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The ability of a material to minimise the transmission ofacoustic energy through the material.
Capability of any boundary to restrict/limitany sound
movement through its boundary is defined as a soundinsulation.
Good sound insulation are obtained with materials whichare reasonable to heavy mass, and are impervious.
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If one wishes to contain water (sound).
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If one wishes to absorb water (sound).
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Noise treatment with sound insulation and soundabsorption
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The difference in average sound pressure levelsbetween two rooms at any given frequency
Sound Insulation = L1L2 dB
where
L1 = SPL in the sound source roomL2 = SPL in the adjacent receiving room
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A sound reduction index quantifying the sound insulationperformance of a material, building element or partition.
TL = L1L2+ 10 log10(S/A2) dB
where TL = Transmission Loss
L1 = SPL in noise source roomL2 = SPL in receiving room
S = Area of partition
A2 = Absorption in receiving room.
Transmission loss performance varies with soundfrequency.
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i. Weight (mass)
ii. Homogeneity and uniformity
iii. Stiffnessiv. Discontinuity or isolation
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The performance is governed by mass and stiffness
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Transmission Loss ataSheet
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TL and SRI are produced from the soundtransmission directly to the partition.
partition
100 dB
10 dB
TL
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However, the sound also transmitted in variouspossible direction to other rooms. This is called asflanking or indirect sound transmission.
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Room B
Room A
Room C
SoundSource
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The sound transmission occurs when the room wall
A or divider, vibrates due to the sound sourcewhich subsequently moves the sound to the otherrooms.
Normally, the sound (energy) transmitted through
the wall (flanking) is insulate-limitedto 50 55 dBbetween the two rooms.
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Sound Pressure Level in A Receiving Room
The resulting reverberant SPL in the adjacent room is given by:
SPL 2= SPL 1TL + 10 log Sp+ 10 log (1/R)
where R = Room constant
Sp = Area of partition
TL = Transmission LossSPL1 = SPL in noise source room
SPL2 = SPL in receiving room
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For a quick and simplified estimate, many practicalapplications could be approximated as
SPL 2= SPL 1TL
Example :
Plant room SPL = 100 dBA
TL of Brick wall = 45 dBAApprox. SPL outside = 100 45 = 55 dBA
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A foremans office is located inside a workshop of 3000m3volume and reverberation time of 2 s. The workshop reverberantsound pressure level was 72 dB. The office is to be constructedwith demountable metal construction pressure level inside theoffice. The office has room dimension of 4m length x 3m width x
2.5 height. The 4m length wall is the common part wall with theworkshop. The average room absorption coefficient ais 0.1
1. Determine the receiving room (office) correction factor
2. Determine a partition transmission area
3. Calculate transmitted noise SPLin receiving room
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20PROF. MADYA DR. SHAHRUDDIN BIN MAHZAN@MOHD ZIN
a) Determine the receiving room (office) correction factor.
S (total surface area) = 2(4x3) + 2(4X2.5) + 2(3x2.5) = 59
V (room volume) = 4 x 3 x 2.5= 30m3
R (room constant) = (59)(0.1)/(1-0.1)= 6.59
Room correction = 10log10(1/R)
= 10log10(1/6.56)= -8
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b) Determine Partition Transmission Area
Sp = 4 x 2.5 = 10 m2
Noise radiation = 10log Sp= = 10log10(10) = +10 dB
c) SPL in receiving Room
SPL 2= SPL 1TL + 10 log Sp+ 10 log (1/R)
SPL 2 = 72 35 + 10 + (-8) = 39 dBA.
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In practice a partition may have a weakerelement (door, window) that would de-rate theoverall sound insulation.
In such case it would be necessary to obtain theoverall performance of the composite
partition.
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A partition has a total area of 10 m2. The partition ismetal, with TL of 35 dB. Of the partition, the door is2 m2with a TL of 20 dB, and glazing of 4 m2, and TL27 dB. Find the overall TL of the partition.
Material Area (m2) TL (dB)Metal 4 35
Door 2 20Glass 4 27
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Ratio Metal : Glass 4 : 4 = 1 : 1Difference in TL 35 - 27 = 8 dB
From graph, loss of insulation = 6 dBCombined Metal+GlassTL =35 6 =29 dBRatio Door : Metal+Glass 2 : 8 = 1 : 4Difference in TL 29 - 20 = 9 dB
From graph, loss of insulation = 4 dBCombined Door+Metal+Glass = 29 4 = 25 dB
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Consider previous example of the Foremans Office inthe factory..
With metal partition only TL metal = 35 dB
With metal + door + glass TL overall = 25 dB
With metal only :
SPL inside office = 72 35 + 10 8 = 39 dBA
With metal plus door and glass :
SPL inside office = 72 25 + 10 8 = 49 dBA
27PROF MADYA DR SHAHRUDDIN BIN MAHZAN@MOHD ZIN