgtQt,ρ,αtdt.Although the left-hand side of (24) is regular in the half-plane Im(α)>Im(-k), the regularity of the right hand side is violated by the presence of simple poles lying at the upper half-plane, namely, at α=αm(Imαm>Imk) with(25)K2αmJ1KmaY1Kmb-J1KmbY1Kma=0,Km=Kαm,m=0,1,2,….In order to provide regularity of the right hand side of (24) in the upper half of the α-plane, these poles must be eliminated by imposing that their residues are zero. This gives (26a)F˙+a,k=1πf0-ikg0a2-b2ab,(26b)F˙+a,αm=1πfm-iαmgmJ12Kma-J12KmbJ1KmaJ1Kmbfor m=0,1,2,…. Here, fm and gm’s are the expansion coefficients of the functions fρ and gρ, respectively, which may be represented trough the following complete sets of orthogonal functions(27)fρgρ=2πbf0g0+∑m=1∞fmgmJ0KmρKmY1Kmb-KmJ1KmbY0Kmρ.Using the continuity relation (4i) together with (24) and taking into account (13) give(28)F˙+a,αK2αJ1KbπMαJ1Ka+a2F-a,α=121K2αMα∫abft-iαgtJ0KtKY1Kb-KJ1KbY0Kttdt-a2eikliα+k,F˙+a,αK2αNα+a2F-a,α=bπaf0-iαg0k2-α2+1π∑m=1∞fm-iαgmαm2-α2J1KmbJ1Kma-a2eikliα+kwhich is the Wiener-Hopf equation to be solved through classical procedures. Here, Nα stands for(29)Nα=J1KbπMαJ1Ka.The final solution of the W-H equation is determined to be(30)F˙+a,αk+αN+α=-aπbf0+ikg0k+αN+k+1π∑m=1∞fm+iαmgm2αmαm+αJ1KmbJ1Kmak+αmN+αm-kaeikliα+kN+k,where N±α are the split functions resulting from the Wiener-Hopf factorization of Nα as(31)Nα=N+αN-α.Their explicit expressions are given in [12] as(32)N+α=πJ1kaJ1kbJ1kaY1kb-J1kbY1ka-1/2e-αχ∏n=0∞1+α/k2-jn/b21+α/k2-jn/a21+α/αn.Here jn’s are the roots of the Bessel function of the first kind (33a)J1jn=0,n=0,1,…,(33b)χ=iπbln⁡b-aln⁡a-b-aln⁡b-awith(33c)N-α=N+-α.


### 2.2. Determination of the Unknown Coefficients

The field in the regiona<ρ<b, z∈0,l can be expressed is terms of the waveguide normal modes as(34)u21ρ,z=∑n=0∞aneiβnz-Pne-iβnzJ0ξnρ-RnY0ξnρwith(35a)Pn=ikη3+iβnikη3-iβn,(35b)Rn=ikη1J0ξna-ξnJ1ξnaikη1Y0ξna-ξnY1ξna=ikη2J0ξnb+ξnJ1ξnbikη2Y0ξnb+ξnY1ξnb.In (34), ξn’s are the roots of the following equation:(36)ikη1J0ξna-ξnJ1ξnaikη1Y0ξna-ξnY1ξna-ikη2J0ξnb+ξnJ1ξnbikη2Y0ξnb+ξnY1ξnb=0while βn’s are defined as(37)βn=k2-ξn2.Taking into account (27) and (34), the continuity relations (4g) and (4h) can be written in the following form:(38)2πbf0+iαg0+∑m=1∞fm+iαgmJ0KmρKmY1Kmb-KmJ1KmbY0Kmρ=i∑n=0∞anα+βneiβnl-Pnα-βne-iβnlJ0ξnρ-RnY0ξnρ.Multiplying (38) by 2ρ/πb and J0(Kmρ)KmY1(Kmb)-KmJ1(Kmb)Y0(Kmρ)ρ, respectively, and then integrating over ρ from a to b read(39a)f0+iαg0=iS0∑n=0∞anα+βneiβnl-Pnα-βne-iβnl△0n,(39b)fm+iαgm=iSm∑n=0∞anα+βneiβnl-Pnα-βne-iβnl△mnwith (40a)△0n=2πξnabJ1ξna-RnY1ξna-J1ξnb-RnY1ξnb,(40b)△mn=2ξnπξn2-Km2J1KmbJ1KmaJ1ξna-RnY1ξna-J1ξnb-RnY1ξnb,where S0 and Sm stand for(41a)S0=2π2a2-b2b2,(41b)Sm=2π2J12Kma-J12KmbJ12Kma.Using the W-H solution (30) together with (26a) and (26b) we obtain a set of linear algebraic equations in terms of the unknown coefficients fm and gm.(42a)π2baf0-ikg0S0N+k=bπaf0+ikg0N+k+kπ∑m=1∞fm+iαmgmαmN+αmJ1KmbJ1Kma-kaeikliN+k,(42b)π2fr-iαrgrJ1KraJ1KrbSrN+αr=bπaf0+ikg0N+k+1πk+αr∑m=1∞fm+iαmgm2αmαm+αrJ1KmbJ1Kmak+αmN+αm-kaeikliN+k,and taking into account (39a) and (39b) we obtain now a set of equations to determine the unknown expansion coefficient an as(43a)-iπ2baN+k∑n=0∞ank-βneiβnl-Pnk+βne-iβnl△0n=bπa1N+kiS0∑n=0∞ank+βneiβnl-Pnk-βne-iβnl△0n+ikπ∑n=0∞∑m=1∞anαm+βneiβnl-Pnαm-βne-iβnl△mnαmN+αmJ1KmbSmJ1Kma-kaeikliN+k,(43b)-iπ2J1KraJ1KrbN+αr∑n=0∞anαr-βneiβnl-Pnαr+βne-iβnl△rn=bπa1N+kiS0∑n=0∞ank+βneiβnl-Pnk-βne-iβnl△0n+iπ∑n=0∞∑m=1∞anαm+βneiβnl-Pnαm-βne-iβnlk+αr△mn2αmαm+αrJ1KmbSmJ1Kmak+αmN+αm-kaeikliN+k.


### 2.3. Reflected and Transmitted Fields

According to (8a), the scattered field in the region 0<ρ<a,thatis,u1(ρ,z), can be obtained by taking the inverse Fourier transform of Fρ,α. By considering (13) we write(44)u1ρ,z=-12π∫LF˙+a,αJ0KρKαJ1Kae-iαz-ldα.The evaluation of this integral for z<l and z>l will give us the reflected wave propagating backward in the inner cylinder and the transmitted wave, respectively.Forz<l, the integral is calculated by closing the contour in the upper half-plane and evaluating the residues contributions from the simple poles occurring at the zeros of J1Ka lying in the upper α-half-plane, namely, at Ka=jn. The reflection coefficient R of the fundamental mode is defined as the complex coefficient multiplying the travelling wave term exp-ikz and is computed from the contribution of the first pole at α=k. The result is(45)R=-ei2klN+k2-iπf0+ikg0kbN+k2eikl+iπeiklaN+k∑m=1∞fm+iαmgmαmN+αmJ1KmbJ1Kma.The first term is the reflection coefficient related to the case where a semi-infinite rigid duct is inserted axially into a larger rigid tube of infinite length [12] whereas the second one is the correction term involving the effect of the impedances of the annular region and the overlap length.Similarly, the transmission coefficientT of the fundamental mode which is defined as the complex coefficient of exp⁡ikz is obtained by evaluating the integral in (44) for z>l. This integral is now computed by closing the contour in the lower half of the complex α-plane. The pole of interest is at α=-k whose contribution gives(46a)-1+Teikz+Oeik2-jn/b2zwith(46b)T=a2b2+ie-iklπkaba-abf0+ikg0.The first term in (46a) cancels out the incident wave in the region ρ<a,  z>l, while the second is the transmission coefficient of the fundamental mode.


## 2.1. The Wiener-Hopf Equations

Consider the Fourier transform of the Helmholtz equation satisfied by the scattered fieldu1ρ,z in the region ρ<a for z∈(-∞,∞); namely,(7)1ρ∂∂ρρ∂∂ρ+K2αFρ,α=0,where Fρ,α is the Fourier transform of the field u1ρ,z defined to be(8a)Fρ,α=∫-∞∞u1ρ,zeiαzdz=eiαlF+ρ,α+F-ρ,αwith(8b)F-ρ,α=∫-∞lu1ρ,zeiαz-ldz,(8c)F+ρ,α=∫l∞u1ρ,zeiαz-ldz.Owing to the analytical properties of Fourier integrals, F+(ρ,α) and F-(ρ,α) are regular functions in the upper half-plane Imα>Im(-k) and in the lower half-plane Imα<Imk, respectively. The solution of (7) reads(9)Fρ,α=-AαJ0KρKαJ1Ka,where A(α) is a spectral coefficient to be determined and K(α) is the square-root function(10)Kα=k2-α2which is defined in the complex α-plane cut as shown in Figure 2 such that K(0)=k. Consider now the Fourier transform of (4a); namely,(11)F˙-a,α=0.The differentiation of (9) with respect to ρ and putting ρ=a gives(12)eiαlF˙+a,α=Aα.Substituting (12) into (9) yields(13)F+ρ,α=-F˙+a,αJ0KρKαJ1Ka-F-ρ,α.In the region a<ρ<b the field u22ρ,z satisfies the Helmholtz equation for z∈l,∞ as denoted in (3). The Fourier transform of this equation for the region in question is(14)1ρ∂∂ρρ∂∂ρ+K2αG+ρ,α=fρ-iαgρ,where(15a)fρ=∂∂zu22ρ,l,(15b)gρ=u22ρ,l.In (14), G+ρ,α is a regular function in the upper half of the complex α-plane which is defined as(16)G+ρ,α=∫l∞u22ρ,zeiαz-ldz.Figure 2
Complexα-plane.Particular solutions to (14) can be found easily by using Green’s function which satisfies the Helmholtz equation(17)1ρ∂∂ρρ∂∂ρ+K2αGρ,α=0,ρ≠t,ρ,t∈a,bwith the following conditions:(18a)Gt+0,t,α=Gt-0,t,α,(18b)∂∂ρGt+0,t,α-∂∂ρGt-0,t,α=1t,(18c)∂∂ρGb,t,α=0,(18d)∂∂ρGa,t,α=0.The solution is(19)Gρ,t,α=1K2αMαQρ,t,αwith(20a)Qρ,t,α=π2J0KρKY1Ka-KJ1KaY0KρJ0KtKY1Kb-KJ1KbY0Kt,a≤ρ≤t,J0KρKY1Kb-KJ1KbY0KρJ0KtKY1Ka-KJ1KaY0Kt,t≤ρ≤b,(20b)Mα=J1KaY1Kb-J1KbY1Ka.Together with the boundary condition (4f) the solution can be written as(21)G+ρ,α=1K2αMαBαJ0KρKY1Kb-Y0KρKJ1Kb+∫abft-iαgtQt,ρ,αtdt.Here, B(α) is a spectral coefficient to be determined.The continuity relation in (4j) requires(22)G˙+a,α=F˙+a,α.B(α) can be solved uniquely from (22) as(23)Bα=-F˙+a,α.The substitution of (23) into (21) gives(24)G+ρ,α=1K2αMα-F˙+a,αJ0KρKY1Kb-Y0KρKJ1Kb+∫abft-iαgtQt,ρ,αtdt.Although the left-hand side of (24) is regular in the half-plane Im(α)>Im(-k), the regularity of the right hand side is violated by the presence of simple poles lying at the upper half-plane, namely, at α=αm(Imαm>Imk) with(25)K2αmJ1KmaY1Kmb-J1KmbY1Kma=0,Km=Kαm,m=0,1,2,….In order to provide regularity of the right hand side of (24) in the upper half of the α-plane, these poles must be eliminated by imposing that their residues are zero. This gives (26a)F˙+a,k=1πf0-ikg0a2-b2ab,(26b)F˙+a,αm=1πfm-iαmgmJ12Kma-J12KmbJ1KmaJ1Kmbfor m=0,1,2,…. Here, fm and gm’s are the expansion coefficients of the functions fρ and gρ, respectively, which may be represented trough the following complete sets of orthogonal functions(27)fρgρ=2πbf0g0+∑m=1∞fmgmJ0KmρKmY1Kmb-KmJ1KmbY0Kmρ.Using the continuity relation (4i) together with (24) and taking into account (13) give(28)F˙+a,αK2αJ1KbπMαJ1Ka+a2F-a,α=121K2αMα∫abft-iαgtJ0KtKY1Kb-KJ1KbY0Kttdt-a2eikliα+k,F˙+a,αK2αNα+a2F-a,α=bπaf0-iαg0k2-α2+1π∑m=1∞fm-iαgmαm2-α2J1KmbJ1Kma-a2eikliα+kwhich is the Wiener-Hopf equation to be solved through classical procedures. Here, Nα stands for(29)Nα=J1KbπMαJ1Ka.The final solution of the W-H equation is determined to be(30)F˙+a,αk+αN+α=-aπbf0+ikg0k+αN+k+1π∑m=1∞fm+iαmgm2αmαm+αJ1KmbJ1Kmak+αmN+αm-kaeikliα+kN+k,where N±α are the split functions resulting from the Wiener-Hopf factorization of Nα as(31)Nα=N+αN-α.Their explicit expressions are given in [12] as(32)N+α=πJ1kaJ1kbJ1kaY1kb-J1kbY1ka-1/2e-αχ∏n=0∞1+α/k2-jn/b21+α/k2-jn/a21+α/αn.Here jn’s are the roots of the Bessel function of the first kind (33a)J1jn=0,n=0,1,…,(33b)χ=iπbln⁡b-aln⁡a-b-aln⁡b-awith(33c)N-α=N+-α.


## 2.2. Determination of the Unknown Coefficients

The field in the regiona<ρ<b, z∈0,l can be expressed is terms of the waveguide normal modes as(34)u21ρ,z=∑n=0∞aneiβnz-Pne-iβnzJ0ξnρ-RnY0ξnρwith(35a)Pn=ikη3+iβnikη3-iβn,(35b)Rn=ikη1J0ξna-ξnJ1ξnaikη1Y0ξna-ξnY1ξna=ikη2J0ξnb+ξnJ1ξnbikη2Y0ξnb+ξnY1ξnb.In (34), ξn’s are the roots of the following equation:(36)ikη1J0ξna-ξnJ1ξnaikη1Y0ξna-ξnY1ξna-ikη2J0ξnb+ξnJ1ξnbikη2Y0ξnb+ξnY1ξnb=0while βn’s are defined as(37)βn=k2-ξn2.Taking into account (27) and (34), the continuity relations (4g) and (4h) can be written in the following form:(38)2πbf0+iαg0+∑m=1∞fm+iαgmJ0KmρKmY1Kmb-KmJ1KmbY0Kmρ=i∑n=0∞anα+βneiβnl-Pnα-βne-iβnlJ0ξnρ-RnY0ξnρ.Multiplying (38) by 2ρ/πb and J0(Kmρ)KmY1(Kmb)-KmJ1(Kmb)Y0(Kmρ)ρ, respectively, and then integrating over ρ from a to b read(39a)f0+iαg0=iS0∑n=0∞anα+βneiβnl-Pnα-βne-iβnl△0n,(39b)fm+iαgm=iSm∑n=0∞anα+βneiβnl-Pnα-βne-iβnl△mnwith (40a)△0n=2πξnabJ1ξna-RnY1ξna-J1ξnb-RnY1ξnb,(40b)△mn=2ξnπξn2-Km2J1KmbJ1KmaJ1ξna-RnY1ξna-J1ξnb-RnY1ξnb,where S0 and Sm stand for(41a)S0=2π2a2-b2b2,(41b)Sm=2π2J12Kma-J12KmbJ12Kma.Using the W-H solution (30) together with (26a) and (26b) we obtain a set of linear algebraic equations in terms of the unknown coefficients fm and gm.(42a)π2baf0-ikg0S0N+k=bπaf0+ikg0N+k+kπ∑m=1∞fm+iαmgmαmN+αmJ1KmbJ1Kma-kaeikliN+k,(42b)π2fr-iαrgrJ1KraJ1KrbSrN+αr=bπaf0+ikg0N+k+1πk+αr∑m=1∞fm+iαmgm2αmαm+αrJ1KmbJ1Kmak+αmN+αm-kaeikliN+k,and taking into account (39a) and (39b) we obtain now a set of equations to determine the unknown expansion coefficient an as(43a)-iπ2baN+k∑n=0∞ank-βneiβnl-Pnk+βne-iβnl△0n=bπa1N+kiS0∑n=0∞ank+βneiβnl-Pnk-βne-iβnl△0n+ikπ∑n=0∞∑m=1∞anαm+βneiβnl-Pnαm-βne-iβnl△mnαmN+αmJ1KmbSmJ1Kma-kaeikliN+k,(43b)-iπ2J1KraJ1KrbN+αr∑n=0∞anαr-βneiβnl-Pnαr+βne-iβnl△rn=bπa1N+kiS0∑n=0∞ank+βneiβnl-Pnk-βne-iβnl△0n+iπ∑n=0∞∑m=1∞anαm+βneiβnl-Pnαm-βne-iβnlk+αr△mn2αmαm+αrJ1KmbSmJ1Kmak+αmN+αm-kaeikliN+k.


## 2.3. Reflected and Transmitted Fields

According to (8a), the scattered field in the region 0<ρ<a,thatis,u1(ρ,z), can be obtained by taking the inverse Fourier transform of Fρ,α. By considering (13) we write(44)u1ρ,z=-12π∫LF˙+a,αJ0KρKαJ1Kae-iαz-ldα.The evaluation of this integral for z<l and z>l will give us the reflected wave propagating backward in the inner cylinder and the transmitted wave, respectively.Forz<l, the integral is calculated by closing the contour in the upper half-plane and evaluating the residues contributions from the simple poles occurring at the zeros of J1Ka lying in the upper α-half-plane, namely, at Ka=jn. The reflection coefficient R of the fundamental mode is defined as the complex coefficient multiplying the travelling wave term exp-ikz and is computed from the contribution of the first pole at α=k. The result is(45)R=-ei2klN+k2-iπf0+ikg0kbN+k2eikl+iπeiklaN+k∑m=1∞fm+iαmgmαmN+αmJ1KmbJ1Kma.The first term is the reflection coefficient related to the case where a semi-infinite rigid duct is inserted axially into a larger rigid tube of infinite length [12] whereas the second one is the correction term involving the effect of the impedances of the annular region and the overlap length.Similarly, the transmission coefficientT of the fundamental mode which is defined as the complex coefficient of exp⁡ikz is obtained by evaluating the integral in (44) for z>l. This integral is now computed by closing the contour in the lower half of the complex α-plane. The pole of interest is at α=-k whose contribution gives(46a)-1+Teikz+Oeik2-jn/b2zwith(46b)T=a2b2+ie-iklπkaba-abf0+ikg0.The first term in (46a) cancels out the incident wave in the region ρ<a,  z>l, while the second is the transmission coefficient of the fundamental mode.


## 3. Results and Discussion

In this section in order to show the effects of the parameters like the length of the extended inletl and the surface admittance η1,2,3 on the transmitted field, some numerical results showing variation of the transmission coefficient T with different parameters are presented. In all numerical calculations the solution of the infinite system of algebraic equations is obtained by truncating the infinite series at N=5, since the transmission coefficient becomes insensitive for N>5. We also limit ourselves with only imaginary values of surface admittance η1,2 for simplicity.In Figures3 and 4, while the admittance X2>0 of the lateral wall of the expanding duct increases the transmitted field is ascending until some value of X2; then it starts to attenuate gradually. But for negative values of X2 the attenuation is more visible especially around -0.5<X2<0. For different values of X1 not much but some decrease in the transmitted field is observed.Figure 3
Transmission coefficientT versus the surface admittance η2=iX2(X2>0) for different values of η1=iX1.Figure 4
Transmission coefficientT versus the surface admittance η2=iX2(X2<0) for different values of η1=iX1.In Figure5, an oscillatory behaviour is seen for increasing values of the extended inlet length kl, but this behaviour is broken for negative values of X1 and X2. From Figure 6 it is observed that the transmission does not alter as the real part of η3 increased. But for small positive values of Reη3 imaginary part Im(η3) becomes effective. The most reduction on the sound transmission is seen for the negative value of Im(η3).Figure 5
Transmission coefficientT versus the extended inlet length kl for different values of η1=iX1 and η2=iX2.Figure 6
Transmission coefficientT versus the real part of η3 for different values of Im(η3).Figure7 shows an excellent agreement between the present paper (for the case of extended inlet length kl→0) and the previous study [9] of the author (for the case of expansion chamber length kl→∞ and surface admittance is taken to be zero). In this comparison, transmission coefficient is calculated as though it is in a rigid-walled duct with sudden area expansion (without extended inlet).Figure 7
Transmission coefficientT versus the expansion chamber radius kb.


## 4. Conclusions

This paper examines the transmission of sound waves in an extended tube resonator whose walls in overlapping region, where extended inlet and expanding duct walls overlap, are treated by acoustically absorbing materials of finite length. In the present work the lined region of the inner surface is assumed to be finite which makes the problem more complicated. To overcome the additional difficulty caused by the impedance discontinuity a hybrid method of formulation consisting of expressing the total field in terms of complete sets of orthogonal waveguide modes where available and using the Fourier transform elsewhere is adopted. The mixed boundary value problem is reduced to a Wiener-Hopf equation whose solution involves infinitely many expansion coefficients satisfying an infinite system of linear algebraic equations. These equations are solved numerically and the effects of various parameters on transmitted field such as the extended inlet length and the surface admittance of the lined section are displayed graphically. As a future work a similar problem now with an extended outlet will be studied following the same method used here.



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*Source: 9485163-2016-11-28.xml*