Engineering Acoustics Lecture 3

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    Chapter 2 . . .

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    Frequency analysis of sound

    Most sounds contain a combination of many

    different frequencies.

    The frequency analysis of sound is essential for

    noise control. Because sound absorption is

    frequency dependent.

    i.e. The same material absorb different amounts

    of sound energy at different frequencies.

    e.g. To choose the proper kind of absorber.

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    Frequency analysis of sound . . .

    Frequency analysis is performed by measuring theoutput of a sound level meter through a band filter,

    which passes only a particular frequency range

    between f1 and f2. This is called the bandwidth f

    (or pass band).

    f = f1 f2, if f1 > f2

    where f1, f2 cut-off frequencies

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    Frequency analysis of sound . . .

    The centre frequency

    fm =f1f2

    It is usually convenient to measure and analyze sound in

    ranges of frequencies such as the octave. An octave bandis the range of frequencies between any one frequency

    and double that frequency. (i.e.f1=2 f2)

    e.g. 75 150 Hz, 150 300 Hz, 300 600 Hz, 600 1200 Hz, 1200

    2400 Hz, 2400 4800 Hz, 4800 9600 Hz

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    Frequency analysis of sound . . .

    Mostly it is sufficient to know the magnitude of the

    sound contains within the octave bands. The

    preferred center frequencies of acoustic

    measurements are;

    31.5 Hz, 63 Hz, 125 Hz, 250 Hz, 500 Hz, 1000 Hz, 2000 Hz, 4000Hz, 8000 Hz for octaves.

    Frequency analysis of sound is performed using

    frequency analyzers such as octaveband analyzer

    and 1/3 octave-band analyzer.

    Note: 1/3 octave band is obtained by dividing the octave

    bandwidth into 3 equal parts.

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    Example

    A certain noise was analyzed into octave bands. The sound

    levels measured in each center frequency are given below.Calculate the combined sound level?

    Center frequency (Hz) sound level dB (A)

    31.5 60

    63 60

    125 65

    250 70

    500 65

    1000 65

    2000 45

    4000 40

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    Answer

    L = 10 log(106.0+106.0+106.5+107.0+106.5+106.5+104.5+104.0)

    = 67.9

    = 68 dB (A)

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    Measurement conditions

    The measurement of sound is done under the following

    standard conditions.

    1) Free field

    2) Reverberant field

    3) Semi reverberant field

    4) Anechoic field

    5) Semi anechoic field

    6) Diffuse sound field

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    Measurement conditions . . .

    1) Free field

    This is completely open space where there are no sound

    reflections or other modifying factors present.

    ReceiverSource

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    Measurement conditions . . .

    2) Reverberant field

    In a reverberant field the sound energy at any pointis the sum of that directly radiated from the sourceand sound levels reflected from adjacent surfaces.

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    Measurement conditions . . .

    Ei = Er + Et + Ea

    In a fully reverberant field all the sound energy

    striking the bounding surfaces is reflected without

    loss. This simplifies that the bounding surfaces

    should be highly reflective.

    Ei Er

    Et

    Ea

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    Measurement conditions . . .

    3) Semi reverberant field

    In a semi reverberant field the prevailing

    conditions may be anywhere between free field

    and reverberant field conditions.

    4) Anechoic field

    All the sound measured comes directly from thesource. (All incident energy striking the walls is fully

    absorbed)

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    Measurement conditions . . .

    5) Semi anechoic field

    In a semi anechoic field the sound source is mounted

    above a hard reflective surface.

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    Measurement conditions . . .

    Note:

    The measurements taken outdoors can be considered

    to approximate to free field condition. And show

    reasonable agreement with anechoic measurements

    provided there is no reflective surfaces nearby.

    Measurements taken indoors can be considered as

    approximating to diffuse field condition. And show

    reasonable agreement with reverberant fieldmeasurements.

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    Measurement conditions . . .

    6) Diffuse Sound Field

    A room is assumed to be completely diffused (Any

    closed space is referred to as a room). This means:

    1) the acoustical energy is uniformly distributed

    throughout the entire room

    2) at any point the sound propagation is uniform

    in all directions

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    Noise Rating

    1) Steady noise

    2) Time varying noise

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    Noise Rating1) Steady noise

    When a sound level meter reading fluctuates within arange of less than 5 dB when using the weighting S

    then the noise can be treated as steady.

    For the measurement of steady noise A-weightingsimple sound level meter is used.

    The average value is taken as the reading,

    ; n total number of readings

    }{0

    10Ln

    10L2

    10L1

    0

    In

    10................10(10I10log

    L

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    Noise Rating . . .

    2) Time Varying Noise

    The level of many noise varies with time.

    e.g. traffic noise, impulse noiseetc

    For the measurement of such noise an integrating

    sound level meter is used.

    It automatically calculates and indicates the value ofequivalent continuous sound level (Leq) for a given time

    interval T together with the value of T.

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    Equivalent continuous sound level

    LAeq,T -> Equivalent Continuous A weighted sound level

    within a specified time interval, T.

    This is defined as

    Where P(t) is the instantaneous A weighted soundpressure.

    T = t2 t1

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    Equivalent continuous sound level . . .

    This is the sound level in dB (A) of a

    continuous steady sound which has the same

    A-weighted sound energy within a specified

    time interval , T as the fluctuating sound

    being measured.

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    Equivalent continuous sound level . . .

    e.g.

    Suppose;

    sound level L1 acts during time t1,

    sound level L2 acts during t2,

    sound level L3 acts during t3,

    ..

    sound level Ln acts during tn.

    Let T = Total time over which the LAeq,T is required.

    T = t1 + t2 + t3+ + tn

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    Equivalent continuous sound level . . .

    10L1x10II

    01

    Let

    =>

    )(0

    11

    I

    I10logL

    }{T

    10t................10t10t10log

    10Ln

    n

    10L2

    2

    10L1

    1

    }{TI

    tI........tItI10log

    0

    nn2211TAeq,

    L

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    Example

    Calculate LAeq,8 over an eight hour period for a

    worker exposed to the following noise levels and

    duration.

    Noise Level dB (A) Duration (hours)

    94 2

    89 3

    98 0.583 2.5

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    Answer

    LAeq,8 = 91.4 dB (A)

    }{8

    2.5x100.5x103x102x1010log1083109810891094

    Aeq,8L

    }{T

    10t................10t10t10log10

    Ln

    n

    10L2

    2

    10L1

    1TAeq,

    L

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    Personal Daily Noise Exposure Level

    The equivalent continuous A weighted sound level

    over an 8 hour period is called personal daily noiseexposure level, according to the Noise at Work

    Regulations, 1989.

    This is used for assessing noise exposure in the work

    place.

    In the regulations recommended level foroccupational noise is a personal daily noise

    exposure level of 85 dB (A).

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    Personal Daily Noise Exposure Level . . .

    i.e. 1st action level is 85 dB (A)

    This is taken as the industrial first action levelfor occupational noise.

    The second action level is 90 dB (A).

    Above this value it is possibly hazardous and earprotection (muff or plugs) must be provided to the

    workers or change their work shifts.

    A 12 hour LAeq,T value of 75 dB (A) is the common

    limit for construction site noise.

    Above this level site operations can be stopped by

    legal action

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    Example

    What is the maximum time for which an employee

    may spend in a particular work shop where thenoise level is 106 dB (A) without using ear

    protection if his noise dose is not to exceed an

    equivalent continuous noise level of 90 dB (A) over

    the period of 8 hour work shift?

    Assume that for the rest of the shift the employee is

    subjected to a constant sound level of 85 dB (A).

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    Answer

    Let t be the required time.

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    Example

    The noise of a construction site is caused by the following.

    Source dB (A) Duration (hrs)

    Compressor 89 8

    Excavator 85 2

    Truck 78 6

    Pump 76 7

    Calculate the equivalent continuous sound level

    over a 12 hour working day of a worker exposed to

    the above noise levels.

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    Answer

    L = 10 log

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    Noise exposure from single discrete events

    In many situations the total noise exposure over

    a period of time is made up from a number of

    different individual events such as passing of an

    air craft over head or a train near by or a short

    bursts of machinery noise.

    The measurements of noise from different

    events will be made over different durations.

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    Noise exposure from single discrete events

    For comparison of different types of events itwould be convenient if the equivalent

    continuous sound level is averaged over the

    same duration.

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    Reference book:

    Acoustics and noise control

    2nd edition

    B J Smith, R J Peters and S Owen

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    Practical schedule

    3 Practical

    2 - Outdoors

    1 Industrial visit

    Assignments:

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

    3 for performance

    7 for assignment