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8.3 THE HELMHOLTZ RESONATOR

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The primary element of the side-branch muffler is a resonator volume

(Helmholtz resonator). We need to become familiar with the characteristics

of this acoustic system before considering the application in silencer design.

In addition, the analysis of the Helmholtz resonator illustrates the principles

of the lumped-parameter analysis for acoustic systems (Howe, 1976).

8.3.1 Helmholtz Resonator System

The Helmholtz resonator and the analogous mechanical and electrical systems

are shown in Fig. 8-8. The system consists of an acoustic resistance RA,

an acoustic mass MA, and an acoustic compliance CA. The elements are in

‘‘series,’’ such that the volumetric flow rate is the same for each element. In

the acoustic system, the pressure is analogous to the mechanical force or the

electrical voltage, and the volumetric flow rate is analogous to the velocity of

a mass or the electrical current.

Silencer Design 341

FIGURE 8-8 Equivalent circuit diagram for the Helmholtz resonator. The pressure

at one terminal of the acoustic compliance is atmospheric pressure, or the acoustic

pressure is zero. The acoustic pressure at the other end of the compliance terminal is

pC. The Helmholtz resonator is analogous to the mechanical system with a spring,

damper, and mass.

8.3.2 Resonance for the Helmholtz Resonator

The resonant frequency for the Helmholtz resonator is similar in principle to

the natural frequency for vibration of a spring–mass system. For an electrical

system, resonance occurs at a frequency such that the electrical reactance

is zero. The resonant frequency for the Helmholtz resonator is given

by the following expression:

fo ј

2_рMACAЮ1=2 (8-43)

This relationship may be used for design purposes to determine the resonator

frequency for given resonator dimensions, or the required dimensions

may be found to achieve a given resonant frequency.

Example 8-1. A Helmholtz resonator is to be constructed of a cylinder

with the diameter equal to the length, D ј L1. The opening to the resonator

is an orifice with a radius a ј 10mm (0.394 in) and a thickness L ј 1mm

(0.039 in). It is desired to select the resonator dimensions such that the

resonant frequency is 250 Hz. The gas in the resonator volume is air at

218C (708F) and 101.3 kPa (14.7 psia), for which the density is 1.200 kg/

m3 (0.0749 lbm=ft3) and the sonic velocity is 343.8 m/s (1128 ft/sec).

Both ends of the hole in the resonator wall are ‘‘flanged,’’ so the

equivalent length of the resonator inlet may be found from Eqs (8-8) and

(8-12):

Le ј 1:00 ю р2Юр8=3_Юр10:0Ю ј 17:98mm

The acoustic mass may be calculated from Eq. (8-11):

MA ј

_oLe

_a2 ј р1:200Юр0:01798Ю

р_Юр0:010Ю2 ј 68:68 kg=m4

The required acoustic compliance for the resonator may be found

from Eq. (8-43):

CA ј

4_2f 2

oMA ј

р4_2Юр250Ю2р68:68Ю ј 5:901 _ 10_9 m5=N ј

_oc2

The resonator volume required to achieve a resonant frequency of 250 Hz

may be determined:

V ј р5:901Юр10_9Юр1:200Юр343:8Ю2 ј 0:837 _ 10_3 m3 ј 0:837dm3

The cylindrical volume (for D ј L1) is given by the following:

V ј 1

4 _D2L1 ј 1

4 _D3

342 Chapter 8

The resonator dimensions may be determined:

D ј L1 ј Ѕр4=_Юр0:837Юр10_3Ю_1=3 ј 0:1021m ј 102:1mm р4:02 inЮ

The wavelength of the sound wave in the resonator is as follows:

_ ј

fo ј

343:8

250 ј 1:375m _ L1 ј 0:1021m

The wavelength is much larger than the resonator dimensions, so the

lumped-parameter approach is valid.

8.3.3 Acoustic Impedance for the Helmholtz

Resonator

For steady-state operation, the acoustic impedance for the Helmholtz resonator

with all elements in series may be written in the following complex

form:

ZA ј

U ј RA ю j!MA _

!CA

(8-44)

ZA ј RA ю jMA ! _

!MACA

_ _

ј RA ю jMA ! _

(8-45)

The quantity !o is the circular resonant frequency:

!o ј 2_fo ј

рMACAЮ1=2 (8-46)

The acoustic impedance for the Helmholtz resonator may be written in

the following alternative form:

ZA ј RA 1 ю j

2_foMA

fo _

_ _ __

(8-47)

Let us define the acoustic quality factor QA by the following relationship:

QA ј

2_foMA

(8-48)

The acoustic quality factor gives a measure of the quality or sharpness of

tuning for the Helmholtz resonator (Nilsson, 1983). A large value for the

acoustic quality factor usually implies that there is a small acoustic resistance

present. The quality factor in the acoustic system is analogous to the

damping ratio (or, actually, the reciprocal of the damping ratio) for the

mechanical vibratory system. The acoustic impedance for the Helmholtz

resonator may be written in terms of the acoustic quality factor:

Silencer Design 343

ZA ј RA 1 ю jQA

fo _

_ _ __

(8-49)

The magnitude of the acoustic impedance may be found from Eq.

(8-49):

jZAj ј рRe2 ю Im2Ю1=2 ј RAЅ1 ю Q2

Aр f =fo _ fo=f Ю2_1=2 (8-50)

The phase angle between the incident acoustic pressure and the volume

flow rate through the necktube of the resonator may also be determined

from Eq. (8-49):

tan ј Im=Re ј QAЅр f =foЮ _ р fo=f Ю_ (8-51)

8.3.4 Half-Power Bandwidth

The acoustic quality factor gives a measure of how sharply the resonator

‘‘resonates.’’ Another qualitative measure of the sharpness of tuning is the

acoustic half-power bandwidth. The average acoustic power delivered to the

resonator is equal to the power dissipated in the resistance element, because

the capacitive and mass elements only momentarily store energy and do not

dissipate energy:

W ј U2RA ј рpin=jZAjЮ2RA (8-52)

The magnitude of the acoustic impedance may be found from Eq. (8-50):

W ј р p2

in=RAЮ

1 ю Q2

AЅр f =foЮ _ рfo=f Ю_2 (8-53)

The power delivered to the resonator at the resonant frequency, f ј fo, is

given by the following:

Wo ј

(8-54)

Equation (8-53) may be written in dimensionless form as follows:

Wo ј

1 ю Q2

AЅр f =foЮ _ рfo=f Ю_2 (8-55)

The ratio of acoustic power delivered to the resonator may also be expressed

in ‘‘level’’ form as follows:

_LW ј LWo _ LW ј 10 log10f1 ю Q2

AЅр f =foЮ _ р fo=f Ю_2g (8-56)

344 Chapter 8

A plot of the acoustic power delivered to the resonator as a function of

frequency is shown in Fig. 8-9. We may observe several facts from this

figure. First, the acoustic power is maximum when the frequency is equal

to the resonant frequency. Secondly, the curve is ‘‘spread out’’ for small

values of the acoustic quality factor (large damping), whereas it has a more

sharp peak for large values of the acoustic quality factor (small damping).

For frequency ratios larger than about р f =foЮ ј 10, the sound power level

decreases at a rate of about 6 dB/octave.

A quantitative measure of the sharpness of the peak in the power curve

for the Helmholtz resonator is given by the acoustic half-power bandwidth.

This quantity is defined as the difference between the two frequencies f1 and

f2 at which the acoustic power delivered to the resonator is one-half of the

power delivered at resonance. This condition corresponds to the frequencies

at which рW=WoЮ ј 1

2 in Eq. (8-55):

W=Wo ј 1

2 ј f1 ю Q2

AЅр f1;2=foЮ _ р fo=f1;2Ю_2g_1 (8-57)

Silencer Design 345

FIGURE 8-9 Graph of the ratio of the acoustic power delivered to a Helmholtz

resonator at any frequency f to the power delivered at the resonant frequency fo. The

plot is made for various values of the acoustic quality factor QA.

The two solutions for Eq. (8-57) are as follows:

fo ј р1 ю 4Q2

AЮ1=2 _ 1

2QA

(lower frequency) (8-58)

fo ј р1 ю 4Q2

AЮ1=2 ю 1

2QA

(upper frequency) (8-59)

The acoustic half-power bandwidth is found by taking the difference

of the two frequencies given by Eqs (8-58) and (8-59):

f2 _ f1

fo ј

(8-60)

If we multiply Eq. (8-58) by Eq. (8-59), we obtain the following:

f1f2

f 2

o ј р1 ю 4Q2

AЮ _ 1

4Q2

A ј 1

fo ј рf1 f2Ю1=2 (8-61)

The resonant frequency is the geometric mean of the upper and lower frequencies

of the acoustic half-power bandwidth.

Example 8-2. Consider the resonator in Example 8-1, for which the resonant

frequency was 250 Hz. Determine the magnitude of the acoustic impedance

at 250 Hz and at 125 Hz, the acoustic half-power bandwidth, and the

acoustic power delivered at 250 Hz and at 125 Hz. The incident sound pressure

level is 80 dB. The viscosity of air at 218C is ј 15:35 mPa-s.

First, let us determine the acoustic resistance for the opening of the

resonator. The parameter from Eq. (8-32) may be calculated for the resonant

frequency:

rv ј aр2_ fo_o=Ю1=2 ј р0:010ЮЅр2_Юр250Юр1:200Ю=р15:35Юр10_6Ю_1=2

rv ј 110:8 > 5:66

For a frequency f ј 125 Hz, we find rv ј 78:3 > 5:66. Both cases fall in the

‘‘intermediate tube’’ range, so the acoustic resistance may be calculated from

Eq. (8-34). The acoustic resistance at the resonant frequency, fo ј 250 Hz, is

as follows:

RA ј Ѕр4_Юр250Юр1:200Юр15:35Юр10_6Ю_1=2

р_Юр0:010Ю2

10 ю 2

_ _

ј 1608 Pa-s/m3

Similarly, we find the following value for the acoustic resistance at a frequency

f ј 125 Hz:

346 Chapter 8

RA ј р1608Юр125=250Ю1=2 ј 1137 Pa-s/m3

The acoustic quality factor, using the acoustic resistance at resonance,

is found from Eq. (8-48):

QA ј р2_Юр250Юр68:68Ю

р1608Ю ј 67:09

Similarly, the acoustic quality factor, using the acoustic resistance at 125 Hz,

is as follows:

QA ј р67:09Юр1608=1137Ю ј 94:88

The acoustic impedance at resonance р f ј foЮ is equal to the acoustic

resistance, as seen from Eq. (8-50). At a frequency of 250 Hz, the acoustic

impedance is as follows:

jZAj ј RA ј 1608 Pa-s/m3 ј 1:608 kPa-s/m3

The acoustic impedance at a frequency of 125 Hz may be calculated from

Eq. (8-50), using the quality factor corresponding to 125 Hz:

jZAj ј р1137Юf1 ю р94:88Ю2Ѕр250=125Ю _ р125=250Ю_2g1=2

jZAj ј р1137Юр142:3Ю ј 161:82 _ 103 Pa-s/m3 ј 161:82 kPa-s/m3

The phase angle between the incident acoustic pressure and the volumetric

flow rate is ј 0 at resonance. The phase angle at 125 Hz may be

found from Eq. (8-51):

tan ј р94:88ЮЅр125=250Ю _ р250=125Ю_ ј _142:3

ј _89:58 ј _1:564 rad

The phase angle for 125 Hz is almost _908 or _1

2 _ rad.

The acoustic half-power bandwidth may be found from Eq. (8-60):

_f ј

QA ј

250

67:09 ј 3:73 Hz ј f2 _ f1

The lower frequency f1 is found from Eq. (8-58):

fo ј Ѕ1 ю р4Юр67:09Ю2_1=2 _ 1

р2Юр67:09Ю ј 0:9926

f1 ј р0:9926Юр250Ю ј 248:1Hz

The upper frequency f2 is found as follows:

f2 ј f1 ю _f ј 248:1 ю 3:7 ј 251:8Hz

Silencer Design 347

The incident sound pressure is found from the sound pressure level:

pin ј р20Юр10_6Ю 1080=20 ј 0:200 Pa

The acoustic power delivered to the resonator at the resonant frequency is

found from Eq. (8-54):

Wo ј

RA ј р0:200Ю2

р1608Ю ј 24:86 _ 10_6W ј 24:86 mW

The acoustic power delivered to the resonator at 125 Hz is found from Eq.

(8-55):

Wo ј

1 ю р94:88Ю2Ѕр125=250Ю _ р250=125Ю_2 ј 0:04937 _ 10_3

W ј р24:86Юр10_6Юр0:04937Юр10_3Ю ј 1:227 _ 10_9W ј 1:227nW

We observe that the resonator readily accepts energy around the resonant

frequency; however, the energy delivered to the resonator rapidly

decreases as the frequency is increased (or decreased) from the resonant

value. This characteristic is important when considering the use of a sidebranch

muffler, which is based on the use of a Helmholtz resonator element.

8.3.5 Sound Pressure Level Gain

Another parameter of interest for the Helmholtz resonator is the difference

between the incident sound pressure level and the sound pressure level

within the resonator volume. The sound pressure level gain LG for the

resonator is defined as follows:

LG ј Lp;c _ Lp;in ј 20 log10р pc=pinЮ (8-62)

The quantity pc is the acoustic pressure in the cavity or resonator volume

and pin is the acoustic pressure incident on the resonator system.

The magnitude of the acoustic pressure within the cavity may be found

from Eq. (8-28) for the capacitive element:

j pcj ј

!CA

(8-63)

The volumetric flow rate and the incident pressure are related through the

system impedance, given by Eq. (8-50):

j pinj ј jZAjjUj (8-64)

We may combine Eqs (8-63) and (8-64) to obtain the expression for the

acoustic pressure ratio:

348 Chapter 8

j pcj

j pinj ј

!CAjZAj ј

MA=RA

!CAMAрjZAj=RAЮ

(8-65)

If we make the substitution for the resonant frequency from Eq. (8-43) and

the acoustic impedance from Eq. (8-50), we may write Eq. (8-65) in the

following form:

j pcj

j pinj ј р fo=f ЮQA

Ѕ1юQ2

Aр f =fo _fo=f Ю2_1=2 (8-66)

This expression has the same mathematical form as that for the mechanical

transmissibility, which we will examine in Chapter 9. It is noted that, at

resonance f =fo ј 1, and the pressure ratio is equal to the acoustic quality

factor QA. The sound pressure level at resonance is given by Eq. (8-62):

LGo ј 20 log10 QA (at resonance) (8-67)

Example 8-3. Determine the sound pressure level gain for the Helmholtz

resonator given in Example 8-2 at resonance (250 Hz) and at 125 Hz.

The sound pressure level gain at resonance may be found from Eq.

(8-67):

LGo ј 20 log10р67:09Ю ј 36:5dB

We note that the acoustic pressure and acoustic pressure level in the resonator

volume for resonant condition are as follows:

pc ј QA pin ј р67:09Юр0:200Ю ј 13:42 Pa

Lp;c ј Lp;in ю LG ј 80 ю 36:5 ј 116:5dB

The pressure ratio for a frequency of 125 Hz may be found from Eq.

(8-66):

j pcj

j pinj ј р250=125Юр94:88Ю

f1 ю р94:88Ю2Ѕр125=250Ю _ р125=250Ю_2g1=2 ј 0:1481 _ 10_3

The sound pressure level gain at 125 Hz is as follows:

LG ј 20 log10р0:1481 _ 10_3Ю ј _76:6dB

The sound pressure level in the resonator cavity for a frequency of 125 Hz is

much less than the sound pressure level of the incident sound wave:

Lp;c ј Lp;in ю LG ј 80 ю р_76:6Ю ј 3:4dB

We observe that the Helmholtz resonator in this example serves as a

good amplifier of sound only in the vicinity of the resonant frequency. This

Silencer Design 349

characteristic may be utilized when Helmholtz resonators are used in connection

with stereo systems to emphasize certain frequency ranges.

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