industrial noise control - silencers

Sound in Ducts

Sound in Ducts

Sound in ducts is an important part of the study of acoustics.

We are usually interested in reducing the noise in ducts, or,

Preventing the noise from occurring in the first place!

Reducing the noise when it does occur may rely on installing some form of silencer.

They are neither trivial nor simple to design!

Today, we will begin to understand the basic theory, and the design process.

Sound in Ducts

The Silencers that are used in ducted systems are of 2 basic types – “reactive”, or “resistive”.

Resistive silencers use sound absorbing material to dissipate the acoustic energy.

Reactive silencers work by reflecting the sound.

We shall consider the former type first – reactive silencers. They are widely used in engine exhaust systems.

Simple Expansion Chamber silencer

Side-branchResonator

Sound in Ducts

Sound in Pipes – a quick reminder

l

tiAe

)klt(iAe

tiBe

)klt(iBe

0x

p

t

u

(Linearised 1 D momentum equation)

titiBeAe

c

1u

titi

BeAep

At x = 0 (and still assuming we only consider the real part)

)kxt(iAe

)kxt(iBe

x

0

(We attach physical significance only to the real part.)

Sound in Ducts

Expansion Chambers and Side Branch Resonators.

We shall see that they are good at blocking low frequency sound, but?

Their performance is highly frequency dependent.

The Quarter-Wave Resonator

ti

1eA

ti

1eB

ti

3eA

ti

2eA ti

2eB

l

)klt(i

2eA )klt(i

2eB

Perfect reflection:

)klt(i

2eB )klt(i

2eA

ikl2

22 eAB

)kl2t(i

2

ti

2 eAeB

o

We would like to know finally the amplitude of the transmitted wave, A3 , compared with the amplitude of the incident wave, A1 .

We start at the end!

We assume for the moment that there is no reflected wavein the downstream pipe.

Thus:


ti

1eA

ti

1eB

ti

3eA

ti

2eA

l

)klt(i

2eA )klt(i

2eB

)kl2t(i

2eA

=+tωi

11 e)BA(pressureEquating

We now focus our attention on the area of the joint.

Provided the wavelength is largecompared with the pipe diameter:

We can assume that the sound pressure in this region is essentially uniform, (regardless of which pipe we consider) - but it is of course varying in time.

We need another equation?

2kl)-tω(i2

tωi2 eAeA +

tωi3eA=

=+ )BA( 112ikl-

22 eAA + 3A=

We simplify:


ti

1eA

ti

1eB

ti

3eA

ti

2eA

l

)klt(i

2eA )klt(i

2eB

)kl2t(i

2eA

Continuity of Volume Velocity -- i.e. particle velocity x pipe area

tωi3

3tωilki2-22

2tωi11

1eA

cρ

Se)eA -A(

cρ

Se)B-A(

cρ

S+=

31

lki2

22111 AS)e1(AS)BA(S

Assume S1 = S3,

and simplify:

Area =S1

Area =S2

Area =S3


ti

1eA

ti

1eB

ti

3eA

ti

2eA

l

)klt(i

2eA )klt(i

2eB

)kl2t(i

2eA

31

lki2

22111 AS)e1(AS)BA(S

3

lki2

2211 AeAABA

From the first eqn:

lki2

32

e1

AA

And also from the first eqn:

31111 AS)BA(S

31

lki2

22111 AS)e1(AS)BA(S

Add:

Recall the second eqn

The 3 equations so far;

We would still like to find the transmitted wave A3 in terms of the incident wave, A1


ti

1eA

ti

1eB

ti

3eA

ti

2eA

l

)klt(i

2eA )klt(i

2eB

)kl2t(i

2eA

3131

lki2

2211 ASAS)e1(ASAS2

Substitute for A2:

31

lki2

lki2

3211 AS2)e1(

)e1(

ASAS2

ikl

ikl

e

e

S

SAA

2

2

1

231

1

1

21

)kltan(

S2

Si1AA

1

231

iklikl

iklikl

1

231

ee

ee

S2

S1AA


ti

1eA

ti

1eB

ti

3eA

ti

2eA

l

)klt(i

2eA )klt(i

2eB

)kl2t(i

2eA

Take the modulus – to find the ratio of the amplitudes:

2

1

2

2

3

1 )kltan(S2

S1

A

A

Hence calculate the Transmission Loss

dBA

Alog10TL

2

3

1

)kltan(

S2

Si1AA

1

231

tan(kl) can vary from zero to infinity!!

Thus A3 could equal A1 , or A3 could equal zero!!


ti

1eA

ti

3eA

l

Quater-Wave Resonator

0

5

10

15

20

25

30

35

40

45

0 2 4 6 8 10

Normalised Frequency (kl)

Tra

nsm

issi

on

Lo

ss

dB

First max when: 2

kl

2

l2

ie

4l

4

3l

4

5l

2kl

2

3kl

2

5kl


ti

1eA

ti

3eA

l

In real life, there is usually areflected wave at the outlet of the open tail pipe.

The actual behaviour is thus also influenced by the length of this tail pipe. But that is another story!

Even for the present simplified silencer, the attenuation may not be too helpful if we have many frequencies present simultaneously!

A few frequencies have large attenuations, but most have very little.

Are there other options?

Sound in Ducts

Expansion Chambers - a simplified analysis

Cross sectional area is S1l Cross sectional area is S2

(no reflected wave)for some reason?

1A2A

1B 2B

lki

2eA

lki

2eB

ikleA

3

What is the relationship between A3 (the transmitted wave) and A1 (the incident wave)?

(The usual terms involving will be cancelled.)tie

Sound in Ducts

At the second discontinuity:

Cross sectional area is S2

lki

2eA

lki

2eB

Continuity of pressure: lkilkilki eAeBeA 322

Continuity of Volume flow: ikllkilki eASeBeAS 31222 )(

Multiply the first eqn by S2, and add and subtract

lki

lki

e

eA

S

SSA

3

2

212

2

lki

lki

e

eA

S

SSB

3

2

122

2Cross sectional area is S1

ikleA

3

Sound in Ducts

At the first discontinuity:


1A2A

1B 2B

Continuity of pressure: 2211 BABA

Continuity of Volume flow: )BA(S)BA(S 222111

Multiply the first eqn by S1, and add:

22122111 B)SS(A)SS(AS2

Substitute now for A2, and for B2.


Sound in Ducts

22122111 B)SS(A)SS(AS2

ikl

ikl

ikl

ikl

e

eA

S

SSSS

e

eA

S

SSSSAS

3

2

1221

3

2

212111

2)(

2)(2

Collecting terms, and replacing by cos(kl)+i sin(kl) we obtain:ikle

lki

lki

e

eA

S

SSA

3

2

212

2 lki

lki

e

eA

S

SSB

3

2

122

2

Sound in Ducts




1A

1B

3A

1

2

2

1

3

1

2

)sin()cos(sincos

S

S

S

Skliklklikl

A

A

Take the modulus – to find the ratio of the amplitudes

2

1

2

2

1

22

2

3

1

S

S

S

S

4

)kl(sin)kl(cos

A

A

Sound in Ducts




1A

1B

3A

2

1

2

2

1

22

2

3

1

S

S

S

S

4

)kl(sin)kl(cos

A

A

Note that if sin(kl) = 0.0, cos(kl) will be 1, and thus A1 = A3

Thus the transmitted wave is of the SAME amplitude as the incident wave

Bad luck! What happens at other frequencies?2

nlornl2


l

(no reflected wave)for some reason?1A

1B

3A

2kl

2

3kl

kl

4l

4

3l

l

1A

1B

3A

If you calculate the energy (using p2) you will find that

A12 = B1

2 + A32

That means that no energy is dissipated/absorbed. The silencer works by reducing the energy transmitted and increasing the reflected wave


l

1A

1B

3A

But life is usually not this simple.

Again there is usually a reflected wave in the downstream pipe. The length of that pipe, L, has an effect on the overall

performance.

L

The “radiation” condition is well understood for a circular pipe, and the reflected wave in the pipe (and the

radiated wave) can be calculated.

The new equations can be solved.

Other Silencers?

We have discovered so far, that even for our simplified analysis, reactive silencers, while they can provide large attenuations, their performance is very dependent on

frequency.

If we are to design useful silencers of this type, they will be much more complicated than the present simple

expansion chamber, or the simple quarter wave resonator.

But they do exist and they do work.

Something more complicated?

A combination of six quarter wave resonators (tuned to different frequencies),

Looks like it might work, but it has to be properly designed.

and three expansion chambers.

Resistive Silencers.

Resistive silencers operate by dissipating energy through the use ofsound absorbing material. If they are to be effective, the sound must make good contact with the sound absorbing material. A typical cross section through a resistive silencer is shown below.

Sound

and air?

The Sound Absorbing material

Perforated metal lining is usually usedto protect the sound absorbing material from air erosion.

These types of silencers are usedextensively in air-conditioning systems.


There are many equations that can be used. The most accurate ones make use of the “acoustic impedance” of the sound absorbing

material.

Sound

and air?

The acoustic impedance is the ratio of the sound pressure to the particle velocity at the surface of the material, when both pressure and velocity are expressed in complex number notation.

The acoustic impedance is a more comprehensive way of defining sound absorbing materials, than is the absorption coefficient.


But reasonable results can be obtained using the following equation:

4.1)/(

S

PmdBnAttenuatio

Sound

and air?



Where:

• P is the internal perimeter of the duct (m) in which the air flows,

• S is the cross-sectional area of the duct (m2) in which the air flows,

• alpha is the absorption coefficient of the sound absorbing material (and lining). [Alpha is a function of frequency.]

4.1)/(

S

PmdBnAttenuatio



Some limitations:

• Accurate generally to about 10% (in either direction)

• Frequency range from about 250 Hz to 2000 Hz

• Absorption coefficient less than about 0.8

• For circular ducts, D > 0.15 m

• For rectangular ducts, width or height to be < 1m, and

• Aspect ratio (i.e. width/height) in range from 0.5 to about 2.0

•Area of duct not exceeding 0.3 m2 (more nearly square then 2:1)

4.1)/(

S

PmdBnAttenuatio


Some typical values for attenuation:

For example :

232.04.08.0 mS

0.8 m0.4 m

4.1)/(

S

PmdBnAttenuatio

63 0.1 0.3

125 0.2 0.5

250 0.3 1.4

500 0.5 2.8

1000 0.6 3.7

2000 0.75 5.0

Absorption

CoefficientFrequency

Attenuation

dB/m

mP 4.24.08.04.08.0

63 0.1 0.3

125 0.2 0.5

250 0.3 1.4

500 0.5 2.8

1000 0.6 3.7

2000 0.75 5.0

Absorption

CoefficientFrequency

Attenuation

dB/m


Resistive Silencer

0

1

2

3

4

5

6

10 100 1000 10000

Frequency (Hz)

Att

en

ua

tio

n d

B/m


Note that the attenuation is more uniform now, but that it is difficult to attenuate the low frequency sound.

• Increasing the ratio of P/S, – say by using “splitters”, and/or,

• Increasing the absorption coefficient – increasing thickness perhaps?

4.1)/(

S

PmdBnAttenuatio

How can we generally improve the attenuation?

Another approach for Square Lined Ducts.

Duct size (mm)

Thicker Sound Absorbing Material.

Duct size (mm)

industrial noise control - silencers

Documents