lec 13 sound insulation

8/13/2019 Lec 13 Sound Insulation

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1PROF. MADYA DR. SHAHRUDDIN BIN MAHZAN@MOHD ZIN


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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

PROF. MADYA DR. SHAHRUDDIN BIN MAHZAN@MOHD ZIN


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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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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

lec 13 sound insulation

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