PRACTICE PROBLEM ONE: From Previous lecturePRACTICE PROBLEM ONE: From Previous lecture

A 16’ x 20’ x 9’ room has absorptive coefficients A 16’ x 20’ x 9’ room has absorptive coefficients as follows: Ignore doors & windows.as follows: Ignore doors & windows.

WallsWalls .30.30FloorFloor .25.25CeilingCeiling .40.40

A. Find total room absorptionA. Find total room absorptionB. Find reverberation timeB. Find reverberation time

The finishes then change to the following The finishes then change to the following coefficients:coefficients:

WallsWalls .46.46FloorFloor .40.40CeilingCeiling .86.86

C. Find the noise reduction due to the new C. Find the noise reduction due to the new finishesfinishesD. Find the difference in reverberation time.D. Find the difference in reverberation time.

PROBLEM SOLUTIONPROBLEM SOLUTION Wall area = Wall area = 648 x .30648 x .30 == 194.4194.4 Floor area = Floor area = 320 x .25320 x .25 == 80.0 80.0 Ceiling area = Ceiling area = 320 x .40320 x .40 == 128.0128.0 total = 402.4 sabinstotal = 402.4 sabins

T = .05 x (2880 / 402.4) = T = .05 x (2880 / 402.4) = .36 seconds.36 seconds

NEWNEW

Wall area =Wall area = 648 x 648 x .46.46 == 298.08298.08Floor area =Floor area = 320 x 320 x .40.40 == 128.0128.0Ceiling area = Ceiling area = 320 x 320 x .86.86 == 275.2275.2

total = total = 701.28701.28Noise Reduction = 10log (701.28/402.4) = 10 log Noise Reduction = 10log (701.28/402.4) = 10 log

1.7427 =1.7427 =10 x .2412 = 2.41 db. T = .05 x 4.1068 = .205 10 x .2412 = 2.41 db. T = .05 x 4.1068 = .205 secsec

Time difference = .36 - .205 = .1550 sec, = 43%Time difference = .36 - .205 = .1550 sec, = 43%

PRACTICE PROBLEM TWO:PRACTICE PROBLEM TWO:

SOUND PROJECTION SOUND PROJECTION Say a fire alarm horn Say a fire alarm horn sounds at a loudness of sounds at a loudness of 100 decibels100 decibels at a at a distance of distance of 12 feet12 feet. At what intervals must horns . At what intervals must horns be placed so that a minimum of be placed so that a minimum of 90 decibels90 decibels can can be heard. Find the distance, x, from the horn be heard. Find the distance, x, from the horn that the sound be only 90 decibels. Consider no that the sound be only 90 decibels. Consider no help from sound reflection.help from sound reflection.

Solution: Solution: FirstFirst, find the amount of sound , find the amount of sound energy in watts/cmenergy in watts/cm2 2 , created at the distance of , created at the distance of 12 feet. 12 feet. SecondSecond, find the amount of energy , find the amount of energy required to produce 90 decibels. required to produce 90 decibels. ThirdThird, find the , find the distance where the 90 db sound energy is distance where the 90 db sound energy is produced. produced.

FirstFirst: EQUATION 1: 100 db = 10 log ( I.E.: EQUATION 1: 100 db = 10 log ( I.E.1212 / 10 / 10--

1616 ) simplify the equation by dividing both sides ) simplify the equation by dividing both sides by 10; soby 10; so

10 db = log (I.E.10 db = log (I.E.1212 / 10 / 10-16-16 ) now, since you ) now, since you cannot solve for I.E.cannot solve for I.E.1212 inside the log function, inside the log function, take the anti-log of each side of the equation to take the anti-log of each side of the equation to get rid of the log function; anti-log of 10 = get rid of the log function; anti-log of 10 = 10101010 ; ;

anti-log of the right side, log (I.E.anti-log of the right side, log (I.E.1212 / 10 / 10-16-16 ) is ) is simply, simply,

I.E.I.E.1212 / 10 / 10-16-16 , so the resulting equation is , so the resulting equation is

10101010 = I.E. = I.E.1212 / 10 / 10-16-16, rearrange and solve for I.E., rearrange and solve for I.E.1212; ;

I.E.I.E.1212 = 10 = 101010 x 10 x 10-16-16 = = 1010-6-6 watts/cm2 watts/cm2

SecondSecond: EQUATION 1 Find the energy : EQUATION 1 Find the energy required to produce 90 decibels. Follow the required to produce 90 decibels. Follow the same procedure as previous: same procedure as previous:

90 = 10 log ( I.E.90 = 10 log ( I.E.xx / 10 / 10-16-16 ) ; divide both sides ) ; divide both sides by 10, and 9 = log ( I.E.by 10, and 9 = log ( I.E.xx / 10 / 10-16-16 ) )

Take anti-log of both sides, and 10Take anti-log of both sides, and 1099 = I.E. = I.E.xx / / 1010-16-16 ; rearrange and I.E. ; rearrange and I.E.xx = 10 = 1099 x 16 x 16-16-16 ; ;

so the amount of energy required to produce 90 so the amount of energy required to produce 90 decibels at the x distance; decibels at the x distance; I.E.I.E.xx = 10 = 10-7-7 watts/cmwatts/cm22

Realize here that 10Realize here that 10-7-7 is a smaller number than is a smaller number than 1010-6-6

Third:Third: EQUATION 2 With the Inverse Square EQUATION 2 With the Inverse Square Law Formula, find the x distance; Law Formula, find the x distance;

I.E.I.E.1212 / I.E. / I.E.xx = ( x / 12 ) = ( x / 12 )22 ; 10 ; 10-6-6 / 10 / 10-7-7 = x = x22 / / 144; rearrange the formula to solve for x144; rearrange the formula to solve for x22 , then , then x ; x ;

divide the left side, so 10divide the left side, so 10-6-6 x 10 x 1077 = x = x22 / 144 ; / 144 ;

so 10 = xso 10 = x22 / 144 ; and solving for x / 144 ; and solving for x22 ; x ; x22 = 10 = 10 times 144 = 1440, so times 144 = 1440, so

x = √1440 , and x = 38 feet. x = √1440 , and x = 38 feet.

So the Interval of fire horns is two times that So the Interval of fire horns is two times that distance, or = distance, or = 2 x 38’ = 76 feet2 x 38’ = 76 feet

FORMULAS FOR SOUND PROJECTION, FORMULAS FOR SOUND PROJECTION, ABSORPTION, AND ISOLATION:ABSORPTION, AND ISOLATION:

IL = 10 log IE / 10IL = 10 log IE / 10-16 -16 SOUND PROJECTION SOUND PROJECTION

IEIE1 1 / IE/ IE22 = [ d = [ d22 / d / d11 ] ]2 2 SOUND ENERGY & DISTANCE SOUND ENERGY & DISTANCE

NR = 10 log [ aNR = 10 log [ a22 / a / a11 ] SOUND REDUCTION BY ] SOUND REDUCTION BY ABSORPTION ON ABSORPTION ON SURFACESSURFACES

NR = TL+10 log ( a NR = TL+10 log ( a R R / S ) SOUND REDUCTION BY / S ) SOUND REDUCTION BY AN ISOLATION AN ISOLATION

BARRIERBARRIER

SOUND REDUCTION SOUND REDUCTION

BY A PHYSICAL BARRIERBY A PHYSICAL BARRIER

SOUND ISOLATIONSOUND ISOLATION

Sound isolation is the restriction of sound from Sound isolation is the restriction of sound from areas where it is not desirable. Examples areas where it is not desirable. Examples include:include:

Privacy within spacesPrivacy within spaces

SecuritySecurity

Annoyance in task areasAnnoyance in task areas

Annoyance due to vibration & sound of equipmentAnnoyance due to vibration & sound of equipment

Isolation is accomplished by:Isolation is accomplished by:

Sound reduction by Isolation Noise ReductionSound reduction by Isolation Noise Reduction

Sealing of openings and cracksSealing of openings and cracks

Rerouting and absorptionRerouting and absorption

Vibration absorptionVibration absorption

Masking by other sounds Masking by other sounds

Sound isolation by use of a barrier that prevents Sound isolation by use of a barrier that prevents transmission of sound is effective in providing transmission of sound is effective in providing privacy and security of speech. privacy and security of speech.

All construction assemblies used as sound All construction assemblies used as sound barriers have been tested for their integrity in barriers have been tested for their integrity in reducing transmission of sound energy. reducing transmission of sound energy.

Tested assemblies are given a rating, over the Tested assemblies are given a rating, over the six sound frequencies, called “Sound six sound frequencies, called “Sound Transmission Class.” TL, or Transmission Loss is Transmission Class.” TL, or Transmission Loss is the number of decibels loudness the assembly the number of decibels loudness the assembly will prevent from passing through.will prevent from passing through.

Continuous sounds which manage to travel Continuous sounds which manage to travel through an assembly will reverberate within an through an assembly will reverberate within an adjacent space if the surfaces are reflective, and adjacent space if the surfaces are reflective, and reinforce the sound level within that space. reinforce the sound level within that space.

NOISE REDUCTION NOISE REDUCTION BY A BARRIERBY A BARRIER is measured by is measured by the formula, the formula,

NR = TL + 10 log ( a NR = TL + 10 log ( a R R / S )/ S ) where where

NR = noise reduction in decibelsNR = noise reduction in decibels

TL = sound transmission loss of the barrier in TL = sound transmission loss of the barrier in decibels, found from the chartdecibels, found from the chart

a a RR = the room absorption of the receiving room = the room absorption of the receiving room

S = surface area of the barrier in square feetS = surface area of the barrier in square feet

Noise Reduction becomes the Noise Reduction becomes the DIFFERENCEDIFFERENCE in in the loudness of the original sound, and the the loudness of the original sound, and the loudness of the sound in the receiving room.loudness of the sound in the receiving room.

Do not confuse Noise Reduction with Intensity Do not confuse Noise Reduction with Intensity Loudness. Loudness.

60 decibelsemerges at 8 db

2" x 4" studs@ 16" oc

@ 2'-0" ochat channels

fiberglass insulation3" sound attenuation

Initial sound of

5/8"gypsum bdboth sides

decibels8

WALL DESIGN NUMBER 5 TRANSMISSION LOSS EQUALS 52 DECIBELS AT FREQUENCY OF 500 HZ

But realize that sound that gets through, or But realize that sound that gets through, or around a barrier into an adjacent room is around a barrier into an adjacent room is available to reverberate within the receiving available to reverberate within the receiving room and build up because of reflection.room and build up because of reflection.

Short duration sounds, such as a door closing do Short duration sounds, such as a door closing do not last long enough to increase the noise in a not last long enough to increase the noise in a receiving room. But steady sound such as that receiving room. But steady sound such as that from the operation of machinery can build up in a from the operation of machinery can build up in a receiving room if the surfaces are reflective, and receiving room if the surfaces are reflective, and the noise level will be louder than when it first the noise level will be louder than when it first enters the space. enters the space.

For that reason, room absorption in the reduction For that reason, room absorption in the reduction of sound is important in addition to noise of sound is important in addition to noise reduction by a barrier. reduction by a barrier.

Example Problem:Example Problem:

Say an office is 20’ x 20’ x 10’ ceiling height. One Say an office is 20’ x 20’ x 10’ ceiling height. One wall separates the room from a mechanical space wall separates the room from a mechanical space where machinery produces 75 decibels of steady where machinery produces 75 decibels of steady sound. The barrier wall has a Transmission Loss of 50 sound. The barrier wall has a Transmission Loss of 50 decibels. decibels.

Find the noise level in the room if the room Find the noise level in the room if the room absorption equals 300 sabins;absorption equals 300 sabins;

NR = TL+10 log ( a NR = TL+10 log ( a R R / S ) / S )

TL = 50, a TL = 50, a R R = 300, S = 200 sq.ft.= 300, S = 200 sq.ft.Initially, the sound enters the room at a level of 25 Initially, the sound enters the room at a level of 25 decibels, because 50 decibels are stopped by the decibels, because 50 decibels are stopped by the wall, sowall, so

NR = 50 + 10 log ( 300/200) = 50 + 10 log (1.5)NR = 50 + 10 log ( 300/200) = 50 + 10 log (1.5)NR = 50 + 10(.1761) = 50 + 1.76 51.76 NR = 50 + 10(.1761) = 50 + 1.76 51.76

decibels decibels

Noise in the room is 75 – 51.76 = 23.24 Noise in the room is 75 – 51.76 = 23.24 decibelsdecibels

As an alternate to this problem,As an alternate to this problem,Say that we tripled the absorption of the room by Say that we tripled the absorption of the room by adding more absorptive surfaces, and aadding more absorptive surfaces, and arr = 900 = 900 sabinssabins

Then Noise Reduction = 50 + 10 log Then Noise Reduction = 50 + 10 log (900/200)(900/200)

NR = 50 + 10 log (4.5) = 50 + 10 (.6532)NR = 50 + 10 log (4.5) = 50 + 10 (.6532)

NR = 50 + 6.532 = 56.53 decibels,NR = 50 + 6.532 = 56.53 decibels,

So the noise level in the room equals 75 – 56.53 = So the noise level in the room equals 75 – 56.53 = 18.47 decibels.18.47 decibels.

which represents a reduction of about which represents a reduction of about 20%.20%.

NOW AS A NOVEL EXAMPLE, CONSIDER A NOW AS A NOVEL EXAMPLE, CONSIDER A SITUATION THAT INVOLVES A COMBINATION OF SITUATION THAT INVOLVES A COMBINATION OF

SOUND PROJECTIONSOUND PROJECTIONSOUND ABSORPTIONSOUND ABSORPTIONSOUND ISOLATIONSOUND ISOLATION

C

TL=52

wall

16'

wall

TL=12

one only through walls C and Dsound from motorcycle gets into rooms two and

walls .90

D

Room Oneceiling ht. 8'

.36

.44

coefficients:absorption

floorceiling32

'

Room Two12' ceiling ht.

24'

.22

coefficients:

.18

.26ceilingwalls

floor

absorption

absorbent coefficients of doors same as walls

A motorcycle at Bbackfires with a soundlevel of 110 decibelsat point A

280'

B

40'

A

EXAMPLE PROBLEM A motorcycle at point B backfires with a Loudness Intensity of 110 decibels at point A. Find the Intensity Loudness in Room One.

Ignore distances of sound travel in rooms one and Two.

PROCEDURE: Find IE at wall C, then IL at wall C, then IL in Room Two, then IL in Room One . . .

roadway

IL = 10 log IE / 10IL = 10 log IE / 10-16-16

110 = 10log IE110 = 10log IE4040 / / 1010-16-16

11 = log IE11 = log IE4040 / / 1010-16 -16

10101111 = = IEIE4040 / / 1010-16-16 ; ; IEIE40 40 = 10 = 10 -5-5

The intensity energy at point AThe intensity energy at point A

IEIE1 1 / IE/ IE22 = [ d = [ d22 / d / d11 ] ]22

Intensity ENERGY at C:Intensity ENERGY at C: 10 10 -5 -5 / IE/ IECC = [280/40] = [280/40] 2 2 = 49 = 49 IEIECC = 1/49 x 10 = 1/49 x 10 -5-5 = 2.04 x 10 = 2.04 x 10 -2-2 x 10 x 10 -5-5 = = 2.04 x 10 2.04 x 10 --

77

The amount of Intensity Energy at wall CThe amount of Intensity Energy at wall C

Then find Intensity Loudness at point C:Then find Intensity Loudness at point C: IL IL CC = 10log [ = 10log [2.04 x 10 2.04 x 10 -7 -7 / / 1010-16-16 ] = ] = 10log [ 2.04 x 10log [ 2.04 x

10 10 99 ] ] IL IL CC = 10 [ 9.309 ] = = 10 [ 9.309 ] = 93 decibels at wall C.93 decibels at wall C.

TL=52

wallC

280'

A motorcycle at B

at point A

backfires with a soundlevel of 110 decibels

B A

40'

TL=12

Dwall

.22floor

Room Two12' ceiling ht.

TL=52

24'

ceilingwalls .18

.26

absorptioncoefficients: C

wall

ABSORPTION OF ROOM TWO:

Floor = 24 x 32 x .22 = 169

Walls = (48+64)12 x .18 = 242

Ceiling = 24x32x.26 = 200

ROOM ABSORPTION = 611 sabins

NOISE REDUCTION IN RM 2:

NR = TL + 10log [ 611 / 12x32]

NR = 52 + 10log 1.59

NR = 52 + 10 ( .20 )

NR = 52 + 2.0 = 54 decibels

SOUND INTENSITY LEVEL IN

ROOM TWO EQUALS

93 – 54 = 39 DECIBELS

32’

Find absorption of Rm 2 in order

To find intensity loudness in Rm 2

coefficients:

TL=12

32

'

wallsceilingfloor

.36

.90

.44 Dwall

ceiling ht. 8'Room One

absorption

16'

TOTAL ROOM ABSORPTION IN ROOM ONE:

Floor: 16 x 32 x .44 = 225

Walls: (32 + 64) x 8 x .90 = 691

Ceiling: 16 x 32 x .36 = 184

ROOM ABSORPTION = 1100 SABINS

NOISE REDUCTION IN ROOM ONE:

NR = TL + 10log [ 1100 / 8 x 32 )

NR = 12 + 10log 4.30 = 12 + 10 x .633

NR = 12 + 6.33 = 18.33 decibels

SOUND INTENSITY LEVEL IN ROOM ONE = 39 – 18.33

Equals 20.67 DECIBELS

Other considerations must be given Other considerations must be given to components within a building to assure to components within a building to assure restriction of sound. Sound travels restriction of sound. Sound travels through openings as does water through a through openings as does water through a leak. leak.

Imagine if you filled a room with Imagine if you filled a room with water, and it leaks, then sound will also water, and it leaks, then sound will also leak out. leak out.

SOUND ENERGY THROUGH OPENINGSSOUND ENERGY THROUGH OPENINGS

Since sound energy emanates much like light Since sound energy emanates much like light energy, openings through a barrier are similar to energy, openings through a barrier are similar to a water leak – sound and light get out through a water leak – sound and light get out through openings in barriers and through cracks between openings in barriers and through cracks between doors and frames. doors and frames.

Diffracted sound increases as a percentage as Diffracted sound increases as a percentage as the opening size is reduced. The smallest the opening size is reduced. The smallest opening has the largest percentage of diffracted opening has the largest percentage of diffracted sound.sound.

Thru-Wall Outlet

Two Separate Outlets

Where two adjacent rooms have electrical or telephone outlets, thru-wall devices should be avoided because they create an opening in the wall.

Some codes, and hotel franchise companies will require back to back outlets to be separated by a sound barrier to stop sound and maintain the fire integrity of the wall.

A device called an “automatic door bottom” has a sound insulating strip that drops down when the door is closed, and retracts when the door is opened.

MASKING OF SOUNDMASKING OF SOUND

Sounds heard, that have no message of interest Sounds heard, that have no message of interest can be blocked out by the brain, as no handle is can be blocked out by the brain, as no handle is present to garner attention. Background sounds present to garner attention. Background sounds such as instrumental music can mask sounds that such as instrumental music can mask sounds that have a variation of pitch and loudness if the have a variation of pitch and loudness if the loudness is minimal, and there is no message. loudness is minimal, and there is no message.

But conversation that has a subject of the But conversation that has a subject of the listener’s interest, will be picked up by the brain, listener’s interest, will be picked up by the brain, and those sounds will mask the background, and those sounds will mask the background, simply because of the interest in the message. simply because of the interest in the message.

Mechanical equipment that grows unbalanced in Mechanical equipment that grows unbalanced in its operation often causes vibrations of movement its operation often causes vibrations of movement and sound, and can be of great annoyance to some and sound, and can be of great annoyance to some people, while with others it will not.people, while with others it will not.

Sounds that can be electronically produced called Sounds that can be electronically produced called “pink noise” and “white noise” are useful in “pink noise” and “white noise” are useful in masking certain irritating sounds for some people. masking certain irritating sounds for some people.

Privacy and security of sound are essentially one in Privacy and security of sound are essentially one in the same, except the circumstances for solution the same, except the circumstances for solution might not be the same. For instance, masking of might not be the same. For instance, masking of sound with other types of sound can solve privacy sound with other types of sound can solve privacy concerns, but masking is not necessarily sufficient concerns, but masking is not necessarily sufficient for security. for security.

Annoyance of unwanted sound in task areas may be Annoyance of unwanted sound in task areas may be relative to the unwanted sound. Disturbance during relative to the unwanted sound. Disturbance during tasks can be not only an annoyance for the ear, but tasks can be not only an annoyance for the ear, but also for the brain, particularly when unwanted also for the brain, particularly when unwanted sounds contain messages that changes one’s level of sounds contain messages that changes one’s level of concentration.concentration.

The human mind cannot concentrate on more than The human mind cannot concentrate on more than one thing simultaneously. one thing simultaneously. Regardless of how well Regardless of how well one thinks he or she can study in a crowded one thinks he or she can study in a crowded restaurant with a Chai tea latte and laptop computer, restaurant with a Chai tea latte and laptop computer, the audible sounds and the visual activity act doubly the audible sounds and the visual activity act doubly to compete with your level of concentration on a to compete with your level of concentration on a particular subject. Think about it.particular subject. Think about it.

Background sounds with no message of interest Background sounds with no message of interest can help to mask distractions can help to mask distractions if the background if the background sounds hold no interestsounds hold no interest - - - such as Tchaikovsky’s - - - such as Tchaikovsky’s third symphony. third symphony.

PRACTICE PROBLEM:PRACTICE PROBLEM:

A 16’ x 20’ x 9’ supply room has absorptive A 16’ x 20’ x 9’ supply room has absorptive coefficients as follows:coefficients as follows:

WallsWalls .46.46FloorFloor .40.40CeilingCeiling .86.86

The 20’ wall side of the room is adjacent to The 20’ wall side of the room is adjacent to another room, in which mechanical equipment another room, in which mechanical equipment sounds with a steady hum of 70 decibels. The sounds with a steady hum of 70 decibels. The separating wall has a Transmission Loss of 46 separating wall has a Transmission Loss of 46 decibels.decibels.

Find the level of sound in the supply room.Find the level of sound in the supply room.

SOLUTION:SOLUTION:

ROOM ABSORPTIONROOM ABSORPTIONWalls: 648 x .46 = 298Walls: 648 x .46 = 298Floor: 320 x .40 = 128Floor: 320 x .40 = 128Ceiling: 320 x .86 = 275Ceiling: 320 x .86 = 275 total = 701total = 701

NR = TL + 10log (701/180) = 46 + 10 log 3.89 =NR = TL + 10log (701/180) = 46 + 10 log 3.89 =46 + 10 ( .59) = 46 + 5.9 = 51.9 decibels46 + 10 ( .59) = 46 + 5.9 = 51.9 decibels

Sound in receiving room = 70 – 51.9 = 18.1 Sound in receiving room = 70 – 51.9 = 18.1 decibels decibels