7.2 STEADY-STATE SOUND LEVEL IN A ROOM

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When a source of sound is turned on in a room, some of the energy emitted

from the source is absorbed at the surfaces of the room and some energy is

reflected back into the room. Steady-state conditions are usually achieved

274 Chapter 7

TABLE 7-1 Subjective Acoustic Characteristics of

Rooms

Room characteristic Average absorption coefficient

Dead room 0.40

Medium-dead room 0.25

Average room 0.15

Medium-live room 0.10

Live room 0.05

Copyright © 2003 Marcel Dekker, Inc.

after at time on the order of 0.25 seconds for most rooms. For a room

having floor dimensions of 8m _ 8m (26.2 ft _ 26.2 ft), a sound wave

will cross the room in (8 m)/(347 m/s) ј 0:023 seconds. The sound wave

will cross the room р0:25Ю=р0:023Ю ј 10ю times in 0.25 seconds. When

steady-state conditions have been achieved, the energy supplied by the

sound source is equal to the energy absorbed by the room surfaces.

For steady-state conditions, we may divide the acoustic field in the

room into two parts, as shown in Fig. 7-4:

1. The direct sound field, which consists of the acoustic energy associated

with sound waves that have not struck surfaces in the

room.

2. The reverberant sound field, which consists of the remainder of

the energy.

The reverberant sound field is associated with all of the sound waves that

have been reflected one or more times from the various surfaces in the room.

The acoustic energy density associated with the direct sound field is

given by Eqs (2-21) and (2-27):

DD ј

ID

c ј

QW

4_r2c

(7-3)

Room Acoustics 275

FIGURE 7-4 The reverberant and the direct sound fields.

Copyright © 2003 Marcel Dekker, Inc.

The quantity Q is the directivity factor for the sound source, W is the

acoustic power radiated by the sound source into the room, r is the distance

between the source of sound and the receiver, and c is the speed of sound in

the air in the room.

To calculate the reverberant sound field, let us consider the element

shown in Fig. 7-5. The acoustic energy contained in the small elemental

volume is equal to the product of the energy per unit volume (acoustic

energy density) and the element volume:

dE ј DR dr dS (7-4)

The sound power radiated from the small area dS toward the surface

element _S is equal to the change in energy of the element per unit time,

where c ј dr=dt.

dW ј

dE

dt ј DR c dS (7-5)

For a uniform reverberant sound field, the intensity of the acoustic

energy radiated from the small area dS is the sound power per unit area:

dI ј

dW

4_r2 ј

DR c dS

4_r2 (7-6)

276 Chapter 7

FIGURE 7-5 Reverberant sound field incident on a small element of the surface of

the room _S.

Copyright © 2003 Marcel Dekker, Inc.

The acoustic power incident on the room surface area increment _S from

the small elemental area dS is the product of the energy per unit area and the

projected area in the direction of the element dS:

dWin ј dIр_S cos _Ю (7-7)

The differential area on the spherical surface is given by the following

expression in spherical coordinates:

dS ј r2 sin _ d_ d’ (7-8)

If we make the substitutions from Eqs (7-6), (7-7), and (7-8) and

integrate, we obtain the total reverberant acoustic power incident on the

small surface _S:

dWin ј

DRc_S

4_r2

р2_

0

р_=2

0

r2 sin _ cos _ d_ d’

dWin ј 1

4DRc_S (7-9)

Assuming a uniform reverberant sound field, we may integrate Eq. (7-9)

over the surface area of the room So to obtain the total reverberant acoustic

power incident on the room surface:

Win ј 1

4DRcSo (7-10)

The acoustic power absorbed by the room surface is given by р__WinЮ,

according to the definition of the surface absorption coefficient in Eq. (7-1).

In steady state, the acoustic power absorbed is equal to the power that is

supplied by the reverberant sound field:

Wр1 _ __Ю ј __Win ј 1

4DRc__So (7-11)

We may solve for the reverberant acoustic energy density from Eq. (7-11):

DR ј

4Wр1 _ __Ю

c__So ј

4W

cR

(7-12)

The quantity R is the room constant, defined by the following expression, for

negligible energy attenuation in the room air:

R ј

__So

1 _ __

(7-13)

The total acoustic energy density in the room in steady state is the sum

of the contributions due to the direct sound field and the reverberant sound

field:

D ј DR ю DD ј

4W

cR ю

QW

4_r2c ј

W

c

4

R ю

Q

4_r2

_ _

ј

p2

_oc2 (7-14)

Room Acoustics 277

Copyright © 2003 Marcel Dekker, Inc.

The steady-state sound pressure may be found from Eq. (7-14):

p2 ј _ocW

4

R ю

Q

4_r2

_ _

(7-15)

Equation (7-15) may be written in an alternative form by introducing the

acoustic reference quantities:

p2

p2

ref ј

W

Wref

4

R ю

Q

4_r2

_ _

_ocWref

p2

ref

(7-16)

This expression may be converted to ‘‘level’’ form by taking log10 of both

sides and multiplying through by 10:

Lp ј LW ю 10 log10

4

R ю

Q

4_r2

_ _

ю 10 log10

_ocWref

p2

ref

_ _

(7-17)

" "

reverberant

sound field

_ _

direct

sound field

_ _

As discussed in Sec. 5.1, the value of the last term in Eq. (7-17) for air at

101.3 kPa and 300K is 0.1 dB. The final form for the expression for the

steady-state sound pressure level in a room may be written as follows:

Lp ј LW ю 10 log10

4

R ю

Q

4_r2

_ _

ю 0:1 (7-18)

The room constant R must be expressed in m2 units, and the distance

between the source and receiver r must be expressed in m units in Eq. (7-18).

An important observation may be made from Eq. (7-18). The term

р4=RЮ is associated with the reverberant sound field, which is dependent on

the absorption characteristics of the room. The term рQ=4_r2Ю is associated

with the sound coming directly from the sound source or the direct sound

field, which is independent of the properties of the room. If the second term

predominates, or if the direct sould field is much larger than the reverberant

sound field, very little reduction in the sound pressure level can be achieved

by adding more acoustic absorptive material to the room surfaces. In fact, it

would be a waste of money and effort to buy and install acoustic material on

the wall or ceiling, for this case. Other noise control techniques, such as using

an acoustic barrier or enclosure, would be required in cases where the direct

sound field is predominating. On the other hand, if the first term predominates,

or if the reverberant sound field is much larger than the direct sound

field, the steady-state sound pressure level can be reduced by adding acoustic

material on the surfaces of the room.

278 Chapter 7

Copyright © 2003 Marcel Dekker, Inc.

Example 7-1. A room has dimensions of 6.20m (20.3 ft) _ 6.00m

(19.7 ft) _ 3.10m (10.2ft) high, as shown in Fig. 7-6. The room has one

solid wood door (1.20m _ 2.20m or 3.94 ft _ 7.22 ft) in the 6-m wall. The

walls are plaster on lath. The ceiling is 1

2-in acoustic tile on hard backing, and

the floor is covered with a 3/8-in carpet on concrete floor, with no pad. A

machine (sound power levels given in Table 7-2) is located in the room, and

the directivity factor for the machine is Q ј 4 for all frequencies. There are

six adults standing in the room. Determine the octave band steady-state

sound pressure level in the room at a distance of 6.00m (19.7ft) from the

machine.

Let us carry out the calculations for the 500Hz octave band in detail.

The results for the other octave bands are given in Table 7-2. The surface

absorption coefficients and surface areas are found as follows:

1. Walls:

_1 ј 0:06

S1 ј р2Юр6:20ю6:00Юр3:10Ю _ р1:20Юр2:20Ю ј 75:64_2:64

ј 73:00m2

RoomAcoustics 279

FIGURE 7-6 Diagram for Example 7-1.

Copyright © 2003 Marcel Dekker, Inc.

2. Door:

_2 ј 0:05

S2 ј р1:20Юр2:20Ю ј 2:64m2

3. Ceiling:

_3 ј 0:55

S3 ј р6:20Юр6:00Ю ј 37:20m2

4. Floor:

_4 ј 0:21

S4 ј 37:20m2

5. People:

р_SЮ ј р6 peopleЮр0:44Ю ј 2:64m2

The total surface area of the room is:

So ј 75:64 ю р2Юр37:20Ю ј 150:04m2 р1615 f t2Ю

The average surface absorption coefficient at 500 Hz may be found

from Eq. (7-2):

__ ј р0:06Юр73:00Ю ю р0:05Юр2:64Ю ю р0:55Юр37:20Ю ю р0:21Юр37:20Ю ю 2:64

р150:04Ю

__ ј

4:38 ю 0:132 ю 20:46 ю 7:812 ю 2:64

150:04 ј

35:424

150:04 ј 0:2361

280 Chapter 7

TABLE 7-2 Solution for Example 7-1

Item

Octave band center frequency, Hz

125 250 500 1,000 2,000 4,000

LW, dB (given) 52 57 60 56 50 43

__ 0.1709 0.1784 0.2361 0.4034 0.4057 0.4113

R, m2 30.93 32.58 46.37 101.5 102.4 104.8

10 log10

4

R

ю Q

4_r2

_ _

_8.6 _8.8 _10.2 _13.2 _13.2 _13.3

Lp(OB), dB 43.5 48.3 49.9 42.9 36.9 29.8

Copyright © 2003 Marcel Dekker, Inc.

The room constant at 500Hz is calculated from Eq. (7-13):

R ј р150:04Юр0:2361Ю

р1_0:2361Ю ј 46:372m2

The steady-state sound pressure level in the 500Hz octave band is

found from Eq. (7-18):

Lp ј 60ю10log10

4

46:372 ю р4:0Ю

р4_Юр6:00Ю2

_ _

ю0:1

Lp ј 60ю10log10р0:08626ю0:00884Юю0:1 ј 60юр_10:2Юю0:1

Lp ј 49:9dB

It is noted that the sound pressure level could be decreased by adding

sound absorption material on the walls, because the reverberant sound field

contribution (0.08626) is much larger (almost 10 times larger) than the direct

sound field contribution (0.00884).

If we extrapolate the sound pressure spectrum to estimate

Lpр63HzЮ ј 37dB, the overall sound pressure level due to the source is

found as follows:

Lp ј 10 log10р103:70 ю104:35 ю104:83 ю_ _ _ю102:98Ю ј 53:4 dB

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