Chapter 6

Room Acoustics

Room Acoustics

A room has good acoustics if,

Background noise is low enough and desired sound loud enough

Sound field is well diffused

No echoes or distortions

Reverberation time suits the purpose

Room Acoustics . . .

Diffuse sound field:

the acoustic energy is uniformly distributed through out the entire room

At any point the sound propagation is uniform in all directions

The sound field in a room is assumed to be completely diffused.

Reverberant Sound

The sound received at a point in a room from a noise source in that room may be direct (without having been reflected from any surface) or due to reflections.

Listener

Source

Reverberant Sound . . .The total sound due to direct reflections and inter-reflections is called the reverberant sound.

The region close to the sound source where direct sound predominates is called the direct field.

Further away from the source the reverberant sound may be the dominant component. This region is called the reverberant field.

The point at which direct and reverberant field intensity are the same is called critical distance.

Reverberation TimeThe reverberant sound decays with a reverberation time.

The reverberation time is defined as the time taken for a sound to decay by 60dB from its original level.

Reverberation TimeDifferent activities require reverberation time in the following ranges.

Speech 0.5 – 1s Music 1 – 2s

Short reverberation times are necessary for clarity of speech.

Longer reverberation times enhance the quality of music or fullness of voice.

Reverberation TimeWhy short reverberation time for speech?

If the reverberation time is long, the continuing presence of reverberant sound will mask the next spoken syllable and cause speech to be blurred.

Short reverberation time increases the sound intelligibility.

Reverberation Time . . .Reverberation times can be calculated by using any of the following formula.

Formula by Stephens & Bate

Sabine’s formula

Stephens & Bate formulaThis formula is suggested by Stephens & Bate using empirical methods to calculate the ideal reverberation time. Reverberation time,

where, V – volume of room (m3) r = 4 for speech 5 for orchestra 6 for choir

0.107} {0.012Vr T 31

R

Example

Calculate the reverberation time of a lecture room designed for 150 people if the optimum volume per person is 2.8 m3 and r = 4.

Answer Volume V = 2.8 x 150

= 420 m3

Reverberation time = 4 x (0.012 x 4201/3 + 0.107)

= 0.787 s

Reverberation Time . . .Optimum volume per person for various types of room,

Type of room Optimum (m3) Maximum (m3)

Room for speech 2.8 4.9

Cinema 3.1 4.2

Churches 7.1 – 9.9 11.9

Sabine’s formulaThe actual reverberation time is calculated using Sabine’s formula,Reverberation time

where V – volume of room (m3) A – total absorption (m2) k = 0.16 s / m

Total absorption,; s – surface area

α – absorption coefficient

AkV TR

iiSα A

ExampleA lecture hall with dimensions 30m x 10m x 5m is designed for 200 persons and has the following surface finishes, areas and absorption coefficients at 500 Hz.

Surfaces Area (m2) αWalls 400 0.02Ceiling 300 0.05Floor 300 0.02Glass 43 0.18Wood 12 0.10

Calculate the reverberation time of this hall at 500Hz(a) When occupied by 200 people(b) When 50 people leave the fully occupied lecture

hall

Assume that the absorption per person is 0.43m2 and the shading of the floor by the audience effectively reduces its absorption by 40% at 500Hz. Take the absorption per seat of unoccupied, fully upholstered seats at 500Hz is 0.28m2.

Answer a) When occupied by 200 people

volume V = 30 x 10 x 5 = 1500 m3

Total absorption,A = 400 x 0.02 + 300 x 0.05 + 300 x 0.02 x 0.6

43 x 0.18 + 12 x 0.10 + 200 x 0.43

= 121.54 m2

Answer Reverberation time

= kV/A= 0.16 x 1500 / 121.54= 1.974 2 s

Answer a) When occupied by 150 people

volume V = 30 x 10 x 5 = 1500 m3

Total absorption,A = 400 x 0.02 + 300 x 0.05 + 300 x 0.02 x 0.6

+ 43 x 0.18 + 12 x 0.10 + 150 x 0.43+ 50 x 0.28

= 114.04 m2

Answer Reverberation time

= kV/A= 0.16 x 1500 / 114.04 2.1 s

Ex: If the reverberation time required for this lecture hall is 0.8s calculate the extra absorption needed.

Answer Total absorption A’ = (121.54 + a ) m2

Reverberation time 0.8 = kV/A0.8 = 0.16 x 1500 / (121.54 + a)

Extra absorption needed,a 178.46 m2

Measurement of absorption coefficient (r)Reverberation chamber method:The value of obtained by measurement in a reverberation chamber is called the reverberation absorption coefficient, r.

This is the random incident absorption Coefficient ().

Reverberation chamberThe reverberation chamber is a room constructed with highly reflecting walls to maximize the internal reflections to provide a completely diffused sound field.

The inside wall surface must be hard, smooth finish using painted concrete, polished stone (e.g.terrazzo) or mosaic tiles.

The surface should have and irregular room shapes without any parallel walls to prevent flutter echo.

0.06 α

Measurement of absorption coefficient (r)Flutter echo:

Repetitive inter reflection of sound energy between opposing parallel or concave sound reflecting surfaces causes flutter echo.

It occurs in small spaces and can be prevented by reshaping to avoid parallel surfaces, providing deep sound absorbing treatment or breaking up smooth surfaces with splayed or scalloped elements of one of the parallel walls.

Reverberation Chamber method . . . Reverberation time without specimen,

V – volume of the reverberation chamberS – total surface area of the reverberation chamber

- average absorption coefficient of the room( between 1% and 2% )

When a specimen of area is placed,

is the random incidence absorption coefficient.

SkV T0

α } T1 -

T1 {

skV α α

α s) - (S α skV T

0mr

rm

ISO 354This provides the following instructions, The reverberation chamber should have a minimum

volume of 180 m3; preferably more than 200 m3.

Specimens should cover an area of about 10 m2 when placed on the floor.

Diffusion is to be provided by using many diffusing panels.

.

Reference book:

Acoustics and noise control2nd editionB J Smith, R J Peters and S Owen

Practical schedule

3 Practical 2 - Outdoors 1 – Industrial visit

Assignments: Three (3) in-class assignments, each carry 10 marks.

3 – for performance7 – for assignment