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Miroslav KuMiroslav KueraeraCTU in CTU in PraguePrague
Department ofDepartment of Environmental EngineeringEnvironmental Engineering((roomroom no. 215)no. 215)
Email: miroslav.Email: miroslav.kucerakucera@fs.@fs.cvutcvut..czcz
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.
method of calculation
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.
method of calculation
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
method of calculation
the resulting values
=
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).
method of calculation
the resulting values
( )( )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
= + +
NoiseNoise to to thethe ExteriorExterior
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
= + + +
NoiseNoise to to thethe ExteriorExterior
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.
method of calculation
the resulting values
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=
NoiseNoise to to thethe ExteriorExterior
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.
method of calculation
the resulting values
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
SoundSound in in RoomRoom
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
SoundSound in in RoomRoom
acoustic field in the room
Decrease with distance
Room constant
( )
=m
mSR
1
Reveberent field
SoundSound in in RoomRoom
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
=
SoundSound in in RoomRoom
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
SoundSound in in RoomRoom
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
SoundSound in in RoomRoom
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
method of calculation
the resulting values
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
Sound Propagation Through the 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
Sound Propagation Through the Wall
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:
method of calculation
the resulting values
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.
method of calculation
the resulting values
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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