7.5 NOISE FROMAN ADJACENT ROOM

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In the previous material, we have considered the sound field produced by a

source of sound within the room. There are situations where the noise

source may be located in an adjacent room, such as the case of a conference

room adjoining a machinery room. In this case, the acoustic properties of

the adjacent room will influence the steady-state sound pressure level in the

room to be considered. We would like to examine the case of a noise source

in an adjacent room in this section.

7.5.1 Sound Source Covering One Wall

For the case of sound transmitted through a wall into a room, we may

consider the wall itself as a source of sound, as far as the interior of the

room is concerned. The reverberant sound field is independent of the location

of the sound source, if the sound field may be considered to be diffuse.

The acoustic energy density associated with the reverberant sound field is

given by Eq. (7-12):

DR ј

4W

cR

(7-54)

where quantityW is the acoustic power radiated from the wall, as shown in

Fig. 7-8, R is the room constant, and c is the sonic velocity in the air.

For small distances from the wall, or for r < рSw=2_Ю1=2, the sound

waves leaving the wall are practically plane waves, where Sw is the surface

area of the wall. For plane sound waves, the acoustic energy density associated

with the direct sound field is given by the following expression:

Room Acoustics 293

Copyright © 2003 Marcel Dekker, Inc.

DD ј

ID

c ј

W

Swc

(7-55)

The acoustic intensity ID is the acoustic power per unit area, W=Sw.

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

the reverberant and direct contributions:

D ј DD ю DR ј

W

c

4

R ю

1

Sw

_ _

ј

p2

_oc2 (7-56)

We may solve for the steady-state sound pressure (squared) from Eq. (7-56)

for the case of the receiver located near the wall source:

p2 ј _ocW

4

R ю

1

Sw

_ _

[for r < рSw=2_Ю1=2_ (7-57)

On the other hand, when the receiver is located a large distance from

the wall, or for r > рSw=2_Ю1=2, the source of sound (the wall) acts as a

source with a directivity factor Q ј 2. In this case, the acoustic energy

associated with the direct sound field is given by the following expression:

DD ј

ID

c ј

QW

4_r2c ј

W

2_r2c

(7-58)

The total acoustic energy density for this case is given as follows:

D ј

W

c

4

R ю

1

2_r2

_ _

ј

p2

_oc

[for r > рSw=2_Ю1=2_ (7-59)

294 Chapter 7

FIGURE 7-8 A wall of area Sw acting as a source of sound.

Copyright © 2003 Marcel Dekker, Inc.

The corresponding steady-state sound pressure (squared) may be found

from Eq. (7-59):

p2 ј _ocW

4

Rю

1

2_r2

_ _

[for r > рSw=2_Ю1=2_ (7-60)

Both Eqs (7-57) and (7-60) may be converted to ‘‘level’’ form by

introducing the reference pressure and reference power, taking log10 of

both sides, and multiplying by 10. The final result is as follows:

For r < рSw=2_Ю1=2 :

Lp ј LW ю10 log10

4

Rю

1

Sw

_ _

ю0:1 (7-61)

For r > рSw=2_Ю1=2 :

Lp ј LW ю10 log10

4

Rю

1

2_r2

_ _

ю0:1 (7-62)

In these equations, we have taken the value of the constant term as follows:

10log10

_ocWref

p2

ref

_ _

ј 0:1dB

We note that the values for the term associated with the direct sound field

are equal when r ј рSw=2_Ю1=2 ј r_.

7.5.2 Sound Transmission froman Adjacent

Room

The result obtained in the previous section may be applied to the case of

sound being generated in one room (by noisy equipment, for example) and

transmitted through a wall of area Sw into an adjacent room, as shown in

Fig. 7-9. Let us denote the room in which the source of sound is located as

Room (1), and the room in which the receiver (people, for example) is

located as Room (2). The steady-state sound pressure (squared) in the

room with the source of sound may be determined from Eq. (7-15):

p21

ј _ocW

4

R1 ю

Q

4_r21

_ _

(7-63)

Similarly, we have shown that the steady-state sound pressure

(squared) in the other room due to sound transmitted through the wall is

given by Eq. (7-57) or Eq. (7-60):

Room Acoustics 295

Copyright © 2003 Marcel Dekker, Inc.

p22

ј _ocW2

4

R2 ю

1

Sw

_ _

[for r2 < рSw=2_Ю1=2_ (7-64)

p22

ј _ocW2

4

R2 ю

1

2_r22

_ _

[for r2 > рSw=2_Ю1=2_ (7-65)

The sound power transmission coefficient at for the wall is defined by

the following expression:

at ј

W2

Win ј

W2

IinSw ј

W2R1

SwW

(7-66)

The acoustic power transmitted into the second room through the wall

between the rooms may be found from Eq. (7-66):

W2 ј atSwW=R1 (7-67)

If we substitute this result from Eq. (7-67) into Eqs (7-64) and (7-65),

we obtain the following expressions for the sound pressure (squared):

p22

ј

_ccW

R1

4Sw

R2 ю 1

_ _

at [for r2 < рSw=2_Ю1=2_ (7-68)

p22

ј

_ocW

R1

4Sw

R2 ю

Sw

2_r22

_ _

at [for r2 > рSw=2_Ю1=2_ (7-69)

Note that the transmission loss for the wall is defined by Eq. (4-90):

TL ј 10 log10р1=atЮ (7-70)

296 Chapter 7

FIGURE 7-9 Sound transmitted from an adjacent room through an interior wall.

Copyright © 2003 Marcel Dekker, Inc.

Equations (7-68) and (7-69) may be converted to ‘‘level’’ form as follows:

Lp2 јLW _ 10 log10рR1Ю ю 10 log10

4Sw

R2 ю 1

_ _

_ TL ю 0:1

Ѕfor r2 < рSw=2_Ю1=2_ (7-71)

Lp2 јLW _ 10 log10рR1Ю ю 10 log10

4Sw

R2 ю

Sw

2_r22

_ _

_ TL ю 0:1

Ѕfor r2 > рSw=2_Ю1=2_ (7-72)

The sound pressure level in the first room, the room with the noise

source, is given by Eq. (7-18):

Lp1 ј LW ю 10 log10

4

R1 ю

Q

4_r21

_ _

ю 0:1 (7-73)

The designer has several choices to produce a noise reduction in the

room adjacent to the room with the noise source, including:

1. Increase the room constant R1 for the room containing the sound

source. This approach reduces noise in the adjacent room

(Room 2) by reducing the reverberant noise before the sound

is transmitted through the wall. If all other factors remain

unchanged, increasing the average surface absorption coefficient,

for example, from 0.10 to 0.20 in Room 1, will increase the room

constant by a factor of 2.25 and decrease the sound pressure level

in Room 2 by about 3.5 dB.

2. Increase the room constant R2 for the adjacent room by adding

acoustic treatment to the room surfaces, for example. If all other

factors remain unchanged, increasing the average surface absorption

coefficient from 0.10 to 0.20 in Room 2 will increase the

room constant by a factor of 2.25 and decrease the sound pressure

level in Room 2 by about 1.5 dB near the separating wall

and up to about 3.5 dB far from the wall.

3. Increase the transmission loss for the wall between the two

rooms. In many situations, this approach results in the most

significant noise reduction. For example, if the wall thickness is

doubled, the transmission loss will be increased by 6 dB and the

sound pressure level in Room 2 will be decreased by 6 dB.

4. The sound pressure level in Room 2 can also be reduced by

reducing the sound power level for the sound source in Room

1. This approach may not always be feasible.

Room Acoustics 297

Copyright © 2003 Marcel Dekker, Inc.

Example 7-6. A Jordan refiner used in a paper mill is located in a room

having a total surface area of 900m2 (9688 ft2) and an average absorption

coefficient of 0.05. The refiner has a sound power level of 105 dB and a

directivity factor of 2.0. The refiner is located 4m (13.1 ft) from the wall

of the operator’s room. The operator’s room has a total surface area of

100m2 (1076 ft2) and an average surface absorption coefficient of 0.35.

The transmission loss for the wall between the refiner room and the operator’s

room is 30 dB. The surface area of the wall between the two rooms is

16m2 (172 ft2). The operator is located 1.5m (4.9 ft) from the wall.

Determine the steady-state sound pressure level in the refiner room and in

the operator’s room, neglecting the effect of air absorption.

The room constant for the refiner room may be calculated from Eq.

(7-13):

R1 ј

__S1

1 _ __1 ј р0:05Юр900Ю

1 _ 0:05 ј 47:37m2

The steady-state sound pressure level in the refiner room is found from

Eq. (7-73):

Lp1 ј 105 ю 10 log10

4

47:37 ю

2:0

р4_Юр4:0Ю2

_ _

ю 0:1

Lp1 ј 105 ю 10 log10р0:08444 ю 0:00995Ю ю 0:1 ј 105 ю р_10:3Ю ю 0:1

Lp1 ј 94:8dB

The room constant for the operator’s room is as follows:

R2 ј

__2S2

1 _ __2 ј р0:35Юр100Ю

1 _ 0:35 ј 53:85m2

Let us calculate the following quantity:

r_ ј рSw=2_Ю1=2 ј р16=2_Ю1=2 ј 1:596m р5:24 ftЮ

The distance between the operator (receiver) and the wall is as follows:

r2 ј 1:50m < 1:596m ј r_

The sound pressure level in the operator’s room is found from Eq.

(7-71):

Lp2 ј 105 _ 10 log10р47:37Ю ю 10 log10 р4Юр16Ю

р53:85Ю ю 1

_ _

_ 30 ю 0:1

Lp2 ј 105 _ 16:8 ю 3:4 _ 30 ю 0:1 ј 61:7dB

298 Chapter 7

Copyright © 2003 Marcel Dekker, Inc.

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