158th Meeting of the Acoustical Society of America San Antonio, Texas Acoustic radiation of a vibrating wall covered by a porous layer Transfer impedance concept and effect of compression Nicolas DAUCHEZ Supméca – Institut Supérieur de Mécanique de Paris, Saint Ouen, France Olivier DOUTRES, Jean-Michel GENEVAUX Laboratoire d’Acoustique UMR CNRS 6613 Université du Maine, Le Mans, France 29 october 2009
Context • Porous materials used in industrial applications ( automotive, aeronautics ,…) Introduction for noise reduction Part I Part II Part III Conclusion � sound absorption (trim panel, floor , …) � vibration damping (fuselage) � sound insulation (fuselage) 2
Context • Porous material attached to a vibrating structure Influence of a porous layer on the acoustic radiation of a plate ? Introduction Part I Method Part II Part III • Analytical model using transfert Conclusion impedance concept • Experimental validation � Porous layer impedance applied to a moving wall: Application to the radiation of a covered piston, Doutres, Dauchez, Genevaux, J. Acoust. Soc. Am. 121 (1), 2007 3
Introduction 1. Transfert impedance concept Introduction 2. Acoustic radiation efficiency Part I 2.1. Infinite plate Part II 2.2. Flat piston Part III 2.3. Circular plate Conclusion 3. Application to multilayer Conclusion 4
Problem to be solved Introduction Part I Part II Part III Ω ∇ + = p k p Conclusion in , ² ² 0 Source at ∂ p Γ + = A p B h r boundary at , ( ) ∂ n Dirichlet Neumann (imposed pressure) (imposed velocity) 5
Problem divided into 2 cases : Introduction Part I Part II Part III Conclusion a) Acoustic excitation : amplitude of reflected wave ? 6
Problem divided into 2 cases : Introduction Part I Part II Part III Conclusion a) Acoustic excitation : amplitude of reflected wave ? b) Vibratory excitation : amplitude of transmitted wave ? 7
Surface Impedance Z S Porous layer characterized by p 1 ( 0 ) Z S = v ( 0 ) Introduction R = 1 + − jkz jkz p z e R e ( ) Part I ( ) 1 Part II = jkz − − jkz v z e R e ( ) Z Part III 0 Conclusion The reflected wave is function of surface impedance Z S : − Z Z = ρ = Z c R S 0 with + Z Z 0 0 0 S 0 Z S is measured in a Kundt tube 8
Transfert impedance Z T Porous layer characterized by p T ( 0 ) 1 = Z T − v v ( 0 ) p Introduction Non-porous massless coating : Part I Part II Bulk modulus Z T = j ω Part III Thickness Conclusion The transmitted wave is function of transfert impedance Z T : Z = − = jkz p z T e T T Z v ( ) gives p + Z Z 0 T 0 Can Z T be measured in a Kundt tube ? Is Z T equivalent to Z S ? 9
A simple coating : Spring Static law: K − − = p v v ( 0 ) ( ) 0 p ω j K Introduction = v 0 Case a: Part I p p Part II ( 0 ) = = ω Z S K j / Part III v p Conclusion ( 0 ) = = ω Z K j / Case b: T − v v p Elastic layer K Bulk modulus = = = Z Z T S ω ω j j L 10
Adding a layer defined by its mass M s Dynamic law: K ω = − − j M v p v v ( 0 ) ( ) S p ω j K Introduction = v 0 Case a: Part I p p Part II ( 0 ) = = ω + ω Z j M K j / S s Part III v Conclusion K Z p ( 0 ) ⇒ = Z = Z 0 Case b: T ω − ω j Z j M v 0 s 0 ω → Z T ≠ Z S excepted when 0 : Elastic layer at low K Bulk modulus = = = Z Z frequency T S 11 ω ω j j L
Monophasic continuous layer Transfert matrix: − p a b p L ( 0 ) ( ) = − v c d v L ( 0 ) ( ) Z e , k kd jZ kd a b cos sin e = Introduction c d j Z kd kd / sin cos Part I e a Part II = ⇒ Z S = v Case a: 0 p Part III c Z p ( 0 ) ⇒ = Conclusion Z T = Z 0 Case b: − − a cZ v 0 1 0 kd ω → → → a c j Z T ≠ Z S excepted when 0 : 1 and Z e Z Monophasic Bulk modulus = = = Z Z j e layer at low T S ω kd j L 12 frequency
Thickness 20 mm Bulk modulus 140(1+j0.1) kPa Impedance curves 30 kg.m -3 Density Real part Imaginary part 14000 8000 Introduction 6000 12000 4000 Part I 10000 imaginary part of impedance 2000 real part of impedance 8000 Part II 0 6000 -2000 Part III 4000 -4000 2000 Conclusion -6000 0 -8000 -2000 -10000 -4000 -12000 2 3 4 2 3 4 10 10 10 10 10 10 Frequency (Hz) Frequency (Hz) Z T transfert impedance ≠ Z S surface impedance 13
Poroelastic layer: Surface impedance Z S Biot-Allard theory in 1D in the porous layer : 2 waves travelling in each direction Two waves in fluid medium Boundary conditions : Continuity of Stress and displacement Introduction s = f = v v ( 0 ) ( 0 ) 0 at x=0 Part I = − φ s + φ f v d v v ( ) ( 1 ) ( 0 ) ( 0 ) at x=d Part II σ f = − φ σ s = − − φ d p d d p d ( ) ( ) and ( ) ( 1 ) ( ) Part III Conclusion Z 2 ,k 2 1 2 p d ( ) A Z S = v(d) B v d ( ) E C F Excitation D p(d) 14 0 x d
Poroelastic layer: Transfert impedance Z T Biot-Allard theory in 1D in the porous layer : 2 waves travelling in each direction p d ( ) ⇒ = Z One waves in fluid medium v d 0 ( ) Boundary conditions : Continuity of Stress and displacement Introduction s = f = v v v Part I ( 0 ) ( 0 ) at x=0 p = − φ s + φ f Part II v d v v ( ) ( 1 ) ( 0 ) ( 0 ) at x=d Part III σ f = − φ σ s = − − φ d p d d p d ( ) ( ) and ( ) ( 1 ) ( ) Conclusion Z 2 ,k 2 1 2 Excitation A v P v(d) p d ( ) B = Z E T − v v d ( ) C p D 15 0 x d
Impedance curves Real part Imaginary part Introduction Part I Part II Part III Conclusion Z T transfert impedance ≠ Z S surface impedance 16
Boundary conditions at excitation interface Introduction Part I Part II Part III Conclusion Fluid-porous interface : Wall-porous interface : ω = = s f ω = − φ + φ v j u u s f v j u u / ( 0 ) / ( 1 ) p ⇒ The skeleton is much more excited in case b) 17
Relative skeleton velocity for both excitation at fluid-porous interface vibratory acoustical Introduction Part I Strong Part II influence of frame borne Part III No strain wave Conclusion in skeleton Skeleton almost at rest 18
Conclusion of Part I Z S Z T Introduction Part I Part II Part III Conclusion Z ≠ Z Excepted at low frequency, T S Z T can not be measured in a Kundt tube 19
1 impedance for each problem Introduction Part I Part II Part III Conclusion ∂ Z p + z = jk p 0 ( a ) 0 ∂ Z S ∂ Z p − = ωρ jk p j v 0 ( b ) p ∂ Z z 0 20 T
Introduction 1. Transfert impedance concept Introduction 2. Acoustic radiation efficiency Part I 2.1. Infinite plate Part II 2.2. Flat piston Part III 2.3. Circular plate Conclusion 3. Application to multilayer Conclusion 21
2. Acoustic radiation efficiency Radiated acoustic Π power σ = Π a R Introduction v Injected vibratory Part I power Part II 2 p ∫∫ 2 sin Part III Π = < > ϑ ϑ ϕ < >= I r d d I with: a ρ c 2 Conclusion S 0 0 � 2 w r ( ) ∫∫ Π = ρ c dS v 0 0 2 S 22
2.1. Acoustic radiation efficiency 2 2 T Z σ = = T Infinite plate at normal incidence (1D) : R + Z Z Z v 2 T p 0 0 Introduction Part I Part II (0 dB) Part III Conclusion with Z S with Z T no effect increase decrease 23
2.2. Acoustic radiation of a flat piston in semi-infinite field Calculation of the far field pressure: Rayleigh Integral Z T Z 0 Introduction Reflective Part I baffle with Part II Part III Conclusion For a flat piston of radius a in far field : Z 0 24
2.2. Acoustic radiation of a flat piston in semi-infinite field Measurement : Introduction Part I Part II Part III Conclusion 2 materials : A : polymer foam B : soft fibrous 25
2.2. Acoustic radiation of a flat piston in semi-infinite field A : polymer foam with porous Introduction Part I Part II Part III Conclusion 26
2.2. Acoustic radiation of a flat piston in semi-infinite field B : soft fibrous Introduction with porous Part I Part II Part III Conclusion 27
Introduction 1. Transfert impedance concept Introduction 2. Acoustic radiation efficiency Part I 2.1. Infinite plate Part II 2.2. Flat piston Part III 2.3. Circular plate Conclusion 3. Application to multilayer Conclusion 28
2.3. Radiation efficiency of a circular plate ρ ω h 2 ∇ − = w w 4 Plate equation 0 D β J a ( ) = β − β w r J r n I r 0 0 ( ) ( ) ( ) Introduction Axisymetrical modes n n n β 0 0 I a 0 0 ( ) n 0 0 Part I F w r w r ( ) ( ) ∑ Part II = w r Modal synthesis n s n ( ) β − β π Λ D a 4 4 2 ( ) Part III n n n Conclusion Rayleigh ρ h integration β = ω 4 2 with: 0 n n D Acoustic Vibratory power power 29
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