Download - 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:
The Quarter-Wave Resonator
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:
The Quarter-Wave Resonator
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
The Quarter-Wave Resonator
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
The Quarter-Wave Resonator
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
The Quarter-Wave Resonator
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!!
The Quarter-Wave Resonator
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
The Quarter-Wave Resonator
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:
Cross sectional area is S1
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.
Cross sectional area is S2
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
Expansion Chambers - a simplified analysis
Cross sectional area is S1l Cross sectional area is S2
(no reflected wave)for some reason?
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
Expansion Chambers - a simplified analysis
Cross sectional area is S1l Cross sectional area is S2
(no reflected wave)for some reason?
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
Cross sectional area is S1
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
Cross sectional area is S1
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.
Resistive Silencers.
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.
Resistive Silencers.
But reasonable results can be obtained using the following equation:
4.1)/(
S
PmdBnAttenuatio
Sound
and air?
Resistive Silencers.
But reasonable results can be obtained using the following equation:
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
Resistive Silencers.
But reasonable results can be obtained using the following equation:
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
Resistive Silencers.
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 Silencers.
Resistive Silencer
0
1
2
3
4
5
6
10 100 1000 10000
Frequency (Hz)
Att
en
ua
tio
n d
B/m
Resistive Silencers.
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)