Fundamentals of Noise Reduction1 - cvut.czzmrhavla/EE/11_EE_Noise Reduction.pdf · Fundamentals of...

Miroslav KuMiroslav KueraeraCTU in CTU in PraguePrague

Department ofDepartment of Environmental EngineeringEnvironmental Engineering((roomroom no. 215)no. 215)

Email: miroslav.Email: miroslav.kucerakucera@[email protected]

Fundamentals of Noise Reduction Fundamentals of Noise Reduction Propagation of noise in spacePropagation of noise in space

NoiseNoise andand peoplepeople

The effect of noise is determined primarily by the intensity of sound I (W/m2)

Sound hits the drum, continues middle earto the inner ear, where it is filtered

Weber - Fechner physiological law

0

log.IIkonstLN =

cpI

=

2

10=konst

The relationship between stimulus and perception is logarithmic. This logarithmic relationship means that if a stimulus varies as a geometric progression, the corresponding perception is altered in an arithmetic progression.

][Ph

PPerceptionerception of the of the HHumanuman EEarar

Smallest noise that human ear percieves is given by valueThreshold of hearing p0 =2.10-5Pa

Ear perceives frequencies from 20Hz to 20 000Hz

perception ear describes this graph

Curves still volume

frequency

Wavelength

cf = Trc = The easiest way is harmonic signals - Sinus

wavelength is the length traveled by a wave during one period

Speed of sound in the air

distance

Example 1

Calculate the wavelength of the signal for the specified frequencies.

we need to know the velocity air c=340 m/swater c=1440 m/ssteel c= 5750 m/s

f [Hz] 63 125 250 500 1000 2000 4000 8000

air

water

steel

Example 1

Calculate the wavelength of the signal for the specified frequencies.

f [Hz] 63 125 250 500 1000 2000 4000 8000

air 5,4 2,7 1,4 0,7 0,3 0,17 0,09 0,04

water 22,9 11,5 5,8 2,9 1,4 0,7 0,36 0,2

steel 91,3 46 23 11,5 5,8 2,9 1,4 0,7

the resulting values

method of calculation

cf = Trc =

Period of Oscillation

Harmonic Signal - Sinus

Circular Frequency Tf /22 ==

cfT == 1

time

Sound Pressure

Relation for harmonic signal

barometric pressure

sound pressure

sound pressure

Acoustic power

Decrease of sound intensity

I decreases with increasing r inversely

.W konst I S= =

1 1 2 2. .I S I S=

1.p konstr

=2

1.I konstr

=

surface of a sphere

2

2

1

2

1

2

1

2

2

1

=

==

pp

rr

SS

II

DirectivityDirectivity FactorFactor

Definition: The ratio of the intensity on a designated axis of a soundradiator at a specific distance from the source to the intensity that wouldbe produced at the same location by a spherical source radiating the sametotal acoustic energy.

22 2

44 4m m mW Q WW I S I r I I

r r

= = = =

WrQ

24=

Q=1in space

Q=2at the wall

Q=4two walls

Q=8in the corner

Spherical surface

ExampleExample 22

Calculate the sound intensity for sound pressure value

p = 2.10-3 Pa, pB =101325 Pa, r = 287,1 J/kg, t = 20C.

2/??? mWI =

ExampleExample 22

Calculate the sound intensity for sound pressure value

p = 2.10-3 Pa, pB =101325 Pa, r = 287,1 J/kg, t = 20C.


the resulting value

cpI

=

2

TrpB =

Equation of state

29 /10.6,9 mWI =

TrTr

pc B

=

ExampleExample 33

Calculate the sound power. Source has a sound pressure in the control point p = 0,01 Pa. The distance between the control point and the source is 5 m. Density of air 1,22 kg/m3, speed of sound 340m/s. Wave has a spherical shape.

WW ???=

Source Control point

W=? p=0.01Pa

Acoustic wave

ExampleExample 33

Calculate the sound power. Source has a sound pressure in the control point p = 0,01 Pa. The distance between the control point and the source is 5 m. Density of air 1,22 kg/m3, speed of sound 340m/s. Wave has a spherical shape.


the resulting value

22

4 rc

pSIW

==

WW 51054,7 =

Decibel Scale

0

2 0 . lo gppLp

=

0

1 0 . lo gIILI

=

Sound pressure [Pa] Sound pressure level [dB]

p0 = 2.10 -5 (Pa)

Acoustic intensity [W/m2] Sound intensity level [dB]

I0 = 10 -12 (W/m2)

recalculation

Reference value

Reference value

Decibel Scale

Sound pressure Sound pressure levelComparison of scale

Decibel Scale

Acoustic power [W] Sound power level [dB]

W0 = 10 -12 (W)

0

10.logWWLW

=

recalculation

Reference value

Interrelationships

W

I

S

LI & Lp

2,0log10log10log10log10 0020

2

00

20

2

0

=

+=

== pI Lcc

pp

cp

cp

IIL

for normal atmospheric conditions

cpI

=

2

Interrelationships

W

I

S

0 0 0

10 log 10 log 10 logW pW I SL L SW I S

= = = +

Source

LW & Lp

Combination of Sound Sources

0,1

110.log 10 i

n L

iL

==

0

0,5

1

1,5

2

2,5

3

0 2 4 6 8 10 12 14

Lp1 - Lp2

DLp

( )10,11 2 110.log 2.10 3LL L L L dB= = = +1 2 16 1L L dB L L dB = = +

1 2 110 0,4L L dB L L dB = = +

When we have more resources

logarithmic sum

graphical method

I add to higher value L1 or L2

Combination of Sound Sources

Example:

calculation

dBL 71)1010101010log(10 621,0681,0601,0631,0631,0 =++++=

ExampleExample 44

Open plan office. Calculate the sound pressure level. In the control point is Lp7=53,5dB (from same 7 sources).

a) Determine the sound pressure level from one source.b) How much is necessary to reduce the number of resources

when the control point may be Lpmax = 50 dB.

ExampleExample 44

Open plan office. Calculate the sound pressure level. In the control point is Lp7=53,5dB (from same 7 sources).

a) Determine the sound pressure level from one source.b) How much is necessary to reduce the number of resources

when the control point may be Lpmax = 50 dB.

a) b)

from 7 do 3



=

nL

pL

p

71,0

110log10

=

1

50

1,0

1,0

1010

p

p

L

L

n

dBLp 451 = 16,3=n

Spectrum

Dependence on the frequency of the monitored variables.

musical instruments

machinery, equipment

individual tones

a continuous spectrum

no periodicperiodic

Spectrum

Example: Relative spectrum of Fan

presented as Line chart

presented as Bar chart

0

10

20

30

40

50

60

70

80

90

100

31,5 63 125 250 500 1000 2000 4000 8000 16000

f [Hz]

Lw

[dB

]

Example: Spectrum of vacuum cleaner

Octave Band

.m h df f f=

Octave

Center frequency of the band

fm (Hz) division of frequency axis

2hd

ff=

31,5 63 125 250 500 31,5 63 125 250 500 10001000 2000 40002000 4000 8000 160008000 16000

audible band is divided into octaves

example

210002000

==d

h

ff

specific values hearing band

Thirds of Octave Band

.m h df f f=

1/3 Octave

Center frequency of the band

fm (Hz) division of frequency axis

3 2hd

ff=

25 25 31,531,5 40 50 40 50 6363 80 100 80 100 125125 160 200 160 200 250250 315 400 315 400 500500 630 800 630 800 1000 1000 1250 1600 1250 1600 20002000 2500 3150 2500 3150 40004000 5000 6300 5000 6300 80008000 10000 1250 10000 1250 1600016000

specific values hearing band

Detailed analysis shows:

Weighted Sound LevelsAA-- weightingweighting

The A scale was designed to correspond to the response of thehuman ear for a sound pressure level at all frequencies.

( )( )ipi ALpAL += 0110log10

f[Hz] A [dB]

31,5 -39,4

63 -26,2

125 -16,1

250 -8,6

500 -3,2

1000 0

2000 1,2

4000 1

8000 -1,1

16000 -6,6

Equivalent Sound Level LAeq

0,1

0

110.log 10 .pAT L

AeqL dT =

20

25

30

35

40

45

50

55

60

0 50 100 150 200as (s)

LpA

(dB

)

Time (s)

For the evaluation of the noise at time

LpAeq has the same effect on humans as a variable value LpA in the same interval.

T

ExampleExample 55

Calculate the total sound pressure level and total soundpressure level A in octave band. In the control point isLp7=53,5 dB (from same 7 sources).

f [Hz] 31,5 63 125 250 500 1000 2000 4000 8000 16000

Lpi 97 99 83 65 62 63 61 60 55 51

Ai -39,4 -26,2 -16,1 -8,6 -3,2 0 1,2 1 -1,1 -6,6

Lpi+Ai 56,7 72,8

dBLpA ???=dBLpc ???=

ExampleExample 55

Calculate the total sound pressure level and total soundpressure level A in octave band. In the control point isLp7=53,5dB (from same 7 sources).



( )( )ipi ALpAL += 0110log10( )( )piLpcL = 0110log10

dBLpA 9,74=dBLpc 2,101=

NoiseNoise to to thethe ExteriorExterior

acoustic field of a point source

210log

4. .p WQL L

r = +

12 1

2

20logp prL Lr

= +

6 / 2.L dB r =Propagation of noise to exterior

Example

Lp=50dB in 5m

and in 10 m is Lp=44dB

This big decrease is good at small distances.

but when

Lp=50dB in 50 m

Is Lp=44dB until in 100 m

acoustic field of a line source

to dist. a = d decrease of 3 dBfrom dist. 2a = d decrease of 6 dB

12 1

2

10logp prL Lr

= +

3 / 2.L dB r =

10log 10log4p WaL L arctg add

= + +


distance

lenght

Surface is cylindernot sphere

-typical examplesare road, manufacturing line

acoustic field of a plane area source

-typical example is facade of the house

to dist. a = d decrease 0 dB

0 / 2.L dB r =

2 2 20 0

110log2

b a

p W

dxL L dyx y d

= + + +


Result is the sum of small areas surface

AAttenuationttenuation with with DDistanceistance for for VVariousarious SSourcesourcesComparisonComparison

0 dB3 dB

6 dB

ExampleExample 66

Determine the noise propagating from 2 sources to controlpoint. Distance x = 20 m.

1. source LW = 120 dB, Q = 2, r1 = 50 m2. Source Lp = 73 dB at a distance r = 40 m.

dBLpc ???=

ExampleExample 66

Determine the noise propagating from 2 sources to controlpoint. Distance x = 20 m.

1. source LW = 120 dB, Q = 2, r1 = 50 m2. Source Lp = 73 dB in the distance of r = 40 m.



1 21

10 log4p W

QL Lr

= + 402 402

20 logp prL Lr

= +

( )1 20,1 0,110 log 10 10p pL LpcL = +

dBLpc 7,78=


Noise Attenuation Barriersdecrease of sound intensity - attenuation by bending

( ) iWipiD

baQLL +

+= 24log10

the attenuation over the top of the screen

Acousticsshadow

OutdoorOutdoor NoiseNoise

attenuation by bending D [dB]

( )2 ).

a bfq hc a b

+=0,39614, 22.D q=

General scheme for endless barrier

one possibility for calculating

speed of sound in the airefective height

frequency

ExampleExample 77

Calculate the attenuation of bending over an obstacle. Determine the sound pressure level in control point.

Where: Q = 2, L1 = 4 m, Hz = 1 m, L2 = 20,4 m, Hb = 4,1 m,LK = 2 m, HK = 12 m a = 5 m, b = 20 m, h = 1 m from geometric calculation

f [Hz] 31,5 63 125 250 500 1000 2000 4000 8000

Lwi [dB] 106,5 105,7 102,2 97,4 92,2 87,0 82,1 77,4 72,3

q

DiLpi

ExampleExample 77

Calculate the attenuation of bending over an obstacle. Determine the sound pressure level in control point.



f [Hz] 31,5 63 125 250 500 1000 2000 4000 8000

Lwi [dB] 106,5 105,7 102,2 97,4 92,2 87,0 82,1 77,4 72,3

q 0,22 0,3 0,43 0,61 0,86 1,21 1,71 2,42 3,43

Di 7,7 8,9 10,2 11,7 13,4 15,3 17,6 20,2 23,2

Lpi 62,8 60,9 56,1 49,8 42,8 35,7 28,6 21,3 13,2

( ) iWipiD

baQLL +

+= 24log10

396,022,14 qDi =( )

abba

cfh

bahq +=

+=

2112

OutdoorOutdoor NoiseNoise

attenuation by bending

++=

=

321 1,01,01,01,03

1

1010101log10

10

1log10iii

inDDD

DicD

real obstacle has length, so three paths of propagation

Total attenuation is given by relation

SoundSound in in RoomRoom

acoustic field in the room is divided into two parts: direct and reflected field

Here is an acoustics situation the same and depends on the absorption coefficient


acoustic field in the room

Direct field

Reverberant field

The total objective sound field in room

210.log

4. .p WQL L

r = +

( )4 110.log.

mp W

m

L LS

= +

( )2

4 110.log

4. . .m

p Wm

QL Lr S

= + +

the same as outside

here is efect absorption coefficient


acoustic field in the room

Decrease with distance

Room constant

( )

=m

mSR

1

Reveberent field


The boundary between the field of direct and reflected waves

Reducing noise exposure of workers in the field of reflected waves

increase

( )2

4 14. . .

m

m

Qr S

=

( ). .

16. . 1m

m

QSr

=

1

1

.i im

i

SS

=

2

2

.i im

i

SS

=

( )( )

( )( )

1 2

2 1

. 110.log

. 1m m

pm m

SL

S

=

soundincidentsoundabsorption

=


Absorption coefficient

wall

Incident waveI0

ReflectionI1

AbsorptionI2

TransmissionI3

>

AbsorptionAbsorption coefficientcoefficient

Materials for absorption

Concrete

wood-paneled wall

window glass

brick wall

carpet

person in the room onetwo


reverberation time T [s]/absorption coefficient a [-]

0,164m

VTS

=

is used for assessment of noise in room

relation between T & a

time of decrease on 60 dB is T

room volume

wall area

mean

SoundSound in in RoomRoomreverberation time -examples


DirectivityDirectivity FactorFactor Q Q [[--]] forfor distributiondistribution elementselements: : gridgrid, , swirlswirl diffuserdiffuser etcetc..

Q = 8 in the corner

Q = 2 at the wall

Q = 1 in space

Q = 4 two walls

frequency and size of the grid

ExampleExample 88

Calculate the total sound pressure level in the control point afrom sources 1 and 2. Room has reverberation time T = 0,7s.

1. source LW1A = 95 dB, Q = 12. source LW2A = 97 dB, Q = 1

dBLpAc ???=

ExampleExample 88

Calculate the total sound pressure level in the control point afrom sources 1 and 2. Room has reverberation time T = 0,7s.

1. source LWA1 = 95 dB, Q = 12. source LWA2 = 97 dB, Q = 1



0,164 0,164mm

V VTS S T

= =

2 2 2r x y z= + +

23,0=m

mr a 5,12 =

dBLpA 7,821 = dBLpA 862 = dBLpAc 7,87=

( )( )pAiLpAcL = 0110log10( )

++=

m

m

aiWAipAi Sr

QLL

14

4log10 2

spatial distance mr a 29,25,012222

1 =++=

Sound Propagation Through the Wall

Room absorption

Transmission loss R - graph

21 2 10 logp okt p okt

AD L L RS

= = +

ii SA =

attenuation through the wall

quality of the wall is given by transmission loss R [dB]

surfaceof dividing wall


when we calculate the wall thickness h [m]

"" AA

m fmf

="" B

B

m fmf

="mh

=

general procedure for determining the transmission loss

two possibility

We must move to left or right than to touching calculated curves with polygon line.

We know Lp1, Lp2, A2and material constants.

Surface density


when we calculate the sound pressure level behind the wall Lp2 [dB]

"m h= "

"A

Am ff

m

= ""

BB

m ffm

=We know Lp1,h, A2and material constants.

the width of one octave

ExampleExample 99

Calculate the sound pressure level Lp2 in receiving room.

Walls thickness between rooms h = 250 mm. Room absorption A2 = 10m2. Dimensions of the dividing wall S = 26,25m2.

Material constants for the brick wall:

2

2

3

/58000"/12600"

35/2000

mkgHzfmmkgHzfm

dBRmkg

B

A

A

=

=

==

f [Hz] 31,5 63 125 250 500 1000 2000 4000 8000

Lp1 [dB] 92 95 99 97 88 83 76 75 71

R [dB]

Lp2 [dB]

from the material tables

ExampleExample 99

Calculate the sound pressure level Lp2 in receiving room.Walls thickness between rooms h = 250 mm. Material constants for the brick wall:



2

2

3

/58000"/12600"

35/2000

mkgHzfmmkgHzfm

dBRmkg

B

A

A

=

=

==

f [Hz] 31,5 63 125 250 500 1000 2000 4000 8000

Lp1 [dB] 92 95 99 97 88 83 76 75 71

R [dB] 35 35 37 45,5 49,5 54 58,5 63 67,5

Lp2 [dB]

2/500" mkghm ==

Hzm

fmf AA 2,25""

== Hzm

fmf BB 116""

== Go to chart Rin real scale draw a graph

ExampleExample 99

Calculate the sound pressure level Lp2 in receiving room.

Room absorption A2 = 10m2. Dimensions of the dividing wall S = 26,25m2.



f [Hz] 31,5 63 125 250 500 1000 2000 4000 8000

Lp1 [dB] 92 95 99 97 88 83 76 75 71

R [dB] 35 35 37 45,5 49,5 54 58,5 63 67,5

Lp2 [dB] 61,3 64,3 66,3 55,8 42,8 33,3 21,8 16,3 7,8

SARLL oktpoktp 212 log10=

Thank you for your attention

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