Chap.4 Duct Acoustics - Seoul National...

Aeroacoustics 2019 - 1 -

Chap.4 Duct Acoustics

About “Duct acoustics”

Industrial use

Efficiency

Predictability

Complexity

Ducts are able to efficiently transmit sound over large distances!



Wave dynamics

vi i

i i

Ht x x

∂ ∂∂+ − =

∂ ∂ ∂

( )

i

i i j ij

i

uu u p

e p u

ρρ δ

ρ

= + +

0

vi ij

k ik iu qτ

τ

= +

iue

ρ =

1. Navier-Stokes equations

Basic assumption for homogeneous linear formulation• Waves are free of non-linear effects and propagate with speed of sound

• Temperature change (heat transfer) is negligible

• Viscous effects are sufficiently diminished in free space

• No external force (or source)

0iq =

0 ′= +

0ijτ =

0H =



Wave dynamics

0t x x

∂ ∂ ∂+ + =

∂ ∂ ∂

0 0

0 0

0

0 0

u uu u p

u vu p p u

ρ ρρ

γ

′ ′+ ′ ′+ = ′

′ ′+

uvp

ρ′ ′ =

′ ′

2. Linearized Euler equations (2-D)

• Fourier-Laplace transformation

0 0 0 0

0 0 0

0 0 0

0 0 0 0

00 0

00 00

x y x y

x y x

x y y

x y x y

k u k v k kk u k v k u

k u k v k vk p k p k u k v p

ω ρ ρ ρω ρ

ω ργ γ ω

− − − − ′ − − − ′ = − − − ′ − − − − ′

( )( ) ( )( ) ( )

1 2 0 0

1/22 23 0 0

1/22 24 0 0

x y

x y x y

x y x y

k u k v

k u k v c k k

k u k v c k k

λ λ ω

λ ω

λ ω

= = − −

= − − + +

= − − − +

0 0

0

0 0

0 0

v vv v

v u pv p p v

ρ ρ

ργ

′ ′+ ′ = ′ ′+

′ ′+

• Dispersion relation (Relationship between frequency and wavenumber)

( )( ) ( )

1 2 0 0

2 2 2 23 4 0 0

0

0

x y

x y x y

k u k v

k u k v c k k

λ λ ω

λ λ ω

= = − − =

= − − − + =

: Entropy and vorticity waves

: Acoustic waves2 2 2 0c kω − = : Acoustic waves w/o mean flow c

kω

= :phase velocity



Wave dynamics

3. Wave Equations

• Momentum conservation

0 0 0u ut x xρ ρρ′ ′ ′∂ ∂ ∂+ + =

∂ ∂ ∂ 00

1 0u u put x xρ′ ′ ′∂ ∂ ∂+ + =

∂ ∂ ∂

• Mass conservation

• Assuming there is no mean flow:2 0p

t xρ′ ′∂ ∂

− =∂ ∂

22

0 0 0 2

1( ) ( )2

dp d pp pd d

ρ ρ ρ ρρ ρ

= + − + − +

• Assuming pressure is a function of density alone,

20 0( ) dpp p p c

dρ ρ ρ

ρ′ ′− = − ⇔ =

2 22 0p c p

t′∂ ′− ∇ =

∂: Only works when there is no mean flow!

2 22 0c p

tρ′∂ ′− ∇ =∂

,in 3D:



Duct acoustic problems• Ventilation ducts• Exhaust ducts• Automotive silencers• Shallow water channels and surface ducts in deep water• Turbofan engine ducts

How to Solve?

• Modelling to simple geometry (circular or rectangular ducts)• Appropriate assumptions (harmonic solution, no mean flow, linearization and so on..)• Dispersion relations• Mathematical functions• Boundary conditions

Analytical approach

ExperimentNumerical simulation



Cylindrical duct• Modelling to simple geometry: simple cylindrical duct with hard wall

• Appropriate assumptions: assume harmonic separable solution

rθ

x

2 22 0p c p

t∂

− ∇ =∂

2 22

2 2 2

1 1rx r r r r∂ ∂ ∂ ∂∂ ∂ ∂ ∂θ

∇ = + +

, Cylindrical Laplacian:


• From dispersion relation of acoustic waves,


• assuming harmonic separable solution,

• If we substitute derivatives over functions into some independent variables,

2 2 2 2 2 2 2( ) 0r c k m rω′′ ′+ + − − =r R rR R

( )2 2 2 2 0mnc kω µ− + =

2 2 2 0c kω − =

2R nR′′= −

2 2 2, mn m nk kµ = +

2 2 2 2( ) 0mnr mµ′′ ′+ + − =r R rR R

( )( ) i kx m tp R r e θ ω+ −=Thus, general solution is implied as



mR(r)= J (r)

2 2 2 2( ) 0mnr mµ′′ ′+ + − =r R rR R

• Final equation resembles Bessel function

αy = J (x)

α

x• Neumann B.C. at the wall

( )

r a

R rr =

dd

( ) 0m mnaµ′ =J

mnr rµ=2 2 2( ) 0r m′′ ′+ + − = r R rR R

• General solution of first derivative of Bessel function:

[Mode shapes of Bessel function]

( ) 0m mnα′ =J

mnα

mnmn a

αµ =


• From dispersion relation axial wavenumber become,2

22 mnk

cω µ= −

22

2mn

mnkc a

αω = −

• General Solution of Duct acoustics

Cut-off frequency


pmn

ckω

= :Axial phase speed



• Modes and mode shape functions

In seeking a solution for the pressure field in a duct we obtained, not a single unique solution, but a family of solutions. The general solution is a linear superposition of these ‘eigenfunction’ solutions:

( ) ( )∑∑∞

=

−∞

−∞=

Φ=1

,,,n

kximnmn

m

mnerpxrp αθθ

( ) ( )Φmn rmn m rmnimk r J k r e= θ

The resultant acoustic pressure in the duct is the weighted sum of fixed pressure patterns across the duct cross section. Each of which propagate axially along the duct at their characteristic axial phase speeds.



• Mode shape functions of cylindrical duct

The first roots of the derivative of the Bessel function

• Example

Consider a duct with radius a=0.5m, c=340m/s, sound frequency=3000rpm.Which modes will propagate?

, ( ) 0mn m mnα α′∀ =J


Rectangular duct

• 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



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

• Similarly function g is derived

• 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



• 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




• Modes shape functions of rectangular duct

( ) [ ] tixikmn

xikmn eeBeA

axn

axmtp mnmn ωππ

3321 coscos, +

=′ −x

0, 0m n= = 0, 1m n= = 1, 1m n= = 1, 2m n= =

( )222

2

2

2

nmac

kmn +−=πω

Chap.4 Duct Acoustics - Seoul National...

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