SOUND INSULATION

Transmission Loss, R When sound energy is transmitted through a panel (wall or floor) one way of expressing the proportion of energy transmitted is by giving the reduction in energy flow in decibels. This is the Transmission Loss, R Sometimes also given as the “Sound Reduction Index”, SRI

Diffuse Field Sound Transmission Loss, R “Coincidence Controlled “ - flexural wave resonance LOW FREQUENCIES “Stiffness Controlled” HIGHER FREQUENCIES “Mass Controlled” TRANSMISSION LOSS. R (dB) FREQUENCY (Hz) Slope 6 dB per doubling of frequency First panel resonance Coincidence Critical frequency

Equation of Motion for a Panel Force on wall = Incident sound pressure (pressure = force / area) Force = P0 sin(2ft) Force required to vibrate the wall F = Mass x Acceleration F = Damping x Velocity F = Stiffness x Displacement Force = mx + dx +kx m is mass, d is damping factor, k is stiffness x is displacement, x is velocity, x is acceleration +  .. . . ..

Simple theoretical model - mass region only .. . Equating we get, Po sin(2ft) = mx + dx + kx This 2nd order differential equation can be solved to find the vibration of the wall and hence sound transmitted from the opposite side into an adjoining space. Difficulties Based on single angle of incidence Stiffness and damping are difficult to predict Simple theoretical model - mass region only Luckily this is usually the most important .. .

Transmission Loss at Angle θ Solving the equation of motion for a single angle and converting this into a decibel reduction we get:-  P A R T I O N

Theoretical Mass Law Rearranging etc. we get, R0 = 20lg(mf) – 42 For normal incidence, θ = 0 so Cosθ = 1 we can replace:- ω by 2πf, ρc by 135 rayls, and the superficial density ρs by m (mass/m2) Rearranging etc. we get, R0 = 20lg(mf) – 42 Since RField = R0 – 5dB the THEORETICAL MASS LAW is RField = 20lg(mf) – 47 dB

Example A thin panel has a surface mass of 20 kg.m-2 find the transmission loss, R @ a frequency of 500Hz R = 20 lg(mf) – 47 dB substitute in for m and f we get, R = 20 lg(20x500) – 47 dB R = 33 dB Repeating this for other frequencies gives, 27 dB @ 250Hz 39 dB @ 1kHz Note that this is a 6 dB improvement for each doubling of frequency and also a 6 dB improvement for each doubling of mass R (dB) Slope 6 dB/ doubling of frequency Frequency (Hz)

Mass .v. Layers From the previous example Double the mass surface mass 20 kg.m-2 R = 33 @ 500Hz Double the mass surface mass 40 kg.m-2 R = 39 @ 500Hz Create two separate isolated layers surface mass 20 kg.m-2 per layer potential R = 33+33 = 66 @ 500Hz Two thin layers are better than one thick layer R=33dB R=39dB R=66dB

PRACTICAL SOUND TRANSMISSION A wall (or floor) is a barrier to airborne sound energy Sound Transmission Coefficient,  (Tau)  = transmitted energy incident energy For instance, if 1% of the energy is transmitted then  = 0.01 5% transmitted  = 0.05 100% transmitted  = 1.0 air Incident Sound energy wall Source room Receiving room transmitted

Transmission Loss, R In practical terms this is the inverse of the transmission coefficient and expressed in dB For instance if 1% of the energy is transmitted then R = 20 dB if 5% of the energy is transmitted then R = 13 dB if 10% of the energy is transmitted then R = 10 dB if 100% of the energy is transmitted then R = 0 dB

SOUND TRANSMISSION EQUATION Mic L1 L2 The level, L2 in the receiving room depends on:- Transmission Loss, (R) of the wall being tested The area, (Se) of the separating wall The amount of absorption, (A) in the receiving room

Measurement of R in a Laboratory Spring isolation to prevent flanking sound transmission L1 L2 Sending Room Receiving Room, volume V Loudspeaker Mic Wall construction being tested with area, Se Isolation pads Sound Transmission Suite Measure L1 & L2, and the reverberation time,T in room 2 Find the room absorption from Sabine formula Note: No flanking sound

MEASUREMENT OF ROOM ABSORPTION The reverberation time can be measured if the room exists Absorption area found by transforming the Sabine formula What if the building hasn’t been built – the design stage! PREDICTION OF ROOM ABSORPTION Room absorption can be calculated by adding all the room surface areas x absorption coefficients. A = s A = s11 + s22 + 

Example Calculation It is proposed to convert an old warehouse building into sheltered accommodation for the elderly. One of the flats will be next to the boiler room and you have been asked to determine whether the proposed partition wall between the boiler room and the flat will provide sufficient sound reduction. Flat Boiler room

The target level in the receiving room is NR35. Flat Boiler room Data The noise level in the boiler room is 87 dB in the dominant 500 Hz frequency band. The separating wall is to be of plastered timber stud construction 3m high x 5m long. For a wall 50mm x 100mm studs, 9mm plasterboard, 12mm plaster the R values are (Woods Practical Guide to Noise Control):- The absorption in the receiving room @ 500 Hz has been calculated by adding together the all the surface areas x absorption coefficients giving a total absorption of 11 m2 Sabine units. The target level in the receiving room is NR35. 63 125 250 500 1k 2k 4k 8k Hz 20 25 28 34 47 39 50 56 R 39 dB

Solution The level in the boiler room L1 = 87 dB @ 500 Hz Transmission Loss, R at 500 Hz is 34 dB (table) The areas of the wall, Se is 3x5 = 15 m2 Total absorption, A is 11m2 sabins Substituting these values into the transmission equation we get, L2 = L1 - R + 10lg(Se/A) L2 = 87 - 34 +10lg(15/11) L2 = 54 dB NR35 requires a level of 39 dB @500 Hz, therefore the target has not been achieved. Action -- Reduce noise level in boiler room by 15 dB Choose a better wall (double masonry wall)

Composite Panels (walls) Let’s consider the sound transmission through the wall between the performance space and the control room

Triple leaf concrete block wall Partition Details Triple leaf concrete block wall Double glazed unit Wall 3m x 3m has R values of:- Glazing 1.5m x 1m has R values of:- What is the composite R value @ say, 500 Hz? Triple leaf cavity wall, 112/215/112mm with 50mm cavities between leaves Freq 63 125 250 500 1k 2k 4k 8k R 53 56 62 69 78 84 6/200/10 double glazing Freq 63 125 250 500 1k 2k 4k 8k R 35 46 56 65

Composite R for panels with different values of R in different areas This is based on calculating the composite sound transmission coefficient, composite composite = 1x Area1 + 2 x Area2 + +n x Arean Total area of the panel Individual ,s are found from R = 10lg(1/ )  = 1/10(R/10) Rcomposite = 10lg(1/ composite)

similarly for the wall (R=69dB) Solution (for 500 Hz)  = 1/10(R/10) glazed panel (R=46dB)  Glazing = 1/10(46/10) = 0.000025118 similarly for the wall (R=69dB) Wall = 1/10(69/10) = 0.000000125 composite = 1x Area1 + 2 x Area2 + +n x Arean Total area of the panel Total area = 9m2 , Glazed area = 1.5m2 , Wall area = 7.5m2 composite = 0.000025118 x 1.5 + 0.000000125 x 7.5 = 0.00000429 9 R composite = 10lg(1/ composite ) = 53.7 dB (54 dB) Note comparison with --- wall = 69 dB, Glaze = 46 dB

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