chapter 8 sound in waveguides

Sound in Waveguides

Slides to accompany lectures in ME 610:

Engineering Acoustics© 2002 by A. F. Seybert

Department of Mechanical EngineeringUniversity of Kentucky

Lexington, KY 40506-0108Tel: 859-257-6336 x 80645

Fax: [email protected]

Dept. of Mech. EngineeringUniversity of Kentucky

8 - 2

Chapter 8: Sound in Waveguides

ME 610 Engineering Acoustics

Introduction

1. Walls are assumed rigid2. Waveguide can bend gradually (R > λ)3. Flow is negligible (M<<1)4. Sound propagation is planar (λ > 1.7d for circular

waveguides or λ > 2w, 2h for rectangular waveguides).

Plane waves

Oblique waves

Plane waves( )( )( )

Spherical waves become planar within a few diameters of discontinuities and bends if λ > 1.7d


8 - 3



Applications/Solutions

1. Intake/Exhaust systems of engines, compressors2. Refrigeration, hydraulic, steam, and water lines3. HVAC systems for buildings, vehicles4. Industrial processes, exhaust stacks

Noise Control Solutions:1. Apply sound absorbing material to walls of ductwork (very

efficient for attenuating oblique waves and plane waves at high frequencies).

2. Install a muffler or silencer.• Dissipative: sound energy converted to heat by

absorption• Reactive: sound cancelled by reflection


8 - 4



Types of Mufflers

Dissipative muffler (silencer):

Sound is attenuated due to absorption

(conversion to heat)

Sound absorbing material(e.g., duct liner)

Duct or pipe


8 - 5



Types of Mufflers (cont.)

Reactive muffler:

Sound is attenuated by reflection and “cancellation” of sound waves


8 - 6



Types of Mufflers (cont.)

Combination reactive and dissipative muffler:

Sound is attenuated by reflection and “cancellation” of sound waves +

absorption of sound

Sound absorbing material

Perforated tubes


8 - 7



Bio-Acoustic Applications


8 - 8



Zeffort variableflow variable

W = effort variable x flow variable

≡ SYSTEM

Flow variable

Effort variable

+

_

The Concept of Impedance

Generalized impedance of a system and instantaneous power supplied W :

pAu AS = p2 AS /zp/uParticle Velocity (u)Pressure (p)Acoustic

T/Q’

P/Q

f/v

e/i

Impedance Z*

PAQ = P2/ZVolume Flow Rate (Q)Pressure (P)Fluid

TAQ’ = T2/ZHeat Flow Rate/Deg (Q’)Temperature (T)Thermal

fAv = f2/ZVelocity (v)Force (f)Mechanical

eAi = e2/ZCurrent (i)Voltage (e)Electrical

Power supplied W*

Flow VariableEffort VariableSystem Type

* For acoustic systems multiply by area S to get W; z is the specific acoustic impedance.


8 - 9



Specific Acoustic Impedance

The specific acoustic impedance z is the ratio of the complex sound pressure at a point in the fluid to that of the associated particle velocity.

units: (for Lord Rayleigh)Pa

m sPa s m rayl

/= ⋅ ⋅ ≡−1

z p u r jx= = +~ ~ ~p

~u

wavefront


8 - 10



( )~( ) ~ ~( ) ( )~

~ ~p x Ae u x j jk Ae

z p u c

jkxo

jkx

o

= = − −

= =

− −1 ω ρ

ρ

For a plane traveling wave in the +x direction (B=0):

Specific Acoustic Impedance Ratio

Define the dimensionless specific acoustic impedance ratio (or normalized specific acoustic impedance) as:

z z c r c j x c r jxo o o/ / /= = + = +ρ ρ ρ

(this is called the characteristic impedance)


8 - 11



1-D Wave Equation - Review

∇ = + + =22

2

2

2

2

2 2

2

2

1p

px

py

pz c

pt

∂∂

∂∂

∂∂

∂∂

Consider harmonic wave propagation in only the x direction:

p p x t p x e j t= =( , ) ~( ) ω ~( ) ~( )p x p x e j= φ

Plane waves

Rigid walls

x


8 - 12



Where it is understood that the actual sound pressure is the real part:

( )Re{~( ) } ~( ) cosp x e p x tj tω ω φ= +

= complex amplitude of the sound pressure at x.~( )p x

Substituting into the one-dimensional wave equation:

d p xdx

k p x kc

2

22 0

~( ) ~( )+ = =ω

1-D Wave Equation - Review


8 - 13



Plane Harmonic Waves in One-Dimension

d p xdx

k p x kc

2

22 0

~( ) ~( )+ = =ω

~, ~A B are determined by the boundary conditions at the two ends of the waveguide.

~eA jkx−~eB jkx

( ) ),(),(~~),(

~~)(~

txptxpeeBeAtxp

e

eBeAxp

tjjkxjkx

tj

jkxjkx

−+−

−

+=+=

+

+=

ω

ω :waves traveling x- and x represent these with


8 - 14



Boundary Conditions for 1-D Waves

~~)(~ eBeAxp jkxjkx− +=x = 0 x = L

Source termination

Source B.C.’s:• u(0) specified (positive displacement pump)• p(0) specified (SPL output of a fan)

Common termination B.C.’s:• open end (tailpipe)• closed end (hydraulic line with rigid termination)• anechoic end (long steam line)• absorbing end (porous material)


8 - 15



Example – Waveguide Closed at x = L

x = 0 x = L

Rigid termination u(L)=0

Source u(0)= Uo

~( )~ ~

~( ) ( ~ ~ )

~( )sin

cos ( )

p x Ae Be

u xj

jk Ae BeU x

x L

p xjz U

kLk L x

jkx jkx

o

jkx jkx o

o o

= +

= − − + ==

=

= −

−

−

−1 0

0ω ρ


8 - 16



Waveguide having Arbitrary Termination Impedance

z p ut = ~( ) ~( )0 0

x=0x=-L

z(-L)

( )( )

( )( )

′ − =+

−=

+

−

′ = ′ =+

−=

′ −′ +

′ − =′ ++ ′

−

−

−

−z L

Ae Be

Ae Be

B A e

B A e

z zB A

B AB A

zz

z Lz j kL

jz kL

jkL jkL

jkL jkL

j kL

j kL

tt

t

t

t

( )~ ~~ ~

~ ~

~ ~

( )~ ~

~ ~~ ~

( )tantan

1

1

01

1

11

1

2

2


8 - 17



Sound Pressure at any Position x

~( )p 0

x=0x=-L

~( )p L−

~( )p x

( )[ ]( )

( )

~( ) ~ ~ ~ ~ ~

~( ) ~( )~ ~

~ ~

~ ~

p x Ae Be A e B A e

p x p Le B A e

e B A e

B Azz

jkx jkx jkx jkx

jkx jkx

jkL jkL

t

t

= + = +

= −+

+

=′ −′ +

− −

−

−

11


8 - 18



Some Common Termination Impedances

A anechoic

closed

Open(unbaffled)

Open(baffled)

z ct o= ρ

zt = ∞

[ ]z c ka j kat o= +ρ π( ) ( )2 2 8 3

[ ]z c ka j kat o= +ρ ( ) ( . )2 0 62a

2a

(ka<<1)

(ka<<1)


8 - 19



Example – Open Unbaffled Pipe

0 100 200 300 400 5000

0.05

0.1

0.15

0.2

0.25

0.3

0.35Open pipe termination impedance ratio

Imag.

Real

Frequency (Hz)

Impe

danc

e ra

tio(unbaffled pipe)

(for a = 0.05 m, c = 343 m/s)


8 - 20



Sound Power Injected into Waveguide

W S p L u L

S p Lp Lz L

S p Lz L

Sp L

z Lz L

Sc

p L

z Lz L

in

o

= − −

= −−−

= −−

=−

−− =

−

′ −′ −

1212

12

1

12

12

2

2

2

2

2

Re{ ~( )~ * ( )}

Re{ ~( )~ * ( )

* ( )} ~( ) Re{

* ( )}

~( )

( )Re{ ( )}

~( )

( )Re{ ( )}

ρ

z p ut = ~( ) ~( )0 0

x=0x=-L

z L p L u L( ) ~( ) ~( )− = − −


8 - 21



Impedance Discontinuities in Ducts

• Any change in the impedance seen by a wave generates a reflected wave

• Less energy is transmitted due to the reflection• Mufflers maximize this effect

Impedance discontinuity

AB C


8 - 22



Example: Area Change

Continuity of pressure (momentum) and volume velocity (mass) at x = 0:

( )( ) ( )

( )2

2

1

2

2

2112

21

14

22

12

12

mm

cAScCS

SSmm

mSS

AC

CcSBAcS

CBA

oo

oo

+=

=

=+

=+

=

=−

=+

ρρτ

ρρ

~~

~~

~~~

~~~

jkxeA −~jkxeB~

jkxeC −~

x = 0

S1 S2

The last quantity is the power transmission coefficient

Anechoic: z = ρoc


8 - 23



Attenuation of Sound by an Area Change

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

S1/S2

Tra

nsm

issi

on

an

d R

efle

ctio

n C

oef

fici

ents

Power Transmission CoefficientPower Reflection Coefficient


8 - 24



Example: Expansion Chamber Muffler

jkxeA −~jkxeB~

jkxeC −~

x = 0

S1 S2jkxeD~

jkxeE −~S1

x = L

kLmmkLAB 2212

2

44

1sin)(cos

~~−++

=−=τ

Apply continuity of pressure and volume velocity at x = 0 and x = L; solve for power reflection coefficient, then subtract from unity to get the power transmission coefficient:

The Transmission Loss (TL) in decibels (dB) is defined:

)(loglog dBWWTL ti τ11010 1010 ==


8 - 25



TL of an Expansion Chamber Muffler

0

5

10

15

0 1 2 3 4 5 6 7 8 9 10

kL = 2π(L/λ)

TL (d

B)

Area Ratio = 2 Area Ratio = 5 Area Ratio = 10


8 - 26



Side Branch (T-Junction) Muffler

Volume velocity continuity

( )βββ SzZZZZ

QQQ

cbj

cbj

=+=

+=

111

~~~

S

Sb

x = 0

A

Bjp~

Assumption: that the frequency is sufficiently low so that the junction sound pressure is constant everywhere in the junction

jp~

==

j

jjj z

pSuSQ

~~~

==

c

jcc z

pSuSQ

~~~

==

b

jbbbb z

pSuSQ

~~~

impedance. acoustical the is that recall

as defined impedance acoustical the is Here

specificz

SzQpuSpZ

Z

=== ~~~

Pressure continuity:

jxpBAxp ~~~)(~

0=+=

=


8 - 27



Side Branch (T-Junction) Muffler

1/

//

///

+=

=

===

+=

b

bj

oc

o

cc

o

bb

o

jj

cb

cbj

ZZ

Z

ScZSc

ZZ

ScZ

ZSc

ZZ

ZZZZ

Z

(anechoic) let simplicity for

follows as impedances normalized define

ρρρρ

Zj

Zb

Zc

{ } { }2

/

/

22

2

/

2

//

/

124

*1Re~~21*1Re~

21

121

~

~

121

1

1~

~

+==

+==

+==

+−

=+

−=

b

b

i

tc

ccjt

c

br

bj

j

ZZ

WW

ZBAZpW

Z

ZA

B

ZZ

Z

A

B

τ

α

tcoefficien ontransmissi power sound the

is into dtransmitte power sound The

tcoefficien reflection power sound the so

tcoefficien reflection pressure the recall


8 - 28



The Helmholtz Resonator

Mx Kx p S x j u xuj

j MK

u p S

zp

uj

SM

K KM

cSLV

j b bb

b j b

bj

b b

b

on

&& &&+ = = =

−

=

= =

−

→ = = =

when

ωω

ωω

ωω

ω ω1

0

Kc SV

M S L

o b

o

o b

=

=

ρ

ρ

2 2

∆F = pjSb

x

Vo

Sb

L


8 - 29



TL of a Helmholtz Resonator

Vo = 0.001 m3

L = 25 mmSb = 2 x 10-4 m2

S = 8 x 10-4 m2

fn = 154 Hz

Anechoic termination

0

5

10

15

20

25

30

0 100 200 300

Frequency (Hz)

TL (d

B)

TL dB

cL

SSn

b

n

( ) log= +

−

10 1

12

110

2

2

2

ωω ω

ωω

90 Hz


8 - 30



Network Interpretation: Electrical to Acoustical

zu

p

source loadZi

e

source load

Electrical System Acoustical System


8 - 31



Source u Any acoustic system

u

p (sound pressure

reaction)

zt

zpu

r jx= = + zpu

r xtt

tt t= = +

Input or load impedance

Termination impedance

Acoustic System Components


8 - 32



{ }

{ } = { }

=

WS

p f u fS

z(f)u f u f

S u fz f S r f u f

=

=

2 2

212

22

Re ~( )~ * ( ) Re ~( )~ * ( )

~( )Re ( ) ( ) ~( )

Sound Power and Impedance

zu

p

source load

W


8 - 33



Performance Measures – Transmission Loss

Transmission loss (TL) of the muffler:

Wi

Wr

Wt AnechoicTerminationMuffler

TL dB LogWW

i

t

( ) = 10 10


8 - 34



Sound Intensity and Sound Power

{ }I p t u t p f u ft

= =( ) ( ) Re ~( )~ * ( )12

{ }W IdS I SS

p f u fS

= = ⋅ =∫ 2Re ~( )~ * ( )

power worktime

(force distance)time N m

sJs W

powerarea intensity = pressure velocity

W m2

= =⋅ ⋅

= =

= ⋅

Time-average active intensity:

u

pS

W


8 - 35



Performance Measures – Insertion Loss

IL (dB) = SPL1 – SPL2

Insertion loss depends on : • TL of muffler• Lengths of pipes• Termination (baffled vs. unbaffled)• Source impedance

Muffler

SPL1

SPL2


8 - 36



Example – TL and IL

24”12”

12”2” 6”Source

-50

-40

-30

-20

-10

0

10

20

0 200 400 600 800 1000

Frequency (Hz)

TL

an

d IL

(d

B)

Insertion Loss

Transmission Loss

Pipe resonances

Inlet Pipe Outlet Pipe

Expansion Chamber Muffler


8 - 37



Example – Branch Impedance

z zt

zB

zz z

z zB t

B t

=+

Can we make zB zero?

zt u

p

zB

z

(any system)


8 - 38



The Quarter-Wave Resonator

zB

L

z j c L c L c n nn c

L

fncL

Lnc

fn

B o= − = = =

=

= = =

ρ ω ω π

ωπ

λ

cot( / ) / / , , ...0 2 13 5

2

4 4 4

when

or

(produces a sound pressure node at junction)


8 - 39



Aside: Mechanical and Acoustic Impedance

M C

K x, u

~Fe j tω

( )( )

Mx Cx Kx Fe

x ue x j ue x u j e

Z F u C j M K

j t

j t j t j t

m

&& & ~

& ~ && ~ ~

~ ~

+ + =

= = =

= = + −

ω

ω ω ωω ω

ω ω

Because the applied force is split among all three elements, while all experience the same velocity, we say the three elements are in series (element impedances add).


8 - 40



Aside: Mechanical and Acoustic Impedance

Average power supplied by the force to the system:

W F t u t Fu uZ u

W u Z C u

W C F Z

m m

m m

m m

= = =

= =

=

( ) ( ) Re{~~*} Re{(~ )~*}

~ Re{ } ~

~

12

12

12

12

12

2 2

2 2or

10 1 10 2 10 3 0

0.00

0.0

0.015

C

3C

Frequency (1/s)

Ave

rage

Pow

er (

W)

Resonance occurs when the average power is maximum:

Im{ } ( )Z M K

KM

m

n

= − =

= =

ω ω

ω ω

0


8 - 41



Example

An engine running at 2400 rpm has a fundamental “firing” frequency of 80 Hz. An exhaust pipe 2 m long and 0.1 m diameter is connected to the exhaust manifold. The sound pressure amplitude at the entrance to the exhaust pipe is 100 Pa at 80 Hz. How much sound power is injected into the exhaust pipe and what is the sound pressure amplitude at the termination of the pipe?

100 Pa

p = ?


8 - 42



Example (cont.)

x=0x=-L=-2 m

p(-L)=100 Pa@ 80 Hz

Open end(unbaffled)p(0) ?

a=0.05 m

ka

kL S

z ka j ka jt

= ⋅ ⋅ = ×

= ⋅ ⋅ = = = ×

′ = + = × + ×

−

−

− −

( ) . / .

( ) / . ( . ) .

( ) . . .

2 80 0 05 343 7 33 10

2 80 2 343 2 93 0 05 7 85 10

0 6 537 10 4 40 10

2

2 3

2 3 2

π

π π m2


8 - 43



Example (cont.)

( )( )

′ − =′ ++ ′

= × − ×

=′ −

′ − = ×

=− +

+=

′ −′ +

= − × + ×

= −

− −

−

−

− −

z Lz j kL

jz kLj

WS p

c z Lz L

pp L B A

e B A eB A

zz

j

p j

t

t

in

o

jkL jkL

t

t

( )tantan

. .

~

( )Re{ ( )} .

~( )~( ) ~ / ~

~ / ~~

/~

. .

~( ) . .

1551 10 168 10

2184 10

01 1

19 90 10 8 68 10

0 26 3 4 06

3 1

2

22

1 2

ρ W

Pa

Pa (119.5 dB)~( ) .p 0 26 6=


8 - 44



Input Impedance and Sound Power

80 Hz

0 100 200 300 400 50010

-5

10-4

10-3

10-2

10-1

100

101

Frequency (Hz)

Sound power injected into pipe

(100 Pa at each frequency)

Soun

d Po

wer

(W)

Resonances: Re(z’) minimum

0 100 200 300 400 500-10

-8

-6

-4

-2

0

2

4

6

8

10

Frequency (Hz)

Impe

danc

e ra

tio

Impedance ratio at x = -L

Real Part

Imag Part

0 100 200 300 400 500-10

-8

-6

-4

-2

0

2

4

6

8

10

Frequency (Hz)

Impe

danc

e ra

tio

Impedance ratio at x = -L

Real Part

Imag Part

80 Hz


8 - 45



Termination (Radiation) Impedance

zpu

r x

cS

Dc

f jDc

f

cS

Dc

f jDc

f

( D)

tt

tt t

o

o

= = +

≈

+

≈

+

>

(unbaffled termination)

(baffled termination)

ρ π π

ρ π π

λ

20 6

2

22

0 852

17

22

22

.

.

.

Baffle (flange)

ut

pt

D

chapter 8 sound in waveguides

Documents