7.4 EFFECT OF ENERGY ABSORPTION IN THE AIR

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As was discussed in Sec. 4.12, the effect of dissipation of acoustic energy in

the air through which the sound wave is moving is generally important only

for high-frequency sound and for sound transmitted over large distances.

The same behavior is observed for sound transmitted in rooms. In this

section, we will consider the effect of dissipation of energy in the air in a

room on the steady-state sound level and on the reverberation time.

7.4.1 Steady-State Sound Level with Absorption

in the Air

In addition to the energy absorbed at the walls, there is an attenuation of the

sound due to absorption in the volume of air in the room. This effect also

modifies the energy from the reverberant field. Equation (7-11) may be

written in the following form to include these effects:

Wр1 _ __Ю e_md ј __Win ю р1 _ e_md ЮWin ј 1

4DRcSoр1 ю __ _ e_md Ю

(7-39)

The quantity d is the mean free path of the sound wave, given by Eq. (7-23),

and m is the energy attenuation coefficient. The first term on the left side of

Eq. (7-39) represents the energy delivered to the reverberant field or the

energy that has not been absorbed after striking the walls or passing through

the air before striking the walls. We may solve for the reverberant acoustic

energy density, including the effect of energy absorption in the air:

DR ј

4Wр1 _ __Ю e_md

cSoр1 ю __ _ e_md Ю ј

4W

cR

(7-40)

The room constant, including the effects of air attenuation, is given by the

following expression from Eq. (7-40):

R ј

Soр1 ю __ _ e_md Ю

р1 _ __Ю e_md (7-41)

Room Acoustics 289

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In general, the term md is usually small for rooms. As an example, the

energy attenuation coefficient for air at 258C (778F) and 60% relative

humidity at 2000Hz is m ј 2:16km_1 ј 0:00216m_1, from Table 4-8. Let

us consider a room having dimensions 30m _ 15m _ 5m high (98:4 ft_ 49:2 ft _ 16:4 ftЮ. The volume is V ј 2250m3, and the surface area is

So ј 1350m2, so the mean free path is d ј 4V=So ј 6:667 m. The quantity

md ј р0:00216Юр6:667Ю ј 0:0144. For small vlaues of md, we may expand

the exponential expression and approximate the exponential by the first two

terms in the series:

e_md ј 1 _ md __ _ _ _ 1 _ 4mV=So (7-42)

If we make the substitution from Eq. (7-42) for small md (or md _ 0:20Ю into Eq. (7-41), we obtain the following expression for the room constant:

R ј

Soр__ ю 4mV=SoЮ

1 _ __ _ р4mV=SoЮ

(7-43)

The direct sound field is also affected by the attenuation in the air in

the room. The modification of Eq. (7-3) to include energy attenuation effects

is as follows:

DD ј

QW e_mr

4_cr2 (7-44)

If the value of the parameter mr is less than about 0.10, the exponential in

Eq. (7-44) is approximately unity and may be neglected. In the previous

example, this would correspond to a distance between the sound source

and receiver of r ј 0:10=0:00216 ј 46:3m (152 ft).

The total acoustic energy density is the sum of the reverberant field,

Eq. (7-40), and the direct field, Eq. (7-44):

D ј DR ю DD ј

W

c

4

R ю

Qe_mr

4_r2

_ _

ј

p2

_oc2 (7-45)

The corresponding sound pressure level may be found by introducing the

reference pressure and reference power, then taking log10 of both sides of the

resulting expression and multiplying by 10:

Lp ј LW ю 10 log10

4

R ю

Qe_mr

4_r2

_ _

ю 0:1 (7-46)

Equation (7-43) must be used to evaluate the room constant, if air attenuation

effects are to be included.

Example 7-4. Determine the octave band sound pressure levels for the

500 Hz and 4000 Hz octave bands for the room given in Example 7-1, if

290 Chapter 7

Copyright © 2003 Marcel Dekker, Inc.

air attenuation is considered. The air in the roomis at 218C (708F) and 50%

relative humidity.

From Table 4-8, we find the following values for the energy attenuation

coefficient:

m ј 0:39 km_1 at 500 Hz and m ј 6:11 km_1 at 4000 Hz.

For a frequency of 500 Hz, the dimensionless parameter is as follows:

4mV=So ј р4Юр0:39Юр10_3Юр115:32Ю=р150:04Ю ј 1:20 _ 10_3

The room constant is found from Eq (7.43):

R ј р150:04Юр0:2361 ю 0:0012Ю

1 _ 0:2361 _ 0:0012 ј 46:68m2

The exponential factor in the direct sound field expression is as follows:

e_mr ј expЅ_р0:39Юр10_3Юр6Ю_ ј expр_0:00234Ю ј 0:9977

The octave band sound pressure level for the 500 Hz octave band is as

follows:

Lp ј 60 ю 10 log10

4

46:68 ю р4Юр0:9977Ю

р4_Юр6Ю2

_ _

ю 0:1

Lpр500 HzЮ ј 60 ю р_10:2Ю ю 0:1 ј 49:9dB

If we repeat the calculations for the 4000 Hz octave band, we obtain

the following values:

4mV=So ј 0:01878

R ј 113:23m2

e_mr ј 0:9640

Lpр4000 HzЮ ј 43 ю р_13:6Ю ю 0:1 ј 29:5dB

It is observed that the effect of air attenuation is negligible (less than

0.1 dB) in the 500 Hz octave band and is essentially negligible (about 0.3 dB

difference) for the 4000 Hz octave band, in this example. If all the room

dimensions were increased by a factor of 10, then the air attenuation would

be more significant, particularly in the 4 kHz octave band.

7.4.2 Reverberation Time with Absorption in the

Air

The general effect of air attenuation is to decrease the reverberation time,

since there is an additional mechanism (air attenuation) present to remove

energy from the acoustic field. If we include the effect of air attenuation, Eq.

Room Acoustics 291

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(7-19) for the acoustic energy density after the first reflection would be

modified as follows:

D1 ј Doр1 _ __Ю e_md (7-47)

The acoustic energy density after the second reflection is found similarly:

D2 ј D1р1 _ __Ю e_md ј Doр1 _ __Ю2 e_2md (7-48)

The acoustic energy density after n reflections is given by the following

expression:

Dn ј Doр1 _ __Юn e_nmd (7-49)

If we substitute the number of reflections n from Eq. (7-26) in Eq.

(7-49), the following relationship is obtained for the time dependence of the

acoustic energy density, considering air attenuation:

D ј Doр1 _ __ЮрSoc=4VЮt e_mct (7-50)

This expression may be written in the following alternative form:

D ј Do exp _

cSot

4V

ln

1

1 _ __

_ _

_ mct

_ _

(7-51)

If we introduce the number of absorption units a from Eq. (7-30), Eq. (7-51)

may be written in the following form:

D ј Do exp _

ct

4V рa ю 4mVЮ

h i

(7-52)

The acoustic energy density in terms of the acoustic pressure is given

by Eq. (7-31). If we set the difference between the original sound pressure

level and the sound pressure level after a time Tr (the reverberation time)

equal to 60 dB, the following expression is obtained for the reverberation

time, including the effect of attenuation in the air:

Tr ј

55:26V

cрa ю 4mVЮ

(7-53)

According to Eq. (7-53), the effect of air attenuation increases the absorption

from a to рa ю 4mVЮ. The effect of air attenuation is more pronounced

for large rooms (large volumes) than for small rooms.

Example 7-5. Determine the reverberation time in Problem 7-2 if the effect

of air attenuation were to be considered. The air in the room is at 218C

(708F) and 50% relative humidity. The energy attenuation coefficient at

500 Hz is m ј 0:39km_1 and the number of absorption units is a ј 18:89m2, according to the Fitzroy relationship.

292 Chapter 7

Copyright © 2003 Marcel Dekker, Inc.

The value of the parameter associated with air attenuation effects is as

follows:

4mV ј р4Юр0:39Юр10_3Юр115:32Ю ј 0:18m2

The total absorption may be calculated:

aю4mV ј 18:89ю0:18 ј 19:07m2

The reverberation time, including air attenuation, is found from Eq. (7-53):

Tr ј р55:26Юр115:32Ю

р343:8Юр19:07Ю ј 0:972 s

The reverberation time calculated in Example 7-2, neglecting air attenuation,

was 0.981 s, so the effect of air attenuation in this example is to

decrease the reverberation time by about 1% or about 0.010 s.

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