Sound Insulation

· Sound insulation is the most useful method for controlling noise in buildings.

· Insulation is the principal method of controlling both airborne sound and impact sound in the buildings.

· Two types of sound insulation are to be dealt with in building construction:

i. Airborne Sound Insulation:: the insulation against noise originating in air, e.g. voices, music, motor traffic, wind.

ii. Impact Sound Insulation: the insulation against noise originating directly on a structure by blows or vibration e.g. footsteps above, furniture being moved, drilling and hammering the structure.

· The Building Regulations 1991 call for ‘an adequate resistance to the transmission of airborne sound’ for walls and for airborne and impact sound in the case of floors. · Means of complying the Regulations are set out in Approved Document E: 1992 Airborne and impact sound. · The standard of airborne sound reduction are specified in term of; ® weight of walls and floors and ® soft coverings and floating layers for impact sound, ® together with special requirements for joints and the elimination of air passengers.

Construction for sound insulation

A. Walls

Heavyweight wall

· Heavyweight walls generally provide high levels of airborne sound insulation.

· Construction of cavity in the heavyweight wall can increase sound insulation.

Lightweight wall

· Lightweight wall can provide adequate levels of airborne sound insulation with airtightness.

· Each side of the wall is built on separate timber frames and an absorbent blanket of mineral fibre is used to provide acoustical isolation between the sides of the wall.

B. Floors

· The insulation of a floor must be maintained at all junctions with the surrounding walls in order to prevent flanking transmission.

· The separation of 2 parts of a floating floor must continue around all edges by the use of resilient materials and airtight techniques.

Concrete floor

· The natural mass concrete floors provide insulation against airborne sound but concrete can transfer impact sound.

· Therefore a resilient layer (mineral wool) is also needed to provide insulation against impact sound.

Timber floor

· Thick layer of composite floorboard or multiple layers of plasterboard is used in timber floor construction to increase the mass of timber floor for better sound insulation.

C. Windows

· Glass has relatively high density so increasing the thickness of glass provides increased of sound insulation.

· Air cavity can be provided between 2 panels of glass for sound insulation.

· Window must design to shut with a good seal in order to provide air tightness.

· The acoustic insulation of insulating glass can be achieved by using glasses of asymmetrical thickness, as well as laminated glass with special PVB (Polyvinyl-Butyral) layer which is highly considerably reduces noise.

CALCULATION

1. A particular sound wave has frequency of 440Hz and a velocity of 340m/s. Calculate the wavelength of this sound.

Answer:

Given, v = 340m/s, f = 440 Hz

Wave Length, = v (Speed of Sound)

f (Frequency)

= 340

= 0.7727m

2. A particular sound wave has frequency of 440Hz and a wavelength of 1m. Calculate the speed of this sound.

Answer:

Velocity (Speed of sound), v = f (Frequency) x (wavelength)

= 440 x 1

= 440m/s

3. A lecture hall with a volume of 1500m³ has the following surfaces finishes, areas and the total absorption of the lecture hall surfaces is 99 m² sabin. Calculate the reverberation time of this hall.

Using the sabine formula;

Reverberation Time, t = 0.16 V = 0.16 x 1500m³

A 99

Therefore, the reverberation time = 2.42 s

4. A hall has a volume of 5000m² and a reverberation time of 1.6 s. Calculate the amount of extra absorption required to obtain a reverberation time of 1 s.

Answer:

Given, t₁ = 1.6 s, A₁ =?, t₂ = 1.0 s, A₂ =?, V = 5000m³

Using the sabine formula;

Reverberation Time, t = 0.16 V

A

So A₁ = 0.16 V = 0.16 x 5000 = 500 m² sabins

t¹ 1.6

A₂ = 0.16 V = 0.16 x 5000 = 800 m² sabins

t² 1.0

Therefore the extra absorption needed = A₂ - A₁ = 800 – 500 = 300m² sabin

TUTORIAL QUESTIONS:

1. A particular sound wave has frequency of 500Hz and a velocity of 380m/s. Calculate the wavelength of this sound.

Answer:

Given, v = 380m/s, f = 500 Hz

Wave Length, = v (Speed of Sound)

f (Frequency)

= 380

= 0.76m

2. A particular sound wave has frequency of 520Hz and a wavelength of 1.5m. Calculate the speed of this sound.

Answer:

Velocity (Speed of sound), v = f (Frequency) x (wavelength)

= 520 x 1.5

= 780m/s

3. A room of 900m² volumes has a reverberation time of 1.2 s. Calculate the amount of extra absorption required to obtain a reverberation time of 0.8 s.

Answer:

Given, t₁ = 1.2 s, A₁ =?, t₂ = 0.8 s, A₂ =?, V = 900m³

Using the sabine formula;

Reverberation Time, t = 0.16 V

A

So A₁ = 0.16 V = 0.16 x 900 = 120 m² sabins

t¹ 1.2

A₂ = 0.16 V = 0.16 x 900 = 180 m² sabins

t² 0.8

Therefore the extra absorption needed = A₂ - A₁ = 180 – 120 = 60m² sabin

4. Calculate the actual reverberation time for the above hall with a volume of 5000m², given the following data for a frequency of 500Hz.

Surface Area	Absorption coefficient
500m^2 brickwork	0.03
600m^2 plaster on solid	0.02
100m^2 acoustic board	0.70
300m^2 carpet	0.30
70m^2	0.40
400 seats	0.30 units each

Answer:

Surface Area	Absorption coefficient	Total absorption
500m^2 brickwork	0.03
600m^2 plaster on solid	0.02
100m^2 acoustic board	0.70
300m^2 carpet	0.30
70m^2	0.40
400 seats	0.30 units each
Total

Using the sabine formula;

Reverberation Time, t = 0.16 V

A

t = 0.16 V = 0.16 x 5000 = 2.4 s

A 335

Therefore, the reverberation time = 2.4 s

5. A large cathedral has a volume of 120 000m3. When the space is empty the reverberation time is 9 s. With a certain number of people present the reverberation time is reduced to 6 s. Calculate the number of people present, if each person provides an absorption of 0.46m³ sabins.

Using the sabine formula;

Reverberation Time, t = 0.16 V = 0.16 V

A N x ά

Given, t₁ = 9s, V = 120 000, ά = 0.46

N₁ = 0.16 V = 0.16 x 120 000 = 4638 persons

t₁ x ά 9 x 0.46

t₂ = 6s,

N₂ = 0.16 V = 0.16 x 120 000 = 6957 persons

t₂ x ά 6 x 0.46

Therefore, the number of people present = N₂ - N₁ = 6957 – 4638 = 2319 number of people present.

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