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8.4 SIDE-BRANCH MUFFLERS

Back

The side-branch muffler is one type of silencer used to reduce noise emission

in a restricted frequency range from a mechanical system. The side-branch

muffler consists of a Helmholtz resonator connected to the main tube

through which the sound is transmitted. The silencer acts to reduce sound

transmission primarily by reflecting acoustic energy back to the source, so it

is classed as a reactive silencer; however, some energy is dissipated within the

acoustic resistance element of the silencer.

Some typical configurations for the side-branch muffler are shown in

Fig. 8-10.

350 Chapter 8

FIGURE 8-10 Configurations for side-branch mufflers: (a) resonator connected by

a tube or tubes and (b) resonator connected through orifices.

8.4.1 Transmission Loss for a Side-Branch Mu\er

The acoustic impedance for the Helmholtz resonator in the side-branch

muffler is given by Eq. (8-44):

ZAb ј RA юjXA (8-68)

The acoustic reactance XA of the resonator is given by the following expression:

XA ј 2_fMA _

2_fCA

(8-69)

The quantities MA and CA are the acoustic mass and acoustic compliance

for the resonator.

The instantaneous acoustic pressure p1рtЮ in the main tube upstreamof

the junction or side branch may be written as follows:

p1рtЮ ј A1 ejр!t_kxЮ юB1 ejр!tюkxЮ (8-70)

The first term represents the incident sound wave at the junction, and the

second term represents the sound wave reflected back toward the source.

Similarly, the instantaneous acoustic pressure downstreamof the junction or

side branch may be written in the following form, assuming that there is

negligible energy reflected beyond the junction:

p2рtЮ ј A2 ejр!t_kxЮ (8-71)

The acoustic pressure at the branch may be written in terms of the acoustic

impedance of the resonator:

pbрtЮ ј UbрtЮZAb ј UbрtЮрRA юjXAЮ (8-72)

As indicated in Fig. 8-11, the acoustic pressure at the junction рx ј 0Ю is the same for the three elements:

p1рx ј 0Ю ј p2рx ј 0Ю ј pb (8-73)

A1 ю B1 ј A2 ј UbрRA ю jXAЮ (8-74)

The volumetric flow rate upstream of the junction or side branch may

be written in terms of the acoustic velocity u1рtЮ and the cross-sectional area

S of the main tube:

U1рtЮ ј Su1рtЮ ј рS=_ocЮЅA1 e jр!t_kxЮ _ B1 e jр!tюkxЮ_ (8-75)

The volumetric flow rate downstream of the junction may be written in a

similar fashion:

U2рtЮ ј Su2рtЮ ј рS=_ocЮA2 e jр!t_kxЮ (8-76)

Silencer Design 351

At the junction рx ј 0Ю, we have the following condition for the volume flow

rate:

U1рx ј 0Ю ј U2рx ј 0Ю ю Ub (8-77)

A1 _ B1 ј A2 ю р_oc=SЮрA1 ю B1Ю

RA ю jXA

(8-78)

If we eliminate the term B1 between Eqs (8-74) and (8-78), we obtain

the following expression for the ratio of the coefficients A1 (incident sound

wave) and A2 (transmitted sound wave):

A2 ј р_oc=2SЮ ю RA ю jXA

RA ю jXA

(8-79)

The square of the magnitude of this ratio is as follows:

____

ј ЅRA ю р_oc=2SЮ_2 ю X2A

A ю X2A

(8-80)

The sound power transmission coefficient at for the side-branch muffler

is defined as the ratio of the sound power transmitted to the sound

power incident on the junction:

at ј

Wtr

Win ј j ptrj2

j pinj2 ј

____

A ю X2A

ЅRA ю р_oc=2SЮ_2 ю X2A

(8-81)

The transmission loss TL for the muffler is related to the sound power

transmission coefficient:

352 Chapter 8

FIGURE 8-11 Physical conditions at the junction of a side branch. Subscript 1

denotes upstream conditions, subscript 2 denotes downstream conditions, and subscript

b denotes conditions for the side branch.

TL ј 10 log10р1=atЮ (8-82)

If more than one tube is used to connect the resonator volume and the

main tube, the total acoustic mass for Nt tubes having the same length and

diameter is given by the following expression:

MA ј

_oрL ю _La ю _LbЮ

_a2Nt ј

_oLe

_a2Nt

(8-83)

The quantity L is the length of each connecting tube, and _La and _Lb are

the equivalent lengths to account for end effects at each end of the connecting

tube. If the end of the connecting tube is flush with the main tube wall or

the wall of the resonator volume (flanged end), use Eq. (8-8) for the end

correction. If the connecting tube protrudes into the main tube or the resonator

volume (free end), use Eq. (8-10) for the end correction.

The expression for the acoustic compliance CA for the resonator is

given by Eq. (8-26).

In many design cases, additional acoustic resistance, in the form of

screens and other elements, must be added to achieve the required total

acoustic resistance. The total acoustic resistance RA is related to the specific

acoustic resistance RS (resistance for a unit area) by the following:

RA ј

_a2Nt

(8-84)

The specific acoustic resistance for one screen layer RS1 is given in Table 8-1

for several screen sizes. The specific acoustic resistance for NS layers of

screens is RS ј NSRS1.

Silencer Design 353

TABLE 8-1 Specific Acoustic Resistance for

Wire Screen

Mesh size,

wires/inch

Screen thickness

RS1, rayl, N-s/m3 mm inch

30 0.66 0.026 5.67

50 0.44 0.0173 5.88

65 0.33 0.0129 6.40

100 0.23 0.0091 9.10

120 0.184 0.0072 13.5

200 0.114 0.0045 24.6

325 0.073 0.0029 49.1

The acoustic reactance for the resonator, using Eq. (8-69) with the

resonant frequency given by Eq. (8-43), may be written in the following

form:

XA ј 2_fMAЅ1 _ рfo=f Ю2_ (8-85)

The acoustic quality factor may be introduced from Eq. (8-48):

XA ј RAQAЅр f =foЮ _ рfo=f Ю_ (8-86)

The final expression for the sound power transmission coefficient (or

the reciprocal) for the side-branch muffler is obtained by making the substitution

from Eq. (8-86) into Eq. (8-81):

at ј

_2 ю Q2

AЅр f =foЮ _ рfo=f Ю_2

1 ю Q2

AЅр f =foЮ _ р fo=f Ю_2 (8-87)

The quantity _ is defined as follows:

_ ј 1 ю

_oc

2SRA

(8-88)

The quantity S is the cross-sectional area of the main tube from the source

of sound, and _o and c are the density and speed of sound, respectively, for

the gas in the main tube.

At resonance р f ј foЮ, we see from Eq. (8-87) that the sound power

transmission coefficient is a minimum (or the reciprocal is a maximum) and

has the following value:

р1=atЮmax ј _2 (8-89)

The transmission loss for the muffler at resonance is a maximum:

TLo ј TLmax ј 10 log10р_2Ю ј 20 log10 _ р8-90)

The half-power bandwidth frequencies for the side-branch muffler

may be determined by using Eqs (8-87) and (8-89):

1=at

р1=atЮmax ј 0:500 ј

_2 ю Q2

AЅр f =foЮ _ рfo=f Ю_2

_2f1 ю Q2

AЅр f =foЮ _ рfo=f Ю_2g

(8-91)

If we solve the quadratic equation, Eq. (8-91), for the two solutions, we

obtain the lower frequency р f1) and the upper frequency р f2Ю of the halfpower

frequency band:

f1=f0 ј _fЅ1 ю р1=_2Ю_1=2 _ 1g (8-92)

f2=fo ј _fЅ1 ю р1=_2Ю_1=2 ю 1g (8-93)

354 Chapter 8

The quantity _ is defined by the following expression:

_ ј

2QAр_2 _ 2Ю1=2 (8-94)

We observe from Eq. (8-94) that the value of the quantity _ must meet the

condition _ >

ffiffiffi2p if the side-branch muffler is to be used.

The sound power transmission coefficients at the upper and lower

frequencies of the half-power band are related to the transmission loss at

resonance:

TL1 ј TL2 ј 10 log10Ѕ0:500р1=atЮmax_ ј TLo _ 3:0 dB (8-95)

The transmission loss at resonance is given by Eq. (8-90).

Example 8-4. A side-branch muffler has a resonator volume of 3.00dm3

(0.1059 ft3). Three side-branch tubes are used, and each tube has a diameter

of 24mm (0.945 in) and a length of 75mm (2.95 in). The side-branch tubes

are flush with the main tube surface at one end and with the resonator

container wall at the other end. The acoustic resistance is provided by one

layer of 100-mesh screen in each tube. The diameter of the main tube is

150mm (5.91 in). The fluid flowing in the main tube is air at 3258C (6178F)

and 110 kPa (15.96 psia), for which the sonic velocity c ј 490 m/s (1608 ft/

sec) and density _o ј 0:641 kg=m3 (0.040 lbm=ft3). Determine the transmission

loss for the muffler at a frequency of 125 Hz.

The equivalent length for the side-branch tubes (flanged ends) is as

follows:

Le ј L ю 2р8_=3Юa ј 75 ю р2Юр8_=3Юр12Ю ј 276mm

The acoustic mass for the side-branch tubes may be calculated from Eq.

(8-83):

MA ј р0:641Юр0:276Ю

р_Юр0:012Ю2р3Ю ј 130:4kg=m4

The acoustic compliance for the resonator may be found from Eq. (8-26):

CA ј р3:00Юр10_3Ю

р0:641Юр490Ю2 ј 19:49 _ 10_9 Pa-s/m3

The resonant frequency for the silencer is found from Eq. (8-43):

fo ј

р2_ЮЅр130:4Юр19:49Юр10_9Ю_1=2 ј 99:8Hz

Silencer Design 355

The specific acoustic resistance for one layer of screen is RS1 ј 9:10

N-s/m3 from Table 8-1. The acoustic resistance for the muffler may be

calculated from Eq. (8-84):

RA ј р1Юр9:10Ю

р3Юр_Юр0:012Ю2 ј 6705 Pa-s=m3 ј 6:705 kPa-s/m3

The acoustic quality factor for the muffler is given by Eq. (8-48):

QA ј р2_Юр99:8Юр130:4Ю

р6705Ю ј 12:20

The cross-sectional area of the main tube is as follows:

S ј р1

4 _Юр0:150Ю2 ј 0:01767m2

The _-parameter may be calculated from Eq. (8-88):

_ ј 1 ю р0:641Юр490Ю

р2Юр0:01767Юр6705Ю ј 1 ю 1:326 ј 2:326

The sound power transmission coefficient may be found from Eq.

(8-87) for a frequency of 125 Hz:

at ј р2:326Ю2 ю р12:20Ю2Ѕр125=99:8Ю _ р99:8=125Ю_2

1 ю р12:20Ю2Ѕр125=99:8Ю _ р99:8=125Ю_2 ј 1:139

The transmission loss at a frequency of 125 Hz is found from Eq. (8-82):

TL ј 10 log10р1:139Ю ј 0:6dB

The transmission loss at the resonant frequency р fo ј 99:8 Hz) is

found from Eq. (8-90).

TLo ј 20 log10р2:326Ю ј 7:3dB

We note that there is a considerable difference between the transmission loss

at resonance and at the higher frequency of 125 Hz.

The parameter _ given by Eq. (8-94) may be calculated:

_ ј р2:326Ю

р2Юр12:20ЮЅр2:326Ю2 _ 2_1=2 ј 0:0516

The upper and lower frequencies for the half-power band may be calculated

from Eqs (8-92) and (8-93):

f1=fo ј р0:0516ЮfЅ1 ю р1=0:0516Ю2_1=2 _ 1g ј 0:9497

f1 ј р99:8Юр0:9497Ю ј 94:8Hz

356 Chapter 8

f2=fo ј р0:0516ЮfЅ1 ю р1=0:0516Ю2_1=2 ю 1g ј 1:0529

f2 ј р99:8Юр1:0529Ю ј 105:1Hz

The half-power bandwidth is relatively small (about 5%) for the muffler in

this example.

8.4.2 Directed Design Procedure for Side-Branch

Mu\ers

In many silencer design situations, the following parameters are known or

specified: (a) minimum acceptable transmission loss, TLmin; (b) primary

range of frequencies for the silencer operation, f1 and f2; (c) cross-sectional

area of the main tube, S; and (d) the type of gas in the silencer, so that the

density _o and speed of sound c are known. There are often more

‘‘unknowns’’ than design equations; therefore, there are several possible

solutions for a particular set of design parameters. The design equations

may be organized, however, in a form that allows a more directed design

procedure. This directed design procedure is outlined in the following

material for a side-branch muffler.

D1. The design resonant frequency for the resonator may be determined

from the primary range of frequencies for the silencer and Eq. (8-61):

fo ј рf1 f2Ю1=2 (8-96)

D2. The maximum transmission loss for the muffler or the transmission

loss at resonance may be determined from the minimum acceptable

transmission loss and Eq. (8-95):

TLo ј TLmin ю 3 dB (8-97)

D3. The dimensionless parameter _ may be determined from the value

of the transmission loss at resonance and Eq. (8-90):

_ ј 10TLo=20 (8-98)

D4. The required acoustic resistance for the muffler may be determined

from the value of the _-parameter and Eq. (8-88):

RA ј

_oc

2р_ _ 1ЮS ј

_a2Nt

(8-99)

The quantity a is the radius of a connecting tube, Nt is the number of

connecting tubes, and RS is the specific acoustic resistance for the connecting

tubes.

D5. The acoustic quantity factor may be determined from the frequencies

and the _-parameter with Eqs (8-92), (8-93), and (8-94). If we subtract

Silencer Design 357

Eq. (8-92) from Eq. (8-93), we obtain the following relationship for the

parameter _:

_ ј

f2 _f1

2fo

(8-100)

Using Eq. (8-94) to eliminate the parameter _, we obtain the following

relationship for the acoustic quality factor:

QA ј

fo_

р f2 _f1Юр_2 _2Ю1=2 (8-101)

D6. The required acoustic mass for the muffler may be determined

from the acoustic quality factor and Eq. (8-48):

MA ј

QARA

2_fo

(8-102)

D7. The required acoustic compliance may be determined from the

acoustic mass, the resonant frequency, and Eq. (8-43):

CA ј

4_2f2o

MA ј

_oc2 (8-103)

D8. Practical choices must be made for two of the following quantities

to completely specify the design: specific acoustic resistance, RS; number of

side-branch tubes, Nt; side-branch tube radius, a; resonator volume, V; and

length of the side-branch tube, L. The remaining three quantities may be

calculated from the previously determined values of the acoustic mass, MA;

the acoustic resistance, RA; and the acoustic compliance, CA.

This design procedure is illustrated in the following example.

Example 8-5. A side-branch muffler is to be constructed by placing a

300mm (11.81 in) inside diameter cylinder around the 250mm (9.84in)

inside diameter circular duct used in an air conditioning system, as shown

in Fig. 8-12. The resonator volume is connected to the main duct through

circular holes covered with wire screen. The thickness of the duct is 1.20mm

(0.047 in). The muffler is to have a minimum transmission loss of 6 dB over

the frequency range from 177 Hz to 354 Hz (the 250 Hz octave band). the

fluid flowing through the main duct is air at 108C (508F) and 105 kPa

(15.23 psia), for which the density _o ј 1:292 kg=m3 (0.0807 lbm=ft3) and

the sonic velocity c ј 337:3 m/s (1107 ft/sec). Determine the muffler dimensions.

The design resonant frequency is found from Eq. (8-96), if we set the

design frequency range as the half-power band range:

358 Chapter 8

fo ј рf1 f2Ю1=2 ј Ѕр177Юр354Ю_1=2 ј 250 Hz

The maximum transmission loss, which occurs at the resonant frequency, is

as follows:

TLo ј TLmin ю 3:0dB ј 6:0 ю 3:0 ј 9:0dB

The damping parameter _ is found from Eq. (8-98):

_ ј 109:0=20 ј 2:8216

The cross-sectional area for the main duct is as follows:

S ј 1

4 _

р0:250Ю2 ј 0:0491m2

The required acoustic resistance may be calculated from Eq. (8-99):

RA ј р1:292Юр337:3Ю

р2Юр2:8216 _ 1Юр0:0491Ю ј 2436 Pa-s/m3

The acoustic quality factor for the resonator may be found from Eq. (8-101):

QA ј р250Юр2:8216Ю

р354 _ 177ЮЅр2:8216Ю2 _ 2_1=2 ј 1:6322

The required acoustic mass may be determined from Eq. (8-102):

MA ј р1:6322Юр2436Ю

р2_Юр250Ю ј 2:531 kg=m4

The required acoustic compliance for the resonator is given by Eq. (8-103):

CA ј

р4_2Юр250Ю2р2:531Ю ј 0:1601 _ 10_6 m5=N

Silencer Design 359

FIGURE 8-12 Figure for Example 8-5. The main tube diameter D1 ј 250mm and

the resonator volume OD is D2 ј 300 mm.

If we use one layer of screen over each of Nt holes, the acoustic

resistance and acoustic mass are given by Eqs (8-99), with RS ј RS1, and

(8-83) with flanged-end tubes:

RA ј

_a2Nt

MA ј

_oЅL ю р16a=3_Ю_

_a2Nt

Solving for the ratio of the acoustic mass and acoustic resistance, we obtain

the following expression:

RA ј

_oЅL ю р16a=3_Ю_

Let us try a 200-mesh wire screen, for which RS1 ј 24:6 N-s/m3, for the

acoustic resistance. If we solve for the required hole radius a, we obtain:

a ј

MARS

RA_o _ L

_ _

р2:531Юр24:6Ю

р2436Юр1:292Ю _ 0:0012

_ _

a ј 0:0109m ј 10:9mm р0:429 inЮ

The hole diameter is:

2a ј 21:8mm р0:858 inЮ

The required number of holes may be found as follows:

Nt ј

_a2RA ј р24:6Ю

р_Юр0:0109Ю2р2436Ю ј 27:0 holes

There are other possible solutions. For example, if we use 120-mesh screens,

we would find that the hole radius would be a ј 5:7mm (0.224 in) and the

number of holes would be Nt ј 55 holes.

The required volume of the resonator is found from the acoustic compliance:

V ј CA_oc2 ј р0:1601Юр10_6Юр1:292Юр337:3Ю2 ј 0:02353m3 р0:831 ft3Ю

The required length of the resonator is as follows:

L2 ј р0:02353Ю 1

4 _

fр0:300Ю2 _ Ѕ0:250 ю р2Юр0:0012Ю_2g ј

0:02353

0:02065

L2 ј 1:139m р44:86 inЮ

The summary of the design for the side-branch muffler is as follows:

Muffler length, L2 ј 1:139m (44.86 in)

360 Chapter 8

Hole diameter, 2a ј 21:8mm (0.858 in)

Resistance element, 1 layer of 200-mesh wire screen

The sound power transmission coefficient for a frequency f ј 63 Hz

may be found from Eq. (8-87):

at ј р2:8216Ю2 ю р1:6322Ю2Ѕр63=250Ю _ р250=63Ю_2

1 ю р1:6322Ю2Ѕр63=250Ю _ р250=63Ю_2 ј 1:1842

The transmission loss for the muffler at 63 Hz is as follows:

TL р63 HzЮ ј 10 log10р1:1842Ю ј 0:7dB

8.4.3 Closed Tube as a Side-Branch Mu\er

For a side-branch muffler with a resonator volume attached, the muffler

operates quite well around the single resonant frequency of the resonator

volume. A long tube, however, has several resonant frequencies (theoretically,

an infinite number of resonant frequencies). The resonant frequencies

for the closed tube shown in Fig. 8-13 are given by the following expression

(Kinsler et al., 1982):

fo ј

n _ 1

2Le рn ј 1; 2; 3; . . .Ю (8-104)

Silencer Design 361

FIGURE 8-13 Closed tube of radius a and length L as a side branch.

The equivalent length of the closed tube is given by Eqs (8-8) and (8-12) for

the open end:

Le ј L ю р8=3_Юa (8-105)

The quantity L is the length of the tube and a is the radius of the tube.

The acoustic reactance for the closed tube is given by the following

relationship, valid for ka < 1 (Reynolds, 1981):

XA ј _

_oc

_a2 cotрkLeЮ (8-106)

The quantity k is the wave number:

k ј

2_f

(8-107)

The total acoustic resistance for the tube involves the energy dissipated

due to fluid friction within the tube, as expressed by the attenuation coefficient

, and the resistance provided by a screen or other element at the inlet

of the tube, if such an element is used:

RA ј

_ocL ю RS

_a2 (8-108)

The quantity RS is the specific acoustic resistance for the screens or other

elements. The attenuation coefficient may be found from the following

expression:

ј р_fe=_oЮ1=2

(8-109)

The quantity e is the effective viscosity for the gas, which includes the effect

of heat conduction (Rayleigh, 1929):

e ј 1 ю р_ _ 1Ю

р_PrЮ1=2

_ _

(8-110)

The quantity is the viscosity of the gas, _ is the specific heat ratio, and Pr

is the Prandtl number for the gas.

The sound power transmission coefficient for the closed-tube sidebranch

muffler may be obtained by substituting the expressions for the

acoustic resistance and acoustic reactance given by Eqs (8-106) and

(8-108) into Eq. (8-81):

at ј ЅL ю рRS=_ocЮ ю р_a2=2SЮ_2 tan2рkLeЮ ю 1

ЅL ю рRS=_ocЮ_2 tan2рkLeЮ ю 1

(8-111)

362 Chapter 8

The sound power transmission coefficient reaches a minimum at the

resonant frequency given by Eq. (8-104). The value (which is a maximum

for 1=atЮ is given by the following expression:

р1=atЮmax ј 1 ю р_a2=2SЮ

L ю рRS=_ocЮ

" #2

(8-112)

From Eq. (8-111), we note that for the frequencies for which

kLe ј 2_fLe=c ј n_ рn ј 1; 2; 3; . . .Ю (8-113)

the term tanрkLeЮ ј 0, and р1=atЮ ј 1. The frequencies given by Eq. (8-113)

correspond to frequencies at which sound is transmitted through the system

without attenuation.

Example 8-6. A tube having an inside diameter of 54.8mm (2.157 in) is

attached to a pipe having an inside diameter of 161.5mm (6.357 in) through

which noise is transmitted. It is desired to select a closed-tube side-branch

muffler that has a resonant frequency of 250 Hz. The gas in the tube is air at

218C (708F) and 101.3 kPa (14.7 psia), for which the sonic velocity

c ј 343:8 m/s (1128 ft/sec), density _o ј 1:200 kg=m3 (0.075 lbm=ft3Ю, viscosity

ј 18:17 mPa-s, Prandtl number Pr ј 0:710, and specific heat ratio

_ ј 1:400. Determine the required length of the attached tube and the

transmission loss at 250 Hz and 177 Hz.

The equivalent length of the closed tube may be found from Eq.

(8-104), using n ј 1, because this value results in the shortest tube length:

Le ј

4fo ј р343:8Ю

р4Юр250Ю ј 0:3438m р13:54 inЮ

The required length of the closed tube is calculated from Eq. (8-105):

L ј Le _ р8=3_Юa ј 0:3438 _ р8=3_Юр0:0274Ю ј 0:3205m р12:62 inЮ

The effective viscosity for the gas is calculated from Eq. (8-110):

e ј р18:17Ю 1 ю р1:40 _ 1Ю

Ѕр1:40Юр0:710Ю_1=2

_ _

ј р18:17Юр1:401Ю ј 25:46 mPa-s

The attenuation coefficient at 250 Hz is found from Eq. (8-109):

ј Ѕр_Юр250Юр25:46Юр10_6Ю=р1:200Ю_1=2

р0:0274Юр343:8Ю ј 0:01370m_1

L ј р0:01370Юр0:3205Ю ј 0:00439

Silencer Design 363

The cross-sectional area of the main tube is as follows:

S ј 1

4 _

р0:1615Ю2 ј 0:02048m2

The sound power transmission coefficient at the resonant frequency

fo ј 250 Hz is found from Eq. (8-112):

р1=atЮ ј 1 ю р_Юр0:0274Ю2

р2Юр0:02048Юр0:00439Ю

" #2

ј р14:117Ю2 ј 199:28

The transmission loss at 250 Hz is as follows:

TLo ј 10 log10р199:28Ю ј 23:0dB

The following values for the parameters for 177 Hz may be calculated:

kLe ј

2_fLe

c ј р2_Юр177Юр0:3438Ю

р343:8Ю ј 1:111 rad

ј Ѕр_Юр177Юр25:46Юр10_6Ю=р1:200Ю_1=2

р0:0274Юр343:8Ю ј 0:01153m_1

L ј р0:01153Юр0:3205Ю ј 0:003695

_a2

2S ј р_Юр0:0274Ю2

р2Юр0:02048Ю ј 0:05758

The sound power transmission coefficient at 177 Hz is calculated from Eq.

(8-111):

at ј р0:003695 ю 0:05758Ю2 tan2р1:111Ю ю 1

р0:003695Ю2 tan2р1:111Ю ю 1 ј

1:0153

1:0000557 ј 1:0153

The transmission loss at 177 Hz is as follows:

TLр177 HzЮ ј 10 log10р1:0153Ю ј 0:066 dB _ 0:1dB

Suppose one layer of 200-mesh screen рRS ј 24:6 Pa-s/m3) is placed

over the open end of the tube:

_oc ј р24:6Ю

р1:20Юр343:8Ю ј 0:05963

The sound power transmission coefficient at resonance (250 Hz) with the

screen is as follows:

р1=atЮmax ј 1 ю р0:0578Ю

р0:00439 ю 0:05963Ю

_ _2

ј р1:899Ю2 ј 3:608

364 Chapter 8

The addition of the screen results in a lower value of the transmission loss at

resonance:

TLo ј 10 log10р3:608Ю ј 5:6dB

The sound power transmission coefficient at 177 Hz may be calculated as

follows:

1=at ј р0:003655 ю 0:5963 ю 0:05758Ю2 tan2р1:111Ю ю 1

р0:003695 ю 0:05963Ю2 tan2р1:111Ю ю 1 ј

1:0596

1:01635

1=at ј 1:0426

The transmission loss at 177 Hz is slightly higher with the addition of the

screen:

TLр177 HzЮ ј 10 log10р1:0426Ю ј 0:18 dB _ 0:2dB

8.4.4 Open Tube (Ori￠ce) as a Side Branch

Let us consider the case in which an open tube or orifice (a short, open tube)

is used as the side branch. The reactance for a tube open to the surroundings

(CA ј1) is equal to the acoustic mass for the tube:

XA ј 2_fMA ј

2f _oLe

a2 ј

_ockLe

_a2 (8-114)

The quantity k ј 2_f =c is the wave number. The lumped-parameter analysis

is valid only for small values of ka, so the acoustic resistance for the open

end (valid for ka < 1:4) is given by Eq. (8.35a):

RA ј

_ock2

2_ ј

2_f 2_o

(8-115)

The sound power transmission coefficient (or the reciprocal) for the

open-tube side-branch muffler may be found by making the substitutions

from Eqs (8-114) and (8-115) into Eq. (8-81):

at ј

_ock2

2_ ю

_oc

_ockLe

_a2

_ _2

_ock2

_ockLe

_a2

_ _2 (8-116)

This relationship may be simplified by canceling the term р_ocЮ and dividing

through by the last term in the numerator and denominator:

at ј Ѕрka2=2LeЮ ю р_a2=2SkLeЮ_2 ю 1

рka2=2LeЮ2 ю 1

(8-117)

Silencer Design 365

If we substitute for the wave number k in terms of the frequency f, we obtain

the alternative expression for the sound power transmission coefficient:

at ј f_р fLe=cЮрa=LeЮ2 ю Ѕрa2=4SЮ=р fLe=cЮ_g2 ю 1

Ѕ_р fLe=cЮрa=LeЮ2_2 ю 1

(8-118)

For small frequencies (or, for f < 0:1c=S1=2Ю, Eq. (8-118) approaches

the limiting expression, as follows:

at ! рa2=4SЮ

р fLe=cЮ

" #2

ю1 (8-119)

On the other hand, for large frequencies (or, for f > c=S1=2Ю, the sound

power transmission coefficient given by Eq. (8-118) approaches a limiting

value of unity, or the transmission loss approaches zero at large frequencies.

If an additional acoustic resistance, such as a screen, is placed over one

end of the short tube, the sound power transmission coefficient expression is

modified as follows:

at ј f_р fLe=cЮрa=LeЮ2 ю ЅрRS=2__ocЮ=р fLe=cЮ_ ю Ѕрa2=4SЮ=р fLe=cЮ_g2 ю 1

fЅ_р fLe=cЮрa=LeЮ2_ ю ЅрRS=2__ocЮ=р fLe=cЮ_g2 ю 1

(8-120)

The quantity RS is the specific acoustic resistance of the screens. The lowfrequency

limiting value of the sound power transmission coefficient with

the use of screen resistances is as follows:

at ! рRS=2__ocЮ ю рa2=4SЮ

р fLe=cЮ

" #2

ю1 (8-121)

The use of a screen resistance increases the low-frequency transmission loss.

At high frequencies, the transmission loss approaches zero, whether screens

are used or not used.

Example 8-7. A hole having an inside diameter of 54.8mm (2.157 in) is

placed in a pipe having an inside diameter of 161.5mm (6.357 in) and wall

thickness of 3.40mm (0.134 in) through which noise is transmitted. The gas

in the tube is air at 218C (708F) and 101.3 kPa (14.7 psia), for which the

sonic velocity c ј 343:8 m/s (1128 ft/sec), and density _o ј 1:200 kg=m3

(0.075 lbm=ft3Ю. Determine the transmission loss at 250 Hz and 125 Hz.

First, let us check the validity of the lumped-parameter model:

ka ј

2_fa

c ј р2_Юр250Юр0:0274Ю

р343:8Ю ј 0:125 < 1

366 Chapter 8

The lumped-parameter model does apply for this example. Let us also calculate

the following parameter:

S1=2 ј р343:8Ю

р0:02048Ю1=2 ј 2402 Hz

The frequency f ј 250 Hz is near р0:1Юр2402Ю ј 240:2 Hz, so the lowfrequency

approximation, Eq. (8-119), does apply; however, let us use the

general expression for this frequency.

The equivalent length of the hole, which has ‘‘flanged ends’’ at both

inlet and outlet, is given by Eqs (8-8) and (8-12):

Le ј L ю 2р8=3_Юa ј 3:40 ю р2Юр8=3_Юр27:4Ю ј 49:92mm

Let us calculate the following dimensionless parameters for a frequency

of f ј 250 Hz:

fLe

c ј р250Юр0:04992Ю

р343:8Ю ј 0:0363

4S ј р0:0274Ю2

р4Юр0:02048Ю ј 0:00916

The sound power transmission coefficient for a frequency of 250 Hz may be

found from Eq. (8-118):

at ј Ѕр_Юр0:0363Юр27:4=49:92Ю2 ю р0:00916=0:0363Ю_2 ю 1

Ѕр_Юр0:0363Юр274=49:92Ю2_2 ю 1

at ј р0:0344 ю 0:2525Ю2 ю 1

р0:0344Ю2 ю 1 ј

1:0823

1:0012 ј 1:081

The transmission loss is as follows:

TL ј 10 log10р1:081Ю ј 0:3 dB (for f ј 250 HzЮ

The frequency f ј 125 Hz falls in the range for which the low-frequency

approximation applies, because 125 Hz < р0:1Юр2402Ю ј 240:2 Hz. For

125 Hz, the frequency parameter has the following value:

fLe

c ј р125Юр0:04992Ю

р343:8Ю ј 0:01815

The sound power transmission coefficient may be evaluated from Eq.

(8-119):

at ј 1 ю

0:00916

0:01815

_ _2

ј 1 ю р0:505Ю2 ј 1:255

Silencer Design 367

The transmission loss is as follows:

TL ј 10log10р1:255Ю ј 1:0dB (for f ј 125HzЮ

The transmission loss for the orifice side-branch muffler is considerably

smaller than that for the closed-tube side-branch muffler at 250 Hz. On

the other hand, the orifice has somewhat better performance at lower frequencies,

although the transmission loss is not very large at a frequency of

125 Hz.

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