chap.4 duct acoustics

Aeroacoustics 2008 - 1 -

Chap.4 Duct Acoustics

Duct Acoustics Plane wave

• A sound propagation in pipes with different cross-sectional area • If the wavelength of sound is large in comparison with the

diameter of the pipe the sound propagates as an one-dimensional wave ( λ>>d → 1-d wave)

( ) ( )

( ) 0in 0in

/

//

>=

<+=′−

+−

xTexeRIep

cxti

cxticxti

ω

ωω

0=x

0>x0<x

IR

T1A Area 2A Area

투과반사입사 :,:, : TRI


Transmission & Reflection of Plane Waves

Duct Acoustics • The mass flux into the junction must equal the mass flux out

• The velocity must equal at both sides of the junction

• Energy flux)in = Energy flux)out

• The pressure of both sides of junction is continuous

220110 uAuA ρρ =

( ) Tc

ARIc

A

0

2

0

1

ρρ=−

TRI =+

2'221

'11 upAupA =



Duct Acoustics • The amplitudes of other wave, R and T ,are can be solve from

above the relations

• The transmission loss , LT is symmetric in A1 and A2

( )

+=

=

=

21

221

1022

21

10

10

4log10 log10

power dtransmittepowerincident log10

AAAA

TAIA

LT

IAA

ATIAAAAR

21

1

21

21 2 ,+

=+−

=



Duct Acoustics A single expansion-chamber ‘silencer’

• The simple muffler that is a used in car ‘silencer’ consists of inlet and outlet pipes with cross-sectional area A1, and expansion chamber between them of cross-sectional area A2 and length l

0=x lx =

0<x

T

1A Area

2A AreaI

R

B

C

1A Area

l



Duct Acoustics • The first area change occurs at x=0 and the second occurs at x=l.

• The condition of continuity of mass flux,

• The condition of continuity of pressure

( ) ( )

( ) ( )

( ) xlTelxCeBe

xeRIep

cxti

cxticxti

cxticxti

<=

<<+=

<+=′

−

+−

+−

in 0in

0in

/

//

//

ω

ωω

ωω

( ) ( )( ) lxCeBeATeA

xCBARIAcliclicli =−=

=−=−−−− at

0at //

2/

1

21ωωω

lxCeBeTexI

cliclicli =+=

=+=+−−− at

0at CBR/// ωωω



Duct Acoustics • The algebraic equation when solved for R and T

• However, the simple ‘silencer’ does not reduce the total energy of sound in the system.

• Reducing the acoustic energy of transmitted wave → Increasing in the reflected wave • Sound absorbing material → reduce the acoustic energy by converting it into heat or vibration

cl

AA

AAicl

clIi

AA

AA

Rωω

ω

sin/cos2

sin

1

2

2

1

1

2

2

1

++

−

=

cl

AA

AAicl

IeTcli

ωω

ω

sin/cos2

2

1

2

2

1

/

++

=

222 ITR =+



Duct Acoustics • The transmission loss , LT is

• The transmission loss is maximum at frequencies for which

• The effect of expansion ratio

= 2

2

10log10TI

LT

sin411log10

2

1

2

2

110

−+=

cl

AA

AALT

ω

etc...2

5,2

3,2

i.e. 1/sinlc

lc

lccl πππωω =±=

1

2A

Am =

LT

(dB)

frequency

m increase



Duct Acoustics Note

• ‘Tuning’ for dominant frequencies of noise

• Theory work for only λ≫d “Low frequency wave only”

• High frequency waves behave like 3-D

• Also, the geometrical shape of the duct is not important (provided the area change occurs in a distance short in comparison with the wavelength



Duct Acoustics

Effect of expansion chamber ratio Effect of expansion chamber shape

Ref. “Theoretical and experimental investigation of mufflers with comments on engine-exhaust muffler design”, Davis et al. NACA 1192(1954)



Higher order modes

• As an illustration, the sound of frequency ω in a rigid walled duct of square cross-section with sides of length a is considered

• With substitution for p′ into the wave equation,

a

a

2x

1x

3x

( ) ( ) ( ) ( ) tiexhxgxftp ω321, =′ x

22

2

αω−=−

′′−′′

−=′′

chh

gg

ff



Higher order modes • Since a wall boundary condition is applied, function f is derived

like this

• Similarly function g is derived like this

• Finally, function h is derived to the propagation form

• The axial phase speed, cp=ω/kmn is now a function of the mode number and the propagation of a group of waves will cause them to disperse.

( ) ma

xmAxf integer somefor , cos 111

=

π

( ) naxnAxg integer somefor , cos 1

22

=π

( ) 333

xikmn

xikmn

mnmn eBeAxh += − ( )222

2

2

2

nmac

kmn +−=πω



Higher order modes • The pressure perturbation in the (m,n) mode has the form

• When kmn is real, the pressure perturbation equation represents that waves are propagating down the x3 axis with phase speed.

• When kmn is purely imaginary, i.e. exceeds the cut-off frequency, the strength of mode varies exponentially with distance along the pipe. Such disturbances are evanescent

( ) [ ] tixikmn

xikmn eeBeA

axn

axmtp mnmn ωππ

3321 coscos, +

=′ −x



Pipes of varying cross-section Wave equation

• If the pipe diameter is small in comparison with both the acoustic wavelength and the length scale over which the cross-sectional area change, most particle motions are longitudinal.

• Conservation of mass

• Linearized momentum equation is

• Modified wave equation

( )xAx

( )uAxt

A∂∂

−=∂∂

0ρρ

xp

tu

∂′∂

−=∂∂

0ρ

∂′∂

∂∂

=∂

′∂xpA

xtp

cA

2

2

2



Pipes of varying cross-section Application to the ‘exponential horn’

• Evaluation of the case of ‘exponential horn’ which cross-sectional area defined as, A(x)=A0eαx

• For such an area variation of wave equation simplifies to

• The pressure perturbation in sound waves of frequency ω then has the form

• Disturbance with ω > αc/2 propagates and the pressure but not the energy flux attenuates during propagation, while lower frequency modes are ‘cut-off’

xp

xp

tp

c ∂′∂

+∂

′∂=

∂′∂ α2

2

2

2

2

1

( ) ( ) ( ){ }kxtikxtix BeAeetxp +−− +=′ ωωα 2,



Normal transmission Physics at the interface

• When a sound wave crosses an interface between two different fluids some of the acoustic energy is usually reflected.

• There are two boundary conditions The pressure on the two sides of the boundary must be equal The particle velocities normal to the interface must be equal

( )0cxtiIe −ω

( )0 cxtieR +ω( )1cxtiTe −ω

00 , cρ 11 , cρ( )0 interface =x

ϖπλ c

fccT 2===

ϖπλ 0

02 c

=ϖπλ 1

12 c

=



Normal transmission • The pressure must be equal at the interface : I+R=T • The particle velocities normal to the interface must be equal

• The result pressure coefficients, R and T , are determined with I

• Velocity Transmission Coefficient :

• The energy flux of the incident wave per unit cross sectional area is equal to that of the reflected and transmitted waves

110000 cT

cR

cI

ρρρ=−

IccccR

+−

=0011

0011

ρρρρ I

cccT

+=

0011

112ρρ

ρ

( )( ) ( ) 00

2

112

0011

221

21

002

0011

220011

11

2

00

2 4c

Iccc

Icccc

Iccc

Tc

Rρρρρ

ρρρρ

ρρρρ

=+

++−

=+

0011

00

00

11 2//

ccc

cIcT

ρρρ

ρρ

+=


Normal transmission Reflection from a high and low impedance fluid

• A typical example is aerial sound waves incident onto a water surface. (ρ0c0 ≪ ρ1c1 )

• Velocity transmission coefficient

so, the transmission wave carries negligible energy


ITIR 2 , ==

02

0011

00 ≈+

=cc

cρρ

ρ



Normal transmission Reflection from a high and low impedance fluid

• In the opposite case, for sound in water incident onto a free surface with air, the reflected and transmitted waves are

• The acoustic energy is totally reflected

0 , =−= TIR



Sound propagation through walls Effect of a wall in transmission

• A sound wave normally incident on a plane material layer partitioning a fluid which has uniform acoustic properties, ρ0c0

• Some sound will be reflected from the layer and some will be transmitted through the wall



Sound propagation through walls • There are two boundary conditions that must be satisfied at all

times and points The velocity of the wall must be equal to wave of each side A pressure difference across the wall in order to provide the

force necessary to accelerate unit area of the surface of material

• By continuity the velocity of wall,

• The pressure difference is the net force of mass per unit area of the wall

( ) titi

ec

Tc

eRIu ωω

ρρ 0000

=−=

( ) titi ecTim

tumeTRI ωω

ρω

00

=∂∂

=−+( )

ti

ti

TepeRIp

ω

ω

=

+='2

'1



Sound propagation through walls • The result pressure coefficients, R and T , are determined with I

• Surface Impedance

• Energy transmitted

Imic

miR

+=

ωρω

002I

miccT

+=

ωρρ

00

00

22

( )( )( ) mic

TRIc

up

mimic

RIRIc

up

ωρρ

ωωρρ

+=+

=

−+

=−+

=

0000

'2

0000

'1

11

00

2

2220

20

20

20

00

2

44

cI

mcc

cT

ρωρρ

ρ +=



Sound propagation through walls • The transmission loss is dependent on the frequency ω.

• For high frequency(ωm≫ρ0c0), the sound waves mostly reflected

• For low frequency(ωm≪ρ0c0), the sound waves mostly travels through the wall with very little attenuation

• “Low frequency waves get through a massive wall easily, while high frequency waves are effectively stopped”

1

2220

20

20

20

10 44log10

−

+

=mc

cLT ωρρ



Sound propagation through walls Example) Attenuation by a wall

• Transmission loss

250 mkgm =

sec00 2410 mkgc =ρ

Hzffor 10=

1

2220

20

20

20

10 44log10

−

+

=mc

cLT ωρρ

= 2

2

10log10TI

LT

dBLT 12≈

kHzffor 1= dBLT 50≈

I

chap.4 duct acoustics

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