noise handout

ME30033/50155 – 1. Revised 2014

ME30033/ME50155

Mechanical Vibrations and Noise

Handout 1: Noise

Lecture 1 Sound levels, Decibels or 2+2=5

1.1 Introduction

Noise is often defined as ‘unwanted sound’. Sound takes the form of waves travelling through the air (or any medium, which may be a gas, liquid or solid). These waves consist of pressure fluctuations and an associated vibration or velocity fluctuation of the medium. The science of sound is called ‘Acoustics’. The behaviour of sound waves in a 3-dimensional space can be extremely complex, and acoustics can be very highly mathematical. In practice, empirical approximations are heavily used.

1.2 Sound pressure level

Sound Pressure Level (SPL) is a measure of the sound level detected by an observer at a particular location.

(a) Free-field (anechoic) conditions (b) Diffuse (reverberant) conditions

Figure 1.1 Different sound fields

SPL detected by observer

Directly radiated sound waves

Reflected sound waves

SPL

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The loudness of sounds detected by the human ear depends on the magnitude of the pressure fluctuations (it also depends on the frequency, which we’ll come on to later). A suitable measure of the magnitude is the root mean square (rms) average of the pressure fluctuation:

2

2

21T

T

rms dttpT

p

The human ear can detect sounds with rms pressure fluctuations from about 10 Pa (10-10 bar)

to about 103 Pa (0.01bar). Because of this wide range, it is normal to compress the numbers into a convenient logarithmic scale. The sound pressure level (SPL or LP) is defined using the decibel (dB) scale as:

SPL or ref

rms

ref

rmsP

p

p

p

pL 102

2

10 log20log10

where pref is defined as 20Pa, which is approximately the threshold of hearing for a young person. Some typical SPLs are listed in table 1.1.

The Decibel scale is nicely compressed to a convenient range of numbers. Note that a factor of 10 increase in prms corresponds to a 20dB increase in SPL. The smallest noticeable change in sound level is approximately 1dB.

Typical source prms (Pa) SPL (dB) Effect

Threshold of hearing 0.000 02 0

Faint rustling of leaves 0.000 2 20

Quiet residential area

at night

0.002 40

Conversation 0.02 60

Vacuum cleaner 0.2 80 > 85dB: Hearing damage due to

long-term exposure Pneumatic drill 2 100

Rock concert, close to

speakers

20 120 > 120dB: Hearing damage due to

short-term exposure

Jet engine at 10m 200 140 130dB: Threshold of pain

Rocket engine at 10m 2000 160

Table 1.1 Typical sound pressure levels

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1.3 Sound intensity

This is a measure of the power flow through a unit area.

Figure 1.2 Sound intensity

It is given by the equation

2

2

1T

T

dttutpT

I

This derives from the power flow being given by force velocity, and pressure being force per unit area. This is time-averaged. I is often taken to be the energy passing through an imaginary surface enclosing a sound source, in which case u is the velocity at a point on that surface, normal to the surface.

Sound intensity has the dimensions of W/m2. It is usually expressed in dB:

SIL or ref

II

IL 10log10 where

1210refI W/m2.

For a wave with no reflections as in figure 1.1(a), the velocity is proportional to the pressure.

c

tptu

where c is the ‘characteristic impedance’.

For air at sea level and 20C, 1.2 kg/m3 and c 344 m/s.

4133442.1 c kg/m2/s (or Ns/m3).

So

2

2

22

413

11T

T

rmspdttpTc

I

So 10

2

10

2

101013.4

log10413

log10

rms

ref

rmsI

p

I

pL

The SPL for the same sound is given by

10

2

1025

2

102

2

10104

log10102

log10log10

rmsrms

ref

rmsP

pp

p

pL

Intensity = power per unit area passing through imaginary surface

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So the sound intensity level and sound pressure level from a point source in reflectionless (anechoic) conditions are (virtually) the same in dB.

However in a reflective sound field (figure 1.1(b)) the sound intensity level in dB is less than the sound pressure level in dB. This is because the intensity relates to the energy flow, and waves travelling in different directions have energy flow in different or opposite directions, so the net energy flow is less. In fact there may be zero energy flow.

1.4 Sound Power

This is a measure of the total sound power produced by a source, or the sound power passing through a given area, in W (Watts). Again it is often expressed as a sound power level (SWL) in dB:

SWL or ref

WW

WL 10log10

where W is the sound power in W and Wref is the reference sound power which is 10-12 W.

(note: ref

ref

refA

WI

where 1refA m2)

Sound powers are often surprisingly small, as most sources are very inefficient at generating sound.

Sound power (W) SWL (dB)

Person speaking loudly

Trumpet

Electric guitar and 100W amplifier

Saturn V space rocket

Table 1.2 Typical sound powers

Note that sound power is a measure of the sound produced by a source, not the sound detected by an observer. It is not related to the distance from the source.

1.5 Addition of sounds

When there are two sound sources, we can calculate the sound level produced at a point by each source individually and then add the result to give the total sound level. However care is needed, and we cannot just add the values in dB. Two noise sources of 60dB each do not produce a combined noise of 120dB! Also we do not add the sound pressures. Instead the total sound is obtained by adding the squares of the pressures (usually! An exception to this rule is given later). This is because the sound energy or power is related to the sound pressure squared.

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Since SPL=

refrefp

p

p

p102

2

10 log20log10 ,

1022 10

SPL

refpp

So we need to take antilogs of the dB values. For n noise sources:

n

i

SPL

tot

i

1

10

10 10log10SPL

In these notes, adding sound levels in dB is represented using the symbol. This is not a standard convention!

1.5.1 Example

Consider two sources, each of which individually produces a SPL of 60dB at a measurement location.

A 3dB increase is equivalent to a doubling of sound power ( 2log10 10 3dB).

Consider two sources which produce SPLs of 60dB and 65dB.

i.e. only a 1dB increase over the louder source’s SPL. In general, when one source produces a SPL more than about 5dB higher than another source, the quieter source has very little effect on the overall SPL. This doesn’t necessarily mean that the quieter source is not significant! It may actually be a more annoying or noticeable sound, depending on its tone and other factors. A rattle in a car may have very little effect on the SPL inside the car, but it can be very noticeable. The overall SPL does not tell the whole story. Noise is rarely as simple or predictable as we would like it to be.

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Finally, add two sounds of 2dB and 2dB.

1.6 Conclusions

Noise is unwanted sound. Sound travels through the air as waves, travelling at the speed of sound.

Sound pressure level (SPL) is a measure of sound level at a location. Sound power level (SWL) is the power of the sound produced by a source. Sound intensity level (SIL) is a measure of power flow. These quantities are usually expressed in decibels.

1.7 Problems

1. Calculate the SPL in dB for these rms pressure levels.

a) 1 Pa

b) 10 Pa

2. Calculate the sound power in Watts for the following:

a) SWL = 65 dB

b) SWL = 75 dB

3. The following values are the SPL produced by machines when running on their own. Calculate the total SPL, to the nearest dB, when the machines are running together.

a) 70 dB, 75 dB, 77 dB

b) 70 dB, 80 dB

c) Four machines, 80 dB each

d) Ten machines, 80 dB each

e) A hundred machines, 80 dB each

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Lecture 2 Wave propagation in one dimension

We shall start off with the much simpler one-dimensional case. Later we shall use practical empirical approximations for 3-D sound.

The aims of this section are to show the behaviour of waves in a pipe both mathematically and qualitatively. The phenomena of standing waves and resonance will be described.

1-D wave propagation applies to several important cases, including organ pipes, wind and brass instruments, hydraulic lines, water pipes etc. Imagine a pipeline containing a fluid (gas or liquid), with a piston at one end. Now imagine what happens if you move the piston a short distance at a constant velocity (we shall neglect the time taken to accelerate the piston and assume a step change to the new velocity).

Figure 2.1 Wave propagation in a pipe

If the fluid was incompressible, the complete column of fluid would have to move as one, at the same velocity as the piston. However this doesn’t happen because a real fluid is compressible.

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Instead, initially the fluid nearest to the piston becomes compressed and moves at the velocity of the piston. This disturbance propagates along the tube at the speed of sound. The speed of sound is typically about 340 m/s in air, 1500m/s in water. The fluid itself does not move at this velocity – it only moves at the speed of the piston – but the disturbance or wavefront propagates along the tube at that speed (imagine a ‘Mexican wave’ in a stadium. Although each person only moves up and down, the wave seems to move at high speed around the stadium).

For this simple case the pressure can be assumed to be uniform across the cross-section of the tube, and these waves are known as plane waves. Later we shall consider what happens when the wave reaches the end of the tube.

2.1 Analysis

The behaviour of pressure waves in a gas in a rigid pipeline can be analyzed by considering a

short element of length x as shown in Figure 2.2. We shall assume that the diameter is sufficiently small to assume no radial velocity or pressure variations and consider variations in one dimension only. We shall also ignore viscous friction (this is usually a fairly small effect anyway).

Figure 2.2 - Short element of fluid in pipe.

Applying Newton's second law gives:

t

uxA

t

umAp

(1)

For the limit as x tends to 0,

t

u

x

p

(2)

We can also apply continuity of mass.

Rate of change of mass = mass flow in – mass flow out

txA

= AuuuA

AuuAuuuuuA (ignoring u term) (3)

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Hence

uut

x

(4)

or as x tends to 0, x

ux

u

t

(5)

Provided that the particle velocity u is relatively small (u << c), the second term is negligible, so:

x

u

t

. (6)

The density is related to the pressure by the bulk modulus B. This is defined as

pB (7)

By the chain rule for differentiation, t

p

Bt

p

pt

so from equation (6) x

u

t

p

Bt

or (8)

Differentiating (2) w.r.t. x, and (8) w.r.t. t, we get

(9)

(10)

Combining equations (9) and (10), we obtain the partial differential equation

(11)

Define

Bc . It turns out that this is the speed of sound.

2

2

22

2 1

t

p

cx

p

(12)

This is the Wave Equation. It also has the same form for velocity: 2

2

22

2 1

t

u

cx

u

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It can be shown that two possible solutions of the wave equation are

c

xtftxp , , and c

xtgtxp , . (13)

The first solution represents a wave travelling in the positive x direction at speed c, and the second solution a wave travelling in the negative x direction.

Figure 2.3 A pressure pulse travelling in the positive x direction, shown at two positions, showing the delay in the pulse at a position x (Fahy and Walker 1.6)

Similar equations also apply for velocity u:

c

xtfc

txu

1, , and

cxtg

ctxu

1, (14)

The velocity wave is proportional to the pressure wave and the ratio is the characteristic impedance c :

c

txptxu

,,

More generally, we may have waves travelling in both directions:

c

xtgc

xtftxp , (15)

c

xtgc

xtfc

txu

1, (16)

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2.2 Speed of sound in air

The speed of sound depends on the bulk modulus. To determine this we shall assume that isentropic compression applies, because we are dealing with high frequencies and rapid

changes, so pV constant where is the ratio of specific heats. Considering rapid

compression from 0p to p:

0

0pp (17)

Differentiating w.r.t. density: (18)

Evaluating this at 0pp and 0 , we get (19)

pB so 0pB . (20)

The pressure-density relationship for air is M

RTp

so

M

RTc 0 where , R and M are

constants.

For dry air at 20C, the speed of sound is 343m/s. It increases with temperature but is independent of pressure.

2.3 Solution for sinusoidal pressure variation

Sound is usually considered in terms of its frequency components, largely because this is how the ear interprets it, but also because it is often easier to analyse this way. A single frequency is represented by a sinusoidal pressure variation.

tjexPxtxAtxp Recos, where xjexAxP . (21)

The exponential form is more convenient for this analysis. P is the complex amplitude, which contains both amplitude and phase information. We can differentiate the exponential form w.r.t. x and t:

2

2

x

p (22)

2

2

t

p (23)

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Substituting these into the wave equation (12) 01

2

2

22

2

t

p

cx

p:

(24)

so (25)

For this to be true for all t, the contents of the square brackets must equal zero.

02

2

2

2

xP

cx

P (26)

This is known as the one-dimensional Helmholtz Equation. It is in fact the Fourier Transform of

the wave equation

j

t.

General solutions to this equation are:

c

xj

c

xj

GeFeP

(27)

c

GeFeU

c

xj

c

xj

(28)

F and G are complex numbers and depend on the boundary conditions. F represents a

pressure wave travelling in the positive x direction. The

c

xj

e

term is effectively a phase lag or delay which increases linearly with increasing x and represents the motion of the waves along the line in the direction of increasing x . G is a pressure wave travelling in the negative x

direction and

c

xj

e

is a phase lag which increases with decreasing x.

The pressure P is equal to the sum of the two pressure waves. The velocity U is equal to the difference between the two waves as they are travelling in opposite directions.

The wavelength is given by f

c. For example, for a mid-range frequency f of 1000Hz in air with

c = 340m/s, = 0.34m. For f = 100Hz, = 3.4m. Wavelength decreases with frequency.

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2.4 Conclusions

We have determined equations for wave behaviour in the time domain and the frequency domain. The frequency domain solution will be useful when analysing reflections and resonance.

2.5 Problems

Calculate the speed of sound at sea level, at a typical airliner cruise altitude of 10,000 m, and at Concorde’s cruise altitude of 18,000 m, given the following data.

Sea level temperature 20C (293 K)

Lapse rate (drop in temperature with altitude): 6.5 K per 1000 m up to 11,000 m

Temperature above 11,000 m: -56.5C

Pressure/density relationship for air M

RTp

,

where R = 8.3145 J/mol/K, M = 0.02896 kg/mol and T = temperature in K.

Ratio of specific heats = 1.4.

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Lecture 3 Reflections and Standing Waves

3.1 Closed ends

Consider a simple pipe with a piston at one end moving sinusoidally, and the other end closed as shown in figure 3.1. The piston generates a pressure and velocity wave G in the pipe. This travels along the pipe at the speed of sound. The wave is reflected at the end producing a wave F travelling in the opposite direction. Further reflection occurs at the piston, and so on.

Figure 3.1 Pressure and velocity amplitude along the length of a closed-ended pipe

The fluid velocity at the piston is equal to the piston velocity v:

tLUtAv coscos

A is the piston velocity amplitude and LU is the fluid velocity amplitude at Lx .

Reflections occur at the pipe ends. If the pipe end is closed at 0x , the velocity is zero, so U(0)=0. Thus from Equation (28), at 0x

(29)

so the wave F is reflected back with no change in its amplitude or phase. Substituting into (27):

(30)

x

x=0 x=L

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Similarly by substituting into (28), (31)

The pressure amplitude xP varies along the pipe in the form of a ‘rectified sinewave’

( xsin or xcos ). The troughs (minimum amplitude points) are called ‘nodes’ and the peaks

‘anti-nodes’. Nodes are spaced a half wavelength apart, as are anti-nodes. Velocity nodes also occur at the positions of the pressure anti-nodes, and velocity anti-nodes at the pressure nodes. A pressure anti-node and a velocity node appear at the pipe end. Also, the phase does not vary along the length of the pipe, so there is no apparent movement of the waves along the pipe. This is called a standing wave.

3.2 Open ends

Perhaps surprisingly, reflections also occur at an open end. Consider figure 3.2. Here P(0) = 0 and from Equation (27):

(32)

i.e. the reflected wave is inverted or has a phase shift of 180. The complex amplitudes of the pressure and flow at a distance x are given by

(33)

(34)

Figure 3.2 Pressure and velocity amplitude along the length of an open-ended pipe

x

x=0 x=L

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Q: What is the sound intensity in a blocked pipe with a sinusoidal input?

Other boundary conditions may give phase shifted or attenuated reflections.

An example of measured pressure amplitude in a pipe is shown in figure 3.3, for two frequencies. Note that the wavelength, and the spacing between nodes, is shorter at the higher frequency.

Figure 3.3 Measured standing wave in a pipe.

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3.3 Resonance

Fluid in a pipeline will oscillate at its resonant frequencies following a disturbance. In practice there will always be some damping so that the oscillations will die away. If a sinusoidal flow or pressure input is applied at one end, high amplitudes may be generated at or near resonance.

For this analysis, we shall assume no damping.

3.3.1 Case 1: ‘closed-closed’ tube

Consider a pipe, length L, with both ends closed (at x=0 and x=L). Equation (31) is

c

x

c

jGxU

sin

2

The velocity amplitude at x=L is 0sin2

c

L

c

jGLU

.

This equation is satisfied either by 0G or

nc

L . The second condition defines the natural

or resonant frequencies, at which the gas will oscillate following some disturbance.

The resonant (or natural) frequencies are where

3,2,,0c

L etc, or

L

nc

L

c

L

cf

2...,,

2

2,

2 .

These are sometimes called ‘closed-closed’ or ‘fixed-fixed’ modes, as both ends of the pipe are closed (fixed).

In practice there will be some damping due to viscous friction, and there will need to be some source of excitation. This could take the form of a piston oscillating at one end of the tube. With a piston oscillating sinusoidally at the right-hand end, the pressure amplitude at point x is given by

c

L

c

xcAj

xP

sin

cos

where A is the piston velocity amplitude. According to this equation, if the piston oscillates at a resonant frequency the pressure amplitude will be infinite. In practice the amplitude would be finite because of damping and non-linearities.

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3.3.2 Case 2: ‘closed-open’ or ‘open-closed’ tube

Similar equations can be derived for an open-ended tube, giving us the ‘closed-open’ or ‘fixed-free’ modes. Assuming that the end is open at x = 0, the velocity amplitude is given by the equation

c

x

c

GxU

cos

2

Assuming that the end is closed at x=L,

0cos2

c

L

c

GLU

This is satisfied by G = 0 or

c

Lcos = 0. The latter condition defines resonance, where

2

12

n

c

L or

L

cn

L

c

L

cf

4

12...,,

4

3,

4

3.3.3 Case 3: ‘open-open’ tube

Assuming that the end is open at x = 0, from equation (33)

c

xjGxP

sin2

Also assuming that the end is open at x = L, 0sin2

c

LjGLP

This is satisfied by G = 0 or

c

Lsin =0. The resonant frequencies are

L

nc

L

c

L

cf

2...,,

2

2,

2 , the

same as for the closed-closed case.

3.4 A pipe as a musical instrument

Each of these three cases has a series of resonant frequencies at integer multiples of the first frequency (although for the closed-open or open-closed case the series consists of odd-numbered multiples). These are known as the fundamental frequency (or first harmonic),

Fundamental frequency

Harmonic series Frequency of nth frequency

Closed-closed

L

c

2

1,2,3,4…

L

nc

2

Closed-open, open-closed L

c

4

1,3,5,7…

L

cn

4

)12(

Open-open

L

c

2

1,2,3,4…

L

nc

2

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second harmonic, third harmonic etc. A musical note is generally composed of a whole series of these harmonics. The tone or timbre of the sound is governed by the balance between the amplitudes of these harmonics. The fundamental frequency is inversely proportional to the length of pipe.

Wind and brass instruments produce notes at the resonant (natural) frequencies of the column of air in the tube, generated by a vibration source at the blown end.

The open-open case is effectively what occurs in a flute, recorder or organ pipe, where air flow across a sharp edge produces von Karman vortices and pressure fluctuations at one end. In a flute or recorder, the effective length L depends on the position and number of open holes (although it is not as simple as one single open end). In an organ, pipes of different length are used for different notes.

Most reed instruments (clarinet, saxophone etc) act as closed-open pipes. The vibrating reed acts as the vibration source, and the open holes act as the open end although in a more complicated way, as there may be several small open holes. Brass instruments also work in this way, but the pitch is changed in two ways: by changing the tube length using valves or a slider, or by changing the frequency of the vibration source at the player’s lips to generate different tones (overtones) in the harmonic series. However there is an added complication: the tube is usually conical or tapered in some way (except in a clarinet, which is mainly cylindrical). This can be shown to give a harmonic series equivalent to an open-open tube.

Note that if we were to block the end of an organ pipe we would change it from open-open to open-closed. This would reduce its fundamental frequency by a factor of two, shifting it down one octave. It would also change its tone, largely because only the odd harmonics would be generated (note the form of the frequency series). This is one reason why flutes (open-open) sound different to clarinets (closed-open), though there are many other reasons.

3.5 Conclusions

Reflections occur at closed and open ends of tubes. These cause standing waves to occur. Resonances and anti-resonances occur at certain frequencies. These occur at multiples of the fundamental frequency.

Musical instruments rely on resonance to produce musical notes. The different tones of different instruments are the result of different end conditions and the shape of the tube, among other factors.

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3.6 Problems

1. Calculate the speed of sound in air, where

Density = 1.2 kg/m3

Ratio of specific heats = 1.4

Pressure = 1 bar

Hence calculate the first natural frequency of:

a. an organ pipe of length 1.1m, with two open ends

b. an organ pipe of length 1.1m, with an open and a closed end

c. an organ pipe of length 0.55m, with an open and a closed end

2. A 10m long pipe is full of air and open to the atmosphere at one end. At the other end is a piston moving sinusoidally with a displacement amplitude of 10mm (i.e. 20mm peak-peak). The piston has the same diameter as the pipe. The air density is 1.2 kg/m3 and speed of sound of 342 m/s. Determine the maximum pressure amplitude and velocity amplitude and the location(s) along the pipe at which these occur, for a piston frequency of

(a) 20Hz

(b) 25Hz.

Hint: You first need to calculate the piston velocity amplitude. Assume that the air velocity is equal to the piston velocity at that end. Then, using the equations

c

xj

c

xj

GeFeP

, c

GeFeU

c

xj

c

xj

,

apply the open-end boundary condition at 0x and then the piston boundary condition at Lx to find F and G.

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Lecture 4 Wave propagation in three dimensions

The wave equation and the Helmholtz equation can be extended to three dimensions:

2

2

22

2

2

2

2

2 1

t

p

cz

p

y

p

x

p

and 0

2

2

2

2

2

2

2

2

P

cz

P

y

P

x

P

or 2

2

2

2 1

t

p

cp

and 0

2

22 P

cP

4.1 Natural frequencies of an enclosed box

It can be shown (Fahy & Walker) that a solution of the wave equation in a rigid-walled cuboidal enclosure of size (Lx, Ly, Lz) is

tL

zn

L

yn

L

xnAtzyxp n

z

z

y

y

x

x

coscoscoscos,,,

where

222

z

z

y

y

x

xn

L

n

L

n

L

nc

For a cube,

√

Different combinations of integers nx, ny and nz, give different natural frequencies and different mode shapes. Unlike in the 1-D case, the natural frequencies are not equally spaced and get closer and closer together at higher frequencies.

4.2 Refraction

Sound waves have many similar properties to light waves. They can be reflected from a surface, and the reflection occurs at the same angle relative to the surface as the incident wave. Refraction can occur when sound travels into a region with a different speed of sound, at an

oblique angle. As the frequency stays the same, the wavelength must change (f

c ) so the

direction changes as shown in figure 4.1.

Figure 4.1 Refraction of sound waves

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Sound travels at a speed that is independent of movement of the source, but is relative to movement of the air. That is, sound travels faster downwind.

Whilst sound is directional, it tends to be less directional than light. This is because sound has much longer wavelengths. Diffraction around obstacles occurs where the wavelength is long relative to the obstruction. High frequency sound is more directional than low; this is why a sub-woofer can be placed in any convenient place in a room, whilst the position of the mid-to-high frequency speakers is important for the spatial image of the sound.

4.2.1 Questions

1. Why can distant sounds across a lake at night sometimes be heard unexpectedly loudly?

2. Why does a distant sound seem louder when its source is upwind of the observer?

4.3 Effect of distance

The sound power produced by a source is equal to the total sound power passing through a surface S enclosing the source. It is related to the sound intensity by the equation

S

IdSW

where I is the component of the sound intensity vector in the direction normal to the surface.

We can use this to estimate the SPL at a distance from a source. The equations that follow are approximate as they assume that the source radiates equally in all directions and that there are anechoic conditions (except for perfect reflection off the floor in cases 4.3.2, 3, 5 and 6).

4.3.1 Point source, sound radiation in all directions

Imagine the idealised case where we have a point source of sound suspended freely in mid-air with no reflections (it is valid to assume a point source if the dimensions of the source are much less than the wavelength of the sound).

The sound intensity I normal to the surface of a sphere surrounding the source at radius r will be uniform over the sphere, so we can say that

24. rIIAW

or 24 r

WI

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Doubling the radius results in a reduction in sound intensity I by a factor of 4. This is the inverse square law.

Also, as conditions are assumed to be anechoic, the sound pressure level is equal to the sound intensity level in dB:

SPL = SIL =

2104

log10rI

W

ref

(note: ref

ref

refA

WI

where 1refA m2)

The SPL and SIL reduce by 6dB when the distance from the source is doubled.

4.3.2 Point source close to reflective surface, hemispherical radiation

This more realistic case occurs when a sound source is located close to the ground or a wall.

Sound travelling downwards is reflected off the surface. Provided that the source is close to the ground, the sound radiation is uniform.

SPL = SIL = rrI

W

ref

10210 log208SWL2

log10

That is, a 3dB increase on the spherical case, section 4.3.1.

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4.3.3 Point source far from reflective surface

If the distance between the source and surface is large (

, say), the hemispherical

radiation assumption does not apply. In this case we can consider a true source and a virtual reflected source, and add the sound from the two.

4.3.4 Line source

Consider an infinitely long line source, suspended in air with anechoic conditions.

For a unit length, where W and SWL are the sound power per unit metre, consider the sound produced from one point on the line. The overall sound level can be found by integration along the length of the source.

The sound intensity I produced by an infinitesimal length x , where x is the distance along the

source line from the nearest point to the observer, is 224 xr

xWI

, since the distance from

that point on the source to the observer is 22 xr .

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Since there are anechoic conditions, the mean square pressure can be related to the intensity:

Icprms 2

Integrating,

So the sound falls off by only 3dB per doubling of the distance.

For a finite length source, this model holds for points close to the line (r<<L). If r>>L the behaviour tends towards that of a point source.

4.3.5 Line source close to reflective surface

Reflections off the surface result in a 3dB increase in SPL.

r10log103SWLSPL

This is roughly equivalent to a busy road, provided that

br , where b is the average

spacing between cars. If

br , the noise is dominated by the nearest car and hemispherical

radiation (case 4.3.2) applies.

For several lines of traffic, calculate the SPL from each line individually, then add.

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4.3.5.1 Estimation of SPL per m for a line of traffic

4.3.6 Line source far from reflective surface

If the distance between the source and surface is large (

, say), consider a true source

and a virtual reflected source, and add the sound from the two.

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4.3.7 Example

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4.4 Conclusions

3D wave propagation is more complex than 1D. Resonances occur in a box or room, but these are usually not considered in its analysis. Empirical/statistical approximations are used instead.

We have derived estimates for the sound pressure level at a distance from some simple sound sources.

4.5 Problems

1. Calculate the first 10 natural frequencies of air in a box of internal dimensions 1m x 0.7m x 0.6m. (note: you need to try different combinations of small values of nx, ny and nz , including zero).

2. Traffic on an urban road is modelled as four lines of noise sources. Lane 1 consists of lorries, 30m apart, each with a SWL of 110dBA. Lane 2 consists of cars 20m apart, each with a SWL of 100dBA. Traffic travelling in the opposite direction is identical, so lane 3 is identical to lanes 2 and lane 4 to lane 1. Each lane is 5m apart.

Estimate the SPL at a distance of 30m from the nearest lane.

3. Traffic on a single lane one-way road is approximated by cars uniformly spaced 30m apart, each with a SWL of 100dBA. An observer is 10m from the road. Assuming uniform hemispherical radiation from each car, and by considering a sufficient number of cars, estimate the SPL at the observer:

(a) when a car is passing the observer;

(b) when the nearest two cars are equidistant from the observer;

(c) assuming a uniform line source;

(d) assuming a uniform line source and assuming that the cars and the observer are 2m above the surface.

30m

10m

30m

10m

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Lecture 5 Sound Measurement

5.1 The Human Ear’s Response

The human ear is an extremely sensitive pressure transducer, but it is more sensitive to some frequencies than others. It has a poor sensitivity at low frequencies and is most sensitive to mid-to-high frequencies. Sounds with frequencies below about 20Hz or above 15-20kHz cannot be heard (although high SPL sounds below 20Hz may be felt). Perceived loudness is sometimes measured in phon. Figure 5.1 shows contours of equal loudness. The bottom line shows the threshold of hearing. It shows that the threshold for frequencies of about 500Hz-5000Hz is about 0dB, but the threshold rises rapidly below 100Hz, meaning that low frequency sounds of low SPL cannot be detected. The ear has a non-linear response; at higher loudness levels the curves level off a bit, so the ear becomes more sensitive to low frequencies at high SPL. If music is played loudly, the bass becomes more audible.

Figure 5.1 Relationship between frequency, SPL and loudness (from http://en.wikipedia.org/wiki/Equal-loudness_contour)

http://en.wikipedia.org/wiki/Equal-loudness_contour

http://en.wikipedia.org/wiki/Image:Lindos1.svg

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Figure 5.2 The human ear

Often noise measurements are weighted to follow the response of the human ear. The most common is ‘A’ weighting. Measurements are quoted in dB(A) or dBA.

Figure 5.3 ‘A’ weighting curve (from http://www.diracdelta.co.uk/science/source/a/w/aweighting/source.html)

http://www.diracdelta.co.uk/science/source/a/w/aweighting/source.html

//upload.wikimedia.org/wikipedia/commons/d/d2/Anatomy_of_the_Human_Ear.svg

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For example, a sound consists of three components:

70dB at 100Hz

65dB at 500Hz

60dB at 1000Hz

To find the overall sound level in dB(A), first add the weighting value for each frequency to the dB level at that frequency. Then add the antilogs of the weighted values.

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5.2 Measurement of Sound Spectrum

The fundamental transducer for air-borne noise measurement is the microphone. Portable sound level meters are available which contain a microphone, filters and meter giving the measurement in dB. A typical device is shown in Figure 5.4.

An instrumentation microphone will generally have a flat response to all frequencies in the audible range from 20 Hz to 20 kHz. The complete signal may be used to represent the overall noise level.

Figure 5.4 Sound level meter

Often an overall dB measurement is too crude to be of much use and more information can be obtained by breaking the sound down into its frequency spectrum. There are two ways to do this: using narrow band filters and using a Fast Fourier Transform (FFT).

5.2.1 Narrow-band filters

The microphone output may be filtered using analog (electronic) or digital filters to give the noise level in a selected frequency band. The frequency bands are spaced equally on a logarithmic scale. In this way the complete noise spectrum can be built up, typically in one octave, 1/3 octave, 1/12 octave or 1/24 octave bands. An octave is a 2:1 frequency ratio; thus an octave band may be from 100 Hz to 200 Hz, for example. A third octave is a 21/3:1 frequency ratio (a major third, musically speaking). A twelfth octave is equivalent to one semitone in the musical scale.

A typical 1/24 octave band measurement is shown in figure 5.5. By analysing the frequency of the peaks it may be possible to determine the source of the noise. This spectrum was measured in a car with the engine running at 840 rev/min. The engine harmonics are at multiples of this frequency (in Hz). The power steering pump has 10 vanes and runs at the same speed as the engine. The dominant engine and power steering harmonics can be seen in the spectrum.

(note: each band is 241

2 or 1.0293 times the frequency of the previous band).

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Figure 5.5 Typical 1/24 octave band spectrum measured in a car

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5.2.2 FFT analysis

This is usually performed using a digital spectrum analyser. The signal is sampled digitally and a digital Fourier transform is performed on it to obtain its spectrum. A fast Fourier transform (FFT) is a very efficient algorithm for this purpose. The spectrum provides great detail with narrow frequency bands uniformly spaced on a linear frequency scale. It can be used to pinpoint accurately the harmonic components of the noise. This detailed information can often be used to pinpoint the source of the noise by relating the frequency of the predominant noise harmonics to the operating speed of the machinery. An example of a measured amplitude spectrum of pump noise is shown in Figure 5.6. This is a twelve-vane pump running at 840 rev/min. The fundamental

frequency is equal to (speed in rev/sec) (number of vanes) = 1401060

840 Hz. The

fundamental frequency and harmonics produced by the pump can clearly be seen at multiples of

140 Hz. Smaller harmonics at multiples of shaft speed ( 1460

840 Hz) can also be seen.

Figure 5.6 Amplitude spectrum of pump noise

5.3 Conclusions

The ear is an extremely sensitive transducer, but has an uneven or non-linear response. Its response varies with frequency and amplitude.

Microphones are used to measure sound pressure levels.

Measurements may be weighted to correspond to the ear’s response.

Sound power level is difficult to measure, and requires special acoustic chambers or specialist instrumentation. This will be considered next.

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5.4 Problems

A sound has been measured using an octave band analyser and split into bands with the following centre frequencies:

125Hz: 80dB 250Hz: 73dB 500Hz: 70dB 1000Hz: 65dB 2000Hz: 61dB 4000Hz: 60dB

By estimating the A weighting factor at the centre of each band, estimate the overall A weighted SPL.

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Lecture 6 Measurement of Sound Power, and More Addition of Sounds

It is necessary to be able to measure the sound power of machinery. However the sound power of a source cannot easily be measured directly. There are several methods, most of which determine it indirectly from sound pressure measurements, taking account of the characteristics of the room.

6.1 Anechoic chamber

In anechoic conditions, the sound pressure level (SPL) and sound intensity level (SIL) are equal. SPL is measured using a microphone moved over a regular grid of points surrounding the source. Remember that the sound power is equal to the integral over a surface surrounding the source of the intensity normal to that source:

S

IdSW

The SWL can be estimated from the average of the measured sound pressures times the area of the imaginary grid surface:

n

i

rmsn

i c

p

n

AI

n

AW

1

2

1

n

i

SPLi

n

ASWL

1

1010 10log10

A possible arrangement for this test is shown in figure 6.1. The microphone is moved over the hemispherical surface. A disadvantage is that an anechoic chamber is required; this is not feasible for large machinery or a jet engine, for example. A large open space may be a possible alternative.

Figure 6.1 SWL measurement in anechoic chamber

Reflecting plane

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6.2 Reverberant chamber

In an enclosed chamber the sound energy is dissipated as heat in the walls, so the sound power is equal to the rate of energy dissipation in the walls. This is given by the equation

4log10 10

CRSPLSWL where RC is the room constant with units of m2 (see lecture 9).

The SPL should be measured at several locations within the chamber and averaged:

n

i

SPLC

i

n

RSWL

1

1010 10

4log10

Ideally the reverberant chamber should be of an irregular shape to avoid room resonances. The value of RC can be determined by measurement of the reverberation decay time in response to an impulsive sound, using the equation:

T

VRC

16.0

where V is the room volume (m3) and T is the time for the sound energy to decay by 60dB.

Figure 6.2 Reverberation decay (from http://www.kettering.edu/~drussell/demos.html)

6.3 Sound intensity measurement

The sound power can be estimated from n sound intensity measurements over an imaginary surface of area A enclosing the source:

S

IdSW or

n

i

iIn

AW

1

This is true if there are reflections, and even if there are other noise sources outside of the surface (the net effect will cancel out as the sound energy passes in and out through the surface). This means that this method can be used in a normal working environment.

Without carpet

With carpet

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The main problem is how to measure the intensity I, as a microphone only measures sound pressure. It can be measured using a special intensity probe, consisting of two closely spaced

microphones x apart, as shown in figure 6.3.

2

2

1T

T

pudtT

I

Since t

u

x

p

(Newton’s second law),

dtx

ppdt

x

pu

2111

A problem with this method is that the microphones have to be very closely matched in their

response, otherwise the approximation x

pp

x

p

21 will be seriously wrong. It requires

expensive, high quality equipment and careful calibration.

Figure 6.3 Intensity probe, with two microphones 1cm apart

6.4 dB algebra

When adding or subtracting noise sources or frequency components: take anti-logs.

101010 1010log10dBdB

YX

YX

101010 1010log10dBdB

YX

YX

Each noise component must have the same weighting or scaling.

dBAdBAdBA

dBdBdB

dBdBA

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When applying weightings, attenuation, gain, transmission loss: just add or subtract. This is

equivalent to a multiplication factor.

dBdBdB

dBAdBdBA

dBAdBA

dBdBAdBA

dBAdB

6.5 Addition of sounds with the same frequency

If two sources are synchronised and produce noise at exactly the same frequency, the previous rule for adding sounds doesn’t apply. The instantaneous pressure is given by the sum of the instantaneous pressure produced by each source alone:

tptptp 21

Let tPp cos11 . If p2 is in phase with p1, tPp cos22

so tPPtp cos21 and the amplitude is the sum of the individual amplitudes.

However p2 is more likely to be out of phase with p1, tPp cos22 , in which case the

overall amplitude is given by cos2 21

2

2

2

1

2 PPPPP (cosine rule). The overall amplitude

may be significantly less than that when they’re in phase, and cancellation may occur with a

minimum of 21 PP if .

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6.6 Beating

If two sources produce noise at slightly different frequencies, an effect called ‘beating’ may occur as the two waves drift in and out of phase. For example consider the combination of two sinusoidal sounds

tPp cos11 , tPp cos22

tPtPp coscos 21 = ttPPttPP sinsincoscos 2121

or if P2=P1,

ttPtp coscos2 1

This appears as a sound of a single frequency , increasing and decreasing in amplitude over a

period of

seconds.

Figure 6.4 Beating produced by two sounds at 98Hz and 102Hz, with equal and unequal amplitudes

This beating effect can be quite unpleasant. It is sometimes heard on aircraft when two engines run at slightly different speeds.

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6.7 Conclusions

Sound power cannot be measured directly but is usually calculated indirectly from sound pressure measurements in special conditions.

Sounds with the same frequency may reinforce or cancel each other. Sounds with slightly different frequency may cause an unpleasant beating effect.

6.8 Problems

1. Which of these are valid calculations? If valid, write the answer with the correct units and give an example where this equation might apply.

a. 70 dB – 10 dB b. 60 dBA + 5 dB c. 60 dBA + 50 dBA d. 60 dBA - 50 dBA e. 33 dB 30 dB f. 70 dB 80 dBA g. 70 dBA 80 dBA 75 dBA

2. Two loudspeakers each produce a pure 1000Hz tone. One on its own produces an SPL of 90 dB and the other produces an SPL of 95 dB. Calculate the maximum and minimum SPL that could be produced by the two together if the signals are in-phase and out-of-phase.

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Lecture 7 Noise control

7.1 Introduction

High noise levels in the workplace are a common and serious problem. Recent statistics suggest that 17% of the UK population have significant hearing loss. 1.7 million UK workers are regularly exposed to noise levels above 85 dB(A), and 0.6 million UK workers to levels above 90 dB(A). Prolonged exposure to these noise levels is known to be potentially damaging to hearing. Stringent UK and EC legislation is in place to control workplace noise.

Medium noise levels can cause annoyance, loss of productivity, communication problems and complaints. A noisy vehicle or other product may lose sales or suffer warranty claims, and customers may choose a quieter alternative.

Noise transmission can be reduced by barriers and enclosures. However it is often better to reduce the noise at source by choosing a quieter machine, where possible.

7.2 European Union Noise at Work Regulations: EU Physical Agents (Noise) Directive 2003

Limits are set for workers’ exposure to noise based on a daily average noise level and a instantaneous peak pressure. Two action levels are defined.

First action level: LEX,8H = 80 dB(A) and peak pressure = 112 Pa

Above this level, employees have the right to demand suitable hearing protectors

Second action level: LEX,8H = 85 dB(A) and peak pressure = 200 Pa

Above this level, employees must be provided with suitable hearing protection which reduces risk below exposures at the second action level. Clearly marked ear protection zones must be implemented into which no employees must enter unless wearing hearing protection.

The regulations state that:

every employer shall ensure that a noise assessment is made when any employee is likely to be exposed at or above any action level;

records must be kept of noise assessments;

employers shall reduce risk of hearing damage to the lowest level practicable;

noise exposure must be reduced other than by provision of hearing protection if the second or peak action levels are reached;

ensure that hearing protection is maintained and used;

provide information to employees if any action level is exceeded.

The daily noise exposure level LEX,8H is a time-weighted average of the noise exposure level over a nominal 8-hour working day as follows, where T = 8 hours.

[

∫

]

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If the noise levels vary for different periods of the day, the daily noise exposure level can be calculated by summing the equivalent level for each period. For n sound pressure levels Lpi

each of duration ti:

[

∑

( )

]

Alternatively the equivalent level of each period can be calculated:

[

]

The total exposure can be calculated by adding the levels.

Example 1. Calculate the daily noise exposure level LEX,8H for 4 hrs at 84 dBA , followed by 3 hrs at 91 dBA , and then 1 hr at 94 dBA.

Example 2. If workers experience 4 hours at 79 dBA and 2.5 hours at 87 dBA, what level is allowable over the final 1.5 hours if LEX,8H of 85 dBA is not to be exceeded?

Hearing damage can be caused by sudden very loud sounds as well as by prolonged exposure to medium-level noise. The regulations also limit the peak pressure at any instant. The peak pressure limits correspond to a very much higher level than the daily limits. For example the second action level limit is 200Pa. This corresponds to a sound of roughly 137dB.

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7.3 Open-topped barriers

Open-topped barriers, walls or partitions are used outdoors to reduce road and rail traffic noise, or indoors in offices and shop floors. They can block sound by reflection or absorption. However they are only partially effective; although a barrier can completely block a line of sight, sound can bend around the barrier by diffraction. An approximate calculation of attenuation can be obtained using the Fresnel number

dBAL

N

22

where L = increase in sound path from source to observer caused by barrier.

Figure 7.1 Open-topped barrier

The attenuation produced by the barrier is given approximately by the equations

3.0log1012dB 10 NAtt for a point source

3.0log108dB 10 NAtt for a line source. This can be used for a road barrier where the

stream of vehicles can be approximated to a line source.

These are shown in figure 7.2.

The actual performance depends on many factors including: shape of barrier; barrier material; type of ground surface; reflections from buildings and other objects (including the opposite barrier). It also depends on frequency.

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Figure 7.2 Attenuation of a barrier

7.3.1 Example

In the diagram below, the car produces a SWL of 100dBA. Estimate the SPL heard by the policeman: (a) without the barrier; (b) with the barrier. Assume the sound is at 1000 Hz.

1.5m

1m

4m 6m

2.5m

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7.3.2 Multiple paths

There may be several possible noise transmission paths, e.g. over barrier, under barrier, around end(s) of barrier, through barrier.

To calculate the noise reduction produced by a barrier, where there are several noise transmission paths:

Assume nominal value of SPL (or calculate SPL without barrier);

Calculate SPL due to each path;

Add the SPLs together (remember how to add dB values).

Compare (subtract) SPL without barrier.

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7.3.3 Example

A barrier is 1.5m high and 4m long. It is calculated by the Fresnel method to give the following attenuations:

14dB over the top;

17dB around one end;

21dB around the other end.

It is also calculated by the ‘mass law’ (see next section) to attenuate sound travelling through it by 25dB (i.e. transmission loss or TL = 25dB).

Calculate the overall attenuation.

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7.4 Conclusions

There are strict rules for workplace noise to try and prevent hearing damage. Noise can be reduced by using quieter machinery, by using enclosures or barriers, or by using ear defenders.

There are also rules for environmental noise which we haven’t covered here.

Open-topped barriers block the direct transmission of sound, but sound can diffract over the top. They are commonly used in offices, with absorbent surfaces to absorb some of the sound.

High walls can help reduce road noise. They are less effective at low frequency.

7.5 Problems

1. A worker receives 1.5 hours at 96dBA, 0.5 hours at 100dBA and 6 hours at 80dBA. Calculate the noise exposure over the 8 hour period.

2. A worker receives 2.5 hours exposure at 89dBA. Calculate the noise level which the worker can then experience for the remaining 5.5 hours without exceeding the maximum permitted exposure of 85dBA.

3. Forestry workers are to be protected from the noise of a mechanical saw by means of a barrier 2.5m high erected 1m from the machine. The noise of concern is at 1000Hz.

The transmission loss of the barrier at this frequency is 23dB. The attenuation round one end is 27dB and round the other end 35dB. Noise cannot go under the barrier. Calculate the attenuation of sound travelling over the barrier from G to H using the Fresnel method. By adding the sound transmitted via the various routes (through, over, around the barrier) calculate the overall attenuation due to the barrier. Estimate the SPL at H due to a SWL at G of 100dB.

1.5m 0.5m

1m 2.5m

G

H

2.5m

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Lecture 8 Noise control: Partitions and enclosures

A solid partition can be used to block the transmission of sound. Most of the sound will be reflected back towards the source, some will be absorbed, and some will be transmitted through the partition. The partition is likely to have the unwanted side-effect of increasing the sound level at the source. Small gaps can greatly reduce the performance of the partition.

The Sound Transmission Coefficient is defined as

powerwaveincident

powerwavedtransmitte

The Transmission Loss (TL) or Sound Reduction Index (SRI) = 10log10

For a partition, 10dB is poor, 30 or 40dB are good.

For anechoic conditions outside, the SPL close to the outside of the barrier is given by

OUTSPL transmitted wave = incident wave TL

For a reflective partition, the reflected wave amplitude is almost the same as the incident wave amplitude. The pressure amplitude at the inside face is approximately the sum of the incident wave amplitude and the reflected wave amplitude (neglecting any phase shift). Therefore the incident wave is approximately half the pressure amplitude at the inside face.

Incident wave 6 INSPL dB

So TLSPLSPL INOUT 6

However, if reverberant conditions exist outside, the SPL will be higher than this. An equation for this will be given later.

Transmission loss can be measured using two adjoining chambers with an opening between them. The chambers can be anechoic or reverberant. The measured barrier is fitted and tightly sealed in the opening. Sound is generated in one chamber and the sound level measured on both sides of the barrier.

Some measured transmission loss results are shown in figure 8.1.

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Figure 8.1 Transmission loss measurements (from Crocker: Noise and Vibration Control, Wiley)

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8.1 Mass law

Except at very low frequencies, most attenuation from a solid partition is due to its mass, not its stiffness or damping. The transmission loss is given approximately by the mass law:

47log20TL 10 f dB

where is the mass per unit area. Some typical densities are given in table 8.1.

Small holes or gaps may have a large effect.

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8.2 Coincidence

The mass law neglects the effects of stiffness or damping. Stiffness of the material may lessen the transmission loss because bending waves can occur in it.

Sound waves oblique to a plate may set up bending waves in the plate. If the wavelength of the bending waves is similar to the wavelength of the sound waves in the air, a form of resonance is set up, called coincidence.

The wavespeed for bending waves, Bc , increases with frequency. The speed of sound in the air,

c , is independent of frequency. The critical frequency is where the wavelength in the air and

in the plate are equal, which occurs when ccB . Coincidence occurs around and above the

critical frequency where ccB .

A high critical frequency is given by thin, heavy, low stiffness plates. This is perhaps the opposite of what would be expected, as high mass and low stiffness generally gives a low natural frequency.

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The transmission loss is reduced by up to 12dB at the critical frequency. The reduction starts about an octave below the critical frequency and continues for about two octaves above. A very approximate equation for transmission loss above the critical frequency is

57log20TL 10 f dB

For a given material, the product of thickness and critical frequency is a constant. Typical values are given in table 8.1.

Material Density Critical frequency (Hz)

thickness (m) tfC

Concrete ~2300 kg/m3 19 Hz.m

Brick ~1900 kg/m3 22 Hz.m

Glass ~2600 kg/m3 12.7 Hz.m

Plasterboard ~800 kg/m3 40 Hz.m

Plywood ~700 kg/m3 20 Hz.m

Steel 7800 kg/m3 12.4 Hz.m

Table 8.1. Typical densities and critical frequencies

Figure 8.2 shows the coincidence effect in glass, and how its frequency reduces with thickness whilst the transmission loss increases with thickness.

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Figure 8.2 Transmission loss measurements for glass, showing how the TL increases with thickness and the coincidence frequency reduces with thickness (from

http://blog.gjames.com/2013/01/24/reducing-noise-with-the-right-windows/)

Panels with low damping, such as glass or steel, exhibit the strongest coincidence effect. The coincidence effect can be reduced by replacing plain glass with laminated glass (figure 8.3) or adding a damping sheet to a metal panel (figure 8.4). Coincidence is less marked for materials with high damping or a fibrous structure, such as wood (figure 8.1).

http://blog.gjames.com/2013/01/24/reducing-noise-with-the-right-windows/

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Figure 8.3 The effect of lamination on the sound insulation of glass (from http://blog.gjames.com/2013/01/24/reducing-noise-with-the-right-windows/)

Figure 8.4 Transmission loss of steel panel (from http://www.hmmh.com/cmsdocuments/Vibrations-Noise_Locomotive-Ross.pdf)

http://blog.gjames.com/2013/01/24/reducing-noise-with-the-right-windows/

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8.2.1 Example

Estimate the transmission loss for 100mm concrete and 16mm plasterboard, for a range of frequencies. Plot these on figure 8.1.

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8.3 Enclosures with multiple surfaces

If there are areas with different transmission losses, we can calculate the overall sound transmission coefficient by a weighted average of the sound transmission coefficient.

2211 AAA

where A is the total area and A1, A2 etc are the areas of materials 1, 2…

8.3.1 Example

Consider a 50mm thick concrete wall of dimensions 5m x 3m, containing a 4mm thick glass window of area 1m2. Calculate the overall transmission loss at 1000Hz.

A hole can be considered as an area with zero transmission loss 1 .

For example, in the above wall there is a hole of 0.1m2. Calculate the transmission loss now.

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Enclosing the source may cause an increase in the SPL inside the enclosure, as the sound is being trapped inside, especially if the enclosure is made from hard, reflective materials. This reduces the effectiveness of the enclosure. This can be avoided by adding an absorbent layer (foam, fibre) inside the enclosure.

8.4 Conclusions

Solid enclosures, partitions and walls reduce sound transmission by reflecting and absorbing the sound energy. The effectiveness of the wall depends on its mass. A stiff material may be less effective than a very flexible material, as bending waves can occur in the material causing coincidence.

Approximate equations are available for the transmission loss.

8.5 Problems

1. The noise levels impinging on a 120mm thick brick wall are as specified below. Calculate the overall A weighted SPL (a) on the near side of the wall, and (b) on the other side, using the ‘mass law’.

Density of brick 1900 kg/m3

Critical frequency thickness 22 Hz.m

Freq. (Hz) 125 250 500 1000 2000 4000 8000

SPL (dB) 95 99 104 107 111 109 102

2. A partition wall 6m 3m contains a door 2m 0.9m made of the same materials as the wall. There is a 10mm gap around the entire door. The wall also contains a window 2m

1m. Estimate the overall TL of the partition at (a) 500Hz; (b) 2000Hz.

Wall: density = 800 kg/m3; thickness 30mm; critical frequency thickness = 20 Hz.m

Window: density = 2600 kg/m3; thickness 4mm; critical frequency thickness = 13 Hz.m

Gap: sound transmission coefficient = 0.35 at 500Hz, 0.11 at 2000Hz

What difference does the gap make at each frequency?

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Lecture 9 Enclosures and Absorbent Materials (I)

9.1 Absorption

Most materials that form effective sound barriers work by reflecting the sound. They do not absorb the sound energy and can significantly increase the sound level on the source side.

To absorb sound energy a material must contain many small air pockets or capillaries. Air moving in the material will dissipate energy as heat due to viscous friction. Most such materials are light and would be poor as barriers to sound transmission (with a transmission loss of 10dB or so). However they can be used inside a room or chamber for reducing reflections and hence reducing reverberant noise. This can reduce sound levels and improve the acoustic quality or clarity for music and speech.

The sound absorption coefficient is defined as

itonincidentpowerwave

surfacethebyabsorbedpowerwave

This depends on frequency and usually increases with frequency.

For several surfaces, the average absorption coefficient is given by the weighted average according to the equation

332211 AAAA

Some typical absorption coefficients are listed in table 9.1.

Material

Glass 0.04

Plastered solid wall 0.06

Plasterboard 0.1

Concrete floor 0.05

Polyurethane foam ceiling tile 0.6

Rubber floor tile 0.1

Heavy pleated curtains 0.5

Person A 1.1m2

Table 9.1 Typical absorption coefficients (note: these may vary in a wide range, and will vary with frequency)

The Room Constant RC is defined as

1

ARC

where A is the total surface area in the room, and is the average absorption coefficient over those surfaces.

If = 0, the room is fully reverberant.

If = 1, the room is anechoic.

ME30033/50155 – 1. Revised 2014

Neither of these extreme values can be achieved fully.

The sound pressure level in a reverberant room can be expressed approximately by the equation

CRR

Q 4

4log10SWLSPL

210

This represents direct noise

24 R

Q

and reverberant noise

CR

4.

Q = 1 for spherical uniform radiation

Q = 2 for hemispherical uniform radiation

In practice the reverberant noise is usually greater than the direct noise, except very close to the source. The reverberant noise also varies with position in the room; the equation only gives the average value.

9.1.1 Example

An unfurnished office of dimensions 10m10m4m has plastered solid walls, 40m2 of windows, plasterboard ceiling and concrete floor.

(a) Calculate the overall sound absorption coefficient. Estimate the SPL 5m from a source of SWL 100dB.

(b) Estimate the overall sound absorption coefficient and SPL for the same room containing a polyurethane foam ceiling, 10m2 of heavy curtains and 20 people.

ME30033/50155 – 1. Revised 2014

ME30033/50155 – 1. Revised 2014

9.2 Use of absorbent materials with partitions or enclosures

A sound absorbent layer may be used to increase . The sound energy is dissipated as heat inside the material. The best materials are porous; air movement inside the holes dissipates energy. These materials are typically lightweight and would be poor as sound insulators or barriers. They need to be used in conjunction with a heavy insulating barrier. A sound absorbent layer is most effective where the acoustic velocities are high, i.e. near an anti-node. Near a wall the acoustic velocity approaches zero, so they are best placed a small distance from the wall – about a quarter-wavelength. In practice there will be multiple frequencies and multiple wavelengths so the exact positioning is not critical.

An absorbent layer is best placed on the inside of an enclosure or on the source side of a partition. It should have the effect of reducing the reverberant sound level on the source side. As the sound level outside the enclosure depends on the sound level inside, a similar reduction in the outside the enclosure should be achieved.

An absorbent layer on the outside of the enclosure may have little effect as it won’t change the SPL inside, and will have little effect on the transmission loss of the barrier.

Generally an absorbent material is most effective at higher frequencies, and a thicker layer is needed for good low frequency performance. Cones and wedges of absorbent material are extremely effective and are used in anechoic chambers.

ME30033/50155 – 1. Revised 2014

9.3 Problems

1. Estimate the room constant from the measurements in figure 6.2 of the notes, with and without carpet. Hence estimate the SPL if there is a source with a constant SWL of 90 dB. Neglect the effect of frequency.

Dimensions of room: 9.25 m long, 7.65 m wide, 3.15 m high

2. The Stereophonics are due to play at the Millennium Stadium in Cardiff. Calculate the total amplifier power in Watts needed to produce a peak SPL of 108dB at a typical seat, assuming reverberant conditions.

People in audience: 50,000;

area of one person and seat: 2 m2;

absorption of one person and seat A = 1.1m2.

Area of concrete: 60,000m2; absorption coefficient = 0.02

Area of steel: 25,000m2; absorption coefficient = 0.01

The roof is closed.

The turf has been removed and the pitch area can be assumed to be concrete.

Average loudspeaker acoustic efficiency (Sound power out / amplifier power in) = 1%

3. The Millennium Stadium is packed to watch Wales win the Grand Slam . Calculate the SPL produced by 40,000 ecstatic fans singing Cwm Rhondda, assuming reverberant conditions.

Assume an SWL of 95dB per singer.

The roof is open with an opening area of 6,000m2 (assume = 1 for the open roof).

Total number of people: 74,000;

area of one person and seat: 2 m2;

absorption of one person and seat A = 1.1m2.

Area of concrete: 50,000m2; absorption coefficient = 0.02

Area of steel: 20,000m2; absorption coefficient = 0.01

Area of grass: 8,000 m2; absorption coefficient = 0.5

ME30033/50155 – 1. Revised 2014

4. An office 15m 12m 4m has windows of area 14m 2m in one wall and 10m 2m at each end. The ceiling is covered with 25mm thick acoustic tiles, and the floor is tiled. There are 30 chairs and desks.

Absorption coefficients :

Frequency (Hz)

250 500 1000 2000

Ceiling 0.35 0.70 0.75 0.65

Walls 0.05 0.06 0.08 0.04

Windows 0.06 0.04 0.03 0.02

Floor 0.005 0.01 0.02 0.045

Absorption A (m2)

Frequency (Hz)

250 500 1000 2000

Desk & chair 0.05 0.05 0.10 0.15

Person 0.75 1.10 1.30 1.40

Desk & chair area 2 m2, person area 2 m2.

Calculate the average absorption coefficient for the room at each frequency when 20 people are present.

Calculate the reverberation times (see section 6.2).

The office contains a printing machine producing the following sound power level (SWL):

Frequency (Hz)

250 500 1000 2000

SWL (dB) 69 70 68 68

Calculate the reverberant SPL (dB) in the room due to this machine.

ME30033/50155 – 1. Revised 2014

Lecture 10 Enclosures and absorbent materials (II)

10.1 Enclosure or partition in a reverberant room

If a sound source is enclosed by an enclosure (box) or partition, the SWL emitted by the enclosure or partition is given by the equation, where AP is the surface area of the enclosure or partition:

PINOUT ATLSPLSWL 10log106

If the room is reverberant, the SPL is given approximately by the equation

C

PIN

C

OUTOUTR

ATLSPL

RSWLSPL

4log106

4log10 1010

Or OUTCPINOUT RATLSPLSPL _1010 log10log10

where OUTCR _ is the room constant for the room outside of the enclosure or partition,

including the surface of the enclosure/partition.

10.1.1 Example

(a) A compressor produces a SWL of 100dB at 250Hz. It is located in a room 10m 5m 4m high. The average absorption coefficient for the room 1.0 . Estimate the reverberant SPL.

(b) A hardboard enclosure 1m 1.5m 1m high is placed over the compressor. The properties of the hardboard are: TL = 12dB; 15.0 . Estimate the SPL in the room now.

ME30033/50155 – 1. Revised 2014

ME30033/50155 – 1. Revised 2014

10.2 Experiment: Noise control using an enclosure

Outline of experiment

The effect of enclosing a sound source using a small enclosure is to be investigated. The effect of foam lining part of the inside of the enclosure is also to be investigated. Measured and calculated sound levels will be compared and transmission and absorption equations tested.

Apparatus

Loudspeaker connected to a laptop. Band-limited random noise generated in 700 – 1400Hz band (i.e. octave band centred at 1kHz).

Plywood box, 0.4m cube, 12mm thick, with small hole to accept microphone of sound level meter.

Foam, 40mm thick, 0.3 x 0.38m (2 pieces), plus small piece to isolate speaker from floor.

Sound level meter.

Tasks

1. Measure background SPL ..................................................................................................................

2. Measure SPL:

2m from source 0.3m from source (just

outside box, if

present)

0.2m from source

(inside box, if present)

(a) Without box

(b) With box

(c) With box and foam inside

3. How much does the box reduce the SPL at a distance? ....................................................................

4. How much does the box increase the SPL inside the box? ................................................................

5. How much does the foam reduce the SPL inside the box and at a distance? ...................................

6. Estimate SWL from case (b) (with box) using

C

INR

SWLSPL4

log10 10 , assuming 1.0

for wood.

ME30033/50155 – 1. Revised 2014

7. Estimate the transmission loss of the plywood and estimate the SPL outside the box. Compare with the measured SPL.

ME30033/50155 – 1. Revised 2014

10.3 Conclusions

Hard materials reduce transmission by reflecting the sound energy. They have the unwanted effect of increasing the sound level on the source side.

Absorbent materials can be used to absorb noise and reduce reverberation, and improve sound quality. They are not generally very effective at blocking the transmission of sound. They can be used in conjunction with a hard partition or enclosure. The absorbent material should generally be on the source side.

Many equations and data are available for calculating sound levels with barriers, enclosures, partitions and absorbent materials. However they are very approximate because many unknown factors are involved.

10.4 Problems

A pump and motor in a concrete lined plant room 8m 6m 5m high produce the following

reverberant noise levels.

f (Hz) 250 500 1000

SPL (dB) 80 82 80

Absorption coefficients are given below.

f (Hz) 250 500 1000

, concrete 0.03 0.05 0.05

, plywood 0.10 0.15 0.25

, plywood + foam 0.30 0.65 0.75

Estimate the overall sound power level (SWL) in dB.

The pump and motor are then enclosed, on top and sides, by a 2m 3m 2m high enclosure

made of plywood (density = 800 kg/m3; thickness 20mm; critical frequency thickness = 18 Hz.m).

At each frequency calculate the reverberant SPL in the enclosure and in the room. Calculate the overall SPL in dB, and deduce the overall reduction in SPL achieved by the enclosure.

A sound absorbent foam layer is now fitted inside the plywood enclosure. Calculate the SPL in the room at each frequency, the overall SPL in dB, and the reduction in SPL (dB) achieved by using the foam.

ME30033/50155 – 1. Revised 2014

Answers to problems

(fuller solutions will be given online later)

Lecture 1

1. 94 dB, 114 dB

2. 3.2 µW, 32 W 3. (a) 80 dB

(b) 80 dB, (c) 86 dB (a 6dB increase), (d) 90 dB (a 10 dB increase) (e) 100 dB (a 20 dB increase)

Lecture 2

343m/s, 303m/s, 295m/s

Lecture 3

1. 155.5Hz (Eb below middle C), 77.7Hz (Eb one octave lower), 155.5Hz (Eb below middle C).

2. (a) Pmax = 599 Pa at 4.27 m from open end, Umax = 1.46 m/s at 0 m and 8.55 m from open end (b) Pmax = 5410 Pa at 3.42 m from open end, Umax = 13.2 m/s at 0 m and 6.84 m from open end

Lecture 4

1. 171.0, 244.3, 285.0, 298.2, 332.4, 342.0, 375.4, 412.5, 420.3, 445.2Hz

2. 80dBA

3. 73dBA, 71dBA, 72 dBA, 72 dBA

Lecture 5

1. 72dBA

Lecture 6

1.

(a) valid, 60 dB (e.g. sound of level 70 dB with a barrier of attenuation 10 dB)

(b) valid, 65 dBA (e.g. sound of level 60 dBA amplified by a gain of 5 dB)

(c) invalid (double A weighting)

(d) valid, 10 dB (e.g. SPL without barrier minus SPL with barrier gives attenuation of barrier)

(e) valid, 35 dB (two sounds added together)

(f) invalid (adding sounds of mixed weightings)

(g) valid, 81.5 dBA (three sounds added together)

2. 99 dB, 88 dB

ME30033/50155 – 1. Revised 2014

Lecture 7

1. 92 dBA

2. 80 dBA

3. 81 dB without barrier, 18dB attenuation, 63 dB with barrier

Lecture 8

1. 115 dBA, 56 dBA

2. With gap: 28 dB at 500 Hz, 33 dB at 2000 Hz.

Without gap: 33 dB at 500 Hz, 37 dB at 2000 Hz. Difference = 5 dB, 4 dB.

Lecture 9

1. SPL = [83, 86] dB.

2. 128,000 W

3. 95 dB

4. = [0.14, 0.24, 0.27, 0.24]

T = [1.1, 0.54, 0.46, 0.53] s

SPL = 58 dB

Lecture 10

SWL = 90 dB

SPL in enclosure = [84 86 82] dB, total = 89 dB

SPL in room = [63 57 56] dB, total = 65 dB

Reduction obtained using box = 21 dB

With foam, SPL in room = [58 48 48] dB, total = 58 dB

Reduction obtained using foam = 7 dB

ME30033/50155 – 1. Revised 2014

Formulae

Natural frequencies of an open-open or closed-closed tube: L

ncf n

2

Natural frequencies of an open-closed or closed-open tube: L

cnfn

4

)12(

Trigonometric/exponential relations: xjxe jx sincos ,

j

eex

jxjx

2sin

,

2cos

jxjx eex

Sound pressure level: SPL or ref

rms

ref

rmsP

p

p

p

pL 102

2

10 log20log10 , where pref = 20Pa

Sound intensity level: SIL or ref

II

IL 10log10 where Iref = 10-12W/m2.

Sound power level: SWL or ref

WW

WL 10log10 where Wref = 10-12W.

Point source, uniform spherical radiation at distance r: rSWLSPL 10log2011

Point source, uniform hemispherical radiation: rSWLSPL 10log208

Uniform line source: rmperSWLSPL 10log106

Uniform line source on reflective surface: rmperSWLSPL 10log103

Addition of uncorrelated sounds:

1010

10

21

1010log10

SPLSPL

totSPL

A weighting coefficients

Freq. (Hz) 125 250 500 1000 2000 4000 8000

A weighting (dB) -16 -9 -3 0 1 1 -1

Noise exposure limits

First action level: LEX,8H = 80 dBA and peak pressure = 112 Pa

Second action level: LEX,8H = 85 dBA and peak pressure = 200 Pa

Exposure limit value: LEX,8H = 87 dBA and peak pressure = 200 Pa

Equivalent level of a period of time t: T

tLL PEQ 10log10 where T = 8 hours

ME30033/50155 – 1. Revised 2014

Attenuation of a barrier

LN

2 where L = increase in sound path caused by barrier

3.0log1012dB 10 NAtt for a point source

3.0log108dB 10 NAtt for a line source.

AttSPLSPL BARRIERWITHOUTBARRIERWITH (or –TL)

Mass law: 47log20 10 fTL dB below coincidence

57log20 10 fTL dB above coincidence

Sound transmission coefficient:

1010

TL

or log10TL

Average sound transmission coefficient or absorption coefficient of surface with multiple materials:

332211 AAAA ,

332211 AAAA

Room constant:

1

ARC

Reverberation time for decay by 60dB: CR

VT

16.0

SPL in a reverberant room:

CRR

QSWLSPL

4

4log10

210

or, if directly radiated sound is neglected,

CRSWLSPL

4log10 10

Sound power emitted by a partition or enclosure: PINOUT ATLSPLSWL 10log106

Sound pressure outside a partition or enclosure:

Anechoic conditions outside: TLSPLSPL INOUT 6

(SPLOUT is defined near the outside surface)

Reverberant conditions outside:

OUTCPINOUT RATLSPLSPL _1010 log10log10

noise handout

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