Chapter 10 Sound in Ducts - University of...

Slides to accompany lectures in

Vibro-Acoustic Design in Mechanical Systems © 2012 by D. W. Herrin

Department of Mechanical Engineering University of Kentucky

Lexington, KY 40506-0503 Tel: 859-218-0609

[email protected]

Chapter 10 – Sound in Ducts

ME 510 Vibro-Acoustic Design 2 Dept. of Mech. Engineering University of Kentucky

1. Dissipative (absorptive) silencer:

Sound is attenuated due to absorption (conversion to

heat)

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

Duct or pipe

Types of Mufflers


2. Reactive muffler:

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

Compressor discharge details

40 mm

Types of Mufflers


3. Combination reactive and dissipative muffler:

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

Sound absorbing material

Perforated tubes

Types of Mufflers


Transmission loss (TL) of the muffler:

Wi

Wr

Wt Anechoic Termination Muffler

Performance Measures Transmission Loss

( )t

i

WWTL 10log10dB =


IL (dB) = SPL1 – SPL2

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

Muffler

SPL1

SPL2

Note: TL is a property of the muffler; IL is a “system” performance measure.

Performance Measures Insertion Loss


24” 12”

12” 2” 6” Source

-50

-40

-30

-20

-10

0

10

20

0 200 400 600 800 1000

Frequency (Hz)

TL a

nd IL

(dB

)

Insertion LossTransmission Loss

Pipe resonances

Inlet Pipe Outlet Pipe

Expansion Chamber Muffler

Example TL and IL


Source Su Any acoustic

system Su

P (sound pressure

reaction)

Zt

Input or load impedance

Termination impedance z = P

Su= r + jx zt =

PtSut

= rt + jxt

Acoustic System Components


•  Dissipative mufflers attenuate sound by converting sound energy to heat via viscosity and flow resistance – this process is called sound absorption.

•  Common sound absorbing mechanisms used in

dissipative mufflers are porous or fibrous materials or perforated tubes.

•  Reactive mufflers attenuate sound by reflecting a portion

of the incident sound waves back toward the source. This process is frequency selective and may result in unwanted resonances.

•  Impedance concepts may be used to interpret reactive muffler behavior.

Summary 1


Named for: Hermann von Helmholtz, 1821-1894, German physicist, physician, anatomist, and physiologist. Major work: Book, On the Sensations of Tone as a Physiological Basis for the Theory of Music, 1862.

von Helmholtz, 1848

The Helmholtz Resonator


F = PSB

x

V

SB

L L’ is the equivalent length of the neck (some air on either end also moves).

Damping due to viscosity in the neck are neglected

(resonance frequency of the Helmholtz resonator)

Helmholtz Resonator Model

Mx +Kx = PSB x = jωuB x = uBjω

j ωM −Kω

"

#$

%

&'uB = PSB

zB =PSBuB

= j 1SB2

"

#$

%

&' ωM −

Kω

"

#$

%

&'

VScK Bo22ρ

=

LSM Bo ʹ′= ρ

VLSc

MKz B

B ʹ′==→ ωwhen0


A 12-oz (355 ml) bottle has a 2 cm diameter neck that is 8 cm long. What is the resonance frequency?

Helmholtz Resonator Example

( )( )( )

Hz1821035508.0402.0

2343

2 6

2

=

×=

ʹ′=

−

n

Bn

fVLScf π

ππ


V = 0.001 m3

L = 25 mm SB = 2 x 10-4 m2

S = 8 x 10-4 m2

fn = 154 Hz

Anechoic termination

0

5

10

15

20

0 50 100 150 200 250 300

Frequency (Hz)

TL (d

B)

35 Hz

Helmholtz Resonator as a Side Branch

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

−ʹ′+=

2

21021log10dB

VcSLScTL

B ωω


Can we make ZB zero?

zA V

P

zB

z

z zA

zB

(any system)

(Produces a short circuit and P is theoretically zero.)

Network Interpretation

AB

AB

zzzzz+

=

zB =PSBuB

= j 1SB2

!

"#

$

%& ωM −

Kω

!

"#

$

%&

VLSc

MKz B


ME 510 Vibro-Acoustic Design

A Tuned Dynamic Absorber

K1

M1 x F

K1

M1 x F

K2

M2

Original System

ω/ω1

|x/F|

Original system

Tuned dynamic absorber M2/M1=0.5

K2

M2

=K1M1

tune

Tuned Dynamic Absorber

15 Dept. of Mech. Engineering University of Kentucky


Resonances in an Open Pipe

P = 1 Pa

Lp= 1 m source

First mode

Second Mode

etc.

λ1 = 2Lp =cf1→ f1 =

3432 1( )

=171.5 Hz

λ2 = Lp =cf2→ f2 =

3431 1( )

= 343 Hz



SPL at Pipe Opening – No Resonator



Example – HR Used as a Side Branch*

V = 750 cm3

L = 2.5 cm (L’= 6.75 cm) DB = 5 cm (SB= 19.6 cm2) D = 10 cm (S = 78.5 cm2)

fn = 340 Hz

Anechoic termination

_____ * e.g., engine intake systems

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

−ʹ′+=

2

21021log10dB

VcSLScTL

B ωω



SPL at Pipe Opening – with Resonator



The Quarter-Wave Resonator has an effect similar to the Helmholtz Resonator:

zB

L

S

SB

The Quarter Wave Resonator

( ) ( )( ) ⎟

⎟⎠

⎞⎜⎜⎝

⎛ += 2

22

10 44tanlog10B

B

SSSSklTL

zB = −jρocSB

cot ωL c( ) = 0 when ωL c = nπ 2 n =1,3, 5...

ωn =nπc2L

fn =nc4L

or L = nc4 f

= n λ4"

#$

%

&'


•  The side-branch resonator is analogous to the tuned dynamic absorber.

•  Resonators used as side branches attenuate sound

in the main duct or pipe.

•  The transmission loss is confined over a relatively narrow band of frequencies centered at the natural frequency of the resonator.

Summary 2


18”

2” 2” 6”

where m is the expansion ratio (chamber area/pipe area) = 9 in this example and L is the length of the chamber.

The Simple Expansion Chamber

( ) ( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ ++= klm

mklTL 22

210 sin1cos441log10

0

5

10

15

20

25

30

0 100 200 300 400 500 600 700 800

Frequency (Hz)

TL (d

B)


2”

9” 18”

2” 2” 6”

Quarter Wave Tube + Helmholtz Resonator

0

5

10

15

20

25

30

0 100 200 300 400 500 600 700 800

Frequency (Hz)

TL (d

B)


18”

2” 2” 6” 9”

(same for extended outlet)

Extended Inlet Muffler

0

5

10

15

20

25

30

0 100 200 300 400 500 600 700 800

Frequency (Hz)

TL (d

B)


9” 9”

4” 6”

Two-Chamber Muffler

0

10

20

30

40

50

0 100 200 300 400 500 600 700 800

Frequency (Hz)

TL (d

B)


Source

Engine Pump Compressor (intake or exhaust)

Area change

Expansion chamber

Helmholtz Resonator

Quarter-wave resonator

termination

We would like to predict the sound pressure level at the termination.

Complex System Modeling


The sound pressure p and the particle velocity v are the acoustic state variables

any acoustic component

1

2

p1, u1

p2, u2

For any passive, linear component:

Transfer, transmission, or four-pole matrix (A, B, C, and D depend on the component)

The Basic Idea

p1 = Ap2 +BS2u2S1u1 =Cp2 +DS2u2

p1S1u1

!"#

$#

%&#

'#= A B

C D

(

)*

+

,-

p2S2u2

!"#

$#

%&#

'#

or



p1, u1 p2 ,u2

S

L

A B

(x = 0) (x = L)

Solve for A, B in terms of p1, u1 then put into equations for p2, u2.

(note that the determinant A1D1-B1C1 = 1)

must have plane waves

The Straight Tube

p x( ) = Ae− jkx +Be+ jkx u x( ) = −1jkρoc

dpdx

p 0( ) = p1 = A+B

u 0( ) = u1 =A−Bρoc

p L( ) = p2 = Ae− jkL +Be+ jkL

u L( ) = u2 =Ae− jkL −Be+ jkL

ρocp1 = p2 cos kL( )+u2 jρoc( )sin kL( )u1 = p2 j ρoc( )sin kL( )+u2 cos kL( )

p1S1u1

"#$

%$

&'$

($=

cos kL( ) jρocS2

sin kL( )

jS1ρoc

sin kL( ) S1S2cos kL( )

)

*

+++++

,

-

.

.

.

.

.

p2S2u2

"#$

%$

&'$

($



Combining Component Transfer Matrices

[ ]22×

⎥⎦

⎤⎢⎣

⎡=

ii

iii DC

BAT Transfer matrix of ith component

[ ] [ ] [ ][ ][ ] [ ]⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

1

1system

1

1123 v

pT

vp

TTTTTvp

inn

n

[ ]22systemsystem

systemsystemsystem

×

⎥⎦

⎤⎢⎣

⎡=

DCBA

T


L

k’,zc

(complex wave number and complex characteristic impedance)

Straight Tube with Absorptive Material

p1S1u1

!"#

$#

%&#

'#=

cos k 'L( ) jzcS2sin k 'L( )

jS1zcsin k 'L( ) S1

S2cos k 'L( )

(

)

*****

+

,

-----

p2S2u2

!"#

$#

%&#

'#



S1 S2

1 2

Area Change

p1 = p2S1u1 = S2u2

p1S1u1

!"#

$#

%&#

'#= 1 0

0 1

(

)*

+

,-

p2S2u2

!"#

$#

%&#

'#


L

S S S’ straight

tube

area changes


T[ ] = 1 00 1

!

"#

$

%&

cos kL( ) jρocS '

sin kL( )

jS 'ρoc

sin kL( ) cos kL( )

!

"

#####

$

%

&&&&&

1 00 1

!

"#

$

%&

T[ ] =cos kL( ) jρoc

S 'sin kL( )

jS 'ρoc

sin kL( ) cos kL( )

!

"

#####

$

%

&&&&&


18”

2” 2” 6”


S 'S= 9


SB

S

1 2

Transfer Matrix of a Side Branch

p1Su1

!"#

$#

%&#

'#=

1 01 zB 1

(

)**

+

,--

p2Su2

!"#

$#

%&#

'#

p1 = p2 = pBSu1 = SBuB + Su2zB = pB SBuB = p2 SBuBSu1 = p2 zB( )+ Su2


F = PSB

x

V

SB

L L’ is the equivalent length of the neck (some air on either end also moves).

Damping due to viscosity in the neck are neglected

(resonance frequency of the Helmholtz resonator)

Helmholtz Resonator Model

VScK Bo22ρ

=

LSM Bo ʹ′= ρ

VLSc

MKz B


Mx +Kx = PSB x = jωuB x = uBjω

j ωM −Kω

"

#$

%

&'uB = PSB

zB =PSBuB

= j 1SB2

"

#$

%

&' ωM −

Kω

"

#$

%

&'


Transmission loss (TL) of the muffler:

Wi

Wr

Wt Anechoic Termination

1 2

⎥⎦

⎤⎢⎣

⎡

DCBA

( )t

i

WWTL 10log10dB =

TL =10 log10Sin4Sout

A+ SoutBρc

+ρcCSin

+SoutSin

D2!

"#

$#

%&#

'#

Performance Measures Transmission Loss


Wi

Wr

Wt Anechoic Termination

1 2

⎥⎦

⎤⎢⎣

⎡

DCBA

Derivation Transmission Loss

p1S1u1

!"#

$#

%&#

'#= A B

C D

(

)*

+

,-

p2S2u2

!"#

$#

%&#

'#

Wi =p+a2

ρcS1

Wt =p+b2

ρcS2

TL =10 logWi

Wt

p1 = p+a + p−a

u1 =p+a − p−aρc

p2 = p+b

u2 =p+bρc

Express p1, p2, u1 and u2 in terms of incident reflected waves


IL = 20 log10A ZS +B ZTZS +C +D ZT

A0 ZS +B0 ZTZS +C0 +D0 ZT

!"#

$#

%&#

'#

Performance Measures Insertion Loss

Muffler

SPL1

SPL2

ZS ZT

T0[ ] =A0 B0C0 D0

!

"##

$

%&&

ZS ZT

T[ ] = A BC D

!

"#

$

%&


Sound Wave Reflections in Engines

Muffler

Engine

Waves leaving engine

Reflected from muffler

Reflected from engine

Waves leaving muffler

Reflected from open end

Reflected from muffler

Resonances can form in the exhaust and tail pipes as well as within the muffler.


Source Impedance

Source

ps pL

Load

zs

zL

pszs + zL

=pLzL

Acoustic Source

Waves Leaving Source

Reflected from Attenuating Element

Attenuating Element

(i.e. Load)

Reflected from Source

uL



Transfer Impedance

1p 2p

21 uu =

ztr

ztr =p1 − p2Su

Incident Wave

Reflected Wave

Transmitted Wave

1p 2pu



Source/Load Concept

L1

Source zs , ps

ps

IL = f TL, zs, zt( )pt = f TL, zs, zt, ps( )pL

L2

zt , pt

Load zL , pL

Muffler

zs

zL



Insertion Loss Prediction

-30

-20

-10

0

10

20

30

40

50

60

0 200 400 600 800 1000Frequency (Hz)

IL (d

B)

Actual source impedancePressure source (Zs=0)Velocity source (Zs=infinite)Anechoic source (Zs=rho*c)



Source

ps pL

Load

zs

zL

uL

pszs + zL

=pLzL= SuL

ps = SuLzs + pL

zs =ps − pLSuL

Source Impedance Series Impedance



Source

us pL

Load

zs zL

uL

SuLzL = SuszszLzs + zL

= pL

zs =pL

S us −uL( )

Source Impedance Parallel Impedance



Derivation Insertion Loss

Muffler SPL2

ZS ZT

p1S1u1

!"#

$#

%&#

'#= A B

C D

(

)*

+

,-

p2S2u2

!"#

$#

%&#

'#

zS =ps − p1S1u1

⇒ p1 = ps − S1u1zs

zT =p2S2u2

⇒ S2u2 =p2zT

1 2

p1 = Ap2 +BzTp2

p1 = ps − S1u1zS

= ps − zS Cp2 +Dp2zT

"

#$

%

&'

Ap2 +BzTp2 = ps − zS Cp2 +D

p2zT

"

#$

%

&'

p2ps=

1

A+ 1zTB+ zSC +

zSzTD



p2ps=

1

A0 +1zTB0 + zSC0 +

zSzTD0

SPL1

ZS ZT

T0[ ] =A0 B0C0 D0

!

"##

$

%&&

Determined in same manner as prior slide



IL = 20 logp2,nomufflerp2,muffler

p2,nomufflerps

=1

A0 +1zTB0 + zSC0 +

zSzTD0

p2,mufflerps

=1

A+ 1zTB+ zSC +

zSzTD

IL = 20 log10A ZS +B ZTZS +C +D ZT

A0 ZS +B0 ZTZS +C0 +D0 ZT

!"#

$#

%&#

'#


•  The transfer matrix method is based on plane wave (1-D) acoustic behavior (at component junctions).

•  The transfer matrix method can be used to determine the

system behavior from component “transfer matrices.” •  Applicability is limited to cascaded (series) components and

simple branch components (not applicable to successive branching and parallel systems).

Summary 3

Chapter 10 Sound in Ducts - University of...

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