basics of sound and noise - college of engineeringdwherr01/bandung/01_introduction.pdf · msound...

Basics of Sound and Noise David Herrin, Ph.D., P.E. University of Kentucky

Department of Mechanical Engineering

Introductions

Noise and Vibration Short Course

Dept. of Mech. Engineering University of Kentucky

2

University of Kentucky

Ø  Public University Ø  16 Colleges Ø  93 Undergraduate Programs Ø  99 M.S. Programs Ø  66 Ph.D. Programs Ø  28,000 Students

Introductions


Faculty Ø  D. W. Herrin, Ph.D., P.E. Ø  T. W. Wu, Ph.D. Students Ø  Yitian Zhang (Ph.D.) Ø  Gong Cheng (Ph.D.) Ø  Kangping Ruan (Ph.D.) Ø  Peng Wang (Ph.D.) Ø  Keyu Chen (Ph.D.) Ø  Weiyun Liu (Ph.D.) Ø  Wanlu Li (M.S.) Ø  Shishuo Sun (M.S.) Ø  Huangxing Chen (M.S.)

Vibro-Acoustics Group

3 Dept. of Mech. Engineering University of Kentucky

Small Engines / Generator Sets Ø  BASCO Ø  Kohler Corp. Sound Absorbing Materials Ø  3M Company Ø  American Acoustical Products Ø  Blachford Inc. Ø  Commercial Vehicle Group Ø  Federal Foam Technologies Ø  Insul-Coustic Corp. Ø  Technicon Acoustics Inc. Others Ø  Bechtel Marine Propulsion Corporation Ø  Ebco Inc. Ø  General Electric Appliances Ø  Lexmark International

HVAC and Refrigeration Industry Ø  Emerson Climate Ø  Ingersoll Rand Trane Ø  JCI York Ø  Transicold Carrier Heavy Equipment Industry Diesel Engines Ø  Caterpillar Inc. Ø  Cummins Inc. Ø  Deere and Company Ø  Southwest Research Institute Ø  Universal Silencers Automotive Supplier / Motorcycle Ø  Dana Corp. Ø  Eaton Corp. Ø  Harley-Davidson Motor Co. Ø  Mann+Hummel Group

Vibro-Acoustics Consortium

Introductions



5

Structural Dynamics Lab

Introductions



6

Hemi-Anechoic Chamber

Introductions



7

Acoustic Materials Characterization

ABSORPTION COEFFICIENT FOR SINGLE AND DOUBLE LAYER ABSORBING MATERIALS

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 200 400 600 800 1000 1200 1400 1600 1800 2000FREQUENCY (Hz)

ALP

HA

Double layerSingle layer

Introductions


8

Sound and Vibration Fields

Sound and vibration waves are mechanical elastic waves, and thus the conditions for their existence are that the medium possess mass and elasticity (i.e., stiffness). If a mass particle is displaced from its equilibrium position, the elastic forces will seek to return it to its original position. The particle influences the surrounding particles and in this way, a disturbance (i.e., wave) propagates through the medium.

Wallin et al. Longitudinal Transverse


Wallin, Carlsson, Åbom, Bodén, and Glav, Sound and Vibration, 2nd Edition, Marcus Wallenberg Laboratoriet, Stockholm, 2011.

Introductions



9

Wave Motion – Some Basics

  Sound waves are pressure disturbances in fluids, such as air, caused by vibration, turbulence, explosions, etc.

  These disturbances propagate at the speed of sound c (c = 343 m/s or 1125 ft/s in air at room temperature)

  The wavelength λ = c/f. For f = 1 kHz, the wavelength is approximately 0.34 m or 1.13 ft.

  As a sound wave passes a point, the fluid particles are displaced but return to their original position until the next wave passes.

Wave animation

Introductions


10

Sound and Vibration Fields

An acoustic field implies a small disturbance. Sound pressure disturbances are only on the order of 1 Pa for 94 dB.


Wallin, Carlsson, Åbom, Bodén, and Glav, Sound and Vibration, 2nd Edition, Marcus Wallenberg Laboratoriet, Stockholm, 2011.

Introductions



11

Particle Motion

  Particles oscillate (but no net flow)   Waves move much faster than particles   Surface displacement determines particle displacement and resulting sound pressure, as well as frequency

Particle displacement amplitude D

ftDtd π2sin)( =

Introductions



12

Particle Velocity

Particle displacement amplitude D

( ) ftfDtu ππ 2cos2)( =

Particle velocity amplitude (m/s)

  u increases with frequency for a constant displacement

  Particle velocity is like current, sound pressure like voltage

Introductions



13

Field Quantities

un = velocity of surface in normal direction – must be known

p, u = sound pressure and particle velocity in the field.

How do we determine

these?

Numerical Acoustics

Introductions



14

Sound Intensity and Power pu ,

  Sound intensity is the sound power radiated per unit area

  To get sound power, we integrate the normal component of the sound intensity over a closed surface

puI =

dSIWS n∫=

I

(watts)

Introductions



15

An Analogy

Like temperature, the sound pressure depends on the source power level AND the environment in which the source is placed.

Introductions



16

Another Analogy

A light bulb produces the same optical power (in watts) regardless of its environment – big or small room – but the intensity of light depends on the environment (reflectance of the walls) and the distance from the light bulb.

A sound source produces the same sound power (in watts) regardless of its environment* – big or small room – but the intensity of sound and the sound pressure depend on the environment (reflectance of the walls) and the distance from the source.

_____ * There are some notable exceptions to this (exhaust noise, close fitting enclosures)

Introductions



17

Plane waves in a duct

λ Oscillating Piston un

Special Cases 1. Plane Waves with no reflection

( ) uzucpuu

oo

n

==

+=

ρ

shift phase a zo = characteristic impedance

ISWcp

cpppuI

oo

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛==

ρρ

2

Introductions



18

2. In the far field* of a source in a free field

p, u

Special Cases

* The far field is where the SPL decreases by 6 dB for a doubling of the distance to the source

cp

cppI

cup

oo

o

ρρ

ρ2

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

(like plane waves in a duct except the sound pressure decreases with distance)

Introductions


Addition of Sound Sources

Combining Acoustic Sources

ptot t( ) = pn t( )n=1

N

∑

19

For Two Sources

ptot2 =

1T

ptot2 t( )dt

0

T

∫ =1T

p1 t( )+ p2 t( )( )2dt

0

T

∫ =

= p12 + p2

2 +2T

p1 t( ) p2 t( )dt0

T

∫


Introductions


Uncorrelated Sources

20

ptot2 = p1

2 + p22 ++ pn

2

Addition by Sound Pressure Level

p2 = pref2 10Lp /10

ptot2 = p1

2 + p22 = pref

2 10Lp1 /10 +10Lp2 /10( )Lptot

=10 log 10Lp1 /10 +10Lp2 /10( )In general

Lptot=10 log 10Lpn /10

n=1

N

∑


Introductions


Example

21

A sound source causes a sound pressure level of Lp1 at a certain point. What increase in SPL is provided by a second source, equal in strength to, but uncorrelated to, the first?

Lptot=10 log 10Lp1/10 +10Lp1/10( ) =10 log 2 10Lp1 /10( )= Lp1 +10 log 2( ) = Lp1 +3 dB


Introductions


Example

22

A machine causes a SPL of 90 dB at a certain point while the background noise is 83 dB. What is the SPL of the machine with the background noise removed.

ptot2 = pM

2 + pB2

pM2 = ptot

2 − pB2

LpM=10 log 10Ltot /10 −10LpB /10( ) = 89 dB


Introductions


Adding Frequency Components

23

ptot2 = pn

2

n=1

N

∑

Lptot=10 log 10Lpn /10

n=1

N

∑

The 1000 Hz octave band includes the 800, 1000, and 1250 Hz third-octave bands. Determine the octave band level if the third-octave band levels are 79, 86 and 84 dB, respectively.

Lptot=10 log 107.9 +108.6 +108.4( )

n=1

N

∑


Introductions



24

Sound Power Level: watts101 log10)dB( 1210

−×== refref

w WWWL

Sound Pressure Level:

Those Amazing dB’s

The main thing to remember is that 100 dB sound pressure level and 100 dB sound power level are completely different!

To avoid confusion, use the reference values:

100 dB (re 20 µPa) sound pressure level 100 dB (re 1x10-12 W) sound power level

Pa 20 log10)dB( ref

2

ref

rms10 µ=⎟⎟

⎠

⎞⎜⎜⎝

⎛= p

ppLp

Introductions



25

r

r

But they are related…

I

(no reflections)

SLL Wp 10log10−= S in m2

24 rS π= (Spherical source) 22 rS π= (Hemi-spherical source)

S = cross-sectional area (duct)

Introductions



26

A source has a sound power level of 90 dB (re 10-12 W). What is the sound pressure level at a distance of 10 m in (a) a free field, (b) in a hemispherical free field, and (c) in a duct of cross-sectional area 1 m2?

An Example

( ) Pa)20(redB59104log90 210 µπ =−=pL

The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.

a.

b.

c.

Introductions



27

Two Vacuums

Shopvac Bosch

Introductions



28

Bosch Vacuum

Exhaust flows through foam

Introductions



29

Shopvac Vacuum

Introductions



30

Sound Power

(Watts) SIW Σ=)(W/mIntensity Sound Avg 2=I

)(m Area Surface 2=S

Exhaust Deflected Upward

Introductions



31

Time and Frequency Domains

Introductions



32

Sound Power Comparison – Bosch, All Sides

10

20

30

40

50

60

70

80

0 1000 2000 3000 4000 5000

Frequency (Hz)

Sou

nd P

ower

(dB

)

Front LeftBack RightTop Total

Introductions



33

Sound Power Comparison – Narrow Band

10

20

30

40

50

60

70

80

0 1000 2000 3000 4000 5000Frequency (Hz)

A-W

eigh

ted

Soun

d Po

wer

(dB)

Bosch w/ Foam (77.6 dBA)

Bosch w/o Foam (81.2 dBA)

Shopvac (85.0 dBA)

Introductions



34

Sound Power Comparison at Low Frequency

Low Frequency Tone

10

20

30

40

50

60

70

80

100 200 300 400 500

Frequency (Hz)

A-W

eigh

ted

Sou

nd P

ower

(dB

)

Bosch w/ Foam

Bosch w/o Foam

Shopvac

Introductions



35

Sound Power Comparison – 1/3 Octave

40

45

50

55

60

65

70

75

80

100 1000 10000

Frequency (Hz)

A-W

eigh

ted

Soun

d Po

wer

(dB)

Bosch w/ Foam (77.6 dBA)

Bosch w/o Foam (81.2 dBA)

Shopvac (85.0 dBA)

Introductions



36

Sound Quality

Bosch (original) Bosch (w/o foam)

Shopvac Bosch (w/o 1st peak)

Introductions



37

Sound Quality: Jury Test

Note: Rate each vacuum on a scale from 1 to 10 where 1 is “very quiet” and 10 is “very loud.”

5.74

7.70 8.00

4.61

0123456789

10

BOSCH Foam BOSCH NoFoam

Other Vacuum BOSCH No 1stPeak

Ave

rage

Rat

ing

Introductions



38

Foam Inside Bosch Vacuum

Introductions



39

Sound Absorption Coefficient

20 1

incidentenergy soundabsorbedenergy sound R−==α

Introductions



40

Sound Absorption Coefficient of Foam

0

0.2

0.4

0.6

0.8

1

0 1000 2000 3000 4000Frequency (Hz)

Abso

rptio

n C

oeffi

cien

t

Introductions



41

Sound Intensity (Shopvac)

Introductions



42

Sound Intensity (Bosch)

Introductions



43

Sound Intensity (Bosch)

Introductions



44

Summary

q  Absorption of foam in BOSCH significantly reduces sound power

q  Sound exhaust is better directed on BOSCH

q  Recommend altering design to reduce/shift first peak

basics of sound and noise - college of engineeringdwherr01/bandung/01_introduction.pdf · msound...

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