Solutions to ExampleProblems in Engineering Noise Control, 2 nd Edn. A companion to "Engineering Noise Control", 3 rd Edn Colin H. Hansen Department of Mechanical Engineering University of Adelaide South Australia 5005 AUSTRALIA FAX: +61-8-8303-4367 e-mail:
[email protected] Published by Colin H Hansen, Adelaide, South Australia Apart from fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Australian copyright Act, 1968, this publication may not be reproduced, stored, or transmitted, in any form, or by any means, without the prior permission in writing of the author. The author makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. A catalogue record for this book is available from the Australian National Library. © 2003 Colin H Hansen Published by Colin H Hansen ISBN 0-9751704-0-6 This version updated August 4, 2006 Contents All page, equation and table references are to the third edition of the textbook, Engineering Noise Control by DA Bies and CH Hansen 1. Solutions to problems in fundamentals 1 2. Solutions to problems relating to the human ear 65 3. Solutions to problems relating to noise measurement and instrumentation 71 4. Solutions to problems relating to criteria 83 5. Solutions to problems related to sound sources and outdoor sound propagation 99 6. Solutions to problems related to sound power, its use and measurement 130 7. Solutions to problems related to sound in enclosed spaces 143 8. Solutions to problems related to sound transmission loss, acoustic enclosures and barriers 199 9. Solutions to problems related to muffling devices 259 10. Solutions to problems in vibration isolation 298 11. Solutions to problems in active noise control 312 12. Errata in Third Edition of Engineering Noise Control 314 Preface This book provides detailed and instructive solutions to the book of problems on acoustics and noise control which is intended as a companion to the 3 rd edition of the book "Engineering Noise Control" by David A. Bies and Colin H. Hansen. The problems and solutions cover chapters 1 to 10 and 12 in that text. Some of the problems and solutions are formulated to illustrate the physics underlying the acoustical concepts and others are based on actual practical problems. Many of the solutions extend the discussion in the text and illustrate the more difficult concepts by example, thus acting as a valuable source and understanding for the consultant and student alike. Although most of the problems and solutions have been tested on students, it is highly likely that there exist errors of which I am unaware. I would dearly like to hear from any readers who may discover any errors, no matter how minor. C.H.H., September, 2003 e-mail:
[email protected] Acknowledgments The author would like to thank the many people (too numerous to mention) who were responsible for providing ideas for problems. The author would also like to thank his undergraduate and graduate students who provided an excellent opportunity for fine tuning the solutions to the problems in this book. Finally, the author would like to express his deep appreciation to his family, particularly his wife Susan and daughters Kristy and Laura for their patience and support during the many nights and weekends needed to assemble the problems and solutions in a form suitable for publication. This book is dedicated to Susan, to Kristy and to Laura. c air c He ' M He γ He × γ air M air 1/2 ' 4 5/3 1/2 1.4 29 1/2 ' 0.34 1 Solutions to problems in Fundamentals Unless otherwise stated an air temperature of 20EC corresponding to an air density of 1.206kg/m 3 and a speed of sound of 343m/s has been assumed. Problem 1.1 • undertake an assessment of the current environment where there appears to be a problem, including the preparation of noise level contours where required; • establish the noise control objectives or criteria to be met; • identify noise transmission paths and generation mechanisms; • rank order noise sources contributing to any excessive levels; • formulate a noise control program and implementation schedule; • carry out the program; and • verify the achievement of the objectives of the program. See pages 8-10 in the third edition of the text for more details. Problem 1.2 (a) Speed of sound given by equation 1.4 in text. The only variables which are different for the two gases are γ and M. Thus: (b) The wavelength of sound emitted from ones mouth is a function of the vocal cord properties which remain unchanged by the presence of helium. As fλ = c and λ is fixed and c is faster in helium, the sound emanating from your mouth will be higher in pitch. Solutions to problems 2 h ' 4240 101.4×10 3 × 95 100 ' 0.0397 c dry ' γRT/ M ' 1.4×8.314×303.2/ 0.029 ' 348.8m/s c wet ' 348.8(1 % 0.0397×0.16) ' 351.0m/s 0 PV ' 0 m M RT 0 V ' 250,000 24 × 3,600 ' 2.894m 3 /s 0 m ' P 0 VM RT ' 101,400 × 0.029 × 2.8935 8.314 × 288.2 ' 3.550kg/s Problem 1.3 Using the relation for h given in the question and knowing that atmospheric pressure is 101.4kPa, we can write: The speed of sound in dry air at 30EC is given by: The speed of sound in wet air is then: Problem 1.4 Gas volume flow rate = 250,000 m 3 per day at STP Gas Law also applies to a moving fluid, so: P and T are the static pressure and temperature respectively and (a) The mass flow rate is given by: (b) Density of gas in pipe is: ρ ' 0 m 0 V ' PM RT ' 8 × 10 6 × 0.029 8.314 × 393.2 ' 71.0kg/m 3 Fundamentals 3 U ' 4 πd 2 0 V ' 4 πd 2 0 m ρ ' 4 π × 0.01 × 3.551 71.0 ' 6.37m/s c ' γP ρ 1/2 ' 1.4 × 8 × 10 6 / 71.0 ' 397m/s PV ' m M RT or R M ' PV mT c ' 1.35 × 101400 × 1273 1.4 × 273 1/2 ' 675 m/ s dP dV ' m M R V dT dV & m M RT V 2 or dp dV ' P T dT dV & P V (c) Gas flow speed in discharge pipe is: Speed of sound (relative to fluid) is: (d) Speed of sound relative to the pipe is thus: 397.2 + 6.4 = 404 m/s. Problem 1.5 The speed of sound is given by: and for any gas, (R/M) is fixed. c ' γRT/ M We can find (R/M) by using the properties at 0EC and the expression: As (m/V) = 1.4 kg/m 3 , and thus: R M ' 101400 1.4 × 273 Problem 1.6 The Universal gas law may be written as . PV ' m M RT or P ' m M RT V Differentiating gives: which may be rewritten as: Solutions to problems 4 dP P % dV V ' dT T dV V ' & 1 γ dP P dP P 1 & 1 γ ' dT T I ' ¯ p 2 2ρc ' I ref 10 L I / 10 and ¯ p ' pressure amplitude ¯ p P 1 & 1 γ ' ¯ τ T ¯ τ ' 1.617(1 & 1/ 1.4) 101400 × 298 ' 1.4 × 10 &3 EC ρc ' γP γRT/ M ' 1.4 × 101400 1.4 × 8.314 × 313.2/ 0.029 ' 400.4 Thus, ¯ p ' [ 2 × 1.205 × 343 × 10 &12 × 10 9.5 ] 1/2 ' 1.62 Pa Another gas property associated with the adiabatic expansion and contraction of a sound wave is , which leads to: PV γ ' const Combining the above two equations gives: The acoustic pressure associated with a sound wave of intensity 95dB is calculated as follows: Thus, We can write the results of the preceding analysis as: Substituting in given values and rearranging the equation to give τ, we obtain: which is the amplitude of the temperature fluctuations. Problem 1.7 Using equations 1.4a and b in the text, it can be shown that: Fundamentals 5 λ ' c/ f ' 789/ 40 ' 19.7 m f ' 250 ' c/ 2L ' c/ 2.4. Thus, c ' 600 m/ s c ' γRT M ' 1.4 × 8.314 × T 0.035 ρ ' γP c 2 ' 1.4 × 101.4 × 10 3 600 2 ' 0.39 kg/ m 3 c ' γRT M ' 1.4 × 8.314 × 1873 0.035 ' 789 m/ s ρ ' γP c 2 ' 1.4 × 101.4 × 10 3 789 2 ' 0.23 kg/ m 3 T ' 600 2 × 0.035 1.4 × 8.314 ' 1083EK ' 810EC Problem 1.8 f = c/4L = 343/(4×4) = 21.4Hz Problem 1.9 (a) Assume that the tail pipe is effectively open at each end as it follows a muffler. Thus: (b) Assumption is that the gas in the pipe is at atmospheric pressure at sea level. Problem 1.10 (a) (b) (c) Solutions to problems 6 D ' &V MV Mp &1 γ dV g V g % dP P ' 0 L w ' L p & 10log 10 4(1 & ¯ α ) S¯ α &10log 10 ρc 400 ' 120 & 10log 10 4 × 0.98 149 × 0.02 & 10log 10 789 × 0.23 400 ' 122.3 dB W ' 10 &12 × 10 122.3/ 10 ' 1.7 W If we treat the furnace like a closed end tube, , so f ' Nc/ 2L L ' c/ 2f ' 789/80 ' 9.9 m So we can expect one dimension of the furnace to be 9.9 m. (d) Surface area, S = πdL + 2πr 2 = π × 4 × 9.86 + 2π × 4 = 149 m 2 (e) Second burner has twice the sound power level, so increase in sound pressure level will be 10 log 10 (3) or an increase of 4.8 dB. So the new SPL = 124.8 dB. Problem 1.11 (a) The bulk modulus of the fluid is given by equation 1.2 in the text as: Consider a unit volume of fluid (V = 1), let the proportion (and volume V g ) of gas be x and the proportion (and volume V L ) of liquid is then (1 - x). Assume a pressure increment of Δp, and a corresponding volume increment ΔV. From the above equation, we can write, ΔV L = -Δp(1 - x)/D Assuming adiabatic compression of gas, . Differentiating P with respect to V PV γ ' const or P ' const × V &γ and rearranging gives: Fundamentals 7 ΔV g ' & xΔP γP ΔV ' & 1 & x D & x γP ΔP D eff ' & ΔV ΔP &1 ' & 1 & x D & x γP &1 ρ eff ' (1 & x) ρ L % xρ g c ' 1 1 & x D % x γP (1 & x) ρ L % xρ g Substituting x for V g and rearranging gives: Thus, the total volume change for a change Δp in pressure = ΔV L + ΔV g = ΔV. Thus: Thus the effective bulk modulus of the fluid is given by: The effective density of the fluid is equal to the total mass for unit volume of fluid and is given by: The speed of sound is given by . Substituting in the effective c ' D/ ρ values calculated for D and ρ above we obtain: where x is the proportion of gas in the fluid. (b) As x approaches 0, using the result of (a) above gives: = which is the speed of sound in the liquid. c ' 1 (1/ D)ρ L D/ ρ L As x approaches 1, using the result of (a) above gives: which is the speed of sound in the gas. c ' 1 (1/ γP)ρ g ' γP ρ g Solutions to problems 8 φ ' A r e j(ωt & kr) p u ' jωρ 1/r % jk p u ' ρc jkr(1 & jkr) 1 % k 2 r 2 ' ρckr(j % kr) 1 % k 2 r 2 ' ρckr 1 % k 2 r 2 × j % kr 1 % k 2 r 2 p u ' ρccosβ(jsinβ % cosβ) ' ρccosβe jβ u ' p ρccosβ e &jβ Problem 1.12 Equation 1.40c in the text is the harmonic solution to the spherical wave equation. That is: For spherical waves, equation 1.6 may be written as and thus the u ' & Mφ Mr particle velocity may be written as in equation 1.42. Using equations 1.7 and 1.40, the acoustic pressure may be written as in equation 1.41b. Using equations 1.41b and 1.42, we may write: Using ω = kc and multiplying the numerator and denominator of the above equation by r gives equation 1.43 in the text. Multiplying the numerator and denominator of equation 1.43 by (1 - jkr) gives: Defining the phase β = tan -1 (1/kr) as the phase by which the pressure leads the particle velocity, the preceding equation may be written as: which is the same as equation 1.72. Harmonic intensity is defined as I = 0.5Re{ } where the bar denotes the ¯ p ¯ u ( complex amplitude. Using equation 1.72, we may write: Thus, , which is the plane wave 0.5 × Re{¯ p ¯ u} ' 0.5 ¯ p 2 cosβ ρccosβ ' ¢p 2 ¦ ρc expression. Fundamentals 9 u ' & Mφ Mr ' Ak ρrω e j(ωt & kr) % A jρr 2 ω e j(ωt & kr) U ' A ρr 0 ω e j(ωt & kr 0 ) k % 1 jr 0 W ' 2πω 2 r 4 0 ρU 2 0 / c ' 2π(2π × 100) 2 × 0.05 4 × 1.205 × 2 2 / 344 ' 0.22W Problem 1.13 (a) At any location, . The velocity potential p ' A r e jω(t & r/ c) ' A r e j(ωt & kr) is then and the particle velocity is φ ' 1 ρ m pdt ' A jρrω e j(ωt & kr) (b) When r = r 0 , u = U and the particle velocity may be written as: But U = U 0 e jωt , so . U 0 ' A ρr 0 ω k % 1 jr 0 e &jkr 0 Thus, A ' ρr 0 ωU 0 k % 1 jr 0 e jkr 0 However, e jkr 0 ' coskr 0 % jsinkr 0 . 1 % jkr 0 for small kr 0 Thus, A ' ρr 0 ωU 0 k % 1 jr 0 1 % jkr 0 ' jρr 2 0 ωU 0 (c) ω = 200π, ρ = 1.206, U 0 = 2, r 0 = 0.05 Power, W = IS = (p 2 /ρc)4πr 2 = and *p* 2 2ρc 4πr 2 ' *A* 2 2ρcr 2 4πr 2 . Thus: *A* 2 ' ω 2 r 4 0 ρ 2 U 2 0 Solutions to problems 10 p ' A r e j(ωt & kr) φ ' A jωρr e j(ωt & kr) u ' & Mφ Mr ' jkA jrωρ e j(ωt & kr) % A jr 2 ωρ e j(ωt & kr) ' A rρc e j(ωt & kr) 1 & j kr ' p ρc 1 & j kr *u* ' *p* ρc *1 & j kr * Problem 1.14 Spherical wave solution to the wave equation is: Using equation 1.7 in the text, the velocity potential is: c Using the one dimensional form of equation 1.6 in the text, the particle velocity is: Thus: Amplitude is twice when or . *p*/ ρc *1 & j / kr* ' 2 1 % (kr) &2 ' 4 Thus kr ' 1/ 3 ' 0.58 Problem 1.15 (a) Plane wave: p = ρcu = 1.206 × 343 × 0.2 L p ' 20log 10 1.206 × 343 × 0.2 2 × 10 &5 ' 132dB (b) L p ' 20log 10 988 × 1486 × 0.2 10 &6 ' 229dB Non-linear effects are definitely important as the level in air is above 130dB Fundamentals 11 L p ' 20log 10 p rms 10 &6 ' 20log 10 106.07 × 10 3 10 &6 ' 220.5dB *u* ' *p* ρc *1 & j kr * ' 150 × 10 3 988 × 1481 1 % 1481 2π × 1000 × 1 2 ' 105mm/s and the level in water is also very high. Cavitation in water would occur if the sound pressure amplitude exceeds the mean pressure which would happen if the water were less than a certain depth. The depth is calculated by calculating the weight of a column of water, 1 m 2 in cross section. Weight = 988 × 9.81 = 9692.3 N, so the water pressure is 9692.3 Pa per meter depth. The acoustic pressure amplitude is 1.41 × 988 × 1486 × 0.2 = 0.414 MPa. So the depth of water this corresponds to (allowing for atmospheric pressure) is: h = (0.414 × 10 6 - 101400)/9692.3 = 32.2m deep. Problem 1.16 (a) The instantaneous total pressure will become negative if the acoustic pressure amplitude exceeds the mean pressure. This corresponds to an r.m.s. acoustic pressure of 150//2 = 106.07 kPa. The sound pressure level is then: (b) The particle velocity is given by . Thus the amplitude of the u ' p/ ρc particle velocity is u = 150 × 10 3 /(988 × 1481) = 103 mm/s (c) Using the analysis of problem 1.14, we can show that for a spherical wave, the particle velocity amplitude is: Solutions to problems 12 p ' A r e j(ωt & kr) φ ' A jωρr e j(ωt & kr) u ' & Mφ Mr ' jkA jrωρ e j(ωt & kr) % A jr 2 ωρ e j(ωt & kr) ' A rρc e j(ωt & kr) 1 & j kr ' p ρc 1 & j kr *u* ' *p* ρc *1 & j kr * *p* ' 2 × p ref × 10 L p / 20 ' 2 × 2 × 10 &5 × 10 110/ 20 ' 8.94 Pa *u* ' 8.94 413.6 1 % 0.546 2 ' 25 mm/ s Problem 1.17 (a) The sound pressure level at 10m will be 20log 10 (10/1) less than at 1m, which translates to 90dB. (b) Spherical wave solution to the wave equation is: Using equation 1.7 in the text, the velocity potential is: Using the one dimensional form of equation 1.6 in the text, the particle velocity is: Thus: ρc = 1.206 × 343 = 413.7; k = 2πf/c = 2π × 100/343 = 1.832 At 1m, 1/kr = 0.546; at 10m, 1/kr = 0.0546. The acoustic pressure amplitude at 1m is: Thus at 1m, the particle velocity amplitude is: The acoustic pressure amplitude at 10m is: Fundamentals 13 *p* ' 2 × p ref × 10 L p / 20 ' 2 × 2 × 10 &5 × 10 90/ 20 ' 0.894 Pa *u* ' 0.894 413.6 1 % 0.0546 2 ' 2.2 mm/s p ' A r e j(ωt & kr) φ ' A jωρr e j(ωt & kr) u ' & Mφ Mr ' jkA jrωρ e j(ωt & kr) % A jr 2 ωρ e j(ωt & kr) ' A rρc e j(ωt & kr) 1 & j kr ' p ρc 1 & j kr Z ' ρc 1 & j kr &1 *Z* ' ρckr 1 % k 2 r 2 1 % k 2 r 2 ' ρckr 1 % k 2 r 2 ' ρc/ 2 4k 2 r 2 ' 1 % k 2 r 2 or k 2 r 2 ' 0.3333 And at 10m, the particle velocity amplitude is: Problem 1.18 (a) Spherical wave solution to the wave equation is: Using equation 1.7 in the text, the velocity potential is: Using the one dimensional form of equation 1.6 in the text, the particle velocity is: (b) Specific acoustic impedance, Z = p/u. Thus: (c) The modulus of the impedance of the spherical wave is half that of a plane wave (ρc) when Thus: Solutions to problems 14 Thus, r ' (λ/ 2π) 0.3333 ' 0.092λ Problem 1.19 (a) r.m.s. sound pressure p rms ' Iρc ' Wρc 4πr 2 ' 1 × 1.206 × 343 4π × 0.3 2 ' 19.12Pa (b) SPL = 20log 10 19.1 2 × 10 &5 ' 119.6dB (c) r.m.s. particle velocity (see equation 1.43 in text): *u r * ' *p r * 1 % k 2 r 2 krρc ' 19.025 1 % k 2 r 2 1/2 413.6kr kr ' 2π × 1000 343 × 0.3 ' 5.50 Thus, u ' 19.125[1 % 5.50 2 ] 1/2 413.7 × 5.50 ' 0.047m/s (d) Phase between pressure and particle velocity given by equation 1.73 as: , and the acoustic pressure β ' tan &1 [1/(kr)] ' tan &1 (1/5.50) ' 10.3E leads the particle velocity. (e) As shown on p35 in the text, spherical wave intensity is . p 2 rms / (ρc) Substituting in the value for p rms calculated in (a) above gives: I ' 19.125 2 / (1.205 × 343) ' 0.885 W/m 2 (f) From equation 1.74 in the text, the amplitude of the reactive intensity is: I r ' p 2 rms ρckr ' 19.125 2 1.205 × 343 × 5.50 ' 0.161 W/ m 2 Fundamentals 15 L oct ' 10log 10 10 7.8 % 10 7.3 % 10 8 ' 80.8dB p 2 rms ' 1.206 × 343 4π 10 100 % 20 200 % 15 100 ' 11.52Pa 2 tanθ ' 0.1A(1 % 1/ 2) 0.1A(1.5 % 1/ 2 ' 0.7734 A B C D 10 m 10 m (g) The sound intensity level is: L I ' 10log 10 0.885 % 120 ' 119.5dB (h) 1500Hz is a different frequency to 1000Hz so the mean square pressures add. Thus, . p rms ' 2 × 19.125 ' 27.0Pa Problem 1.20 Problem 1.21 (a) For a spherical source: . p 2 rms ' ρcI ' ρcW/ S ' ρcW/ 4πr 2 As the sources are uncorrelated we may add p 2 for each. For source A, r = 10. For source B, r = 10/2. For source C, r = 10. Thus the total p 2 at location D is: Thus, the sound pressure level is: L p = 10log 10 11.52 (2 × 10 &5 ) 2 ' 104.6 dB (b) Intensity, I % p 2 . Thus, I % W/r 2 = AW/r 2 . The resultant intensity can thus be calculated using the figure on the next page. Solutions to problems 16 45 o 45 o 45 o 0.1A 0.1A 0.1A 0.1A 0.15A 2 2 u u Thus the direction of the intensity vector is 37.7E below the horizontal. (c) Sound intensity is a measure of the net flow of energy in an acoustic disturbance. For a single frequency field the sound intensity is the product of the acoustic pressure with the in-phase component of the particle velocity. For a broadband field it is the time average product of the acoustic pressure and particle velocity. Thus its measurement requires a knowledge of the acoustic pressure and particle velocity. Equations 1.6 and 1.7 in the text indicate that the particle velocity is proportional to the pressure gradient which can be approximated by subtracting the measurements from two microphones and dividing the result by the separation distance. In the far field of the source, the pressure and particle velocity are in phase and related by p = ρcu. Thus, in the far field, only one microphone is necessary as the pressure gradient need not be calculated. (d) For the determination of sound power, sound intensity measurements can still give accurate results in the presence of reflecting surfaces or other nearby noisy equipment, whereas results obtained using sound pressure measurements are likely to be seriously in error. Problem 1.22 (a) Specific acoustic impedance is the ratio of acoustic pressure to particle velocity at any location in the medium containing the acoustic disturbance. It is a function of the type of disturbance as well as of the acoustic medium. (b) Characteristic impedance is a material or acoustic medium property, is equal to ρc and is the specific acoustic impedance of a plane wave in an infinitely extending medium. Fundamentals 17 10log 10 10 98/ 10 & 10 95/ 10 ' 95.0dB 10log 10 10 102/ 10 & 10 98/ 10 ' 99.8dB 10log 10 10 96/ 10 & 10 94/ 10 ' 91.7dB (c) Interference describes the interaction between two or more sound waves of the same frequency such that regions of reinforcement (increased sound pressure) and regions of cancellation (reduced sound pressure) are formed. (d) Phase speed is the speed at which a single frequency sound wave propagates and is proportional to the rate of change of phase experienced by a stationary observer as the sound wave propagates past. (e) Sound power is a measure of the total rate of energy emission by an acoustic source and has the units of watts. (f) Particle velocity describe the oscillatory motion of particles in an acoustic medium during propagation of an acoustic disturbance. Problem 1.23 A flat spectrum level implies equal energy in each 1Hz wide frequency band. As the bandwidth of an octave band doubles from one band to the next, the energy level will increase by 3dB each time the octave band centre frequency is doubled. Similarly, the level will increase by 1dB for one third octave bands, each time the band centre frequency is stepped up. Problem 1.24 (a) The noise levels due to the machine only at each of the three locations are: The L eq at 500Hz is: 10log 10 10 95/ 10 % 10 99.8/ 10 % 10 91.7/ 10 ' 101.5dB Solutions to problems 18 p 2 1 ' p 2 2 ' p 2 ref 10 L p /10 and p 2 t ' 4 × p 2 1 p 1 ' p 2 ' p ref 10 85/20 p 2 t ' p 2 ref 10 8.5 % 10 8.5 % 2×10 8.5 × cos30 L p t ' 10log 10 p 2 t p 2 ref ' 90.7dB dB reduction ' &10log 10 1 3 10 &1.5 % 10 &2 % 10 &2.3 ' 18.1 dB (b) Problem 1.25 Use equation 1.90 in text, 3 rd edn. as the two signals will be coherent. If the two speakers operate together, the phase difference between the two signals is zero as the speakers are identical and driven by the same amplifier. Thus cosθ = 1. Thus, L p t ' L p 1 % 10log 10 4 ' 85 % 6 ' 91dB If a 45E phase shift were introduced, then cosθ = 0.707 and . p 2 t ' (2 % 2 × 0.707)p 2 1 ' 3.414p 2 1 Thus: . L p t ' L p 1 % 10log 10 3.414 ' 85 % 5.3 ' 90.3dB Problem 1.26 Use equation 1.90 in text, as the signals to be added are coherent. For signals 180E out of phase, cos180E = -1 and thus p t 2 = 0 which means that L pt = - 4. In practice the measured level would be greater than this due to electronic instrumentation noise. Fundamentals 19 p rms ' A 2 / 2 % A 2 / 2 % A 2 cosφ ' A (1 % cosφ) ΔL ' 10log 10 p rms p ) rms 2 ' 10log 10 (2 % 2cosφ) ' 4.8dB β 2 & β 1 ' 2πf c c × c 2f c ' π ¢ p 2 t ¦ ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ & 2¢ p 1 p 2 ¦ ' ¢ ( p 1 & p 2 ) 2 ¦, p 1 > p 2 Problem 1.27 (a) Each signal has an amplitude of A, an r.m.s. value of A//2 and the relative phase between them is φ radians. (b) As the signals are at the same frequency and are shifted in phase by a constant amount, they are coherent, so equation 1.90 in the text is used to add them together. Therefore the r.m.s. value of the combined signal is given by: With just a single source, . The difference in dB between p ) rms ' A/ 2 the two is thus: Thus for a value of φ = 60E, ΔL = 4.8dB Problem 1.28 (a) As the waves are from the same source, one may be described by and the other by , where x 1 and x 2 are p 1 ' P 1 e j (ωt & kx 1 ) p 2 ' P 2 e j (ωt & kx 2 ) the path lengths of the two waves. The phase difference is thus , where f c is the centre frequency of β 2 & β 1 ' k(x 1 & x 2 ) ' 2πf c c (x 1 & x 2 ) the band of noise. (b) If (x 1 - x 2 ) = λ/2 = c/2f c , the phase difference is: Substituting this result into equation 1.90 in the text gives: Solutions to problems 20 β 2 & β 1 ' 2πf c c × c f c ' 2π ¢ p 2 t ¦ ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ % 2¢ p 1 p 2 ¦ ' ¢ ( p 1 % p 2 ) 2 ¦ ¢ p 2 t ¦ ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ % 2¢ p 1 p 2 ¦ 1 2π m 2π 0 cosα dα ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ % 2¢ p 1 p 2 ¦ 1 2π sinα 2π 0 ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ ¢ p 2 t ¦ ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ % 2¢ p 1 p 2 ¦ cos(β 1 & β 2 ) ¢ p 2 t ¦ ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ % 2¢ p 1 p 2 ¦ If (x 1 - x 2 ) = λ = c/f c , the phase difference is: Substituting this result into equation 1.90 in the text gives: (c) If all phases are present the total pressure is given by equation 1.90 averaged over all phases. Thus: which is the result for the incoherent case. Problem 1.29 Following example 1.4 on p49 in the text, the level due to the "first" signal alone is given by . 10log 10 10 7.5 & 10 6.9 ' 73.7dB Problem 1.30 (a) The phenomenon is the superposition of acoustic waves of the same frequency and fixed phase. Thus the total pressure, , is given by: ¢ p 2 t ¦ If the phase difference between the two waves is zero, then: Fundamentals 21 ¢ p 2 t ¦ ' ¢ p 2 1 ¦ % ¢ p 2 2 ¦ & 2¢ p 1 p 2 ¦ 20log 10 2p 1 2 × 10 &5 ' 60 20log 10 p 1 2 × 10 &5 ' 60 & 6 ' 54dB 10log 10 10 9.7 & 10 9.4 . 94dB If p 1 = p 2 , then ¢ p 2 t ¦ ' 4¢ p 2 1 ¦ When the two waves are 180E out of phase: If p 1 = p 2 , then ¢ p 2 t ¦ ' 0 The noise level could be reduced substantially by using a control system which ensured that when one pump was turned on, the other was turned on at such a time that it was 180E out of phase with the first pump. (b) When the problem is noticed, the sound pressure level at the house is 60dB. This would occur when the two pumps are in phase. Thus: If one pump only were operating, then the level should be: If the level which is measured with one pump operating is closer to 54dB than 57dB, then the theory of in-phase addition of sound waves would be verified. However if the level with one pump operating were closer to 57dB, then incoherent addition would be suggested and the problem would need further investigation. Problem 1.31 Following example 1.4 in the text, the level due to the machine alone is equal to: Thus the machine is in compliance with specifications. Solutions to problems 22 10log 10 10 9.5 & 10 9.1 . 92.8dB NR ' 10log 10 [10 0/10 % 10 &5/10 ] & 10log 10 [10 &8/10 % 10 &13/10 % 10 &13/10 % 10 &8/10 % 2( 10 &18/10 % 10 &12/10 )] ' 1.2 & (&2.4) ' 3.6dB Problem 1.32 Following example 1.4 in the text, the level due to the machine alone is equal to: Problem 1.33 See pages 49 and 50 in text, "combining level reductions". The difference level with the barrier removed can be calculated by adding the barrier noise reduction to 60dB(A). The noise reduction is calculated using equation 1.97 and is: Thus the level with the barrier removed is 60 + 3.6 = 63.6dB. Problem 1.34 (a) Level at the receiver due to both waves = 75dB. Reflected signal has suffered a 5dB loss. Let the signal due to the direct wave = xdB. Then: 75 ' 10log 10 10 x/ 10 % 10 (x & 5) / 10 or 75 ' 10log 10 10 x/ 10 % 10log 10 1 % 10 (& 5/ 10) Thus, x = 75 - 1.2 = 73.8dB (b) Contributions to the total level from various paths are: Path A: 73.8 - 4 - 7 = 62.8 Path B: 73.8 - 5 - 5 = 63.8 Path C: 73.8 - 4 = 69.8 Sound pressure level at receiver = 10log 10 10 6.28 % 10 6.38 % 10 6.98 ' 71.4dB (c) The answer can be found by calculating what the direct field contribution Fundamentals 23 u tot ' & 1 ρ M Mx m p dt ' & A jωρ M Mx e jωt e &jkx % 0.25e jkx ' kA ωρ e jωt e &jkx & 0.25e jkx I ' 1 2 Re{¯ p ¯ u ( } ' 1 2 Re A e &jkx % 0.25e jkx kA ωρ e jkx & 0.25e &jkx I ' 1 2 Re kA 2 ωρ 1.0 & 0.25 2 % 0.25e 2jkx & 0.25e &2jkx ' 0.47 kA 2 ωρ is, using the level with the barrier in place (note incoherent addition with the barrier in place). Let the direct sound field = xdB. The total level with the barrier in place is 70dB. Thus: 70 ' 10log 10 10 (x & 11) / 10 % 10 (x & 10))10 % 10 (x & 4) / 10 or 70 ' 10log 10 10 x/ 10 % 10log 10 10 &11/ 10 % 10 &10/ 10 % 10 &4/ 10 Thus, x = 70.0 + 2.4 = 72.4dB Reduction due to destructive interference = 72.4 - 65 = 7.4dB Problem 1.35 The positive going wave may be represented as and the p i ' Ae j(ωt & kx) negative going wave as . At x = 0, the phase between the two (A/4) e j(ωt % kx % θ) waves is 0. Thus, θ = 0. The total pressure field may then be written as: p tot ' Ae j(ωt & kx) % (A/4)e j(ωt % kx) ' Ae jωt e &jkx % 0.25e jkx . Combining equations 1.6 and 1.7 in the text gives for the particle velocity: The active acoustic intensity is then: The preceding equation may be rearranged to give: Alternatively, the intensity of each of the two waves could have been calculated separately and combined vectorially. That is, for the positive going Solutions to problems 24 I i ' ¢p 2 i ¦ ρc ' A 2 2ρc I i ' ¢p 2 r ¦ ρc ' A 2 4 2 ×2ρc I tot ' I i & I r ' A 2 2ρc & A 2 32ρc ' 0.47 A 2 ρc ' 0.47 kA 2 ωρ plane wave: and for the negative going plane wave: The total intensity is then: If the two waves had the same amplitude it is clear from the preceding equations that the active intensity would be zero. Problem 1.36 (a) Higher order mode cut-on frequency is: f co = 0.586c/d, where d is the tube diameter (see p456 in text). Thus f co = 0.586 × 343/0.05 = 4020Hz Frequency range for plane waves = 0 to 4020Hz. (b) For plane waves, the acoustic power is: , where S is the duct cross sectional area. W ' I S ' S¢ p 2 ¦ ρc ' ρcS¢ u 2 ¦ ¢ u 2 ¦ ' ω 2 ξ 2 0 / 2 ' (2π × 500 × 0.0001) 2 / 2 ' 0.04935(m/s) 2 ρc = 1.206 × 343 = 413.7, S = (π/4) × 0.05 2 Thus, W ' 413.7 × 1.964 × 10 &3 × 0.04935 ' 0.04watts (c) As power is proportional to the square of the cone velocity, the cone velocity squared should be kept constant which means that the displacement squared of the speaker cone should vary inversely with frequency. That is, the displacement should vary inversely with the square root of the frequency. Fundamentals 25 u(x, t) ' & 1 jωρ Mp(x, t) Mx ' p(x, t) ρc ξ(x, t) ' 1 ρc 5 500 e j(500t & k 1 x & π/ 2) % 3 200 e j(200t & k 2 x & π/ 2) ' 1 343×1.206 5 500 e j(500t & 2500/ 343 & π/ 2) % 3 200 e j(200t & 1000/343 & π/ 2) ' 1 413.7 1 100 e j(500t & 7.29 & π/ 2) % 3 200 e j(200t & 2.91 & π/ 2) Problem 1.37 (a) The acoustic pressure is given by: p(x, t) ' 5e j(500t & k 1 x) % 3e j(200t & k 2 x) The particle velocity can be obtained using equation 1.7 and the one dimensional form of equation 1.6 as follows: Using the above expression for acoustic pressure and k 1 = 500/343 k 2 = 200/343 and x = 5, the acoustic particle velocity can be written as: u(x, t) ' 1 ρc 5e j(500t & k 1 x) % 3e j(200t & k 2 x) ' 1 343×1.206 5e j(500t & 2500/ 343) % 3e j(200t & 1000/343) ' 1 413.7 5e j(500t & 7.29) % 3e j(200t & 2.92) The displacement is given by . Thus: ξ(x, t) ' m u dt Solutions to problems 26 u(x, t) ' & 1 jωρ M(p R (x, t) & p L (x, t)) Mx ' p R (x, t) & p L (x, t) ρc (b) r.m.s. values are: u rms ' 1 2 × 1 413.6 5 2 % 3 2 1/2 ' 0.010m/ s ξ rms ' 1 2 × 1 413.6 1/ 100 2 % 3/ 200 2 1/2 ' 3.1 ×10 &5 m (c) Active intensity. As we have a plane wave propagating in only one direction, the sound intensity is given by equation 1.70 in the text. Also note that the mean square pressures for two different frequencies add. Thus: I a ' p 2 rms ρc ' 0.5 5 2 % 3 2 413.7 ' 0.041 W/ m 2 (d) Reactive intensity. This is undefined as we have more than one frequency component in the wave. (e) The acoustic pressure for each wave is given by: p R (x, t) ' 5e j(500t & k 1 x) % 3e j(200t & k 2 x) p L (x, t) ' 4e j(500t % k 1 x) % 2e j(200t % k 2 x) Total acoustic pressure, p = p R + p L . Thus: p = + 5e j(500t & k 1 x) % 3e j(200t & k 2 x) 4e j(500t % k 1 x) % 2e j(200t % k 2 x) The total acoustic particle velocity is then: Thus: u(x, t) ' 1 1.206 × 343 5e j(500t & k 1 x) % 3e j(200t & k 2 x) & 4e j(500t % k 1 x) & 2e j(200t % k 2 x) Fundamentals 27 I ' 0.5 1.206 × 343 Re 5e &jk 1 x % 4e jk 1 x 5e jk 1 x & 4e &jk 1 x % 0.5 1.206 × 343 Re 3e &jk 2 x % 2e jk 2 x 3e jk 2 x & 2e &jk 2 x Z ' p(x, t) u(x, t) ' ρc Ae j(ωt & kx) Ae j(ωt & kx) ' ρc x 0 Z = p u T T / -L The active intensity is given by I = 0.5Re{ } where the bar denotes ¯ p ¯ u ( the complex amplitude. The amplitude of the reactive component is not defined as there is more than one frequency present. The active intensity is calculated at each frequency and the results added together as follows: Thus, I = 1.20 × 10 -3 (9 + 5) = 0.017 W/m 2 Problem 1.38 (a) The acoustic pressure may be written as . Using p(x, t) ' Ae j(ωt & kx) equations 1.6 and 1.7 in the text, the acoustic particle velocity may be written as: . u(x, t) ' kA ρω e j(ωt & kx) ' A ρc e j(ωt & kx) (b) Particle velocity is the magnitude of the motion of the particles disturbed during the passage of an acoustic wave, whereas the speed of sound refers to the speed at which the disturbance propagates. Acoustic particle velocity is a function of the loudness of the noise, whereas the speed of sound is independent of loudness. (c) The specific acoustic impedance is the ratio of acoustic pressure to particle velocity. Using the preceding equations we obtain: (d) Solutions to problems 28 p T ' A e j(ωt & kx) % e j(ωt % kx) u T ' A ρc e j(ωt & kx) & e j(ωt % kx) Z ρc ' e &j kx % e j kx e &j kx & e j kx ' cos(kx) &j sin(kx) ' j cot(kx) p T ' A e j(ωt & kx) % e j(ωt % kx % 2kL) x 0 Z = -p u T T / -L rigid termination speaker To simplify the algebra, set the origin of the coordinate system at the rigid end of the tube as shown in the figure. As the tube is terminated non-anechoically, the pressure will include a contribution from the reflected wave. For a rigid termination, the phase shift on reflection is 0E and the amplitude of the reflected wave is equal to the amplitude of the incident wave. As the origin, x = 0 is at the point of reflection, the phase of the two waves must be the same when x = 0. Of course if the origin were elsewhere, this would not be true and the following expressions would have to include an additional term (equal to the distance from the origin to the point of reflection) in the exponent of the reflected wave. With the origin at the point of reflection, the total acoustic pressure and particle velocity at any point in the tube may be written as: and The specific acoustic impedance is then: Problem 1.39 The coordinate system is as shown in the figure at right. (a) For a rigid termination, the phase shift on reflection is 0E and the amplitude of the reflected wave is equal to the amplitude of the incident wave. As the origin, x = 0 is at the loudspeaker location, the phase of the two waves must be the same when x = -L. Thus, the total acoustic pressure at any point in the tube may be written as: The velocity potential and acoustic particle velocity may be derived from Fundamentals 29 φ T ' A jρω e j(ωt & kx) % e j(ωt % kx % 2kL) u T ' A ρc e j(ωt & kx) & e j(ωt % kx % 2kL) p T ' U 0 ρc 1 & e j2kL e j(ωt & kx) % e j(ωt % kx % 2kL) u T ' U 0 1 & e j2kL e j(ωt & kx) & e j(ωt % kx % 2kL) I ' 1 2 Re ¯ p T ¯ u ( T ' ρcU 2 0 Re e &j kx % e j(kx % 2kL) × e j kx & e &j(kx % 2kL) 2 1 & e j2kL × 1 & e &j2kL ' ρcU 2 0 Re e &j kx % e j(kx % 2kL) × e j kx & e &j(kx % 2kL) 2 2 & e j2kL & e &2jkL ' ρcU 2 0 2 2 & 2cos(2kL) Re 1 & 1 % e 2jk(x % L) & e &2jk(x % L) ' 0 the above expression as: (b) At x = 0, u T = U 0 e jωt , thus and so U 0 ' A ρc 1 & e j2kL A ' U 0 ρc 1 & e j2kL (c) Rewriting the expressions of (a) in terms of U 0 , we obtain: and The real part of the acoustic intensity (where the bar denotes the complex amplitude which is time independent) is: The amplitude of the imaginary part of the acoustic intensity can be derived in a similar way and from the last line in the above equation, it Solutions to problems 30 I ' 1 2 Im ¯ p T ¯ u ( T ' ρcU 2 0 2 & 2cos(2kL) sin[2k(x % L)] p i ' Ae j(ωt & kx) and p r ' Be j(ωt % kx % θ) p T ' Ae j(ωt & kx) % Be j(ωt % kx % θ) x speaker 1 speaker 2 0 is: (d) The acoustic intensity is a vector quantity and as the amplitudes of the two waves travelling in opposite directions are the same, their intensities will vectorially add to zero. Problem 1.40 (a) Incident wave and reflected wave pressures may be written as: The total pressure is thus: The maximum pressure amplitude occurs when the left and right going waves are in-phase which is at location x, such that the phase, θ = -2kx, giving a pressure amplitude of (A + B). The minimum pressure amplitude occurs when the left and right going waves are π radians out of phase, at the location x, such that the phase θ = -2kx + π, with a corresponding amplitude of A - B. Thus, the ratio of maximum to minimum pressure is Fundamentals 31 π & 2kx min ' &2kx max x min & x max ' π 2k ' λ 4 ' c 4f f ' c 4(x min & x max ) ' 343 4(0.09 &0.03) ' 1430 Hz u T ' 1 ρc (p i & p r ) u T ' 1 ρc Ae j(ωt & kx) & Be j(ωt % kx % θ) ¯ u T ' 1 ρc A & Be jθ θ ' &0.06k ' &0.06 π 2(x min & x max ) ' &π/ 2 * ¯ u T * ' 1 413.7 A 2 % B 2 1/2 (A + B)/(A - B) and the standing wave ratio is 20log 10 [(A %B) / (A & B)] The location of the minimum closest to the end where x = 0 (left end) is 0.09m. Thus θ = -0.18k + π. As θ is a constant: So: and thus: (b) The total particle velocity can be calculated using equations 1.6 and 1.7 in the text as: Thus: The complex particle velocity amplitude at x = 0 is then: The phase angle θ is given by part (a) as: Using the above 2 equations, the particle velocity amplitude can be written as: The standing wave ratio is given by: Solutions to problems 32 * A % B A & B * ' 10 (100 & 96.5) / 20 ' 1.496 A % B ' 2 × 10 &5 × 1.414 × 10 100/ 20 ' 2.828 Z m ' ρcS A % Be j(2kL % θ) A & Be j(2kL % θ) Z m ' 413.7 × 0.001 × 5.032 % e jπ/ 2 5.032 & e jπ/ 2 ' 0.4136 × 5.032 % j 5.032 & j ' 0.01571 × (5.032 % j) 2 ' 0.382 % 0.158j Thus, A = 5.03B. However, the maximum pressure amplitude is A + B. So: From the preceding two equations, we have A = 2.36 and B = 0.469. Thus the velocity amplitude at x = 0 is: . * ¯ u T * ' 1 413.7 2.36 2 % 0.469 2 1/2 ' 5.8mm/s Thus the volume velocity amplitude is 5.9 × 10 -6 m 3 /s (0.001m 2 area) which is equivalent to an r.m.s volume velocity of 4.2 × 10 -6 m 3 /s. (c) The mechanical impedance of the second loudspeaker is given by the cross-sectional area multiplied by the ratio of the pressure and particle velocity at the surface of the loudspeaker. Thus Z m = pS/u. Using the relationships derived in parts (a) and (b), we have: We previously found that θ = -π/2 and k = -θ/0.06. Also L = 0.3 and 2kL = (2π/0.12)×0.3 = 5π. Thus Z m may be written as: Problem 1.41 (a) It is sufficient to show that adding two waves of the same frequency but shifted in phase will give a third wave of the same frequency but shifted Fundamentals 33 p 1 ' A 1 e jωt and p 2 ' A 2 e jωt % β p 1 ' A 1 cosωt and p 2 ' A 2 cos(ω % Δω)t u et | p 2 p 3 p 1 Im Re in phase. Assuming plane wave propagation, let the two waves to be added be described as: and the third wave as . To find A 3 and θ in terms of A 1 , A 2 p 3 ' A 3 e jωt % θ and β, it is easiest to express p 1 , p 2 and p 3 as rotating vectors and use the cosine rule as shown in the figure. From the cosine rule: A 2 3 ' A 2 1 % A 2 2 % 2A 1 A 2 cosβ which is the same as equation 1.90 in the text. The phase angle θ is: θ ' tan &1 A 2 sinβ A 1 % A 2 cosβ (b) It is sufficient to show that adding together two plane waves of slightly different frequency with the same amplitude result in a third wave. Let the two waves to be added be described as: The sum of the two pressures written above may be expressed as: Solutions to problems 34 p 1 % p 2 ' A(cosωt % cos(ω % Δω)t) ' 2A cos t 2 (ω % ω % Δω) cos t 2 (ω & ω & Δω) ' 2Acos Δω 2 t cos ω % Δω 2 t p i ' Ae j(ωt % kx) and p r ' Be j(ωt & kx % θ) p T ' Ae j(ωt % kx) % Be j(ωt & kx % θ) x 0 Z = -p /Su a T T L which is a sine wave of frequency (ω + Δω) modulated by a frequency Δω/2 (c) If Δω is small we obtain the familiar beating phenomenon (see page 46, fig 1.9 in text which shows a beating phenomenon where the two waves are slightly different in amplitude resulting in incomplete cancellation at the null points). Problem 1.42 (a) As the origin is at the left end of the tube, the incident wave will be travelling in the negative x direction. Assuming a phase shift between the incident and reflected waves of θ at x = 0, the incident wave and reflected wave pressures may be written as: The total pressure is thus: The total particle velocity can be calculated using equations 1.6 and 1.7 in the text as: Fundamentals 35 u T ' 1 ρc (p r & p i ) u T ' 1 ρc Be j(ωt & kx % θ) & Ae j(ωt % kx) P 0 ' A % Be j θ and ρcU 0 ' Be j θ & A A ' 0.5(P 0 & ρcU 0 ) and Be jθ ' 0.5(P 0 % ρcU 0 ) p T ' 0.5(P 0 & ρcU 0 ) e j(ωt % kx) % 0.5(P 0 % ρcU 0 ) e j(ωt & kx) u T ' 0.5 ρc (P 0 % ρcU 0 ) e j(ωt & kx) & (P 0 & ρcU 0 ) e j(ωt % kx) Z a ' & p T Su T ' & ρc S (P 0 & ρcU 0 ) e jkx % (P 0 % ρcU 0 ) e &jkx (P 0 % ρcU 0 ) e &jkx & (P 0 & ρcU 0 ) e jkx Z a ' ρc S jρcU 0 sin(kx) & P 0 cos(kx) ρcU 0 cos(kx) & jP 0 sin(kx) Thus: At x = 0, p T = P 0 e jωt and u T = U 0 e jωt . Thus: Thus: and the total acoustic pressure and particle velocity may be written as: and The acoustic impedance looking towards the left in the negative x- direction is the negative ratio of the total acoustic pressure to the product of the duct cross-sectional area and the total acoustic particle velocity (see the preceding figure). Thus: As , the e jkx ' cos(kx) % jsin(kx) and e &jkx ' cos(kx) & jsin(kx) impedance may be written as: Solutions to problems 36 Z a ' ρc S P 0 cos(kx) & jρcU 0 sin(kx) ρcU 0 cos(kx) & jP 0 sin(kx) Z a ' ρc S Z 0 cos(kL) % jρc sin(kL) ρc cos(kL) % jZ 0 sin(kL) 2m (b) Using a similar analysis to that outlined above, the following expression is obtained for the acoustic impedance looking to the right in the positive x direction: The expressions in parts (a) and (b) for the impedance can be shown to be equal if evaluated at the open end of the tube (x = L in part (a) and x = -L in part (b). In addition, the quantity -P 0 /U 0 is equal to the termination impedance Z 0 in part (a), while in part (b), the quantity P 0 /U 0 is equal to the termination impedance Z 0 . Making the appropriate substitutions, both expressions give the following for the impedance at the open end of the tube. Problem 1.43 (a) Set the origin, x = 0 at the surface of the sample. Then the entrance of the tube is at x = -2.0. As the origin is at the surface of the sample, the incident wave will be travelling in the positive x direction. Assuming a phase shift between the incident and reflected waves of θ at x = 0, the incident wave and Fundamentals 37 p i ' Ae j(ωt & kx) and p r ' Be j(ωt % kx % θ) p T ' Ae j(ωt & kx) % Be j(ωt % kx % θ) u T ' 1 ρc Ae j(ωt & kx) & Be j(ωt % kx % θ) Z ' p T u T ' ρc Ae &j kx % Be j kx % jθ Ae &j kx & Be j kx % jθ ' ρc A % Be j(2kx % θ) A & Be j(2kx % θ) ' ρc A/ B % cos(2kx % θ) % jsin(2kx % θ) A/ B & cos(2kx % θ) &jsin(2kx % θ) R p ' (B/ A) e jθ ' 0.5 % 0.5j reflected wave pressures may be written as: The total pressure is thus: The total particle velocity can be calculated using equations 1.6 and 1.7 in the text as: The pressure reflection coefficient, R p , is defined as p r /p i . Thus: The reflection coefficient amplitude is B/A and the phase is θ. Thus, B/A = 0.707 and θ = 45E = 0.7854 radians. Thus the specific acoustic impedance at any point in the tube may be written as: = 1.832 and x = -2. Thus, 2kx + θ = -6.542. k ' ω/ c ' 2π × 100/ 343 . cos(2kx % θ) ' 0.9667 and sin(2kx % θ) ' &0.25587 and A/ B ' 1.414 Thus, Z = 414 × (2.3807 - j0.25587)/(0.4475 + j0.25587) = 1558(1 - j0.7237) = 1560 - j1130 (b) The absorption coefficient is defined as , so α = 0.5 α ' 1 & *R p * 2 Solutions to problems 38 p i ' Ae j(ωt % kx) and p r ' Be j(ωt & kx % θ) p T ' Ae j(ωt % kx) % Be j(ωt & kx % θ) 10 L 0 / 20 ' A % B A & B x 0 Z = -p u T T / -L speaker sample Problem 1.44 (a) To simplify the algebra, assume that the tube is horizontal with the left end at x = 0 containing the sample of material whose absorption coefficient is to be determined, as shown in the figure. As the origin is at the left end of the tube, the incident wave will be travelling in the negative x direction. Assuming a phase shift between the incident and reflected waves of θ at x = 0, the incident wave and reflected wave pressures may be written as: The total pressure is thus: The maximum pressure will occur when θ = 2kx, and the minimum will occur when θ = 2kx + π. Thus: and p max ' e jkx A % B p min ' e jkx A & B and the ratio of maximum to minimum pressures is (A + B)/(A - B) The standing wave ratio, L 0 , is defined as: Thus the ratio (B/A) is: Fundamentals 39 B A ' 10 L 0 /20 &1 10 L 0 /20 %1 *R p * 2 ' 10 L 0 /20 &1 10 L 0 /20 %1 2 α ' 1 & 10 15/ 20 & 1 10 15/ 20 % 1 2 ' 0.51 u T ' 1 ρc (p r & p i ) u T ' 1 ρc Be j(ωt & kx % θ) & Ae j(ωt % kx) Z ' & p T u T ' &ρc Ae j kx % Be &j kx % jθ Be &j kx % jθ & Ae j kx ' &ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A Z s ρc ' & p T ρcu T ' A % Be j θ A & Be j θ Z s ' A/ B % cosθ % j sinθ A/ B & cosθ & j sinθ ' (A/ B) 2 & 1 % (2A/ B) j sinθ (A/ B) 2 % 1 & (2A/ B) cosθ The amplitude of the pressure reflection coefficient squared is which can be written in terms of L 0 (= 95 - 80) as: *R p * 2 ' (B/ A) 2 The absorption coefficient is defined as , so: α ' 1 & *R p * 2 (b) The total particle velocity can be calculated using the equation in part (a) for the acoustic pressure and equations 1.6 and 1.7 in the text as: Thus: Thus the specific acoustic impedance at any point in the tube may be written as: At x = 0, the specific acoustic impedance is the normal impedance, Z s , of the surface of the sample. Thus: The above impedance equation may be expanded to give: Solutions to problems 40 *Z s * ' (A/ B) 2 & 1 2 % (2A/ B) 2 sin 2 θ (A/ B) 2 % 1 & (2A/ B)cosθ β ' tan &1 2(A/ B) sinθ (A/ B) 2 & 1 *Z s * ρc ' (1.433 2 & 1) 2 % (2.866 2 × 0.966 2 ) 1.433 2 % 1 & 2.866 × 0.258 ' 1.28 β ' tan &1 2 × 1.433 × (&0.966) 1.433 2 & 1 ' &69.2E α st ' 8 cosβ ξ 1 & cosβ ξ log e (1 % 2ξ cosβ % ξ 2 ) % cos(2β) ξ sinβ tan &1 ξ sinβ 1 % ξ cosβ The modulus of the impedance is then: and the phase is given by: Using the previous analysis, . A B ' 10 L 0 / 20 % 1 10 L 0 / 20 & 1 ' 1.433 At the pressure minimum, θ = 2kx - π, where k = 2π/λ. x = 0.2m and f = 250Hz, thus k = 2πf/c = 4.58. Thus, θ = 2 × 4.58 × 0.2 - π = -1.31 radians cosθ = 0.258 and sinθ = -0.966 Thus from the preceding equations: and the phase is: (c) The statistical absorption coefficient is given by equation C.37 in the text as Substituting in the modulus and phase of the impedance, we obtain” Fundamentals 41 α st ' 8 × cos(&69.17) 1.28 1 & cos(&69.17) 1.28 × × log e 1 % 2 × 1.28 × cos(&69.17) % 1.28 2 % cos(&138.4) 1.28sin(&69.17) tan &1 1.28 × sin(&69.17) 1 % 1.28 × cos(&69.17) ' 2.22(1 & 0.278 × 1.266 % 0.625 × (&0.688) ' 2.22(1 & 0.352 & 0.430) ' 0.485 W ' S 2 Re ¯ p T ¯ u ( T ' S* ¯ u T * 2 2 Re Z x 0 Z = -p /u T T L Problem 1.45 (a) Assume a horizontal tube with the left end containing the termination impedance Z 0 at x = 0. As the origin is at the left end of the tube, the incident wave will be travelling in the negative x direction. The power radiated by a source at the other end of the tube is related to the specific acoustic impedance, Z, it "sees" as follows: where the bar represents the complex amplitude and S is the tube cross- sectional area. Assuming a phase shift between the incident and reflected waves of θ at x = 0, the incident wave and reflected wave pressures may be written as: Solutions to problems 42 p i ' Ae j(ωt % kx) and p r ' Be j(ωt & kx % θ) p T ' Ae j(ωt % kx) % Be j(ωt & kx % θ) u T ' 1 ρc (p r & p i ) u T ' 1 ρc Be j(ωt & kx % θ) & Ae j(ωt % kx) Z ' & p T u T ' &ρc Ae j kx % Be &j kx % jθ Be &j kx % jθ & Ae j kx ' &ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A Z 0 ρc ' & P 0 ρcU 0 ' & A % Be j θ Be j θ & A A ' 0.5(P 0 & ρcU 0 ) and Be jθ ' 0.5(P 0 % ρcU 0 ) p T ' 0.5(P 0 & ρcU 0 ) e j(ωt % kx) % 0.5(P 0 % ρcU 0 ) e j(ωt & kx) u T ' 0.5 ρc (P 0 % ρcU 0 ) e j(ωt & kx) & (P 0 & ρcU 0 ) e j(ωt % kx) The total pressure is thus: The total particle velocity can be calculated using equations 1.6 and 1.7 in the text as: Thus: Thus the specific acoustic impedance at any point in the tube may be written as: At x = 0, Z = Z 0 . For convenience also set p T = P 0 e jωt and u T = U 0 e jωt Thus: Thus: and the total acoustic pressure and particle velocity may be written as: and Fundamentals 43 Z ' &ρc (P 0 & ρcU 0 ) e jkx % (P 0 % ρcU 0 ) e &jkx (P 0 % ρcU 0 ) e &jkx & (P 0 & ρcU 0 ) e jkx Z ' ρc jρcU 0 sin(kx) & P 0 cos(kx) ρcU 0 cos(kx) & jP 0 sin(kx) Z ' ρc Z 0 / ρc % j tankL 1 % j(Z 0 / ρc)tankL ' ρc (R 0 % jX 0 ) / ρc % j tankL 1 % j((R 0 % jX 0 ) / ρc)tankL Re{Z} ' ρc (R 0 / ρc) ( 1 % tan 2 kL) (1 & X 0 / ρc) 2 tan 2 kL % (R 0 / ρc) 2 tan 2 kL W ' SρcU 2 L 2 (R 0 / ρc) ( 1 % tan 2 kL) (1 & X 0 / ρc) 2 tan 2 kL % (R 0 / ρc) 2 tan 2 kL The specific acoustic impedance looking towards the left in the negative x direction may then be written as: As , the e jkx ' cos(kx) % jsin(kx) and e &jkx ' cos(kx) & jsin(kx) impedance may be written as: Dividing through by ρcU 0 cos(kx), replacing x with L and replacing gives: & P 0 U 0 with Z 0 Thus: and the power is then: (b) It can be seen from the equation derived in part (a) that when R 0 = 0, the power will be zero. (c) If all losses are zero, the impedance presented to the loudspeaker will be given by the previously derived expression for Z with R 0 = 0. In this case: Solutions to problems 44 Z ' ρc jX 0 / ρc % j tankL 1 & ( X 0 / ρc)tankL ' j ρc X 0 / ρc % tankL 1 & ( X 0 / ρc)tankL A ' 0.5(P 0 & ρcU 0 ) and Be jθ ' 0.5(P 0 % ρcU 0 ) R p ' Be jθ A R p ' Be jθ A ' (P 0 % ρcU 0 ) (P 0 & ρcU 0 ) R p ' &ρc & jX 0 % ρc &ρc & jX 0 & ρc ' j X 0 2ρc % jX 0 *R p * ' X 0 [ 4ρ 2 c 2 % X 2 0 ] &1/2 which is imaginary. Thus there will be no real power generated; only imaginary power which represents non-propagating energy stored in the near field. (d) As no real power is generated, the acoustic pressure and particle velocity must be 90E out of phase. (e) The pressure amplitude reflection coefficient is given by: From part (a): Thus: Dividing numerator and denominator by U 0 and putting , we obtain: Z 0 ' ρc % jX 0 ' &P 0 / U 0 Thus: Fundamentals 45 10 SWR/ 20 ' A % B A & B p i ' Ae j(ωt % kx) and p r ' Be j(ωt & kx % θ) p T ' Ae j(ωt % kx) % Be j(ωt & kx % θ) p T ' A e j(ωt % kx) % Re j(ωt & kx) x 0 Z = -p u T T / L sample speaker Problem 1.46 (a) As given in the problem, the tube is assumed to be horizontal with the left end at x = 0 containing the sample of material whose impedance is to be determined, as shown in the figure. As the origin is at the left end of the tube, the incident wave will be travelling in the negative x direction. Assuming a phase shift between the incident and reflected waves of θ at x = 0, the incident wave and reflected wave pressures may be written as: The total pressure is thus: At the surface of the sample, the pressure amplitude reflection coefficient is thus and . Thus the total pressure at any R ' (B/ A)e jθ B ' (RA)e &jθ location, x, in the tube may be written as: (b) Returning to the first expression for the total pressure of part (a), the maximum pressure will occur when θ = 2kx, and the minimum will occur when θ = 2kx + π. Thus , and the p max ' e jkx A % B p min ' e jkx A & B ratio of maximum to minimum pressures is (A + B)/(A - B) (c) The standing wave ratio (SWR) which is L 0 is defined as: Solutions to problems 46 B A ' 10 L 0 /20 &1 10 L 0 /20 %1 *R p * 2 ' 10 L 0 /20 &1 10 L 0 /20 %1 2 u T ' 1 ρc (p r & p i ) u T ' 1 ρc Be j(ωt & kx % θ) & Ae j(ωt % kx) Z ' & p T u T ' &ρc Ae j kx % Be &j kx % jθ Be &j kx % jθ & Ae j kx ' &ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A Z ρc ' & p T ρcu T ' A % Be j θ A & Be j θ ' 1 % R p 1 & R p Thus the ratio (B/A) is: From part (a), the amplitude of the pressure reflection coefficient squared is which can be written in terms of L 0 as: *R p * 2 ' (B/ A) 2 (d) The absorption coefficient is defined as , which on α ' 1 & *R p * 2 substituting the equation derived in part (c) for , gives the required *R p * 2 result. (e) The total particle velocity can be calculated using equations 1.6 and 1.7 in the text as: Thus: Thus the specific acoustic impedance at any point in the tube may be written as: At x = 0, the specific acoustic impedance is the normal impedance, Z s , of the surface of the sample. Thus: Fundamentals 47 p i ' Ae j(ωt % kx) and p r ' Be j(ωt & kx % θ) p T ' Ae j(ωt % kx) % Be j(ωt & kx % θ) u T ' 1 ρc (p r & p i ) u T ' 1 ρc Be j(ωt & kx % θ) & Ae j(ωt % kx) Z ' & p T u T ' &ρc Ae j kx % Be &j kx % jθ Be &j kx % jθ & Ae j kx ' &ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A x 0 Z = -p /u T T L Problem 1.47 (a) Assume a horizontal tube with the left end containing the termination impedance Z L at x = 0. As the origin is at the left end of the tube, the incident wave will be travelling in the negative x direction. Assuming a phase shift between the incident and reflected waves of θ at x = 0, the incident wave and reflected wave pressures may be written as: The total pressure is thus: The total particle velocity can be calculated using equations 1.6 and 1.7 in the text as: Thus: Thus the specific acoustic impedance at any point in the tube may be written as: Solutions to problems 48 Z L ρc ' & P 0 ρcU 0 ' & A % Be j θ Be j θ & A A ' 0.5(P 0 & ρcU 0 ) and Be jθ ' 0.5(P 0 % ρcU 0 ) p T ' 0.5(P 0 & ρcU 0 ) e j(ωt % kx) % 0.5(P 0 % ρcU 0 ) e j(ωt & kx) u T ' 0.5 ρc (P 0 % ρcU 0 ) e j(ωt & kx) & (P 0 & ρcU 0 ) e j(ωt % kx) Z ' &ρc (P 0 & ρcU 0 ) e jkx % (P 0 % ρcU 0 ) e &jkx (P 0 % ρcU 0 ) e &jkx & (P 0 & ρcU 0 ) e jkx Z ' ρc jρcU 0 sin(kx) & P 0 cos(kx) ρcU 0 cos(kx) & jP 0 sin(kx) Z ' ρc Z L / ρc % j tankL 1 % j(Z L / ρc)tankL At x = 0, Z = Z L . For convenience also set p T = P 0 e jωt and u T = U 0 e jωt Thus: Thus: and the total acoustic pressure and particle velocity may be written as: and: The specific acoustic impedance looking towards the left in the negative x direction may then be written as: As , the e jkx ' cos(kx) % jsin(kx) and e &jkx ' cos(kx) & jsin(kx) impedance may be written as: Dividing through by ρcU 0 cos(kx), replacing x with L and replacing gives: & P 0 U 0 with Z L The same result can be obtained by putting the open end of the tube to the left as shown in the figure above. Fundamentals 49 p T ' Ae j(ωt & kx) % Be j(ωt % kx % θ) u T ' 1 ρc &Be j(ωt % kx % θ) % Ae j(ωt & kx) Z ' p T u T ' ρc Ae j kx % Be &j kx % jθ &Be &j kx % jθ % Ae j kx ' ρc A % Be j(&2kx % θ) A & Be j(&2kx % θ) R p ' Be jθ A Z L ρc ' A % Be j θ A & Be j θ ' 1 % Be j θ / A 1 & Be j θ / A ' 1 % R p 1 & R p x 0 Z = p u T T / -L The same result can be obtained by putting the open end of the tube to the left as shown in the figure below. The total pressure is thus: and the total particle velocity is thus: The specific acoustic impedance at any point in the tube is then: The remaining part of the analysis is the same as before. (b) The reflection coefficient, R p , is defined as the ratio of the complex amplitudes of the reflected to incident waves at x = 0. Thus, from part (a): At x = 0: Thus the pressure reflection coefficient may be written as: Solutions to problems 50 R p ' Z L / ρc & 1 Z L / ρc % 1 Z L ' Z H S T S H Z L ρc ' A % Be j θ A & Be j θ ' 1 % Be j θ / A 1 & Be j θ / A ' 1 % R p 1 & R p R p ' Z L / ρc & 1 Z L / ρc % 1 ' j(S T / S H )tankR & 1 j(S T / S H )tankR % 1 Z L ' j(S T / S H )ρctan[kR(1 & M)] % R a which is equal to zero when Z L = ρc. (c) Assuming continuity of acoustic pressure and volume velocity (uS, where S is the tube cross sectional area) implies that the acoustic impedance is continuous across the hole and thus the specific impedance in the tube at x = 0 is related to the specific acoustic impedance, Z H of the hole by: where S H is the cross sectional area of the hole and S T is the cross sectional area of the tube. From equation 9.14 in the text we have Z H = jρctankR, where R is the effective length of the hole. From part (a), we have at x = 0: Thus the pressure reflection coefficient may be written as: (d) Using equation 9.8 in the text and the condition of continuity of acoustic pressure and acoustic volume velocity at the hole, the equality Z L /S T = Z H /S H holds. Thus: (e) For good absorption, R p = 0, which as we showed in part (b), means that Z L = ρc. For this to be true, the first term in the equation derived in (d) must be large compared to the other terms. Thus, the product kR/M must be much less than 1. For a thin plate, the effective length of the hole is made up of two parts, one corresponding to the side of the hole in the tube and the other corresponding to the side looking into free space. Fundamentals 51 Z ' ρc Z L / ρc % j tankL 1 % j(Z L / ρc)tankL Z ' ρc 4/ ρc % j kL 1 % j(4/ ρc) kL ' ρc jkL ' &jρc kL p ξ ' jωp u ' ρcω kL m ¨ ξ % Kξ ' f or &ω 2 mξ % Kξ ' f Z ' ρc 0 % j kL 1 % j0kL ' jρckL Using equations 9.16 and 9.19 in the text, assuming that the hole diameter is very small compared to the tube diameter, it can be shown that for small M, the effective length of equation 9.7 is R = 0.73d H . Thus the condition that kR/M << 1 implies that 2πf/c × 0.73d H << M, or fd H << 70M. (f) The specific acoustic impedance looking into a tube was shown in part (a) to be: In the limit of small L and large λ (small k), tan(kL) = kL. Also for a rigidly terminated tube, Z L = 4. Thus the preceding expression becomes: The ratio of the pressure to the particle displacement is then: which indicates that the pressure is in phase with the particle displacement. Considering the analogy of a single degree of freedom spring-mass system the equation of motion is: where K is the spring stiffness and m is the mass. It can be seen from the above equation that for a stiffness only, the exciting force will be in phase with the displacement and for a mass only it will be 180E out of phase with the displacement. Thus it is clear from this analogy that the previous case of the pressure and particle displacement in-phase represents a stiffness. For an open ended tube the impedance, Z L = 0 and the impedance of the tube becomes: Solutions to problems 52 p ξ ' jωp u ' &ρcωkL Z 0 ρc ' Z 0 / ρc % j tankL 1 % j(Z 0 / ρc)tankL Z 0 ρc % j Z 0 ρc 2 tankL ' Z 0 ρc % j tankL The ratio of the pressure to the particle displacement is then: which indicates that the acoustic pressure is 180E out of phase with the particle displacement. Thus by the preceding argument, this represents a lumped mass. (g) Using the equation given in the question and using the hint for maximum power we may substitute Z 0 for Z L and Z to give kL at maximum power. Thus: Rearranging gives: For the left side to equal the right side, tan(kL) = 0. Thus kL = nπ, and the optimum length is given by L = nλ/2, where n = 1, 2, 3, ....... (h) The above result suggests that the frequency response will be characterised by a number of resonant peaks and so will be very poor and non-uniform. The frequency response could be made smoother by adding some absorptive porous acoustic material to the tube. This would have the effect of damping the resonances, thus reducing the difference between the peaks and troughs in the frequency response. (i) The resonances are less damped at low frequencies, thus resulting in bigger differences between the peaks and troughs in the frequency response. Problem 1.48 Using the result derived in the answer to 1.45(a), the impedance seen by the loudspeaker may be written as: Fundamentals 53 Z ' ρc Z r / ρc % j tankL 1 % j(Z r / ρc)tankL Z ρc ' R r / ρc % j X r / ρc % tan(kL) 1 & (X r / ρc)tan(kL) & j (R r / ρc) tan(kL) 1 & (X r / ρc)tan(kL) 2 % (R r / ρc) tan(kL) 2 R ρc ' R r / ρc 1 % tan 2 (kL) 1 & (X r / ρc)tan(kL) 2 % (R r / ρc) tan(kL) 2 R ρc ' 1.04 × 10 &3 1 % tan 2 (kL) 1 & 0.00486tan(kL) 2 % 1.04 × 10 &3 tan(kL) 2 which on substituting Z r = R r + jX r may be rewritten as: The power output of the loudspeaker will vary with tube length because the power output of a source is dependent on impedance presented to it and the impedance it is presented is dependent on the tube length. The maximum power output will occur when the impedance presented to the loudspeaker is equal to the internal impedance of the loudspeaker which is infinite. Thus the maximum power output will occur when the denominator in the above equation is zero. Of course the real power output is only dependent on the resistive impedance while the imaginary power output is dependent on the reactive impedance. Thus the specific resistive impedance is: At 250Hz, k = 2πf/c = 2π × 250/343 = 4.58 As a = 0.075, ka = 4.58 × 0.075 = 0.343, πa 2 = 0.0177 Thus, R r /ρc = 0.0177 × 0.343 2 /2 = 1.04 × 10 -3 and X r /ρc = 0.0177 × 0.343 × 0.8 = 0.00486 Substituting in the expression given for R/ρc, we obtain: By trial and error it can be shown that the maximum value of the above expression occurs when tan(kL) = 3.636, or kL = 1.262. Thus optimum L for maximum power out = 1.262/4.58 = 276mm. Problem 1.49 (a) A standing wave tube is used to determine the normal specific impedance Solutions to problems 54 p i ' Ae j(ωt % kx) and p r ' Be j(ωt & kx % θ) p T ' Ae j(ωt % kx) % Be j(ωt & kx % θ) u T ' 1 ρc (p r & p i ) u T ' 1 ρc Be j(ωt & kx % θ) & Ae j(ωt % kx) x 0 Z = -p /u T T sample L of a solid by placing the sample in one end of the tube which is then rigidly closed, and then generating a pure tone sound field in the tube with a loudspeaker placed at the other end of the tube. The ratio of maximum to minimum acoustic pressure in the standing wave generated in the tube is measured, as is the distance of the first minimum from the surface of the sample. To simplify the algebra, a horizontal tube with the left end at x = 0 containing the solid whose impedance is to be determined will be assumed as shown in the figure. As the origin is at the left end of the tube, the incident wave will be travelling in the negative x direction. Assuming a phase shift between the incident and reflected waves of θ at x = 0, the incident wave and reflected wave pressures may be written as: The total pressure is thus: The total particle velocity can be calculated using equations 1.6 and 1.7 in the text as: Thus: Thus the specific acoustic impedance at any point in the tube may be written as: Fundamentals 55 Z ' & p T u T ' &ρc Ae j kx % Be &j kx % jθ Be &j kx % jθ & Ae j kx ' &ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A Z s ρc ' & p T ρcu T ' A % Be j θ A & Be j θ Z s ρc ' (A/ B) % e jθ (A/ B) & e jθ 10 SWR/ 20 ' A % B A & B At x = 0, the specific acoustic impedance is the normal impedance, Z s , of the surface of the sample. Thus: To calculate the impedance, it is necessary to evaluate the constants A and B, and the phase angle, θ. Returning to the above expression for the total acoustic pressure, it can be seen that the maximum sound pressure in the tube will occur when θ = 2kx, and the amplitude will be (A + B). The minimum pressure amplitude occurs at the location, where θ = 2kx - π, with a corresponding amplitude of A - B. Thus, the ratio of maximum to minimum pressure is (A + B)/(A - B) and the standing wave ratio is ,which is the difference in dB between 20log 10 [(A %B) / (A & B)] the maximum and minimum sound pressure levels in the tube. The phase angle θ is determined by the distance, x of the first minimum sound pressure level from the face of the sample by using θ = 2kx - π. This is equivalent to the equation for θ, given on p623 the text, where k = 2π/λ = 2πf/c. The first minimum is used because the effect of any losses due to non rigid tube walls will be minimised. The minimum rather than the maximum is used because its location is much more sharply defined. The equation for the impedance may be rewritten as: The standing wave ratio (SWR) which can be measured is defined as: Thus, and the impedance may be calculated. A B ' 10 SWR/ 20 % 1 10 SWR/ 20 & 1 Solutions to problems 56 Z s ρc ' A/ B % cosθ % j sinθ A/ B & cosθ & j sinθ ' (A/ B) 2 & 1 % (2A/ B)j sinθ (A/ B) 2 % 1 & (2A/ B)cosθ *Z s * ρc ' (A/ B) 2 & 1 2 % (2A/ B) 2 sin 2 θ (A/ B) 2 % 1 & (2A/ B)cosθ β ' tan &1 2(A/ B) sinθ (A/ B) 2 & 1 *Z s * ρc ' (4.216 2 & 1) 2 % (8.432 2 × 0.951 2 ) 4.216 2 % 1 % 8.432 × 0.309 ' 0.87 β ' tan &1 2 × 4.216 × 0.951 4.216 2 & 1 ' 25.5E This is done by expanding the above impedance equation to give: The modulus of the impedance is then: and the phase is given by: (b) As can be seen in the above derivation, the amplitude of the pressure reflection coefficient is . The sound power *R p * ' B A ' 10 SWR/ 20 & 1 10 SWR/ 20 % 1 reflection coefficient is defined as and the absorption coefficient *R p * 2 is given by . α ' 1 & *R p * 2 (i) From part (a), = . A B ' 10 SWR/ 20 % 1 10 SWR/ 20 & 1 10 0.21 % 1 10 0.21 & 1 ' 4.216 θ = 2kx - π, where k = 2π/λ. x = 0.4λ, thus θ = 4π × 0.4 - π = 0.6π cosθ = -0.309 and sinθ = 0.951 Thus from the preceding equations: and the phase is: Fundamentals 57 I ' 1 2 Re p T u ( T ' 1 2ρc Re (A % Be jθ ) × (Be &jθ & A) ' 1 2ρc Re B 2 & A 2 & 2jABsinθ ' 1 2ρc (B 2 & A 2 ) A % B ' 2p ref 10 L p / 20 ' 0.0894 (ii) The normal incidence sound power reflection coefficient is given by: *R p * 2 ' * 10 SWR/ 20 & 1 10 SWR/ 20 % 1 * 2 ' 4.216 &2 ' 0.056 (iii) The absorption coefficient is given by α ' 1 & *R p * 2 ' 0.94 (iv) The intensity is given by: The maximum sound pressure level is 70dB. Thus: Using A = 4.216B, we have 5.216B = 0.0894. Thus B = 0.0171 and A = 0.0723. Thus the sound intensity is: I = (2×413.7) -1 × (0.0171 2 - 0.0723 2 ) = -5.96×10 -6 W/m 2 . The negative sign indicates that the net intensity is in the negative x- direction. The intensity will not vary along a lossless tube as the sound intensity is a vector quantity which in this case is the vector sum of the intensities in the left and right going waves which is independent of tube location for a lossless tube. This can be verified by using the expressions for p T and u T derived in part (a). (v) Let successive minima be located at x 1 and x 2 . Then using the equation derived in part (a) for the total pressure, setting θ = 2kx, and equating the pressures at x 1 and x 2 we obtain . For this to (A % B)e jkx 1 ' (A % B)e jkx 2 be true, . Thus , and , which e jk(x 2 & x 1 ) ' 1 k(x 2 &x 1 ) ' &π x 1 & x 2 ' λ/ 2 implies that the minima are separated by half a wavelength. At 200Hz, this is equal to 343/(2 × 200) = 0.86m. Problem 1.50 (a) Assuming no losses in the tube, the intensity of a single plane wave Solutions to problems 58 I ' 1 ρc ¢p 2 % ¦ & ¢ p 2 & ¦ u ' & 1 ρ M Mx m p dt propagating in any one direction is , where the amplitude of p as ¢ p 2 ¦ / ρc well as its r.m.s. value is independent of axial location. Thus the intensity of the positive going wave may be written as and the ¢ p 2 % ¦ / ρc negative going wave as . As intensity is a vector quantity, the ¢ p 2 & ¦ / ρc intensities of the positive and negative going waves can be combined by adding the intensity of the positive going wave to the negative value of the negative going wave to give: which is independent of location. (b) The intensity cannot be measured with a single microphone because it cannot distinguish between the pressures associated with the two wave components travelling in opposite directions. (c) Two identical microphones can be used because intensity is also the time averaged product of the acoustic pressure and particle velocity and from equations 1.6 and 1.7 in the text we can show that the acoustic particle velocity is related to the acoustic pressure gradient by: where p 1 and p 2 are the pressures measured by microphones 1 and 2 respectively. The above equation may be rewritten as equation 3.21 in the text. Equations 3.21 and 3.22 may then be used to write equation 3.23 which is an expression for the intensity as a function of the measurements made by microphones 1 and 2. Equation 3.23 needs a slight modification as in this case (replacement of n with 1) the intensity is in the positive x-direction down the tube rather than in an arbitrary direction n. Fundamentals 59 R p ' Z s & ρc Z s % ρc α ' 1 & *R 2 p * ' 1 & *Z s & ρc* 2 *Z s % ρc* 2 ' *Z s % ρc* 2 & *Z s & ρc* 2 *Z s % ρc* 2 α ' [ Re{Z s } % ρc] 2 % [ Im{Z s } ] 2 & [ Re{Z s } & ρc] 2 & [ Im{Z s } ] 2 *Z s % ρc* 2 ' [ Re{Z s } ] 2 % 2ρcRe{Z s } % (ρc) 2 & [ Re{Z s } ] 2 % 2ρcRe{Z s } & (ρc) 2 *Z s % ρc* 2 ' 4ρcRe{Z s } *Z s % ρc* 2 Problem 1.51 (a) and Rearranging gives: (b) From the above expression it can be seen that the maximum value of α is 1 which would occur when Z s = ρc. Problem 1.52 (a) The sound power reflection coefficient is simply the square of the modulus of the pressure amplitude reflection coefficient. Referring to equation 5.129 in the text, the required result can be obtained by allowing θ = 0 (normal incidence) and substituting ρ 2 c 2 for Z m and ρ 1 c 1 for ρc. Note that if θ = 0, then sinθ = 0 and from equation 5.130, cosψ = 1. (b) Again, referring to equation 5.129 in the text, the required result can be obtained by substituting ρ 2 c 2 for Z m and ρ 1 c 1 for ρc. The angle ψ is defined in equation 5.130, with the substitutions, k 1 = k and k 2 = k m . Solutions to problems 60 p T ' p I % p R ' A I e jωt e &j(k x x & k y y) & e &j(k x x % k y y) ' A I e jωt e &jk x x e jk y y & e &jk y y ' 2j A I sinky e jωt R p ' ρ w c w & ρc ρ w c w % ρc ' ρ w c w ρc & 1 ρ w c w ρc % 1 Problem 1.53 (a) As a result of the property of zero pressure at the pressure release boundary, we have at the boundary (y = 0), p R = -p I . Thus using equations 5.118 and 5.119 in the text and setting y = 0, we obtain A R = - A I . As the wave is propagating along the y-axis, θ = 0, k y = k and k x = 0. Thus the total pressure (adding the time dependent term, ) is given by: e jωt (b) For normally incident sound (and with Z m = ρ w c w ), equation 5.129 in the text can be written as: From the preceding equation it can be seen that if ρ w c w >> ρc, then R p = 1 and the reflected wave amplitude will be equal to the incident wave amplitude. For an air/water interface, ρ w c w = 1026 × 1500 = 1.54 × 10 6 which is much greater than ρc = 413.7 thus satisfying the required condition. (c) Using equations 1.6.and 1.7 in the text and the result of part (a), the acoustic particle velocity amplitude is 2A i /ρc. Using equations 1.6, 1.7 and 5.118, the acoustic particle velocity amplitude (normally incident wave) is A i /ρc. Thus the total particle velocity amplitude is twice the incident wave particle velocity. Problem 1.54 (a) Referring to the analysis on pages 210 – 212 in the text, for normal Fundamentals 61 R p ' 830 & ρc 830 % ρc ' 830 & 413.6 830 % 413.6 ' 0.335 P i ' 2 × 10 &5 × 10 60/ 20 ' 0.02Pa ρ 2 c 2 cosθ & ρccosψ ' ρ 2 c 2 cosθ % ρccosψ cosψ ' 1 & c 2 2 c 2 sin 2 θ 1/2 ρ 2 c 2 cosθ & ρc 1 & c 2 2 c 2 sin 2 θ 1/2 ' ρ 2 c 2 cosθ % ρc 1 & c 2 2 c 2 sin 2 θ 1/2 incidence, θ = 0. Thus, from equation 5.130, ψ = 0, and equation 5.129 becomes (with Z m = ρ 2 c 2 = 830): The r.m.s. amplitude of the incident wave is given by: Thus the amplitude of the reflected wave is: P r ' 0.335P i ' 0.0067Pa (b) When all the energy is reflected, , Thus: *R p * ' 1 Using Snell's Law, . Thus: sinψ ' c 2 sinθ c Substituting this result into the previous equation gives: This is true only if (c 2 /c)sinθ = 1, or θ = sin -1 (c/c 2 ) which is the angle above which all incident energy will be reflected. Problem 1.55 (a) Referring to the analysis on pages 210 – 212 in the text and substituting Solutions to problems 62 p T p I ' A T A I ' 2cosθ/ (ρc) cosθ ρc % cosψ ρ w c w ' 2ρ w c w cosθ ρ w c w cosθ % ρccosψ τ ' 4ρcρ w c w cos 2 θ (ρ w c w cosθ % ρccosψ) 2 ' (1 & R 2 p ) cosθ/ cosψ u u ¢ µc µ w w c y x r R r I r T incident wave reflected wave transmitted wave air water ρ w c w for Z m in equation 5.129, the required result is obtained. (b) The transmission coefficient, τ, is an energy related quantity and is defined by: τ ' *p T * 2 *p I * 2 ρc ρ w c w ' *A T * 2 *A I * 2 ρc ρ w c w Using equation 5.121 and 5.127 in the text and substituting Z 1 = ρc and Z 2 = ρ w c w , we obtain: Thus: (c) For air, ρc = 413.6 and for sea water ρ w c w = 1.026 × 10 3 × 1500 = 1539 × 10 3 . For θ = 10E, cosθ = 0.9848 and sinθ = 0.1736. Using Snell's Law, Fundamentals 63 R p ' 1.539 × 10 6 × 0.9848 & 413.6 × 0.6506 1.539 × 10 6 × 0.9848 % 413.6 × 0.6506 ' 0.99964 τ ' (1 & R 2 p )cosθ/ cosψ ' 0.0011 ρ w c w cosθ & ρccosψ ' ρ w c w cosθ % ρccosψ cosψ ' 1 & c 2 w c 2 sin 2 θ 1/2 ρ w c w cosθ & ρc 1 & c 2 w c 2 sin 2 θ 1/2 ' ρ w c w cosθ % ρc 1 & c 2 w c 2 sin 2 θ 1/2 . Thus ψ = 49.4E and sinψ ' c w sinθ c ' 1500 × 0.1736/ 343 ' 0.7594 cosψ = 0.6506. Substituting these values into the result of part (a) gives: (d) When all the energy is reflected, , Thus *R p * ' 1 Using Snell's Law, . Thus, sinψ ' c w sinθ c Substituting this result into the previous equation gives This is true only if (c w /c)sinθ = 1, or θ = 13E, which is the angle above which all incident energy will be reflected. (e) If the sound source were a point source, the amount of sound power entering the water would be independent of altitude (except for atmospheric losses) as the same amount of power is contained in any specified included angle. However, the power would be distributed over an ever increasing area of ocean surface. A distributed sound source Solutions to problems 64 ¯ u I ' A I ρc ( sinθ & cosθ) e &jk 1x x ¯ u R ' A R ρc ( sinθ % cosθ) e &jk 1x x ¯ u R ¯ u I ' A R A I ( sinθ % cosθ) ( sinθ & cosθ) p u ' ρc jkr 1 % jkr p u . ρcjkr p ' ρc 0 ukr ω ' ρ 0 ur such as a helicopter would exhibit similar behaviour. (f) Velocity reflection coefficient. Using equations 1.6, 1.7, 5.115, 5.116, 5.118 and 5.119 in the text, the acoustic velocity amplitude of the incident wave at y = 0 is: and the velocity amplitude of the reflected wave is: The velocity amplitude reflection coefficient is the ratio of reflected to incident velocity amplitudes. Thus: Problem 1.56 (a) Assuming that the bubble is a spherical source, the specific acoustic impedance is given by equation 1.43 in the text as: At the bubble surface, kr << 1, so: Acceleration is defined as , so: 0 u ' jωu Fundamentals 65 γ dV V % dP P ' 0 p P ' &γ dV V ρ 0 ur P ' &γ dV V 0 u ' & γP ρr dV V 0 u ' juω dV ' 4πr 2 u jω ; V ' 4 3 πr 3 j uω ' & γP4πr 2 u ρrjω(4/ 3)πr 3 ' & 3γPu ρr 2 jω ω 2 ' 3γP ρr 2 ' 3c 2 r 2 f res ' ω 2π ' c 3 2πr (b) Adiabatic compression PV γ ' const or P ' const × V &γ Differentiating P with respect to V and rearranging gives: dP . p, where p is the acoustic pressure. Thus: Substituting the result for p from part (a) gives: or (c) Resonance frequency is the frequency at which the bubble prefers to vibrate given the physical parameters. Substituting for dV and V in the expression of part (b), we obtain: Rearranging gives: Thus: (d) At resonance, the bubble screen should act like a Helmholtz resonator (see Ch. 9) and remove considerable energy from the sound field. ± 28 × 10 &9 / 9.81 ' ±2.87 × 10 &9 m 2 Solutions to problems relating to the Human Ear Problem 2.1 (a) Weight of a column of water equivalent to the weight of a column of atmosphere of cross sectional area 1m 2 is equal to 101.4kN, which is equal to 101400/9.81 kg. Density of water = 1000kg/m 3 , thus volume of water = 101.4/9.81 m 3 and thus height of water column = volume/(1×1)= 10.34m. Minimum audible sound = 0dB = 20 × 10 -6 Pa r.m.s = 28.2 × 10 -6 Pa peak. This corresponds to a variation in water height of (b) 120dB represents an increase in pressure by a factor of 10 6 over 0dB, so variation in column height would be . ±2.9mm (c) Figure 2.9 indicates an increase of 60dB for 31.5Hz sound which represents a factor of 10 60/20 = 1000. (d) See p56, text. Overall mechanical advantage = 15:1 = 23.5dB sound pressure level. Linkage mechanical advantage = 3:1 = 9.5dB. Problem 2.2 (a) A scaling of physical dimensions by a factor of 10 would mean that the mouse's range of hearing is 200Hz to 180kHz. (b) Figure 3.4 and equations 3.14 and 3.16 indicate that the sensitivity is proportional to d 4 , so the mouse's ear should be 4 orders of magnitude (or 40dB) less sensitive. The human ear 67 (c) The mouse's transduction mechanism must be 4 orders of magnitude more sensitive than the human mechanism. (d) Yes, because the differences would be necessary to make the mouse transduction mechanism more sensitive. (e) The differences could take the form of a larger mechanical advantage in the mouse's middle ear as well as differences in the relative physical dimensions of the inner ear. Referring to possible inner ear differences, a critical component of the inner ear in regard to sensitivity is the hinge mechanism of the tectorial membrane. The sensing of sound by the inner and outer hair cells is by a shearing action imposed on the hair cell stereocelia caused by relative movement between the tectorial membrane and the rods of Corti. To increase the sensitivity, it would be necessary to increase this relative movement which could be achieved by lengthening the rods of Corti. Referring to the middle ear differences, a much smaller oval window and a different arrangement of the bone linkage could account for a large sensitivity increase. Problem 2.3 (a) Sound introduced to the ear using ear muffs will be fairly reverberant, having frontal as well as lateral components; thus, we would expect the MAP to be lower than the MAF. If the earmuffs distort the pinna sufficiently, the MAP may approach the MAF in magnitude. (b) As the sound field within the ear muffs is similar to a diffuse field, one would expect the MDF to be similar to the MAP and less than the MAF. Solutions to problems 68 Problem 2.4 The first symptoms of noise induced hearing loss are difficulties with understanding conversation in a noisy environment, in focussing on the speaker and in localising noise sources. These symptoms are caused by the breaking off of stereocelia on the outer hair cells leaving only the inner hair cell stereocelia functional. The loss usually occurs first in the 4kHz range and then extends to higher and lower frequencies. A conventional hearing aid which amplifies all frequencies by the same amount will not be much help as it will only amplify the "noise" experienced and not help with the symptoms mentioned above. A frequency selective hearing aid will do a little better by amplifying the signal at the frequencies most affected but it is difficult to see how even this type of hearing aid will ameliorate the above mentioned symptoms of early noise induced hearing loss. Problem 2.5 Not necessarily. As stated in the text, repeated exposures to noise which results in a temporary threshold shift will result in permanent damage. On the other hand, there is evidence that damage is accumulative, so even one exposure will contribute to the eventual permanent damage. Problem 2.6 Pitch: Determined by location on the basilar membrane which responds most to the noise. Also the hair cells are tuned to maximum output at frequencies corresponding to the resonance frequencies of the parts of the basilar membrane to which they are attached. However, at low frequencies and loud noise, some neurons will fire for hair cells all along the basilar membrane and a second mechanism by which neurons fire in locked phase with the acoustic signal (at acoustic signal maxima or once per cycle) dominates the pitch determination. As the frequency increases, the localisation of the basilar membrane excitation (and associated resonant hair cells) becomes increasingly important as the method for determining pitch until at 5000Hz the locked phase phenomenon of neuron firing ceases altogether and the neurons fire randomly. The human ear 69 Loudness: Determined by the rate of neuron firing, which is controlled by the motion of the hair cell stereocelia which in turn is controlled by the basilar membrane. There is also a hair cell feedback mechanism whereby the voltage generated by the hair cells causes them to deform, thus increasing the movement of the tectorial membrane. This feedback mechanism effectively increases the dynamic range of the hearing mechanism and also results in improved pitch resolution. Problem 2.7 (a) Two signals may have exactly the same spectral content and thus sound the same, but they may have entirely different phase relationships between the particular spectral components making up the signal. Thus the time histories as seen on an oscilloscope could look quite different. Both amplitude and relative phase of the spectral components of a signal are necessary to reconstruct a signal uniquely. However, the ear discards phase information. (b) In 0.05 seconds sound will travel a distance of 17.15m. This corresponds to a wavelength at 20Hz. Thus for a 20Hz tone, the peak sound pressures will occur at intervals of 0.05 seconds. Below this frequency there will be greater intervals between the peak sound pressures and the sound will be heard as a sequence of auditory events. (c) The dimensions of an auditorium must be such that the sound arriving at any location after being reflected from a wall, ceiling or floor must not arrive longer than 0.05 seconds after the direct sound. This means the difference between direct and reflected paths should be less than 17m. Problem 2.8 (a) A low frequency warning device would be more effective mainly because it is not as easily masked by the 500Hz noise as higher frequencies would be but also because it would diffract more effectively around obstacles to create a more uniform coverage. (b) See fig 2.10(a) in text and read off values as accurately as possible. Solutions to problems 70 Octave band centre frequency 63 125 250 500 1k 2k 4k 8k SPL forward 40 44 49 54 56 55 45 33 SPL side 39 43 47 51 53 49 39 27 SPL rear 38 42 44 48 49 44 27 15 Sones forward 0 0.6 1.7 2.8 3.8 4.3 2.8 1.6 Sones side 0.0 5 0.5 5 1.4 2.4 3.2 3 2 1.1 Sones rear 0.0 2 0.5 1.1 2 2.5 2.2 0.9 0.3 Phons forward 0 33 48 55 59 61 55 46.8 Phons side 0 33 45 53 57 55.8 50 41.4 Phons rear 0 30 41 50 53 51.4 39 22.6 Overall Levels: Sones Forward = 4.3 + 0.3[0.0+0.6+1.7+2.8+3.8+2.8+1.6] = 8.3 Sones Side = 3.2 + 0.3[0.05+0.55+1.4+2.4+3.0+2.0+1.1] = 6.3 Sones Rear = 2.5 + 0.3[0.02+0.5+1.1+2.0+2.2+0.9+0.3] = 4.7 Using equation 2.1, Phons = 40 + (10log 10 S)/(log 10 2). Thus, Phons forward = 70.5 Phons side = 66.6 Phons rear = 62.3 The human ear 71 (c) Yes, a person with no hearing loss would be able to hear, as the levels in the bands important for speech recognition are well above the hearing threshold level or MAF. (d) If the distance increases to 10m from 2m, the sound pressure level in all bands will decrease by 20log 10 (10/2) = 14dB (assuming free field conditions). The MAF from figure 2.9 and the new sound pressure levels are in the table below. Octave band centre frequency (Hz) 63 125 250 500 1k 2k 4k 8k MAF 39 22 15 9 4 0 0 12 New SPL forward 26 30 35 40 42 41 31 19 New SPL side 25 29 33 37 39 35 25 13 New SPL rear 24 28 30 34 35 30 13 1 Difficulty in hearing sound in the 63Hz and 8kHz bands would be encountered but as these bands are not important for speech, it is expected that there would be no difficulty in understanding speech for any head orientation. (e) Speech recognition is just possible for S = 8 Sones. A speaker speaking twice as loudly will increase the rear level from 4 to 8 sones, making speech recognition just possible again, so the situation will be improved. Problem 2.9 See figure 2.9(a). Masking tone is 800Hz at 60dB. From the figure, the threshold of detection of an 630Hz tone would be increased by 20dB. Figure 2.5 indicates that the MAF at 630Hz is approximately 5dB, so the sound level would need to be 25dB in order to be heard. 3 Solutions to problems relating to instrumentation and measurement Problem 3.1 (a) Pressure response: microphone response when subjected to a uniform pressure field which is accomplished in practice electrostatically. Free-field response: microphone response when subjected to a sound wave coming from a specified direction in an otherwise free field (free of reflected sound). Random incidence response: microphone response averaged over all possible angles of sound wave incidence. Microphones are designed so that the roll-off in pressure response at high frequencies is just compensated for by the increase in free field pressure due to diffraction of a normally incident wave (normal incidence microphone) or by the increase in pressure due incident waves averaged over all possible directions of incidence (random incidence microphone). (b) Random incidence microphones are used to measure noise in reverberant test chambers, reverberation times in auditoria and industrial noise when the direction of origin is uncertain or there are a number of sources located in various directions from the observer. Free field microphones are used in anechoic test chambers and in industrial noise measurement cases where the origin of the noise is from a single direction or a narrow direction angle. Instrumentation and measurement 73 L p ' 20log 10 E & S % 94 ' &110 % 26 % 94 ' 10dB L p ' 10log 10 10 1.3 & 10 1.0 ' 10dB Problem 3.2 (a) Using equation 3.16, the sound level represented by the noise floor on the instrument is: If the sound level meter actually reads 13dB, then the contribution due to the actual noise is: (b) The ear can hear 0dB of frontally incident sound at 2kHz and it can probably discern signals just above its noise floor. So the equivalent sensitivity of the sound level meter would be approximately 10dB, implying that the ear is 10dB more sensitive. Problem 3.3 (a) Sensitivity = -25dB re 1V per Pa. Using equation 3.15, we have: -25 = 20log 10 [E/p] which gives a sensitivity of 10 (-25/20) volts/Pa which = 56.2 mV/Pa. (b) The pressure response of the microphone would have to be 4.5dB less at 10kHz than at 250Hz for the overall response to remain flat; that is, - 29.5dB re 1V per Pa. (c) The microphone overall sensitivity for a 0E incident field at 10kHz is -29.5 + 5 = -24.5dB. (d) The microphone overall sensitivity for a reverberant field at 10kHz is -29.5 + 1.5 = -28.0dB. (e) At 250Hz, it can be seen from figure 3.3 in the text that the correction for a 180E angle of incidence would be 0dB. Thus the overall microphone sensitivity at 250Hz would be equal to -24.5dB, the pressure sensitivity. At 10kHz, the overall sensitivity is: -29.5 - 0.2 = -29.7dB. Solutions to problems 74 Problem 3.4 The MAF is the minimum audible field to frontally incident sound and would correspond to the free field calibration of a microphone. Similarly the MDF would correspond to the diffuse field calibration of a microphone. The pressure calibration of a microphone would correspond to the MAP which is the minimum pressure audible at the tympanic membrane of the ear. Problem 3.5 The random incidence microphone is used in cases where one is not sure of the direction from which the sound is coming or if it is coming from a number of directions simultaneously. To minimise measurement error, the microphone would be held vertically, thus resulting in most sound being incident at angles close to 90E. The difference between the microphone diffuse field response and the 90E response is much smaller than the error resulting from pointing a free field microphone in the wrong direction (see figure 3.3 in text). In both cases measured sound pressure levels will be less than actual levels. Problem 3.6 This would be similar to connecting the microphone to ground through a low impedance and this would seriously reduce the microphone sensitivity, resulting in large measurement errors. Problem 3.7 The A-weighted sound pressure level is related to loudness perception of low level environmental noise as well as to hearing damage (although not to loudness perception of loud industrial noise), and regulations are written in terms of this quantity due to its ease of measurement by unskilled people. It is a reasonably valid measure of noise exposure as there is a direct correspondence between A-weighted sound level and hearing loss suffered by noise exposed people. Instrumentation and measurement 75 The advantages of the A-weighted level for characterising equipment and workplaces are its direct relationship to noise exposure and its single number simplicity. However it does not provide the frequency content information necessary for effective noise control measures to be specified and this is the main disadvantage. Problem 3.8 (a) The A-weighted levels are calculated by adding the A-weighting corrections (most are negative) to each octave band level and then logarithmically adding the results as described on page 48 of the text. The overall linear level is calculated by adding the values given in the problem together logarithmically as described on page 48 of the text. The answers are: A-weighted = 89.4dB(A) and Linear = 98.5dB. (b) The main source of error is a result of the assumption that all frequencies in each octave band can be weighted by a single quantity (the weighting corresponding to the band centre frequency) when in fact the A- weighting is a smoothly varying function of frequency. The maximum possible error can be estimated by comparing the results using the A- weighting corresponding to the band centre frequency with results obtained bu using the A-weighting corresponding to the upper and lower band limit frequencies. Problem 3.9 Plot out the values given in Table 3.1 on graph paper using a logarithmic frequency scale. Join all of the points by a straight line which is as good as a smooth curve for the present purposes. Then plot the upper and lower octave band frequency limits from table 1.2 and read off from your graph the A-weighting corrections at these frequencies The difference between these values and the octave band centre frequency A-weighted values represent the largest errors which could occur if all of the energy just happened to be at frequencies at the edge of each octave band. To find the overall possible variation calculate the dB(A) levels for each extreme and compare them to the overall level calculated using band centre frequency corrections as illustrated in the table on the next page. Solutions to problems 76 Combining the band levels together as described on p 38 in the text gives: Upper limit = 84.7dB(A) Lower limit = 83.3dB(A) Centre value = 84.1dB(A) (corresponding to A-weighted corrections at band centre frequency. The A-weighted levels peak at high frequencies so the noise would sound a little "hissy". Octave band centre freq. (Hz) 63 125 250 500 1000 2000 4000 8000 Upper f limit 88 176 353 707 1,414 2,825 5,650 11,300 Lower f limit 44 88 176 353 707 1,414 2,825 5,650 Upper dB(A) adj. -21 -12.2 -5.8 -1.2 1.2 1.3 1.1 0.0 Centre dB(A) adj. -26.2 -16.1 -8.6 -3.2 0.0 1.2 1.0 -1.1 Lower dB(A) adj. -35 -21 -12.2 -5.8 -1.2 1.1 0.0 -3.5 SPL 76 71 68 70 73 76 79 80 Upper dB(A) 55 58.8 62.2 68.8 74.2 77.3 80.1 80.0 Centre dB(A) 49.8 54.9 59.4 66.8 73 77.2 80 78.9 Lower dB(A) 41 50 55.8 64.2 71.8 77.1 80.0 76.5 Instrumentation and measurement 77 Problem 3.10 First remove the background noise contribution from each octave band measurement as described in example 1.4, p.49 in the text. Then arithmetically add (some numbers are negative) the A-weighted corrections to each octave band level as was done for problem 3.9. Then add the A- weighted octave band levels together logarithmically as described on p.48 in the text. The final answer is 89.0dB(A). Problem 3.11 (a) A-weighted level = 86.8dB(A) (b) It is not a good way to calculate overall weighted levels because errors arise from the inherent assumption that the A-weighting is uniform across any particular octave band. Problem 3.12 To begin, assume an arbitrary level of 50dB in the 63Hz octave band, calculate the A-weighted levels in all bands, the overall A-weighted level and thus the amount to add to each band level to reach an overall level of 105dB(A). The calculations are summarised in the following table, where it is noted that a constant spectrum level is reflected in octave band levels increasing at the rate of 3dB per octave (reflecting a doubling of bandwidth per octave). Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 total Un-weighted level 50 53 56 59 62 65 68 71 A-weighted level 23.8 36.9 47.4 55.8 62 66.2 69 69.9 73.8 Adjustment needed 31.2 31.2 31.2 31.2 31.2 31.2 31.2 31.2 Unweighted level for 105dB(A) 81.2 84.2 87.2 90.2 93.2 96.2 99.2 102.2 105.0 Solutions to problems 78 Problem 3.13 Although the signal levels in the experiments to determine equal loudness contours varied substantially, the A-weighted scale corresponds approximately to the loudness contour of 60dB. As industrial noise is usually much louder than this and equal loudness contours for high sound levels do not have the same shape as those at 60dB, it is unlikely that the A-weighted scale will indicate correct loudness levels for most industrial noise. Problem 3.14 (a) If the observer is too close to the microphone when noise measurements are being taken, then reflections from the observer can affect the noise levels being measured. In tonal sound fields, the reflections could result in an increase in the measured sound level of up to 5dB (and also decreases) and in broadband sound fields, the increase measured could be up to 2dB. The reasons for the above numbers not being 6 and 3dB respectively is because the observer will absorb and scatter some of the noise while reflecting it. (b) The fast response (0.1s time constant) of the sound level meter approximates the way the ear hears but the slow response (1.0s time constant) is useful for determining L Aeq levels (average of the upper swings of the needle on an analog meter) and L 90 levels (average of lower meter swings). However most modern instrumentation allows direct digital readout of these quantities. In addition, most legislation is written in terms of measurements taken using the "slow" response as some researchers suppose that this is more representative of the hearing damage caused by the noise. (c) The "frontal/diffuse" control is used for selecting the microphone characteristic most suitable for the measurement being undertaken. Note that the microphone remains unchanged - the electronics in the sound level meter effectively change its characteristics to compensate for the different diffraction effects of the microphone grid for each of the two types of field. Instrumentation and measurement 79 ' 1 8 8 5 5 % 4 × 8 4 % 24 × 8 3 % 32 × 8 2 % 16 × 8 × 10 &4 ' 0.4675 L eq ' 10log 10 p 2 p 2 ref ' 10log 10 0.4675 4 × 10 &10 ' 90.7dB L Aeq ' 10log 10 1 1/4 % 2 % 2 % 1/12 % 4 × × 0.25 × 10 8 % 2 × 10 7 % 2 × 10 9 % (1/12) × 10 9.9 % 4 × 10 7.5 Problem 3.15 , thus p ' (t 2 % 8t % 4) × 10 &2 p 2 ' (t 4 % 16t 3 % 72t 2 % 64t % 16) × 10 &4 Average p 2 ' m 8 0 (t 4 % 16t 3 % 72t 2 % 64t % 16) × 10 &4 dt A-weighting at 250Hz is -8.6dB, so L Aeq = 82.1dB(A). Problem 3.16 L Aeq is generally used to describe noise as it is an A-weighted energy average which seems to be related to loudness perception of low level environmental noise as well as to hearing damage (although not to loudness perception of loud industrial noise), and regulations are written in terms of this quantity due to its ease of measurement by unskilled people. = 85.3dB(A). Problem 3.17 A sound level meter on site is preferable, as a tape recorder is not sufficiently accurate for legal disputes (see table 3.3 in the text). Problem 3.18 Sources of measurement error: Solutions to problems 80 1m 1m 1m possible wind screen configuration shade cloth Microphone vibration; SLM vibration; background noise; overloading input amplifier when taking octave or 1/3 octave band measurements with an old SLM; too cold or too hot; moisture or dust on microphone diaphragm; reflections from nearby surfaces; and wind noise. Can minimise effects of wind noise by placing a foam wind shield on the microphone AND placing the mic in an enclosure made using shade cloth as shown in the figure below. Problem 3.19 (a) See text, p114 B 120. (b) (i) At low frequencies, errors arise because the phase difference between the two microphone signals (due to the spacing being small compared to a wavelength) is not sufficiently large compared to the phase accuracy of the microphones. (ii) At high frequencies, errors arise because the microphone spacing becomes significant compared to a wavelength Instrumentation and measurement 81 p rms ' p ref 10 L p / 20 ' 2 × 10 &5 × 10 95/ 20 ' 1.12Pa causing the finite difference approximation for the pressure gradient to be inaccurate. (iii) In very reactive sound fields, the phase between the acoustic pressure and particle velocity is close to 90E, so any phase errors translate to a large error in the intensity as it is proportional to the cosine of the phase angle. (iv) In the presence of external noise sources, sound power measurements could exhibit significant errors if the sound pressure level of the external noise is sufficiently high (usually about 10dB or more above the level from the noise source being measured). This is because the power measurement relies on averaging normal intensity measurements and the result of the external source will be to create a situation where small differences between large numbers will dominate the result. Problem 3.20 (a) See part (a) in previous question. (b) See part (b) in previous question. (c) Applications include: sound power measurement; localisation and identification of noise sources; sound transmission loss measurement; determining the importance of flanking sound transmission paths in noise control applications; Problem 3.21 (a) Sound pressure associated with 95dB sound level is Force on microphone diaphragm is Solutions to problems 82 F rms ' 1.12 × π × (0.012) 2 / 4 ' 1.27 × 10 &4 N p uA ' & jρc 2 Vω p rms ' 2 × 10 &5 × 10 65/ 20 ' 3.557 × 10 &2 u rms ' pωV ρc 2 S ; ( S is the area of the microphone diaphragm) ' 3.557 × 10 &2 × 2π × 500 × 0.01 × 0.02 2 × (π/ 4) 1.206 × 343 2 × 0.012 2 × (π/ 4) ' 2.19 × 10 &5 m/ s Vol.displ.ampl ' 2u rms S ω ' 2 × 2.188 × 10 &5 × π × 0.012 2 4 × 2 × π × 500 ' 1.11 × 10 &12 m 3 (b) r.m.s. velocity of the diaphragm is the same as the air particle velocity. From equation 9.35 in the text, the ratio of the sound pressure to particle velocity in a cavity of dimensions much smaller than a wavelength is The sound pressure measured by the monitoring microphone is 65dB which corresponds to an r.m.s.pressure of Thus the particle velocity is (c) The volume displacement in the cavity corresponding to a sound pressure level of 65dB can be calculated using the same equation as used in part (b). Thus the volume displacement is (d) Mechanical input impedance, Z m = F/u. Thus Instrumentation and measurement 83 Z m ' 1.27 × 10 &4 2.188 × 10 &5 ' 5.8 N&s/ m (e) The volume displacement of the monitoring microphone will be 30dB (95 - 65) below the displacement of the test microphone. This represents a percentage difference of 100/10 1.5 = 3% which will not affect the sound pressure sensed by the test microphone significantly. (f) Upper test frequency is limited by the onset of resonant cavity modes. As the largest dimension is the radius, cross modes will occur before axial modes. This is explained in Chapter 7 in the text. T a ' 8 × 2 &(99 & 90) / 3 ' 1 hour T a ' 8 × 2 &(99 & 90) / 5 ' 2.3 hours 4 Solutions to problems relating to criteria Problem 4.1 (a) Using equation 4.42 in the text, the allowable exposure time using European criteria is (b) The allowable exposure using USA criteria is Problem 4.2 (a) A-weighted SPL is given by: L pA ' 10log 10 ( 10 (9.5 & 0.86) % 10 (9.7 & 0.32) % 10 9.9 ) ' 100 dB(A) (b) Allowed daily exposure time in Australia is: T a ' 8 × 2 &( 100 & 90) / 3 ' 0.8 hours Allowed daily exposure time in USA is: T a ' 8 × 2 &( 100 & 90) / 5 ' 2 hours Problem 4.3 Fan noise = 91dB(A) Saw idling noise = 88dB(A) Saw cutting noise = 93dB(A) Let required fan noise for L Aeq,8h = 90dB(A) be x dB(A). Criteria 85 90 ' 10log 10 1 8 6.4 10 88/ 10 % 10 x/ 10 % 1.6 10 93/ 10 % 10 x/ 10 10 x/ 10 ' 9.618 × 10 7 90 ' 16.667log 10 1 8 6.4 10 0.3 × [ 10log 10 (10 8.8 % 10 x/ 10 ) & 90] / 5 % 1.6 10 0.3 × [ 10log 10 (10 9.3 % 10 x/ 10 ) & 90] / 5 % 90 (a) European criteria Using equation 4.3 or 4.39 with L B = 90 and L = 3: Solving for x gives: Thus, x = 79.8dB(A) The required fan noise reduction is then 91 - 79.8 = 11.2dB(A) (b) USA criteria Using equation 4.39 with the integral replaced with a sum and with with L B = 90 and L = 5, we obtain: In solving for x, we must remember the proviso that combined fan and saw noise levels of less than 90dB(A) at any time do not contribute to the noise exposure results in a value of x as close to 90dB(A) as possible. Assuming a precision of 0.1dB(A), the allowed fan noise plus saw idle noise is 89.9dB(A). This is because if the fan noise plus saw idle noise is greater than 90dB(A), the overall is greater than 90dB(A). Thus L ) Aeq the maximum allowed fan noise is . Thus the required fan noise x ' 10log 10 10 8.99 & 10 8.8 ' 85.4dB(A) reduction is 91 - 85.4 = 5.6dB(A). Problem 4.4 (a) L Aeq,8h = 10Log 10 (1/8)[2 ×10 95/10 + 6 ×10 70/10 ] = 89.0 dB(A) (b) E A,8 ' 32 × 10 (89 & 100) / 10 ' 2.54 Pa 2 @h (c) We may assume that the 70 dB(A) does not contribute significantly, so we Solutions to problems 86 HDI ' 10log 10 m t 0 10 L p / 20 dt 59.5 ' 10log 10 10 110/ 20 × T ' 55 % 10log 10 T L Aeq,8h ' 10log 10 1 8 4.5 × 10 10.5 % 1.5 × 10 9.5 ' 102.6dB(A) L ) Aeq,8h ' 16.667 × log 10 ( 1/ 8) 10 0.3 × (105 & 90) / 5 × 4.5 % 10 0.3 × (95 & 90) / 5 × 1.5 % 90 ' 101.4dB(A) (USA criteria) T a ' 6 × 2 &(102.6 & 90) / 3 ' 0.33 hours need to find the allowed exposure to 95 dB(A). This is given by T a ' 8 × 2 &( 95 & 90) / 3 ' 2.52 hours (d) SPL due to machine only is: 10log 10 [ 10 9.5 & 10 9.1 ] ' 92.8 dB(A) Problem 4.5 In this case, where T is the number of years to cross the hearing loss criterion. Thus, T . 3 years, and he will be 23 years old before he joins the old folks (assuming that he is in the 20% more sensitive part of the population). Problem 4.6 (a) Using equation 4.3 in the text: (European criteria) Using equation 4.41 in the text with the integral replaced with a summation sign: (b) European criteria Using equation 4.42 in the text, Criteria 87 T a ' 6 × 2 &(101.43 & 90) / 5 ' 1.23 hours HDI ' 10log 10 j i T i × 10 L pi / 20 '10log 10 5 × 10 8.5/ 2 % 3 × 10 9/ 2 % 6 × 10 9.5/ 2 % 1 × 10 10/ 2 % 10 × 10 8/ 2 ' 58.6 L ) Aeq,8h ' 16.667 × log 10 1 8 2 × 10 0.3 × ( 91 & 90) / 5 % 2 × 10 0.3 × (96 & 90) / 5 % 90 ' 88.9dB L Aeq ' 10log 10 1 8 2.4 × 10 85/ 10 % 1.6 × 10 88/ 10 % 2 × 10 91/ 10 % 2 × 10 96/ 10 ' 91.8dB(A) USA criteria Problem 4.7 From figure 4.4(b) in the text, there is a 22% risk of developing a 22dB handicap. Problem 4.8 (a) USA criteria Using equation 4.41 with L B = 90 and L = 5, we obtain: Daily noise dose = 2 (L ) Aeq & 90) / 5 ' 0.86 No reduction in exposure time is necessary. (b) European criteria Using equation 4.3 or 4.39 with L B = 90 and L = 3, Solutions to problems 88 T a ' 8 2 (91.8 & 90) / 3 ' 5.2hours Daily noise dose = 2 (L Aeq & 90) / 3 ' 1.53 Allowable exposure time: Thus a reduction of 2.8 hours is required. Problem 4.9 Number of impacts per day = 80 × 60 × 8 × 0.6 = 23,040 B-duration = 100 msec B-duration × number of impacts = BN = 2.3 × 10 6 Peak SPL = 125dB Allowable level for BN = 2.3 × 10 6 is obtained from fig 4.5 in the text. European criteria L a = 112.5dB, and noise dose = 2 (125 - 112.5)/3 = 18 U.S.A. criteria L a = 121dB, and noise dose = 2 (125 - 120.5)/5 = 1.9 Allowable BN for 125dB peak - see fig 4.5 in text. European criteria BN = 1.40 × 10 5 , and noise dose = 23.04/1.4 = 16.4 U.S.A. criteria BN = 1.29 × 10 6 , and noise dose = 2.304/1.29 = 1.8 The small differences in results obtained using the two methods (allowable level vs allowable BN) are due to difficulties in reading the figure any more accurately. The operator is overexposed according to both criteria. Required work day decrease European criteria Assuming that the press accounts entirely for the exposure of the employee, Criteria 89 then from figure 4.6, the allowed BN product is 1.4 × 10 5 . In terms of hours, this is equal to: 1.4 × 10 5 /(80 × 60 × 100) = 0.29 hours of press operation. Accounting for the background noise, the exposure will be controlled by this for a minimum of 7.71 hours (8 - 0.29). This corresponds to a noise dose of (7.71/8) × 2 (85 - 90)/3 = 0.3. Thus the allowable time of press operation is 0.7 × 0.29 = 0.20 hours. [Iterating again does not affect the result significantly]. Thus required workday decrease = 4.8 - 0.2 = 4.6 hours of press operation. USA criteria Background noise of 85dB(A) does not contribute From figure 4.6, allowable BN product = 1.2 × 10 6 . In terms of operating hours, this is equal to: 1.29 × 10 6 /(80 × 60 × 100) = 2.69 hours of press operation. The background noise when the press is not operating does not contribute to the exposure according to USA criteria. Thus required work day decrease = 4.8 - 2.7 = 2.1 hours of press operation. Problem 4.10 Number of impacts per day = 40,000 B-duration = 60 msec B-duration × number of impacts = BN = 2.4 × 10 6 Peak SPL = 135dB Allowable level for BN = 2.4 × 10 6 is obtained from fig 4.6 in the text. European criteria L a = 112dB, and noise dose = 2 (135 - 112)/3 = 200. U.S.A. criteria L a = 121dB, and noise dose = 2 (135 - 121)/5 = 7. Allowable BN for 135dB peak - see fig 4.6 in text. European criteria BN = 1.2 × 10 4 and noise dose = 240/1.2 = 200 Solutions to problems 90 U.S.A. criteria BN = 3.4 × 10 5 and noise dose = 240/34 = 7 Allowable number of impacts USA criteria The background noise of 87dB(A) contributes nothing to the daily noise dose because it is less than 90dB(A). Thus the allowed number of impacts = 3.4 × 10 5 /60 = 5700. Thus required decrease = 40,000 - 5,700 = 34,300. European criteria Assuming that the press accounts entirely for the exposure of the employee, then from figure 4.6, the allowed BN product is 1.2 × 10 4 . The allowable number of impacts is then 1.2 × 10 4 /60 = 200. However this represents such a small part of the 8-hour day, that the 87dB(A) background can be considered to dominate the exposure for almost 8 hours, resulting in a noise dose of 0.5 due to this alone. Thus the allowable dose due to the impact noise is 0.5, which corresponds to 100 impacts. Problem 4.11 Use figure 4.7 in the text. Expected level - "very loud voice to shout" Required level - "shout". Problem 4.12 See figure 4.7 in text. The answers are: No. No. No. Too Loudly. Problem 4.13 The one third octave band levels would have to be first combined into octave band levels by logarithmically adding together three third octave bands for Criteria 91 L p250 ' 10log 10 10 60/ 10 % 10 65/ 10 % 10 63/ 10 ' 67.9 dB each octave band result. The three bands to add would be one with a centre frequency the same as the octave band and one band above and one below that one. For example, if the 200Hz, 250Hz and 315Hz one third octave band levels were 60dB, 65dB, and 63dB respectively, then the 250Hz octave band level would be: Problem 4.14 A-weighted overall sound levels are inadequate for noise level specification and control because they give no indication of the frequency content of the noise which is necessary for assessing annoyance and determining the type of noise control approach which may be feasible. It is preferable to have data as octave band levels for control purposes, although often for specification purposes, NR, NC or RC numbers are adequate as they take into account the spectral content of the noise. However overall dB(A) numbers are adequate for the purposes of assessing hearing damage risk and for comparing noise (either occupational or environmental) with permitted levels according to local regulations. Problem 4.15 (a) A-weighted levels are calculated and tabulated below Frequency (Hz) 63 125 250 500 1k 2k 4k 8k L p (dB re 20µPa) 100 101 97 91 90 88 86 81 A-weighting -26.2 -16.1 -8.6 -3.2 0.0 1.2 1.0 -1.1 A-weighted level 73.8 84.9 88.4 87.8 90 89.2 87 79.9 Solutions to problems 92 L p ' 10log 10 10 7.38 % 10 8.49 % 10 8.84 % 10 8.78 % 10 9 % 10 8.92 % 10 8.7 % 10 7.99 ' 96.1 dB(A) T a ' 8 × 2 &(96.1 & 90) / 3 . 1.95 hours L ' 38 % 0.3[ 28.5 % 35.3 % 28.5 % 33 % 35.3 % 38 % 33] ' 107.5 sones The overall A-weighted level is: (b) Using equation 4.42 in the text, the allowable number of hours is: (c) NR level of noise - plot levels on NR curves as shown below, where it may be seen that NR = 91. (d) Loudness level Frequency (Hz) 63 125 250 500 1k 2k 4k 8k L p (dB re 20µPa) 100 101 97 91 90 88 86 81 Sones 28.5 38 35.3 28.5 33 35.3 38 33 The overall level in sones is calculated using equation 2.33. Thus: Criteria 93 P ' 40 % 10 log 10 S log 10 2 ' 107.5 phons x ' 10log 10 10 96.1/ 10 & 10 ( 96.1 & 1.5) / 10 ' 90.8dB(A) 0 10 20 30 40 50 45 35 15 MAF 63 125 250 500 1k 2k 4k 8k Octave band center frequency (Hz) O c t a v e b a n d s o u n d p r e s s u r e l e v e l ( d B r e 2 0 P a ) µ 25 From equation 2.32: Note that the above equation is inaccurate for levels above 100 phons. (e) Following example 1.4, the contribution of the machine to the overall level is x dB(A) where x is defined as: Problem 4.16 (a) The levels are first plotted on NC and NCB curves The result is NC = 33 and NCB = (38 + 30 + 20 + 16)/4 = 26 Solutions to problems 94 0 10 20 30 40 50 45 35 15 10 0 63 16 125 31.5 250 500 1k 2k 4k 8k Octave band center frequency (Hz) O c t a v e b a n d s o u n d p r e s s u r e l e v e l ( d B r e 2 0 P a ) µ 25 NCB The system is sufficiently quiet for churches holding less than 250 people (see table 4.8 in the third edition of the textbook and table 4.2 in the first edition or look up AS2107-1987), however for larger churches, the level should be about 5dB lower. (b) This NCB level is exceeded by more than 3dB in 125Hz, 250Hz and 500Hz bands so it will sound rumbly. Note that RC criteria would result in a neutral classification (not rumbly or hissy). Best fit between 125Hz and 500Hz is NCB = 33. No levels in the octave bands between 1000Hz and 8000Hz are above NCB = 33 so sound is not hissy. This conclusion can be checked by plotting on RC curves. As can be seen from the following RC plot, the noise will be neutral (not rumbly or hissy -see 3 rd . Edn. text, page 159) as no levels in bands below 500 Hz exceed the RC level by more than 5 dB. Note RC = (38+30+20)/3 = 29. Although this conclusion is different to that drawn using NCB curves, it can be seen that the RC classification is close to rumbly. Criteria 95 L A ' 10log 10 10 53/ 10 & 10 48/ 10 ' 51.3dB 45 35 25 30 40 50 (RC) 16 31.5 63 125 250 500 1k 2k 4k 8k 10 20 30 40 50 O c t a v e b a n d s o u n d p r e s s u r e l e v e l ( d B r e 2 0 P a ) µ Octave band center frequency (Hz) (c) Optimum spectrum levels of added masking noise would be equal to the RC-33 levels (as this corresponds to the highest existing spectrum levels) with the existing levels logarithmically subtracted from it. For example, in the 63Hz band, the RC-33 value is 53dB and the desired added level is: Desired spectrum levels of masking noise are listed in the table below. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Sound pressure level (dB) 48 48 43 38 30 20 16 12 RC-30 values 50 48 43 35 33 28 23 Desired added levels 51.3 0 0 0 30 27 22 Solutions to problems 96 L A ' 10log 10 10 (60 & 26.2) / 10 % 10 (55 & 16.1) / 10 % 10 (55 & 8.6) / 10 % 10 (50 & 3.2) / 10 % 10 (55 & 0) / 10 % 10 (55 % 1.2) / 10 % 10 (50 % 1.0/ 10 % 10 (45 & 1.1) / 10 ' 59.9dB(A) 0 10 20 30 40 50 45 35 15 10 0 63 16 125 31.5 250 500 1k 2k 4k 8k Octave band center frequency (Hz) O c t a v e b a n d s o u n d p r e s s u r e l e v e l ( d B r e 2 0 P a ) µ 25 NCB Problem 4.17 (a) The NCB value for the noise is (37 + 33 + 33 + 32)/4 = 34 (b) The line of best fit for the NCB curve between 125 Hz and 500 Hz is 32 or 33 NCB. Three high frequency bands exceed this curve so the noise sounds hissy. No bands below 500 Hz exceed 34 NCB so noise is not rumbly. See following figure. Problem 4.18 (a) A-weighted level is: Criteria 97 45 35 25 30 40 50 (RC) 16 31.5 63 125 250 500 1k 2k 4k 8k 10 20 30 40 50 60 O c t a v e b a n d s o u n d p r e s s u r e l e v e l ( d B r e 2 0 P a ) µ Octave band center frequency (Hz) NR level from following figure (NR curves) = 58. (b) The levels are plotted on RC curves below, where it can be seen that the spectrum would sound hissy. (c) The noise level is 59.9dB(A). The dB(A) adjustments to the base level of 40dB(A), the resulting allowable noise levels and the expected public reactions are: Solutions to problems 98 day: +20 -2 = 58dB(A) (marginal public reaction) evening: +20 -5 -2 = 53dB(A) (little public reaction) night: +20 -10 -2 = 48dB(A) (medium public reaction) See table 4.11 in the text for public reaction estimates. (d) The noise reductions between inside and outside and the resulting inside levels are in the table below. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Exterior sound pressure levels 60 55 55 50 55 55 50 45 Expected noise reduction (dB) 5 5 8 10 14 16 20 21 Interior sound pressure levels 55 50 47 40 41 39 30 24 The spectrum in the last line of the table is plotted in the figure below and represents an NR value of 42. Criteria 99 From Table 4.12 in the text, daytime base level = NR 30 and nighttime base level = NR 25. Daytime adjustments = +5 +10, which give an allowable NR = 45 nighttime adjustment = +10, which gives an allowable NR = 35 Thus we would expect complaints during the evening and night but not during the day if the windows are closed. When the windows are open, 5dB is added to levels in all octave bands and the NR value of the interior noise becomes NR 47. This would result in a few complaints during the day and an increase in the number of nighttime complaints. (e) Factory could be built provided it only operated during the day. If the noise occurred only 25% of the time: . L Aeq ' 10log 10 0.25 × 10 59.9/ 10 ' 53.9dB(A) From the results of (c) above, it can be seen that the factory could now operate in the evening as well, but not at night. However from the results of (d) above, and table 4.11 in text, the NR criteria would indicate that it should still only operate during the day. Problem 4.19 From table 4.10 in the text, the acceptable level for nighttime operation is L Aeq = 40 - 10 + 15 = 45dB(A). From table 4.11 in the text, the expected community response would be widespread complaints. φ ' A r e j( ωt & kr) p ' j ωρ A r e j( ωt & kr) u ' A r 2 e j( ωt & kr) % jkA r e j( ωt & kr) ' A r e j( ωt & kr) 1 r % jk ' p jωρ 1 r % jk ' p ρc 1 & j kr 5 Solutions to problems relating to sound sources and outdoor sound propagation Problem 5.1 (a) Intensity, I = W/S = 10 -2 /(4π × 0.5 2 ) Watts/m 2 = 3.18 mWatts/m 2 (b) and thus the p rms ' Iρc ' 3.18 × 1.206 × 343 × 10 &3 ' 1.15Pa pressure amplitude = p rms /2 = 1.62 Pa. (c) For outwardly travelling spherical waves in free space, equation 1.40c may be written as: Using equations 1.6 and 1.7, the acoustic pressure and particle velocity may be written respectively as: and Sound sources and outdoor sound propagation 101 *u* ' *p* ρc 1 % c 2πf r 2 ' 1.62 1.206 × 343 1 % 343 2 × π × 400 × 0.5 2 ' 4.06mm/ s Q ' 4πW k 2 ρc ' Wc πf 2 ρ ' 0.01 × 343 π × 400 2 × 1.206 ' 2.38 × 10 &3 m 3 / s Thus: *u* ' *p* ρc *1 & j kr * or, (d) L p ' 20log 10 p rms % 94 ' 95.2dB (e) L w ' 10log 10 W % 120 ' 100dB (f) Source strength, Q, may be calculated using equation 5.12 in the text. Thus: Problem 5.2 (a) From equation 5.13(b) in the text, it can be seen that for a simple source the r.m.s. pressure 2 is inversely proportional to the distance 2 from the source. Thus the sound pressure level at 10m would be 110 - 20log 10 (10/1) = 90dB. (b) We can use the result of 5.1(c) above. The acoustic pressure amplitude at 1m is and the amplitude at ¯ p ' 2 × 10 &5 × 2 × 10 110/ 20 ' 8.945 Pa 10m is . ¯ p ' 2 × 10 &5 × 2 × 10 90/ 20 ' 0.8945 Pa At 1m, kr = 2πf/c = 2π × 100/343 = 1.832. At 10m, kr = 18.32. The Solutions to problems 102 ¯ u ' ¯ p ρc 1 % 1 (kr) 2 β ' &tan &1 1 kr ' &tan &1 1 1.832 ' &28.6E β ' &tan &1 1 kr ' &tan &1 1 18.32 ' &3.1E W ' Q 2 k 2 ρc 4π ' 6.28 2 × 10 &8 × 1.832 2 × 1.206 × 343 4π ' 43.6µWatts W ' Q 2 k 2 ρc 4π ' 6.28 2 × 10 &8 × 14.655 2 × 1.206 × 343 4π ' 2.79 milli Watts L w ' 10log 10 W % 120 dB result of 5.1(c) may be written as: Thus at 1m, and the phase ¯ u ' 8.945 1.206 × 343 1 % 1 1.832 2 ' 24.6mm/ s relative to the acoustic pressure is: At 10m, and the phase ¯ u ' 0.8945 1.206 × 343 1 % 1 18.32 2 ' 2.2mm/ s relative to the acoustic pressure is: Problem 5.3 Source volume velocity = 4πr 2 u rms = 4π × 0.01 2 × 0.5 = 6.28 × 10 -4 At 100Hz, k = 2π × 100/343 = 1.832 m -1 and at 800Hz, k = 14.655 m -1 . Using equation 5.12 in the text, the acoustic power radiated at 100Hz is: and at 800Hz, the acoustic power radiated is: The corresponding sound power levels are calculated using: Thus at 100Hz, L w = 76.4dB and at 800Hz, L w = 94.5dB. Sound sources and outdoor sound propagation 103 ¯ Q ' 4 π r 2 ¯ u ' 4π × 0.01 2 × 0.1 ' 1.257 × 10 &4 m 3 / s ¯ p ' Qkρc 4πr ' 1.257 × 10 &4 × 0.916 × 1.206 × 343 4π × 10 ' 3.79 × 10 &4 Pa β ' &tan &1 1 kr φ ' A r e j( ωt & kr) Problem 5.4 (a) The amplitude of the pressure fluctuations can be calculated using equation 5.13(a) in the text. The amplitude of the volume velocity, , ¯ Q is given by: At 50Hz, k = 2πf/c = 2π × 50/343 = 0.916, and the amplitude of the pressure fluctuations is then: (b) Using the equation from problem 5.2(c), the phase of the pressure minus the phase of the particle velocity is given by: At r = 0.5m, the above expression gives β = 65.4E and at r = 10m, β = 6.2E. This indicates that close to the source the acoustic pressure field is dominated by near field effects, whereas at 10m from the source, the near field effects will be small and the field may be approximated as a propagating plane wave. Problem 5.5 The radiation impedance per unit area is equivalent to the specific acoustic impedance, Z which is simply p/u. For outwardly travelling spherical waves in free space, equation 1.40c may be written as: Solutions to problems 104 p ' j ωρ A r e j( ωt & kr) u ' A r 2 e j(ωt & kr) % jkA r e j( ωt & kr) ' A r e j(ωt & kr) 1 r % jk ' p jωρ 1 r % jk ' p ρc 1 & j kr p u ' ρc 1 & j ka &1 ' ρc 1 % j ka 1 % 1 (ka) 2 ' ρc (ka) 2 % jka (ka) 2 % 1 ' ρc ω 2 a 2 % jωac c 2 % ω 2 a 2 1 r 2 M Mr r 2 M Mr % 1 r 2 sinθ M Mθ sinθ M Mθ % 1 r 2 sin 2 θ M 2 Mψ 2 φ & 1 c 2 M 2 φ Mt 2 ' 0 Using equations 1.6 and 1.7, the acoustic pressure and particle velocity may be written respectively as: and Thus, setting r = a, Problem 5.6 Follow the analysis in the text described by equations 5.1 to 5.12. In equation 5.12, replace the mean square volume velocity Q 2 with the product of half the velocity amplitude and the surface area of the pulsating sphere; that is, . The required result is then obtained. Q 2 ' *U* 2 2 ( 4πa 2 ) 2 Problem 5.7 The wave equation is: Sound sources and outdoor sound propagation 105 φ ' 2f ) (ct & r) (h/ r) cosθ r 2 Mφ Mr ' &2f )) (ct & r) (hr) cosθ & 2f ) (ct & r) hcosθ 1 r 2 M Mr r 2 Mφ Mr ' &2f ))) (ct & r) (h/ r) cosθ & 2f )) (ct & r) (h/ r 2 ) cosθ % 2f )) (ct & r) (h/ r 2 ) cosθ sinθ Mφ Mθ ' &2f ) (ct & r) (h/ r) sin 2 θ 1 r 2 sinθ M Mθ sinθ Mφ Mθ ' &2f ) (ct & r) (h/ r 3 ) 2cosθ 1 r 2 sin 2 θ M 2 Mψ 2 ' 0 & 1 c 2 M 2 φ Mt 2 ' &2f ))) (ct & r) (h/ r) cosθ &4f ) (ct & r) (h/ r 3 ) cosθ p ' ρ Mφ Mt ' A cosθ kr 1 & j (kr) e j(ωt & kr) u r ' &Lφ ' A cosθ krρc 1 & 2 (kr) 2 & 2j kr e j(ωt & kr) (a) The solution given by equation 5.25 in the text is: Substituting the solution into the various terms in the wave equation gives: Adding all the above terms together gives: which must equal zero to satisfy the wave equation. This is true provided that r is very large compared to h. (b) Equations 5.32 and 5.33 in the text may be written as: Solutions to problems 106 φ ' 1 ρ m T 0 p dt ' & jAcosθ ρck 2 r 1 & j kr e j( ωt & kr) & Mφ Mr ' & jAcosθ ρck 2 r 2 1 & j kr e j( ωt & kr) % jAcosθ ρck 2 r j kr 2 e j( ωt & kr) % jAcosθ ρck 2 r 1 & j kr (&jk) e j(ωt & kr) r 2 Mφ Mr ' Arcosθ ρck 1 & 2j kr & 2 (kr) 2 e j( ωt & kr) 1 r 2 M Mr r 2 Mφ Mr ' & Acosθ r 2 ρck 1 & 2j kr & 2 (kr) 2 e j( ωt & kr) & Arcosθ r 2 ρck 2j kr 2 % 4 k 2 r 3 e j( ωt & kr) & Arcosθ r 2 ρck 1 & 2j kr & 2 (kr) 2 (&jk) e j( ωt & kr) ' & Acosθ rρck & 1 r % 2 k 2 r 3 & jk % 2j kr 2 e j( ωt & kr) Using the first of the above equations and omitting the integration constant: Checking the expression for u, by evaluating -Lφ, we obtain: To verify that the expression obtained above for φ is a solution to the wave equation we substitute it into the wave equation and calculate the result term by term as follows: Sound sources and outdoor sound propagation 107 sinθ Mφ Mθ ' jAsin 2 θ ρck 2 r 1 & j kr e j(ωt & kr) 1 r 2 sinθ M Mθ sinθ Mφ Mθ ' 2jAcosθ ρck 2 r 3 1 & j kr e j(ωt & kr) 1 r 2 sin 2 θ M 2 Mψ 2 ' 0 & 1 c 2 M 2 φ Mt 2 ' ω 2 c 2 φ ' & j k 2 Acosθ ρck 2 r 1 & j kr e j( ωt & kr) jAcosθ ρcr & j kr % 2j (kr) 3 % 1 & 2 (kr) 2 % 2 (kr) 2 & 2j (kr) 3 & 1 % j kr e j( ωt & kr) φ ' f (ct & r) r r 2 Mφ Mr ' &f (ct & r) & rf ) (ct & r) Adding all the above terms together gives: which is equal to zero. Thus the solutions given by equations 5.32 and 5.33 satisfy the spherical wave equation exactly. Problem 5.8 Equation 1.35 in the text is: To verify that the expression obtained above for φ is a solution to the wave equation we substitute it into the wave equation and calculate the result term by term. Solutions to problems 108 1 r 2 M Mr r 2 Mφ Mr ' f ) (ct & r) r 2 % f )) (ct & r) r & f ) (ct & r) r 2 1 r 2 sinθ M Mθ sinθ Mφ Mθ ' 0 1 r 2 sin 2 θ M 2 Mψ 2 ' 0 & 1 c 2 M 2 φ Mt 2 ' & f )) (ct & r) r Q ' 4πW k 2 ρc ' 4π × 0.5 4.58 2 × 1.206 × 343 ' 2.69 × 10 &2 m 3 / s W D ' ρc k 4 h 2 Q 2 3π ' 1.205 × 343 × 4.58 4 × (0.08/ 2) 2 × 2.691 2 × 10 &4 3π ' 0.0224watts Adding all the above terms together gives 0, so equation 1.35 in the text is a solution. Problem 5.9 The wavenumber, k, is equal to (2πf/c) = (2π × 250/343) = 4.58. The source strength, Q, of the monopole may be calculated using equation 5.12 in the text. That is: The dipole acoustic power, W D , may be calculated using equation 5.29 in the text to give: Sound sources and outdoor sound propagation 109 p(r) ' jωρqe &jkr 4πr p(r) ' p 1 % p 2 ' jωρ 4π q 1 r 1 e &jkr 1 % q 2 r 2 e &jkr 2 r 1 . r % h cosθ and r 2 . r & h cosθ p(r, θ) ' jωρ 4πr e &j kr q 1 e &jkhcosθ % q 2 e jkhcosθ ' p 1 (r, θ) 1 % q 2 q 1 e 2jkhcosθ u Sound power level = 10log 10 W + 120 = 103.5dB. Problem 5.10 For a single source: For 2 sources separated by a distance 2h, the total pressure, p, is the sum of the pressures p 1 and p 2 from each source. Thus: As shown on page 179 in the text, for phase accuracy purposes, the following approximations are adequate: where it has been assumed that h << r. Noting that for amplitude purposes, r 1 . r 2 . r The above equation may be rewritten as: When θ = θ 0 , p = 0, or . 1 % q 2 q 1 e 2jkhcosθ 0 ' 0 Thus, . q 2 q 1 ' &e &2jkhcosθ 0 Substituting this into the preceding equation for p gives: Solutions to problems 110 p(r, θ) ' p 1 (r, θ) 1 & e &2jkh(cosθ 0 & cosθ) *p* ' 2*p 1 * 1 & cos(2khcosθ) Q ' 4πW k 2 ρc ' 4π × 0.01 9.16 2 × 1.206 × 343 ' 1.903 × 10 &3 m 3 / s I D ' ρc k 4 h 2 Q 2 (2πr) 2 cos 2 θ ' 1.205 × 343 × 9.16 4 × (0.005/ 2) 2 × 1.903 2 ×10 &6 (2π × 0.5) 2 × 0.707 2 ' 3.34µWatts/ m 2 ¢ p 2 ¦ ' ρcI ' 1.205 × 343 × 3.33 × 10 &6 ' 1.38 × 10 &3 Pa 2 If θ 0 = 90E, then p(r, θ) ' p 1 (r, θ) 1 & e 2jkhcosθ Taking the modulus of the preceding equation gives: This function contains the directivity information and is plotted in the figure above. Problem 5.11 The wavenumber, k, is equal to (2πf/c) = (2π × 500/343) = 9.16. The source strength, Q, of each monopole making up the dipole source may be calculated using equation 5.12 in the text. That is: (a) The dipole intensity at θ = 45E is given by equation 5.28 in the text and is: (b) Sound sources and outdoor sound propagation 111 L p ' 10log 10 ¢ p 2 ¦ p 2 ref ' 65.4dB F rms ' 4πaA 3k 2 1 % 1 (ka) 2 A 2 ' ρchQk 3 2π A 2 ' 1.206 × 343 × (0.005/ 2) × 1.903 × 10 &3 × 9.16 3 2π ' 0.241 F rms ' A 2 4π 3k 2 ' 0.241 × 4π 3 × 9.16 2 ' 12.0 mN Q 2 ' W M 4π ρck 2 u R h = L =0.05m 0.1m 0.1m 1W 0.5W 0.5W 20m (c) Required driving force can be calculated by taking the mean square value calculated using equation 5.39 in the text. Thus: From equation 5.35 in the text: Substituting in the previously calculated values for k and Q gives: As ka is very small, equation 5.40 in the text may be used. Thus: Problem 5.12 The arrangement is illustrated in the figure. As R is 1m off the floor, the distance to it is 20 2 % 1 2 ' 20.02m (a) The strength of each source may be calculated using equation 5.12 in the text: Solutions to problems 112 Q 2 ' 0.5 × 4π 1.206 × 343 × 2.29 2 ' 2.897 × 10 &3 ¢ p 2 ¦ ' 5ρcW long cos 4 θ 4πr 2 × 2 ' Q 2 ρck 3 hLcos 2 θ πr 2 × 2 ' 2.897 × 10 &3 1.206 × 343 × 2.29 3 × 0.05 2 × cos 2 (30) π × 20.02 2 × 2 ' 1.271 × 10 &4 Pa 2 L p ' 10log 10 1.271 × 10 &4 4 × 10 &10 ' 55.0dB L p ' 121.8 & 10log 10 (2π × 20.02 2 ) & 10log 10 (400/ 413.66) ' 87.9dB In this case, W M = 0.5W and k = 2π × 125/343 = 2.290. Thus: The arrangement shown is a longitudinal quadrupole and the mean square sound pressure at any location may be calculated using equations 5.54 and 5.55 in the text. Note that the equation for the mean square pressure is multiplied by 2 in this case because the radiation is into half space. Thus: The sound pressure level is then: (b) 125Hz random noise will make the sources act independently and the power radiated will be the arithmetic sum of the individual sources. Thus W = 1.5W and . L w ' 10log 10 (1.5/ 10 &12 ) ' 121.8dB Using equation 5.108 in the text, the sound pressure level may be calculated using S = 2πr 2 and ρc = 413.6. Thus: Sound sources and outdoor sound propagation 113 ¢ p 2 D ¦ ¢ p 2 M ¦ ' 4(kh) 2 cos 2 (90 & 1.145) ' 4(4.58 × 0.1) 2 (0.02) 2 ' 3.36 × 10 &4 Reduction ' 10log 10 3.36 × 10 &4 &1 ' 34.7dB | 0.1m hole speaker 5m O 0.2m Problem 5.13 (a) T h e a r r a n g e m e n t approximates a simple dipole and is illustrated in the figure. The angle β = sin -1 0.1/5 = 1.145E. The azimuthal angle given in the problem is irrelevant. The wavenumber, k = (2π × 250)/343 = 4.58. The sound pressure levels radiated by the hole alone (monopole) and hole + speaker (dipole) may be calculated using equations 5.13a, 5.30b and 5.29 in the text. Using these equations, the ratio of the mean square pressures (dipole/monopole) may be written as: Thus the reduction in sound pressure level due to the presence of the speaker is: (b) If a speaker is placed below the hole as well, a longitudinal quadrupole is formed with L = h = 0.1. In this case, β = 0, and as can be seen from equation 5.55 in the text (where θ = 90 - β), the theoretical mean square sound pressure will be zero, implying a reduction of infinity dB. Problem 5.14 In equation 5.62, (W/b) is effectively the power per unit length of source (as W is the power of each source separated by b) and in equation 5.70, W/D is the same quantity. Thus the difference between the finite length and infinite Solutions to problems 114 ¢ p 2 ¦ ' ρc W 2π 2 r 0 D [α u & α l ] L p ' L w & 10log 10 (4πr 0 D) % 10log 10 ( α u & α l ) % 10log 10 (ρc/ 400) ' 130 & 10log 10 (4π × 80 × 20) % 10log 10 (0.248) % 10log 10 (1.0342) ' 81.1dB o u o l 20m 80m A pipe length source is the quantity which is the ratio of the angle ( α u & α l ) / π subtended by the source at the observer in each case. Thus, by logical argument, equation 5.65 can be rewritten as follows for a finite coherent line source. Problem 5.15 The arrangement is as shown in the figure. α u ' &α l ' tan &1 (10/ 80) ' 0.124 c Finite length pipe, D = 20m, r 0 = 80m and L w = 130dB. Turbulent flow, so assume an incoherent source. Also assume incoherent addition of the direct and ground reflected waves. Taking logs of equation 5.70 in the text gives for the direct wave: The ground reflected wave level is then 81.1 - 3 = 78.1dB. Thus the total Sound sources and outdoor sound propagation 115 L p ' 10log 10 10 8.11 % 10 7.81 ' 82.9dB ¢ p 2 ¦ ' [ Wρc/ 4πr 0 D] [ α u & α l ] × 2 ¢ p 2 ¦ ' [ 2 × 1.206 × 343/ ( 4 π × 200 × 50) ] × 0.249 × 2 ' 3.275 × 10 &3 Pa 2 L p ' 10log 10 3.275 × 10 &3 4 × 10 &10 ' 69.1dB o u o l 50m 200m A pipe level at the receiver is: Problem 5.16 The situation is as shown in the figure below. Equation 5.70 in the text may be used to calculate the sound pressure level. The equation must be multiplied by the directivity factor (2 in this case). Thus: . α u ' α l ' tan &1 25 200 ' 0.124 radians W = 2, r 0 = 200, D = 50. Thus: Solutions to problems 116 A g ' &10log 10 [ 1 % 10 &2/ 10 ] ' &2.1dB ¢ p 2 1 ¦ ' ρc W 4πr 2 0 × DF ¢ p 2 2 ¦ ' ρc W 4br 0 × DF ¢ p 2 1 ¦ ¢ p 2 2 ¦ ' br 2 π × r 2 1 ' 6 × 50 π ' 95.5 From table 5.3 in the text, A a = 19.3dB per 1000m, so for 200m, A a = 19.3/5 = 3.9dB. Sound intensity loss due to ground reflection is 2dB. Thus the ground effect is given by equation 5.175b in the text as: Thus, A g + A a = 1.8dB and the sound pressure level at the receiver is: L p = 69.1 - 1.8 = 67.3dB Problem 5.17 The traffic may be treated as an infinite line source. At 1m r 0 < b/π and the mean square sound pressure is related to the source sound power, W, of one vehicle by: where DF is the directivity factor for the source/ground combination. At 50m, the sound pressure is related to the sound power, W, of one vehicle by: Thus: The level at 50m = level at 1m - 10log 10 = 88 - 19.8 = 68.2dB(A). ¢ p 2 1 ¦ ¢ p 2 2 ¦ Sound sources and outdoor sound propagation 117 I ' (¯ p 2 o / 3ρcr 2 ) (2 % cosθ) u r ¢ p 2 ¦ ' ρc W 4br 0 ' 413 × 2 4 × 7 × 250 ' 0.118 Pa 2 L p ' 10log 10 0.118 ( 2 × 10 &5 ) 2 ' 84.7 dB L p ' 84.7 & 0.6 % 3.0 ' 87 dB (%6, &3 dB) Problem 5.18 From Equation 5.62: Concrete ground, so ground effect is A g = -3 dB. Assume 20EC, air absorption ranges from 2.6 to 2.8 dB per 1000 m. For 250 m air absorption - 0.7 dB. From table 5.3, meteorological influence is +6, -3 dB. Assume no obstacles blocking line of sight to the road from the residence. So the sound pressure level at the residence is: Problem 5.19 (a) The arrangement is shown in the figure. We are given: The sound power is obtained by integrating the intensity over an imaginary hemispherical surface centred at the centre of the speaker. The sound power is then: Solutions to problems 118 W ' m S I dS ' ¯ p 0 2 3ρc m 2π 0 dψ m π/ 2 0 (2 % cosθ) r 2 sinθ r 2 dθ ' ¯ p 2 0 2π 3ρc m π/ 2 0 (2sinθ % cosθ sinθ) dθ ' ¯ p 2 0 2π 3ρc . π/ 2 0 & 2cosθ & 0.25 cos2θ ' ¯ p 2 0 2π 3ρc 2 % 1/ 2 ' 0.0127¯ p 2 0 Watts (b)& (c) As the speaker only radiates into a hemispherical space, the presence of a baffle will have no influence on the power radiated, regardless of whether the source is constant volume or constant pressure. Problem 5.20 (a) If the piston is assumed to be made up of an infinite number of point monopole sources, all pulsating in phase, then the sound pressure at any location can be calculated by summing the contributions from each source. This effectively means that an expression for the sound pressure at some distance, r, due to a monopole on the piston surface must be integrated over the piston surface. If a baffle is present, the monopole source must be replaced with a hemispherical source which radiates twice the pressure. (b) The function 2J 1 (x)/x vs x is plotted out in the figure below, where x = ka sinθ. Sound sources and outdoor sound propagation 119 0 5 10 15 20 0.5 1.0 2J ( ) 1 x x x At 500Hz, ka = 2π × 500 × 0.1/343 = 0.91. Nodes in the radiation pattern occur when 0.91 sinθ = 3.8, 7, 10.1, etc. That is, there are no nodes. At 2,500Hz, ka = 2π × 2,500 × 0.1/343 = 4.58. Nodes in the radiation pattern occur when 4.58 sinθ = 3.8, 7, 10.1, etc. That is, there is only one node at θ = 60E. At 10,000Hz, ka = 2π × 10,000 × 0.1/343 = 18.31. Nodes in the radiation pattern occur when 18.31 sinθ = 3.8, 7, 10.1, 13.3, 16.4, and 19.6. That is, there are 5 nodes at θ = 12E, 23E, 33E, 47E and 64E. Directivity patterns are shown for each of these cases in the figures on the next page. The side lobes for the 10kHz case have been expanded for clarity. For sketching purposes, the figure shown at the beginning of part (b) (figure 5.7 in the text) may be used with the x-axis crossings representing the nodal locations of each lobe and the peaks in the curve representing the relative amplitude of each lobe. [Note that fig 5.6 will only provide information for the first three lobes so it must be extended for the 10kHz case.] Solutions to problems 120 R R ' ρcπa 2 × (2ka) 2 / 8 ' ρc(πa 2 ) 2 4ω 2 / (8πc 2 ) ' ρcS 2 ω 2 / (2πc 2 ) σ ' R R ρcS ' πa 2 ω 2 2πc 2 ' (ka) 2 / 2 W ' R R πa 2 ρcU 2 / 2 500Hz 2.5kHz 10kHz Problem 5.21 (a) Using equations 5.95b and 5.96 in the text, the radiation resistance for a piston can be written as: where it has been assumed that ka is sufficiently small that all but the first term of equation 5.96 in the text is negligible. (b) The radiation efficiency at low frequencies is: which described the solid line in figure 5.9 in the text for ka < 0.8. (c) See fig 5.9 in the text. Problem 5.22 (a) Piston radiating from an infinite baffle. k = 2πa/λ = 2 and a = 0.1m. From equation 5.98b in the text: Sound sources and outdoor sound propagation 121 U ' ¯ ξω ' 2 ¯ ξc/ λ ' 2 ¯ ξc/ a ' 2 × 0.0002 × 343/ 0.1 ' 1.372m/ s W ' π × 0.1 2 × 1.206 × 343 × 1.372 2 / 2 ' 12.2W I ' ρck 2 8π 2 r 2 F 2 (w) I ' ρch 2 U 2 π 2 a 4 8π 2 r 2 ' ρcU 2 a 2 8r 2 ' 1.206 × 343 × 1.372 2 × 0.01 2 × 2 2 ' 0.97W/ m 2 L p ' 10log 10 ¢ p 2 ¦ p 2 ref ' 10log 10 ρcI 4 × 10 &10 ' 120.0dB and from fig 5.8, for ka = 2, R R = 1. The piston velocity amplitude, U, is given by: Thus the radiated power is: (b) From equation 5.84 in the text, the on-axis intensity is: where w = kasinθ = 0. Thus, F(0) = Uπa 2 . Thus: (c) Radiation mass loading = πa 2 ρc[X(2ka)], where X(2ka) = 0.55 (see fig 5.8 in the text). Thus the mass loading is π × 0.01 × 1.206 × 343 × 0.55 ' 7.1 kg/ s (d) Sound pressure level at 2m: Problem 5.23 (a) The arrangement is shown in the figure on the next page where it can be seen that h = L = 102.5mm. Solutions to problems 122 W M ' ( Q 2 H ρck 2 / 4π) × 2 W long ' [ ( 2k 3 hLQ L ) 2 ρc/ 5π] × 2 W long W M ' 4k 6 h 2 L 2 Q 2 H ρc4π 20πQ 2 H ρck 2 ' 4 5 k 4 h 2 L 2 ' 4 5 (2π) 4 f 4 (0.1025) 4 343 4 ' 9.94 × 10 &12 f 4 hole speaker speaker 200mm 205mm (b) Power radiated by original opening (assuming a constant volume velocity source in an infinite baffle) is (see equation 5.12 in the text and multiply by 2 to account for radiation into half space): The power radiated by the longitudinal quadrupole may be calculated using equation 5.54 in the text. Again the equation in the text must be multiplied by 2. Thus: where Q L = Q H /2. Thus: Frequency, Hz dB reduction (-10log 10 (W long /W m ) 63 125 250 500 38 26 14 2 Sound sources and outdoor sound propagation 123 ¢ p 2 long ¦ ¢ p 2 M ¦ ' [ 5ρcW long cos 4 θ/ 4πr 2 ] × 2 [ W M ρc/ 4πr 2 ] × 2 ' 5cos 4 θ W long W M ΔL p ' &10log 10 [5cos 4 θW long / W M ] L p ' L w & 10log 10 [4πr 2 ] ' 120 & 10log 10 [2π × 100] ' 92dB (c) The ratio of the mean square pressures may be obtained using equations 5.13b and 5.55 in the text. Thus: The reduction in sound pressure level is then: Values of sound pressure reduction (dB re 20µPa) are tabulated below for the required values of θ. frequency (Hz) θ = 0 θ = π/4 θ = π/2 63 125 31 19 37 25 4 4 When the speakers are turned on, the sound field amplitude distribution has 2 lobes with maxima at θ = 0, π and minima at θ = π/2, 3π/2. Problem 5.24 (a) The directivity index due to the hard floor is 3dB. (b) The expected sound pressure level due to an omni directional source on a hard floor is: The actual sound pressure level is 110dB so the directivity due to the source characteristics is -18dB. Solutions to problems 124 ¢ p 2 ¦ ' 2ρcW πHL tan &1 HL 2r H 2 % L 2 % 4r 2 ' 2 × 1.206 × 343 × 0.01 × S S × π tan &1 0.25 2 × 25 0.25 % 0.25 % 4 × 25 2 ' 2.63 × 10 &4 Pa 2 L p ' 10log 10 2 × 2.633 × 10 &4 4 × 10 &10 ' 61.2dB hole 2m 25m Problem 5.25 The radiated sound power is W = IS = 0.01 × S = 0.01S Watts. The arrangement is shown in the figure. With no ground reflection, the on-axis sound pressure level may be calculated using equation 5.105 in the text (as we may assume that the opening behaves like an incoherent plane source). Here, H = L = 0.5, S = HL and r = 25. Thus: Assuming a similar travel distance for the ground reflected wave, the total sound pressure level at the receiver is: Interestingly, in this case, the receiver is sufficiently far from the source for the source to appear as a point source and the same result would have been obtained if equation 5.106 had been used instead of equation 5.105. Sound sources and outdoor sound propagation 125 ρf R 1 ' 1.206 × 2000 2.25 × 10 5 ' 0.011 β R 1 ρf 1/ 2 ' 1.72 1 0.011 1/2 ' 16.6 A g ' &10log 10 1 % 10 &A R / 10 ' &2.4dB hole 3m 150m 1.5m | | x y R Problem 5.26 (a) Sound power level, L w = 10log 10 W + 120 = 123dB. (b) The arrangement for calculating the ground effect is shown in the figure. From similar triangles, x = 2y. Thus x = 100 and y = 50. tanβ = 3/100, so β = 1.72E. For grass covered ground, R 1 = 2.25 × 10 5 (middle of range). Thus: From figure 5.20 in the text, A R = 1.3dB. The ground effect is then: Thus the effect of the ground is to increase the level at the receiver by 2.4dB. (c) Loss due to atmospheric absorption. From table 5.3 in the text, A a = 15.5dB per 1000m (25% RH and 20EC). So for a distance of 150m, A a = 15.5 × 0.15 = 2.3dB. Solutions to problems 126 ¢ p 2 ¦ ' 1.206 × 343 × 2 2π × 150 2 ' 5.852 × 10 &3 Pa 2 L p ' 10log 10 5.852 × 10 &3 4 × 10 &10 ' 71.7dB 5 90 10000 10000 - x x r 1 r 2 r (d) The opening should be treated as an incoherent plane source and equation 5.105 in the text used to calculate the sound pressure level. However as we found in the previous problem, the receiver is far enough from the source for it to appear as a point source (see figure 5.11) and we may use equation 5.106. Thus: The sound pressure level is then: A a + A g = -0.1, thus the sound pressure level at the receiver is equal to 71.8dB. (e) Adding a second opening will add 3dB to the sound pressure levels at the receiver as it may be assumed that the sound fields from the two sources are incoherent. Thus the sound pressure level at the community location of (b) above would be 74.8dB. Problem 5.27 The arrangement is illustrated in the figure. Using similar triangles: Sound sources and outdoor sound propagation 127 x 5 ' 10000 & x 90 90x ' 50000 & 5x x ' 50000/ 95 ' 526.31579m r 1 ' 5 2 % (526.315) 2 ' 526.33954 r 2 ' 90 2 % (10000 & 526.31579) 2 ' 9474.111701 r ' 85 2 % 10000 2 ' 10000.36124 r 1 % r 2 & r ' 0.0900m f ' c λ ' 1500 2 × 0.0900 ' 8.3kHz Destructive interference will occur if λ/2 = 0.0900m. This occurs at a frequency, f, given by: Problem 5.28 We would NOT expect to measure 84dB(A) at the operator's position due to reflected energy from the nearby wall. As the sound is predominantly in the 500Hz to 2000Hz band, the operator is not in the hydrodynamic near field of the machine (although he/she could be in the geometric near field). Also the machine is far enough from the wall for its sound power to be unaffected by the wall. Assuming incoherent addition of direct and reflected waves, assuming that geometric near field effects are negligible assuming that the machine does not act as a barrier to the reflected sound and assuming that the loss on reflection from the wall is negligible, the level at the operator's position may be calculated as the logarithmic sum of the direct and reflected waves. The reflected wave path length is 6m and the direct path length is 2m. Thus the sound pressure level of the reflected wave is Solutions to problems 128 L p ' 10log 10 10 8.4 % 10 7.45 ' 84.5dB(A) R p ' (Z s / ρc) & 1 (Z s / ρc) % 1 R p ' (Z s / ρc)cosθ & cosψ (Z s / ρc)cosθ % cosψ *R p * ' *(Z s / ρc) cosθ & 1* *(Z s / ρc) cosθ % 1* . Thus the total sound pressure level expected 84 & 20log 10 (6/ 2) ' 74.5dB at the operator's position is: Problem 5.29 (a) Specific acoustic impedance is a complex quantity characterised by an amplitude and a phase and is the complex ratio of acoustic pressure to acoustic particle velocity at any point in an acoustic medium, including the interface between two different media. Characteristic impedance is equal to ρc. For an acoustic medium where the viscous and thermal losses are small (such as air or water), it is a real quantity and is equal to the specific acoustic impedance of a plane wave propagating in an acoustic medium of infinite size. (b) The absorption coefficient (assuming plane incident waves) is defined in terms of the reflection coefficient, R p , as α = 1 - *R p * 2 . Also, the normal specific acoustic impedance of the surface of an acoustic medium of infinite extent is the characteristic impedance of the medium, for an infinitely thick medium. Thus equation 5.129 in the text may be used with θ = 0 to give: If θ is not equal to 0, then If it is assumed that the wavenumber in the material is much larger than that in air, cosψ = 1 and: The maximum absorption coefficient will occur when the modulus squared of the reflection coefficient is a minimum; that is, when Sound sources and outdoor sound propagation 129 *R p * 2 ' (2cosθ & 1) 2 % 9cos 2 θ (2cosθ % 1) 2 % 9cos 2 θ ' 13cos 2 θ & 4cosθ % 1 13cos 2 θ % 4cosθ % 1 α ' 1 & *R p * 2 ' 1 & 13 × 0.0769 & 4 × 0.2774 % 1 13 × 0.0769 % 4 × 0.2774 % 1 ' 1 & 0.286 ' 0.71 ¯ p ¯ p 0 ' e &αx Decay rate ' 20log 10 e α ' 20 × 0.4343ln e α ' 8.686α (dB/ m) L p ' L w & 10log 10 ( 2πr 2 ) & A g & A a & A m '110 & 65.5 % 3 & 2.7 × 0.75 &A m ' 45.5 dB (%8, &6 dB) is a mimimum. Differentiating the above expression wrt cos2 using the chain rule, and setting the result equal to zero, we obtain the minimum value when cosθ = 0.2774 or θ = 74E. The corresponding maximum value of the absorption coefficient is given by: Problem 5.30 Power radiated by window = 0.1 W = 110 dB Concrete ground, so A g = -3 dB Assume 20 EC temperature, so A a ranges from 2.6 to 2.8. Use A a = 2.7. Range due to meteorological conditions is: (+8, -6 dB) from Table 5.10. Assume no barriers between the source and receiver. The receiver is far enough away for the window to be treated as a point source in a baffle. Thus: Problem 5.31 The pressure amplitude at any location x is given by: where is the amplitude at x = 0. The decibel decay rate per unit distance ¯ p 0 is 20log 10 of the reciprocal of the above expression when x = 1 and is thus given by: Important factors are air temperature and humidity. Solutions to problems 130 A 2 600 r - 600 r Problem 5.32 The situation is illustrated in the figure below. We need to find the radius of curvature, r, of the wave and hence the distance, d. We may use the aircraft as the reference frame. Thus we assume a coordinate system moving horizontally at the speed of the aircraft and then later on calculate the distance that the aircraft travels during the time it takes for the sound to reach the ground (with an assumed aircraft speed). The wave which hits the ground at grazing incidence will be the one heard first. The total sonic gradient is 1/10 = 0.1s -1 . Using equation 5.190 in the text, the radius of curvature of the wave is thus 343 × 10 = 3430m. The distance d is then: Sound sources and outdoor sound propagation 131 d ' r 2 & (r & 600) 2 ' 3430 2 & 2830 2 ' 1940m θ 0 ' cos &1 3430 & 600 3430 ' 34.4E t ' m θ 0 0 r dθ c 0 & 0.1h ' m θ 0 0 dθ (c 0 / r) & 0.1 % 0.1cosθ ' 10 × m θ 0 0 dθ cosθ ' 10 log e [ secθ % tanθ] 34.4E 0 ' 6.4seconds We now need to take into account the speed of the aircraft and the distance it will travel in the time the sound wave travels to the observer. Let the angle subtended at the centre of the circular arc shown in the previous figure be θ 0 . The speed of sound as a function of θ (with θ = 0 corresponding to ground level and θ = θ 0 corresponding to the aircraft level) is given by c θ = c 0 - 0.1h, where h = height above the ground and c 0 = 343m/s is the speed of sound at ground level. The value of θ 0 is given by Thus the time taken for the sound to travel from the aircraft to the ground is If the aircraft travels at 400km/hour (not given in the question), then it would travel 710m in 6.4 seconds. Thus the aircraft emerges from ground shadow (1940m - 710m) = 1230m from the observer. 6 Solutions to problems relating to sound power, its use and measurement Problem 6.1 (a) This is discussed in detail on p246–247 in the text. (b) For a constant power source in the corner of the room, the radiated power would be concentrated over a one eighth sphere instead of a sphere and the sound pressure level in the direct field would thus be increased by 9dB. However, there would be no change in the reverberant field sound pressure level. For a constant volume velocity source in the corner of the room, the radiated power would be increased by a factor of 8 due to there being 3 reflecting surfaces and in addition the power would be concentrated over a one eighth sphere instead of a sphere and the direct field sound pressure level would thus be increased by 18dB. However, the reverberant field sound pressure level would be increased by only 9dB, corresponding to the power increase. For a constant pressure source in the corner of the room, the direct field radiated sound pressure level would be unchanged, but the reverberant field sound pressure level would be reduced by 9dB, corresponding to a reduction of 9dB in the radiated power. (c) A good approximation would be a constant pressure source model because the noise is originally generated by a fluctuating pressure, the amplitude of which is controlled by the aerodynamics and not the acoustics of the problem. Sound sources and outdoor sound propagation 133 10log 10 W ' 10log 10 ¢ p 2 ¦ % 10log 10 S & 10log 10 (ρc) 10log 10 W W ref ' 10log 10 ¢ p 2 ¦ p 2 ref % 10log 10 S % 10log 10 (400) & 10log 10 (ρc) L w ' L p % 10log 10 S L w ' L p % 10log 10 S ' 85 % 10log 10 (2π × 2 2 ) ' 85 % 14 ' 99dB Problem 6.2 (a) Sound power level is a measure of the rate of total energy radiated by an acoustic source while sound pressure level is a measure of the fluctuating sound pressure at a particular location. Sound power level is a source property whereas sound pressure level depends on the measurement location as well as the strength and size of the source. (b) Beginning with and taking logs of both sides gives: W ' ¢ p 2 ¦ S/ ρc Dividing both sides by 10 -12 and remembering that the reference sound power level is 10 -12 W and the reference sound pressure level is 2 × 10 -5 Pa, the preceding equation may be written as: If the quantity ρc is approximated as 400, then the preceding equation becomes: (c) Using the above equation, the sound power level would be: Assumptions: ρc = 400 and the sound pressure level measurements were made in the acoustic far field of the machine. (d) The desirable quantity is usually sound power level as it indicates the amount of acoustic energy which will be added to an environment and allows the increase in sound pressure level to be calculated at Solutions to problems 134 r > 3λ/ 2π; r > 3R; r > 3πR 2 / 2λ any location in the far field of the source as a result of the introduction of the machine (provided the sound radiation is omnidirectional or directivity information is also given). Sound power is also independent of the presence of nearby reflecting surfaces, provided that these are more that a quarter of a wavelength from the acoustic centre of the machine or noise source. On the other hand, the sound pressure level at a specified location is affected by the presence of reflecting surfaces and also does not necessarily allow the sound pressure level at other locations to be calculated. However, if only the noise exposure of the operator of a machine is of concern, then perhaps it is better to specify sound pressure level at the operator's location measured in the presence of specified reflecting surfaces than sound power level as the operator may not be in the acoustic far field of the source. Problem 6.3 (a) This is discussed in detail on pages 249-251 in the text. (b) Assuming that "negligible" is a factor of 10, the frequency would be given by (see figure 6.1 in the text). γ ' 10/ κ and therefore2r/ R ' 10λ/ (πR ) ' 10 × 343/ (πR f ) f ' 10 × 343/ (2πr) ' 3430/ (2 × π × 1) . 550Hz (c) r = R = 1, so γ = 2 and κ = π × 550/343 . 5. From figure 6.1, for the far field to be dominant, γ = 15. Thus r = 7.5m is the distance at which the far field would be dominant. Problem 6.4 (a) The anechoic room should be sufficiently large that sound pressure measurements can be made in the far field of the source. Thus the following criteria should be satisfied (see eq. 6.5, p.250 in text). The lowest frequency of interest is 44Hz (see table 1.2 on p43 in the text) and the highest frequency is 11,300Hz. Sound power use and measurement 135 2 × 6 (4.5 % 0.75 % 7.8/ 4) by (3.72 % 0.2 % 7.8/ 4) > by (3.72 % 0.6 % 7.8/ 4) > ' 14.4m× 11.7m× 6.3m 1.3λ 3 ' 1.3 343 63 3 ' 210m 3 W ' ¢ p 2 ¦ S ρc ' ¢ p 2 ¦ 2πr 2 ρc frequency (Hz) λ (m) 3λ/2π 3R 3πR 2 /2λ R = 1.5 R = 0.4 R = 0.6 44 11,300 7.8 0.03 3.72 0.015 4.5, 1.2, 1.8 1.36 349 0.1 24.8 0.22 55.9 The last criterion in the above table for r represents the transition from the geometric near field to the far field. Thus for high frequencies it is not practical to take measurements in the far field; the geometric near field will suffice. As all measurements must be taken at least λ/4 from the room walls, the following minimum interior room dimensions are needed. As standard measurements use a hemispherical surface, the required room dimensions are 14.4m × 14.4m × 7.2m high. (b) For measurements in octave bands, the required room volume is: Optimum dimensions are in the ratios 2:3:5. Thus 2x × 3x × 5x = 210, or x = 1.91m. Thus the required dimensions are 3.8m × 5.7m × 9.6m. Problem 6.5 The sound power is related to the average mean square pressure by: or in terms of sound power level, L w : Solutions to problems 136 L w ' L p % 10log 10 (2πr 2 ) % 10log 10 400/ ρc ' 70 % 10log 10 (8π) % 10log 10 400/ 413.6 ' 83.9dB W ' 10 &12 × 10 L w / 10 ' 10 &12 % 8.39 ' 0.245mW 3λ/ 2π; 3R; and 3πR 2 / 2λ ¢ p 2 ¦ ' (Qkρc) 2 (4πr) 2 The sound power, W, is thus: Problem 6.6 (a) r.m.s. acoustic pressure: p rms ' 2 × 10 &5 × 10 L p / 20 ' 2 × 10 3.3 & 5 ' 0.04Pa (b) Source dimension, R = 0.1m; wavelength, λ = 1481/250 = 5.92m; r = 2m. The distance r must be larger than the following quantities for the location to be in the far field: Substituting values for λ and l into the preceding expressions and evaluating gives 2.82, 0.3 and 0.008 respectively, so the location is not in the far field but rather in the transition between the hydrodynamic near field and the far field. (c) Intensity: I ' p 2 rms ρc ' 0.04 2 998 × 1481 ' 1.08nano&watts/ m 2 (d) Power, W ' I S ' 1.08 × 10 &9 × 2 × π × 2 2 ' 0.027µW (e) The mean square sound pressure is related to the source volume velocity, Q, for a spherical source by equation 5.13a in the text as: Sound power use and measurement 137 d ' 2Q 2πf × 4πa 2 ' 1.791Q/ f d ' 3.582r p rms ρf 2 d ' 3.582 × 2 × 0.04 998 × 250 2 ' 4.6 × 10 &9 m L w ' L p % 10log 10 V & 10log 10 T 60 % 10log 10 1 % Sλ 8V & 13.9dB In this case, the source is a hemisphere and it is radiating into hemispherical space. Thus the same result is obtained as for a spherical source radiating into spherical space. The volume velocity, Q, is related to the surface displacement, d, by: Combining the above two equations gives: Substituting values for the variables gives: Problem 6.7 Equation 6.13 in the text is: From problem 6.4 the room size is 3.8m × 5.7m × 9.6m. The volume, V, is 207.9m 3 and the surface area, S, is 2(3.8 × 5.7 + 3.8 × 9.6 + 5.7 × 9.6) = 225.7m 2 . Using the above equation, the following table may be constructed. f L p 10log 10 T 60 λ 10log 10 (1+Sλ/8V) L w 63 125 250 500 1000 2000 4000 8000 85 105 100 90 95 98 90 88 9.1 8.8 7.4 6.5 5.4 4.0 3.0 1.8 5.44 2.74 1.37 0.69 0.34 0.17 0.086 0.043 2.4 1.4 0.7 0.4 0.2 0.1 0.1 0.0 87.6 106.9 102.6 93.2 99.1 103.4 96.4 95.5 Overall 110.3 Solutions to problems 138 L p (av) ' 10log 10 1 N j i 10 L pi / 10 ' 86.8dB L w ' L p % 10log 10 S & Δ 1 & Δ 2 Problem 6.8 Average L p = 10 log 10 {(1/5)[10 8.5 + 10 8.3 + 10 8.0 + 10 8.7 + 10 8.6 ]} = 84.8dB Area of measurement surface = 2(2 × 4 + 2 × 3) + 3 × 4 = 40m 2 . Thus radiated power, (eq. 6.25 in text). L w ' L p % 10log 10 S & Δ 1 & Δ 2 Ratio of area of measurement surface to machine surface = 40/8 = 5, so Δ 2 = 0. Assuming that the machine is in a large enclosure, Δ 1 = 0 as well. Thus L w = 84.8 + 10log 10 40 = 100.8dB. Problem 6.9 The data may be used to construct the following table. meas- ured 85 88 86 90 84 85 87 88 89 90 90 88 87 88 89 85 back- ground 80 82 80 81 80 79 81 79 80 81 83 83 82 80 80 79 machine only 83. 3 86.7 84.7 89.4 81.8 83.7 85.7 87.4 88.4 89.4 89 86.3 85.3 87 88.4 83.7 The sound power level may be calculated using equation 6.25 in the text which is: where S = 40m 2 , and the factory volume V = 20 × 20 × 5 = 2000m 2 . Thus V/S = 50 and from table 6.4, p. 266 in the text, Δ 1 = 2.5dB. Machine surface area is the area of a cube of dimensions 1m smaller than the test cube. Let the side of the test cube = x. Then 5x 2 = 40 and x = 2.828m. Thus the machine size is 0.828m × 0.828m × 1.828m high (as the machine is resting on the floor). The machine surface area is then S m = 1.828 × 0.828 × 4 + 0.828 × 0.828 = Sound power use and measurement 139 L w ' 86.8 % 10log 10 40 & 2.5 ' 100.3dB re 10 &12 W L w ' 86.8 % 10log 10 40 & 3 . 100dB re 10 &12 W 4 R ' (1/ 188) & (1/ 428) [10 2/ 10 ] 10 2/ 10 & 1 ' 2.763 × 10 &3 L w ' L p2 & 10 log 10 [ S &1 1 & S &1 2 ] % 10 log 10 [ 10 ( L p1 & L p2 ) / 10 & 1] & 10log 10 ρc 400 L w ' 84 & 10log 10 188 &1 & 428 &1 % 10log 10 10 2/10 & 1 & 10log 10 413.7 400 ' 84 %25.25 & 2.33 & 0.15 ' 106.8dB re 10 &12 W 6.7m 2 . Thus S/S m = 40/6.7 = 5.9 and from table 6.3, p266 in the text, Δ 2 = 0. Assuming ρc = 400: Problem 6.10 Second surface average = 86.8 - 2 = 84.8dB. Area ratio (surface1 to surface2) = 40/120 = 0.3333. From figure 6.3 or equation 6.27 in the text, Δ 1 = 2.6dB and from problem 6.8, Δ 2 = 0. Thus: Problem 6.11 Machine surface area = 2(8 × 3 + 4 × 3) + 8 × 4 = 104m 2 . Area of test surface 1m from machine = 2(10 × 4 + 6 × 4) + 10 × 6 = 188m 2 = S 1 . Area of test surface 3m from machine = 2(14 × 6 + 10 × 6) + 14 × 10 = 428m 2 = S 2 . Using equation 6.22 in the text: Thus R = 1448 m 2 . Sound power level, L w is calculated using equation 6.24 in the text which is: Thus: Solutions to problems 140 L p2 ' 10log 10 1 11 10 8.7 % 10 8.75 % 10 8.6 % 10 8.5 % 10 8.65 % 10 8.8 % 10 8.68 % 10 8.72 % 10 8.6 % 10 8.58 % 10 8.53 ' 86.6dB L p1 ' 10log 10 10 90/ 10 & 10 80/ 10 ' 89.5dB L p2 ' 10log 10 10 86.6/ 10 & 10 80/ 10 ' 85.5dB Δ 1 ' 89.5 & 85.5 & 10log 10 [ 10 0.4 & 1] % 10log 10 [ 1 & 96/ 280] ' 0.4dB However, this excludes the near field correction term of equation (6.25) in the text, which for this case is -1 dB. So if we used equation 6.25 and also applied the correction due to ρc not equal to400, the result would be L w = 105.8 dB. Problem 6.12 (a) The average noise level measured on the larger surface is calculated by logarithmically averaging the given values. Thus: The background level may be subtracted from the overall averaged levels as this will give the same result as subtracting it from the individual levels and then averaging. Thus the noise level on test surface 1 due only to the machine is: and the noise on test surface 2 due only to the machine is: (b) Reverberant field correction, Δ 1 , use equation 6.27 in the text. First calculate the areas of the test surfaces. Surface 1, S 1 = 2(6 × 2.5 × 2) + 6 × 6 = 96m 2 . Surface 2, S 2 = 2(10 × 4.5 × 2) + 10 × 10 = 280m 2 . Thus: (c) Correction for non-normal sound propagation is Δ 2 . Machine surface area, S m = 2(5 × 2 × 2) + 5 × 5 = 65m 2 . S 1 /S m = 96/65 = 1.48. Thus from table 6.3 in the text, Δ 2 = 1. Sound power use and measurement 141 L w ' L p % 10log 10 S & Δ 1 & Δ 2 ' 89.5 % 10log 10 (96) & 0.4 & 1 ' 107.9dB L w ' L p % 10log 10 (2πr 2 ) % 10log 10 400 ρc ' 75 % 17.5 & 0.1 ' 92.4dB(A) L p ' 10log 10 10 85/ 10 & 10 82/ 10 ' 82dB(A) L w ' 87.4 & 10log 10 3 ' 82.6dB(A) (d) Sound power level, L w is calculated using equation 6.25 in the text which is: Problem 6.13 (a) sound power level: (b) Sound power level of existing machinery is 92.4 + 82 - 87 = 87.4dB(A). (c) Maximum allowable total reverberant sound pressure level generated by the three new machines is: Thus the maximum allowable sound power level is 87.4dB(A). (d) If all new machines emit the same sound power level and the total allowed is 87.4dB(A), the upper bound on the level generated by each machine is: Problem 6.14 (a) Advantages of sound intensity for sound power measurement: ! reduces errors arising from presence of reflecting surfaces; ! reduces errors from near field effects resulting from measurements taken close to the source; ! reduces errors caused by background noise generated by sources other than the one under test; ! Allows good results to be obtained at low frequencies. Disadvantages of sound intensity for sound power measurement: Solutions to problems 142 σ ' W Sρc¢ v 2 ¦ ! Usually more time consuming; ! instrumentation is more expensive; ! When the radiated sound field is complex, sound intensity measurements can provide too much data which is time consuming to analyse and can be confusing. (b) Advantages of sound intensity for transmission loss measurement: ! reduces errors arising from flanking path transmission; ! only requires a single reverberant room rather than two; ! Allows good results to be obtained at low frequencies. Disadvantages of sound intensity for transmission loss measurement: ! Usually more time consuming; ! instrumentation is more expensive; (c) Advantages of sound intensity for localisation and identification of noise sources: ! more reliable than a directional microphone due to greater spatial resolution and the ability to measure very close to a source; ! thus contamination from other nearby sources is reduced; Disadvantages of sound intensity for localisation and identification of noise sources: ! instrumentation is more expensive; Problem 6.15 (a) "radiation efficiency" is a measure of the amount of sound power radiated by a vibrating surface compared to that carried by a plane wave having the same mean square acoustic velocity (as the vibrating surface) and equal area. It can be expressed as: (b) Referring to the above equation, it can be seen that the sound power radiated by a surface of area S, mean square velocity and radiation ¢ v 2 ¦ efficiency σ is: Sound power use and measurement 143 W ' σSρc¢ v 2 ¦ L w ' 10log 10 ¢ v 2 ¦ % 10log 10 S % 10log 10 σ % 146 dB re 10 &12 W To identify possible paths of sound transmission between rooms, the radiation efficiencies of all walls ceiling and floor can be used together with their measured mean square velocities and the above equation to calculate the relative contributions of each surface to the overall sound power transmitted into the room, thus allowing the flanking paths to be identified and ranked. (c) The radiation efficiency of a surface is close to one when the bending wavelength of bending waves in the surface is less than or equal to the wavelength of the radiated acoustic waves; that is, at frequencies equal to or above (and in practice just below, for finite size surfaces) the critical frequency of the surface. In this case there will always be some radiation angle at which the bending waves on the surface will match the trace acoustic wavelength in the surrounding medium with a resulting strong coupling between the surface vibration field and the radiated acoustic field. Problem 6.16 (a) Equation 6.31 in the text is: The critical frequency, f c ' 0.551c 2 / (c L h) ' 0.551 × 343 2 /(5400 × 0.003) ' 4001 Hz 10log 10 S = 0. Ph/S = 4 × 0.003/1 = 0.012. Solutions to problems 144 L p ' L w & 10log 10 (2πr 2 ) % 10log 10 ρc 400 L p ' 10log 10 10 4.43 % 10 5.39 % 10 4.81 ' 55.3dB re 20µPa L pA ' 10log 10 10 (4.43 & 0.86) % 10 (5.39 & 0.3) % 10 4.81 ' 52.8dB(A) Thus the following table may be constructed. Octave band centre frequency (Hz) f/f c 10log 10 σ 10log 10 ¢ v 2 ¦ L w (dB re 10 -12 W) 250 500 1000 0.06 0.12 0.24 -19.8 -18.2 -16.0 -54.0 -46.0 -54.0 72.2 81.8 76.0 (b) From figure 5.11 in the text we can see that for , the panel r/ HL ' 10 will appear as a point source and the sound pressure level is given by: The quantity 10log 10 2πr 2 = 28 and 10log 10 (ρc/400) = 0.1. Thus the following table can be constructed. Octave band centre frequency (Hz) L w (dB re 10 -12 W) L p (dB re 20µPa) 250 500 1000 72.2 81.8 76.0 44.3 53.9 48.1 The overall sound pressure level is: (c) The A-weighted sound pressure level is: (d) r.m.s. acceleration levels (approximate). Sound power use and measurement 145 a = 2πfv (see following table) f (Hz) v (mm/s) a (m/s 2 ) 250 500 1000 2 5 2 3.1 15.7 12.6 M ' (20 % 25 % 30 % 32) / 177 ' 0.6 7 Solutions to problems relating to sound in enclosed spaces Problem 7.1 Direct field The direct field of a sound source is defined as that part of the sound field which has not suffered any reflection from any room surfaces or obstacles. Reverberant field The reverberant field of a source is defined as that part of the sound field radiated by a source which has experienced at least one reflection from a boundary of the room or enclosure containing the source. Problem 7.2 250Hz octave band, bandwidth from table 1.2 on p43 in the text is 353 - 176 = 177Hz. Thus the modal overlap, M, is (from equation 7.24 in the text) given by: Problem 7.3 (a) If n z = 0, the pressure distribution in the room is uniform and the monopole source will excite the mode. For n z = 1 and n z = 3, there will be a node at L z /2 and the monopole will not excite the mode. For n z = 2, there will be an antinode at L z /2 and the mode will be excited by a monopole source. (b) A dipole will be ineffective at exciting the n z = 0 and n z = 2 modes Sound in enclosed spaces 147 f 1,1,1 ' 343 2 1 10 2 % 1 5 2 % 1 2 2 ' 93.9Hz because the phase of each part of the dipole is opposite to the other and the modal response is not. On the other hand, one would expect good excitation of the n z = 3 mode because the phase on one side of a node is 180E different to that on the other side and this can be matched by the dipole. (c) At a rigid wall, the particle velocity is zero. The required derivation is described on pages 278 and 279 in the text. (d) The required shape is shown in the figure below where the nodes are represented by lines and the relative phases of acoustic pressure are shown by plus and minus signs. Using equation 7.17 in the text: Problem 7.4 (a) Cut-on frequency is the frequency at which the wavenumber, κ becomes real and the mode begins to propagate down the duct without decaying in amplitude (assuming a rigid, hard walled duct). (b) From the equation given in the problem, for a single mode the acoustic Solutions to problems 148 κ 3,2 ' (ω/ c) 2 & [ 45π 2 / L 2 z ] ω ' c[ 45π/ L z ] κ 3,2 ' [5π 2 / L 2 z ] & [ 45π 2 / L 2 z ] ' ± j 40π/ L z ' c 1 & 45π 2 c 2 / (ω 2 L 2 z ) ' c 1 & const / ω 2 c 32 f c f co pressure amplitude is given by . Thus the dB decay per ¯ p ' ¯ p 0 e &j κ mn x unit distance is: , where is the sound pressure Δ ' 20log 10 ¯ p 0 ¯ p 1 ' 20 × 0.4343 × jκ mn ¯ p 1 amplitude at 1m. For m = 3, n = 2, and L y = L z /3: Cut-on frequency is thus given by: Thus at 1/3 of the cut-on frequency, κ 3,2 is given by: Thus Δ = 172/L z (dB/m). (c) Phase speed, c 3,2 = ω/κ 3,2 = ω (ω 2 / c 2 ) & 45π 2 / L 2 z Sound in enclosed spaces 149 f ' c 2 1 6.2 ' 343 2 × 1 6.2 ' 27.7Hz ψ ' ¢ p 2 ¦ ρc 2 ¢ p 2 max ¦ ' ( 2 × 10 &5 ) 2 ×10 80/ 10 ' 0.04Pa 2 ¢ p 2 ¦ ' 0.04 × cos 2 πx L The variation of c 3,2 as a function of frequency is shown in the figure. Near the mode cut-on frequency it can be seen that the phase speed approaches infinity and at high frequencies it approaches the speed of sound in free space. Problem 7.5 (a) Room dimensions are 4.6m × 6.2m × 3.5m. The lowest resonance frequency is the axial mode corresponding to the longest room dimension. Thus: (b) Pressure distribution follows a half cosine wave in the 6.2m direction as shown in the figure and calculated from equation 7.19 in the text. The pressure is uniform across a given cross section defined by the 4.6m × 3.5m dimensions. Energy density is given by equation 7.32 in the text as: The sound pressure level in a room corner is 80dB and this corresponds to the maximum sound pressure. Thus: The sound pressure at any other location is (from equation 7.19 in the text): Thus the energy in the room is given by: Solutions to problems 150 E ' m V ψ dV ' 4.6 × 3.5 m L 0 ¢ p 2 ¦ ρc 2 dL ' 4.6 × 3.5 × 0.04 1.206 × 343 2 m L 0 cos 2 πx L dx ' 4.54 × 10 &6 x 2 % Lsin(2πx/ L) 4π L 0 dE E ' & Sc¯ α V dt m E E 0 dE E ' & m t 0 Sc¯α V dt log e E & log e E 0 ' & Sc¯αt V E ' E 0 e &Sc¯ αt / V ¢ p 2 ¦ ' ¢ p 2 0 ¦ e &Sc¯ αt / V L p &L p0 ' 4.343c¯αtS/ V ' 4.343c¯αt / L Substituting L = 6.2 gives E = 1.4 × 10 -5 Joules. (c) The analysis procedure used on p290 and 291 in the text may be used here. Alternatively, we may start with the given equation and write: Integrating gives: which gives: or As E % +p 2 ,, we can write: Thus: Sound in enclosed spaces 151 60 ' 4.343c¯ αT 60 / L ¯α ' 60 4.343 × L cT 60 ' 60 4.343 × 6.2 343 × T 60 ' 0.250 T 60 ¢ p 2 ¦ 4ρc ' I W a ' 2 × 0.04 4 × 1.206 × 343 × 0.05 × 16.1 ' 38.9µW W a ' dE dt ' ESc¯α V ' 1.407 × 10 &5 × 343 × 0.05 6.2 ' 38.9µW f ' 343 2 × 0.5861 5.5 ' 18.3Hz and Therefore: (d) T 60 = 5 seconds, so = 0.25/5 = 0.05. ¯α Power consumption, . and S w = 4.6 × 3.5 = 16.1 m 2 . The W a ' 2I¯αS w quantity, I, is the sound intensity in the direction of one wall. As shown on p287 in the text, the effective intensity, I, in one direction is related to the acoustic pressure in a diffuse field by: Substituting 0.04Pa 2 for the maximum mean square pressure and the above equation into the equation for W a gives for the absorbed power: Alternatively, the expression given in part (c) may be used to give: Problem 7.6 (a) Substituting L = 2.7, c = 343, a = 5.5 and the values of the characteristic function ψ given in the problem table into the equation for f given in the problem gives for the lowest order resonance frequency (n = 0, m = 1 and n z = 0) gives: Solutions to problems 152 f 1,1,1 ' 343 2 1 2.7 2 % 1.697 5.5 2 ' 82.7Hz (b) The lowest order mode pressure distribution is shown in the figure at right, where the nodal plane which runs the full length of the cylinder is indicated. The sound field is in opposite phase from one side of the nodal line to the other. The sound pressure will be at a maximum along two axial lines at the surface of the cylinder furthest from the nodal plane. (c) The air particles are oscillating with a velocity in-quadrature with the local acoustic pressure as the mode is characterised by a standing wave generated by the interference of two acoustic waves travelling in opposite directions across the nodal plane. (d) This problem may be answered by inspection of the given table and equation. The modes are listed in the table below in order of ascending frequency, column 1 followed by column 2 etc. 1,0 2,0 0,1 3,0 4,0 1,1 5,0 2,1 0,2 6,0 3,1 1,2 7,0 4,1 2,2 0,3 5,1 3,2 1,3 6,1 4,2 2,3 7,1 0,4 5,2 3,3 1,4 6,2 4,3 2,4 7,2 5,3 3,4 6,3 4,4 7,3 5,4 6,4 7,4 (e) f = 343/(2 × 2.7) = 63.5Hz. (f) Axial modes have 2 zeroes in the subscripts n z , n, and m. Tangential modes have one zero and oblique modes have none. These criteria can be used to list any number of axial, tangential and oblique modes. The resonance frequency of the first oblique mode is: Modes with resonances below this with n z = 1 are found from the table in the problem as those with ψ less than the value of ψ corresponding to Sound in enclosed spaces 153 ψ < 82.7 × 2 × 5.5/ 343 ' 2.651 area of circular region area of square region ' πr 2 4r 2 ' π 4 r n = 1 and m = 1. For n z = 0, the value of ψ must be below that given by: The modes with resonance frequencies below the (1,1,1) mode are listed below in the form (n z , m, n). 1,0,0 1,1,0 1,2,0 1,3,0 1,4,0 1,0,1 0,1,0 0,2,0 0,3,0 0,4,0 0,5,0 0,6,0 0,0,1 0,1,1 0,2,1 0,3,1 0,0,2 That is, there are 17 modes with resonance frequencies below that of the first oblique mode. Of these, 9 are axial and 8 are tangential modes. Problem 7.7 (a) A 2-D space would be one where the dimensions in 2 directions were very much larger than in the third direction. An example would be a large factory with a low ceiling in the frequency range below the first floor/ceiling axial resonance frequency. Following the procedure for a 3-D space described on p285-287 in the text, a 2-D square region enclosing a circle is considered as shown in the figure. Time for sound to travel through the circular region (length of encompassing square) = 2r/c. Energy in the square region of length 2r (and unit thickness) as a result of a wave travelling normally to any of the sides is , where I is the incident wave intensity. The energy per unit I 2r c perimeter in the circular region as a result of an incident wave 2r wide is: Solutions to problems 154 ΔE ' π 4 I 2r c E ' m 2πr ΔE dx ' m 2π 0 ΔEr dθ ' π 2 Ir 2 c m 2π 0 dθ ' I r 2 π 2 c ψ ' E S ' I r 2 π 2 cπr 2 ' πI c I ' ψc π ψ ' I c ' ¢ p 2 ¦ ρc 2 ψ ' πI c ' ¢ p 2 ¦ ρc 2 I ' ¢ p 2 ¦ πρc 1 1 A The total energy in the circular region due to waves from all directions is found by integrating the preceding expression over the perimeter of the circle. Thus: Energy density, ψ, is given by: Thus: Consider a plane wave travelling unit distance. The energy in unit area is the intensity multiplied by the time it is present which is, E = I/c. The energy is also equal to the energy density multiplied by unit area. Thus ψ = E. Thus for a plane wave: Equation 7.32 in the text also gives the relation between sound pressure and energy density. Thus for a diffuse 2-D field: Rearranging gives: (b) An example of a 1-D field would be the inside of a rigid tube closed at both ends at frequencies below the first higher order mode cut-on Sound in enclosed spaces 155 I ' ¢ p 2 ¦ 2ρc W a ' I P¯α ' ψc π P¯α W ' S Mψ Mt ' W 0 & ψc π P¯ α X ' [ πW 0 / Pc¯ α] & ψ dX dt ' & Mψ Mt ' & 1 S W 0 & ψc π P¯ α 1 X ' P¯ αc π W 0 & ψc π P¯ α &1 L 1 A frequency. Assuming unit cross-sectional area and a wave travelling from left to right over a distance of L. The time for a wave to travel a distance, L, is L/c. The energy in the wave travelling from left to right is IL/c. The total energy (due to left and right travelling waves) is thus 2IL/c. The total energy in the volume is also ψL. Thus, I = ψc/2 and ψ = 2I/c. Using this relation and equation 7.32 in the text gives for a 1-D field: (c) Let S = area of 2-D room of unit height as shown in the figure. Rate of energy absorbed, W a , around the perimeter, of length P and absorption coefficient is given by: ¯α Rate of change of energy in the reverberant field = rate of supply, W 0 - rate absorbed, W a . Thus: Introducing the dummy variable: into the preceding equation, we may write: and Solutions to problems 156 1 X dX dt ' & P¯ αc πS m X X 0 dX X ' m t 0 & P¯αc πS log e X & log e X 0 ' & P¯ αct πS X ' X 0 e &P¯ αct / πS ψ ' ψ 0 e &P¯ αct / πS ¢ p 2 ¦ ' ¢ p 2 0 ¦ e &P¯ αct / πS W a ' 2I ¯α ' 2 ψc 2 ¯α L Thus: Integrating gives: Thus: and For a decaying sound field, W 0 = 0 at t = 0. Using the preceding equation, X = X 0 when t = 0. Also the definition of the dummy variable, X, above can be used to show that when W 0 = 0, X = X 0 = -ψ 0 . The same equation may be used to show that as W 0 = 0, then at any time t, X = -ψ. From the preceding discussion and the equation above we may write: We know that ψ % +p 2 ,. Thus: (d) Let L be the length of the 1-D tube of unit cross sectional area. Rate of energy absorbed, W a , at the ends of the tube of length L and absorption coefficient is given by: ¯α Rate of change of energy in the reverberant field = rate of supply, W 0 - Sound in enclosed spaces 157 W ' L Mψ Mt ' W 0 & ψc¯ α X ' [ W 0 / c¯α] & ψ dX dt ' & Mψ Mt ' & 1 L W 0 & ψc¯ α 1 X ' ¯αc W 0 & ψc¯α &1 1 X dX dt ' & ¯αc L m X X 0 dX X ' m t 0 & ¯ αc L log e X & log e X 0 ' & ¯αct L X ' X 0 e &¯ αct / L rate absorbed, W a . Thus: Introducing the dummy variable: into the preceding equation, we may write: and Thus: Integrating gives: Thus: and For a decaying sound field, W 0 = 0 at t = 0. Using the preceding equation, X = X 0 when t = 0. Also the definition of the dummy variable, X, above can be used to show that when W 0 = 0, X = X 0 = -ψ 0 . The same equation may be used to show that as W 0 = 0, then at any time t, X = -ψ. From the preceding discussion and the equation above we may write: Solutions to problems 158 ψ ' ψ 0 e &¯ αct / L ¢ p 2 ¦ ' ¢ p 2 0 ¦ e &¯ αct / L c x ' ccosθ and c y ' csinθ N r ' ccosθ L x % csinθ L y N av ' 1 π/ 2 m π/ 2 0 N r dθ ' 2c π m π/ 2 0 cosθ L x % sinθ L y ' 2c π sinθ L x & cosθ L y π/ 2 0 ' 2c π 1 L x % 1 L y 0 L y L x y x We know that ψ % +p 2 ,. Thus: (e) Mean free path, 2-D space. Sound propagation in any direction in the 2- D plane is equally likely (see figure). Due to symmetry, we need to consider only one of the 4 quadrants and find an average path length between reflections for sound propagating in these directions. Consider a sound wave propagating with speed c, in an angular direction of ψ from the x-axis as shown in the figure. Resolving into x and y components, we have: The number of reflections per unit time for a wave travelling in the θ direction is given by: Averaging over all directions in one quadrant (the average for the other three quadrants would be the same) we find that the average number of reflections per unit time is: Sound in enclosed spaces 159 Λ ' π 2 L x L y L x % L y ' πS P L w ' L p % 10log 10 V & 10log 10 T 60 % 10log 10 (1 % Sλ/8V) & 13.9 ' 95%10log 10 (105) &10log 10 (2.5) %10log 10 1% 142 × 0.686 8 × 105 &13.9 ' 95 % 20.2 & 4.0 % 0.5 & 13.9 ' 97.8dB re 10 &12 W η ' acoustic power electrical power ' 10 &12 × 10 97.8/ 10 10 ' 6 × 10 &4 ' 0.06% c sinu c cosu u But N av = c/Λ, where Λ is the mean free path. Thus: For a 1-D space (see figure), the mean free path, which is t he di st ance bet ween reflections is equal to L, the length of the space. Problem 7.8 (a) Room size = 7m × 5m × 3m. Volume = 105m 3 , surface area = 2(7 × 5 + 7 × 3 + 5 × 3) = 142m 2 . Wavelength of sound, λ = 343/500 = 0.686m. The sound power in the reverberation room is related to the sound pressure level by equation 6.13 in the text. Thus: Alternatively, we could assume that the correction term, (1 + Sλ/8V) is already included in L p . In this case, L w = 97.8 - 0.5 = 97.3dB. Power conversion efficiency is given by: (b) Fire box dimensions are: 10 × 12 × 20m. Axial resonances - first mode in each direction given by: Solutions to problems 160 f ' c 2L ' 864 2 × 10 , 864 2 × 12 , 864 2 × 20 ' 43.2Hz, 36Hz, 21.6Hz ¯ p ' 2p rms ' 2 2 × 10 &5 × 10 155/ 20 ' 1.59kPa I ' p 2 i ρc ' ( 1.125 × 10 3 ) 2 ( 1 % 1 & ¯α ) 2 ρc (i) As the frequency of instability is 36Hz, it seems likely that it is associated with the axial mode across the width in the 12m direction. (ii) The acoustic pressure will be largest at the two side walls normal to the 12m dimension, because on reflection the pressure amplitude is doubled. (iii) Amplitude of cyclic force acting on walls. Acoustic pressure is 155dB. Amplitude is then: Side wall area = 10 × 20 = 200m 2 . Force on each wall = 1.59 × 200 = 318kN. (iv) Power absorbed by 2 side walls. Pressure at wall is the sum of the incident and reflected pressures. Thus, . However, . (p i % p r ) rms ' 1.125 × 10 3 p r ' ( 1 & ¯α) p i So, [ 1 % ( 1 & ¯α) ] p i ' 1.125 × 10 3 The absorbed power is controlled by the incident sound intensity which is: The absorbed power is , where A w = area of one wall. W a ' 2¯αI A w For a 1-D sound field, . Thus: ¢ p 2 ¦ ' ¢ p 2 0 ¦ e &c¯αt / L L p0 & L p ' 4.343c¯αt / L and . But T 60 ' ( 60/ 4.343) × L c¯α Sound in enclosed spaces 161 T 60 ' 2.2Q f n ' 2.2 × 30 36 ' 1.833secs ¯α ' ( 60/ 4.343) × 12 864 × 1.833 ' 0.105 W a ' 2 × 0.1047 × ( 1.1246 × 10 3 ) 2 × 200 (1 % 1 & 0.1047) 2 × 1.16 × 864 ' 14kW η ' 13953 800 × 10 3 ' 1.7% L p0 & L p ' 4.343c¯αt / L 60 ' 4.343 × 343 × 0.05 × T 60 / 5 L p0 & L p ' 4.343P¯ αct / (πS) 60 ' 4.343 × 20 × 0.25 × 343 × T 60 / (π × 25) Thus: The absorbed power is then: (v) Power conversion efficiency is: Problem 7.9 (a) Taking logs of the equation given in the problem for a 1-D field, we have: Thus: which gives, T 60 = 4.0 seconds. (b) Again taking logs of the equation given in the problem gives: P = 4 × 5 = 20m and S = 5 × 5 = 25m 2 . Thus: which gives, T 60 = 0.63 seconds. (c) S = 2(5 × 5 + 5 × 5 + 5 × 5) = 150m 2 Solutions to problems 162 60 ' 1.086 × 150 × 343 × 0.1833 × T 60 / 125 W ' IA ' ¢ p 2 ¦ 4ρc A ' 1.265 4 × 413 × 30 ' 0.023 W ' 103.6 dB re 10 &12 W V = 5 × 5 × 5 = 125m 3 ¯ α ' 50 × 0.05 % 100 × 0.25 150 ' 0.183 Using equation 7.50 in the text, we obtain: which gives, T 60 = 0.73 seconds. Problem 7.10 In a reverberant field the sound intensity in any particular direction is equal to that in any other direction resulting in a net active intensity (averaged over all directions) of zero and a large reactive intensity. Thus in the situation under consideration here, the active intensity will only be contributed to by the direct field of the source and the reactive intensity field will dominate the active field by a large amount, especially at large distances from the source. Due to the dominance by the reactive field together with limitations on the phase accuracy between the two microphones of any measurement system, accurate measurements of the active field will not be generally feasible in a reverberant room. Problem 7.11 (a) Energy density in a reverberant field is and the given SPL is 95 ¢ p 2 ¦ / ρc 2 dB. Thus, and the energy density is ¢ p 2 ¦ ' (2 × 10 &5 ) 2 × 10 9.5 ' 1.265 Pa 2 then: Energy density = 1.265/(413 × 343) = 8.9 × 10 -6 J/m 3 (b) Sound power incident on a wall is given by: Sound in enclosed spaces 163 T 60 ' 55.25 × 400 343 × 360 × 0.1 ' 1.8seconds W ' 360 × 0.1 × 4 × 10 &4 4 × 1.206 × 343 × (1 & 0.1) ' 9.67µW I ' 4 × 10 &4 4 × 1.206 × 343 ' 2.42 × 10 &7 W/ m 2 r ' S¯ α 8π( 1 & ¯α) 1/2 ' 36 8π × 0.9 1/2 ' 1.26m Problem 7.12 (a) Room 10m × 10m × 4m, S = 2(10 × 10 + 10 × 4 × 2) = 360m 2 , V = 10 × 10 × 4 = 400m 3 . Using equation 7.51 in the text for reverberation time, we may write: (b) L p = 60dB corresponds to +p 2 , = 4 × 10 -10 × 10 60/10 = 4 × 10 -4 Pa 2 . Using equation 7.41 in the text we obtain: Sound power level, L w = 10log 10 (9.67 × 10 -6 ) + 120 = 69.9dB re 10 -12 W. (c) Using equation 7.33 in the text: (d) From equations 7.40 to 7.42 in the text, the reverberant field level is equal to the direct field level when D θ /(4πr 2 ) = 4/R = . 4( 1 & ¯α)/( S¯α) Assuming the acoustic centre of the source is within a quarter wavelength from the hard floor, D θ = 2 and the distance, r, at which the fields are equal is: Problem 7.13 (a) Room 3.05m × 6.1m × 15.24m, S = 2(3.05 × 6.1 + 3.05 × 15.24 + 6.1 × 15.24) = 316.1m 2 , V = 3.05 × 6.10 × 15.24 = 283.54m 3 . Using equation 7.51 in the text for reverberation time, we may write: Solutions to problems 164 ¯α ' 55.25V ScT 60 ' 55.25 × 283.54 316.1 × 343 × 2 ' 0.072 L w ' L p % 10log 10 4(1 & ¯ α) S¯ α % 10log 10 ρc 400 ' 74 & 10log 10 4 × 0.928 316.1 × 0.072 & 0.15 ' 81.7dB re 10 &12 W 316.1¯ α ' 246.1 & 246.1¯ α T 60 ' 55.25V Sc¯ α ' 55.25 × 283.54 316.1 × 343 × 0.438 ' 0.33seconds c ' γRT M ' 1.4 × 8.314 × 1473 0.035 ' 700 m/ s Using equation 7.42, the sound power output may be written as: (b) To lower the reverberant field by 10dB, we need to increase R by a factor of 10. Old = 24.61m 2 , thus required new = S¯α/ (1 & ¯α) S¯α/ (1 & ¯α) 246.1m 2 . Thus in the new situation, expanding the above equation gives: which results in a new required value of = 0.438. The old value of ¯α ¯α is 0.0722, so the increase in absorption needed is: = 316.1(0.438 - 0.0722) = 115.5m 2 . Δ S¯α (c) Using the value of = 0.438 obtained above and equation 7.51 in the ¯α text, we obtain: Problem 7.14 (a) Sound power level = 10 log 10 W + 120 = 10 log 10 3.1 + 120 = 124.9 dB re 10 -12 W (b) Sound in enclosed spaces 165 R b ' S¯α (1 & ¯α) ' 400 × 0.08 0.92 ' 34.78m 2 ¯ α n ' 100 × 0.08 % 300 × 0.5 400 ' 0.395 L p ' L w % 10log 10 4(1 & ¯ α ) S¯ α %10log 10 ρc 400 ' 124.9 % 10log 10 4 × 0.95 100 × 0.05 % 10log 10 700 × 0.29 400 ' 120.8 dB ρ ' γP c 2 ' 1.4 × 101.4 × 10 3 700 2 ' 0.29 kg/ m 3 Surface area, S = πdL + πd 2 /2 = π × 4 × 6 + π × 16/2 = 32π = 100 m 2 Problem 7.15 L p ' 10log 10 ( 10 9.5 % 10 9.7 % 10 9.9 ) ' 102.1 dB(A) L p ' L w % 10log 10 Q 4πr 2 % 4(1 & ¯ α) S¯ α Thus, 102.1 & 110 ' 10log 10 2 4πr 2 % 3.64 320 × 0.09 and, 10 &7.9/ 10 & 3.64 320 × 0.09 ' 1 6.28r 2 Thus, r ' 1 6.28 × 0.0358 ' 2.1 m Problem 7.16 Room 10m × 10m × 5m, S = 2(10 × 10 + 10 × 5 × 2) = 400m 2 , V = 10 × 10 × 5 = 500m 3 . The room constant before treatment is Adding sound absorbing treatment to the walls and ceiling results in a new mean absorption coefficient calculated as follows: Solutions to problems 166 R n ' S¯α (1 & ¯α) ' 400 × 0.395 0.605 ' 261.2m 2 Δ L p ' 10log 10 D θ 4πr 2 % 4 R old & 10log 10 D θ 4πr 2 % 4 R new ' 10log 10 2 4π3 2 % 4 34.78 & 10log 10 2 4π3 2 % 4 261.2 ' 6.0dB ¯ α ' 48 × 0.15 % 132 × 0.05 180 ' 0.0767 ¢ p 2 R ¦ ' 4 × 25 × 10 &3 × 1.206 × 343 (1 & 0.0767) 180 × 0.0767 ' 2.766 Pa 2 L pR ' 10log 10 (2.766) % 94 ' 98.4dB Thus the new room constant after treatment is: Using equation 7.42 in the text allows the difference in sound pressure level (assuming that the sound power is constant) to be calculated as follows (assuming the acoustic centre of the machine is within a quarter wavelength of the hard floor): Problem 7.17 (a) Room 8m × 6m × 3m, S = 2(8 × 6 + 8 × 3 + 6 × 3) = 180m 2 . The mean Sabine absorption coefficient may be calculated using equation 7.78 in the text to give: Using equation 7.41, we can write: The reverberant field sound pressure level is then: (b) Direct and reverberant fields equal (see equation 7.42 in the text) when Sound in enclosed spaces 167 r ' S¯ α 16π(1 & ¯ α) 1/2 ' 180 × 0.0767 16π × (1 & 0.0767) 1/2 ' 0.55m 10log 10 ¢ p 2 ¦ ' 10log 10 Wρc πa 2 & 8 L p(reverb) ' L w % 10log 10 ρc 400 & 10log 10 ( πa 2 ) & 8 ' 130 % 0.14 & 18.95 & 8 ' 103.2 dB L p(direct) ' L w % 10log 10 ρc 400 & 10log 10 ( 4πr 2 ) ' 130 % 0.14 & 24.97 ' 105.2 dB . Assuming that the acoustic centre of the source is well D θ / 4πr 2 ' 4/ R above the floor, D θ = 1 and the distance, r, at which the two fields are equal is: Problem 7.18 (a) This is a flat room as on pages 314-319 in the text. Pressure reflection coefficient amplitude = 0.7, β = 0.7 2 and ¯ α ' 1 & 0.7 2 ' 0.51 Room height, a = 5 m and distance r = 5 m. Thus r/a = 1. From the figure, the reverberant field sound pressure is given by: Thus the reverberant field sound pressure level is: The direct field sound pressure level is: (b) Sabine room, area, and volume, S ' 2[10 × 10 % 10 × 5 × 2] ' 400 m 2 V ' 10 × 10 × 5 ' 500 m 2 The total sound pressure level (direct plus reverberant) in the room is: Solutions to problems 168 L p ' L w % 10log 10 Q 4πr 2 % 4(1 & ¯α) S¯α % 0.15 C ' 10log 10 0.3 % S E ( 1 & ¯ α) S i ¯ α ' 10log 10 0.3 % 300 × 0.49 400 × 0.51 ' 0.1 dB L p ' 87.4 & 24.9 ' 62.5 dB L p ' L w % 10log 10 D 4πr 2 % 4( 1 & ¯ α) S¯ α ' 130 % 10log 10 1 4π × 25 % 4 × 0.49 400 × 0.51 ' 111 dB r ' DS¯ α 4π × 4( 1 & ¯ α) ' 400 × 0.51 16π × 0.49 ' 2.9 m T 60 ' 55.25V Sc¯ α ' 55.25 × 500 400 × 343 × 0.51 ' 0.39 seconds L p ' L w % 0.14 & 10log 10 2πr 2 ' 130 % 0.14 & 42.8 ' 87.4 dB (c) Direct and reverberant fields equal when . D/ 4πr 2 ' 4(1 & ¯ α) / S¯ α Thus: (d) (e) Treat the room like an enclosure. Thus, the sound level at the receiver without the enclosure at a distance of 50 + 5 m is: The enclosure noise reduction is given by NR = TL - C, where Thus the noise reduction = 24.9 dB and the SPL at 50 m is: Problem 7.19 Q = 2, r = 1, α = 0.08, L w = 95dB, S = 400. Thus: Sound in enclosed spaces 169 L p ' 95 % 10log 10 2 4 × π % 4(1&0.08) 400 × 0.08 % 0.15 ' 89.5dB L w ' L p % 20log 10 r % 10log 10 (4π) & 3 % 10log 10 400 ρc ' 80 % 9.54 % 11.00 & 3.01 & 0.15 ' 97.4dB L p ' L w % 10log 10 (4/ R) % 0.15 10log 10 R ' 97.38 & 85 % 0.15 % 6.02 ' 18.55 328¯ α (1 & ¯α) ' 71.62 Problem 7.20 (a) Reference source on hard asphalt (DI = 3), L p = 80dB at 3m. Sound power level is: (b) Assuming a negligible direct field, the room constant may be calculated using equation 7.42 in the text (with a 0.15dB correction for ρc different from 400). Thus: Substituting in values for the parameters gives: Thus, R = 71.6. (c) Room 14m × 6m × 4m, S = 2(14 × 6 + 14 × 4 + 6 × 4) = 328m 2 . Using equation 7.43 in the text gives: Thus, = 0.18. ¯α (d) Existing L p = 75dB and the allowable total = 80dB. Thus the allowable contribution from the three new line printers is . The allowable contribution from each 10log 10 (10 8 & 10 7.5 ) ' 78.3dB line printer is then . The allowable sound 10log 10 10 7.83 3 ' 73.5dB power level is the obtained using equation 7.42 in the text (using only the reverberant field part) as follows: Solutions to problems 170 L w ' 73.6 & 10log 10 4 71.6 & 0.15 ' 86.0dB R ' 328 × 0.261 (1 & 0.261) ' 116m 2 Δ L p ' 10log 10 116 71.6 ' 2.1dB L w ' 74.29 & 10log 10 4 116 & 0.15 ' 88.8dB (e) Area of ceiling = 14 × 6 = 84m 2 , corresponding = 0.5. ¯α Area of floor and walls = 328 - 84 = 244m 2 , corresponding = 0.179. ¯α Average . Thus: ¯ α ' 0.5 × 84 % 0.179 × 244 328 ' 0.261 Using equation 7.120 in the text, the reduction in sound pressure level is thus: Thus the level in the room before addition of the new printers will now be 75 - 2.1 = 72.9dB. The allowable sound pressure level of the three new line printers together is now . The 10log 10 (10 8 & 10 7.29 ) ' 79.1dB allowable contribution from each line printer is then . The allowable sound power level is the 10log 10 10 7.91 3 ' 74.3dB obtained using equation 7.42 in the text (using only the reverberant field part) as follows: Problem 7.21 (a) Considering only the reverberant field of the machine we may use equation 7.41 in the text. The surface area of the factory is S = 2(10 × 10 × 3) = 600m 2 . An L p of 83dB corresponds to a +p 2 , of . Using equation 7.41, the required 4 × 10 &10 × 10 83/ 10 ' 7.98 × 10 &2 Pa 2 mean absorption coefficient is given by: Sound in enclosed spaces 171 ¯α (1 & ¯α) ' 4 × 0.01 × 1.206 × 343 7.981 × 10 &2 × 600 ' 0.346 600 × 0.257 ' 100 × 0.01 % 500x 90 ' 100 % 10log 10 2 4πr 2 % 4 208 Thus . However this would be the required absorption ¯α ' 0.257 coefficient if the floor were lined as well. If we assume that the concrete floor has an absorption coefficient of 0.01, and we let the required absorption coefficient of the walls and ceiling be x, then we can use equation 7.78 in the text to write: which gives x = 0.306. Thus the required absorption coefficient for the walls and ceiling is 0.31. (b) The room constant, R, is from part (a) 0.346 × 600 = 208m 2 . The sound power level, L w = 10log 10 (0.01) + 120 = 100dB. For a total L p of 90dB we may write (assuming a directivity factor, D θ = 2 as the source is assumed close to a hard floor and other surfaces are more absorptive): Solving the above gives r = 1.40m as the radius around the machine within which the sound pressure level will exceed 90dB. Problem 7.22 (a) From Table 4.10 in the text, the allowable community noise level to ensure minimal risk of complaints is 40 + 15 - 10 = 45dB(A). If the only noise is in the 500Hz octave band then the -3.2dB A-weighting at this frequency, results in an allowable level of 48.2dB in that band. The existing level is 44dB so the allowed increase is 4.2dB. A reverberant field sound pressure level corresponds to a sound power level (see equation 7.42 in the text) of: Solutions to problems 172 L w ' 88 & 10log 10 4(1 & ¯α) S¯α & 0.15 dB ¯α ' 55.25 × 2500 343 × 2.1 × 1450 ' 0.132 L w ' 88 & 10log 10 4(1 & 0.132) 1450 × 0.132 & 0.15 ' 105.3dB L w ' 10log 10 10 109.5/ 10 & 10 105.3/ 10 ' 107.4dB S¯α ' 500 × 0.5 % 950 × 0.132 ' 375.4m 2 Room 25m × 20m × 5m, S = 2(25 × 20 + 25 × 5 + 20 × 5) = 1450m 2 . V = 25 × 20 × 5 = 2500m 3 T 60 = 2.1 secs. From equation 7.51 in the text: Thus: This is the sound power level of the existing equipment. Thus the allowable total sound power of existing + new equipment is 105.3 + 4.2 = 109.5dB. Thus the allowed power level for the new equipment is: As there are 5 new machines, the allowed power level for each is L w = 107.4 - 10log 10 (5) = 100.4dB Assumptions ! Only absorption is due to floor, walls and ceiling ! Air temperature of 20EC. ! The relative contribution of direct and reverberant sound energy to the community noise levels will be the same for the new machines as for the old machines. (b) If ceiling tile with = 0.5 were added then the new is: ¯α S¯α corresponding to = 0.259 ¯α Old R = and (1450 × 0.132) / (1 & 0.132) ' 220.5m 2 new R = . (1450 × 0.259) / (1 & 0.259) ' 506.8m 2 Thus the allowed increase in sound power level for the same reverberant Sound in enclosed spaces 173 Δ L w ' 10log 10 (506.8) & 10log 10 (220.5) ' 3.6dB W a ' ψcS¯ α/ 4 ' 0.282 × 343 × 0.144 × 0.1/ 4 ' 0.35W L p ' L w % 10log 10 4(1 & ¯α) S¯α %10log 10 ρc 400 L p ' 84 % 10log 10 4 × 0.85 216 × 0.15 % 0.15 ' 74.4dB D θ 4πr 2 ' 1 4π × 1.5 2 ' 0.035 field sound pressure level is: Assuming that the reverberant field dominates the direct field, the new allowed power level of each machine is 100.4 + 3.6 = 104dB. Problem 7.23 (a) Energy density, ψ = ¢ p 2 ¦ / ρc 2 Thus ψ = (2 × 10 &5 ) 2 × 10 140/ 10 / (1.205 × 343 2 ) ' 0.28J / m 3 (b) Enclosure surface area, S = 2(0.2 × 0.15 + 0.2 × 0.12 + 0.15 × 0.12) = 0.144m 2 . Power flow into walls (equation 7.39a) is given by (c) The power generated is equal to the power absorbed by the walls. Thus the power required to drive the source is 0.348/0.2 = 1.7W. Problem 7.24 (a) Reverberant field L p from taking logs of equation 7.41 in the text. S = 2(10 × 6 + 10 × 3 + 6 × 3) = 216m 2 , = 0.15, so: ¯α (b) Compare As the acoustic centre of the source is D θ 4πr 2 with 4(1 & α) Sα well above the hard floor, D θ = 1; so: Solutions to problems 174 4(1 & α) S¯ α ' 4 × 0.85 216 × 0.15 ' 0.105 L p ' 84 % 10log 10 0.0354 % 4 × 0.85 216 × 0.15 % 0.15 ' 75.6dB(A) L p ' 84 % 10log 10 0.0354 % 4 × 0.5 216 × 0.5 % 0.15 ' 71.5dB(A) r ' 216 × 0.15 4 × 4 × π × 0.85 ' 0.87m L p ' 90.2 % 10log 10 4 × 0.85 216 × 0.15 % 0.15 ' 80.6dB(A) L p ' 80.6 ' 84 & 10log 10 (4πr 2 ) % 0.15 Thus the reverberant field dominates. (c) Using equation 7.42 in the text (and allowing for ρc not equal to 400) the sound pressure level corresponding to = 0.15 is: ¯α and the sound pressure level corresponding to = 0.5 is: ¯α This corresponds to a reduction of 4.1dB. (d) From equation 7.42 in the text, when each machine is running separately, the fields are equal when . Thus for each machine: D θ 4πr 2 ' 4(1 & α) Sα When the machines are running together, the reverberant field contribution will be the sum of the two reverberant fields originating from each machine. The total sound power output is . The reverberant field sound pressure 10log 10 10 8.9 % 10 8.4 ' 90.2dB level is then: Thus the required distance is that at which the direct field L p is equal to 80.6dB(A). For the original machine: Sound in enclosed spaces 175 r ' 216 × 0.5 4 × 4 × π × 0.5 ' 2.07m L p ' 90.2 % 10log 10 4 × 0.5 216 × 0.5 % 0.15 ' 73.0dB(A) L p ' 73.0 ' 84 & 10log 10 (4πr 2 ) % 0.15 L p ' 73.0 ' 89 & 10log 10 (4πr 2 ) % 0.15 Thus r = 0.42m (distance from original machine at which original machine direct field = reverberant field with both machines running). For the new machine, L p ' 80.6 ' 89 & 10log 10 (4πr 2 ) % 0.15 Thus r = 0.75m (distance from new machine at which new machine direct field = reverberant field with both machines running). (e) Increase α to 0.5. From equation 7.42 in the text, when each machine is running separately, the fields are equal when . Thus for each machine, D θ 4πr 2 ' 4(1 & α) Sα The reverberant field sound pressure level when both machines are running is, Thus the required distance is that at which the direct field L p is equal to 73.0dB(A). For the original machine, Thus r = 1.02m (distance from original machine at which original machine direct field = reverberant field with both machines running). For the new machine, Thus r = 1.81m (distance from new machine at which new machine direct field = reverberant field with both machines running). (f) Assumptions: ! Each machine is radiating omni-directional sound. ! Both machines exhibit similar frequency spectra. ! The frequency averaged absorption coefficient is obtained using a sound source with a frequency spectrum similar to that of the machines under test. Solutions to problems 176 T 60 ' 55.25 × 900 600 × 343 × 0.1 ' 2.4secs R ' 600 × 0.1 1 & 0.1 ' 66.67m 2 L p ' L w % 10log 10 (4/ R) % 10log 10 4 % 0.15 ' 94 % 10log 10 (4/ 66.7) % 10log 10 (4) % 0.15 ' 94 & 12.22 % 6.02 % 0.15 ' 88.0dB (87.95) L p ' 10log 10 10 7.5 % 10 8.80 ' 88.2dB (88.16) L p ' 10log 10 10 7.5 % 10 7.795 ' 79.7dB ΔL p ' 88.16 & 85 ' 3.16dB ' 10log 10 R f / R i Problem 7.25 (a) Room 10m × 15m × 6m, S = 2(10 × 15 + 10 × 6 + 15 × 6) = 600m 2 . V = 10 × 15 × 6 = 900m 3 . The room reverberation time can be calculated using equation 7.51 in the text. Thus: (b) Room constant from equation 7.43 in the text: (c) For each machine, L w = 94dB. Thus for 4 new machines, L w = 94 + 10log 10 4. Thus the reverberant field sound pressure level due to the 4 new machines is: The existing reverberant field level is 75dB prior to installation of the new machines. Thus the total level after installation is: (d) When quiet machines are installed, the sound power and the sound pressure level contribution due to the new machines is reduced by 10dB to 94 and 78.0 respectively. Thus the total reverberant field level after installation of quiet machines is: (e) The design goal is 85dB. Thus the required reduction due to ceiling tile if untreated machines are used is: Sound in enclosed spaces 177 0.187 ' 0.1(600 & x) % 0.6x 600 D 4πr 2 ' 4( 1 & ¯ α) S¯ α or r ' DS¯ α 16π( 1 & ¯ α) ¯ α ' (1450 & 500) × 0.1 % 500 × 0.5 1450 ' 0.238 Thus . The initial room constant is 66.67m 2 . R f / R i ' 10 3.16/ 10 ' 2.07 Thus the new requirement is R = 138m 2 . The required average absorption coefficient is then given by . Thus . To achieve this, let x 600 × ¯ α ' 138 & 138 × ¯α ¯α ' 0.187 square metres of room surface be covered by tile. Then: which gives x = 105m 2 . Area of ceiling = 10 × 15 = 150m 2 . So covering the ceiling with ceiling tile would be adequate. This would cost $50 + 3 × 150 = $500 which is much less than the machine noise control and is thus the preferred option. Problem 7.26 (a) Distance at which direct and reverberant fields are equal is given by: The room surface area, S, is given by S ' 2( 20 ×25 % 20 ×5 % 25 × 5) ' 1450m 2 So r ' 2 × 1450 × 0.1 16π × 0.9 ' 2.5m (b) In this case, D=4 instead of 2 and r = 2.53 × 2 1/2 = 3.6m (c) If the ceiling were covered with ceiling tiles, the new sabine absorption coefficient would be: So r ' 2 × 1450 × 0.238 16π × 0.762 ' 4.2m Solutions to problems 178 L w ' L p & 10log 10 (4/ R) & 0.15 L p1 & L p2 ' 10log 10 D 4πr 2 % 4(1 & ¯α 1 ) S¯α 1 & 10log 10 D 4πr 2 % 4(1 & ¯ α 2 ) S¯ α 2 ' 10log 10 2 4π × 0.5 2 % 4 × 0.9 1450 × 0.1 & 10log 10 2 4π × 0.5 2 % 4 × 0.762 1450 × 0.238 ' &1.795 % 1.901 ' 0.1dB (d) The operator is only 0.5m from the machine so he/she is in the direct field. The reverberant field contribution at this distance is small so the ceiling tiles will have only a very small effect. Could calculate this (but not necessary). Problem 7.27 (a) Room 5m × 5.5m × 3m, S = 2(5 × 5.5 + 5 × 3 + 5.5 × 3) = 118m 2 . V = 5 × 5.5 × 3 = 82.5m 3 . Perimeter, L = 4(5.5 + 5 + 3) = 54m. Assuming that measurements of L p are made far enough from the machine that the direct field is negligible compared to the reverberant field, the sound power level is given by: This allows the following table to be constructed. Octave band centre frequency (Hz) L p ¯α R L w 63 250 1000 4000 75 85 84 70 0.01 0.02 0.02 0.03 1.192 2.408 2.408 3.649 69.6 82.6 81.6 69.5 Overall 85.4 (b) The reverberation time may be calculated using equation 7.51 in the text and the modal density may be calculated using equation 7.21. Equations Sound in enclosed spaces 179 2.569 × 10 &5 f 2 % 1.575 × 10 &3 f % 0.0197 ' 7.673 f ' & 1.575 × 10 &3 2 × 2.569 × 10 &5 % ( 1.575 × 10 &3 ) 2 % 4 × 2.569 × 10 &5 × 7.673 2 × 2.569 × 10 &5 ' &31 % 547 ' 516Hz d ' 5.5 2 2 % 5 2 2 % 3 2 2 ' 4.01m 7.23 and 7.24 are then used to calculate the modal overlap. The following table may be constructed. Octave band centre frequency (Hz) ¯α T 60 Δf dN df M 63 250 1000 4000 0.01 0.02 0.02 0.03 11.26 5.63 5.63 3.75 0.195 0.391 0.391 0.590 0.22 2.0 27.3 417.4 0.04 0.79 10.6 246.3 The reverberation time may be assumed constant between 250 and 1000Hz, so we need to calculate the frequency where dN/df = 3/0.391 = 7.67. That is: Thus: Thus the modal overlap is greater than 3 for frequencies above 520Hz. (c) The effect of acoustic tile may be calculated using equation 7.42. The quantity L w will remain the same in each case so we need to calculate the change in the second term in the equation. Distance from room corner to centre is given by: Assuming that the acoustic centre of the source is within a quarter of a wavelength from the hard corner, D θ = 8 and so Solutions to problems 180 10log 10 10 7.5 % 10 8.5 % 10 8.4 % 10 7 & 10log 10 10 (7.5 & 0.39) % 10 (8.5 & 0.38) % 10 (8.4 & 0.46) & 10 (7 & 0.41) ' 87.8 & 83.7 ' 4.1dB . The following two tables (one with no D θ 4πr 2 ' 8 4π × 4.01 2 ' 0.0396 tile and the other with 25m 2 of acoustic tile) may be constructed. Octave band centre frequency (Hz) ¯α (no tile) R D θ 4πr 2 % 4 R 10log 10 D θ 4πr 2 % 4 R 63 250 1000 4000 0.01 0.02 0.02 0.03 1.192 2.408 2.408 3.649 3.4 1.7 1.7 1.14 5.3 2.3 2.3 0.6 In the following table (which includes the effects of acoustic tile, the overall absorption coefficient is calculated using equation 7.78 in the text. Octave band centre frequency (Hz) ¯α (wall) ¯α (tile) ¯α (overall) R D θ 4πr 2 % 4 R 10log 10 D θ 4πr 2 % 4 R Improve- ment (dB) 63 250 1000 4000 0.01 0.02 0.02 0.03 0.08 0.15 0.20 0.25 0.025 0.048 0.058 0.077 3.0 5.95 7.27 9.84 1.37 0.72 0.59 0.45 1.4 -1.5 -2.3 -3.5 3.9 3.7 4.6 4.1 The noise reductions in each octave band are given by the last column in the above table. (d) Assuming that the overall space average level will be reduced by the same amount as the level at the centre of the room, the difference between new and old overall levels is: Sound in enclosed spaces 181 L w ' 95 & 8.3 % 22.5 % 0.2 & 13.9 ' 95.5dB re 10 &12 W L p ' 95.5 % 10log 10 2 4π × 0.5 2 % 4(1 & 0.022) 193.2 × 0.022 % 0.15 ' 97.6dB re 20µPa T 60 ' 55.25V Sc¯ α ' 55.25 × 179.7 193.2 × 343 × ¯ α ' 0.150 ¯ α Problem 7.28 (a) Using equation 6.13 in the text and substituting in the appropriate values gives: (b) Using equation 7.42 in the text, (and allowing for ρc … 400) we can write: Problem 7.29 (a) Room 6.84m × 5.565m × 4.72m, S = 2(6.84 × 5.565 + 6.84 × 4.72 + 5.565 × 4.72) = 193.2m 2 . V = 6.84 × 5.565 × 4.72 = 179.7m 3 . The room reverberation time can be calculated using equation 7.51 in the text. Thus: The results for each third octave band are given in the following table. Solutions to problems 182 λ ' 179.7/ 4.6 1/ 3 ' 3.393m λ ' 179.7/ 1.3 1/ 3 ' 5.171m One third octave band centre frequency (Hz) ¯α T 60 63 80 100 125 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000 6300 8000 0.010 0.010 0.011 0.011 0.013 0.015 0.017 0.017 0.018 0.018 0.019 0.020 0.022 0.025 0.028 0.031 0.034 0.037 0.040 0.044 0.047 0.050 15.0 15.0 13.6 13.6 11.5 10.0 8.8 8.8 8.3 8.3 7.9 7.5 6.8 6.0 5.4 4.8 4.4 4.0 3.7 3.4 3.2 3.0 (b) Lowest 1/3 octave band given by V = 4.6λ 3 (see p259 in text). Thus: This corresponds to a frequency, f = 343/3.393 = 101Hz. Thus the room is suitable for measurements down to and including the 100Hz 1/3 octave band. (c) Lowest octave band given by V = 1.3λ 3 (see p259 in text). Thus: This corresponds to a frequency, f = 343/5.171 = 66Hz. Thus the room may just be suitable for measurements down to and including the 63Hz Sound in enclosed spaces 183 L w ' L p & 10log 10 (4/ R) % 10log 10 (400/ ρc) 92.5 ' 87 & 10log 10 (4/ R) & 0.15 (4/ R) ' 10 (87 & 92.5 & 0.15) / 10 ' 0.272 octave band. (d) For pure tone noise, the lowest acceptable frequency is given by (see p259 in text). We need to solve for the frequency by f ' T 60 / V 1/ 2 trial and error as illustrated in the table below. One third octave band centre frequency (Hz) T 60 corresponding lowest acceptable frequency 63 80 100 125 160 200 250 315 400 500 630 15.0 15.0 13.6 13.6 11.5 10.0 8.8 8.8 8.3 8.3 7.9 578 578 550 550 506 472 443 443 430 430 419 From the table, it can be seen that the lowest acceptable frequency for tonal noise is 430Hz. Problem 7.30 (a) L w = 92.5dB, L pr = 87dB for reference source. The room constant is calculated using only the reverberant part of equation 7.42 in the text and allowing for ρc = 413.6. Thus: and so R = 14.7m 2 . Solutions to problems 184 L p ' 10log 10 10 8.5 & 10 8.1 & 10log 10 (4) ' 76.8dB(A) re 20µPa L w ' 76.77 & 10log 10 (4/ 14.7) & 0.15 ' 82.3dB(A) re 10 &12 W L w ' 76.77 & 10log 10 (4/ 29.4) & 0.15 ' 85.3dB(A) re 10 &12 W 10log 10 D θ 4πr 2 % 4(1 & ¯ α 1 ) S¯α 1 & 10log 10 D θ 4πr 2 % 4(1 & ¯α 2 ) S¯α 2 (b) Existing L p is 81dB(A). Allowed total L p = 85dB(A). 4 new machines, so allowed L p from each new machine is: The corresponding allowed sound power level of each machine is then: Assumptions: ! Direct field small compared to reverberant field at measurement locations. ! ρc = 413.6 ! Spectral content of noise from new machines is similar to that of existing noise. If not, then calculations should be done in octave bands. (c) Doubling room constant gives the allowed sound power level of Same assumptions as for part (b). Problem 7.31 Room 15m × 15m × 5m, S = 2(15 × 5 × 2 + 15 × 15) = 750m 2 . Using equations 7.42 and 7.43, the noise reduction is given by: The mean absorption coefficient before treatment is calculated using equation 7.78 and is: Sound in enclosed spaces 185 ¯α 1 ' 15 × 15 × 0.01 % 525 × 0.1 750 ' 0.073 ¯α 2 ' 15 × 15 × 0.01 % 525 × 0.7 750 ' 0.493 ΔL p ' 10log 10 2 4 × π × 16 % 4(1 & 0.073) 750 × 0.073 & 10log 10 2 4 × π × 16 % 4(1 & 0.493) 750 × 0.493 ' &11.1 % 18.1 ' 7.0dB and after treatment it is: Assuming that the acoustic centre of the machine is within a quarter of a wavelength of the floor, D θ = 2; thus the noise reduction at the specified location is: Problem 7.32 Room: L w = 105dB, V = 100m 3 , S = 130m 2 , T 60 = 1.5secs. Office: L w = 85dB, V = 80m 3 , S = 100m 2 , T 60 = 0.75secs. Partition area: 15m 2 . For the room and the office, the quantity is calculated using equation 7.51 S¯α in the text and the reverberant sound pressure level existing in each space is calculated by combining equations 7.42 and 7.43 with the direct field term of equation 7.42 omitted. Thus the following table may be constructed. Room Office S¯α 10.739 17.182 ¯α 0.0826 0.1718 L p ' L w % 10log 10 4(1 & ¯α) S¯α % 0.15 100.5dB 78.0dB Solutions to problems 186 L p ' 10log 10 10 7.9 & 10 7.8 ' 72.1dB ¢ p 2 ¦ ' 4 × 10 &10 × 10 116/ 10 ' 159.2Pa 2 1 & ¯ α ¯ α ' 159.2 × 50 4 × 1.206 × 343 × 1 ' 4.812 T 60 ' 55.25 × 30 343 × 50 × 0.172 ' 0.56secs ¯ α ' 10 × 0.8 % 40 × 0.172 50 ' 0.298 ¢ p 2 ¦ ' 4 × 1 × 1.206 × 343 × (1 & 0.298) 50 × 0.298 ' 77.96Pa 2 Allowable level in office = 78 + 1 = 79dB. Allowable level due to new machine is: The level in the room is 100.5dB, so the noise reduction required of the wall is 100.5 - 72.1 = 28.4dB which should be rounded up to 30dB for specification purposes. Problem 7.33 (a) Room volume V = 30m 3 , surface area, S = 50m 2 Reverberant field mean square pressure is: Using equation 7.41 in the text we obtain: Thus = 0.172 ¯α (b) Using equation 7.51 in the text: Thus, T 10 = 0.56/6 = 0.094secs. (c) New is given by: ¯α Using equation 7.41, we obtain for the new mean square sound pressure: Sound in enclosed spaces 187 L p ' 94 % 10log 10 (77.96) ' 112.9dB The corresponding sound pressure level is then: which corresponds to a reduction of 116 - 112.9 = 3.1dB. (d) Adding 3 more sources increases the existing number by a factor of 4. Providing all sources produce the same sound pressure level, the increase in total sound pressure level over that corresponding to one source would be . ΔL p ' 10log 10 (4) ' 6dB Problem 7.34 porous acoustic fibrous material (fibreglass, rockwool, ceramic fibre, steel wool): Absorption mechanisms: viscous friction losses due to difference in velocities of air particles adjacent to fibres and those in the centre of the gap between fibres. Absorption characteristics: good at mid to high frequencies but at low frequencies need a large thickness of material to be effective. Applications: air handling duct mufflers, pipe lagging, reverberant space absorption. Avantages: Inexpensive, high absorption coefficient in mid to high frequency range. Disadvantages: fibre loss can be a health hazard so care has to be taken to properly contain the material; susceptible to powdering in the presence of vibration; not oil, water or chemical resistant. Porous plastic and rubber (polyurethane foam etc.): Absorption mechanisms: viscous friction losses due to difference in velocities of air particles adjacent to capillary walls and those in the centre of the capillaries. Absorption Solutions to problems 188 characteristics: good at mid to high frequencies but at low frequencies need a large thickness of material to be effective. Applications: vehicle cabins, pipe lagging, reverberant space absorption. Avantages: High absorption coefficient in mid to high frequency range; no health risk due to fibres. Disadvantages: not oil, water or chemical resistant; expensive, fire and smoke hazard; will not tolerate high temperatures. Helmholtz resonators: Absorption mechanisms: viscous friction losses around neck of resonator associated with large air particle motion in the centre of the neck and zero motion at the walls of the neck at resonance. Absorption characteristics: good in a narrow band of frequencies around the design frequency. Applications: electrical transformer enclosures, vehicle mufflers, mufflers for tonal noise generated by large industrial fans. Avantages: can be made to be immune to moisture, oil and chemicals. Can withstand high temperatures; no health risk. Disadvantages: Narrow frequency range of effective absorption. Resonant panels Absorption mechanisms: panel and backing cavity damping losses. Absorption characteristics: good in a narrow band of frequencies around the design frequency. Applications: electrical transformer enclosures, auditoria, concert halls, reverberant spaces. Avantages: can be made to be immune to moisture, oil and chemicals. Can withstand high temperatures; no health risk. Disadvantages: Narrow frequency range of effective absorption. Sound in enclosed spaces 189 ¯ α ' j S i ¯α i j S i ' 2 × 6.84 × 5.565 × 0.02 % 2 × 5.565 × 4.72 × 0.05 %2 × 6.84 × 4.72 × 0.06 2 × 6.84 5.565 % 2 × 5.565 × 4.72 % 2 × 6.84 × 4.72 ' 8.023/ 193.2 ' 0.042 Problem 7.35 We require = 0.8 at 125Hz. Referring to figure 7.8 in the text (p.307) it is ¯α clear that we need to use curve "D" which implies the use of sound absorbing material behind the panel. From figure 7.9, it can be seen that the 125Hz line crosses curve "D" when the cavity depth is 110mm and the panel mass is 2kg/m 2 . Thus this is the required design. Note that the guide notes in the caption of figure 7.8 should also be included as design specifications. Problem 7.36 Following equation 7.78 in the text: Problem 7.37 Room volume = π × 7 2 × 2.5 = 384.85m 3 . The optimum reverberation times are calculated using equation 7.121 in the text with K = 5. The calculated values are listed in the following table. Octave band centre frequency (Hz) T 60 125 250 500 1000 2000 4000 8000 1.45 1.06 0.96 0.96 0.96 0.96 0.96 Solutions to problems 190 (b) Using equation 7.51 in the text, the recommended is 65m 2 at 500Hz S¯α and above, 59m 2 at 250Hz and 43m 2 at 125Hz. (c) Area ceiling = π × 49 = 154m 2 . At 125Hz, the Sabine absorption coefficient of tile is 0.2. Thus if all the ceiling were covered the maximum would be 154 × 0.2 = 30.8m 2 (assuming that the floor and S¯α walls contributed a negligible amount). Alternatively, if it is assumed that the floor and walls are of concrete with = 0.01, then the total ¯α = 30.8 + 0.01 × (154 + 2π × 7 × 2.5) = 33.4m 2 . The recommended ¯α S¯α at 125Hz is 43m 2 (from part (b)), so the ceiling tile would NOT be adequate. (d) A compromise would be to use sufficient tiles to achieve as closely as possible the required absorption over the range 500 to 1000Hz and then design a panel absorber with a maximum absorption at 125Hz and no absorption at 500Hz. The required absorption in the range 500 to 4000Hz is 65m 2 . Thus the optimum amount of tile is 65/0.8 = 81m 2 , which will be OK at 500 and 2000Hz, a bit much at 1000Hz and a bit little at 4000Hz, but nevertheless a good compromise. We are then left with an area of ceiling of 154 - 81 = 73m 2 for panel absorbers. The amount of absorption needed at 125Hz is 43 - 3 - 81 × 0.2 = 24m 2 . So the panel absorber must have an absorption coefficient of 24/73 = 0.33 at 125Hz and nothing at 500Hz. Choosing curve H from figure 7.8 in the text and allowing for the fact that the 125Hz band includes the peak, the average for the 125Hz band ¯α is approximately 0.35. The absorption coefficient of the panel at 500 and above is likely to be 0.08 and at 250Hz it will be 0.12. To optimise the required absorption coefficients, we can vary the relative areas. The area of floor and walls is 264m 2 . Thus the amount of absorption due to the floor and walls is 2.6m 2 in the 125, 250 and 500Hz bands and 5.3m 2 in the 1000, 2000 and 4000Hz octave bands. Including this and with an area of 70m 2 of tile and 65m 2 of panel, the amount of absorption in the octave bands from 125Hz to 4000Hz is 41, 53, 64, 70, 67, and 63m 2 which is a little low at 125Hz and 250Hz (optimum is 43 and 59 respectively) and a little high in the other bands (optimum at 500Hz and higher frequencies is 62m 2 ). Choosing the area of panel = 75m 2 and the area of tile = 65m 2 , gives the following amounts of absorption: 42, 51, 61, 67, 63, 60m 2 which is close enough. Note that there are many other adequate solutions to this problem. Sound in enclosed spaces 191 L w ' L p % 10log 10 (2πr 2 ) & 0.15 ' L p % 21.85 dN df ' 4 × π × 125 2 × 30 343 3 % π × 125 × 62 2 × 343 2 % 40 8 × 343 ' 0.264 Problem 7.38 (a) sound power of the source is given by: Thus the following table may be constructed. Octave band centre frequency (Hz) 63 125 250 500 1k 2k L p (dB re 20µPa) 90 85 78 73 70 65 L w (dB re 10 -12 W) 112 107 100 95 92 87 (b) Room dimensions 5m × 3m × 2m. Using equation 7.17 in the text, we have: f 1,0,0 ' 343 2 × 1 5 ' 34.3Hz f 0,1,0 ' 343 2 × 1 3 ' 57.2Hz f 0,0,1 ' 343 2 × 1 2 ' 85.8Hz f 2,0,0 ' 343 2 × 2 5 ' 68.6Hz f 1,1,0 ' 343 2 × (1/ 5) 2 % (1/ 3) 2 ' 66.7Hz Thus the 3 lowest order modes are in the 31.5Hz and 63Hz third octave bands. Room volume, V = 5 × 3 × 2 = 30m 2 Area, S = 2(5 × 3 + 5 × 2 + 3 × 2) = 62m 2 . Perimeter, L = 4(5 + 2 + 3) = 40 The modal density is given by equation 7.21 in the text. Thus: Solutions to problems 192 N ' 4 × π × 141 3 × 30 3 × 343 3 % π × 141 2 × 62 4 × 343 2 % 40 × 141 8 × 343 ' 19modes N ' 4 × π × 113 3 × 30 3 × 343 3 % π × 113 2 × 62 4 × 343 2 % 40 × 113 8 × 343 ' 11.4modes From table 1.2 on p43 in the text, the 125Hz third octave bandwidth is 141 - 113 = 28Hz, so the number of modes in the band is 28 × 0.264 = 7 to 8 modes. Alternatively equation 7.20 could be used to calculate the number of modes occurring below 141Hz and 113Hz and then take the difference. The number of modes below 141Hz is: The number of modes below 113Hz is: Thus the number in the 125Hz octave band is between 7 and 8 modes. (c) Equations 7.51, 7.43 and 7.42 (with a correction for ρc = 413.6) in the text, and the knowledge that S = 62m 2 , may be used to construct the following table. Octave band centre frequency (Hz) 63 125 250 500 1k 2k Overall L w (dB re 10 -12 W) 112 107 100 95 92 87 T 60 5.5 5 4 3 2 1.5 S¯α 0.877 0.964 1.206 1.608 2.411 3.215 ¯α 0.0141 0.015 6 0.019 4 0.025 9 0.038 9 0.0519 R ' S¯ α 1 & ¯ α 0.890 0.979 1.230 1.650 2.509 3.391 L p (reverb) 118.7 113.3 105.3 99.0 94.2 87.9 A-weighting -26.2 -16.1 -8.6 -3.2 0.0 1.2 L p (dB(A)) 92.5 97.2 96.7 95.8 94.2 89.1 102.8 Sound in enclosed spaces 193 Assumptions: ! A-weighting assumed uniform across each octave band when in fact it varies continuously with frequency. ! Direct field contribution assumed negligible. ! Reflections from and absorption of surfaces other than room boundaries is excluded. (d) If 2 more generators were added, the sound pressure level would increase by . 10log 10 (2 % 1) ' 4.8dB(A) (e) Ceiling tiles added with area = 15m 2 . Remaining room surface area = 62 - 15 = 47m 2 = S floor, walls . The following table may be constructed, where: . ¯ α overall ' ¯ α new ' S¯ α floor walls % S¯ α ceiling S floor walls % S ceiling Octave band centre frequency (Hz) 63 125 250 500 1k 2k Over- all L w (dB re 10 -12 W) 112 107 100 95 92 87 S¯α floor walls 0.66 0.73 0.91 1.22 1.83 2.44 ¯α ceiling 0.15 0.25 0.55 0.85 1.0 1.0 S¯α ceiling 2.25 3.75 8.25 12.75 15 15 ¯α overall ' ¯α new 0.0469 0.0722 0.1477 0.2253 0.2715 0.2813 R new ' S¯ α new 1 & ¯ α new 3.05 4.83 10.75 18.03 23.10 24.27 L p (reverb) 113.3 106.3 95.9 88.6 84.5 79.3 114.2 A-weighting -26.2 -16.1 -8.6 -3.2 0.0 1.2 L p (dB(A)) 87.1 90.2 87.3 85.4 84.5 80.5 94.5 Similar assumptions as made in part (c). Solutions to problems 194 α st ' 1 & e &55.25V/ ScT 60 Problem 7.39 Room 20m × 15 × 4m, S = 2(20 × 15 + 20 × 4 + 15 × 4) = 880m 2 . V = 20 × 15 × 4 = 1200m 3 . The desired reverberation times are calculated using equation 7.121 on p. 329 in the text, with 10% increase at 250Hz, 50% increase at 125Hz and 100% increase at 63Hz. The existing mean statistical absorption coefficient may be calculated using equation 7.56 in the text rearranged to give: Thus the following table may be constructed. Octave band centre frequency (Hz) Existing T 60 Existing mean α st Desired T 60 Desired mean α st Required increase in α st 63 125 250 500 1000 2000 4000 8000 3.0 2.6 2.3 2.1 2.0 2.0 2.0 1.8 0.0706 0.0810 0.0911 0.0993 0.1040 0.1040 0.1040 0.1149 1.86 1.40 1.02 0.93 0.93 0.93 0.93 0.93 0.111 0.145 0.194 0.211 0.211 0.211 0.211 0.211 0.040 0.064 0.103 0.112 0.107 0.107 0.107 0.096 There are many possible solutions to achieve the desired mean absorption coefficients. One alternative is to look for the frequency at which the additional absorption required is the largest and choose a material which has a maximum absorption coefficient at this frequency. In this case, the maximum increase in absorption is needed at 500Hz, so a material thickness of 25mm should be chosen. The amount of material required is then calculated on the basis of achieving the optimum reverberation time in the octave bands most important for speech; namely, 500Hz to 2000Hz. In this case, it would seem that the required increase in mean absorption coefficient is 0.107 which would satisfy the requirements at and above 1000Hz, with a compromise of a little less than desired at 500Hz. let x be the fraction of room surface area to be covered with absorbing material. Using equation 7.58, we Sound in enclosed spaces 195 (1 & 0.211) ' (1 & 0.85) x × (1 & 0.104) (1 & x) log 10 (0.79) ' xlog 10 (0.15) % (1 & x)log 10 (0.896) W ' ¢ p 2 i ¦ S 4ρc W ' τ ¢ p 2 i ¦ S 4ρc ¢ p 2 i ¦ ' 4× 10 &10 × 10 88/ 10 ' 0.252Pa 2 W ' 1.995 × 10 &3 × 0.252 × 1.5 4 × 1.206 × 343 ' 0.457µW ¢ p 2 ¦ ' 1.206 × 343 × 4.565 × 10 &7 2 × π × 60 2 ' 8.349 × 10 &9 Pa 2 have: Taking logs of both sides, we obtain: which gives x = 0.0711. So the required area of 25mm thick acoustic material is 63m 2 . Problem 7.40 Assume that the house is approximately in a direction along the normal axis from the window. The power incident on the window is the intensity in the direction of the window multiplied by the area of the window. Thus The power radiated through the window is then where = . T he reverberant sound pressure τ ' 10 &TL/ 10 10 &2.7 ' 1.995 × 10 &3 level is 88dB. Thus The power radiated through the window is then For an incoherent plane source, the on-axis sound pressure at the receiver is given by equation 5.106. The quantity . Thus from r/ HL ' 60/1.5 ' 40 figure 5.11, it is clear that we can treat it as a hemispherically radiating point source producing a sound pressure described by equation 5.106 in the text. Thus: Solutions to problems 196 L p ' 10log 10 (8.349 × 10 &9 ) % 94 ' 13.2dB re 20µPa q ' 1000 × π × 0.002 2 4 × 0.07 ' 6.7mm q ' 1000 × π × 0.002 2 2 × 0.07 × 2 3 ' 5.1mm This corresponds to a sound pressure level of : As the ground is hard asphalt, we may add 3dB to the level to account for the effect of ground reflection. Thus the expected level at the house is 16dB. Problem 7.41 (a) First calculate distance between holes. Could assume parallel or staggered holes as shown in the two figures to follow. Let q be the distance between holes as shown in the figures. Choosing a segment of plate as shown in the figure we can calculate the ratio of holes to total area of the segment and set this equal to 0.07. This gives a hole spacing for the parallel holes of: and for the staggered holes, For the purposes of this problem we will use q = 6.7mm. Using equation E.7 in the 2 nd . edn. of the text on p528, we have: Sound in enclosed spaces 197 f max 2 × π × 0.1 343 tan f max × 2 × π × 0.1 343 ' 7 × 0.1/ 100 0.003 % 0.85 × 0.002 × (1 & 0.22 × 0.002/ 0.0067) 0.00183 × f max tan f max × 0.00183 ' 1.5256 a(X) ' 3.1696(1.63259 & 3.1696) × 0.0659 2 & 0.1663 2 × 1.63259 2 3.1696 2 × 0.0659 2 % 0.1663 2 × 1.63259 2 ' &0.80849 Rewriting gives: Can solve by trial and error, choosing values of f max until the LHS = 1.5256 as illustrated in the table below. f max (Hz) LHS 100 500 600 550 540 543 542.9 0.0339 1.1892 2.1467 1.5902 1.5000 1.5263 1.5254 Thus the frequency of maximum absorption is 543Hz. If we used Equation 7.77 in the 3 rd edn. we would get 691 Hz, but the error is greater than 15% because the condition, fL/c < 0.1 is not satisfied. (b) Specific normal impedance is given by equation C.43. To evaluate this equation we need to use equation C.41 and to evaluate that we need equations C.3 and C.4. Referring to equation C.15, X = 1.206 × 543/10000 = 0.0655. Thus, T 1 = 1.63259, T 2 = 0.06590, T 3 = 3.1696, T 4 = 0.1663. Referring to equation C.9: and Solutions to problems 198 b(X) ' 1.63259 2 × 0.0659 × 0.1663 3.1696 2 × 0.0659 2 % 0.1663 2 × 1.63259 2 ' 0.24893 a(X 1 ) ' 3.0785(1.5415 & 3.07850) × 0.05636 2 & 0.16526 2 × 1.5415 2 3.0785 2 × 0.05636 2 % 0.16526 2 × 1.5415 2 ' &0.84133 b(X 1 ) ' 1.5415 2 × 0.05636 × 0.16526 3.0785 2 × 0.05636 2 % 0.16526 2 × 1.5415 2 ' 0.23297 κ ' ( 1 & 0.4 × (&0.49807 % j0.23297)) &1 ' ( 1.19923 % j0.093188) &1 ρ m ' ( 1 & 0.8085 % j0.2489) &1 ' ( 0.1915 % j0.24893) &1 Z m ρc ' ρ m κ ' 1 (0.1915 % j0.24893) × (1.19923 & j0.093188) ' 1 0.25285 % j0.28068 ' 1.7717 & j1.9667 ' 2.6471e &j0.8375 ' 1.6270e &j0.4188 ' 1.4864 & j0.6616 X 1 = 0.856 × 1.206 × 543/10000 = 0.05605. Thus, T 1 = 1.54150, T 2 = 0.05636, T 3 = 3.07850, T 4 = 0.16526. Referring to equation C.9: and Thus τ = -0.84133×0.592 + j0.23297 and σ = -0.80849 + j0.24893. The quantities κ and ρ m may be calculated using equations C.5 and C.6 as follows: and Using equations C.3 , we obtain: Sound in enclosed spaces 199 k m ' 2πf max c ρ m κ ' 2π × 543 343 1.19923 & j093188 0.1915 % j0.24893 ' 9.9468 (1.19923 & j0.093188) × (0.1915 & j0.24893) 0.098638 ' 9.9468 2.09319 & j3.2074 ' 9.9468 3.8300e &j0.99259 ' 19.4663e &j0.49629 ' 17.1178 & j9.2692 j tan(k m R) ' e 0.9269 cos(1.7118) % jsin(1.7118 & e &0.9269 cos(1.7118 & jsin(1.7118) e 0.9269 cos(1.7118) % jsin(1.7118) % e &0.9269 cos(1.7118) & jsin(1.7118) ' &0.2995 % j2.8934 &0.4107 % j2.1097 ' 1.3480 & j0.1205 tan(k m R) ' &0.1205 & j 1.3480 Z N ρc ' (0.6616 % j1.4864) 0.1205 % j1.3480 ' 2.0834 & j0.7127 1.8316 ' 1.1375 & j0.3891 and using equation C.4b we obtain: Before continuing, it will be useful to evaluate the quantity, tan(k m R). Using the previous result for k m , and setting R = 0.1, we can write: Thus: Using the previous results and equation C.41 (assuming a rigid backing for the porous material), we can write: To calculate the overall impedance, we use equation C.43, but first we need to evaluate R and tan(kR). From equation 9.25 in the text, the effective length of the holes in the perforated sheet is: Solutions to problems 200 R ' 0.003 % 16 × 0.001 3 × π 1 & 0.43 × 0.001/ 0.0067 ' 0.004588 tan(kR) ' tan(2 × π × 543 × 0.004588/ 343) ' 0.04567 t ' 2 × 1.8 × 10 &5 1.206 × 2 × π × 543 ' 9.3538 × 10 &5 Thus: We also need to calculate the acoustic resistance of the holes using equation 9.29 in the text. To evaluate this equation we need the following quantities: , k ' (2 × π × 543) / 343 ' 9.9467 ; A ' π × 0.002 2 / 4 ' 3.1416 × 10 &6 m 2 D ' π × 0.002 ' 0.006283 ε = 1.0 as radiation from a baffle. The quantity, h, is the largest of the half plate thickness or t. Thus: h = w/2 = 0.0015m. The above quantities may be inserted into equation 9.29 to give: Sound in enclosed spaces 201 Z P ρc ' (100/ 7) (0.04567j % 4.1256 × 10 &3 ) 1 % 100 × 1.206 × 343 2 × π × 543 × 21.762 × 7 × 0.04567 & 4.1256j × 10 &3 ' 0.05894 % 0.6524j 1.00363 & 3.2836j × 10 &4 ' 0.05894 % 0.6548j Z N ρc % Z P ρc ' 1.1375 & 0.3891j % 0.05894 % 0.6548j ' 1.1964 % 0.2657j ' 1.2255 e 0.21854j R a A ρc ' 9.9467 × 9.3538 × 10 &5 × 6.283 × 10 &3 × 0.003 2 × 3.1416 × 10 &6 × 1 % (1.4 & 1) 5 3 × 1.4 % 0.288 × 9.9467 × 9.3538 × 10 &5 × log 10 4 × 3.1416 × 10 &6 π × 0.0015 2 % 3.1416 × 10 &6 × 9.9467 2 2 × π ' 4.00923 × 10 &3 % 6.69555 × 10 &5 % 4.9468 × 10 &5 ' 4.1256 × 10 &3 We can now use equation C.43 to evaluate the overall impedance. The second term on the right is the impedance due to the perforated sheet and is: Thus the total impedance is: cosβ = 0.9762; cos2β = 0.9060; sinβ = 0.2168 and ξ = 1.2255. Using the above data, the statistical absorption coefficient may be calculated using equation C.37 in the text as follows: Solutions to problems 202 α st ' 8 × 0.9762 1.2255 1 & 0.9762 1.2255 × log e ( 1 % 2 × 1.2255 × 0.9762 % 1.2255 2 ) % 0.9060 1.2255 × ( 0.2168) × tan &1 1.2255 × (0.2168) 1 % 1.2255 × 0.9762 ' 6.373 × (1 & 0.7966 × 1.5881 % 3.410 × tan &1 [0.12097]) ' 6.373 × (1 & 1.2651 % 0.4105) ' 0.93 NRC ' ¯ α 250 % ¯ α 500 % ¯ α 1000 % ¯ α 2000 4 ' 0.6 % 0.8 % 1.0 % 1.0 4 ' 0.85 Problem 7.42 The NRC is given by: So the material is adequate for the purpose. Problem 7.43 Truck emits 110 dB. This is equal to: . W ' 10 &12 × 10 110/ 10 ' 0.1 watts r/a = 60/6 = 10. From Fig 7.14, reverberant field pressure squared is: 10log 10 ¢ p 2 R ¦ & 10log 10 Wρc πa 2 ' &4 dB Thus, 10log 10 ¢ p 2 R ¦ ' 10log 10 0.1 × 413.6 π × 36 & 4 ' &18.4 dB so reverberant field pressure is, L pR ' &18.4 % 94 ' 75.6 dB Direct field pressure: 10log 10 ¢ p 2 D ¦ ' 10log 10 0.1 × 413.6 4π × 60 2 ' &30.4 dB Sound in enclosed spaces 203 So direct field pressure is, L pD ' &30.4 % 94 ' 63.6 dB Total pressure, L p ' 10log 10 10 75.4/ 10 % 10 63.6/ 10 ' 75.7 dB Assumptions: • Effective acoustic source location is in the centre of the cross section • Specularly reflecting surfaces • Ambient temperature of 20 EC TL ' &10log 10 τ 8 Solutions to problems relating to sound transmission loss, acoustic enclosures and barriers Problem 8.1 (a) The Transmission Loss of a partition is an inverse decibel measure (bigger TL means a smaller amount of transmitted energy) of the amount of incident energy which is transmitted to the space on the side opposite that on which the energy is incident. It is defined in terms of the transmission coefficient, τ, which is the fraction of transmitted to incident energy, as follows: It may be measured using two reverberant rooms with the panel to be measured acting as a partition between the two rooms with the space around the panel of high transmission loss construction so that all of the significant acoustic energy transmitted between the two rooms passes through the panel under test. The test is conducted by exciting the one of the reverberant rooms with 1/3 octave band noise and then measuring the difference in the space averaged sound pressure level in the two rooms. The appropriate mathematical analysis is embodied in equations 8.13 to 8.16 in the text (page 342). (b) Measurements often do not agree with theoretical calculations because the latter do not take into account the size of the panel exactly. Also the experimental determination of space average sound pressure level is often characterised by errors, especially at low frequencies when the sound fields in the two rooms are not sufficiently diffuse. Sometimes, energy is transmitted from one room to the next by way of paths not through the panel (called "flanking"), resulting in Transmission Loss measurements which are too small. Sound transmission loss, acoustic enclosures and barriers 205 TL ' 10log 10 1 % πfm ρc 2 & 5.5 (dB) 4 ' 10log 10 1 % π × f × 7800 × 0.01 988 × 1481 2 & 5.5 f ' 10 0.95 & 1 2.8046 × 10 &8 ' 16,800Hz 45 o 45 o 75 25 25 30 75 75 All dimensions in mm z 0 z n Problem 8.2 Mass Law Transmission Loss is obtained by combining equations 8.34 (with θ = 0E) and equation 8.35b, which gives: Substituting in values for the variables gives: Rearranging gives: The Transmission Loss in air at this frequency is much greater because the impedance of the panel compared to the characteristic impedance in the propagating medium is much larger for air than water. Problem 8.3 (a) First find location of neutral axis by taking moments about an axis through the centre of the angled section and shown as z 0 in the figure. In the following equations b i is the length of the ith section. If the neutral axis is denoted as z n where z is the vertical coordinate on the figure, then: Solutions to problems 206 z 0 & z n ' j i b i z i0 j i b i ' 75 × 15 % 2 × 25 × 2.5 & 25 × 10 % 75 × 15 & 75 × 15 75 % 3 × 25 % 75 % 2 2 × 30 %75 ' 1000 384.9 ' 2.6mm B ' Eh (1 & ν 2 )R j n b n z 2 n % h 2 % b 2 n 24 % h 2 & b 2 n 24 cos2θ n B 1 ' 207 × 10 9 × 0.0012 0.91 × 0.31 0.15 0.0124 2 % 0.0012 2 12 % 0.05 0.0001 2 % 0.025 2 12 % 0.025 0.0126 2 % 0.0012 2 12 % 0.075 0.0176 2 % 0.0012 2 12 % 0.06 2 0.0026 2 % 0.0012 2 % 2 × 0.03 2 24 ' 8.805 × 10 8 ( 2.3082 × 10 &5 % 2.6046 × 10 &6 % 3.9720 × 10 &6 % 2.3241 × 10 &5 % 6.9427 × 10 &6 ) ' 5.27 × 10 4 kg m 2 s &2 Thus the neutral axis is 12.4mm from the centre of the top of the section. The section thickness, h = 1.2mm and the horizontal length, R, before repeating itself is 250 + 60mm = 0.31m. The bending stiffness in the direction along the ribs may be calculated with E = 207GPa and = 0.3 using equation 8.10 in the text, which is ν incorrect in the first printing of the text and should be: Thus: Sound transmission loss, acoustic enclosures and barriers 207 B 2 ' 207 × 10 9 × 0.0012 3 12 × 0.91 × 0.385 0.31 ' 40.7kgm 2 s &2 c B1 ' 5.27 × 10 4 × 4π 2 × 10 6 11.62 1/ 4 ' 650m/ s c B2 ' 40.7 × 4π 2 × 10 6 11.62 1/ 4 ' 108m/ s f c1 ' 343 2 2π 11.62 5.27 × 10 4 1/ 2 ' 278Hz f c2 ' 343 2 2π 11.62 40.7 1/ 2 ' 10,000Hz B ab ' 0.5 5.27 × 10 4 × 0.3 % 40.7 × 0.3 % 207 × 10 9 × 0.0012 3 3 × 2.6 ' 7934 The stiffness in the direction across the ribs may be calculated using equation 8.5 as follows: (b) The bending wavespeed is calculated using equation 8.1 in the text. The surface mass of the panel is m = 7800 × 0.0012 × 0.385/0.31 = 11.62kg m -2 and the frequency is 1000Hz. Thus the lower and upper bending wavespeeds corresponding to waves propagating parallel and perpendicular to the ribs respectively are: (c) The lower and upper critical frequencies for the panel may be calculated using equation 8.3 in the text. Thus: (d) Assuming that the enclosure wall edge condition is simply supported (a good approximation in practice for most enclosures), the first resonance frequency of the panel may be calculated using equation 8.22 in the text with i = n = 1. Thus: and: Solutions to problems 208 f 1,1 ' π 2 11.62 5.27 × 10 4 2 4 % 40.7 2 4 % 7934 2 4 1/ 2 ' 28.4Hz TL A ' 20log 10 (278 × 11.62) & 54 ' 16.2dB TL B ' 20log 10 (278) % 10log 10 (11.62) & 10log 10 (278) & 20log 10 [ log e (4)] & 13.2 ' 19.1dB TL C ' 20log 10 (5000) % 10log 10 (11.62) & 10log 10 (278) & 20log 10 [ log e (20000/278)] & 13.2 ' 34.4dB TL D ' 10log 10 (11.62) % 15log 10 (10000) & 5log 10 (278) & 17 ' 41.4dB TL oct ' &10Log 10 6 (1/3) [10 &TL 1 /10 % 10 &TL 2 /10 % 10 &TL 3 /10 ] > (e) The sound transmission loss of the panel may be calculated using figure 8.8b in the text. Point A is at 139Hz and the corresponding TL is given by: At point B (278Hz), the TL is: At point C (5,000Hz), the TL is: At point D (20,000Hz), the TL is: These points are plotted on the following graph and interpolation is used to find the octave band TL values. Strictly speaking, the curve should only be used to find 1/3 octave values and the octave band levels must then be calculated from the following equation: Sound transmission loss, acoustic enclosures and barriers 209 f c1 0.5f c1 0.5f c2 B C A However, for most practical purposes, the results obtained that way are little different to the results obtained by reading the octave band levels directly from the figure. However, in the case of isotropic panels, care should be taken to avoid errors near the dip in the curve corresponding to the critical frequency. Following the figure, the octave band results are summarised in a table. Octave band centre frequency (Hz) Transmission Loss (dB) 63 125 250 500 1000 2000 4000 8000 9 15 19 21 25 29 33 37 Problem 8.4 (a) Only one half of the sine wave section needs to be considered. y 1 = 20 sin(15π/40) = 18.48mm. Solutions to problems 210 B 1 ' 207 × 10 9 × 0.0016 0.91 × 0.04 × 10 9 23.8 × 2 (18.48/ 2) 2 % 1.6 2 % 23.8 2 24 % 1.6 2 & 23.8 2 24 cos(101.8E) % 10 18.48 2 % 1.6 2 12 ' 8.04 × 10 4 kg m 2 s &2 B 2 ' 207 × 10 9 × 1.6 3 × 10 &9 12 × 0.91 57.6 40 ' 111.8kg m 2 s &2 f c1 ' 343 2 2π 18 8.04 × 10 4 1/ 2 ' 280Hz f c2 ' 343 2 2π 18 111.8 1/ 2 ' 7,500Hz u 1 y 1 15 10 15 b 1 = 23.8mm, θ 1 = 50.9E b 2 = 10mm, θ 2 = 0 b 3 = 23.8mm, θ 3 = 50.9E l = 40, E = 207GPa = 0.3 ν Using the corrected form of equation 8.10 in the text, we can write the following for the bending stiffness for waves travelling parallel to the corrugations: The bending stiffness for waves travelling perpendicular to the corrugations can be calculated using equation 8.11 in the text as: The surface mass is m = 7800 × 0.0016 × (23.8 + 23.8 + 10)/40 = 18.0kg/m 2 . The lower and upper critical frequencies are calculated using equation 8.3 as: Sound transmission loss, acoustic enclosures and barriers 211 f 1,1 ' π 2 18 8.04 × 10 4 3 4 % 111.8 3 4 % 0.5 8.04 × 10 4 × 0.3 3 4 % 111.8 × 0.3 3 4 % 207 × 10 9 × 0.0016 3 3 4 × 3 × 2.6 1/ 2 ' 12.5Hz TL A ' 20log 10 (280 × 18.0) & 54 ' 20.0dB TL B ' 20log 10 (280) % 10log 10 (18.0) & 10log 10 (280) & 20log 10 [ log e (4)] & 13.2 ' 21.0dB TL C ' 20log 10 (3750) % 10log 10 (18.0) & 10log 10 (280) & 20log 10 [ log e (15000/280)] & 13.2 ' 34.4dB TL D ' 10log 10 (18.0) % 15log 10 (7500) & 5log 10 (280) & 17 ' 41.4dB Assuming that the enclosure wall edge condition is simply supported, the first resonance frequency may be calculated using equation 8.22 in the text. Thus: (b) The sound transmission loss of the panel may be calculated using figure 8.8b in the text. Point A is at 140Hz and the corresponding TL is given by: At point B (280Hz), the TL is: At point C (3,750Hz), the TL is: At point D (15,000Hz), the TL is: These points are plotted on the graph below and interpolation is used to find the octave band TL values. Strictly speaking, the curve should only Solutions to problems 212 TL oct ' &10log 10 6 (1/3) [10 &TL 1 /10 % 10 &TL 2 /10 % 10 &TL 3 /10 ] > f c1 0.5f c1 0.5f c2 B C A be used to find 1/3 octave values and the octave band levels must then be calculated from the following equation: However, for most practical purposes, the results obtained that way are little different to the results obtained by reading the octave band levels directly from the figure. However, in the case of isotropic panels, care should be taken to avoid errors near the dip in the curve corresponding to the critical frequency. Following the figure, the octave band results are summarised in a table. Octave band centre frequency (Hz) Transmission Loss (dB) 63 125 250 500 1000 2000 4000 8000 13 19 21 23 26 30 35 38 Sound transmission loss, acoustic enclosures and barriers 213 f c ' 343 2 2π 36 111.8 1/ 2 ' 10,600Hz TL A ' 20log 10 (10,600 × 36) & 54 ' 57.6dB TL ' 20log 10 (10,600 × 36) % 10log 10 (0.1) & 45 ' 56.6dB 0.5f c A (c) With viscoelastic damping the panel may be treated as isotropic with the surface mass, m, now equal to 2 × 18 = 36kg/m 2 . The critical frequency is: On the isotropic panel curve, point A is at 5,300Hz (10,600/2) and the TL is: At point B, the frequency is 10,600Hz and the TL is: The results are plotted on the graph below from which can be read TL values as a function of 1/3 octave band centre frequency. (d) Second panel, . The 1 f c1 ' 0.55 × 343 2 / (2000 × 0.013) ' 2500Hz subscript is used because this critical frequency is smaller than that for the other panel. Surface mass, m = 0.013 × 1000 = 13kg/m 2 . Solutions to problems 214 f 0 ' 80 36 % 13 36 × 13 × 0.1 ' 82Hz TL A ' 20log 10 (36 % 13) % 20log 10 (81.86) & 48 ' 24dB TL B1 ' 24 % 20log 10 (2500/ 82) & 6 ' 47.7dB TL B2 ' 20log 10 (13 × 0.6) % 40log 10 (10600) & 99 ' 80dB TL C ' 80 % 6 % 10log 10 (0.01) ' 66dB Cavity resonance frequency is: and the corresponding TL is: At point B the frequency is 5300 Hz and the TL is: and as there are rubber spacers, one panel may be considered to be point supported. Note that the panel with the higher critical frequency (the damped corrugated panel in this case) must be the one which is point supported to obtain the high TL predicted. The value of TL B2 for line- point support is: Assuming that there is sound absorbing material in the cavity, TL B = 80dB. At point C, the frequency is 10,600Hz and the TL is: At point D, the frequency is f 1 = 55/0.1 = 550Hz. The TL data are plotted in the following figure from which the 1/3 octave values can be read directly. Problem 8.5 A double wall partition may perform more poorly than a single partition of the same weight at resonance frequencies corresponding to acoustic modes in the cavity and also at the critical frequencies of the individual panels (if they are lightly damped). Sound transmission loss, acoustic enclosures and barriers 215 f 0 ' 80.4 1000 × 0.012 % 7800 × 0.0016 0.1 × 1000 × 0.012 × 7800 × 0.0016 ' 103.2Hz TL A ' 20log 10 (12 % 12.48) % 20log 10 (103.2) & 48 ' 20.0dB f c1 ' 0.55 × 343 2 2100 × 0.012 ' 2570Hz and 0.5f c1 ' 1280Hz f c2 ' 0.55 × 343 2 5400 × 0.0016 ' 7490Hz f 0 f R 0.5f c2 A B D C Problem 8.6 Following the procedure on page 360 in the text, we have at point A: and the corresponding Transmission Loss is: The critical frequencies are: Solutions to problems 216 TL B1 ' 20 % 20log 10 (2568/ 102.2) & 6 ' 42dB TL B2 ' 20log 10 (12) % 10log 10 (0.6) % 30log 10 (7490) % 20log 10 1 % 12.48 2570 12 7490 & 78 ' 61.7dB TL C ' 61.7 % 6 % 10log 10 (0.01) ' 47.7dB f 0 f R 0.5f c2 f c2 A B D C The Transmission Loss at point B is the larger of: and: At point C: The frequency f R = 55/0.1 = 550Hz. The Transmission Loss may thus be plotted as shown in the following figure and 1/3 octave (but not octave) values may be read directly from the figure. Sound transmission loss, acoustic enclosures and barriers 217 0.001 ' τ w (18 & 0.25 & 2) % 0.25 × 10 &28/ 10 % 2 × 10 &25/ 10 18 τ ' 0.25 × 10 &28/ 10 % 2 × 10 &25/ 10 18 ' 3.73 × 10 &4 τ ' 7.16 × 10 &4 × 15.75 % 0.025 × 1 × 1 % 0.25 × 10 &2.8 % 1.975 × 10 &2.5 18 ' 0.00238 TL ' &10log 10 (2.38 × 10 &3 ) ' 26.2dB Problem 8.7 (a) TL overall = 30dB, so τ = 10 -30/10 = 0.001. If we let the required transmission coefficient of the wall be τ w , then we can write: From which τ w = 7.161 × 10 -4 . Thus the required wall Transmission Loss is, TL w = -10log 10 (7.161 × 10 -4 ) = 31.5dB. (b) Maximum TL possible corresponds to τ w = 0. Thus: which corresponds to TL = 10log 10 (3.73 × 10 -4 ) = 34.3dB (c) 25mm crack under the door. Effective crack height is 50mm due to reflection in the floor, and τ crack = 1.0. Thus the overall transmission coefficient is: which corresponds to a Transmission Loss of: Problem 8.8 When designing an enclosure, always (if possible) include sound absorptive material on the ceiling and walls, thus resulting in an "average" enclosure with the coefficients, C, as listed in the table below (along with the required noise reduction and corresponding wall TL given by TL = NR + C). Solutions to problems 218 f c (steel) ' 0.55 × 343 2 5400 × 0.003 . 4000Hz ' f c2 f c (plaster) ' 0.55 × 343 2 1600 × 0.025 . 1600Hz ' f c1 m 1 ' 760 × 0.025 ' 19kg/ m 2 and m 2 ' 7800 × 0.003 ' 23.4kg/ m 2 f 0 ' 80 19 % 23.4 0.1 × 19 × 23.4 ' 78Hz TL A ' 20log 10 (19 % 23.4) % 20log 10 (78) & 48 ' 22dB at 78Hz TL 63 ' TL A % 60log 10 (63/ f 0 ) ' 27 20log 10 (m 1 % m 2 ) % 20log 10 f 0 % 60log 10 63 & 60log 10 f 0 ' 27 % 48 Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Required NR 14 18 25 35 50 40 40 40 C 13 11 9 7 5 4 3 3 required wall TL 27 29 34 42 55 44 43 43 100mm studs implies gap, d = 0.1m. Assume that the door and window have the same TL as the walls. To begin, try 3mm steel and 25mm plasterboard. This is insufficient as we need 27dB at 63Hz. It seems that we need to lower f 0 below 63Hz, to put this point on the 18dB/octave slope. That is: Sound transmission loss, acoustic enclosures and barriers 219 20log 10 (m 1 % m 2 ) & 40log 10 f 0 ' &33 (1) 40log 10 f 0 ' 40log 10 80 % 20log 10 (m 1 % m 2 ) & 20log 10 (m 1 × m 2 ) % 20 ' 96 % 20log 10 (m 1 % m 2 ) & 20log 10 (m 1 × m 2 ) 20log 10 (m 1 × m 2 ) ' 63dB or m 1 m 2 ' 1410 f c1 ' 0.55 × 343 2 1600 × 0.05 . 810Hz f c2 ' 0.55 × 343 2 5400 × 0.005 . 2400Hz m 1 ' 760 × 0.05 ' 38kg/ m 2 and m 2 ' 7800 × 0.005 ' 39.0kg/ m 2 f 0 ' 80 38 % 39 0.1 × 38 × 39 ' 58Hz TL A ' 20log 10 (38 % 39) % 20log 10 (58) & 48 ' 25dB at 58Hz TL B1 ' 24.9 % 20log 10 (810/ 57.7) & 6 ' 41.8dB or Using the equation for f 0 on page 357 in the text and taking logs gives: Substituting the above expression into equation (1) above gives: Try 50mm plasterboard (same as gypsum board), m = 760 × 0.05 = 38kg/m 2 . Required steel weight = 1410/38 = 37kg/m 2 , which is 4.7mm thick. That is, use 50mm thick plasterboard and 5mm thick steel plate. Assume a stud spacing of 0.6m and line-line support. Thus: Solutions to problems 220 TL B2 ' 20log 10 (38) % 10log 10 (0.6) % 30log 10 (2400) % 20log 10 1 % 39 × 810 1/ 2 38 × 2400 1/ 2 & 78 ' 57dB TL C ' 57 % 6 % 10log 10 (0.005) ' 40dB f 0 f c2 f c2 f R 0.5f c2 0.5f c2 A B B C C As the cavity is filled with sound absorbing material, TL B = 57dB. Assume a loss factor for the steel of 0.005 (see table on p.609 in text). Then: The frequency at point D is f R = 55/0.1 = 550Hz. The TL for this construction is plotted on the following figure (where the dashed line is the required Transmission Loss and the solid line is the predicted Transmission Loss, calculated using Figure 8.9 in the text), where it can be seen that the design is deficient between 1000Hz and 2500Hz. It is clear that the first critical frequency must be increased or the quantity TL B (and hence TL C ) must be increased. An easy solution to the problem is to use point supports for the steel panel but this may not be practical. Thus try 75mm thick plaster board (m = 760 × 0.075 = 57kg/m 2 . The required thickness of steel panel is such that steel weight = 1410/57 = 25kg/m 2 , which Sound transmission loss, acoustic enclosures and barriers 221 f c1 ' 0.55 × 343 2 1600 × 0.088 . 460Hz f c2 ' 0.55 × 343 2 5400 × 0.003 . 4000Hz m 1 ' 760 × 0.088 ' 66.9kg/ m 2 and m 2 ' 7800 × 0.003 ' 23.4kg/ m 2 f 0 ' 80 66.9 % 23.4 0.1 × 66.9 × 23.4 ' 61Hz TL A ' 20log 10 (66.9 % 23.4) % 20log 10 (61) & 48 ' 27dB at 61Hz TL B1 ' 26.8 % 20log 10 (460/ 61) & 6 ' 38.3dB TL B2 ' 20log 10 (66.9) % 10log 10 (0.6) % 30log 10 (4000) % 20log 10 1 % 23.4 × 460 1/ 2 66.9 × 4000 1/ 2 & 78 ' 65dB TL C ' 65 % 6 % 10log 10 (0.005) ' 48dB is 3.2mm thick. This is an odd thickness, so try 88mm thick plasterboard (3 × 25 + 13) m = 760 × 0.088 = 66.9kg/m 2 . Required steel weight = 1410/66.9 = 21kg/m 2 , which is 2.7mm thick. That is, use 88mm thick plasterboard and 3mm thick steel plate. Assume a stud spacing of 0.6m and line-line support. Thus: As the cavity is filled with sound absorbing material, TL B = 65dB. Assume a loss factor for the steel of 0.005. Then: The frequency at point D is f R = 55/0.1 = 550Hz. The TL for this construction is plotted on the previous figure as the grey line, where it can be seen that the proposed construction easily meets the noise reduction requirements and is even a little too good. Solutions to problems 222 Problem 8.9 There is a 5mm air gap under door. For the purpose of calculating the transmission coefficient, the reflection in the floor effectively doubles the width of the gap. However, once the transmission coefficient has been determined using figure 8.11 in the text, the area of gap in subsequent calculations is determined without doubling the width. Area of walls and ceiling = 2(4 × 2.5 + 3 × 2.5) + 4 × 3 = 47m 2 . Area under door = 0.005 × 1 = 0.005m 2 . Effective gap under door = 0.01m. τ gap is calculated using figure 8.11 in the text. We can construct the following table using equations 8.12, 8.65 and 8.66 in the text. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 TL wall 27 45 51 56.5 60.5 52 49 63 S wall τ wall 0.0938 0.00149 3.73× 10 -4 1.05× 10 -4 4.19× 10 -5 2.97× 10 -4 5.92× 10 -4 2.36× 10 -5 S gap τ gap 0.005 0.005 0.003 0.0016 0.001 5.5× 10 -4 2.5× 10 -4 1.3× 10 -4 ¯τ 2.10× 10 -3 1.38× 10 -4 7.18× 10 -5 3.63× 10 -5 2.22× 10 -5 1.80× 10 -5 1.79× 10 -5 3.26× 10 -6 ¯ T ¯ L 27 39 41 44 47 47 47 55 From the table it can be seen that the effect of the gap under the door is to significantly reduce the effective wall TL and on comparing the results in the above table with the table of problem 8.8, it can be seen that the required enclosure noise reduction will no longer be achieved in the 1000Hz octave band. Problem 8.10 The required air flow may be calculated using equation 8.84 in the text, in which ρ = 1.206kg/m 2 , C p = 1010m 2 s -2 C -1 , ΔT = 3EC and H = 0.05 × 10 4 W. Thus: Sound transmission loss, acoustic enclosures and barriers 223 V ' 0.05 × 10 4 3 × 1010 × 1.206 ' 0.14m 3 / s The required Insertion Loss specifications for the silencer would be the same as the TL of the walls (not the noise reduction required of the enclosure as this excludes reverberant build-up in the enclosure) and this is found in problem 8.8. Problem 8.11 The following steps would need to be taken. 1. Check local noise regulations for allowable levels in the residential area. Measure existing levels in dB(A) on the perimeter of the supermarket property at the closest location to the proposed compressor location over an extended period (about a week with a statistical noise analyser, or if this is impractical use a sound level meter). 2. Choose as the design criterion for the compressor noise in the community the smallest of the regulation level and the lowest measured existing level plus 5dB(A). If existing levels were determined using spot checks with a sound level meter, then the criterion may be more conservative, such as the existing level plus 2dB(A). 3. Use table 4.8 to convert the allowable dB(A) level in the community to an NR level and then use the corresponding NR curve (figure 4.8) to specify the allowable community noise levels in octave bands. 4. Calculate the noise reduction from the compressor site to the nearest community location due to atmospheric absorption, turbulence, ground effects and meteorological influences, or place a loudspeaker at the proposed compressor location and measure the noise reduction as a function of distance from it. 5. Use the sound power data for the compressor and the excess attenuation data together with equation 5.158 in the text to calculate the noise levels due to the compressor at the nearest community location in octave bands. 6. Thus calculate the required enclosure noise reduction and following that, Solutions to problems 224 L p ' L w & 10log 10 (2πr 2 ) ' 105 & 10log 10 (2π × 80 2 ) ' 59dB TL ' 10log 10 1 % πfm ρc 2 & 4 ' 10log 10 1 % π × 500 × 7800 × h 1.206 × 343 2 & 4 dB the corresponding required wall transmission loss. 7. Check compressor cooling requirements and if necessary design lined inlet and outlet ducts (with forced air ventilation) with the same Insertion Loss as the Transmission Loss of the enclosure walls. Problem 8.12 (a) Compressor is 80m from perimeter. The sound level at the receiver with no enclosure (assuming hard ground, zero reflection loss) may be calculated using equation 5.158 in the text with DI M = A E = 0, so that: The required noise level = 38dB, so reduction required = 21dB at 500Hz. Smallest panel of enclosure = 0.6m × 0.6m (stud spacing), so 500Hz is well above the first panel resonance frequency (you can calculate this using equation 8.21 to be sure). Critical frequency is: . So f c ' 0.55 × 343 2 / (5400 × h) ' 12/ h Hz, where h is in metres 500Hz is in the mass law range for all choices of panel. From equation 8.79 in the text, the required panel TL = NR + C, and for an enclosure lined on the inside, C = 7 (see table 8.4, p376 in the text). Thus the required enclosure TL = 21 + 7 = 28dB at 500Hz. (b) The wall TL may be calculated in the mass law range by using equation 8.36 in the text (with the assumption NOT made that fm/ρc > 1), with a constant of 4 instead of 5.5 to account for octave band calculations. Thus: Sound transmission loss, acoustic enclosures and barriers 225 f 0 ' 80 2/ (0.1 × 9.88) ' 114Hz f c1 ' f c2 ' 0.55 × 343 2 1600 × 0.013 ' 3100Hz TL A ' 20log 10 (9.88 × 2) % 20log 10 (113.8) & 48 ' 19.0dB TL B1 ' 19 % 20log 10 (3100/ 114) & 6 ' 42dB When h = 0.001m, TL = 25.4dB which is too low. As we are in mass law range a doubling of panel thickness will increase the TL by 6dB which is a bit high. Trying h = 0.0016m, gives TL = 29.5dB which is OK, so choose a wall thickness of 1.6mm. (c) Should consider the effect of a door and the design of appropriate door seals, the need for cooling air, the introduction of the inlet air, the pipe penetration (to be isolated from the enclosure wall) for the compressed air and the need for vibration isolation of the enclosure from the compressor. Problem 8.13 Equation 8.79 in the text gives TL = NR + C. From Table 8.3, C = 5dB for an enclosure with surfaces lined with sound absorbing material. Thus, TL = 15 + 5 = 20dB Problem 8.14 (a) The performance of a machinery noise enclosure should not be expressed as a single number dB(A) rating because the dB(A) performance will be dependent on the spectrum shape of the noise generated by the enclosed source. (b) We may use equation 8.79 in the text to relate TL and noise reduction and thus define the minimum required TL. The TL due to the panel may be calculated using the procedure on p360 in the text. d = 0.1m, m 1 = m 2 = 760 × 0.013 = 9.88kg/m 2 . Solutions to problems 226 TL B2 ' 20log 10 (9.88) % 10log 10 (0.6) % 30log 10 (3100) % 20log 10 (2) & 78 ' 50dB TL C ' 50.4 % 6 % 15log 10 (0.02) ' 31dB f R ' 55/ 0.1 ' 550Hz TL oct ' &10log 10 6 (1/3) [10 &TL 1 /10 % 10 &TL 2 /10 % 10 &TL 3 /10 ] > f 0 f R 0.5f c2 f c2 A B D C For line-line support: Assuming that there is sound absorbing material in the wall cavity, TL B = 50dB. The corresponding TL curve is plotted in the figure below and the octave band data are listed in the following table, where octave band results are obtained by using 1/3 octave values read from the graph and calculated using: where TL 1 , TL 2 , and TL 3 , are the (1/3) octave band data read from the figure. Sound transmission loss, acoustic enclosures and barriers 227 V ' H ρC p ΔT ' 0.02 × 50000 1.206 × 1010 × 5 ' 0.164m 3 / s L w ' L p1 % TL % 10log 10 S E & C ' L p1 % NR % 10log 10 S E Octave Band Centre Frequency 63 125 250 500 1000 2000 4000 8000 TL (from plot) 14 21 35 41 47 43 36 52 C 13 11 9 7 5 4 3 3 NR 1 10 26 34 42 39 33 49 Required NR 10 15 20 25 30 35 40 13 It can be seen that the enclosure is deficient in the 63Hz, 125Hz, and 4000Hz octave bands, so it is not adequate. (c) ! Use double, staggered stud wall. ! Use point-line or point-point support by placing rubber grommets between the panel and stud at attachment points. ! Use thicker panels. ! Fix additional 13mm thick panels to existing panels with patches of silicone sealant. ! Use panels of different thickness. (d) From equation 8.84 in the text: (e) The sound power level is given by equation 8.70 in the text. Thus: where S E = 2(5 × 2.5 + 4 × 2.5) + 4 × 5 = 65m 2 and 10log 10 S E = 18dB. The power level calculations may be summarised in the following table. Solutions to problems 228 L pd ' L w & 10log 10 S ' L w & 16dB ¢ p 2 ¦ ' ¢ p 2 d ¦ % ¢ p 2 R ¦ Octave Band Centre Frequency 63 125 250 500 1000 2000 4000 8000 L p1 80 83 78 73 70 60 60 60 NR 10 15 20 25 30 35 40 20 L w 108 116 116 116 118 113 118 98 (f) Test surface 1m from machine of dimensions 2m × 1m × 1m, so measurement surface is 4m × 3m × 2m (machine assumed to be resting on the ground), having an area of S = 2(4 × 2 + 3 × 2) + 4 × 3 = 40m 2 (4 sides and 1 top). Machine surface area S m = 2(2 × 1 + 1 × 1) + 2 × 1 = 8m 2 . The sound pressure is related to the sound power by equation 6.25 in the text with Δ 1 = 0 because the measurements are made outdoors. The ratio, S/S m = 5, so the near field correction Δ 2 = 0 and the sound pressure level 1m from the machine surface is given by: The results are tabulated in the following table: Octave Band Centre Frequency 63 125 250 500 1000 2000 4000 8000 L w 108 116 116 116 118 113 118 98 L pd 92 100 100 100 102 97 102 82 (g) With the enclosure in place, the mean square sound pressure is equal to the sum of the direct and reverberant field contributions. Sound pressure levels are converted to mean square sound pressures using equation 1.78 in the text and sound powers are converted to sound power levels and vice versa using equation 1.80 in the text. Thus the mean square sound pressure in the enclosure is: Sound transmission loss, acoustic enclosures and barriers 229 ¢ p 2 R ¦ ' 4Wρc R 4 R ' 4 3.33 10 C/ 10 0.3 × S E & 1 S E The reverberant mean square pressure is: where R is the room constant and W is the source sound power. The enclosure constant C is given in terms of R by equation 8.82 in the text which can be rearranged (with the aid of equation 7.43) to give: The results are summarised in the following table, where for convenience, ρc = 400. Octave Band Centre Frequency 63 125 250 500 1000 2000 4000 8000 C 13 11 9 7 5 4 3 3 L w 108 116 116 116 118 113 118 98 W 6.3× 10 -2 0.40 0.40 0.40 0.63 0.20 0.63 6.3× 10 -3 4/R 1.21 0.76 0.47 0.29 0.18 0.14 0.10 0.10 L pd 92 100 100 100 102 97 102 82 ¢ p 2 d ¦ 0.63 4.0 4.0 4.0 6.3 2.5 6.3 0.063 ¢ p 2 R ¦ 30.5 122 75.2 46.4 45.4 11.2 25.2 0.25 ¢ p 2 d ¦ % ¢ p 2 R ¦ 31 126 79 50 52 14 32 0.31 L pT 109 115 113 111 111 105 109 89 dB(A) correct. -26 -16 -9 -3 0 1 1 -1 dB(A) sound pressure levels 92 100 100 100 102 97 103 88 Solutions to problems 230 L p2 ' 60 % 10log 10 (65) % 10log 10 2 4 × π × 200 2 & A E ' 24 & A E (dB) (h) In the 2000Hz band, L p1 = 60dB and the sound pressure level at distance r from the enclosure is given by equation 8.72 in the text (with D θ = 2 due to the hard ground surface) as: A E is the excess attenuation given by equation 5.165 in the text. A g is included in D θ above as the asphalt is hard resulting in essentially hemispherical spreading. A a = 10.8/5 = 2dB (see table 5.3 on page 225 in the text). A m = +5, -4dB (see table 5.10) Thus the range of variability is 18 to 27dB. Problem 8.15 Noise level inside enclosure = 101dB at 1000Hz. Noise level on surface, 1m from enclosure = 91dB. Noise level at 50m distance = 70dB. (a) Sound power level radiated by enclosure may be calculated using equation 6.25 in the text. Test surface 1 m from enclosure of dimensions 3m × 3m × 3m, so measurement surface is 5m × 5m × 4m (assuming that the enclosure is resting on the ground), having an area of S = 2(5 × 4 × 2) + 5 × 5 = 105m 2 (4 sides and 1 top). Machine surface area, S m = 2(3 × 3 × 2) + 3 × 3 = 45m 2 . The sound pressure is related to the sound power by equation 6.25 in the text with Δ 1 = 0 because the measurements are made outdoors. The ratio, S/S m = 105/45 = 2.3, so the near field correction Δ 2 = 1 and the sound power level of the enclosure is given by: Sound transmission loss, acoustic enclosures and barriers 231 L w ' 91 % 10log 10 (105) & 1 ' 110dB L p ' 110 & 20log 10 (50) & 10log 10 (2π) ' 68dB (b) We may use equations 5.158 and 5.161 with r = 50m and DI M = A E = 0 so that: (c) As the measured noise level is 70dB, the excess attenuation due to atmospheric absorption and meteorological influences is 68 - 70 = -2dB. The excess attenuation due to the ground effect is -3dB, so the total excess attenuation is -5dB. (d) The machine is probably not well vibration isolated from the enclosure, causing the enclosure wall to vibrate and radiate noise. Vibration could be transmitted by way of the floor or by direct connection of parts of the machine (or attached pipework) to parts of the enclosure. Also, pipework or other equipment not included in the enclosure but attached to the noisy machine could radiate noise which was not apparent prior to installation of the enclosure. Problem 8.16 Adequate internal absorption is necessary to prevent the build-up of reverberant energy which will compromise the predicted acoustic performance. Problem 8.17 (a) The enclosure should be vibration isolated to prevent the walls from being excited to vibrate and thus radiate sound which in turn will compromise the enclosure performance. (b) One possible disadvantage associated with vibration isolation from the floor is increased difficulty in producing an adequate acoustic seal around the base of the enclosure. (c) Other factors which could degrade the enclosure performance are ! poor seals around doors and windows Solutions to problems 232 ! inadequate TL performance of doors and windows ! inadequate internal absorption ! pipe penetrations in the enclosure walls not vibration isolated from the walls or poorly sealed acoustically. ! pipework or other equipment not included in the enclosure but attached to the noisy machine could radiate noise which was not apparent prior to installation of the enclosure. Problem 8.18 ! acoustic absorbing material left out of wall cavity or too rigid and touching both walls ! poor seals around windows and doors ! glass in double glazing too thin ! doors of insufficient acoustic performance ! floor vibration transmitted to enclosure walls because of inadequate vibration isolation ! wall stud spacing incorrect ! incorrect wall thickness or wall materials ! poor seal at base of enclosure ! poor bricklaying (if brick walls) leading to gaps in the mortar ! change from original noise sources ! tonal noise from the machine corresponding to a structural resonance or an acoustic resonance (wall cavity or enclosure space). ! pipework or other equipment not included in the enclosure but attached to the noisy machine could radiate noise which was not apparent prior to installation of the enclosure. A test to determine whether the problem was airborne or structure-borne would be to turn the machines off and use loudspeakers in the enclosure to generate the same noise levels inside the enclosure. If the exterior noise levels are then the same with the loudspeakers operating as they were with the machine, then the problem is airborne flanking. If the noise external to the enclosure with the loudspeakers operating is lower, then the problem is likely to be structure-borne vibration or radiation from equipment attached to the noisy machine but not included in the enclosure. Sound transmission loss, acoustic enclosures and barriers 233 L pA ' 10log 10 10 (6.3 & 2.62) % 10 (6.7 & 1.61) % 10 (6.2 & 0.86) % 10 (5.5 & 0.32) % 10 5.2 % 10 (5 % 0.12) % 10 (4.5 % 0.1) % 10 (4.2 & 0.11) ' 59.2dB(A) Problem 8.19 (a) The Noise Rating (NR) is obtained by plotting the un-weighted octave band data on a set of NR curves (see fig 4.7, p118 in text) as illustrated in the following figure. The NR value is 52.5. A-weighted sound level is given by: (b) The acceptable noise level is obtained using table 4.10, p167 in the text and is L p = 40 + 20 - 10 = 50dB(A). The actual level is over 9dB(A) above the allowable level, a situation which is not acceptable and according to table 4.11, it will generate widespread complaints from the community. If the noise only occurs during the hours of 7am and 6pm, then the allowable level is L p = 40 + 20 = 60dB(A) and the existing level is thus acceptable. Solutions to problems 234 0.5 d 1 ' 1.5 40 & d 1 ; d 1 ' 10m β ' tan &1 0.5 10 ' 2.862E ρf R 1 ' 1.206 × 500 2 × 10 5 ' 0.003 | 0.5m S R 1.5m 40 - d 1 d 1 (c) The octave band levels essentially follow the shape of the NR curve (except at 63Hz, where the level is much lower), so the noise will sound neutral. (d) With no barrier, there are two propagation paths; the direct path and the ground reflected path. With a barrier, there are 8 paths as listed below: ! over the top with no ground reflections ! over the top with a ground reflection on the source side ! over the top with a ground reflection on the receiver side ! over the top with ground reflections on both sides ! around each end with no ground reflections ! around each end with a ground reflection (e) Attenuation of ground reflected wave (see figure) Using similar triangles: From table 5.2 on p209, R 1 = 2 × 10 -5 ; ρ = 1.206. Thus at 500Hz: Sound transmission loss, acoustic enclosures and barriers 235 β R 1 ρf 1/ 2 ' 52E d 1 0.5 ' 20 & d 1 3 ; d 1 ' 2.857m β ' tan &1 0.5 2.857 ' 9.93E ; β R 1 ρf 1/ 2 ' 181E 0.5m S | 20m 20m 1.5m 3m d 1 R S 0.5m | R 1.5m 3m 20m 20m d 2 From figure 5.20, the reflection loss on ground reflection is thus 3.9dB. (f) Attenuation due to barrier. First calculate the reflection loss for each path which involves a ground reflection. ! Over the top, source side reflection Using similar triangles: and from figure 5.20, p231 in the text, A R = 7.9dB ! over the top, receiver side reflection Using similar triangles: Solutions to problems 236 d 2 1.5 ' 20 & d 2 3 ; d 2 ' 6.667 β ' tan &1 1.5 6.667 ' 12.7E ; β R 1 ρf 1/ 2 ' 231E d 3 0.5 ' 40 & d 3 1.5 ; d 3 ' 10m 0.5m 3m 1.5m R 20m 20m d 3 S S / A A A 0.5m 20m 20m d 1 1.5m 3m | | B B B S R and from figure 5.20, p231 in the text, A R = 7.0dB ! around each end of the barrier Using similar triangles: which is the same as if there were no barrier, so A R = 3.9dB Now calculate Fresnel numbers for all paths over and around the barrier. ! Over the top (see figure) Sound transmission loss, acoustic enclosures and barriers 237 A ) ' [ 0.5 2 % 2.857 2 ] 1/ 2 ' 2.9004m; A )) ' [ 3 2 % 17.143 2 ] 1/ 2 ' 17.404m B ) ' [ 1.5 2 % 6.667 2 ] 1/ 2 ' 6.834m; B )) ' [ 3 2 % 13.333 2 ] 1/ 2 ' 13.663m d ) ' [ 2 2 % 40 2 ] 1/ 2 ' 40.05m A ' [ 20 2 % 2.5 2 ] 1/ 2 ' 20.156m; B ' [ 20 2 % 1.5 2 ] 1/ 2 ' 20.056m d ' [ 40 2 % 1 2 ] 1/ 2 ' 40.01m N 1 ' (20.156 % 20.056 & 40.01) × 2.92 ' 0.58 N 2 ' (2.900 % 17.404 % 20.056 & 40.05) × 2.92 ' 0.91 N 3 ' (20.156 % 6.834 % 13.663 & 40.05) × 2.92 ' 1.76 N 4 ' (2.900 % 17.404 % 6.834 % 13.663 & 40.01) × 2.92 ' 2.31 Δ b1 ' 11.5 ; Δ b2 ' 12.8 ; Δ b3 ' 15.8 ; Δ b4 ' 17.0dB Top view d B R A S The distance d, between source and receiver depends on the path which is being considered. For example, for waves reflected from the ground on the source side of the barrier only, the value of d (denoted d') below, is that from the image source to the receiver, etc. From figure 8.14, p389 in the text: It can be shown that in this case, the corrections of equation 8.88 are: A b1 = 11.5 + 0.0 = 11.5; A b2 = 12.8 + 0.1 = 12.9; A b3 = 15.8 + 0.1 = 15.9; A b4 = 17.0 + 0.2 = 17.2. ! Around the ends (see figure) Solutions to problems 238 A ' B ' [ 20 2 % 5 2 % (1.5 & 0.5) 2 / 4] 1/ 2 ' 20.622m; d ' 40.01m. N 5,6 ' (20.622 × 2 & 40.01) × 2.92 ' 3.6 ; Δ b5, 6 ' 18.7dB A ) ' A )) ' [ 10 2 % 0.5 2 % 2.5 2 ] 1/ 2 ' 10.32m B ) ' [ 20 2 % 1 2 % 5 2 ] ' 20.64m; d ) ' [ 40 2 % 2 2 ] 1/ 2 ' 40.05m N 7, 8 ' (10.32 × 2 % 20.64 & 40.05) × 2.92' 3.6 ; Δ b7, 8 ' 18.7dB NR ' 10log 10 10 &0/ 10 % 10 &3.9/ 10 & 10log 10 10 &1.15 % 10 &(1.29 % 0.79) % 10 &(1.59 % 0.70) % 10 &(1.72 % 1.49) % 2 × 10 &1.9 % 2 × 10 &(1.90 % 0.39) ' 1.5 % 9.2 ' 10.7dB A A 0.5m 0.5m 20m 20m 1.5m | B S S R 10m 3m Direct waves, no reflection (one each side): From figure 8.14 in the text: Thus A b5,6 ' 18.7 % 0.3 ' 19.0dB Reflected waves (one each side) From figure 8.14 in the text: Thus A b7,8 ' 18.7 % 0.3 ' 19.0dB NR due to barrier is then calculated using equation 1.97 in the text as: Sound transmission loss, acoustic enclosures and barriers 239 A 1.5m B d 48m 1.5m 2m ?? Problem 8.20 NR-50 octave band values may be read from figure 4.8, p155 in the text and the following table may be generated. Octave band centre frequency (Hz) 63 125 250 500 100 0 200 0 400 0 8000 Existing noise 68 77 65 67 63 58 45 40 NR-50 76 67 60 54 50 48 46 44 Required reduction 0 10 5 13 13 10 0 0 Noise reduction is a function of Fresnel number which is directly proportional to frequency. The important frequency is then 125Hz. If this is satisfied, the reduction at 500Hz will be OK. It is left to the reader to verify this. Total of 8 paths (4 over top and 4 around sides) to consider to get an overall reduction of 10dB at 125Hz. Consider first the paths around the barrier edges as the attenuation around these paths is independent of barrier height. Solutions to problems 240 Around edge (elevation) S h g j e f c 1.5m 2m 23m 25m 1.5m ?? Plan 2m 23m 25m b 5m S R As the source and receiver are at the same height, the location of the ground reflection will be mid-way (in a horizontal direction only) between the source and receiver. Path with no ground reflection Path h = (2 2 + 5 2 ) 1/2 = 5.385m Path j = (48 2 + 5 2 ) 1/2 = 48.260m A % B & d ' 5.385 % 48.260 & 50 ' 3.64m At 125Hz, = 343/125 = 2.744m and N = (2/ ) × 3.64 = 2.65. From figure λ λ 8.14 on p389 in the text, Δ b = 17.0dB. The correction term given by equation 8.88 in the text (assuming an omnidirectional source) is , so . 20log 10 (53.64/ 50) ' 0.6dB A b ' 17.0 % 0.6 ' 17.6dB Path with ground reflection From the figure, and using similar triangles, b/5 = 25/48; b = 2.6m Path e = ( 25 2 % 1.5 2 % 2.604 2 ) 1/ 2 ' 25.180m By similar triangles, f/1.5 = 23/25; f = 1.38m Sound transmission loss, acoustic enclosures and barriers 241 A , A 1 3 B , B 1 2 h-1.5 d 2m 48m B , B 3 4 1.5m 1.5m 1.5m h h A , A 2 4 1.5m 2m 48m Path g = ( 23 2 % 1.38 2 % 2.396 2 ) 1/ 2 ' 23.166m Path c = ( 0.12 2 % 2 2 % 5 2 ) 1/ 2 ' 5.387m A % B & d ' 5.387 % 23.166 % 25.180 & [ 50 2 % 3 2 ] 1/ 2 ' 3.64m Thus, N = (2/ ) × 3.64 = 2.65. from figure 8.14 on p389 in the text, λ Δ b = 17.0dB. The correction term given by equation 8.88 in the text (assuming an omnidirectional source) is , so 20log 10 (53.74/ 50.09) ' 0.6dB . Losses due to ground reflection are zero. A b ' 17.0 % 0.6 ' 17.6dB The required barrier height can be found by trial and error. The results are summarised in the following table, where the subscript "1" refers to waves travelling over the barrier with no reflections, the subscript "2" implies a reflection on the source side, the subscript "3" implies a reflection on the receiver side and the subscript "4" implies a reflection on both sides. The distances "A" refer to the source side of the barrier (source to barrier top along the particular path specified by the associated subscript) and the distances "B" refer to distances on the receiver side. We will begin the trial with a barrier height of 3.0m. The correction term given by equation 8.88 in the text (assuming an omnidirectional source) is added to the barrier attenuation Δ b to give A b . Solutions to problems 242 NR ' 10log 10 10 0 % 10 0 & 10log 10 4 × 10 &1.76 % 10 &A b1 / 10 % 10 &A b2 / 10 % 10 &A b3 / 10 % 10 &A b4 / 10 The barrier overall noise reduction is given by equation 1.97 in the text which can be expanded to give: Barrier height 3.0m 3.5m 4.0m A 1 (m) A 2 (m) A 3 (m) A 4 (m) B 1 (m) B 2 (m) B 3 (m) B 4 (m) N 1 N 2 N 3 N 4 Δ b1 Δ b2 Δ b3 Δ b4 A b1 (dB) A b2 (dB) A b3 (dB) A b4 (dB) 2.50 4.92 2.50 4.92 48.02 48.02 48.21 48.21 0.38 2.07 0.45 2.28 10.1 16.1 10.5 17 10.2 16.6 10.6 17.5 2.828 5.385 2.828 5.385 48.042 48.042 48.260 48.260 0.63 2.43 0.73 2.66 11.9 17 12.0 17.1 12.0 17.5 12.2 17.6 3.202 5.852 3.202 5.852 48.065 48.065 48.314 48.314 0.923 2.79 1.04 3.036 12.6 17.3 13.1 18 12.8 17.9 13.3 18.7 NR (dB) 8.3 9.4 10.0 Thus the wall height should be about 4.0m. Problem 8.21 The layout is illustrated in the following figure. Sound transmission loss, acoustic enclosures and barriers 243 A ) ' 20 2 % 2 2 ' 20.0998m A )) ' 30 2 % 3 2 ' 30.1496m B ) ' 6.667 2 % 3 2 ' 7.3109m B )) ' 3.333 2 % 1.5 2 ' 3.6550m A ' 50 2 % 1 2 ' 50.0100m B ' 10 2 % 1.5 2 ' 10.1119m N 1 ' 2 0.686 50.0100 % 10.1119 & (60 2 % 0.5 2 ) 1/ 2 ' 0.35 N 2 ' 2 0.686 20.0998 % 30.1496 % 10.1119 & (60 2 % 3.5 2 ) 1/ 2 ' 0.76 2m d 1 50m 10m d 2 1.5m 3m | 1 | 2 A A' A'' B' B'' B From similar triangles, ; so d 1 = 20.0m d 1 50 & d 1 ' 2 3 ; so d 2 = 3.333m d 2 10 & d 2 ' 1.5 3 λ = 343/500 = 0.686m There are 4 paths which will contribute to the sound level at the receiver so we need to calculate the Fresnel Number corresponding to each. Path 1, A Y B Path 2, A' Y A'' Y B Solutions to problems 244 N 3 ' 2 0.686 50.0100 % 7.3106 % 3.6553 & (60 2 % 3.5 2 ) 1/ 2 ' 2.5 N 4 ' 2 0.686 20.0998 % 30.1496 % 7.3106 % 3.6553 & (60 2 % 0.5 2 ) 1/ 2 ' 3.5 NR ' 10log 10 10 &0/ 10 % 10 &3/ 10 & 10log 10 10 &9.9/ 10 % 10 &15/ 10 % 10 &20/ 10 % 10 &24.5/ 10 ' 1.76 % 8.31 ' 10.1dB Path 3, A Y B' Y B'' Path 4, A' Y A'' Y B` Y B'' As the wall completely surrounds the factory, no sound is diffracted around its edges. The correction term given by equation 8.88 in the text is greatest for path 4 and is equal to 0.09dB which is negligible, so the correction will be ignored. From figure 8.14, the attenuations corresponding to N 1 to N 4 are Δ N1 = 9.9, Δ N2 = 12, Δ N3 = 17, Δ N4 = 18.5. Adding 3dB to all ground reflections results in the following noise reductions corresponding to the 4 paths: Path 1, 9.9dB; Path 2, 15dB; Path 3, 20dB; Path 4, 24.5dB. From equation 1.97 in the text, the noise reduction due to the enclosure is: Problem 8.22 Referring to the following figures, we have for the reflection angles: over top, source side, tanβ 1 ' 4 25 & x ' 0.5 x ; x ' 2.78m, β 1 ' 10.2E over top, receiver side, tanβ 2 ' 4 25 & y ' 1.5 y ; y ' 6.82m, β 2 ' 12.4E Sound transmission loss, acoustic enclosures and barriers 245 Over top (elevation) Plan R S | 1 | 2 x 25m 25m y 25m 25m 7.5m 7.5m 0.5m 4m 1.5m Around edge (elevation) Plan 7.5m 25m R S S S R h 1 1m | 3 25m 25m 7.5m 25m 1.5m 4m z 0.5m Solutions to problems 246 around edge, tanβ 3 ' 1.5 50 & z ' 0.5 z ; z ' 12.5m, β 3 ' 2.3E Flow resistivity of ground = R 1 = 3 × 10 7 MKS rayls/m and ρ = 1.206kg/m 3 . With no barrier, β 4 = β 3 = 2.3E. The reflection loss, A R is calculated using figure 5.20 in the text and the results are tabulated in the table below. Octave band centre frequency (Hz) ρf R 1 R 1 ρf 1/ 2 A R1 A R2 A R3 and A R4 63 125 250 500 1000 2000 4000 8000 2.5×10 -6 5.0×10 -6 1.0×10 -5 2.0×10 -5 4.0×10 -5 8.0×10 -5 1.6×10 -4 3.2×10 -4 630 447 316 223 158 112 79 56 0.2 0.3 0.5 0.7 1.0 1.5 2.3 3.0 0.2 0.3 0.5 0.7 1.0 1.5 2.3 3.0 1.5 2.0 2.5 3.5 5.0 6.0 7.0 6.5 Path 1 - over top of barrier with no ground reflections d = 50.010m, A = 25.244, B = 25.125 and A + B - d = 0.359m Path 2 - over top of barrier with ground reflection on the source side d = 50.040m, B = 25.125 A = (2.78 2 + 0.5 2 ) 1/2 + (4 2 + 22.22 2 ) 1/2 = 25.402m and A + B - d = 0.487m Path 3 - over top of barrier with ground reflection on the receiver side d = 50.010m, A = 25.244 B = (6.82 2 + 1.5 2 ) 1/2 + (4 2 + 18.18 2 ) 1/2 = 25.598m and A + B - d = 0.802m Path 4 - over top of barrier with ground reflection on both sides d = 50.010m, A = 25.402, B = 25.598 and A + B - d = 0.990m Paths 5&6 - around edges with no reflection B = A = (0.5 2 + 25 2 + 7.5 2 ) 1/2 = 26.106m and A + B - d = 2.201m Sound transmission loss, acoustic enclosures and barriers 247 Paths 7&8 - around edges with reflection in ground Intersection height of diffracted wave with barrier edge = 1.5 × 12.5/37.5 = 0.5m. A = 2(0.5 2 + 12.5 2 + (7.5/2) 2 ) 1/2 = 26.120m B = 2(1 2 + 25 2 + 7.5 2 ) 1/2 = 26.120m and A + B - d = 2.20m Fresnel Number, . Values of N for each path are N ' 2f 343 (A % B & d) tabulated below. Octave band centre frequency (Hz) N 1 N 2 N 3 N 4 N 5&6 N 7&8 63 125 250 500 1000 2000 4000 8000 0.13 0.26 0.52 1.05 2.09 4.19 8.37 16.7 0.18 0.36 0.71 1.42 2.84 5.68 11.4 22.7 0.30 0.59 1.17 2.34 4.68 9.35 18.7 37.4 0.36 0.72 1.44 2.89 5.77 11.5 23.1 46.2 0.80 1.60 3.21 6.42 12.8 25.7 51.3 103 0.80 1.61 3.21 6.4 12.8 26 51 103 The noise reductions (NR or Δ b ) corresponding to the above Fresnel Numbers are calculated using figure 8.14, p389 in the text and are tabulated below. The correction term given by equation 8.88 in the text (assuming an omnidirectional source) will be less than 0.1dB overall and will be ignored here. The numbers in brackets indicate the sum of the ground reflection losses and barrier diffraction loss. All quantities are in dB. Solutions to problems 248 Octave band centre frequency (Hz) NR 1 NR 2 NR 3 NR 4 NR 5&6 NR 7&8 63 125 250 500 1000 2000 4000 8000 8.2 9.5 10.8 13.0 16.1 19.2 22.3 24 8.9 (9.1) 10.1 (10.4) 12.0 (12.5) 14.8 (15.5) 17.9 (18.9) 20.9 (22.4) 23.9 (26.1) 24 (27) 9.9 (10.1) 11.8 (12.1) 13.8 (14.3) 17.0 (17.7) 19.8 (20.8) 23.0 (24.5) 24 (26.3) 24 (27) 10.0 (10.4) 12.0 (12.6) 14.5 (15.5) 17.7 (19.1) 20.7 (22.7) 23.8 (26.8) 24 (28.6) 24 (30) 12.3 15.0 18.1 21.2 24 24 24 24 12.3 (13.8) 15.0 (17.0) 18.1 (20.6) 21.2 (24.7) 24 (29) 24 (30) 24 (31) 24 (30.5) The barrier noise reduction is given by equation 1.97 in the text and the results of the calculations are summarised in the table below. The subscript "A" refers to the condition with no barrier and the subscript "B" refers to the condition with barrier. All quantities are in dB. Octave band centre frequency (Hz) 10log 10 j 10 &NR Ai / 10 &10log 10 j 10 &NR Bi / 10 NR SPL at receiver 63 125 250 500 1000 2000 4000 8000 2.3 2.1 1.9 1.6 1.2 1.0 0.8 0.9 1.8 3.7 6.0 8.9 12.1 14.8 16.7 17.3 4.1 5.8 7.9 10.5 13.3 15.8 17.5 18.2 65.9 69.2 64.1 49.5 44.7 40.2 32.5 33.8 The overall A-weighted level is calculated using the octave band levels as described on pages 101 and 102 in the text. The A-weighted level with the barrier is calculated using the numbers in the last column of the preceding table and is 58.2dB(A). The A-weighted level without the barrier is calculated using the octave band levels given in the problem and is 67dB(A). Thus the noise reduction due to the barrier is 9dB(A). Sound transmission loss, acoustic enclosures and barriers 249 ΔC ' Klog 10 ( 2πf b/ 343) Problem 8.23 If the barrier were a building, the additional noise reduction may be calculated using equation 8.98, p394 of the text which may be rewritten as: The values of K are calculated using figure 8.17 in the text and trigonometry is used to calculate the angles, . The results are summarised in the θ and φ two following tables. Values for ΔC are in dB and the subscript on C refers to the path number. Path 1 Path 2 Path 3 Path 4 Paths 5&6 Paths 7&8 θ (degrees) 98 100 98 100 117 117 (degrees) 96 96 102 102 117 117 φ K 1.2 1.5 2.3 2.7 5.6 5.6 Octave band centre frequency (Hz) ΔC 1 ΔC 2 ΔC 3 ΔC 4 ΔC 5,6,7&8 63 125 250 500 1000 2000 4000 8000 0.8 1.2 1.5 1.9 2.2 2.6 3.0 3.3 1.0 1.4 1.9 2.3 2.8 3.2 3.7 4.2 1.5 2.2 2.9 3.6 4.3 5.0 5.7 6.4 1.8 2.6 3.4 4.2 5.0 5.8 6.7 7.4 3.7 5.4 7.1 8.8 10.4 12.1 13.8 15.5 The noise reductions are calculated as before with the attenuation due to the barrier thickness added to each path. The results (in dB) are summarised in the table below. Solutions to problems 250 Octave band centre frequency (Hz) NR 1 NR 2 NR 3 NR 4 NR 5&6 NR 7&8 63 125 250 500 1000 2000 4000 8000 9.0 10.7 12.3 14.9 18.3 21.8 25.3 27.3 10.1 11.8 14.4 17.8 21.7 25.6 29.8 31.2 11.6 14.3 17.2 21.5 25.2 29.5 33.0 33.4 12.2 15.2 18.9 23.3 27.7 32.6 35.4 37.4 16.0 19.4 25.2 30.0 34.4 36.1 37.8 39.5 17.5 22.4 27.8 33.5 39.4 42.1 44.8 46.0 The barrier noise reduction is given by equation 1.97 in the text and the results of the calculations are summarised in the table below. The subscript "A" refers to the condition with no barrier and the subscript "B" refers to the condition with barrier. Octave band centre frequency (Hz) 10log 10 j 10 &NR Ai / 10 &10log 10 j 10 &NR Bi / 10 NR SPL at receiver 63 125 250 500 1000 2000 4000 8000 2.3 2.1 1.9 1.6 1.2 1.0 0.8 0.9 3.6 6.0 8.6 12.0 15.7 19.3 22.9 24.5 5.9 8.1 10.5 13.6 16.9 20.3 23.7 25.4 64.1 66.9 61.5 46.4 41.1 35.7 26.3 26.6 The overall A-weighted level is calculated using the octave band levels as described on pages 101 and 102 in the text. The A-weighted level with the thick building as the barrier is calculated using the numbers in the last column of the preceding table and is 55.6(A) which is 2.6dB(A) less than for the thin barrier. Thus the effect of thickening the barrier is to increase the noise reduction by 2.6dB(A) to a total of approximately 11dB(A). Sound transmission loss, acoustic enclosures and barriers 251 IL ' 10log 10 D θ 4πr 2 % 4 S 0 ¯α 0 & 10log 10 D θ F 4πr 2 % 4K 1 K 2 S(1 & K 1 K 2 ) ' term1 & term2 S A d B R 3m 1.5m 25m 25m 0.5m Problem 8.24 If the barrier is moved closer to the refrigeration unit, the noise reduction will increase because the Fresnel numbers will increase due to the increased path length from source to receiver over the top of the barrier and around the edges. Problem 8.25 An elevation view of the situation is shown in the figure. As the barrier is indoors, equation 8.109 on p402 in the text is appropriate. Thus: Source and receiver are each 25m from barrier. For an omnidirectional source on a hard floor D θ = 2; and . D θ 4πr 2 ' 6.366 × 10 &5 r ' d ' [ 50 2 % 1 2 ] 1/2 ' 50.01m A ' (25 2 % 2.5 2 ) 1/ 2 ' 25.125m B ' (25 2 % 1.5 2 ) 1/ 2 ' 25.045m A + B - d = 0.160m; S = 2 × 50 = 100m 2 = area of gap between barrier and ceiling. S 0 = 2(100 × 5 + 50 × 5 + 100 × 50) = 11,500m 2 ; = 0.08. ¯α 0 Thus, 4/ S 0 ¯α 0 ' 4.348 × 10 &3 Solutions to problems 252 ¯ α 1 ' ¯ α 2 ' 5750 × 0.08 % 150 × 0.15 5900 ' 0.082 S 1 ¯ α 1 ' S 2 ¯ α 2 ' 482.5m 2 and K 1 ' K 2 ' 100 100 % 482.5 ' 0.172 term1 ' 10log 10 6.366 × 10 &5 % 4.348 × 10 &3 ' &23.6dB term2 ' 10log 10 6.366 × 10 &5 F % 4 × 0.172 2 100(1 & 0.172 2 ) ' 10log 10 6.366 × 10 &5 F % 1.29 × 10 &3 S 1 = S 2 = S 0 /2 + 3 × 50 = 5,900m 2 . Thus: Referring to the first equation: There is only one path over the top of the barrier, so F = (3 + 10N) -1 , where N is the Fresnel number for the path, calculated using equation 8.85 and figure 8.13 on page 388 in the text. Thus term 2 can be written as: The calculations which are a function of frequency are summarised in the following table, where the barrier IL is given by IL = term 1 - term 2. Octave band centre frequency (Hz) N = 2 λ (A % B & d) F D θ F 4πr 2 term 2 IL 63 125 250 500 1000 2000 4000 8000 0.059 0.117 0.233 0.466 0.933 1.865 3.732 7.46 0.28 0.240 0.188 0.131 0.081 0.046 0.025 0.013 1.78×10 -5 1.53×10 -5 1.12×10 -5 8.34×10 -6 5.16×10 -6 2.93×10 -6 1.59×10 -6 8.20×10 -7 -29 -29 -29.1 -29.1 -29.1 -29.1 -29.1 -29.1 5.4 5.4 5.5 5.5 5.5 5.5 5.5 5.5 Sound transmission loss, acoustic enclosures and barriers 253 L pd ' L w & 10log 10 (4π × 16) ' L w & 23dB &10log 10 (1 & ¯ α f ) % 10log 10 (1 % 1.2) 2 % 4 2 4 2 ' &10log 10 (1 & ¯α f ) % 1.1 &10log 10 (1 & ¯α c ) % 10log 10 (2 % 1.8) 2 % 4 2 4 2 ' &10log 10 (1 & ¯ α c ) % 2.8 S R C Problem 8.26 (a) The most dominant five paths from source to receiver are shown in the figure. We are to ignore any paths contributing less than 0.2dB to the total. That is, we will ignore paths which contribute 10dB or more lower than the direct path. Of course, the correct way to do this problem is to use equation 7.110 in the text for a flat room with a diffusely reflecting floor (as there is furniture present) and specularly reflecting ceiling. However, here we are happy with an approximate solution. The direct path contribution is given by equations 5.158 and 5.160 in the text with DI M and A E = 0. Thus: The second path (reflection from the floor) will be attenuated by: The third path (reflection from the ceiling) will be attenuated by: The fourth path (reflection from the floor then ceiling) will be attenuated by: Solutions to problems 254 &10log 10 (1 & ¯ α f ) (1 & ¯ α c ) % 10log 10 (1 % 3 % 1.8) 2 % 4 2 4 2 ' &10log 10 (1 & ¯α f ) (1 & ¯α c ) % 4.9 &10log 10 (1 & ¯ α f ) (1 & ¯ α c ) % 10log 10 (1.2 % 3 % 2) 2 % 4 2 4 2 ' &10log 10 (1 & ¯α f ) (1 & ¯α c ) % 5.3 The fifth path (reflection from the ceiling then floor) will be attenuated by: The contributions of each path to the final sound pressure level is obtained by subtracting the attenuations from the direct wave sound pressure level (arithmetically). The total sound pressure level is then obtained by logarithmically combining the contributions from each path (assuming incoherent combination). The results are tabulated in the table below. Note that paths involving more reflections will result in a contribution of less than 0.2dB to the total and are thus not included. Perhaps you could prove this by trial and error. frequency (Hz) 500 1000 2000 L w 70 77 75 L pd 47 54 52 L p2 42.2 47.8 45.5 L p3 38.2 38.2 29.2 L p4 32.4 31 21.7 L p5 31.9 30.5 21.2 L pt 48.8 55.1 52.9 Sound transmission loss, acoustic enclosures and barriers 255 N ' 2 × f c 1.2 2 % 2 2 % 1 2 % 2 2 & 4 ' 3.31 × 10 &3 f A b ' Δ b % 20log 10 2.282 % 2.282 4 ' Δ b % 1.2 L pb ' L pd & A b (b) When the barrier is present, the only contributions to the sound level on the other side will be the wave reflected from the ceiling (L p3 ) and the wave diffracted over the top, as all other paths are blocked by the barrier (do a sketch of the arrangement to prove this to yourself). For the wave diffracted over the top, we can write the following for the Fresnel number: The attenuation, Δ b is read from figure 8.14 in the text (point source) and corrected using equation 8.88 to give: The contribution, L pb of the wave diffracted over the top of the barrier to the sound field at the receiver is calculated using: The results are summarised in the following table. frequency (Hz) 500 1000 2000 L pd 47 54 52 L p3 39.9 33.9 24.4 N 1.65 3.3 6.6 A b 16 19 22 L pb 31 35 30 L pt 39.0 39.9 33.3 reduction due to barrier 10 15 20 Solutions to problems 256 IL ' 10log 10 D θ 4πr 2 % 4 S 0 ¯α 0 & 10log 10 D θ F 4πr 2 % 4K 1 K 2 S(1 & K 1 K 2 ) ' term1 & term2 A B S R 3 m 1.5m 2m 3m 5m 10m 5m 2m 20m a b R S 50m 5m 5m 1m d Assumptions: ! The barrier is sufficiently wide that the contribution at the receiver position due to refraction around the edges is negligible. ! All waves at the receiver combine together incoherently. ! The directivity of the source is uniform in all directions. (c) The noise reduction could be increased by moving the barrier sufficiently close to either the noise source or receiver that the wave reflected from the ceiling will also be blocked. In addition the barrier Fresnel number will be increased, further reducing the energy diffracted over the top of the barrier. Problem 8.27 (a) The situation is illustrated in the figure. As the barrier is indoors, equation 8.109 on p402 in the text is appropriate. Thus: First we calculate the Fresnel Numbers for diffraction over the top and around the edges using equation 8.85 and figure 8.13 on page 388 in the text. Sound transmission loss, acoustic enclosures and barriers 257 S ' 2 × 5 × 3 % 20 × 2 ' 70m 2 S 1 ' S 2 ' 30 % 1350 ' 1380m 2 ¯ α 1 ' ¯ α 2 ' (216/ 2 % 30¯ α b ) / (1350 % 30) Diffraction over the top X R = X S = 5m; Z R = 1.5m; Z S = 1m; Y = 2m; h = 3m Y R = 2 × 5/ (5 % 5) ' 1 d ' [ ( 5 % 5) 2 % (1 % 1) 2 % (1.5 & 1) 2 ] 1/ 2 ' 10.210m A ' [ 5 2 % 1 2 % (3 & 1) 2 ] 1/ 2 ' 5.477m B ' [ 5 2 % 1 2 % (3 & 1.5) 2 ] 1/ 2 ' 5.315m Fresnel Number, N = (2f/343)(A + B - d) = 3.393 × 10 -3 f Diffraction around edge "a" X R = X S = 5m; Z R - Z S = 2m; Y = 0.5m; h - Z S = 5m; h - Z R = 3m Y R = 0.5 × 5/ (5 % 5) ' 0.25 d ' [ ( 5 % 5) 2 % 4 × 0.25 2 % 4] 1/ 2 ' 10.210m A ' [ 5 2 % 0.25 2 % 5 2 ] 1/ 2 ' 7.075m B ' [ 5 2 % 0.25 2 % 3 2 ] 1/ 2 ' 5.836m Fresnel Number, N a = (2f/343)(A + B - d) = 1.575 × 10 -2 f Diffraction around edge "b" X R = X S = 5m; Z R - Z S = 2m; Y = 0.5m; h - Z S = 5m; h - Z R = 7m Y R = 0.5 × 5/ (5 % 5) ' 0.25 d ' [ ( 5 % 5) 2 % 4 × 0.25 2 % 4] 1/ 2 ' 10.210m A ' [ 5 2 % 0.25 2 % 5 2 ] 1/ 2 ' 7.075m B ' [ 5 2 % 0.25 2 % 7 2 ] 1/ 2 ' 8.606m Fresnel Number, N b = (2f/343)(A + B - d) = 3.190 × 10 -2 f S 0 ¯α 0 ' 2(5 × 20 % 5 × 50 % 20 × 50) × 0.08 ' 216.0m 2 r = d = 10.210m; DI = 5, thus D θ = 3.16 and D θ /4πr 2 = 2.51 × 10 -3 Solutions to problems 258 term1 ' 10log 10 D θ 4πr 2 % 4 S 0 ¯ α 0 ' &16.8dB and F ' j a, b, c 1 3 % 10N i D θ F 4πr 2 % 4K 1 K 2 S(1 & K 1 K 2 ) The quantities K 1 and K 2 are calculated using equation 8.111 in the text. They are frequency dependent quantities, but all other variables needed for their calculation have been evaluated above. We now proceed to evaluate the second term of equation 8.109 as a function of octave band centre frequency. Results for the second term as well as the overall Insertion Loss of the barrier are tabulated in the following table. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 N (Top) 0.21 0.42 0.85 1.70 3.39 6.79 13.57 N a 0.99 1.97 3.94 7.88 15.75 31.5 63.0 N b 2.01 3.99 7.98 15.95 31.9 63.8 127.6 F 0.315 0.205 0.123 0.068 0.036 0.019 0.010 ¯α b (table 7.1) 0.08 0.25 0.83 1.0 1.0 1.0 1.0 ¯α 1 , ¯ α 2 0.08 0.08 0.10 0.10 0.10 0.10 0.10 K 1 , K 2 0.388 0.377 0.345 0.337 0.337 0.337 0.337 1.09× 10 -2 9.98× 10 -3 8.02× 10 -3 7.46× 10 -3 7.39× 10 -3 7.34× 10 -3 7.32× 10 -3 Insertion Loss (dB) 2.8 3.2 4.2 4.5 4.5 4.5 4.6 Sound transmission loss, acoustic enclosures and barriers 259 f r ' c L πd ' 5476 π × 0.4022 ' 4,334Hz f c ' 0.55c 2 c L h ' 0.55 × 343 2 5476 × 0.00222 ' 5,323Hz pipe 0.20m 0.40m Jacket c L ' E/ [ρ m ( 1 & ν 2 ) ] ' 71.6/ [2700( 1 & 0.34 2 ) ] ' 5476 m/s (b) Moving the barrier closer to the source will increase the value of N which will reduce the value of F and thus the direct field contribution. The value of K 1 will be reduced and the term involving K 1 K 2 will be reduced slightly. Thus we can expect an increase in the barrier insertion loss. This effect is expected to be only 1 or 2dB but calculations as were done in part (a) are necessary to verify this. (c) Extending the barrier length will increase the barrier Insertion Loss because it will remove the contribution from waves diffracted around the barrier ends and will also reduce the reverberant field contribution because of a smaller gap area and also because of an increase in the effective room absorption. This effect should be slightly larger than the effect in part (b) above but would still be restricted to a few dB. The cost effectiveness of this action can only be assessed by completing the calculations as was done in part (a) of the problem and fully evaluating the benefit of the increased noise reduction vs the inconvenience of restricted passage. Problem 8.28 This is a pipe lagging problem so we follow the procedure on pages 404 and 405 in the text with the errata corrected in equations 8.116, 8.117 and 8.119. The thickness of the jacket is h = 6/2700 = 2.22mm and the diameter of the jacket, d = 0.2 + 2 × 0.1 + 0.0022 = 0.4022m. The quantity, 1000(m/d) 1/2 = 3862. The longitudinal wavespeed is: The ring frequency is: The critical frequency is obtained using the equation on p337 as: Solutions to problems 260 IL ' 40 1 % 0.12/ 0.2 log 10 f 6 × 0.1 132 ' 25log 10 ( f × 0.00587) For the Insertion Loss calculations, we use equation 8.112 on p404 for octave bands of 4000Hz and below and equation 8.115 for the 8000Hz octave band. The results of the calculations are summarised in the table below at octave band centre frequencies. Of course, if three 1/3 octave bands are averaged, then the result would be slightly different. In addition, equation 8.120 has also been used to generate an alternative set of Insertion Loss predictions. With equation 8.120, the Insertion Loss is given by: which is valid for frequencies defined by . f $ 120/ 6 × 0.1 ' 155Hz Octave band centre frequency (Hz) f/f r or f/f c C r or C c X r or X c Insertion Loss eqns. 8.112 - 8.119 Insertion Loss eqn.8.120 63 125 250 500 1000 2000 4000 8000 0.0145 0.0288 0.0577 0.115 0.231 0.461 0.923 1.503 0.1090 0.2163 0.4326 0.8653 1.7305 3.4610 6.9220 15.849 789.5 1307 1731 2120 2479 2717 1991 4289 -18.8 7.6 17.9 25.5 29.4 19.4 22.6 15.9 - - 4.2 11.7 19.2 26.7 34.3 41.8 The Insertion Loss results are somewhat different between the two methods of calculation and as we shall see in the next problem, neither prediction scheme is particularly good. However, it is supposed that the true results lie somewhere between the two. Problem 8.29 (a) The jacket cross section is shown schematically in the following figure. Sound transmission loss, acoustic enclosures and barriers 261 x × 10 &3 × 2700 % (1 & x) × 10 &3 × 11300 ' 6 m eff ' ρ 1 h 1 % ρ 2 h 2 ' 6.00kg/ m 2 y ' (h Pb % h Al / 2)E Al % E Pb h Pb / 2 E Al % E Pb ' (0.384 % 0.616/ 2) × 71.6 % 16.5 × 0.384/ 2 71.6 % 16.5 ' 0.598mm B eff ' E Al h Al 12 h 2 Al % (h Pb % h Al / 2 & y) 2 1 & ν 2 Al % E Pb h pb 12 h 2 Pb % (y & h Pb / 2) 2 1 & ν 2 Pb ' 71.6 × 0.616 12 0.616 2 % 12 × (0.384 % 0.308 & 0.598) 2 1 & 0.34 2 % 16.5 × 0.384 12 0.384 2 % 12 × (0.598 & 0.192) 2 1 & 0.44 2 ' 2.017 % 1.392 ' 3.41kgm 2 s &2 h Pb 1mm h Al Lead Aluminum y neutral axis Let x mm be the thickness of the aluminium part of the jacket. Then: Thus x = 0.616mm = thickness of Aluminium. Thickness of lead = 1 - 0.616 = 0.384mm Effective surface mass is given by: The neutral axis location is given by: The bending stiffness is given by: Solutions to problems 262 c L ' 12 h B eff m eff ' 12 0.001 × 3.409 6 ' 2611m/ s f c ' c 2 2π m B ' 343 2 2π 6 3.409 ' 24.84kHz f r ' c L πd ' 2611 π × 0.251 ' 3311Hz Assume that the effective Poisson's ratio is that of Aluminium and equal to 0.34. Thus the longitudinal wave speed is: The critical frequency is given by equation 8.3 in the text as: and the ring frequency is given on p404 in the text as: The jacket insertion loss may be calculated using either equation 8.120 or equations 8.112 to 8.119 in the text. We will use both methods here as a comparison. The quantities used in the equations are m = 6kg/m 2 , R = 0.05m, d = 0.15 + 2 × 0.05 = 0.25m and D = 0.15m. The results of the calculations at the 1/3 octave band centre frequencies are summarised in the following table. Note that the calculations using equation 8.120 are only valid for frequencies defined by . f $ 120/ 6 × 0.05 ' 220Hz Sound transmission loss, acoustic enclosures and barriers 263 Octave band centre frequency (Hz) X r or X m C r or C c Insertion Loss (dB) Eqs. 8.112 - 8.119 Eq. 8.120 63 80 100 125 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000 6300 8000 10000 1279 1500 1690 1869 2056 2220 2380 2543 2708 2859 3010 3157 3280 3378 3435 3403 3193 2258 2714 3356 4134 5038 5893 0.1137 0.1444 0.1805 0.2256 0.2888 0.3610 0.4512 0.5685 0.7219 0.9024 1.1370 1.4439 1.8048 2.2560 2.8877 3.6096 4.5120 5.6852 1.868 2.335 2.942 3.736 4.670 -2.4 4.1 8.5 12.1 15.7 18.6 21.3 23.9 26.3 28.4 30.2 31.5 31.7 30.1 21.1 24.9 31.4 24.1 30.0 29.5 20.7 30.4 37.0 - - - - - 0.35 2.6 4.9 7.0 9.3 11.6 13.7 15.9 18.3 20.4 22.6 24.8 27.1 29.3 31.5 33.8 36.0 The data in the table are plotted in the following figure where the solid black line represents the theory embodied in equations 8.112-8.119 and the dashed black line represents the theory embodied in equation 8.120. In addition, some experimental data (solid grey line) from the text book (figure 8.18, second edition) are shown for comparison. It can be seen that neither theory provides a very good prediction of the measured data. Unfortunately there are no better theories available at present. Solutions to problems 264 (b) It is clear from the equations used for either calculation method that increasing the mass of the liner will increase the low frequency Insertion Loss. For the first prediction scheme, reducing c L will also result in increased values for the low frequency Insertion Loss. (c) One advantage of porous acoustic foam over rockwool is that the porous foam will support the weight of the jacket indefinitely whereas rockwool will gradually compress and in a high vibration environment, it will turn to powder. A disadvantage of foam is that it is not fireproof and if ignited it emits toxic gas. Another disadvantage is that the foam is much more expensive than rockwool. R V L t B B N holes 9 Solutions to problems relating to muffling devices Problem 9.1 Referring to the figure, we can imagine that each hole in the perforated sheet represents a neck of a Helmholtz resonator with the volume associated with each neck being equal to the total volume behind the perforated sheet divided by the number of holes. Let L be the depth of the backing cavity and lets consider a section of sheet of dimensions B × B with a number of holes, N. The total backing volume is then LB 2 and the volume associated with one hole is LB 2 /N. The percent open area is given by: . P ' 100N B 2 πd 2 4 Thus the effective resonator volume is: , V ' 100L P πd 2 4 ' 100LA P where A is the neck cross-sectional area. Solutions to problems 266 f 0 ' c 2π A LV ' c 2π P 100LR 2R 0 ' 0.85d(1 & 0.22d/ a) p i ' Ae j ( ωt & kx) ; p r ' Be j ( ωt % kx % θ) p T ' Ae j ( ωt & kx) % Be j ( ωt % kx % θ) u T ' 1 ρc Ae j ( ωt & kx) & Be j ( ωt % kx % θ) Z ' p T u T ' ρc Ae &j kx % Be j ( kx % θ) Ae &jkx & Be j ( kx % θ) ' ρc A % Be j ( 2kx % θ) A & Be j ( 2kx % θ) ' ρc A/ B % e j ( 2kx % θ) A/ B & e j ( 2kx % θ) A B ' 10 8/ 20 % 1 10 8/ 20 & 1 ' 3.5119 1.5119 ' 2.323 From equation 9.38 in the text: where R = 2R 0 + t is the effective length of the neck and 2R 0 is the total effective end correction for the hole. Comparing the above equation with equation 7.77 in the text gives the effective end correction of the perforate as: The Helmholtz model is appropriate because the system is effectively a small mass of air vibrating against a stiffness represented by the backing volume. Problem 9.2 (a) Diameter, 0.4m, so higher order mode cut on frequency is . Thus at 200 Hz only plane waves propagate. f ' 0.586c/ 0.4 ' 500 Hz Minimum pressure occurs when θ ' &2kx % π ; x ' 1.8 m At 200 Hz, ; k ' 2πf / c ' 2π ×200/ 343 ' 3.664 = -10.048 c . θ ' &2 × 3.664 × 1.8 % π Adding 4π, gives θ = 2.518 c = 144.2E At x = 2, 2kx = 14.656 and 2kx + θ = 4.608 c = 264 degrees. Thus: Muffling devices 267 R 0 ' 0.61×0.05% 8 ×0.05 3π 1&(1.25×0.25) ×(1&0.639) 2 ' 0.0078 m Z ' 413 2.323 % cos(4.608) % jsin(4.608) 2.323 & cos(4.608) & jsin(4.608) ' 413 2.219 % j(&0.995) 2.427 & j(&0.995) Z ' 413 2.219 & j0.995 2.427 % j0.995 ' 413 (2.219 & j0.995) × (2.427 & j0.995) (2.427 % j0.995) × (2.427 & j0.995) ' 413 4.395 & j4.623 6.88 M ' Re{Z}/ ρc ' 263.8 413 ' 0.64 Multiplying numerator and denominator by complex conjugate of the denominator gives: Thus, Z = 264 - j278 (b) The hole impedance is in parallel with the rigid Re{Z A } ' Re {Z/ A orifice } plate containing it. If the rigid plate impedance is effectively infinity, the combined specific acoustic impedance is that of the hole and this is what is measured by the standing wave. The acoustic impedance of the hole is the specific acoustic impedance divided by the area of the hole only. If cross flow dominates resistance, M ' Re{Z A }A orifice / ρc Thus, (c) End correction = no flow correction × (1 - M) 2 . Hole radius = 0.05 m Thus total end correction is: Problem 9.3 A quarter wave tuning stub works by changing the wall impedance of a duct on which it is mounted so that downstream propagating waves are reflected back upstream. The viscous losses at the entrance to the stub also account for some of the energy loss, because of the large particle motions near the edges of the entrance. The stub can also act by changing the radiation impedance (and thus the amount of power radiated) of the source producing the noise. Solutions to problems 268 0.2 g f e d c b a Z 1 Z b Z d Z f Z a Z 1 Z 2 Z c Z 3 Z e Z g Problem 9.4 (a) The impedance looking into the stub must be zero as we are ignoring the resistive component. The equivalent electrical circuit is shown in the figure below: Impedance looking in at location 3 = Z 3 ' Z e (Z f % Z g ) Z e % Z f % Z g Impedance looking in at location 2 = Z 2 ' Z c (Z d % Z 3 ) Z c % Z d % Z 3 Impedance looking in at location 1 = Z 1 ' Z a (Z b % Z 2 ) Z a % Z b % Z 2 If we neglect resistive impedance, then Z 1 = 0 and Muffling devices 269 . If we now substitute the expression for Z 3 Z b ' &Z 2 ' & Z c (Z d % Z 3 ) Z c % Z d % Z 3 into the preceding equation, we obtain: (1) Z b Z c % Z d % Z e (Z f % Z g ) Z e % Z f % Z g % Z c Z d % Z e (Z f % Z g ) Z e % Z f % Z g ' 0 We now need to substitute in physical dimensions in place of the impedances. and the effective length of the hole Z b ' Z d ' Z f ' j ρωR A is given by: . R ' 2R 0 ' 8d 3π 1 & 1.25 d 0.2 The density, ρ = 1.206, A = πd 2 /4 and ω = 200π. The volumes between the orifices are given by: and V e ' V c ' πD 2 4 L 3 ' 0.01047L . V a ' V g ' 0.5V e ' 0.005236L Required length, L = 0.5 × λ/4 = c/8f = 343/800 = 0.429m. Using equation 9.35, we obtain: . Z c ' Z e ' &j 1.206 × 343 2 0.01047 × 0.4288 × 2 × π × 100 ' &j 5.03 × 10 4 Z a = Z g = 2Z c = -j1.01 × 10 5 . Equation (1) can now be solved by computer for d. The result is d = 44mm. (b) The device with baffles would have a much larger resistive impedance than the one without baffles because of all the cross-sectional changes at which there will be viscous losses. The larger resistive impedance will lower the quality factor and thus the peak attenuation, although the bandwidth of significant attenuation will increase. Solutions to problems 270 R s ' 414 0.00152 0.288 × 1.832 × 2.180 × 10 &4 log 10 4 × 0.00152 π × (2.180 × 10 &4 ) 2 ' 146MKS Rayls IL ' 20log 10 1 % Z d R s Problem 9.5 (a) To calculate the resistive impedance, equation 9.29 on p417 in the text may be used. First we will evaluate the variables used in the equation. ρc = 343 × 1.206 = 414, A = π × 0.044 2 /4 = 0.00152 m 2 , k = 2π × 100/343 = 1.832, t = h = , 2 × 1.8 × 10 &5 / ( 1.206 × 2π × 100) ' 2.180 × 10 &4 w = 0, and M = 0. Substituting these values into equation 9.29, ε ' 0 we obtain: (b) At the design frequency (100Hz), the Insertion Loss is found by substituting R s for Z s in equation 9.46 in the text to give: If the side branch is mounted an odd number of quarter wavelengths from the end of the duct, the reactive part of the impedance Z d will theoretically be infinite (based on equation 9.14) and the side branch Insertion Loss will also be infinite as indicated by the above equation. However, in practice, the quarter wave tube is of finite cross section and Z d cannot therefore be too large. This results in the Insertion Loss being finite and usually limited to 25 to 30dB. Problem 9.6 (a) Closed end side branches have less viscous losses than open ended tubes because of one less cross-sectional change and corresponding edge. Thus the quality factor and therefore the peak attenuation for open ended tubes will be smaller, although the bandwidth of significant attenuation will increase. (b) Equation 9.46 in the text may be used here. The downstream duct impedance for an infinite duct is simply ρc/A. Thus the ratio Z d /Z s in equation 9.46 is given by: Muffling devices 271 Z d Z s ' 1 j(ω & 10 4 / ω & jω × 10 &4 ) ω 2 ' 10 4 and ω ' 100rad/ s IL ' 20log 10 *1 % Z d Z s * ' 20log 10 *1 % 10 4 ω * ' 40dB end correction ' 0.61 × 0.03 % 8 × 0.03 3 π ( 1 & 1.25 × 0.3) ' 0.0342m ω 0 ' c A/ RV ' 343 2.83 × 10 &3 0.0742 × 6.28 × 10 &3 1/ 2 ' 844.6rad/ sec K M C When the Insertion Loss is at its peak, the side branch reactive impedance, Z s is zero. Thus: The Insertion Loss is then: Problem 9.7 The equation is simply a differential equation of the vibration of a mass acting against a spring. The quantity, M, represents the vibrating mass of air in the resonator neck, the quantity, C, represents the viscous damping losses at the entrance and exit of the neck caused by motion of the air particles relative to the edges of the neck and the quantity, K, represents the stiffness of the volume of air in the cavity of the resonator as shown in the figure. Problem 9.8 (a) Helmholtz resonator - cylindrical cavity, 100mm radius, 200mm high. Neck 40mm long with a radius of 30mm. Volume of cylinder, V = π × (0.1) 2 × 0.2 = 6.28 × 10 -3 m 3 Area of neck, A = π × (0.03) 2 = 2.83 × 10 -3 m 2 . Length of neck, R = 0.04m + end corrections. so R = 0.0742m. From equation 9.38 in the text: Solutions to problems 272 Q ' ρc R s R / AV ' 413.66 1000 0.0742 2.83 × 6.28 × 10 &6 1/ 2 ' 26.7 ¢ p 2 ¦ ' 4 × 10 &10 × 10 8 ' 0.04Pa 2 Thus, f 0 = 844.6/2π = 134Hz. (b) The quality factor may be calculated using equation 9.40 in the text. Thus: (c) If we define effectiveness as the frequency range within 3dB of the maximum, then the bandwidth of effectiveness is obtained using equation 7.22 in the text as . Thus we can expect Δf ' f / Q ' 135/ 26.7 ' 5.0Hz the device to be effective between 132 and 137Hz. (d) Incident plane wave of 80dB L p and 135Hz. Power dissipated = = at resonance. ¢ p 2 ¦ *Z s * ¢ p 2 ¦ R s The mean square sound pressure is: Thus the power dissipated = 0.04 1000 ' 40µWatts (e) Power in a plane wave = IA = ¢ p 2 ¦ A ρc ' 40 × 10 &6 Thus area of plane wave, A = 413.6 × 40 × 10 &6 / 0.04 ' 0.4136m 2 (f) The sabine absorption of the resonator, = 0.414m 2 S¯α Cross-sectional area of resonator = . Thus the cross- π × 0.1 2 ' 0.031m 2 sectional area of the resonator volume is 13 times smaller than its Sabine absorption area. Thus to make the wall look anechoic, we would need 1/0.414 = 2.4 resonators per square metre. (g) From equation 1.4 in the text, an ambient temperature variation from - 5EC to 45EC corresponds to a speed of sound change from to , which is a range 1.4 × 8.314 × 268/ 0.029 1.4 × 8.314 × 318/ 0.029 from 328 to 357m/s. From equation 9.38, it can be seen that the resonance frequency of the resonator will vary from 808 to 880 rad/sec which corresponds to a range from 129 to 140Hz which is outside the 3dB range discussed in part (c). The problem could be overcome by Muffling devices 273 p t ' p i % p r ' A i e j(ωt % kx) % A r e j(ωt & kx) u t ' u i % u r ' 1 ρc A i e j(ωt % kx) & A r e j(ωt & kx) ρc S A i % A r A i & A r ' Z L ' Z c A i % A r A i & A r p t Su t ' Z i ' Z c A i e jkL % A r e &jkL A i e jkL & A r e &jkL Z L L x 0 Z i using two additional types of resonator with resonance frequencies of 131.5 and 137.5Hz respectively at 20EC. Problem 9.9 (a) The total acoustic pressure anywhere along the tube is the sum of the incident and end r ef l ect ed pressures and may be written in terms of two complex constants, A i and A r , representing the complex amplitudes of the incident and reflected (from x = 0) waves respectively. Thus: Using equations 1.6, and 1.7, the acoustic particle velocity may be written as: At x = 0, . Thus: p t Su t ' Z L At x = L: Expanding the exponents gives: Solutions to problems 274 Z i ' Z c A i cos(kL) % jA i sin(kL) % A r cos(kL) & jA r sin(kL) A i cos(kL) % jA i sin(kL) & A r cos(kL) % jA r sin(kL) ' Z c A i % jA i tan(kL) % A r & jA r tan(kL) A i % jA i tan(kL) & A r % jA r tan(kL) ' Z c A i % A r % jtan(kL)(A i & A r ) A i & A r % jtan(kL)(A i % A r ) Z i ' A i % A r A i & A r % jtan(kL) 1 % A i % A r A i & A r jtan(kL) ' Z c Z L Z c % jtan(kL) 1 % Z L Z c jtan(kL) Z i ' Z c Z L % jZ c tan(kl) Z c % jZ L tan(kl) Rearranging gives: Rearranging gives: (b) (i) Z c ' ρc S ' 414 0.2 × 0.2 ' 10,400 MKS Rayls (ii) At x = 0, p r /p i = 0.5; thus A r /A i = 0.5 and A i % A r A i & A r ' 1 % 0.5 1 & 0.5 ' 3 Thus, Z L = 3Z c = 31,000 MKS Rayls (iii) Z i ' 10350 31050 % j 10350tan(200π × 10.29/ 343) 10350 % j 31050tan(200π × 10.29/ 343) ' 10350 31050 10350 ' 31000MKS Rayls Muffling devices 275 κ 2 n ' (ω/c) 2 & [(n x π/L x ) 2 % (n y π/ L y ) 2 ] (ω/c) 2 ' [ (n x π/L x ) 2 % (n y π/ L y ) 2 ] c n ' ω 2 ω c 2 & n x π L x 2 & n y π L y 2 1/ 2 ω 2,2 ' 343 4π L y 2 % 2π L y 2 1/ 2 ' 343 L y 16π 2 % 4π 2 1/ 2 ' 4819 L y c n c 0 f nx,ny f Problem 9.10 (a) The wavenumber is: (i) Cut-on when = 0. That is when: κ 2 (ii) Phase speed is defined as . Thus: c n ' ω/ κ n This equation is shown sketched in the figure below, where ω = 2πf. (b) Drop in L p over a distance of L y /4 for 2,2 mode. Cut-on frequency is given by: Excitation frequency = 0.75ω 2,2 = 3614/L y rad/sec. The wavenumber is then: Solutions to problems 276 κ 2 n ' 3614 343L y 2 & 16π 2 % 4π 2 L 2 y ΔL p ' 20log 10 1 e &jκ 2,2 L y / 4 ' 20log 10 1 e &9.294/ 4 ' 20.2dB W t ' 0.5Re{p 2 v ( 2 } ' 0.5Re{Z c v 2 v ( 2 } ' 0.5Re{Z c }*v 2 * 2 τ ' W t W i ' 0.5Re{Z c }*v 2 * 2 0.5Re{Z c }*v* 2 ' * v 2 v * 2 R % jX ' Z s Z c v Z c Z s v 1 Z c v 2 Thus, at the excitation frequency. The acoustic pressure κ 2,2 ' ±j 9.293 L y is related to the distance along the duct as . Thus p(x) % e &j κ n x . Setting x 1 = 0 and x 2 = L y /4, we can write the following p(x 1 ) p(x 2 ) ' e &jκ 2,2 x 1 e &jκ 2,2 x 2 for the reduction in sound pressure level as the wave travels from x 1 to x 2 . Problem 9.11 (a) The transmitted power is given by: A similar expression can be derived for the incident power. Thus the transmission coefficient is given by: The characteristic impedance is defined as Z c = ρc/A. Thus: The circuit equations are v ' v 1 % v 2 and v 1 Z s ' v 2 Z c from which: Muffling devices 277 v ' v 2 1 % Z c Z s and * v 2 v * 2 ' 1 *1 % 1 R s % jX s * 2 τ ' * v 2 v * 2 ' 1 *1 % 1 R s % jX s * 2 ' *R s % jX s * 2 *R s % jX s % 1* 2 ' R 2 s % X 2 s (R s % 1) 2 % X 2 s *R p * 2 ' 1 & R 2 s % X 2 s ( R s % 1) 2 % X 2 s ' (R s % 1) 2 % X 2 s & R 2 s & X 2 s (R s % 1) 2 % X 2 s *R p * 2 ' 2R s % 1 (R s % 1) 2 % X 2 s The transmission coefficient can then be written as: (b) The power reflection coefficient, , is related to the transmission *R p * 2 coefficient, τ, by . Thus: *R p * 2 ' 1 & τ Rearranging gives: Problem 9.12 (a) The Insertion Loss is calculated using equation 9.54 as it is a constant volume velocity source. Assuming that the damping term of Equation (7.33) is negligible, the resonance frequency of the muffler is defined by: . IL ' 10log 10 1 & 2π × 20 ω 0 2 2 ' 10 Thus ω 0 = 61.6rad/sec. To be a little conservative, use ω 0 = 60rad/sec. The required chamber volume is then found using equation 9.38 in the Solutions to problems 278 V ' 343 2 × 0.01767 0.3 × 65 2 ' 1.92m 3 0 PV ' 0 m M RT 0 m ' MP 0 V RT ' 0.029 × 101.4 × 10 3 × 2.894 8.314 × 288 ' 3.55kg/ s 0 V ' 0 mRT MP ' 3.55 × 8.314 × 623 0.029 × 12 × 10 6 ' 0.0529m 3 / s U ' 0.0529 × 4 π × 0.1 2 ' 6.735m/ s ρ ' 0 m 0 V ' 3.54 0.0529 ' 66.9kg/ m 3 text. The cross-sectional area of the inlet pipe is . Thus the required chamber volume is: A ' π × 0.15 2 / 4 ' 0.01767m 2 (b) The attenuating device could be made smaller by using a low pass filter as described on pp.433-438 in the text. Problem 9.13 Low pass acoustic filter (see figure 9.11 in text). Head loss 4 velocity . heads due to tube inlets and exits, so total pressure drop, Δp = 2ρU 2 . We now must calculate the flow speed, U. Assume that the choke tube diameter is the same as the inlet and exit pipes and equal to 0.1m. Flow rate at STP = 250,000m 3 /day = 2.894m 3 /sec. The mass flow rate can be calculated from the Universal Gas Law. Thus: and: At operating conditions, T = 623EK and P = 12 × 10 6 Pa. Thus: The velocity is thus: The gas density is: Thus the pressure drop, Δp is: Muffling devices 279 Δp ' 2 × 67.2 × 6.735 2 ' 6.1kPa c g ' γRT/ M ' (1.3 × 8.314 × 623/ 0.029) 1/ 2 ' 482m/ s f 0 ' c g 2π A c R c 1 V 1 % 1 V 2 1/ 2 ' 482 2π π × 0.1 2 4 × 1.8 1 % 1 1/ 2 ' 7.2Hz which is much less than 0.5% of 12MPa, so it is OK. Speed of sound in the gas is: Following the design procedure on pages 437 and 438 of the text, we have 1. f 0 = 0.6 × 10 = 6Hz 2. Try V 1 = V 2 = 1m 3 4. Choke tube diameter = 0.1m and length = 1.8m (assuming the chambers are cylinders 1m long and 0.64m diameter). 7. Resonance frequency is: which is close enough to 6Hz for now. Choke tube x-sectional area = π × 0.01/4 = 0.007854m 2 Equation 9.71 in the text may be used to calculate the Insertion Loss for the muffler. Substituting the values into this equation gives IL = 30dB which is too much. Thus try changing the volumes to 0.75m 3 and the choke tube length to 1.6m. This gives an Insertion Loss of 17.6dB which is too small. Try changing the volumes to 0.9m 3 and the choke tube length to 1.7m. This gives an Insertion Loss of 25.9dB which is too large. Try changing the choke tube length to 1.4m. This gives an Insertion Loss of 20.8dB which is OK. Thus, the final design is for 2 volumes, each of 0.9m 3 with a 0.1m diameter choke tube, 1.4m long connecting them. In practice, conservatism would usually dictate sticking with the 30dB design if it is practical. Solutions to problems 280 26 ' 343 2 × π π ×0.05 2 3 × 4 × R c (0.03 &1 % 0.03 &1 ) 1/ 2 compressor 1 2 4 3 L Problem 9.14 As the tubes are short, they may be treated as lumped elements and the design procedure outlined on pages 437 and 438 in the text may be used. The impedance of the three tubes is three times that for a single tube. The design frequency is 40Hz. Thus the required resonance is 0.65 × 40 = 26Hz. To simplify matters, use the largest allowable chamber volumes and smallest allowable choke tube diameter. Using equation 9.80, we obtain: Thus and R c = 192mm. 26R 1/ 2 c ' 11.403 The filter thus consists of 2 volumes, each 0.03m 3 , connected by three 0.05m diameter tubes which are approximately 0.2m long. Problem 9.15 Muffling devices 281 v ' v 2 % v 3 % v L v 2 Z 2 ' v 3 (Z 3 % Z 4 ) ' V L Z L v 3 ' v L Z L Z 3 % Z 4 v 2 ' v L Z L Z 2 v ' v L 1 % Z L Z 3 % Z 4 % Z L Z 2 IL ' 20log 10 *1 % Z L Z 3 % Z 4 % Z L Z 2 * Z L ' ρc A ' 413.6 × 4 π × 0.02 2 ' 1.317 × 10 6 Z 2 ' Z 4 ' &j ρc 2 Vω ' &j 1.206 × 343 2 0.2 × 2π × 10 ' &j 1.129 × 10 4 Z 1 Z 2 v 2 v 3 v L Z L Z 4 Z 3 v (a) The equivalent acoustical circuit is shown above. (b) The Insertion Loss of the muffler is given by equation 9.64 in the text. The circuit equations are: Thus the Insertion Loss is given by: (c) For no reflections from the pipe exit: The volume impedances are: Solutions to problems 282 Z 1 ' Z 3 ' j ρc A tan 2πL λ ' j 413.6 × 4 π × 0.02 2 tan 6π 343 ' j7.24 × 10 4 IL ' 20log 10 *1 % 1.317 × 10 6 j(7.24 × 10 4 & 1.129 × 10 4 ) % 1.317 × 10 6 &j1.129 × 10 4 * ' 20log 10 *1 & j21.54 % j116.7* . 39.6dB v ' v 1 % v 2 % v 3 % v 4 v 1 Z a ' (v 2 % v 3 % v 4 ) (Z b / 2) % (v 3 % v 4 ) (Z b / 2) % v 4 Z g v 2 (Z c % Z d ) ' (v 3 % v 4 ) (Z b / 2) % v 4 Z g v 3 Z e ' v 4 Z g c d b e a f g v Z f Z b /2 Z b /2 v 2 v 3 Z c Z d Z a v 1 Z e Z g v4 and the tube impedances are: The Insertion Loss is then: Problem 9.16 (a) Equivalent circuit diagram. (b) System equations: Muffling devices 283 Z i ' Z 3 % 1 (1/ Z 2 ) % (1/ Z 1 ) ' Z 3 % Z 1 Z 2 Z 1 % Z 2 ' Z 3 Z 1 % Z 3 Z 2 % Z 1 Z 2 Z 1 % Z 2 Z 3 ' 4jρc πd 2 3 tan(kL 3 ); Z 2 ' & 4jρc 2 πd 2 2 L 2 ω 0.5m 0.5m 0.5m 0.4m 10mm 10mm Z 1 Z 2 Z 3 Z i Z i Z 3 Z 2 Z 1 (c) Inductive (with resistive part) impedances are Z b , Z c , Z f and Z g . Capacitative impedances are Z a , Z d and Z e . Problem 9.17 (a) Impedance looking into the tube is: (b) Solutions to problems 284 Z 1 ' & 4jρc πd 2 1 cot(kL 1 ) Z i ' & (4jρc) 2 π 2 d 2 1 d 2 3 cot(kL 1 )tan(kL 3 ) & (4jρc) 2 c π 2 d 2 2 d 2 3 L 2 ω tan(kL 3 ) % (4jρc) 2 c π 2 d 2 1 d 2 2 L 2 ω cot(kL 1 ) & 4jρc π cot(kL 1 ) d 2 1 % c d 2 2 L 2 ω & cot(kL 1 ) tan(kL 3 ) d 2 1 d 2 3 & ctan(kL 3 ) d 2 2 d 2 3 L 2 ω % ccot(kL 1 ) d 2 1 d 2 2 L 2 ω ' 0 &ω × 8 × 10 &2 cot(ω/ 686) tan(ω/ 686) & 343 × 10 &4 tan(ω/ 686) % 343 × 10 &4 cot(ω/ 686) ' 0 &ω % 0.429(&tan(ω/ 686) % cot(ω/ 686)) ' 0 and: Thus: (c) Resonance occurs when inductive impedance = capacitative impedance, or Z i = 0. Eliminating (4jρc/π) 2 from previous expression for Z i and setting the result = 0, we obtain: Substituting for k = ω/343, L 1 = L 2 = L 3 = 0.5, d 1 = d 3 = 0.01 and d 2 = 0.4, we obtain: Simplifying gives: Solving by trial and error: Muffling devices 285 reciprocating compressor Z 1 Z 2 Z 3 Z L Z 4 Z 5 Z 4 Z 3 Z 1 Z 5 Z 2 Z L reciprocating compressor v v 1 v 2 v 3 ω Value of expression ω Value of expression 1 10 20 18 293 19.4 -5.3 -1.6 17 17.3 17.15 17.1586 0.3 -0.28 0.017 -0.000045 Thus the resonance frequency = 17.1586/2π = 2.7Hz. (d) At resonance, there will be a pressure maximum at the closed end and a maximum particle velocity at the open end. Problem 9.18 (a) (b) Constant volume velocity source. Thus . The Insertion v ' v 1 % v 2 % v 3 Loss is given by: Solutions to problems 286 IL ' 20log 10 * v v 3 * Z 2 v 1 ' Z 3 (v 2 % v 3 ) % v 2 (Z 4 % Z 5 ) v 1 ' Z 3 v 2 Z 2 % Z 3 v 3 Z 2 % Z 4 v 2 Z 2 % Z 5 v 2 Z 2 v 2 (Z 4 % Z 5 ) ' v 3 Z L ; thus v 2 ' v 3 Z L Z 4 % Z 5 v 1 ' v 3 Z 3 Z L Z 2 (Z 4 % Z 5 ) % Z 3 Z 2 % Z 4 Z L Z 2 (Z 4 % Z 5 ) % Z 5 Z L Z 2 (Z 4 % Z 5 ) IL ' 20log 10 * (v 1 % v 2 % v 3 ) v 3 * ' 20log 10 / 0 0 0 0 1 % Z L Z 4 % Z 5 % Z 3 Z L Z 2 (Z 4 % Z 5 ) % Z 3 Z 2 / 0 0 0 0 % Z 4 Z L Z 2 (Z 4 % Z 5 ) % Z 5 Z L Z 2 (Z 4 % Z 5 ) Examining pressure drops, we may write: Thus: Also: Substituting for v 2 in the expression for v 1 gives: The Insertion Loss is then: (c) The result of (b) is no longer valid if the dimensions of the chambers represented by impedances Z 2 and Z 5 exceed one quarter of a wavelength of sound. Muffling devices 287 1 Z s ' 1 Z v % 1 Z a Z v ' &j ρc 2 Vω and Z a ' j ρωR A 1 Z s ' j Vω ρc 2 & j πa 2 ρωR ' j Vω ρc 2 & j 3π 2 a 8ρω 2 & 2.5a π 4V 1/ 3 Z a Z v cross-sectional area, , of hole A Volume, V v Z v Z a Problem 9.19 (a) Load seen by speaker (ignoring external air load), Z s given by: where: and R is the effective length of the orifice. From equation 9.16, the end correction for the side of the hole in free space is 8a/3π. If we assume that the enclosure is cylindrical of diameter D, then the end correction for the side of the hole in the enclosure is and the total R 0 ' 8a 3π (1 & 2.5a/ D) effective length of the hole is then . If the cylinder R ' 8a 3π (2 & 2.5a/ D) diameter is equal to its length, then and D ' (4V/ π) 1/ 3 . R ' 8a 3π 2 & 2.5a π 4V 1/ 3 Thus: Solutions to problems 288 Z s ' &j 16ρωc 2 (1 & 1.153aV &1/ 3 ) 16Vω 2 (1 & 1.153aV &1/ 3 ) & 3π 2 ac 2 16Vω 2 (1 & 1.153aV &1/ 3 ) ' 3π 2 ac 2 ω 0 ' 3π 2 ac 2 16V(1 & 1.153aV &1/ 3 ) Thus: (b) At low frequencies, the second term in the denominator will be larger than the first and so the phase of the impedance will be +j which means that the acoustic pressure leads the acoustic particle velocity by 90E. As the particle displacement also leads the velocity by 90E, the particle displacement at the orifice and the acoustic pressure will be in phase. As the box is small compared to a wavelength, the acoustic pressure will be in phase with the displacement of the cone; thus the out flow from the orifice will be in the same direction as the cone motion. At higher frequencies, the second term in the denominator will become smaller than the first and the phase of the impedance will be -j, a 180E shift from the lower frequency case. Thus in this case the out flow from the orifice will be in the opposite direction to the cone motion and will thus reinforce the out flow at the cone as shown in the figure. The crossover frequency is thus when the two terms in the denominator are equal, which is when: or: (c) We need to solve the above equation for V, given that ω 0 = 2π × 100. Rearranging the above equation and substituting values for variables, we obtain: Muffling devices 289 V ' 3π 2 ac 2 16ω 2 0 (1 & 1.153aV &1/ 3 ) ' 3π 2 × (0.01/ π) 1/ 2 × 343 2 16 × 4 × π 2 × 10 4 (1 & 1.153 × (0.01/ π) 1/ 2 V &1/ 3 ) ' 0.0311 1 & 0.0651V &1/ 3 a b 1m 3 d 1.5m 3 f 2m 3 0.2m L 0.2m dia 0.1m long 0.2m dia 0.1m long 0.2m dia 10m long c e Equivalent acoustical circuit Z a Z c Z e Z L Z f Z d Z b p v b v d v f v L Solving by trial and error gives V = 0.039m 3 . Problem 9.20 Referring to the figures above and using equation 9.55 in the text for a constant pressure source, we may write the following Insertion Loss and acoustical circuit equations: Solutions to problems 290 IL ' 20log 10 * p v L Z L * (1) p ' (v b % v d % v f % v L )Z a % v b Z b (2) v b Z b ' v d Z d % Z c (v d % v f % v L ) (3) v d Z d ' v f Z f % Z e (v f % v L ) (4) v f Z f ' v L Z L (5) v d ' v f Z f Z d % Z e Z d % v L Z e Z d (6) v d ' v L Z L Z d % Z e Z L Z d Z f % Z e Z d (7) v b ' v d Z d Z b % Z c Z b % v f Z c Z b % v L Z c Z b (8) v b ' v L Z d Z b % Z c Z b Z L Z d % Z e Z L Z d Z f % Z e Z d % Z c Z L Z b Z f % Z c Z b (9) p Z L v L ' Z a % Z b Z L Z d Z b % Z c Z b Z L Z d % Z e Z L Z d Z f % Z e Z d % Z c Z L Z b Z f % Z c Z b % Z a Z L Z d % Z e Z L Z d Z f % Z e Z d % Z L Z f % 1 Using eq. (4), we can write: Using eqs. (5) and (6), we can write: Using eq. (3): Using eqs. (5), (7) and (8): Using eqs. (2), (5), (7) and (9): The Insertion Loss is then: Muffling devices 291 IL ' 20log 10 / 0 0 0 0 Z a % Z b Z L Z d Z b % Z c Z b Z L Z d % Z e Z L Z d Z f % Z e Z d % Z c Z L Z b Z f % Z c Z b / 0 0 0 0 % Z a Z L Z d % Z e Z L Z d Z f % Z e Z d % Z L Z f % 1 Z a ' jρc A a tan(k R a ) % R a ; Z b ' &j ρc 2 V b ω ; Z d ' &j ρc 2 V d ω Z c ' j ρc A c tan(k R c ) % R c ; Z e ' j ρc A e tan(k R e ) % R e ; Z L ' ρc A L R a ' ρc A a ω c tD a w a 2A a 1 % (γ & 1) 5 3γ % 0.288 ω c d log 10 4A a πh 2 a % ε ω 2 c 2 A a 2π % M t ' 2µ/ (ρω) ' 0.00550/ ω ; ( µ ' 1.8 × 10 &5 kgm &1 s &1 ) where the impedances are defined as: From equation 9.29, p.417: R c and R e are defined similarly to R a , except that subscripts c and e are substituted for subscript a respectively. ρc = 1.206 × 343 = 414; A e = A c = A a = π × 0.2 2 /4 = 0.00314m 2 . Pipe end corrections, R 0 = (8 × 0.1)(1 - 1.25 × 0.2/1)/(3 × π) = 0.064m h c = h e = largest of t or 0.05; h a = largest of t or tube inlet radius. w a = 10m, w e = w c = 0.1m, γ = 1.4, V b = 1, V d = 1.5, V f = 2m 3 Assume that M is small enough to neglect. ε a ' ε c ' ε e ' 0 Solutions to problems 292 1 2 Re Δ p Z 0 v ( 1 ' 1 2 Re Z 0 v 1 v ( 1 ' *v 1 * 2 2 Re Z 0 v 1 (Z 1 % Z 0 ) ' v 2 Z 2 ' Δ p v 1 ' Δ p Z 1 % Z 0 W 0 ' *v 1 * 2 2 Re6 Z 0 > ' *Δ p* 2 2 Re6 Z 0 > *Z 1 % Z 0 * 2 Ap Ap Z 1 Z 1 Z 2 Z 2 Z 0 Z 0 v 2 v 1 v 2 v 1 Problem 9.21 (a) Fan and plenum equivalent circuit diagram is shown in the figure below. Power flow through Z 0 is the radiated sound power and is equal to: v 1 is an rms quantity and is the pressure drop across Z 0 . Equating Δ p Z 0 circuit pressure drops gives: At low frequencies, Re{Z 0 } = 0, and so W 0 = 0 (b) For Z 2 , use the analysis leading up to equation 9.35 in the text. For Z 1 use the analysis leading to equation 9.14 for the imaginary part and equation 9.29 for the real part. The expressions are valid over the frequency range from zero up to where the neck diameter or plenum dimensions approach 0.2 of a wavelength. (c) The resistive component consists of viscous friction losses due to air particles vibrating back and forth against the edge of the inlet. (d) Varying the fan position along the duct will have the effect of adding an inductive impedance between the fan and plenum and will also change impedance, Z 1 . Thus the radiated sound power will vary. Muffling devices 293 c g ' γRT M ' 1.3 × 8.314 × (273 % 900) 0.03 1/ 2 ' 650m/ s 25 ' 10log 10 1 & ( ω/ ω 0 ) 2 2 ; Thus, ω 0 ' ω 18.78 ' 2π × 50 18.78 ' 72.5rad/ s 1 1.3 & 18 × 0.1 314.2V & 2 × 0.01 A 2 × 650 2 72.5 2 A 1.5 V 10 π × 8 × 10 &3 × 0.01 > 1 0.77 & 5.73 × 10 &3 V & 4.734 × 10 &8 A 2 3.707 × 10 6 A 1.5 V > 1 A min ' 3 × 4.734 × 10 &8 0.77 1/ 2 ' 4.3 × 10 &4 m 2 V min ' 3 × 5.73 × 10 &3 0.77 ' 0.022m 3 [ 0.769 & 0.143 & 0.051] 4.41 ' 2.53 , which is a bit large Problem 9.22 The speed of flow, U 0 = 0.1m/s. The speed of sound in the gas is: Following the design procedure on p.432 in the text, the desired resonance frequency is given by: Using equation 9.60 in the text, the design equation is: Rewriting gives: To begin, set each term in brackets = (0.77/3). Thus: However, this will not satisfy the minimum requirement for A 1.5 V. Thus increase A and V by a factor of 2. Thus try V = 0.04m 3 and A = 9.6 × 10 -4 m 2 (d = 35mm). Design equation becomes: Solutions to problems 294 [ 0.769 & 0.143 & 0.095] 2.78 ' 1.48 , which is OK R ' 7.06 × 10 &4 0.04 650 72.5 2 ' 1.42m 0.3m 0.6m Try reducing d to 30mm; A = 7.06 × 10 -4 and the design equation is The required tail pipe length, R, is: Design summary (to achieve 25dB at 50Hz) Volume = 40litres Tail pipe diameter = 30mm tail pipe length = 1.42m Could use a smaller tail pipe length and larger tail pipe diameter if a larger volume were chosen. Problem 9.23 Use maximum allowable OD = 0.6m. Inside diameter = 0.3m = 2h Liner thickness, R = 0.15m Thus R/h = 1.0 M = -17/343 = -0.05 At 500Hz, 2h λ ' 0.3 × 500 343 ' 0.437 From Figure 9.15 in the text, for M = 0, R/h = 1.0 and 2h/λ = 0.437, the figure with the highest attenuation is the top right figure corresponding to R 1 R/ρc = 2. For a square duct lined on 4 sides (equivalent to a circular duct), the attenuation would be 6.3dB per length of lined duct equal to the duct radius. From Figure 9.17, top right figure, the attenuation for M = -0.1, is7.1 dB per length of duct equal to the duct radius. As M = -0.05, use an attenuation rate of 0.5(6.3 +7.1) = 6.7dB per length of duct equal to the duct radius. From figure 9.22 in the text, (expansion ratio = (0.6/0.3) 2 = 4), the required liner attenuation for an overall attenuation of 15dB is 10.8dB. Thus the required length of liner = 0.15 × 10.8/6.7 = 0.24m Muffling devices 295 S λ ' π × 0.3 2 4 1/ 2 × 500 343 ' 0.39 450mm 300mm 75mm Additional noise reduction if direction of sound is distributed equally in all directions at the duct inlet (for example, if the duct were venting a machine enclosure) can be found from figure 9.21, p.459 in the text, where: From the figure, the additional attenuation is 7dB. Problem 9.24 The total attenuation is made up of ! entrance losses (figure 9.21, p.459) ! exit losses (table 9.5, p.464, numbers in brackets) ! liner attenuation (figure 9.15, p.449) Mach no., M = 0.0. Area of open duct = 75 × 300 × 10 -6 = 0.0225, σ = 0. To maximise the attenuation choose the curve corresponding to R/h = 4 and R 1 R/ρc = 4 in figure 9.15 which corresponds to M = 0.0. For the large dimension, 2h = 0.3 and duct length is equal to 3h, and for the small dimension, 2h = 0.075 and the duct length is equal to 12h. The following table may now be generated. Octave band centre frequency (Hz) 2h 1 / λ short 2h 2 / λ long attenuation (dB) - lining only short sides long sides all sides 500 1000 2000 4000 0.437 0.875 1.749 3.499 0.109 0.219 0.437 0.875 7.5 4.6 2.4 0.9 19.1 24.2 29.5 24.0 26.6 30.2 31.9 24.9 The total loss may now be calculated with the aid of the following table. Solutions to problems 296 Octave band centre frequency (Hz) S/ λ (S=0.3×0.075) Entrance loss (dB) Exit loss (dB) D ' 4A π ' 0.169 lining loss (dB) total loss (dB) 500 1000 2000 4000 0.219 0.437 0.875 1.749 3.1 7.3 9.6 10.0 4.2 1.6 0.6 0 26.6 30.2 31.9 24.9 33.9 39.1 42.1 34.9 (a) It is clearly possible to achieve 30dB or more in each of the octave bands, 1 2 & 4kHz. In fact a thinner liner would most likely be adequate. (b) The best attenuation possible at 500Hz is 34dB. Problem 9.25 Dissipative muffler - 3 attenuations: inlet, outlet and lined section (no expansion loss as it is mounted on an enclosure). Try beginning with the smallest allowed cross section of duct. Thus, S = 0.25 m 2 and and S/ λ ' f × S/ c ' 1.458 × 10 &3 f 2h/ λ ' 0.5f / c ' 1.458 × 10 &3 f frequency (Hz) inlet S λ ' 2h λ loss outlet loss total atten. needed Liner atten needed 125 0.18 2 5.5 9 1.5 1000 1.46 10 0 15 5 2000 2.91 10 0 15 5 If we use the largest outer cross section allowed, the ratio of liner thickness to half airway width is 1.0. Using curve 3 in Figure 9.16, assuming a flow speed of M=0.1, and using R 1 R/ρc = 8, we obtain the following attenuations for 0.25 m length of duct lined on all 4 sides. 125 Hz 1.2 dB 1000Hz 3.4 dB 2000 Hz 1.0 dB It is clear that the critical frequency is 2000 Hz and to satisfy the requirement of 5 dB at this frequency, we need a length of duct equal to 5 × 0.25/1.0 = Muffling devices 297 45 o 100mm 200mm 300mm a b a b 1.25 m. Note that many other solutions would be acceptable as well. Problem 9.26 (a) From the figure to the right, using similar triangles, it can be seen that the ratio, R/h is 100/200 = 0.5 (as one side only of the airway is lined). It may also be assumed that the flow speed is small enough to ignore. (b) For the ratio R/h = 0.5, an acceptable value of R 1 R/ρc is 2. Thus the required flow resistance of the liner is R 1 = 2 × 413.6/0.0707 = 11,700 MKS Rayls. (c) h = 2R = 200//2 = 141.4mm. Duct cross-sectional area = (0.2//2) × 0.4 = 0.0566m 2 . The effective duct length, L = 300/2 = 424mm = 3h. Wavelength, λ = 343/f. The exit loss is obtained from Table 9.5 in the text and the inlet loss is from figure 9.21 in the text (assuming diffuse field input). The lined duct loss is from figure 9.15 in the text (curve 2, top right figure). The results are summarised in the following table. Solutions to problems 298 r t Octave band centre frequency (Hz) 2h/λ lined duct loss (dB) length, 3h /S/λ Inlet Loss (dB) Exit Loss (dB) Total Atten. (dB) 63 125 250 500 1000 2000 4000 8000 0.052 0.103 0.206 0.412 0.825 1.65 3.30 6.60 0.1 0.6 2.1 6.6 8.2 3.0 0.9 0.3 0.044 0.087 0.17 0.35 0.69 1.39 2.78 5.55 0 0 2 6.1 8.8 10 10 10 12.3 8.2 4.2 1.2 0.1 0 0 0 14 9 9 14 18 13 11 10 Problem 9.27 (a) Insertion Loss is the difference in sound level at the end of the duct with and without the silencer in place. Transmission Loss is the difference in sound pressure level measured at the inlet and outlet of the silencer. (b) Power transmitted down duct with no muffler = 1. Power transmitted with muffler = τ. Thus IL = -10log 10 τ Power incident on muffler = 1. Power transmitted by muffler = τ Thus Transmission Loss, TL = -10log 10 τ = IL (c) Dissipative attenuators absorb energy and contain surfaces lined with sound absorbing material. They are cost effective for high frequency noise. Reactive mufflers change the radiation impedance "seen" by the sound Muffling devices 299 TL ' &10log 10 A R % Acosθ πr 2 R ' S¯ α (1 & ¯ α) ' 2(3.32 × 2.82 % 3.32 × 2.95 % 2.82 × 2.95) × 0.1/ 0.9 ' 6.106m 2 3.32m 2.82m source (tonal noise or system resonances) and reflect energy back to the source. They can also dissipate energy through viscous losses at entrances and exits of small air passage ways. Reactive mufflers are cost effective for low frequency broadband noise. They can also be tuned to attenuate tonal noise at one or more frequencies. Active mufflers act in a similar way to reactive mufflers but the impedance change, sound reflection or absorption is provided by a sound source such as a loudspeaker. These mufflers are cost effective for low frequency tonal noise problems and in many cases they are preferred to reactive mufflers because of their relatively small size and installation convenience. They are also preferred in dirty environments where reactive mufflers can become clogged. Problem 9.28 (a) Room 3.32m long, 2.82m wide and 2.95m high. Doors at each end are 2.06m high, 0.79m wide. Can treat it like a plenum chamber. and the sound power attenuation or transmission loss is: ¯α ' 0.1 A = 2.06 × 0.79 = 1.627m 2 , r = 3.32m Thus: Solutions to problems 300 TL ' &10log 10 1.627 6.106 % 1.627 π × 3.32 2 ' 5.0dB R ' 6.106 × 0.9 0.1 × 0.5 0.5 ' 54.95m 2 TL ' &10log 10 1.627 54.95 % 1.627 π × 3.32 2 ' 11.2dB TL ' &10log 10 1.627 6.106 ' 5.7dB TL ' &10log 10 1.627 54.95 ' 15.3dB W ' ¢ p 2 ¦ 4ρc A ' 4 × 10 &10 × 10 8.5 × 1.627 4 × 413.6 ' 1.244 × 10 &4 W (b) Increasing to 0.5 gives: ¯α Thus, That is, a 6dB improvement. (c) If the direct line of sight were prevented, then the direct field term contribution will be zero. Thus for case (a) above: and for case (b) above, That is, there is little difference in the first case where the reverberant field contribution dominates. (d) Sound power leaving the doorway of the first room is given by the sound intensity directed towards the door multiplied by the door opening area. Using equations 7.33 and 1.75 in the text, we obtain: The sound power level is: L w ' 10log 10 W% 120 ' 81.0dB re 10 &12 W Thus the sound power incident on the second doorway is: 81.0 - 5.0 = 76.0dB. Assuming no reflection from the doorway, this corresponds to a sound pressure level in the doorway given by equation 6.2 in the text as (where Muffling devices 301 L p ' L w & 10log 10 A % 0.15 ' 76.0 & 10log 10 (1.627) % 0.15 ' 74.0dB N s ' fd/ c ' 500 × 1/ 343 ' 1.458 L p ' L w & K %DI M & A E (dB re 20µPa) 100m 90 o R 2m 1 2 the correction for ρc … 400 has been included): Problem 9.29 At a distance of 50m, the angular orientation from the stack axis of the line joining the stack to the observer is approximately 90E. The directivity index may be obtained using figure 9.27 in the text. The Strouhal number is: Thus from figure 9.27 in the text, DI M = -9.6dB. The sound pressure level at the receiver is related to the sound power radiated by the stack using equation 5.158 in the text which may be written as: L w = 135dB K = 10log 10 (2πr 2 ) = 10log 10 (2π × 100 2 ) = 48dB (equation 5.161 in text) A E = A a + A g + A m + A b + A f (equation 5.165 in text) A a = 0.2 - 0.3dB (table 5.3, p. 225 in text) A g = -3dB (attenuation of ground reflected wave = attenuation of direct wave) A m = (+3, -1)dB (table 5.10, p.243 in text) A b = A f = 0 Solutions to problems 302 L p ' 135 & 48 & 9.6 & 0.2 % 3 & (%3, &1) ' 77 & 81dB re 20µPa Thus: If meteorological influences are ignored, L p = 80dB re 20µPa. 10 Solutions to problems in vibration isolation Problem 10.1 (a) The difference in levels between the plant room and apartment for the two cases of the plant operating and the test source operation are obtained by subtracting appropriate rows in the table given in the problem and are given in the table below Octave band centre frequency (Hz) 63 125 250 500 Equipment operating 25 36 38 40 Test source operating 34 37 38 39 Inspection of the above table indicates that in the 63Hz band, the problem is dominated by structure-borne noise whereas at higher frequencies, air- borne noise dominates. Thus improved vibration isolation will only help the low frequency noise problem and after allowing for the A-weighting correction of table 3.1, it can be seen that reducing the 63Hz problem will not significantly reduce the A-weighted noise level in the apartment. Problem 10.2 The single degree of freedom model gives inaccurate estimations of vibration transmission in the audio frequency range because it treats the spring and supported mass as lumped elements and does not include the effects of wave transmission along the spring. The wave transmission effects are negligible at sub-audio frequencies but are often important mechanisms of vibration transmission in the audio frequency range. Solutions to problems 304 T s ' m L 0 1 2 ρS y L 0 x 2 dy ' 1 2 ρS L 2 0 x 2 L 3 3 ' 1 2 ( ρSL) 0 x 2 3 ' 1 2 m s 0 x 2 T tot ' T s % T m ' 1 2 m 3 0 x 2 % 1 2 m 0 x 2 ' 1 2 m % m s 3 0 x 2 T max ' 1 2 m % m s 3 (ωA) 2 V ' 1 2 kx 2 max ' 1 2 kA 2 1 2 m % m s 3 (ωA) 2 ' 1 2 kA 2 m y dy L k, m s 0 x Problem 10.3 (a) Assuming that the displacement of the end of the spring attached to the mass is described by x(t). An intermediate point on the spring, a distance of y from the fixed base will have a displacement of (y/L)x(t). The total kinetic energy of the spring is: The total kinetic energy of the spring and mass is: If x(t) = Asin(ωt), then and: 0 x(t) ' ωAsin( ωt ) The maximum potential energy stored in the spring is: where k is the spring stiffness. Equating T max with V max gives: Vibration isolation 305 ω n ' k m % m s 3 f 0 ' ω n 2π ' 1 2π k m % m s / 3 E ' force/ area spring extension/ spring length ' kx/ S x/ L ' kL S c L ' E ρ ' kL Sρ ' f s λ s ' 4f s L f s ' 1 4 k SρL ' 0.25 k m s M 2 ξ Mx 2 ' 1 c 2 L M 2 ξ Mt 2 ξ ' Ae j (ωt & ( ω/ c L ) x) % Be j (ωt % ( ω/ c L ) x % θ) Thus: and the resonance frequency f 0 is thus: (b) Effective Young's modulus, E, is given by: Longitudinal wave speed is thus given by: Thus the surge frequency is given by: (c) The wave motion in the spring satisfies the one dimensional wave equation given by: where ξ is the spring longitudinal displacement as a function of axial location x. The solution for this equation is the same as for the acoustic case. That is: Solutions to problems 306 ξ ' A e j (ωt & ( ω/ c L ) x) & e j (ωt % ( ω/ c L ) x) m M 2 ξ Mx 2 ' &kL Mξ Mx &mω 2 e j (ωt & ( ω/ c L ) L) & e j (ωt % ( ω/ c L ) L) ' jkL ω c L e j (ωt & ( ω/ c L ) L) % e j (ωt % ( ω/ c L ) L) mω 2 e j ( ω/ c L ) L) & e &j ( ω/ c L ) L) ' jkL ω c L e &j ( ω/ c L ) L) % e j ( ω/ c L ) L) ωL c L tan ωL c L ' ρSL m ' m s m ' 1 N One boundary condition is that at x = 0, ξ = 0. Substituting this into the solution to the wave equation gives . Thus the wave equation Be jθ ' &A solution becomes: The second boundary condition is that the inertia force of the mass at x = L is equal to the spring force. That is: where k is the spring stiffness. Substituting the wave equation solution into this gives: which can be rewritten as: As shown in part (b), kL = c L 2 ρS. Using this relation and rearranging the above equation gives: The surge frequency is 2π × ω, where ω is the solution of the above transcendental equation. As stated in the problem, the upper frequency bound will be 0.9 × this value. Vibration isolation 307 k e ' k 1 k 2 k 1 % k 2 ω 1 ' k 1 k 2 (k 1 %k 2 )m ω 2 ' k 1 m ω 1 ω 2 ' k 2 k 1 % k 2 ' 1 1 % k 1 / k 2 m k 1 k 2 0 x M m M m M i M f M f f m f m f f f f f in f in without isolator with isolator Problem 10.4 As shown in the figure, let the frame stiffness be represented by k 2 and the isolator stiffness by k 1 . The effective stiffness, k e is then given by: Thus the resonance frequency is given by: With a rigid frame, the resonance frequency is given by: The ratio of the two (which is plotted in figure 10.7) is: Problem 10.5 (a) The equivalent mobility electrical circuits are shown in the two figures below. Solutions to problems 308 f m M m ' f f M f f in ' f m % f f ' f f 1 % M f M m ' f f M m %M f M m T 1 ' f f f in ' M m M m % M f f m ' f f M i % M f M m f in ' f m % f f ' f f 1 % M i % M f M m ' f f M m % M i % M f M m T 2 ' f f f in ' M m M m % M f % M i T F ' T 2 T 1 ' M m % M f M m % M f % M i (b) For the circuit without the isolator: and: The force transmissibility is the ratio of f f /f in , which is: For the circuit with the isolator: and: The force transmissibility is the ratio of f f /f in , which is: The force transmissibility with the isolator compared to that without the isolator is then T 2 /T 1 and is: which is the same as equation 10.31. Vibration isolation 309 Problem 10.6 2b = 0.7 m; machine width = 0.9m 2e = 1.0 m; machine depth =1.2m 2h = 0.2 m a = 0.5 - h = 0.4 m; machine height = 2×(0.4-h) = 0.6m vertical dimension of mass, 2d = 2(a - h) = 0.6m k = 4000 N/m Radius of gyration about vertical y-axis (eq. 10.18 in text): δ y ' [ (0.9/2) 2 % (1.2/2) 2 ] / 3 ' 0.4330 m Radius of gyration about horizontal x-axis (eq. 10.18 in text): m δ x ' [ (0.6/2) 2 % (1.2/2) 2 ] / 3 ' 0.3873 Radius of gyration about horizontal z-axis (eq. 10.18 in text): m δ z ' [ (0.6/2) 2 % (0.9/2) 2 ] / 3 ' 0.3122 f 0 ' 1 2π k m ' 1 2π 4000 × 4 50 ' 2.847Hz Now the rocking modes will be calculated, first in the x-y plane (about the z- axis) and then in the z-y plane (about the x-axis): W x ' (δ z / b) k x / k y ' 0.3122 0.35 1/ 4 ' 0.446 M x ' a/ δ z ' 0.4/0.3122 ' 1.281 From figure 10.5 in the text, Ω a = 0.38 and Ω b = 1.19. Thus: and f a ' 0.38 × 2.847 × 0.35/ 0.3122 ' 1.21Hz f b ' 1.19 × 2.847 × 0.35/ 0.3122 ' 3.80Hz W z ' (δ x / e) k z / k y ' 0.3873 0.5 1/ 4 ' 0.3873 M z ' a/ δ x ' 0.4/0.3873 ' 1.033 From figure 10.5 in the text, Ω a = 0.33 and Ω b = 1.09. Thus: Solutions to problems 310 T F ' M m % M f M m % M f % M i T F ' 0.1 % 0.2 0.1 % 0.2 % 1 ' 0.231 k 2 k 1 ' m 1 m 2 ( m 1 % m 2 ) 2 2πf 2 ' k 2 m 2 ζ 2 ' 3( m 2 / m 1 ) 8( 1 % m 2 / m 1 ) 3 and f c ' 0.33 × 2.847 × 0.5/ 0.3873 ' 1.21Hz f d ' 1.09 × 2.847 × 0.5/ 0.3873 ' 4.01Hz The resonance frequency of the rotational mode is calculated using equation 10.17 in the text and is: f y ' 1 π 0.35 2 × 4000 % 0.5 2 × 4000 50 × 0.433 2 ' 4.01Hz Problem 10.7 From equation 10.31, the increase in force transmission is: From the question, M f = 0.2M i = 2M m . Thus: which corresponds to a reduction in force transmission by a factor of 4.3. Problem 10.8 Referring to figure 10.11 in the text and the discussion on pages 496, the optimal absorber is characterised by: The excitation frequency is 3000/60 = 50Hz and this should be equal to the Vibration isolation 311 9.87 × 10 4 × m 2 10 7 ' 1000m 2 (1000 % m 2 ) 2 k 2 ' (100π) 2 × m 2 ' 19.7MN/ m ζ ' 3 × 200/ 1000 8( 1 % 200/ 1000) 3 ' 0.21 resonance frequency of the absorber mass and spring system. Thus or (100π) 2 ' k 2 / m 2 k 2 ' 9.87 × 10 4 × m 2 Using the first equation and substituting the above for k 2 , and the given values for k 1 and m 1 , we obtain: which results in a negative mass m 2 . The text indicates that the damping mass should be as large as possible; thus it seems impractical to try to satisfy the second of the above three design equations. Thus let the design mass m 2 =20% of m 1 = 200kg. Thus the required stiffness is given by: The required damping may be obtained from equation 10.47 as: Problem 10.9 Damping a vibrating surface will only reduce the resonant response. Thus damping will only result in a reduction in sound radiation if the resonant modes are contribution most to the radiated sound field. This is generally the case when the surface is excited mechanically. However, if the surface is excited by an acoustic wave on the side opposite that which is causing the radiation problem, then it is likely that the sound radiation will be dominated by vibration modes which are being forced to vibrate at frequencies well above their resonances. In this case damping the surface will not reduce the sound radiation directly but may reduce it a little because adding mass to the vibrating structure will decrease its mobility for excitation by the incoming sound field. See pages 504-506 in the text. Problem 10.10 (a) 100dB below 1 volt corresponds to a voltage of 1 × 10 -100/20 Volts = Solutions to problems 312 u ' p ρc ' 2 × 10 &5 × 10 80/ 20 413.6 ' 4.836 × 10 &4 m/ s 0 u rms ' 2πf u ' 2π × 1000 × 4.836 × 10 &4 ' 3.04m/ s 2 d rms ' u 2πf ' 4.836 × 10 &4 2π × 1000 ' 0.078µm 10µVolts. Accelerometer mass, m (grams), is approximately equal to sensitivity in mV/g. Smallest detectable acceleration (in g) is the smallest detected voltage in milli-volts divided by m. Thus smallest detectable acceleration is 10 -2 /m "g" which in metres/sec is 0.01 divided by the accelerometer weight in grams. (b) The required relation can be determined by considering the mass loading effect of the accelerometer. To obtain results within 3dB of the correct level, the accelerometer mass must satisfy the requirement that . m < 3.7 × 10 &4 (ρc L h 2 / f ) For steel, ρ = 7800 and c L = 5150/0.954. Thus the condition must be satisfied. mf u < 15,580 h 2 (c) From part (a), if the smallest acceleration to be detected is 0.01g, then the lightest accelerometer which can be used is 1 gram. Substituting m = 1 and h = 1 in the relation of (b) above gives f u = 15.58kHz. Problem 10.11 (a) For a large plate vibrating as a piston, sound will be radiated as plane waves with no near field. Thus at any point the pressure and acoustic particle velocity are related by p = ρcu. As the acoustic particle velocity adjacent to the plate is equal to the plate velocity, the velocity of the plate is given by: The r.m.s. acceleration is thus given by: (b) The r.m.s. displacement is given by: Vibration isolation 313 (c) At high frequencies accelerations are generally large compared to displacements, whereas the opposite is true at very low frequencies. An accelerometer is thus the best means of measuring the acceleration at 1kHz, provided that the accelerometer did not significantly mass load the plate (see equation 10.53 in the text). For thin plates at high frequencies, a measurement of the sound pressure close to the plate may be the best way of determining the plate response. Problem 10.12 (a) Adding damping will reduce the sound radiated by a vibrating surface if the surface vibration modes which are excited are resonant as is usually the case if the structure or surface is excited mechanically (but not acoustically). A full discussion of this concept may be found on pages 504-506 of the text. (b) Adding stiffness to a vibrating surface will only decrease its sound radiation or increase its transmission loss at frequencies below the first resonance frequency of the surface. If the surface is excited at resonance by a tonal excitation source, then adding stiffness will increase its resonance frequencies and if the carefully done will result in a reduction in radiated sound. (c) Adding mass to a vibrating surface will reduce the sound radiation and increase the panel transmission loss at frequencies above the first resonance frequency of the surface and below the surface critical frequency. Problem 10.13 Vibration velocities measured in octave bands on a diesel engine are listed in the following table. Solutions to problems 314 Octave band centre frequency (Hz) 63 125 250 500 1k rms vibration velocity (mm/s) 5 10 5 2 0.5 rms acceleration estimate (v2πf) (m/s 2 ) 1.98 7.85 7.85 6.28 3.14 rms displacement estimate (v/2πf) (µm) 12.6 12.7 3.18 0.64 0.08 (a) Overall rms velocity = (5 2 + 10 2 + 5 2 + 2 2 + 0.5 2 ) 1/2 = 12.4 mm/s (b) Overall velocity in dB re 10 -6 mm/s = 20 log 10 (12.4/10 -6 ) = 142dB re 10 -6 mm/s (c) Estimate of the overall acceleration level in dB re 10 -6 m/s 2 = 20 log 10 (13.29/10 -6 ) = 142dB re 10 -6 m/s 2 (d) Estimate of the overall displacement level in dB re 10 -6 µm = 20 log 10 (18.22/10 -6 ) = 145dB re 10 -6 µm Problem 10.14 (a) I would mount the accelerometer so that its axis was normal to the beam surface and thus parallel to the beam displacement. (b) Longitudinal waves would cause the accelerometer to vibrate normal to its axis and the cross-axis sensitivity of the accelerometer will result in a signal due to the longitudinal wave. For a properly selected and aligned accelerometer the effect can be very small but for any other accelerometer the effect could be significant depending on the relative levels of the bending and longitudinal displacements and the actual cross axis sensitivity of the accelerometer. Problem 10.15 (a) Using the relation, , we obtain f 0 = 11.1Hz f 0 ' 1 2π g d Vibration isolation 315 T F ' 1 % (2×0.05×4.49) 2 (1 & 4.49 2 ) 2 % (2 × 0.05 × 4.49) 2 ' 0.057 M i ' j 2 π × 50 4.905e%6 ' j6.405 × 10 &5 m/s/N T F ' / 0 0 0 / 0 0 0 &3.183 × 10 &6 & 2 × 10 &5 &3.183 × 10 &6 & 2 × 10 &5 % 6.405 × 10 &5 ' 0.567 4πζ 1 & ζ 2 ' 0.5 (b) We may use equation (10.14) in the text. The value of X is 3000/(60×11.1) = 4.49. Thus: Thus the reduction in transmitted vertical force is -20 log 10 (0.057) = 25dB (c) We may use equation (10.31) in the text. First we must calculate the isolator mobility. The overall spring stiffness is found by setting equations (10.2) and (10.3) in the text, equal. The result is k = 1000×9.81/0.002 = 4.905MN/m. The isolator mobility is then calculated using to give: M i ' jω k i The mobility of the supported mass is calculated using M m = 1/j2πfm = -j3.183 × 10 -6 m/s/N. Using equation (10.31), we obtain: Thus the reduction in transmitted vertical force is now -20 log 10 (0.567) = 5dB; thus there is an increase of 20dB. Problem 10.16 The critical damping ratio is given by: Solutions to problems 316 ζ ' 0.25 16π 2 % 0.25 ' 0.04 Thus: and η = 0.02. This loss factor is about 10 to 20 times greater than would be expected from a sheet of steel, so one might conclude that the product would be effective. One application would be for lining of parts bins. 11 Solutions to problems in active noise control Problem 11.1 Acoustic mechanisms associated with active noise control include: 1. Suppression of the primary source by changing its input impedance with a control source 2. Reflection of energy as a result of causing an impedance mismatch at the control source 3. Absorption of energy by the control source 4. Local cancellation at the expense of increased levels elsewhere Applications: (a) Feasible, reference sensor would be tacho signal, control source should be downstream of primary source and remote from turbulence generating parts of the duct system, and error sensor should be downstream of control source (out of source near field). Mechanism involved is suppression of primary source by changing its radiation impedance. (b) Not feasible if the source of noise is the grille. This is because it would be difficult to obtain a causal reference signal. If the source of noise is upstream of the grille, then active control may be feasible. A reference signal could be obtained from a microphone (with a turbulence filter) placed upstream. The control source would be placed at least 1.2m downstream of the reference sensor (depending on the controller time delay) and the error sensor would be placed 0.5 to 1m downstream of the control source, and it may even work better if placed on the room side of the grille. Mechanism would be reflection and absorption of primary energy. (c) Not feasible if global control is needed. Possible to establish zones of local noise reduction. Active noise control 318 (d) Not feasible due to difficulty in obtaining a causal reference signal. (e) Feasible, reference sensor would be tacho signal on rotating shaft related to noise producing machine, control sources should be in factory corner if room is small, otherwise they should be near the locations where noise reduction is needed. Mechanism involved is suppression of primary source by changing its radiation impedance for small room and local cancellation for large room. (f) Feasible, reference sensor would be tacho signal from aircraft engine, control sources would be placed in cabin ceiling or in seat headrests, and error sensors should be as close as possible to the passengers. Mechanism involved is suppression of primary source by changing its radiation impedance, although local cancellation may dominate in some cases (g) Not generally feasible due to complexity of radiated sound field. (h) Feasible. Feedback system is needed. Control sources could be actuators on the fuselage skin or loudspeakers in the passenger headrests. Microphones would be best located in passenger head rests. (i) Feasible. Reference signal would be derived from electricity mains signal, control sources could be shakers on the tank or loudspeakers surrounding the transformer and very close to it. Error sensors would need to surround the transformer and be located further away than the control sources. Mechanism is suppression of primary noise by changing the radiation impedance of the transformer tank. φ ' f (k( ct ± r)) r 12 Errata in the 3rd edition of Engineering Noise Control p xi, Change “Noise Reduction Index (NRI)” to “Noise Reduction Coefficient (NRC)” p xv, change “FWHA” to “FHWA” p xviii In line 19, change “Noise Reduction Index” to “Noise Reduction Coefficient” p16, In line 3, change the equation to (1/ hf) E/ ρ > 2 p16, line 10, change D P = 1.346E to D P = 1.099E p16, Change Eq. (1.3) to p18, In Eq. (1.5), change "332" to "331" p27, Change Eq. 1.40a to p29, 3 lines above Eq, (1.50), change "1.36" to "1.41" p34, Change the reference just above Eq. (1.69) to "Fahy, 1995" p35, First line under Eq. (1.67), change "1.65" to "1.64" p41, 4 lines above Section 1.10., replace "pet" with "per" p45, 2 lines under Eq. (1.89) and in Eq. (1.90), remove the subscript, " t " from p t . D C ' D F 1 % D F E W 2R t % ρ w ρ ν 2 Errata for third edition text book 320 DND ' 2 (L ) Aeq,8h & 90) /L T a ' 8×2 &(91.2 & 90.0) / 3 ' 8/ 2 0.39 ' 6.1 hours p51, Table 1.3, line 3, replace "U " with "u" p51, Table 1.3, line 5, replace "Z d " with "Z A " p51, Heading 1.12.2, replace "Z " with "Z s " p72, line immediately below the figure, add "is the" after the word, "ordinate" p76, Line 13, change "sound" to "sounds" p87, 2 lines above Example 2.1, the text should read, "Figure 2.10(b) is an alternative representation of Figure 2.10(a)" p111, line 4, change "1252" to "61252". p134, 3 rd line, replace H with H ) p142, The number "3" and "0.3" should be replaced by "3.01" and "0.301" respectively in Equations (4.37) to (4.41) inclusive p143, Replace equation 4.43 and the 2 lines preceding it with: The daily noise dose (DND), or "noise exposure", is defined as equal to 8 hours divided by the allowed exposure time, T a with L B set equal to 90. That is: p143, Replace the sentence following equation (4.42) with: "If the number of hours of exposure is different to 8, then to find the actual allowed exposure time to the given noise environment, the "8" in Equation (4.42) is replaced by the actual number of hours of exposure." p144, 3 rd equation down should be: p147, Replace Figure 4.6 with the more accurate figure below. Errata for third edition text book 321 impact and steady state (equal energy) 5 dB / doubling steady state impulse 180 170 160 150 140 130 120 110 100 90 0.1 1 10 10 10 10 10 10 10 10 10 2 3 4 5 6 7 8 9 2 5 5 5 5 5 5 5 5 8-hour dB(A) equivalent B-duration x number of impulses (ms) P e a k s o u n d p r e s s u r e l e v e l ( d B r e 2 0 P a ) µ p147, 4 lines under Figure 4.6, change "1414" to "1474". p149, 5 th and 6 th lines from the top, change "645" to "60645" in four places. p150, 13 lines from the bottom, change Figure 4.6 to Figure 4.7. p153, First line after the headings in Table 4.6, change "0.06" to "0.6". p157, Fig 4.9 caption, add "MAF" = minimum audible field. p160, On y-axis, change label from "dB re 20 mPa" to "dB re 20 µPa" p165, First paragraph in section 4.9, replace "1995" with "1995, 1999". p176, Line above Eq. (5.6), change "r" to "r = a" p176, In Eq (5.6), change "r" to "a" p177, In Eq. (5.7), change "r" to "a" p179, 2 lines under figure 5.2, replace "(x,y)" with "O" and label the observer as O in Figure 5.2 p192, 2 lines above Eq. (5.71), add "each of which has a radius of a i " immediately after "sources" Errata for third edition text book 322 p192, 2 lines above Eq. (5.72), change "a" to "a i " p192, Line above Eq. (5.72), change "ka" to "ka i " p192, In Eq. (5.72), change "a" to "a i " in 5 places p192 Eq. 5.71 and below, change Q to in 4 places ¯ Q p192 last line add "amplitude" immediately after "velocity" p193 Eq 5.73 and below change Q to in 2 places ¯ Q p225, In Table 5.3 caption, change "Sutherland et al., 1974" to "Sutherland and Bass, 1979" p226, 13 lines above Eq. (5.171), change "2613" to "9613". p226, Paragraph beginning "Note that ISO" only applies to overall A- Weighted calculations and should be deleted here. The paragraph following this one should also be deleted as the meteorological effects should not be taken into account in two separate places - either they should be included in the barrier calculations or calculated separately but not both. p229, Interchange the 63 Hz and 2000 Hz labels on the curves in Fig. 5.19. p232, Eq. 5.181, change "-0.09" to "-0.9" p236, In Eq. (5.188) change "10.3" to "10.0" p241, Table 5.9, -3.0<<<+0.5 should be replaced with -3.0<<<-0.5 p244, ISO 9613-2 procedures for calculating ground effects and shielding effects are based on an assumption of downwind propagation from the sound source to the receiver. Thus the only correction term (Equation (5.193)) that is offered by ISO for meteorological effects is a term to reduce the A-weighted calculated sound pressure level for long time averages of several months to a year. Thus section 5.11.12.4.should be deleted and replaced with the paragraph above. p251, In Figure 6.1, in the centre on the right hand side replace γ ' 1/ κ with γ ' κ Errata for third edition text book 323 p253, 2 lines above section 6.6, change "1989" to "1995". p259, The equation numbered "6.12" should be numbered "6.11" p264, The equation numbered "6.25" should be numbered "6.24" p264, 2 lines below Eq. 6.20, replace S 1 with 1/S 1 p264, 3 lines below Eq. 6.20, replace S 2 with 1/S 2 p267, The first equation should be numbered "6.26" p267, In Fig 6.3, there are two curves labelled "4". The lower curve should be labelled "5" p292, 3 lines above Eq. 7.52, change to and add "at time t=0" ¢ p 2 k (t) ¦ ¢ p 2 k (0) ¦ after "mode k" p292, 2 lines above Eq. 7.52, change to ¢ p 2 k (t) ¦ ¢ p 2 k (0) ¦ p292, In Eq. 7.52, change to ¢ p 2 k (t) ¦ ¢ p 2 k (0) ¦ p293, 3 lines above Eq. 7.55, change p k to p k (0) p293 6 lines from the bottom, there should be a minus sign before log e p294, 5 lines from the bottom, change (2000) to (2001) p294, Eq. (7.59), replace 0.16V S with 0.16V S 2 p295, Eq. (7.64), multiply each of the three terms in brackets by -1 p295, 2 lines beneath Eq. (7.62), add "energy" before "reflection" p295, 2 lines above Equation (7.64), change "2001" to "2000" p296, lines 2 and 3, change "S x , S x and S x " to , "S x , S y and S z " p301, In each of the top two lines of the table, add "(m 2 )" after Sα p303, Section 7.7.2, change "NRI" to "NRC" in three places and change "Noise Reduction Index" to "Noise Reduction Coefficient" in two places. Also change Eq. 7.76 to: Errata for third edition text book 324 NRC ' ¯ α 250 % ¯ α 500 % ¯ α 1000 % ¯ α 2000 4 (7.76) p303, 2 lines from bottom, change "20 mm" to "20 µm" p304, Caption of Figure 7.6, line 1, change "porous surface" to "rigidly backed porous material" and in the last line, change "L" to R p310, Immediately following Equation (7.88), add the following: "Note that for square, clamped-edge panels, the fundamental resonance frequency is 1.83 times that calculated using Equation (8.21). For panels with aspect ratios of 1.5, 2, 3, 6, 8 and 10 the factors are 1.89, 1.99, 2.11, 2.23, 2.25 and 2.26 respectively." p310, Equation 7.85 should be: ξ c ' f f c 1/ 2 p311, End of second full paragraph, change "Elbert" to "Elfert" p329, Eq. (7.122), replace T 60u with 1 T 60u p330, 10 th line, change "2000" to "2001" p339, 12 th line from the bottom, change "1973" to "1988" p343, 5 lines above the figure, change "ASTM E90-66T" to "ASTM E413- 87" p347, replace the line immediately above section 8.2.4 and the last word in the line above that with "contour value at 2000 Hz is increased by 1 dB." and add "Note that IIC, R w and STC values are all reported as integers." p352, 3 lines under Equation (8.36), change "below" to "above". p353, change x-axis label to f (Hz) (log scale)" p354, 2 nd and 3 rd lines from the bottom, replace "8.37" with "8.38" p355, 2 nd line after Eq. 8.44, replace f c2 /2 with f c1 /2 Errata for third edition text book 325 20 % 20log 10 (2500/100) & 6 ' 42.0 dB h ' 1 & f f c1 2 2 1 & f f c2 2 2 D ' 2 h if f < 0.9 × f c1 πf c1 8f η 1 η 2 f c2 f if f > 0.9 × f c1 p355, 3 rd line, replace "8.37" with "8.38" p359, In Eq. 8.50, replace 10 log 10 m 1 with 20 log 10 m 1 p360, change x-axis label to "frequency (Hz) (log scale)" p360, on the x-axis of the figure, change "0.5 f c2 " to "0.5 f c1 " p360, first line of item (b) in the caption, change to "LineBpoint support ( f c2 is the critical frequency of the point supported panel)" p360, Under "Point B", item (a), replace "30log 10 f c2 " with "20log 10 f c1 + 10log 10 f c2 " p360, Under "Point B", items (b) and (c), replace "40log 10 f c2 " with "20log 10 f c1 + 20log 10 f c2 " p360, Eq (a) under "Point C", add the term, "20 log 10 (f c2 / f c1 )" to the RHS of the equation p360, last Eqn., change f l to f R p361, replace Eq. 8.55 with: p363, 6 lines from the bottom of the page, change the equation to: p363, 4 lines from the bottom of the page, change "77" to "78" and "61" to "60" in 2 places p363, last line, change "61" to "60" and "52" to "51" Errata for third edition text book 326 p365, Section 8.2.6.2, 5 lines down, replace the sentence beginning with "Alternatively" with the following: "This mechanism can be considered to approximately double the loss factor of the base panels. Alternatively, the panels could be connected together with a layer of visco-elastic material to give a loss factor of about 0.2." p365, Section 8.2.6.2, 9 lines down, after the words "(0.3 to 0.6 m)", add the words, "or connected with a layer of visco-elastic material or even nailed together". p371, In the 500 Hz column, 7 th number from the bottom, replace S1" with "51" p379, 2 lines above "Example 8.4", change "Example 8.7" to "Example 8.8" p380, replace the example table with the following table. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 TL from Table 8.2 30 36 37 40 46 54 57 59 from Table 7.1 0.013 0.013 0.015 0.02 0.03 0.04 0.05 0.06 α w from Table 7.1 0.01 0.01 0.01 0.01 0.02 0.02 0.02 0.03 α f S i (m) 0.463 0.463 0.525 0.68 1.05 1.36 1.67 2.04 α 67 67 59 45.6 29.5 22.8 18.6 15.2 S E / S i α i 10log 10 ( ) 18 18 18 17 15 14 13 12 S E / S i α i NR (dB) 12 18 19 23 31 40 44 47 p381, 3 rd line down, Equation (8.75) should be (8.65), 6 lines down Equation (8.76) should be (8.66) and 8 lines down, Equation (8.6), should be (8.65). p381, 4 th Eq. in section 3, "30.5/30" should be "30.5/31" p391, At the end of the paragraph above the figure, add the following sentences. "When paths involving the ground reflected wave on the source side are considered, the straight line distance, d, used in Equation (8.85) is the distance between the image source and the receiver. The same reasoning applies to paths involving ground reflections on the receiver side." p394, 3 lines following Eq. 8.98, replace "barier" with "barrier". p395, replace the four equations for A b with the following in the same order Errata for third edition text book 327 A b ' 15.8 % 20log 10 [5.8/4.5] ' 18.0 dB; A R ' 1.3 dB; A b % A R ' 19.3 dB A b ' 19.8 % 20log 10 [7.2/ 4] ' 24.9 dB; A R ' 2.6 dB; A b % A R ' 27.5 dB A b ' 19.5 % 20log 10 [7.5/4.5] ' 23.9 dB; A R ' 5 dB; A b % A R ' 28.9 dB A b ' 12.0 % 20log 10 [4.5/4] ' 13.0 dB A b ' 19.8 % 20log 10 [7.2/4] ' 24.9 dB A b % A R ' 19.3 dB p395, 6 lines from the bottom, replace "4.6" with "4.7" p395, Solution, item 1, last line, change "5.18" to "5.20". p396, replace the two equations for A b with the following in the same order. p396, Item 3, lines 2 and 3, change the numbers to 19.3 dB, 19.3 dB, 27.5 dB, 28.9 dB, 28.9 dB, 13 dB, 24.9 dB and 24.9 dB p396, Item 3, line 4, change "5.18" to "5.20". p396, Item 3, line 4, change "10 dB" to "12 dB" p399, Figure 8.19, replace r with R Errata for third edition text book 328 R ) s ' R θ cosα h ) s ' H b & R θ sinα α ' 1 2 (π & θ) & β β ' cos &1 (H b / A) θ ' ±cos &1 [1 & (A 2 / 2R 2 ) ], *R* > A/ 2 N ' ± 2 λ X 2 S % (h b & Z S ) 2 1/ 2 % X 2 R % (h b & Z R ) 2 1/ 2 % b 2 % Y 2 1/ 2 & d Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 0 10 20 30 40 50 O c t a v e b a n d i n s e r t i o n l o s s Figure 8.21 Typical pipe lagging insertion loss for 50 mm glass- fibre, density 70-90 kg/m 3 , covered with a lead / aluminium jacket of surface density, 6 kg/m 2 . The I symbols represent variations in measured values for three pipe diameters (75 mm, 150 mm and 360 mm). p399, Replace Eq. (8.100) with: p400, 1 st paragraph, change "Figure 8.12" to "Figure 8.14" p401, Eq. (8.107) should be: p404, Figure 8.21 is missing (see following figure) Errata for third edition text book 329 C c ' 0.232ξ c R / h (8.117) X c ' [41.6( m/ h) 1/2 ξ c ( 1 & 1/ ξ c ) &1/4 ] & [258h/ R ξ c )] (8.116) X m ' [ 226(m/ h) 1/ 2 ξ c (1 & ξ 2 c ) ] & [258h/ (Rξ c )] (8.119) curve no 1 2 3 4 0.01 0.1 0.5 1 o µh p405, Replace Equations (8.116), (8.117) and (8.119) with the following: p415, lines 6 and 7 under Eq 9.16, replace with, "the end correction. In this case, > = 0. For a" p417, replace the text between Eqs. (9.25) and (9.26) with: "An alternative expression for the effective length, which may give slightly better results than Equation (9.25), for grazing flow across the holes, and which only applies for flow speeds such that , is (Dickey and Selamet, 2001)" u τ / (ωd ) > 0.03 p429, Move Equation (9.52) up one line. p432, Item 5, line 1, Replace "Equation (8.48)" with "Equation (9.52)" p439, line following Equation (9.81), replace µ with f m p444, In Table 9.2, "19" should be "-19" p453, 454, Replace the legend in the figures with p459, Figure 9.21, x-axis label, change "S" to "A" and in the caption add "open" immediately before "duct". p461, In the equation in the centre of the page, change "6" to "5" p461, 4 lines below the equation in the middle of the page, change "5.5" to "7" Errata for third edition text book 330 p461, 8 lines below the equation in the middle of the page, change "12.5" to "13" p462, line 1, change "1.2" to "1.0" p462, Figure 9.23 caption, last line, change "1992" to "1987" p464, Replace Table 9.5 with the following: Octave band centre frequency (Hz) Duct diameter (mm) 63 125 250 500 1000 2000 150 18(20) 13(14) 8(9) 4(5) 1(2) 0(1) 200 16(18) 11(12) 6(7) 2(3) 1(1) 0(0) 250 14(16) 9(11) 5(6) 2(2) 1(1) 0(0) 300 13(14) 8(9) 4(5) 1(2) 0(1) 0(0) 400 10(12) 6(7) 2(3) 1(1) 0(0) 0(0) 510 9(10) 5(6) 2(2) 1(1) 0(0) 0(0) 610 8(9) 4(5) 1(2) 0(1) 0(0) 0(0) 710 7(8) 3(4) 1(1) 0(0) 0(0) 0(0) 810 6(7) 2(3) 1(1) 0(0) 0(0) 0(0) 910 5(6) 2(3) 1(1) 0(0) 0(0) 0(0) 1220 4(5) 1(2) 0(1) 0(0) 0(0) 0(0) 1830 2(3) 1(1) 0(0) 0(0) 0(0) 0(0) p470, Figure 9.27, caption, and Eq. (9.115), replace "D" with "d" p471, Eq. (9.116) and (9.117) and 2 lines below Fig. 9.28, replace "D" with "d" p476, 3 rd and 6 th line of the first paragraph, change "1979" to "1978" p478, line above Equation (10.14), change "1979" to "1978" p479, Figure 10.2, replace the lowest y-axis label (currently 0) with 0.02 p483, In Equation (10.18) and 2 lines above it, replace "e" with "q" to avoid confusion with the distance, e, between spring supports. p484, In Figure 10.6, the force should be shown as acting on mass m 2 , not Errata for third edition text book 331 mass m 1 . p485 In Eqs. (10.25a,b), the left hand side should be squared. p487, line above Equation (10.31), change "1986" to "1988" p495, Equation (10.42), remove the symbol "d" from the right hand side. P496, Equation (10.48), replace "d" with |F|/ k 1 p496, Equation (10.47), the numerator on the RHS should be 3(m 2 /m 1 ) 3 p498, 8 lines from the top of the page, change "1979" to "1978" p513, Table 11.2, 3 rd line in 2000 Hz column should be "25" p513, Table 11.2, the 8000 Hz column should be replaced with 13, 15, 18, 27, 35, 35, 26, 32, 32, 34, 42 and 44 respectively and the BFI column for the two tubeaxial entries should be " 7 " p513, Remove the paragraph containing Equation (11.2) and remove "(11.2)" in the second to bottom line. p514, last Equation, label (11.2) p515, Example 11.1 table, replace "30" with "36" p517, Equation (11.10), change to: L w ' 72 % 13.5log 10 kW (dB re 10 &12 W) p526, last line, change "8.8" to "8.3" p528, replace the values in the table with the following. 0 72 77 80 81 80 76 69 63 60 74 79 82 83 82 78 71 65 120 61 66 69 70 69 65 58 52 180 55 60 63 64 63 59 52 46 Errata for third edition text book 332 p535, 4 lines above Eq.(11.33), and 2 lines after Eq. (11.34), change "534" to "60534". p536, 4 lines from the bottom, change "534" to "60534". p541, Following Eq. 11.64, insert the statement, "If the second term in brackets of Equation (11.64) exceeds 0.3, it is set equal to 0.3". p542, line 3, change "534" to "60534". p542, Immediately before Equation (11.67), add the following: "Note that the final spectrum levels must all be adjusted by adding or subtracting a constant decibel number so that when A-weighted and added together, the result is identical to the A-weighted overall levels form Equations (11.65) and (11.66)." p543, 1 line and 4 lines above Eq. (11.70), change "534" to "60534". p544, Equation 11.73, second term on the right should have the "log 10 " removed and "17.27" replaced with "17.37", so it reads "- 17.37(...........)" p544, Replace the last paragraph with, "The octave band external sound pressure levels may be calculated using Equations (11.73) and (11.76) with octave band sound power levels used in Equation (11.76) instead of overall sound power levels." p552, The constant in Equation (11.89) should be "55", not "53". p558, Replace the paragraph following Table 11.29 with the following: "The road surface or condition correction is taken as zero for either sealed roads at speeds above 75 km/hr or gravel roads. For speeds below 75 km/hr on impervious sealed roads, the correction is -1 dB. For pervious road surfaces, the correction is -3.5 dB. For concrete roads with deep random grooves greater than 5 mm in width, the correction is, C cond = 4 - 0.03P where P is the percentage of heavy vehicles." p559, Replace the nine lines following Eq. 11.102 with the following: "Low barriers such as twin beam metal crash barriers can have less Errata for third edition text book 333 SEL ref ' SEL v % 10 log 10 N % C 2 effect than soft ground. So if these are used with any proportion, P d , of soft ground, their effect should be calculated by looking at the lower noise level (or the most negative correction) resulting from the following two calculations: $ Soft ground correction (0 < P d < 1.0), excluding the barrier correction; and $ hard-ground correction (P d = 0) plus the barrier correction.” p560, Remove the sentence beginning 12 lines from the bottom of the page, "Note that the two values for $ must add up to 180E " p561, In the heading and first line, change "FWHA" to "FHWA" p561, 6 lines from the bottom, add "Menge, et al.," before "1998". p562, 4 lines under Equation (11.108), add "Menge, et al.," before "1998". p563, 5 th line in first paragraph, and 3 lines under Equation (11.109), replace "1995" with "U.K. DOT, 1995a". p563, 3 lines under Equation (11.111), replace "1995" with "U.K. DOT, 1995a,b". p563, p564, Replace the last two lines of page 563 and the top three lines of page 564 with the following: "Note that different vehicle types must be considered as separate trains. For any specific train type consisting of N identical units, the quantity SEL ref is calculated by adding 10log 10 N to SEL v . In addition the track correction, C 2 from Table 11.32 must also be added so that: p564, The second entry of "Freight vehicles, tread braked, 2 axles" should actually be "Freight vehicles, disc braked, 4 axles" p565, Lines 1 and 3, change SEL to SEL ref . Errata for third edition text book 334 p565, table 11.32, add ,C 2 , after "Correction" in the column 2 label. p567, In Equation (11.121), remove the minus sign p568, Add equation numbers, 11.122 and 11.123 to the equations at the top of the page. p580, 10 lines above Equation (12.1), change "1985" to "1986". p609, line 2 in the table for fresh water, change "988" to "998". p609, line in the table for iron, Young’s Modulus = 206, density =7,600, = 4910, E/ ρ 0 = 0.0005 and < = 0.27. p609, line in the table for Nylon, move the "6.6" next to "nylon" and Young’s Modulus = 2, density =1,140, = 1,320. E/ ρ p609, line in table for lead, loss factor = 0.015 p609, line in table for concrete, loss factor = 0.005 - 0.02 p610, the last column of numbers is the density and the 2 nd last column is Young’s modulus. p617, In figure captions, change "C.6" to "C.5" and "C.5" to "C.6". p621, Change number of Eq. 1.36 to C.24. p622, In Equation (C.29), replace Z N with Z N /Dc p623, In Equation (C.30), replace 2 with $ in three places. p645, Missing references. Allard, J.F. and Champoux, Y. (1989). In situ two-microphone technique for the measurement of acoustic surface impedance of materials. Noise Control Engineering Journal , 32, 15-23. Barron, M. (1993). Auditorium acoustics and architectural design. E&FN Spon: London. Errata for third edition text book 335 p646, Missing references. Beranek, L. L. (ed.) (1988). Noise and Vibration Control. Revised edition. Washington D.C: Institute of Noise Control Engineering. Beranek, L.L. (1996). Concert and Opera Halls. How They Sound. Acoustical Society of America: New York. Berglund, B., Lindvall, T. and Schwela, D.H. (1995). Community Noise. Stockholm: Stockholm University and Karolinska Institute. Berglund, B., Lindvall, T. and Schwela, D.H. Eds. (1999). Guidelines for Community Noise. Geneva: World Health Organization. p647, Missing references. Bragg, S.L. (1963). Combustion noise. Journal of the Institute of Fuel, Jan., 12B16. Broner, N. and Leventhall, H.G. (1983). A criterion for predicting the annoyance due to lower level low frequency noise. Journal of Low Frequency Noise and Vibration, 2, 160B168. p648, Missing references. Cazzolato, B.S. (1999). Sensing systems for active control of sound transmission into cavities. PhD thesis, Adelaide University, South Australia. Cazzolato, B.S. and Hansen, C.H. (1999). Structural radiation mode sensing for active control of sound radiation into enclosed spaces. Journal of the Acoustical Society of America, 106, 3732B3735. Chapkis, R.L. (1980). Impact of technical differences between methods of INM and NOISEMAP. In Proceedings of Internoise '80, pp. 831B834. Chapkis, R.L., Blankenship, G.L. and Marsh, A.H. (1981). Comparison of aircraft noise-contour prediction programs. Journal of Aircraft. 18, 926 B 933. p649, Missing references. Davy, J.L. (1993). The sound transmission of cavity walls due to studs. In Proceedings of Internoise '93, pp. 975B978. Davy, J.L. (1998). Problems in the theoretical prediction of sound insulation. In Proceedings of Internoise '98, Paper #44. Errata for third edition text book 336 Davy, J.L. (2000). The regulation of sound insulation in Australia. In Proceedings of Acoustics 2000. Australian Acoustical Society Conference, Western Australia, November 15-17, pp. 155-160. Delaney, M.E., Harland, D.G., Hood, R.A. and Scholes, W.E. (1976). The prediction of noise levels L 10 due to road traffic. Journal of Sound and Vibration, 48, 305-25. p650, Missing references. Dutilleaux, G., Vigran, T.E. and Kristiansen, U.R. (2001). An in situ transfer function technique for the assessment of acoustic absorption of materials in buildings. Applied Acoustics, 62, 555-572. Edge, P.M. Jr. and Cawthorn, J.M. (1976). Selected methods for quantification of community exposure to aircraft noise , NASA TN D-7977. Fahy, F.J. (2001). Foundations of Engineering Acoustics. London: Academic Press. Fahy, F.J. and Walker, J.G. (1998). Fundamentals of Noise and Vibration. London: E&FN Spon. FHWA (1995). Highway Traffic Noise Analysis and Abatement Guide. U.S. Dept. of Transportation, Federal Highway Administration, Washington, D.C. Fitzroy, D. (1959). Reverberation formula which seems to be more accurate with nonuniform distribution of absorption. Journal of the Acoustical Society of America,31, 893-97. Fleming, G.G., Burstein, J., Rapoza, A.S., Senzig, D.A. and Gulding, J.M. (2000). Ground effects in FAA's integrated noise model. Noise Control Engineering Journal, 48, 16B24. p652, Missing references. Hidaka, T., Nishihara, N. and Beranek, L.L. (2001). Relation of acoustical parameters with and without audiences in concert halls and a simple method for simulating the occupied state. Journal of the Acoustical Society of America,109, 1028B1041. Add Wosal, E-M. to the authors of the Hodgson (2002) paper. Howard, C.Q., Cazzolato, B.S. and Hansen, C.H. (2000). Exhaust stack silencer design using finite element analysis. Noise Control Engineering Journal, 48, 113-120. Errata for third edition text book 337 p653, Missing reference Jean, Ph., Rondeau, J.-F. and van Maercke, D. (2001). Numerical models for noise prediction near airports. In Proceedings of the 8 th International Congress on Sound and Vibration, Hong Kong, 2-6 July, pp. 2929B2936. p654, the reference, "Landau, L.D. and Lifsltitz, E.W." should be "Landau, L.D. and Lifshitz, E.W." p654, Missing references. Kuo, S.M. and Morgan, D.R. (1996). Active noise control systems. New York: John Wiley. Kurze, U.J. and Anderson, G.S. (1971). Sound attenuation by barriers. Applied Acoustics, 4, 35B53. Larson, K.M.S. (1994). The present and future of aircraft noise models: a user's perspective. In Proceedings of Noise-Con '94, pp969 B 974. Lee, J-W., Hansen, C.H., Cazzolato, B. and Li, X. (2001). Active vibration control to reduce the low frequency vibration transmission through an existing passive isolation system. In Proceedings of the 8 th International Congress on Sound and Vibration, Hong Kong, 2-6 July. Li, K.M. (1993). On the validity of the heuristic rayBtraceBbased modification to the WeylBVan der Pol formula. Journal of the Acoustical Society of America,93, 1727B1735. Li, K.M. (1994). A high frequency approximation of sound propagation in a stratified moving atmosphere above a porous ground surface. Journal of the Acoustical Society of America, 95, 1840B1852. Li, K.M., Taherzadeh, S. and Attenborough, K. (1998). An improved rayBtracing algorithm for predicting sound propagation outdoors. Journal of the Acoustical Society of America,104, 2077B2083. p655, Missing references. Maidanik, G. (1962). Response of ribbed panels to reverberant acoustic fields. Journal of the Acoustical Society of America, 34, 809B826. Errata for third edition text book 338 Menge, C.W., Rossano, C.F., Anderson, G.S. and Bajdek, C.J. (1998). FHWA Traffic Noise Model, Version 1.0, Technical Manual. U.S. Dept. Transportation, Washington, D.C. p656, Missing references. Neubauer, R.O. (2000). Estimation of reverberation times in non- rectangular rooms with non-uniformly distributed absorption using a modified Fitzroy equation. 7 th International Congress on Sound and Vibration, Garmisch-Partenkirchen, Germany, July, pp. 1709B1716. Neubauer, R.O. (2001). Existing reverberation time formulae - a comparison with computer simulated reverberation times. In Proceedings of the 8 th International Congress on Sound and Vibration, Hong Kong, July, 805-812. Nilsson, A. (2001). Wave propagation and sound transmission in sandwich composite plates. In Proceedings of the Eighth International Congress on Sound and Vibration, Hong Kong, July, pp. 61B70. p657, Missing references. Parkins, J.W. (1998). Active minimization of energy density in a threeBdimensional enclosure. PhD thesis, Pennsylvania State University, USA. Passchier-Vermeer, W. (1968). Hearing Loss Due to Exposure to Steady State Broadband Noise. Report No. 36. Institute for Public Health Eng., The Netherlands. Passchier-Vermeer, W. (1977). Hearing Levels of Non-Noise Exposed Subjects and of Subjects Exposed to Constant Noise During Working Hours. Report B367, Research Institute for Environmental Hygiene, The Netherlands. Plovsing, B. (1999). Outdoor sound propagation over complex ground. In Proceedings of the Sixth International Congress on Sound and Vibration, Copenhagen, Denmark, 685B694. p658, Missing references. Price, A.J. and Crocker, M.J. (1969). Sound transmission through double panels using Statistical energy analysis. Journal of the Acoustical Society of America,47, 154B158. Errata for third edition text book 339 Raney, J.P. and Cawthorn, J.M. (1998). Aircraft noise, Chapter 47 in Handbook of Acoustical Measurements and Noise Control, 3 rd edn. reprint, edited by C.M. Harris, Acoustical Society of America, New York. Raspet, R., L'Esperance, A. and Daigle, G.A. (1995). The effect of realistic ground impedance on the accuracy of ray tracing. Journal of the Acoustical Society of America,97, 683B693. p659, Missing references. Sandberg, U. (2001). Noise Emissions of Road Vehicles: Effect of Regulations. Final Report 01-1 of the I-INCE Working Party on Noise Emissions of Road Vehicles. International Institute of Noise Control Engineering. Saunders, R.E., Samuels, S.E., Leach, R. and Hall, A. (1983). An Evaluation of the U.K. DoE Traffic Noise Prediction Method. Research Report ARR No. 122. Australian Road Research Board, Vermont South, VIC., Australia. Sendra, J.J. (1999). Computational Acoustics in Architecture. Southampton: WIT Press. p659, In the Shepherd reference, change "1985" to "1986" p660, Missing references. Soom, A. and Lee, M. (1983). Optimal design of linear and nonlinear vibration absorbers for damped systems. Journal of Vibration, Acoustics, Stress and Reliability in Design, 105, 112B119. Steele, C. (2001). A critical review of some traffic noise prediction models. Applied Acoustics, 62, 271-287. Sutton, O.G. (1953). Micrometeorology. New York: MGraw-Hill. Tadeu, A.J.B. and Mateus, D.M.R. (2001). Sound transmission through single, double and triple glazing. Experimental evaluation. Applied Acoustics, 62, 307B325. Takagi, K. and Yamamoto, K. Calculation methods for road traffic noise propagation proposed by ASJ. In Proceedings of Internoise ‘94. Yokohama, Japan, pp.289B294. p660, Tse reference, change "1979" to "1978". Errata for third edition text book 340 p661, Missing references. U.K. DOT (1988). Calculation of Road Traffic Noise. Department of Transport. London: HMSO. U.K. DOT (1995a). Calculation of Railway Noise. Department of Transport. London: HMSO. U.K. DOT (1995b). Calculation of Railway Noise. Supplement 1. Department of Transport. London: HMSO. Watters, B.G., Labate, S. and Beranek, L.L. (1955). Acoustical behavior of some engine test cell structures. Journal of the Acoustical Society of America,27, 449B456. Wiener, F.M. and Keast, D.D. (1959). Experimental study of the propagation of sound over ground. Journal of the Acoustical Society of America, 31, 724. Yoshioka, H. (2000). Evaluation and prediction of airport noise in Japan. Journal of the Acoustical Society of Japan (E), 21, 341B344. Zaporozhets, O.I. and Tokarev, V.I. (1998). Aircraft noise modelling for environmental assessment around airports. Applied Acoustics, 55, 99B127. p661, In the Zinoviev reference, replace "In print" with " 269, 535-548." p663, after last line, add, "ANSI S3.6 B 1997. Specification for Audiometers." p667, line 1, replace "E90-99" with "E90-02". p715, Change "Noise Reduction Index" to "Noise Reduction Coefficient". Published by Colin H Hansen, Adelaide, South Australia Apart from fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Australian copyright Act, 1968, this publication may not be reproduced, stored, or transmitted, in any form, or by any means, without the prior permission in writing of the author. The author makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. A catalogue record for this book is available from the Australian National Library. © 2003 Colin H Hansen Published by Colin H Hansen ISBN 0-9751704-0-6 This version updated August 4, 2006 Contents All page, equation and table references are to the third edition of the textbook, Engineering Noise Control by DA Bies and CH Hansen 1. Solutions to problems in fundamentals 2. Solutions to problems relating to the human ear 3. Solutions to problems relating to noise measurement and instrumentation 4. Solutions to problems relating to criteria 1 65 71 83 5. Solutions to problems related to sound sources and outdoor sound propagation 99 6. Solutions to problems related to sound power, its use and measurement 7. Solutions to problems related to sound in enclosed spaces 130 143 8. Solutions to problems related to sound transmission loss, acoustic enclosures and barriers 199 9. Solutions to problems related to muffling devices 10. Solutions to problems in vibration isolation 11. Solutions to problems in active noise control 12. Errata in Third Edition of Engineering Noise Control 259 298 312 314 Preface This book provides detailed and instructive solutions to the book of problems on acoustics and noise control which is intended as a companion to the 3rd edition of the book "Engineering Noise Control" by David A. Bies and Colin H. Hansen. The problems and solutions cover chapters 1 to 10 and 12 in that text. Some of the problems and solutions are formulated to illustrate the physics underlying the acoustical concepts and others are based on actual practical problems. Many of the solutions extend the discussion in the text and illustrate the more difficult concepts by example, thus acting as a valuable source and understanding for the consultant and student alike. Although most of the problems and solutions have been tested on students, it is highly likely that there exist errors of which I am unaware. I would dearly like to hear from any readers who may discover any errors, no matter how minor. C.H.H., September, 2003 e-mail:
[email protected] Acknowledgments The author would like to thank the many people (too numerous to mention) who were responsible for providing ideas for problems. The author would also like to thank his undergraduate and graduate students who provided an excellent opportunity for fine tuning the solutions to the problems in this book. Finally, the author would like to express his deep appreciation to his family, particularly his wife Susan and daughters Kristy and Laura for their patience and support during the many nights and weekends needed to assemble the problems and solutions in a form suitable for publication. This book is dedicated to Susan, to Kristy and to Laura. 4 in text. As fλ = c and λ is fixed and c is faster in helium.4 29 1/2 ' 0. rank order noise sources contributing to any excessive levels. See pages 8-10 in the third edition of the text for more details. The only variables which are different for the two gases are γ and M. formulate a noise control program and implementation schedule. Problem 1. identify noise transmission paths and generation mechanisms. including the preparation of noise level contours where required.206kg/m3 and a speed of sound of 343m/s has been assumed.1 • • • • • • • undertake an assessment of the current environment where there appears to be a problem. the sound emanating from your mouth will be higher in pitch. carry out the program. Problem 1.2 (a) Speed of sound given by equation 1. Thus: cair cHe ' MHe γHe × γair Mair 1/2 ' 4 5/3 1/2 1. . establish the noise control objectives or criteria to be met.34 (b) The wavelength of sound emitted from ones mouth is a function of the vocal cord properties which remain unchanged by the presence of helium.1 Solutions to problems in Fundamentals Unless otherwise stated an air temperature of 20EC corresponding to an air density of 1. and verify the achievement of the objectives of the program. 400 × 0.4 Gas volume flow rate = 250.314 × 288.314×303.000 m3 per day at STP Gas Law also applies to a moving fluid.000 ' 2.8(1 % 0.8 m/s The speed of sound in wet air is then: cwet ' 348.029 × 2.314 × 393.0m/s Problem 1.3 Solutions to problems Using the relation for h given in the question and knowing that atmospheric pressure is 101.16) ' 351.0397×0.8935 ' ' 3.894m 3/s 24 × 3.0397 100 The speed of sound in dry air at 30EC is given by: cdry ' γRT / M ' 1.2 V .2 (b) Density of gas in pipe is: ρ ' m 0 PM 8 × 106 × 0.4kPa. so: m 0 0 RT PV ' M P and T are the static pressure and temperature respectively and 0 V ' 250.4×8. we can write: h ' 4240 101.029 ' ' ' 71.2 / 0.0 kg/m 3 0 RT 8.029 ' 348.2 Problem 1.550 kg/s RT 8.600 (a) The mass flow rate is given by: m ' 0 0 PVM 101.4×10 3 × 95 ' 0. 4 kg/m3.Fundamentals (c) Gas flow speed in discharge pipe is: U ' 4 πd 2 3 0 V ' 4 m 0 4 3.4 × 273 ' 675 m/ s Problem 1.35 × 101400 × 1273 1. Problem 1.4 × 273 1/2 c ' 1. We can find (R/M) by using the properties at 0EC and the expression: PV ' m R PV RT or ' M M mT As (m/V) = 1.2 + 6.0 ' 397 m/s (d) Speed of sound relative to the pipe is thus: 397.01 71.551 ' × ' 6. (R/M) is fixed.37 m/s 2 ρ π × 0. R 101400 ' and thus: M 1.4 × 8 × 106 / 71.4 = 404 m/s. M V .0 πd Speed of sound (relative to fluid) is: c ' γP ρ 1/2 ' 1.5 The speed of sound is given by: c ' γRT / M and for any gas.6 The Universal gas law may be written as PV ' Differentiating gives: dP m R dT m RT ' & dV M V dV M V 2 which may be rewritten as: or dp P dT P ' & dV T dV V m RT M or P ' m RT . 617 (1 & 1 / 1.314 × 313.4 ) × 298 ' 1.4 .029 ' 400. we obtain: τ ' ¯ 1. L / 10 p2 ¯ I ' and p ' pressure amplitude ¯ ' Iref 10 I 2ρc Thus. it can be shown that: ρc ' γP γRT / M ' 1. p ' [ 2 × 1. Problem 1.4 × 10&3 EC 101400 which is the amplitude of the temperature fluctuations.7 Using equations 1.4 × 8.4 Solutions to problems dP dV dT % ' P V T Another gas property associated with the adiabatic expansion and contraction of a sound wave is PV γ ' const .62 Pa ¯ We can write the results of the preceding analysis as: p ¯ 1 1& P γ ' τ ¯ T Substituting in given values and rearranging the equation to give τ. which leads to: dV 1 dP ' & V γ P Combining the above two equations gives: dP 1 1& P γ ' dT T The acoustic pressure associated with a sound wave of intensity 95dB is calculated as follows: Thus.5 ]1/2 ' 1.205 × 343 × 10&12 × 109.2 / 0.4a and b in the text.4 × 101400 1. 4 × 8.39 kg/ m 3 Assumption is that the gas in the pipe is at atmospheric pressure at sea level.314 × T 0. Thus: Thus.4 × 8. Problem 1.035 T ' (b) 6002 × 0. c ' 600 m/ s c ' γRT ' M 1.4Hz 5 Problem 1. f ' 250 ' c / 2L ' c / 2.035 ' 1083EK ' 810EC 1.Fundamentals Problem 1.4 × 103 7892 ' 0.035 (b) ρ ' γP c 2 ' 1.314 × 1873 ' 789 m/ s 0.4 × 101.4.4 × 101.23 kg/ m 3 (c) λ ' c / f ' 789 / 40 ' 19.10 (a) c ' γRT ' M 1.314 ρ ' γP c 2 ' 1.4 × 8.9 (a) Assume that the tail pipe is effectively open at each end as it follows a muffler.4 × 103 600 2 ' 0.7 m .8 f = c/4L = 343/(4×4) = 21. let the proportion (and volume Vg) of gas be x and the proportion (and volume VL) of liquid is then (1 x).8 dB.3 dB 149 × 0. so L ' c / 2f ' 789/80 ' 9. S = πdL + 2πr2 = π × 4 × 9. So the new SPL = 124.9 m So we can expect one dimension of the furnace to be 9.2 in the text as: D ' &V MV Mp &1 Consider a unit volume of fluid (V = 1).6 Solutions to problems If we treat the furnace like a closed end tube. f ' Nc / 2L .x)/D Assuming adiabatic compression of gas. so increase in sound pressure level will be 10 log10(3) or an increase of 4. PV γ ' const or P ' const × V &γ . From the above equation. (d) Surface area. and a corresponding volume increment ΔV. Assume a pressure increment of Δp. Problem 1.11 (a) The bulk modulus of the fluid is given by equation 1. we can write.7 W (e) Second burner has twice the sound power level.9 m. Differentiating P with respect to V and rearranging gives: γ dVg Vg % dP ' 0 P . ΔVL = -Δp(1 .98 789 × 0.86 + 2π × 4 = 149 m2 Lw ' Lp & 10 log10 ' 120 & 10 log10 4(1 & α ) ¯ ρc &10 log10 Sα ¯ 400 4 × 0.02 400 W ' 10 &12 × 10122.8 dB.23 & 10 log10 ' 122.3 / 10 ' 1. . (b) As x approaches 0. Thus: ΔV ' & 1&x x & ΔP D γP Thus the effective bulk modulus of the fluid is given by: Deff ' & ΔV ΔP &1 ' & 1&x x & D γP &1 The effective density of the fluid is equal to the total mass for unit volume of fluid and is given by: ρeff ' (1 & x) ρL % x ρg The speed of sound is given by c ' D / ρ .Fundamentals Substituting x for Vg and rearranging gives: Δ Vg ' & xΔP γP 7 Thus. As x approaches 1. the total volume change for a change Δp in pressure = ΔVL + ΔVg = ΔV. using the result of (a) above gives: c ' 1 ' (1 / γP)ρg γP ρg which is the speed of sound in the gas. Substituting in the effective values calculated for D and ρ above we obtain: c ' 1 1&x x % D γP (1 & x) ρL % x ρg where x is the proportion of gas in the fluid. using the result of (a) above gives: c ' 1 = D / ρL (1 / D)ρL which is the speed of sound in the liquid. 7 and 1.42. 0.8 Problem 1. p 2 cosβ ¯ ¢p 2¦ ' . we may write: p e&jβ u ' ρc cosβ Thus.5 × Re{¯u} ' 0.43 in the text.40. Harmonic intensity is defined as I = 0.6 may be written as u ' & p jωρ ' u 1/r % jk Using ω = kc and multiplying the numerator and denominator of the above equation by r gives equation 1.41b and 1. Using equations 1.12 Solutions to problems Equation 1. the acoustic pressure may be written as in equation 1. which is the plane wave ρc cosβ ρc .41b.jkr) gives: p jkr(1 & jkr) ρckr(j % kr) ' ρc ' ' u 1 % k 2r 2 1 % k 2r 2 ρckr 1%k r 2 2 × j % kr 1 % k 2r 2 Defining the phase β = tan-1(1/kr) as the phase by which the pressure leads the particle velocity. Using equations 1. That is: A φ ' e j(ωt & kr) r Mφ and thus the Mr particle velocity may be written as in equation 1. the preceding equation may be written as: p ' ρc cosβ (jsinβ % cosβ) ' ρc cosβ ejβ u which is the same as equation 1.43 by (1 .72.72.40c in the text is the harmonic solution to the spherical wave equation.42.5 p¯ expression. Using equation 1.5Re{ p u ( } where the bar denotes the ¯¯ complex amplitude. equation 1. we may write: For spherical waves. Multiplying the numerator and denominator of equation 1. ρ = 1. The velocity potential r r 1 A j(ωt & kr) p dt ' e is then φ ' and the particle velocity is ρm jρrω u ' & Mφ Ak j(ωt & kr) A ' e e j(ωt & kr) % 2 Mr ρrω jρr ω (b) When r = r0. ρr0ω jr0 But U = U0ejωt . 1 % jkr0 ρr0ωU0 2 1 % jkr0 ' jρr0 ωU0 1 k% jr0 for small kr0 (c) ω = 200π. r0 = 0. Thus: W ' 2πω2r0 ρU0 / c 4 2 4 2 *p* 2 *A* 2 4πr 2 and 4πr 2 ' 2 2ρc 2ρcr ' 2π(2π × 100)2 × 0. W = IS = (p2/ρc)4πr2 = *A* 2 ' ω2 r0 ρ2 U0 .206. A ' 1 jr0 e jkr0 However. U0 = 2.13 (a) At any location. u = U and the particle velocity may be written as: U ' A j(ωt & kr0) 1 k% e ρr0ω jr0 & jkr A 1 k% e 0.05 Power. e Thus.054 × 1.22 W .205 × 22 / 344 ' 0.Fundamentals Problem 1. p ' 9 A jω(t & r / c) A e ' e j(ωt & kr) . A ' ' coskr0 % jsinkr0 . so U0 ' ρr0ωU0 k% jkr0 Thus. 14 Solutions to problems Spherical wave solution to the wave equation is: A p ' e j(ωt & kr) r Using equation 1.7 in the text.206 × 343 × 0.2 2 × 10&5 988 × 1486 × 0.2 Lp ' 20log10 1.15 (a) Plane wave: p = ρcu = 1.10 Problem 1. the velocity potential is: c A j(ωt & kr) e φ ' jωρr Using the one dimensional form of equation 1. the particle velocity is: u ' & Mφ jkA j(ωt & kr) A j(ωt & kr) ' e % e 2 Mr jrωρ jr ωρ ' p j 1& ρc kr ' A j(ωt & kr) j e 1& rρc kr Thus: *u* ' *p* j *1 & * ρc kr Amplitude is twice * p * / ρc when * 1 & j / kr * ' 2 or 1 % (kr)&2 ' 4 . Thus kr ' 1 / 3 ' 0.2 10&6 ' 132 dB (b) Lp ' 20log10 ' 229 dB Non-linear effects are definitely important as the level in air is above 130dB .6 in the text.58 Problem 1.206 × 343 × 0. s.414 MPa. So the depth of water this corresponds to (allowing for atmospheric pressure) is: h = (0.3 Pa per meter depth.5 dB (b) The particle velocity is given by u ' p / ρc .414 × 106 . The acoustic pressure amplitude is 1.2m deep. so the water pressure is 9692.16 (a) The instantaneous total pressure will become negative if the acoustic pressure amplitude exceeds the mean pressure. we can show that for a spherical wave. the particle velocity amplitude is: *u* ' *p* j 150 × 103 *1 & * ' ρc kr 988 × 1481 ' 105 mm/s 1% 1481 2π × 1000 × 1 2 .81 = 9692. acoustic pressure of 150//2 = 106.Fundamentals and the level in water is also very high.41 × 988 × 1486 × 0.14.07 kPa.07 × 103 10&6 ' 220.3 N. 11 Cavitation in water would occur if the sound pressure amplitude exceeds the mean pressure which would happen if the water were less than a certain depth.101400)/9692. Weight = 988 × 9. Problem 1.2 = 0. The depth is calculated by calculating the weight of a column of water. The sound pressure level is then: Lp ' 20log10 p rms 10&6 ' 20log10 106. 1 m2 in cross section.m. This corresponds to an r. Thus the amplitude of the particle velocity is u = 150 × 103/(988 × 1481) = 103 mm/s (c) Using the analysis of problem 1.3 = 32. 94 1 % 0.6 in the text.206 × 343 = 413. 1/kr = 0.5462 ' 25 mm/ s 413. the particle velocity amplitude is: *u* ' 8.7 in the text.546.94 Pa Thus at 1m.7. The acoustic pressure amplitude at 1m is: *p* ' 2 × p ref × 10 L p / 20 ' 2 × 2 × 10&5 × 10110 / 20 ' 8.12 Problem 1.832 At 1m. the particle velocity is: u ' & Mφ jkA j(ωt & kr) A ' e % e j(ωt & kr) 2 Mr jrωρ jr ωρ ' p j 1& ρc kr ' A j(ωt & kr) j e 1& rρc kr Thus: *u* ' *p* j *1 & * ρc kr ρc = 1. k = 2πf/c = 2π × 100/343 = 1. 1/kr = 0.17 Solutions to problems (a) The sound pressure level at 10m will be 20log10(10/1) less than at 1m. (b) Spherical wave solution to the wave equation is: p ' A j(ωt & kr) e r A j(ωt & kr) e jωρr Using equation 1. at 10m. which translates to 90dB.0546. the velocity potential is: φ ' Using the one dimensional form of equation 1.6 The acoustic pressure amplitude at 10m is: . Z = p/u. Thus: Z ' ρc 1 & j kr (c) The modulus of the impedance of the spherical wave is half that of a plane wave (ρc) when * Z * ' ρckr Thus: 4k 2 r 2 ' 1 % k 2 r 2 or k 2 r 2 ' 0.894 1 % 0.6 in the text. the particle velocity amplitude is: *u* ' 0.Fundamentals *p* ' 2 × p ref × 10 L p / 20 13 ' 2 × 2 × 10&5 × 1090 / 20 ' 0.18 (a) Spherical wave solution to the wave equation is: A p ' e j(ωt & kr) r Using equation 1.894 Pa And at 10m.6 Problem 1.3333 1 % k 2r 2 1%k r 2 2 ' ρckr 1 % k 2r 2 ' ρc / 2 .7 in the text.05462 ' 2.2 mm/s 413. the velocity potential is: φ ' A j(ωt & kr) e jωρr Using the one dimensional form of equation 1. the particle velocity is: Mφ jkA j(ωt & kr) A ' e u ' & % e j(ωt & kr) 2 Mr jrωρ jr ωρ ' A j(ωt & kr) j e 1& rρc kr ' p j 1& ρc kr &1 (b) Specific acoustic impedance. 19 (a) r.6 dB (c) r.161 W / m 2 1.025 1 % k 2r 2 413.206 × 343 4π × 0.74 in the text.m. Substituting in the value for prms calculated in (a) above gives: I ' 19.885 W/m 2 (f) From equation 1.047 m/s 413.32 ' 19. the amplitude of the reactive intensity is: Ir ' prms ρckr 2 2 ' 19. u ' (d) Phase between pressure and particle velocity given by equation 1.12 Pa (b) SPL = 20 log10 ' 119.50) ' 10.502 ] 1/2 ' 0.s.205 × 343) ' 0.1252 ' 0.3333 ' 0.7 × 5.1 2 × 10&5 2 ' 1 × 1.50 Thus.3 ' 5.125 [1 % 5.205 × 343 × 5.1252 / (1.m.092 λ Problem 1. spherical wave intensity is prms / (ρc) .43 in text): * ur * ' kr ' * pr * 1 % k 2r 2 krρc 19.50 .s. (e) As shown on p35 in the text. Solutions to problems r ' (λ / 2π) 0.50 343 19.73 as: β ' tan&1[1/(kr)] ' tan&1(1/5.3E .14 Thus. and the acoustic pressure leads the particle velocity. sound pressure prms ' Iρc ' Wρc 4πr 19. particle velocity (see equation 1.6 k r 1/2 ' 2π × 1000 × 0. prms ' 2 × 19.5 % 1/ 2 ' 0.52 (2 × 10&5)2 ' 104.206 × 343 4π 10 20 15 % % 100 200 100 ' 11.52 Pa 2 Thus. Problem 1.1A (1. 10 m For source B.21 (a) For a spherical source: 2 prms ' ρcI ' ρcW / S ' ρcW / 4πr 2 . D Thus the total p2 at location D is: p rms ' 2 A B C 10 m 1.6 dB (b) Intensity. Thus.125 ' 27. For source C.0 Pa . Thus. r = 10/2. the sound pressure level is: Lp = 10log10 11. r = 10.20 Loct ' 10 log10 107. r = 10.885 % 120 ' 119. tanθ ' 0.Fundamentals (g) The sound intensity level is: LI ' 10log10 0. I % p2.8 dB Problem 1.1A (1 % 1/ 2 ) 0.8 % 107. I % W/r2 = AW/r2. For source A.7734 . As the sources are uncorrelated we may add p2 for each.5 dB 15 (h) 1500Hz is a different frequency to 1000Hz so the mean square pressures add.3 % 108 ' 80. The resultant intensity can thus be calculated using the figure on the next page. whereas results obtained using sound pressure measurements are likely to be seriously in error. Thus.7E below the horizontal.7 in the text indicate that the particle velocity is proportional to the pressure gradient which can be approximated by subtracting the measurements from two microphones and dividing the result by the separation distance. For a single 0.1A 45o frequency field the sound 2 intensity is the product of 45o the acoustic pressure with 0. Equations 1.15A the in-phase component of 2 the particle velocity. In the far field of the source. the pressure and particle velocity are in phase and related by p = ρcu.1A 0. in the far field. only one microphone is necessary as the pressure gradient need not be calculated.6 and 1. sound intensity measurements can still give accurate results in the presence of reflecting surfaces or other nearby noisy equipment. It is a function of the type of disturbance as well as of the acoustic medium. Thus its measurement requires a knowledge of the acoustic pressure and particle velocity. For a broadband field it is the time average product of the acoustic pressure and particle velocity.16 Solutions to problems Thus the direction of the intensity vector is 37.1A .22 (a) Specific acoustic impedance is the ratio of acoustic pressure to particle velocity at any location in the medium containing the acoustic disturbance. is equal to ρc and is the specific acoustic impedance of a plane wave in an infinitely extending medium. 0. (d) For the determination of sound power. (c) Sound intensity is a measure of the net flow of energy in an acoustic 45o disturbance.1A 0. Problem 1. (b) Characteristic impedance is a material or acoustic medium property. Fundamentals 17 (c) Interference describes the interaction between two or more sound waves of the same frequency such that regions of reinforcement (increased sound pressure) and regions of cancellation (reduced sound pressure) are formed.23 A flat spectrum level implies equal energy in each 1Hz wide frequency band. (e) Sound power is a measure of the total rate of energy emission by an acoustic source and has the units of watts.7 / 10 ' 101.5 dB . Problem 1. the energy level will increase by 3dB each time the octave band centre frequency is doubled.8 dB 10log10 1096 / 10 & 1094 / 10 ' 91.0 dB 10log10 10102 / 10 & 1098 / 10 ' 99. Problem 1.24 (a) The noise levels due to the machine only at each of the three locations are: 10log10 1098 / 10 & 1095 / 10 ' 95. the level will increase by 1dB for one third octave bands. each time the band centre frequency is stepped up.8 / 10 % 1091. Similarly. (f) Particle velocity describe the oscillatory motion of particles in an acoustic medium during propagation of an acoustic disturbance. (d) Phase speed is the speed at which a single frequency sound wave propagates and is proportional to the rate of change of phase experienced by a stationary observer as the sound wave propagates past. As the bandwidth of an octave band doubles from one band to the next.7 dB The Leq at 500Hz is: 10log10 1095 / 10 % 1099. p1 ' p2 ' pref 1085/20 pt ' pref 108.25 Use equation 1. In practice the measured level would be greater than this due to electronic instrumentation noise.707 and 2 2 2 pt ' (2 % 2 × 0.4.3 ' 18.5 % 108. then cosθ = 0. Thus: Lp ' Lp % 10 log10 3. .414 p1 .5 % 10& 2 % 10& 2.5 % 2 × 108. 3rd edn. Lp ' Lp % 10 log10 4 ' 85 % 6 ' 91 dB t 1 If a 45E phase shift were introduced.90 in text.3 dB .7 dB For signals 180E out of phase.5× cos30 Lp ' 10 log10 t 2 2 pt 2 pref 2 ' 90.3 ' 90.414 ' 85 % 5. p1 ' p2 ' pref 10 2 2 2 L p/10 and pt ' 4 × p1 2 2 Thus.90 in text.707)p1 ' 3. as the signals to be added are coherent. the phase difference between the two signals is zero as the speakers are identical and driven by the same amplifier.26 Use equation 1. cos180E = -1 and thus pt2 = 0 which means that Lpt = . t 1 Problem 1. If the two speakers operate together.1 dB 3 dB reduction ' &10 log10 Problem 1. Thus cosθ = 1. as the two signals will be coherent.18 (b) Solutions to problems 1 10& 1. m.s. The phase difference is thus 2πfc β2 & β1 ' k(x1 & x2) ' (x1 & x2) .x2) = λ/2 = c/2fc. Therefore the r.8dB Problem 1. The difference in dB between the two is thus: ΔL ' 10 log10 p rms p rms ) 2 ' 10 log10 (2 % 2 cosφ) ' 4.90 in the text gives: ¢ pt ¦ ' ¢ p1 ¦ % ¢ p2 ¦ & 2 ¢ p1 p2 ¦ ' ¢ ( p1 & p2 )2 ¦. (b) As the signals are at the same frequency and are shifted in phase by a constant amount. an r.m.90 in the text is used to add them together. the phase difference is: β2 & β1 ' 2πfc c × c ' π 2fc Substituting this result into equation 1. where x1 and x2 are p1 ' P1e the path lengths of the two waves. ΔL = 4. prms ' A / 2 . 2 2 2 p 1 > p2 . (b) If (x1 . value of A//2 and the relative phase between them is φ radians.Fundamentals Problem 1.8 dB Thus for a value of φ = 60E.28 (a) As the waves are from the same source. one may be described by j (ωt & kx1) j (ωt & kx2) and the other by p2 ' P2e . value of the combined signal is given by: p rms ' A 2 / 2 % A 2 / 2 % A 2 cosφ ' A (1 % cosφ) ) With just a single source. they are coherent. where fc is the centre frequency of c the band of noise.s. so equation 1.27 19 (a) Each signal has an amplitude of A. Problem 1. the level due to the "first" signal alone is given by 10log10 107.x2) = λ = c/fc. is given by: ¢ pt ¦ ' ¢ p1 ¦ % ¢ p2 ¦ % 2 ¢ p1 p2 ¦ cos(β1 & β2) 2 2 2 If the phase difference between the two waves is zero.29 Following example 1.90 averaged over all phases. then: ¢ pt ¦ ' ¢ p1 ¦ % ¢ p2 ¦ % 2 ¢ p1 p2 ¦ 2 2 2 . Thus: ¢ pt ¦ ' ¢ p1 ¦ % ¢ p2 ¦ % 2 ¢ p1 p2 ¦ 2 2 2 1 cosα dα 2π m 0 2π ' ¢ p1 ¦ % ¢ p2 ¦ % 2 ¢ p1 p2 ¦ 2 2 1 sinα 2π 2π 0 ' ¢ p 1 ¦ % ¢ p2 ¦ 2 2 which is the result for the incoherent case.90 in the text gives: ¢ pt ¦ ' ¢ p1 ¦ % ¢ p2 ¦ % 2 ¢ p1 p2 ¦ ' ¢ ( p1 % p2 )2 ¦ 2 2 2 (c) If all phases are present the total pressure is given by equation 1.20 Solutions to problems If (x1 . Problem 1.7 dB .4 on p49 in the text. ¢ pt ¦ .30 (a) The phenomenon is the superposition of acoustic waves of the same 2 frequency and fixed phase.5 & 106. the phase difference is: β2 & β1 ' 2πfc c × c ' 2π fc Substituting this result into equation 1.9 ' 73. Thus the total pressure. 94 dB Thus the machine is in compliance with specifications. then the level should be: ' 60 & 6 ' 54 dB 2 × 10&5 If the level which is measured with one pump operating is closer to 54dB than 57dB. then ¢ pt ¦ ' 0 2 ¢ pt ¦ ' ¢ p1 ¦ % ¢ p2 ¦ & 2 ¢ p1 p2 ¦ The noise level could be reduced substantially by using a control system which ensured that when one pump was turned on.Fundamentals If p1 = p2. the level due to the machine alone is equal to: 10 log10 109. . (b) When the problem is noticed. However if the level with one pump operating were closer to 57dB. the other was turned on at such a time that it was 180E out of phase with the first pump.7 & 109.31 Following example 1. 20 log10 p1 Problem 1. the sound pressure level at the house is 60dB.4 in the text. then incoherent addition would be suggested and the problem would need further investigation. then the theory of in-phase addition of sound waves would be verified.4 . Thus: 20 log10 2 p1 2 × 10&5 ' 60 If one pump only were operating. then ¢ pt ¦ ' 4¢ p1 ¦ 2 2 21 When the two waves are 180E out of phase: 2 2 2 If p1 = p2. This would occur when the two pumps are in phase. 8 Path B: 73.8 .1.8 Path C: 73.4 dB (c) The answer can be found by calculating what the direct field contribution .4) ' 3.5 .33 See pages 49 and 50 in text.8 .98 ' 71.8 dB Problem 1.1 . Then: 75 ' 10 log10 10x / 10 % 10(x & 5) / 10 or 75 ' 10 log10 10x / 10 % 10 log10 1 % 10(& 5 / 10) Thus.97 and is: NR ' 10log10 [100/10 % 10&5/10] & 10log10[10&8/10 % 10&13/10 % 10&13/10 % 10&8/10 % 2 ( 10&18/10 % 10&12/10 )] ' 1.4 .32 Solutions to problems Following example 1. 92. Let the signal due to the direct wave = xdB.6 dB Thus the level with the barrier removed is 60 + 3.7 = 62.5 & 109. Problem 1.38 % 106. x = 75 .22 Problem 1.5 = 63.8dB (b) Contributions to the total level from various paths are: Path A: 73. The noise reduction is calculated using equation 1.34 (a) Level at the receiver due to both waves = 75dB. The difference level with the barrier removed can be calculated by adding the barrier noise reduction to 60dB(A). "combining level reductions".6 = 63.2 & (&2.4 = 69.8 . the level due to the machine alone is equal to: 10 log10 109. Reflected signal has suffered a 5dB loss.28 % 106.4 in the text.6dB.8 Sound pressure level at receiver = 10 log10 106.2 = 73. 25e jkx utot ' & ρ Mx m jωρ Mx ' kA jωt &jkx e e & 0.4 . using the level with the barrier in place (note incoherent addition with the barrier in place).25e&2jkx 2 ωρ ' 0.25e jkx ωρ The active acoustic intensity is then: I ' 1 1 kA jkx Re{p u (} ' Re A e &jkx % 0. The total pressure field may then be written as: ptot ' Ae j(ωt & kx) % (A/4)e j(ωt % kx) ' A e jωt e &jkx % 0.65 = 7.25e&jkx 2 2 ωρ The preceding equation may be rearranged to give: I ' 1 kA 2 Re 1. the phase between the two waves is 0.6 and 1. x = 70.7 in the text gives for the particle velocity: 1 M A M jωt &jkx p dt ' & e e % 0. the intensity of each of the two waves could have been calculated separately and combined vectorially. At x = 0. Thus: 70 ' 10 log10 10(x & 11) / 10 % 10(x & 10))10 % 10(x & 4) / 10 or 70 ' 10 log10 10x / 10 % 10 log10 10& 11 / 10 % 10& 10 / 10 % 10& 4 / 10 Thus. θ = 0.47 kA 2 ωρ Alternatively.4 = 72. for the positive going . Thus.25e 2jkx & 0.35 The positive going wave may be represented as pi ' Ae j(ωt & kx) and the negative going wave as (A/4) e j(ωt % kx % θ) .25e jkx ¯¯ e & 0.4dB Reduction due to destructive interference = 72.0 + 2.0 & 0.Fundamentals 23 is. Combining equations 1.4dB Problem 1.25e jkx . Let the direct sound field = xdB. The total level with the barrier in place is 70dB. That is.252 % 0. W ' IS ' ρc ¢ u 2 ¦ ' ω2 ξ0 / 2 2 ' (2π × 500 × 0.04935 (m/s)2 ρc = 1. Problem 1.7. where S is the duct cross sectional area.04935 ' 0. where d is the tube diameter (see p456 in text).7 × 1.586c/d.47 ' 0. W ' 413.964 × 10&3 × 0.05 = 4020Hz Frequency range for plane waves = 0 to 4020Hz.052 Thus. S = (π/4) × 0. .206 × 343 = 413. the displacement should vary inversely with the square root of the frequency. Thus fco = 0.0001)2 / 2 ' 0. the cone velocity squared should be kept constant which means that the displacement squared of the speaker cone should vary inversely with frequency. the acoustic power is: S¢p 2¦ ' ρcS ¢ u 2 ¦ . That is.36 (a) Higher order mode cut-on frequency is: fco = 0.47 2ρc 32 ρc ρc ωρ If the two waves had the same amplitude it is clear from the preceding equations that the active intensity would be zero.04 watts (c) As power is proportional to the square of the cone velocity. (b) For plane waves.586 × 343/0.24 plane wave: Solutions to problems ¢pi ¦ 2 Ii ' ρc ' A2 2ρc and for the negative going plane wave: 2 ¢pr ¦ A2 ' Ii ' ρc 42 × 2ρc The total intensity is then: Itot ' Ii & Ir ' A2 A2 A2 kA 2 & ' 0. 6 as follows: 1 Mp(x. t) ' & jωρ Mx ρc Using the above expression for acoustic pressure and k1 = 500/343 k2 = 200/343 and x = 5. t) ' ' 5 j(500t & k1x & π / 2) 3 j(200t & k2x & π / 2) e e % 500 200 5 j(500t & 2500 / 343 & π / 2) e 500 1 343×1. t) ' ' ' j(500t & k1x) j(200t & k2x) 1 5e % 3e ρc 1 5e j(500t & 2500 / 343) % 3e j(200t & 1000/343) 343×1. t) ' 5e j(500t & k1x) 25 % 3e j(200t & k2x) The particle velocity can be obtained using equation 1.206 % 3 j(200t & 1000/343 & π / 2) e 200 ' 1 413.206 1 5e j(500t & 7.29) % 3e j(200t & 2. t) ' u(x . t) ' 1 ρc ξ(x .7 and the one dimensional form of equation 1.91 & π / 2) e 200 .37 (a) The acoustic pressure is given by: p(x .92) 413. t) p(x.7 % 1 j(500t & 7.7 u dt .29 & π / 2) e 100 3 j(200t & 2. the acoustic particle velocity can be written as: u(x .Fundamentals Problem 1. Thus: m The displacement is given by ξ(x . Also note that the mean square pressures for two different frequencies add. p = pR + pL. values are: 1 1 urms ' 52 % 32 × 413.s. As we have a plane wave propagating in only one direction. t) & pL(x.5 ' 0. t) ' j(500t & k1x) j(200t & k2x) 1 5e % 3e 1. t)) ' R Mx ρc jωρ & 4e j(500t % k1x) & 2e j(200t % k2x) .6 2 1/2 ' 3.6 2 ξrms ' 1 2 × 1/2 ' 0.m.206 × 343 p (x. t) ' 5e pL(x . t) 1 M(pR(x. t) ' 4e j(500t & k1x) j(500t % k1x) % 3e % 2e j(200t & k2x) j(200t % k2x) Total acoustic pressure.7 (d) Reactive intensity. t) ' & Thus: u(x . This is undefined as we have more than one frequency component in the wave. t) & pL(x. the sound intensity is given by equation 1.010 m/ s 1 1 / 100 2 % 3 / 200 413.041 W / m 2 Ia ' ρc 413.70 in the text. Thus: 2 prms 52 % 32 ' 0. Thus: p = 5e j(500t & k1x) % 3e j(200t & k2x) + 4e j(500t % k1x) % 2e j(200t % k2x) The total acoustic particle velocity is then: u(x .1 × 10&5 m (c) Active intensity. (e) The acoustic pressure for each wave is given by: pR(x .26 Solutions to problems (b) r. 20 × 10-3 (9 + 5) = 0. t) ' A e j(ωt & kx) . ρω ρc (b) Particle velocity is the magnitude of the motion of the particles disturbed during the passage of an acoustic wave. whereas the speed of sound is independent of loudness. the acoustic particle velocity may be k A j(ωt & kx) A j(ωt & kx) e e ' written as: u(x. whereas the speed of sound refers to the speed at which the disturbance propagates. The active intensity is calculated at each frequency and the results added together as follows: I ' % &jk x jk x jk x &jk x 0.6 and 1. Using the preceding equations we obtain: Z ' p(x.206 × 343 &jk x jk x jk x &jk x 0. t) Ae j(ωt & kx) (d) Z = pT/uT -L 0 x .5 Re 3e 2 % 2e 2 3e 2 & 2e 2 1. The amplitude of the reactive component is not defined as there is more than one frequency present. t) ' . I = 1.7 in the text.206 × 343 Thus.38 (a) The acoustic pressure may be written as p(x. (c) The specific acoustic impedance is the ratio of acoustic pressure to particle velocity.Fundamentals 27 The active intensity is given by I = 0.017 W/m2 Problem 1.5Re{ p u ( } where the bar denotes ¯¯ the complex amplitude. t) A e j(ωt & kx) ' ρc ' ρc u(x. Using equations 1. Acoustic particle velocity is a function of the loudness of the noise.5 Re 5e 1 % 4e 1 5e 1 & 4e 1 1. x = 0 is at the loudspeaker location. the phase shift on x 0 reflection is 0E and the -L amplitude of the reflected wave is equal to the amplitude of the incident wave. As the origin. the phase shift on reflection is 0E and the amplitude of the reflected wave is equal to the amplitude of the incident wave. With the origin at the point of reflection. the total acoustic pressure at any point in the tube may be written as: p T ' A e j(ωt & kx) % e j(ωt % kx % 2kL) The velocity potential and acoustic particle velocity may be derived from .39 The coordinate system is as shown in the figure at right.28 Solutions to problems To simplify the algebra. For a rigid termination. Of course if the origin were elsewhere. Thus. set the origin of the coordinate system at the rigid end of the tube as shown in the figure. As the origin. rigid termination Z = -pT/uT speaker (a) For a rigid termination. As the tube is terminated non-anechoically. this would not be true and the following expressions would have to include an additional term (equal to the distance from the origin to the point of reflection) in the exponent of the reflected wave. the phase of the two waves must be the same when x = 0. the phase of the two waves must be the same when x = -L. the total acoustic pressure and particle velocity at any point in the tube may be written as: p T ' A e j(ωt & kx) % e j(ωt % kx) and uT ' A j(ωt & kx) e & e j(ωt % kx) ρc The specific acoustic impedance is then: Z e&j kx % e j kx cos(kx) ' ' j cot(kx) ' &j kx j kx ρc &j sin(kx) e &e Problem 1. the pressure will include a contribution from the reflected wave. x = 0 is at the point of reflection. we obtain: pT ' U0 ρc 1&e j2kL e j(ωt & kx) % e j(ωt % kx % 2kL) and uT ' U0 1 & e j2kL e j(ωt & kx) & e j(ωt % kx % 2kL) The real part of the acoustic intensity (where the bar denotes the complex amplitude which is time independent) is: I ' 1 ( Re p T u T ¯ ¯ 2 2 ' ρcU0 Re 2 e&j kx % e j(kx % 2kL) × e j kx & e&j(kx % 2kL) 2 1 & e j2kL × 1 & e&j2kL e&j kx % e j(kx % 2kL) × e j kx & e&j(kx % 2kL) 2 2 & e j2kL & e& 2jkL 2 ' ρcU0 Re ' ρcU0 2 2 & 2 cos(2kL) Re 1 & 1 % e 2jk(x % L) & e&2jk(x % L) ' 0 The amplitude of the imaginary part of the acoustic intensity can be derived in a similar way and from the last line in the above equation. U0ρc A 1 & e j2kL and so A ' thus U0 ' ρc 1 & e j2kL (c) Rewriting the expressions of (a) in terms of U0. uT = U0ejωt.Fundamentals the above expression as: φT ' uT ' A j(ωt & kx) e % e j(ωt % kx % 2kL) jρω A j(ωt & kx) e & e j(ωt % kx % 2kL) ρc 29 (b) At x = 0. it . Problem 1.B. giving a pressure amplitude of (A + B). The minimum pressure amplitude occurs when the left and right going waves are π radians out of phase. such that the phase.40 speaker 1 speaker 2 0 x pi ' A e j(ωt & kx) and pr ' B e j(ωt % kx % θ) (a) Incident wave and reflected wave pressures may be written as: The total pressure is thus: pT ' A e j(ωt & kx) % B e j(ωt % kx % θ) The maximum pressure amplitude occurs when the left and right going waves are in-phase which is at location x. with a corresponding amplitude of A . at the location x. the ratio of maximum to minimum pressure is . θ = -2kx.30 is: I ' Solutions to problems 1 ( Im p T u T ¯ ¯ 2 ρcU0 2 ' 2 & 2 cos(2kL) sin[2k(x % L)] (d) The acoustic intensity is a vector quantity and as the amplitudes of the two waves travelling in opposite directions are the same. their intensities will vectorially add to zero. such that the phase θ = -2kx + π. Thus. 09 &0.7 The standing wave ratio is given by: .6 and 1. the particle velocity amplitude can be written as: 1 A 2 % B 2 1/2 * uT * ' ¯ 413.09m.06 k ' &0.06 π 2(xmin & xmax) ' &π / 2 Using the above 2 equations.03) π λ c ' ' 2k 4 4f (b) The total particle velocity can be calculated using equations 1. As θ is a constant: π & 2kxmin ' &2kxmax So: xmin & xmax ' and thus: f ' c 343 ' 1430 Hz ' 4(xmin & xmax) 4(0.18k + π. Thus θ = -0.B) and the standing wave ratio is 20log10[(A %B) / (A & B)] The location of the minimum closest to the end where x = 0 (left end) is 0.7 in the text as: 1 (p & pr) uT ' ρc i Thus: uT ' 1 A e j(ωt & kx) & B e j(ωt % kx % θ) ρc The complex particle velocity amplitude at x = 0 is then: 1 uT ' ¯ A & Be jθ ρc The phase angle θ is given by part (a) as: θ ' &0.Fundamentals 31 (A + B)/(A . 414 × 10100 / 20 ' 2.5 ) / 20 ' 1.m. A = 5. we have: Zm ' ρcS A % Be j(2kL % θ) A & Be j(2kL % θ) We previously found that θ = -π/2 and k = -θ/0.32 Solutions to problems A%B * ' 10(100 & 96.4136 × 5.7 × 0.3 and 2kL = (2π/0.158j Problem 1.41 (a) It is sufficient to show that adding two waves of the same frequency but shifted in phase will give a third wave of the same frequency but shifted .2 × 10-6 m3/s.362 % 0. Thus the velocity amplitude at x = 0 is: * uT * ' ¯ 1 2.001 × 5.032 % j)2 ' 0.032 % ejπ / 2 5.03B. the maximum pressure amplitude is A + B.36 and B = 0.032 & j ' 0.s volume velocity of 4.496 A&B * Thus.032 % j 5.032 & e jπ / 2 ' 0. we have A = 2.7 1/2 ' 5. Also L = 0.001m2 area) which is equivalent to an r. (c) The mechanical impedance of the second loudspeaker is given by the cross-sectional area multiplied by the ratio of the pressure and particle velocity at the surface of the loudspeaker. Thus Zm may be written as: Zm ' 413. Thus Zm = pS/u. Using the relationships derived in parts (a) and (b).12)×0.469.8 mm/s . However.9 × 10-6 m3/s (0.3 = 5π. So: A % B ' 2 × 10&5 × 1.828 From the preceding two equations.01571 × (5.4692 413.382 % 0.06. Thus the volume velocity amplitude is 5. A2 and β. let the two waves to be added be described as: p1 ' A1e jωt and p2 ' A2e jωt % β and the third wave as p3 ' A3e jωt % θ . To find A3 and θ in terms of A1.Fundamentals 33 in phase. p2 and p3 as rotating vectors and use the cosine rule as shown in the figure. The phase angle θ is: θ ' tan&1 A2 sinβ A1 % A2 cosβ (b) It is sufficient to show that adding together two plane waves of slightly different frequency with the same amplitude result in a third wave. Im p3 t p1 p2 Re From the cosine rule: A3 ' A1 % A2 % 2A1A2 cosβ 2 2 2 which is the same as equation 1. Assuming plane wave propagation. Let the two waves to be added be described as: p1 ' A1cos ωt and p2 ' A2cos(ω % Δω)t The sum of the two pressures written above may be expressed as: . it is easiest to express p1.90 in the text. 42 (a) Za = -pT/SuT 0 x L As the origin is at the left end of the tube. Assuming a phase shift between the incident and reflected waves of θ at x = 0.9 in text which shows a beating phenomenon where the two waves are slightly different in amplitude resulting in incomplete cancellation at the null points).7 in the text as: . fig 1. the incident wave and reflected wave pressures may be written as: pi ' A e j(ωt % kx) and pr ' B e j(ωt & kx % θ) The total pressure is thus: pT ' A e j(ωt % kx) % B e j(ωt & kx % θ) The total particle velocity can be calculated using equations 1. the incident wave will be travelling in the negative x direction.6 and 1. Problem 1.34 Solutions to problems p1 % p2 ' A(cosωt % cos(ω % Δω)t) t t ' 2A cos (ω % ω % Δω) cos (ω & ω & Δω) 2 2 ' 2A cos Δω 2 t cos ω % Δω 2 t which is a sine wave of frequency (ω + Δω) modulated by a frequency Δω/2 (c) If Δω is small we obtain the familiar beating phenomenon (see page 46. 5(P0 & ρcU0 ) e j(ωt % kx) % 0. Thus: Za ' & pT Su T ' & jkx &jkx ρc (P0 & ρcU0 ) e % (P0 % ρcU0 ) e S (P0 % ρcU0 ) e &jkx & (P0 & ρcU0 ) e jkx As ejkx ' cos(kx) % jsin(kx) and e&jkx ' cos(kx) & jsin(kx) . Thus: P0 ' A % B e j θ and ρcU0 ' B e j θ & A Thus: A ' 0.5(P0 & ρcU0 ) and Bejθ ' 0.Fundamentals uT ' Thus: uT ' 1 B e j(ωt & kx % θ) & A e j(ωt % kx) ρc 1 (p & pi) ρc r 35 At x = 0.5 (P0 % ρcU0 ) e j(ωt & kx) and uT ' 0.5 (P0 % ρcU0 ) and the total acoustic pressure and particle velocity may be written as: p T ' 0. pT = P0ejωt and uT = U0ejωt.5 (P % ρcU0 ) e j(ωt & kx) & (P0 & ρcU0 ) e j(ωt % kx) ρc 0 The acoustic impedance looking towards the left in the negative xdirection is the negative ratio of the total acoustic pressure to the product of the duct cross-sectional area and the total acoustic particle velocity (see the preceding figure). impedance may be written as: Za ' ρc jρcU0 sin(kx) & P0cos(kx) S ρcU0 cos(kx) & jP0sin(kx) the . 36 (b) Solutions to problems Using a similar analysis to that outlined above. while in part (b). the quantity P0/U0 is equal to the termination impedance Z0. As the origin is at the surface of the sample. the following expression is obtained for the acoustic impedance looking to the right in the positive x direction: Za ' ρc P0cos(kx) & jρcU0 sin(kx) S ρcU0 cos(kx) & jP0sin(kx) The expressions in parts (a) and (b) for the impedance can be shown to be equal if evaluated at the open end of the tube (x = L in part (a) and x = -L in part (b). In addition. Making the appropriate substitutions.0.43 (a) 2m Set the origin. Assuming a phase shift between the incident and reflected waves of θ at x = 0. the incident wave and . the quantity -P0/U0 is equal to the termination impedance Z0 in part (a). Za ' ρc Z0cos(kL) % jρc sin(kL) S ρc cos(kL) % jZ0sin(kL) Problem 1. both expressions give the following for the impedance at the open end of the tube. Then the entrance of the tube is at x = -2. x = 0 at the surface of the sample. the incident wave will be travelling in the positive x direction. Fundamentals reflected wave pressures may be written as: pi ' A e j(ωt & kx) and pr ' B e j(ωt % kx % θ) The total pressure is thus: pT ' A e j(ωt & kx) % B e j(ωt % kx % θ) 37 The total particle velocity can be calculated using equations 1.4475 + j0.25587) = 1558(1 . B/A = 0.7 in the text as: 1 uT ' A e j(ωt & kx) & B e j(ωt % kx % θ) ρc The pressure reflection coefficient.6 and 1. Thus the specific acoustic impedance at any point in the tube may be written as: Z ' pT uT ' ρc A e & j kx % B e j kx % jθ A e & j kx & B e j kx % jθ ' ρc A % Bej(2kx % θ) A & Bej(2kx % θ) ' ρc A / B % cos(2kx % θ) % jsin(2kx % θ) A / B & cos(2kx % θ) &jsin(2kx % θ) k ' ω / c ' 2 π × 100 / 343 = 1.j1130 (b) The absorption coefficient is defined as α ' 1 & * Rp * 2 .542.832 and x = -2.414 . cos(2kx % θ) ' 0.7237) = 1560 . Thus. Rp.25587 and A / B ' 1. Thus.3807 . so α = 0.5 .j0.7854 radians.5 % 0. 2kx + θ = -6.707 and θ = 45E = 0. Z = 414 × (2.25587)/(0.9667 and sin(2kx % θ) ' &0. is defined as pr /pi. Thus. Thus: Rp ' (B / A) ejθ ' 0.j0.5j The reflection coefficient amplitude is B/A and the phase is θ. and the minimum will occur when θ = 2kx + π. L0. the incident wave will be travelling in the negative x direction. Thus: sample speaker Z = -pT/uT -L 0 x pmax ' e jkx A % B and pmin ' e jkx A & B and the ratio of maximum to minimum pressures is (A + B)/(A .38 Problem 1. is defined as: 10 L0 / 20 ' A%B A&B Thus the ratio (B/A) is: . As the origin is at the left end of the tube. assume that the tube is horizontal with the left end at x = 0 containing the sample of material whose absorption coefficient is to be determined.B) The standing wave ratio. the incident wave and reflected wave pressures may be written as: pi ' A e j(ωt % kx) and pr ' B e j(ωt & kx % θ) The total pressure is thus: pT ' A e j(ωt % kx) % B e j(ωt & kx % θ) The maximum pressure will occur when θ = 2kx. as shown in the figure.44 Solutions to problems (a) To simplify the algebra. Assuming a phase shift between the incident and reflected waves of θ at x = 0. Zs.7 in the text as: uT ' Thus: uT ' 1 B e j(ωt & kx % θ) & A e j(ωt % kx) ρc 1 (p & pi) ρc r Thus the specific acoustic impedance at any point in the tube may be written as: p A e j kx % B e &j kx % jθ A % Be j(&2kx % θ) Z ' & T ' & ρc ' & ρc &j kx % jθ j kx uT Be & Ae Be j(&2kx % θ) & A At x = 0.Fundamentals B 10 0 & 1 ' L /20 A 10 0 % 1 L /20 39 The amplitude of the pressure reflection coefficient squared is * Rp * 2 ' (B / A)2 which can be written in terms of L0 (= 95 . Thus: Zs ρc A & B e jθ The above impedance equation may be expanded to give: Zs ' A / B % cos θ % j sinθ (A / B)2 & 1 % (2A / B) j sinθ ' A / B & cos θ & j sinθ (A / B)2 % 1 & (2A / B) cosθ ' & pT ρcu T ' A % B e jθ . of the surface of the sample.80) as: * Rp * 2 ' 10 10 L0 /20 L0 /20 &1 %1 2 The absorption coefficient is defined as α ' 1 & * Rp * 2 . so: α ' 1& 1015 / 20 & 1 1015 / 20 % 1 2 ' 0.6 and 1. the specific acoustic impedance is the normal impedance.51 (b) The total particle velocity can be calculated using the equation in part (a) for the acoustic pressure and equations 1. we obtain” .966) 1.966 Thus from the preceding equations: * Zs * ρc ' (1.4332 & 1 ' 1.2E (c) The statistical absorption coefficient is given by equation C.9662) 1. thus k = 2πf/c = 4.258 and sinθ = -0.866 × 0. x = 0. A 10 0 % 1 ' 1. Thus.58. θ = 2 × 4.8662 × 0.2 .2m and f = 250Hz.π.π = -1.4332 % 1 & 2. θ = 2kx .433 × (&0.4332 & 1) 2 % (2. ' L / 20 B 10 0 & 1 At the pressure minimum.37 in the text as 8 cosβ cosβ loge(1 % 2ξ cosβ % ξ2) % αst ' 1 & ξ ξ cos(2β) ξ sinβ tan&1 ξ sinβ 1 % ξ cosβ Substituting in the modulus and phase of the impedance.58 × 0.31 radians cosθ = 0. where k = 2π/λ.40 Solutions to problems The modulus of the impedance is then: * Zs * ' (A / B)2 & 1 2 % (2A / B)2 sin2θ (A / B)2 % 1 & (2A / B)cosθ and the phase is given by: β ' tan&1 2(A / B) sin θ (A / B)2 & 1 L / 20 Using the previous analysis.258 2 × 1.28 and the phase is: β ' tan&1 ' &69.433 . 22(1 & 0.17) 1 % 1.278 × 1.17) ' 2.485 Problem 1. the incident wave and reflected wave pressures may be written as: .22(1 & 0.28 1& cos(&69.28 × cos(&69. the incident wave will be travelling in the negative x direction.266 % 0.625 × (&0.17) % 1.28sin(&69.430) ' 0.17) 1. As the origin is at the left end of the tube. it "sees" as follows: W ' S * u T *2 ¯ S ( Re p T u T ' ¯ ¯ Re Z 2 2 where the bar represents the complex amplitude and S is the tube crosssectional area.4) 1.282 % cos(&138.352 & 0.28 × 41 × loge 1 % 2 × 1. Assuming a phase shift between the incident and reflected waves of θ at Z = -pT/uT 0 x L x = 0.17) tan&1 1.Fundamentals αst ' 8 × cos(&69.45 (a) Assume a horizontal tube with the left end containing the termination impedance Z0 at x = 0.28 × sin(&69. The power radiated by a source at the other end of the tube is related to the specific acoustic impedance.28 × cos(&69.688) ' 2.17) 1. Z. Z = Z0.42 Solutions to problems pi ' A e j(ωt % kx) and pr ' B e j(ωt & kx % θ) The total pressure is thus: pT ' A e j(ωt % kx) % B e j(ωt & kx % θ) The total particle velocity can be calculated using equations 1.5 (P0 % ρcU0 ) e j(ωt & kx) & (P0 & ρcU0 ) e j(ωt % kx) ρc .5 (P0 % ρcU0 ) e j(ωt & kx) and uT ' 0.5(P0 & ρcU0 ) e j(ωt % kx) % 0.5(P0 & ρcU0 ) and Be jθ ' 0.7 in the text as: uT ' Thus: uT ' 1 B e j(ωt & kx % θ) & A e j(ωt % kx) ρc 1 (p & pi) ρc r Thus the specific acoustic impedance at any point in the tube may be written as: Z ' & pT uT ' & ρc A e j kx % B e &j kx % jθ B e &j kx % jθ & A e j kx ' & ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A At x = 0.6 and 1. For convenience also set pT = P0ejωt and uT = U0ejωt Thus: Z0 ρc ' & P0 ρcU0 ' & A % B e jθ B e jθ & A Thus: A ' 0.5 (P0 % ρcU0 ) and the total acoustic pressure and particle velocity may be written as: p T ' 0. the impedance presented to the loudspeaker will be given by the previously derived expression for Z with R0 = 0. the power will be zero. replacing x with L and replacing P & 0 with Z0 gives: U0 Z ' ρc Thus: Re {Z } ' ρc (R0 / ρc) ( 1 % tan2kL) (1 & X0 / ρc)2 tan2kL % (R0 / ρc)2tan2kL Z0 / ρc % j tan kL 1 % j(Z0 / ρc)tan kL ' ρc (R0 % jX0 ) / ρc % j tan kL 1 % j((R0 % jX0) / ρc)tan kL and the power is then: W ' S ρc UL 2 2 (R0 / ρc) ( 1 % tan2kL) (1 & X0 / ρc)2 tan2kL % (R0 / ρc)2tan2kL (b) It can be seen from the equation derived in part (a) that when R0 = 0. impedance may be written as: Z ' ρc jρcU0 sin(kx) & P0 cos(kx) ρcU0 cos(kx) & jP0sin(kx) the Dividing through by ρcU0cos(kx).Fundamentals 43 The specific acoustic impedance looking towards the left in the negative x direction may then be written as: Z ' & ρc (P0 & ρcU0 ) e jkx % (P0 % ρcU0 ) e &jkx (P0 % ρcU0 ) e &jkx & (P0 & ρcU0 ) e jkx As e jkx ' cos(kx) % jsin(kx) and e&jkx ' cos(kx) & jsin(kx) . (c) If all losses are zero. In this case: . (d) As no real power is generated. Thus there will be no real power generated.5 (P0 % ρcU0 ) From part (a): Rp ' B e jθ A Thus: Rp ' (P0 % ρcU0 ) B e jθ ' A (P0 & ρcU0 ) Dividing numerator and denominator by U0 and putting Z0 ' ρc % jX0 ' & P0 / U0 .44 Solutions to problems jX0 / ρc % j tan kL 1 & ( X0 / ρc)tan kL X0 / ρc % tan kL 1 & ( X0 / ρc)tan kL Z ' ρc ' j ρc which is imaginary. we obtain: Rp ' & ρc & jX0 % ρc & ρc & jX0 & ρc ' j X0 2 ρc % jX0 2 Thus: * Rp * ' X0 [ 4ρ2c 2 % X0 ]&1/2 . the acoustic pressure and particle velocity must be 90E out of phase.5(P0 & ρcU0 ) and Be jθ ' 0. (e) The pressure amplitude reflection coefficient is given by: A ' 0. only imaginary power which represents non-propagating energy stored in the near field. in the tube may be written as: pT ' A e j(ωt % kx) % R e j(ωt & kx) (b) Returning to the first expression for the total pressure of part (a). Thus pmax ' e jkx A % B .46 (a) sample speaker 45 Z = -pT/uT L 0 x As given in the problem. Assuming a phase shift between the incident and reflected waves of θ at x = 0. pmin ' e jkx A & B and the ratio of maximum to minimum pressures is (A + B)/(A . x.B) (c) The standing wave ratio (SWR) which is L0 is defined as: 10SWR / 20 ' A%B A&B . the pressure amplitude reflection coefficient is thus R ' (B / A)e jθ and B ' (R A)e&jθ . As the origin is at the left end of the tube. the maximum pressure will occur when θ = 2kx. the incident wave and reflected wave pressures may be written as: pi ' A e j(ωt % kx) and pr ' B e j(ωt & kx % θ) The total pressure is thus: pT ' A e j(ωt % kx) % B e j(ωt & kx % θ) At the surface of the sample. Thus the total pressure at any location. the incident wave will be travelling in the negative x direction. and the minimum will occur when θ = 2kx + π. as shown in the figure. the tube is assumed to be horizontal with the left end at x = 0 containing the sample of material whose impedance is to be determined.Fundamentals Problem 1. of the surface of the sample. gives the required result. Zs.6 and 1.46 Solutions to problems Thus the ratio (B/A) is: B 10 0 & 1 ' L /20 A 10 0 % 1 L /20 From part (a). the specific acoustic impedance is the normal impedance. (e) The total particle velocity can be calculated using equations 1. which on substituting the equation derived in part (c) for * Rp * 2 . Thus: p 1 % Rp Z A % B e jθ ' & T ' ' jθ ρc ρcu T 1 & Rp A & Be . the amplitude of the pressure reflection coefficient squared is * Rp * 2 ' (B / A)2 which can be written in terms of L0 as: * Rp * 2 ' 10 10 L0 /20 L0 /20 &1 %1 2 (d) The absorption coefficient is defined as α ' 1 & * Rp * 2 .7 in the text as: uT ' Thus: uT ' 1 B e j(ωt & kx % θ) & A e j(ωt % kx) ρc 1 (p & pi) ρc r Thus the specific acoustic impedance at any point in the tube may be written as: Z ' & pT uT ' & ρc A e j kx % B e &j kx % jθ Be &j kx % jθ & Ae j kx ' & ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A At x = 0. Assuming a phase shift between the incident and reflected waves of θ at x = 0. the incident wave will be travelling in the negative x direction.6 and 1.47 (a) 47 Z = -pT/uT 0 x L Assume a horizontal tube with the left end containing the termination impedance ZL at x = 0.Fundamentals Problem 1. the incident wave and reflected wave pressures may be written as: pi ' A e j(ωt % kx) and pr ' B e j(ωt & kx % θ) The total pressure is thus: pT ' A e j(ωt % kx) % B e j(ωt & kx % θ) The total particle velocity can be calculated using equations 1.7 in the text as: uT ' Thus: uT ' 1 B e j(ωt & kx % θ) & A e j(ωt % kx) ρc 1 (p & pi) ρc r Thus the specific acoustic impedance at any point in the tube may be written as: p A e j kx % B e &j kx % jθ Z ' & T ' & ρc uT B e &j kx % jθ & A e j kx ' & ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A . As the origin is at the left end of the tube. impedance may be written as: Z ' ρc jρcU0 sin(kx) & P0 cos(kx) ρcU0 cos(kx) & jP0sin(kx) the Dividing through by ρcU0cos(kx). .5 (P0 % ρcU0 ) e j(ωt & kx) & (P0 & ρcU0 ) e j(ωt % kx) ρc The specific acoustic impedance looking towards the left in the negative x direction may then be written as: Z ' & ρc (P0 & ρcU0 ) e jkx % (P0 % ρcU0 ) e &jkx (P0 % ρcU0 ) e &jkx & (P0 & ρcU0 ) e jkx As e jkx ' cos(kx) % jsin(kx) and e&jkx ' cos(kx) & jsin(kx) . replacing x with L and replacing P & 0 with ZL gives: U0 ZL / ρc % j tan kL The same result Z ' be obtained by putting the open end of the tube to can ρc 1 % j(ZL / ρc)tan kL the left as shown in the figure above. For convenience also set pT = P0ejωt and uT = U0ejωt Thus: ZL ρc ' & P0 ρcU0 ' & A % B e jθ B e jθ & A Thus: A ' 0.5 (P0 % ρcU0 ) e j(ωt & kx) and: uT ' 0.5(P0 & ρcU0 ) and Be jθ ' 0. Z = ZL.5(P0 & ρcU0 ) e j(ωt % kx) % 0.48 Solutions to problems At x = 0.5 (P0 % ρcU0 ) and the total acoustic pressure and particle velocity may be written as: p T ' 0. Z = pT/uT 49 -L 0 x The total pressure is thus: pT ' A e j(ωt & kx) % B e j(ωt % kx % θ) and the total particle velocity is thus: uT ' 1 & B e j(ωt % kx % θ) % A e j(ωt & kx) ρc The specific acoustic impedance at any point in the tube is then: Z ' pT uT ' ρc A e j kx % B e &j kx % jθ & B e &j kx % jθ % A e j kx ' ρc A % Be j(&2kx % θ) A & Be j(&2kx % θ) The remaining part of the analysis is the same as before. Thus. is defined as the ratio of the complex amplitudes of the reflected to incident waves at x = 0. (b) The reflection coefficient. from part (a): Rp ' B e jθ A At x = 0: ZL ρc ' A % B e jθ A & Be jθ ' 1 % B e jθ / A 1 & Be /A jθ ' 1 % Rp 1 & Rp Thus the pressure reflection coefficient may be written as: .Fundamentals The same result can be obtained by putting the open end of the tube to the left as shown in the figure below. Rp. Thus. From equation 9.14 in the text we have ZH = jρctankR. the product kR/M must be much less than 1. . For a thin plate. Rp = 0. the equality ZL/ST = ZH/SH holds. where R is the effective length of the hole. the effective length of the hole is made up of two parts. we have at x = 0: ZL ρc ' A % B e jθ A & Be jθ ' 1 % B e jθ / A 1 & Be /A jθ ' 1 % Rp 1 & Rp Thus the pressure reflection coefficient may be written as: Z / ρc & 1 j(S T / S H)tankR & 1 Rp ' L ' ZL / ρc % 1 j(S T / S H)tankR % 1 (d) Using equation 9. means that ZL = ρc. From part (a). (c) Assuming continuity of acoustic pressure and volume velocity (uS. the first term in the equation derived in (d) must be large compared to the other terms. Thus: ZL ' j(S T / S H)ρc tan[kR(1 & M)] % Ra (e) For good absorption. where S is the tube cross sectional area) implies that the acoustic impedance is continuous across the hole and thus the specific impedance in the tube at x = 0 is related to the specific acoustic impedance. For this to be true.50 Solutions to problems ZL / ρc & 1 ZL / ρc % 1 Rp ' which is equal to zero when ZL = ρc. one corresponding to the side of the hole in the tube and the other corresponding to the side looking into free space. ZH of the hole by: ZL ' ZH ST SH where SH is the cross sectional area of the hole and ST is the cross sectional area of the tube. which as we showed in part (b).8 in the text and the condition of continuity of acoustic pressure and acoustic volume velocity at the hole. ZL = 4. Thus the preceding expression becomes: Z ' ρc 4 / ρc % j kL ρc &jρc ' ' 1 % j(4 / ρc ) kL jkL kL The ratio of the pressure to the particle displacement is then: p jωp ρcω ' ' ξ u kL which indicates that the pressure is in phase with the particle displacement. Also for a rigidly terminated tube.73dH.16 and 9. Considering the analogy of a single degree of freedom spring-mass system the equation of motion is: mξ¨ % Kξ ' f or &ω2mξ % Kξ ' f where K is the spring stiffness and m is the mass.19 in the text. For an open ended tube the impedance. ZL = 0 and the impedance of the tube becomes: Z ' ρc 0 % j kL ' jρckL 1 % j0 kL . It can be seen from the above equation that for a stiffness only. tan(kL) = kL. it can be shown that for small M. (f) The specific acoustic impedance looking into a tube was shown in part (a) to be: Z ' ρc ZL / ρc % j tan kL 1 % j(ZL / ρc)tan kL In the limit of small L and large λ (small k). or fdH << 70M. the effective length of equation 9. the exciting force will be in phase with the displacement and for a mass only it will be 180E out of phase with the displacement.73dH << M.Fundamentals 51 Using equations 9. assuming that the hole diameter is very small compared to the tube diameter. Thus the condition that kR/M << 1 implies that 2πf/c × 0. Thus it is clear from this analogy that the previous case of the pressure and particle displacement in-phase represents a stiffness.7 is R = 0. 45(a). (h) The above result suggests that the frequency response will be characterised by a number of resonant peaks and so will be very poor and non-uniform.48 Using the result derived in the answer to 1. the impedance seen by the loudspeaker may be written as: . this represents a lumped mass. (g) Using the equation given in the question and using the hint for maximum power we may substitute Z0 for ZL and Z to give kL at maximum power... 3. This would have the effect of damping the resonances. (i) The resonances are less damped at low frequencies.. Problem 1. The frequency response could be made smoother by adding some absorptive porous acoustic material to the tube. tan(kL) = 0.. Thus by the preceding argument. Thus kL = nπ. where n = 1. Thus: Z0 ρc Rearranging gives: Z0 ρc %j Z0 ρc 2 ' Z0 / ρc % j tan kL 1 % j(Z0 / ρc)tan kL tan kL ' Z0 ρc % j tan kL For the left side to equal the right side. thus reducing the difference between the peaks and troughs in the frequency response. 2.52 Solutions to problems The ratio of the pressure to the particle displacement is then: p jωp ' ' &ρcωkL ξ u which indicates that the acoustic pressure is 180E out of phase with the particle displacement.. . and the optimum length is given by L = nλ/2. thus resulting in bigger differences between the peaks and troughs in the frequency response.. we obtain: R 1.262. Thus the maximum power output will occur when the denominator in the above equation is zero.00486 Substituting in the expression given for R/ρc. Of course the real power output is only dependent on the resistive impedance while the imaginary power output is dependent on the reactive impedance.262/4.04 × 10-3 and Xr /ρc = 0. Thus optimum L for maximum power out = 1.58 As a = 0.0177 Thus.8 = 0.636.04 × 10&3 tan(kL) 2 2 By trial and error it can be shown that the maximum value of the above expression occurs when tan(kL) = 3.343. ka = 4. The maximum power output will occur when the impedance presented to the loudspeaker is equal to the internal impedance of the loudspeaker which is infinite. πa2 = 0.075 = 0. or kL = 1.3432/2 = 1.00486 tan(kL) 2 % 1.Fundamentals Zr / ρc % j tan kL 1 % j(Zr / ρc)tan kL 53 Z ' ρc which on substituting Zr = Rr + jXr may be rewritten as: R / ρc % j Xr / ρc % tan (kL) 1 & (Xr / ρc)tan(kL) & j (Rr / ρc) tan(kL) Z ' r ρc 1 & (Xr / ρc)tan(kL) 2 % (Rr / ρc) tan(kL) 2 The power output of the loudspeaker will vary with tube length because the power output of a source is dependent on impedance presented to it and the impedance it is presented is dependent on the tube length.04 × 10&3 1 % tan2(kL) ' ρc 1 & 0. Problem 1.0177 × 0.58 × 0.58 = 276mm.075.343 × 0. Rr /ρc = 0. k = 2πf/c = 2π × 250/343 = 4.0177 × 0.49 (a) A standing wave tube is used to determine the normal specific impedance . Thus the specific resistive impedance is: Rr / ρc 1 % tan2(kL) R ' ρc 1 & (Xr / ρc)tan(kL) 2 % (Rr / ρc) tan(kL) At 250Hz. the incident wave will be travelling in the negative x direction. and then generating a pure tone sound field in the tube with a loudspeaker placed at the other end of the tube. To simplify the algebra. as is the distance of the first minimum from the surface of the sample.7 in the text as: 1 (p & pi) uT ' ρc r Thus: uT ' 1 B e j(ωt & kx % θ) & A e j(ωt % kx) ρc Thus the specific acoustic impedance at any point in the tube may be written as: . As the origin is at the left end of the tube. the incident wave and reflected wave pressures may be written as: pi ' A e j(ωt % kx) and pr ' B e j(ωt & kx % θ) sample Z = -p T/u T 0 x L The total pressure is thus: pT ' A e j(ωt % kx) % B e j(ωt & kx % θ) The total particle velocity can be calculated using equations 1.6 and 1.54 Solutions to problems of a solid by placing the sample in one end of the tube which is then rigidly closed. Assuming a phase shift between the incident and reflected waves of θ at x = 0. The ratio of maximum to minimum acoustic pressure in the standing wave generated in the tube is measured. a horizontal tube with the left end at x = 0 containing the solid whose impedance is to be determined will be assumed as shown in the figure. Fundamentals Z ' & pT uT ' & ρc A e j kx % B e &j kx % jθ B e &j kx % jθ & A e j kx 55 ' & ρc A % Be j(&2kx % θ) Be j(&2kx % θ) & A At x = 0. where θ = 2kx . B 10SWR / 20 & 1 . The first minimum is used because the effect of any losses due to non rigid tube walls will be minimised. the ratio of maximum to minimum pressure is (A + B)/(A . The minimum rather than the maximum is used because its location is much more sharply defined.B. of the surface of the sample. A 10SWR / 20 % 1 ' and the impedance may be calculated. where k = 2π/λ = 2πf/c. Thus.π. and the phase angle. Zs. The equation for the impedance may be rewritten as: Zs ρc ' (A / B) % e jθ (A / B) & e jθ The standing wave ratio (SWR) which can be measured is defined as: 10SWR / 20 ' A%B A&B Thus. given on p623 the text. x of the first minimum sound pressure level from the face of the sample by using θ = 2kx .which is the difference in dB between the maximum and minimum sound pressure levels in the tube. The phase angle θ is determined by the distance.B) and the standing wave ratio is 20log10[(A %B) / (A & B)] . This is equivalent to the equation for θ.π. Thus: Zs ρc ' & pT ρcu T ' A % B e jθ A & B e jθ To calculate the impedance. it can be seen that the maximum sound pressure in the tube will occur when θ = 2kx. with a corresponding amplitude of A . The minimum pressure amplitude occurs at the location. θ. Returning to the above expression for the total acoustic pressure. and the amplitude will be (A + B). the specific acoustic impedance is the normal impedance. it is necessary to evaluate the constants A and B. where k = 2π/λ.4322 × 0.216 .951 4. the amplitude of the pressure B 10SWR / 20 & 1 ' reflection coefficient is * Rp * ' . thus θ = 4π × 0.2162 % 1 % 8.309 ' 0.6π cosθ = -0.2162 & 1 ' 25.87 and the phase is: β ' tan&1 2 × 4.951 Thus from the preceding equations: * Zs * ρc ' (4.4λ. (i) From part (a).π.309 and sinθ = 0. x = 0.216 × 0.21 % 1 ' = ' 4. A 10SWR / 20 % 1 100.2162 & 1) 2 % (8. B 10SWR / 20 & 1 100. The sound power A 10SWR / 20 % 1 reflection coefficient is defined as * Rp * 2 and the absorption coefficient is given by α ' 1 & * Rp * 2 .5E .432 × 0.π = 0.4 .21 & 1 θ = 2kx .56 Solutions to problems This is done by expanding the above impedance equation to give: Zs ρc ' A / B % cos θ % j sinθ (A / B)2 & 1 % (2A / B)j sinθ ' A / B & cos θ & j sinθ (A / B)2 % 1 & (2A / B)cosθ The modulus of the impedance is then: * Zs * ρc ' (A / B)2 & 1 2 % (2A / B)2 sin2θ (A / B)2 % 1 & (2A / B)cosθ and the phase is given by: β ' tan&1 2(A / B) sin θ (A / B)2 & 1 (b) As can be seen in the above derivation.9512) 4. which implies that the minima are separated by half a wavelength.0723.07232) = -5. (v) Let successive minima be located at x1 and x2.216&2 ' 0. Thus k(x2 &x1) ' &π .0894. For this to be true.216B. At 200Hz. this is equal to 343/(2 × 200) = 0.7)-1 × (0. and x1 & x2 ' λ / 2 . Thus B = 0. This can be verified by using the expressions for pT and uT derived in part (a). Problem 1.056 (iv) The intensity is given by: 1 1 ( Re (A % Be jθ ) × (Be&jθ & A) I ' Re p T u T ' 2 2ρc ' 1 1 Re B 2 & A 2 & 2jAB sinθ ' (B 2 & A 2 ) 2ρc 2ρc The maximum sound pressure level is 70dB. setting θ = 2kx. The intensity will not vary along a lossless tube as the sound intensity is a vector quantity which in this case is the vector sum of the intensities in the left and right going waves which is independent of tube location for a lossless tube.50 (a) Assuming no losses in the tube.0.0894 Using A = 4. Then using the equation derived in part (a) for the total pressure.01712 . The negative sign indicates that the net intensity is in the negative xdirection.96×10-6 W/m2. the intensity of a single plane wave .94 * 2 ' 4. and equating the jkx jkx pressures at x1 and x2 we obtain (A % B)e 1 ' (A % B)e 2 .0171 and A = 0.Fundamentals 57 (ii) The normal incidence sound power reflection coefficient is given by: * Rp *2 ' * 10SWR / 20 & 1 SWR / 20 10 %1 (iii) The absorption coefficient is given by α ' 1 & * Rp * 2 ' 0. e jk(x2 & x1) ' 1 .216B = 0. Thus: A%B ' 2 p ref 10 L p / 20 ' 0.86m. Thus the sound intensity is: I = (2×413. we have 5. where the amplitude of p as well as its r.22 may then be used to write equation 3. The above equation may be rewritten as equation 3.m. Equations 3.21 in the text. (b) The intensity cannot be measured with a single microphone because it cannot distinguish between the pressures associated with the two wave components travelling in opposite directions. .7 in the text we can show that the acoustic particle velocity is related to the acoustic pressure gradient by: u ' & 1 M p dt ρ Mx m where p1 and p2 are the pressures measured by microphones 1 and 2 respectively.6 and 1.21 and 3. Equation 3. Thus the 2 intensity of the positive going wave may be written as ¢ p% ¦ / ρc and the negative going wave as ¢ p& ¦ / ρc . value is independent of axial location.s. (c) Two identical microphones can be used because intensity is also the time averaged product of the acoustic pressure and particle velocity and from equations 1.23 needs a slight modification as in this case (replacement of n with 1) the intensity is in the positive x-direction down the tube rather than in an arbitrary direction n. As intensity is a vector quantity.58 Solutions to problems propagating in any one direction is ¢ p 2 ¦ / ρc .23 which is an expression for the intensity as a function of the measurements made by microphones 1 and 2. the 2 intensities of the positive and negative going waves can be combined by adding the intensity of the positive going wave to the negative value of the negative going wave to give: I ' 1 2 2 ¢p% ¦ & ¢ p& ¦ ρc which is independent of location. . Problem 1. Referring to equation 5. cosψ = 1. Note that if θ = 0. with the substitutions. (b) Again.129 in the text.129 in the text. k1 = k and k2 = km. then sinθ = 0 and from equation 5. the required result can be obtained by allowing θ = 0 (normal incidence) and substituting ρ2c2 for Zm and ρ1c1 for ρc.Fundamentals Problem 1.130. the required result can be obtained by substituting ρ2c2 for Zm and ρ1c1 for ρc.52 (a) The sound power reflection coefficient is simply the square of the modulus of the pressure amplitude reflection coefficient. referring to equation 5.130.51 (a) Rp ' and α ' 1 & * Rp * ' 1 & 2 59 Zs & ρc Zs % ρc * Zs % ρc *2 & * Zs & ρc *2 * Zs % ρc *2 *Zs & ρc*2 *Zs % ρc*2 ' Rearranging gives: α ' [ Re{Zs} % ρc ]2 % [ Im{Zs} ]2 & [ Re{Zs} & ρc ]2 & [ Im{Zs} ]2 * Zs % ρc *2 [ Re{Zs} ]2 % 2ρc Re{Zs} % (ρc)2 & [ Re{Zs} ]2% 2ρc Re{Zs} & (ρc)2 * Zs % ρc *2 4 ρc Re{Zs} * Zs % ρc *2 ' ' (b) From the above expression it can be seen that the maximum value of α is 1 which would occur when Zs = ρc. The angle ψ is defined in equation 5. 54 (a) Referring to the analysis on pages 210 – 212 in the text.118.and 1. As the wave is propagating along the y-axis.119 in the text and setting y = 0.7 in the text and the result of part (a). ky = k and kx = 0. Using equations 1. 1. Thus using equations 5. For an air/water interface.129 in the text can be written as: ρwcw &1 ρwcw & ρc ρc Rp ' ' ρwcw % ρc ρwcw %1 ρc From the preceding equation it can be seen that if ρwcw >> ρc.6. equation 5. (c) Using equations 1. the acoustic particle velocity amplitude (normally incident wave) is Ai/ρc.118 and 5. e jωt ) is given by: p T ' p I % p R ' AI e jωt e ' AI e jωt e &jk xx &j(k xx & k yy) &jk yy &e &j(k xx % k yy) e jk yy &e ' 2j AI sinky e jωt (b) For normally incident sound (and with Zm = ρwcw).60 Problem 1. θ = 0.6. Thus the total pressure (adding the time dependent term. pR = -pI.53 Solutions to problems (a) As a result of the property of zero pressure at the pressure release boundary. we obtain AR = AI. Thus the total particle velocity amplitude is twice the incident wave particle velocity. ρwcw = 1026 × 1500 = 1. the acoustic particle velocity amplitude is 2Ai/ρc. Problem 1. for normal .7 thus satisfying the required condition.7 and 5. then Rp = 1 and the reflected wave amplitude will be equal to the incident wave amplitude.54 × 106 which is much greater than ρc = 413. we have at the boundary (y = 0). 335 830 % ρc 830 % 413.129 becomes (with Zm = ρ2c2 = 830): Rp ' 830 & ρc 830 & 413.6 ' ' 0.55 (a) Referring to the analysis on pages 210 – 212 in the text and substituting . * Rp * ' 1 .335 Pi ' 0. and equation 5.6 The r. Thus: 2 1/2 2 cos ψ ' c2 c2 sin θ Substituting this result into the previous equation gives: ρ2c2 cosθ & ρc 1 & c2 2 1/2 c2 sin2θ c2 2 1/2 ' ρ2c2 cosθ % ρc 1 & c2 sin θ 2 This is true only if (c2/c)sinθ = 1. Thus. sinψ ' c2 sinθ c 1& . ψ = 0. θ = 0.s.Fundamentals 61 incidence.m.02 Pa Thus the amplitude of the reflected wave is: Pr ' 0. from equation 5.130. Problem 1. amplitude of the incident wave is given by: Pi ' 2 × 10&5 × 1060 / 20 ' 0.0067 Pa (b) When all the energy is reflected. Thus: ρ2c2 cosθ & ρc cosψ ' ρ2c2 cosθ % ρc cosψ Using Snell's Law. or θ = sin-1(c/c2) which is the angle above which all incident energy will be reflected. cosθ = 0. Using Snell's Law. (b) The transmission coefficient. ρc = 413. For θ = 10E. we obtain: pT pI ' AT AI ' 2 ρw cw cos θ 2 cos θ / (ρc) ' cos θ cos ψ ρw cw cos θ % ρ c cos ψ % ρc ρw c w Thus: τ ' 4 ρ c ρw cw cos2θ (ρw cw cos θ % ρ c cos ψ ) 2 ' (1 & Rp ) cos θ / cos ψ 2 (c) For air.127 in the text and substituting Z1 = ρc and Z2 = ρwcw.026 × 103 × 1500 = 1539 × 103.62 Solutions to problems ρwcw for Zm in equation 5. the required result is obtained.129.1736. τ. .121 and 5. is an energy related quantity and is defined by: * p T *2 ρc * AT *2 ρc τ ' ' * p I *2 ρw cw * AI *2 ρw cw incident wave rI air c wcw water y reflected wave rR x rT transmitted wave Using equation 5.9848 and sinθ = 0.6 and for sea water ρwcw = 1. the power would be distributed over an ever increasing area of ocean surface.9848 % 413. or θ = 13E. which is the angle above which all incident energy will be reflected.6506 1. Thus ρwcw cosθ & ρc cosψ ' ρwcw cosθ % ρc cosψ Using Snell's Law.Fundamentals sinψ ' cw sinθ 63 c cosψ = 0.9848 & 413. (e) If the sound source were a point source.539 × 106 × 0. the amount of sound power entering the water would be independent of altitude (except for atmospheric losses) as the same amount of power is contained in any specified included angle. ' 1500 × 0. Thus ψ = 49.7594 .4E and Substituting these values into the result of part (a) gives: Rp ' 1.6506 τ ' (1 & Rp )cos θ / cos ψ ' 0. However.1736 / 343 ' 0. sinψ ' cos ψ ' cw sinθ c 1& . A distributed sound source . * Rp * ' 1 .6 × 0.99964 (d) When all the energy is reflected. Thus.6506. 2 1/2 cw c2 sin θ 2 Substituting this result into the previous equation gives ρwcw cosθ & ρc 1 & cw 2 1/2 c2 cw 2 sin2θ 1/2 ' ρwcw cosθ % ρc 1 & c2 sin θ 2 This is true only if (cw/c)sinθ = 1.0011 2 ' 0.539 × 106 × 0.6 × 0. 7. ρcjkr u 0 Acceleration is defined as u ' jω u . the specific acoustic impedance is given by equation 1.118 and 5. the acoustic velocity amplitude of the incident wave at y = 0 is: uI ' ¯ AI ρc ( sinθ & cosθ ) e &jk1x x and the velocity amplitude of the reflected wave is: A &jk x u R ' R ( sinθ % cosθ ) e 1x ¯ ρc The velocity amplitude reflection coefficient is the ratio of reflected to incident velocity amplitudes. 1.6. so: p ' ρc ukr 0 ' ρur 0 ω . Thus: uR ¯ uI ¯ ' AR ( sinθ % cosθ ) AI ( sinθ & cosθ ) Problem 1.43 in the text as: p jkr ' ρc u 1 % jkr At the bubble surface. 5. Using equations 1. kr << 1. 5.56 (a) Assuming that the bubble is a spherical source. 5. so: p . (f) Velocity reflection coefficient.64 Solutions to problems such as a helicopter would exhibit similar behaviour.115.116.119 in the text. u ' juω 0 dV ' 4πr 2 u . p.Fundamentals (b) Adiabatic compression PV γ ' const or P ' const × V & γ Differentiating P with respect to V and rearranging gives: γ dV dP % ' 0 V P 65 dP . where p is the acoustic pressure. Thus: p dV ' &γ P V Substituting the result for p from part (a) gives: ρur 0 dV ' &γ P V or u ' & 0 γ P dV ρr V (c) Resonance frequency is the frequency at which the bubble prefers to vibrate given the physical parameters. the bubble screen should act like a Helmholtz resonator (see Ch. we obtain: juω ' & γP 4πr 2 u ρr jω (4 / 3)πr ω2 ' 3 ' & 3γPu ρr 2jω Rearranging gives: 3γP ρr 2 ' 3c 2 r2 Thus: fres ' ω c 3 ' 2π 2πr (d) At resonance. . jω V ' 4 3 πr 3 Substituting for dV and V in the expression of part (b). 9) and remove considerable energy from the sound field. 14 and 3.5Hz sound which represents a factor of 1060/20 = 1000.34m. which is equal to 101400/9.87 × 10&9 m (b) 120dB represents an increase in pressure by a factor of 106 over 0dB. This corresponds to a variation in water height of ± 28 × 10&9 / 9. thus volume of water = 101.5dB sound pressure level.16 indicate that the sensitivity is proportional to d4.81 ' ± 2.2 × 10-6 Pa peak. Linkage mechanical advantage = 3:1 = 9.2 Solutions to problems relating to the Human Ear Problem 2.4kN.9 indicates an increase of 60dB for 31. (b) Figure 3. so variation in column height would be ± 2.4/9. text.4 and equations 3.s = 28.5dB.1 (a) Weight of a column of water equivalent to the weight of a column of atmosphere of cross sectional area 1m2 is equal to 101. Density of water = 1000kg/m3.m. (c) Figure 2.81 kg. Overall mechanical advantage = 15:1 = 23.81 m3 and thus height of water column = volume/(1×1)= 10. Problem 2. Minimum audible sound = 0dB = 20 × 10-6 Pa r. . (d) See p56. so the mouse's ear should be 4 orders of magnitude (or 40dB) less sensitive.2 (a) A scaling of physical dimensions by a factor of 10 would mean that the mouse's range of hearing is 200Hz to 180kHz.9 mm . Referring to the middle ear differences. (d) Yes. Problem 2.3 (a) Sound introduced to the ear using ear muffs will be fairly reverberant. the MAP may approach the MAF in magnitude. it would be necessary to increase this relative movement which could be achieved by lengthening the rods of Corti. Referring to possible inner ear differences. one would expect the MDF to be similar to the MAP and less than the MAF. . To increase the sensitivity. we would expect the MAP to be lower than the MAF. because the differences would be necessary to make the mouse transduction mechanism more sensitive. (b) As the sound field within the ear muffs is similar to a diffuse field. a critical component of the inner ear in regard to sensitivity is the hinge mechanism of the tectorial membrane.The human ear 67 (c) The mouse's transduction mechanism must be 4 orders of magnitude more sensitive than the human mechanism. thus. If the earmuffs distort the pinna sufficiently. a much smaller oval window and a different arrangement of the bone linkage could account for a large sensitivity increase. having frontal as well as lateral components. (e) The differences could take the form of a larger mechanical advantage in the mouse's middle ear as well as differences in the relative physical dimensions of the inner ear. The sensing of sound by the inner and outer hair cells is by a shearing action imposed on the hair cell stereocelia caused by relative movement between the tectorial membrane and the rods of Corti. A frequency selective hearing aid will do a little better by amplifying the signal at the frequencies most affected but it is difficult to see how even this type of hearing aid will ameliorate the above mentioned symptoms of early noise induced hearing loss. repeated exposures to noise which results in a temporary threshold shift will result in permanent damage. the localisation of the basilar membrane excitation (and associated resonant hair cells) becomes increasingly important as the method for determining pitch until at 5000Hz the locked phase phenomenon of neuron firing ceases altogether and the neurons fire randomly. The loss usually occurs first in the 4kHz range and then extends to higher and lower frequencies. so even one exposure will contribute to the eventual permanent damage.4 Solutions to problems The first symptoms of noise induced hearing loss are difficulties with understanding conversation in a noisy environment. A conventional hearing aid which amplifies all frequencies by the same amount will not be much help as it will only amplify the "noise" experienced and not help with the symptoms mentioned above. there is evidence that damage is accumulative. On the other hand.68 Problem 2. Also the hair cells are tuned to maximum output at frequencies corresponding to the resonance frequencies of the parts of the basilar membrane to which they are attached. As stated in the text. at low frequencies and loud noise.5 Not necessarily. some neurons will fire for hair cells all along the basilar membrane and a second mechanism by which neurons fire in locked phase with the acoustic signal (at acoustic signal maxima or once per cycle) dominates the pitch determination.6 Pitch: Determined by location on the basilar membrane which responds most to the noise. Problem 2. These symptoms are caused by the breaking off of stereocelia on the outer hair cells leaving only the inner hair cell stereocelia functional. Problem 2. in focussing on the speaker and in localising noise sources. . As the frequency increases. However. This corresponds to a wavelength at 20Hz. the ear discards phase information.05 seconds.05 seconds after the direct sound. (b) See fig 2. the peak sound pressures will occur at intervals of 0. However.The human ear 69 Loudness: Determined by the rate of neuron firing. Thus the time histories as seen on an oscilloscope could look quite different. (c) The dimensions of an auditorium must be such that the sound arriving at any location after being reflected from a wall. . but they may have entirely different phase relationships between the particular spectral components making up the signal.05 seconds sound will travel a distance of 17. Thus for a 20Hz tone.8 (a) A low frequency warning device would be more effective mainly because it is not as easily masked by the 500Hz noise as higher frequencies would be but also because it would diffract more effectively around obstacles to create a more uniform coverage. Problem 2. Problem 2. Both amplitude and relative phase of the spectral components of a signal are necessary to reconstruct a signal uniquely.15m.7 (a) Two signals may have exactly the same spectral content and thus sound the same.10(a) in text and read off values as accurately as possible. There is also a hair cell feedback mechanism whereby the voltage generated by the hair cells causes them to deform. This feedback mechanism effectively increases the dynamic range of the hearing mechanism and also results in improved pitch resolution. This means the difference between direct and reflected paths should be less than 17m. which is controlled by the motion of the hair cell stereocelia which in turn is controlled by the basilar membrane. (b) In 0. Below this frequency there will be greater intervals between the peak sound pressures and the sound will be heard as a sequence of auditory events. ceiling or floor must not arrive longer than 0. thus increasing the movement of the tectorial membrane. 4 0 30 41 50 53 51.8 1.55+1.1 1.3 Sones Rear = 2.6+1.4+3.5+1.6 1.7 Using equation 2.0 2 0 0.02+0.8+1.3 33 48 55 59 61 55 46.5 + 0.6] = 8.3[0.0+2.2 0.7 2.4+2.3 + 0.0+1.3] = 4. Phons forward = 70.9+0.5 Phons side = 66.2 + 0.05+0.4 2.3 Sones Side = 3.8+2.5 5 0. Phons = 40 + (10log10S)/(log102).5 1.3[0.6 Phons rear = 62.6 Overall Levels: Sones Forward = 4.1 2 2.9 0.3 2.1+2.8 50 41.1.8 4.6 0.4 3.0 5 0.7+2.0+0.8+3.0+2.2+0.3 .1] = 6. Thus.8 3.2 3 2 1.5 2.8 0 33 45 53 57 55.4 39 22.70 Octave band centre frequency Solutions to problems 63 125 250 500 1k 2k 4k 8k SPL forward SPL side SPL rear Sones forward Sones side Sones rear Phons forward Phons side Phons rear 40 44 49 54 56 55 45 33 39 43 47 51 53 49 39 27 38 42 44 48 49 44 27 15 0 0.3[0. the threshold of detection of an 630Hz tone would be increased by 20dB. A speaker speaking twice as loudly will increase the rear level from 4 to 8 sones. Octave band centre frequency (Hz) MAF New SPL forward New SPL side New SPL rear 63 125 250 500 1k 2k 4k 8k 39 26 25 24 22 30 29 28 15 35 33 30 9 40 37 34 4 42 39 35 0 41 35 30 0 31 25 13 12 19 13 1 Difficulty in hearing sound in the 63Hz and 8kHz bands would be encountered but as these bands are not important for speech. making speech recognition just possible again. From the figure.The human ear 71 (c) Yes. Figure 2.5 indicates that the MAF at 630Hz is approximately 5dB. (e) Speech recognition is just possible for S = 8 Sones.9(a). . so the situation will be improved. the sound pressure level in all bands will decrease by 20log10(10/2) = 14dB (assuming free field conditions). it is expected that there would be no difficulty in understanding speech for any head orientation. as the levels in the bands important for speech recognition are well above the hearing threshold level or MAF. (d) If the distance increases to 10m from 2m. The MAF from figure 2.9 See figure 2. Masking tone is 800Hz at 60dB. a person with no hearing loss would be able to hear.9 and the new sound pressure levels are in the table below. so the sound level would need to be 25dB in order to be heard. Problem 2. Free-field response: microphone response when subjected to a sound wave coming from a specified direction in an otherwise free field (free of reflected sound).1 (a) Pressure response: microphone response when subjected to a uniform pressure field which is accomplished in practice electrostatically. (b) Random incidence microphones are used to measure noise in reverberant test chambers. Microphones are designed so that the roll-off in pressure response at high frequencies is just compensated for by the increase in free field pressure due to diffraction of a normally incident wave (normal incidence microphone) or by the increase in pressure due incident waves averaged over all possible directions of incidence (random incidence microphone). Random incidence response: microphone response averaged over all possible angles of sound wave incidence. .3 Solutions to problems relating to instrumentation and measurement Problem 3. Free field microphones are used in anechoic test chambers and in industrial noise measurement cases where the origin of the noise is from a single direction or a narrow direction angle. reverberation times in auditoria and industrial noise when the direction of origin is uncertain or there are a number of sources located in various directions from the observer. 16.3 (a) Sensitivity = -25dB re 1V per Pa.3 in the text that the correction for a 180E angle of incidence would be 0dB.5dB re 1V per Pa.5 + 5 = -24. (e) At 250Hz. the overall sensitivity is: -29. then the contribution due to the actual noise is: Lp ' 10 log10 101. Using equation 3. that is. it can be seen from figure 3. At 10kHz.2 mV/Pa. Problem 3. the pressure sensitivity. So the equivalent sensitivity of the sound level meter would be approximately 10dB. the sound level represented by the noise floor on the instrument is: Lp ' 20 log10 E & S % 94 ' &110 % 26 % 94 ' 10 dB If the sound level meter actually reads 13dB.2 = -29. (c) The microphone overall sensitivity for a 0E incident field at 10kHz is -29.5dB less at 10kHz than at 250Hz for the overall response to remain flat. we have: -25 = 20log10[E/p] which gives a sensitivity of 10(-25/20) volts/Pa which = 56. .5dB.5 . (d) The microphone overall sensitivity for a reverberant field at 10kHz is -29.7dB.2 73 (a) Using equation 3.0dB.Instrumentation and measurement Problem 3.3 & 101. implying that the ear is 10dB more sensitive. 29.5 + 1.0 ' 10 dB (b) The ear can hear 0dB of frontally incident sound at 2kHz and it can probably discern signals just above its noise floor.0.15. (b) The pressure response of the microphone would have to be 4. Thus the overall microphone sensitivity at 250Hz would be equal to -24.5dB.5 = -28. Problem 3. . the microphone would be held vertically.4 Solutions to problems The MAF is the minimum audible field to frontally incident sound and would correspond to the free field calibration of a microphone. Similarly the MDF would correspond to the diffuse field calibration of a microphone. thus resulting in most sound being incident at angles close to 90E. resulting in large measurement errors.7 The A-weighted sound pressure level is related to loudness perception of low level environmental noise as well as to hearing damage (although not to loudness perception of loud industrial noise).74 Problem 3. Problem 3. Problem 3. The pressure calibration of a microphone would correspond to the MAP which is the minimum pressure audible at the tympanic membrane of the ear.3 in text). It is a reasonably valid measure of noise exposure as there is a direct correspondence between A-weighted sound level and hearing loss suffered by noise exposed people.6 This would be similar to connecting the microphone to ground through a low impedance and this would seriously reduce the microphone sensitivity. and regulations are written in terms of this quantity due to its ease of measurement by unskilled people. In both cases measured sound pressure levels will be less than actual levels. The difference between the microphone diffuse field response and the 90E response is much smaller than the error resulting from pointing a free field microphone in the wrong direction (see figure 3.5 The random incidence microphone is used in cases where one is not sure of the direction from which the sound is coming or if it is coming from a number of directions simultaneously. To minimise measurement error. 4dB(A) and Linear = 98. To find the overall possible variation calculate the dB(A) levels for each extreme and compare them to the overall level calculated using band centre frequency corrections as illustrated in the table on the next page. The answers are: A-weighted = 89. Then plot the upper and lower octave band frequency limits from table 1. . Join all of the points by a straight line which is as good as a smooth curve for the present purposes. (b) The main source of error is a result of the assumption that all frequencies in each octave band can be weighted by a single quantity (the weighting corresponding to the band centre frequency) when in fact the Aweighting is a smoothly varying function of frequency.9 Plot out the values given in Table 3.8 (a) The A-weighted levels are calculated by adding the A-weighting corrections (most are negative) to each octave band level and then logarithmically adding the results as described on page 48 of the text. The maximum possible error can be estimated by comparing the results using the Aweighting corresponding to the band centre frequency with results obtained bu using the A-weighting corresponding to the upper and lower band limit frequencies.2 and read off from your graph the A-weighting corrections at these frequencies The difference between these values and the octave band centre frequency A-weighted values represent the largest errors which could occur if all of the energy just happened to be at frequencies at the edge of each octave band. Problem 3. The overall linear level is calculated by adding the values given in the problem together logarithmically as described on page 48 of the text.1 on graph paper using a logarithmic frequency scale. Problem 3.5dB. However it does not provide the frequency content information necessary for effective noise control measures to be specified and this is the main disadvantage.Instrumentation and measurement 75 The advantages of the A-weighted level for characterising equipment and workplaces are its direct relationship to noise exposure and its single number simplicity. 8 41 71 58.0 1.300 1.414 707 1. (Hz) Upper f limit Lower f limit Upper dB(A) adj.2 77. SPL Upper dB(A) Centre dB(A) Lower dB(A) -35 -21 -8.7dB(A) Lower limit = 83.2 -5.0 80 80.1 0.0 -1.1 -12.4 55.3dB(A) Centre value = 84.0 Centre -26.1 0.8 76 77. The A-weighted levels peak at high frequencies so the noise would sound a little "hissy".76 Solutions to problems Combining the band levels together as described on p 38 in the text gives: Upper limit = 84.8 70 68.2 1.0 78.1 dB(A) adj.3 1.9 76.6 -3.825 5.1dB(A) (corresponding to A-weighted corrections at band centre frequency.5 76 55 49.8 54. Octave band centre freq.2 1.8 707 353 -1. 63 125 250 500 1000 2000 4000 8000 88 44 -21 176 88 -12.414 2.2 73 74. Lower dB(A) adj.8 66.2 73 71.5 .2 353 176 -5.3 77.650 1.650 11.1 80 80.2 59.2 0.2 2.8 64.2 -16.2 1.825 5.0 -3.1 79 80.9 50 68 62.8 -1. 9.8 total 23. calculate the A-weighted levels in all bands.2 68 69 31.Instrumentation and measurement Problem 3.0 .49 in the text.0dB(A).2 31.2 99.2 105.2 81.2 71 69.48 in the text.11 (a) A-weighted level = 86. assume an arbitrary level of 50dB in the 63Hz octave band. The final answer is 89.8dB(A) (b) It is not a good way to calculate overall weighted levels because errors arise from the inherent assumption that the A-weighting is uniform across any particular octave band.12 To begin.2 84. the overall A-weighted level and thus the amount to add to each band level to reach an overall level of 105dB(A).2 31. Then arithmetically add (some numbers are negative) the A-weighted corrections to each octave band level as was done for problem 3.2 96.8 31. Problem 3.8 36. Then add the Aweighted octave band levels together logarithmically as described on p. The calculations are summarised in the following table. p. Octave band centre frequency (Hz) Un-weighted level A-weighted level Adjustment needed Unweighted level for 105dB(A) 63 50 125 53 250 500 1000 2000 4000 8000 56 59 62 62 65 66.2 93.2 102. Problem 3.2 87.9 47.10 77 First remove the background noise contribution from each octave band measurement as described in example 1.4 55. where it is noted that a constant spectrum level is reflected in octave band levels increasing at the rate of 3dB per octave (reflecting a doubling of bandwidth per octave).9 31.2 73.2 31.2 31.2 31.4.2 90. the A-weighted scale corresponds approximately to the loudness contour of 60dB.0s time constant) is useful for determining LAeq levels (average of the upper swings of the needle on an analog meter) and L90 levels (average of lower meter swings). In addition.78 Problem 3. the reflections could result in an increase in the measured sound level of up to 5dB (and also decreases) and in broadband sound fields. most legislation is written in terms of measurements taken using the "slow" response as some researchers suppose that this is more representative of the hearing damage caused by the noise. Note that the microphone remains unchanged .the electronics in the sound level meter effectively change its characteristics to compensate for the different diffraction effects of the microphone grid for each of the two types of field.14 (a) If the observer is too close to the microphone when noise measurements are being taken. (c) The "frontal/diffuse" control is used for selecting the microphone characteristic most suitable for the measurement being undertaken. The reasons for the above numbers not being 6 and 3dB respectively is because the observer will absorb and scatter some of the noise while reflecting it. (b) The fast response (0. Problem 3. In tonal sound fields. As industrial noise is usually much louder than this and equal loudness contours for high sound levels do not have the same shape as those at 60dB.13 Solutions to problems Although the signal levels in the experiments to determine equal loudness contours varied substantially. it is unlikely that the A-weighted scale will indicate correct loudness levels for most industrial noise. . then reflections from the observer can affect the noise levels being measured. the increase measured could be up to 2dB. However most modern instrumentation allows direct digital readout of these quantities.1s time constant) of the sound level meter approximates the way the ear hears but the slow response (1. 17 A sound level meter on site is preferable. so LAeq = 82.16 LAeq is generally used to describe noise as it is an A-weighted energy average which seems to be related to loudness perception of low level environmental noise as well as to hearing damage (although not to loudness perception of loud industrial noise).15 p ' (t 2 % 8t % 4) × 10&2 .6dB.7 dB A-weighting at 250Hz is -8. thus p 2 ' (t 4 % 16t 3 % 72t 2 % 64t % 16) × 10&4 Average p 2 ' (t 4 % 16t 3 % 72t 2 % 64t % 16) × 10&4 dt m 0 8 79 1 85 ' % 4 × 84 % 24 × 83 % 32 × 82 % 16 × 8 × 10&4 ' 0.4675 8 5 Leq ' 10 log10 p2 2 pref ' 10 log10 0.5 = 85.9 % 4 × 107.3 in the text).4675 4 × 10&10 ' 90.3dB(A).25 × 108 % 2 × 107 % 2 × 109 % (1/12) × 109.18 Sources of measurement error: . Problem 3. LAeq ' 10log10 1 × 1/4 % 2 % 2 % 1/12 % 4 × 0. Problem 3. Problem 3.Instrumentation and measurement Problem 3. as a tape recorder is not sufficiently accurate for legal disputes (see table 3. and regulations are written in terms of this quantity due to its ease of measurement by unskilled people.1dB(A). reflections from nearby surfaces. too cold or too hot.19 (a) See text. Problem 3. p114 B 120. overloading input amplifier when taking octave or 1/3 octave band measurements with an old SLM. errors arise because the microphone spacing becomes significant compared to a wavelength (ii) . (b) (i) At low frequencies. At high frequencies.80 Solutions to problems Microphone vibration. and wind noise. SLM vibration. Can minimise effects of wind noise by placing a foam wind shield on the microphone AND placing the mic in an enclosure made using shade cloth as possible wind screen configuration shade cloth 1m 1m 1m shown in the figure below. moisture or dust on microphone diaphragm. errors arise because the phase difference between the two microphone signals (due to the spacing being small compared to a wavelength) is not sufficiently large compared to the phase accuracy of the microphones. background noise. This is because the power measurement relies on averaging normal intensity measurements and the result of the external source will be to create a situation where small differences between large numbers will dominate the result.21 (a) Sound pressure associated with 95dB sound level is p rms ' p ref 10 L p / 20 ' 2 × 10&5 × 1095 / 20 ' 1. (b) See part (b) in previous question. localisation and identification of noise sources. (iii) In very reactive sound fields. sound power measurements could exhibit significant errors if the sound pressure level of the external noise is sufficiently high (usually about 10dB or more above the level from the noise source being measured). sound transmission loss measurement. so any phase errors translate to a large error in the intensity as it is proportional to the cosine of the phase angle. Problem 3.Instrumentation and measurement 81 causing the finite difference approximation for the pressure gradient to be inaccurate. the phase between the acoustic pressure and particle velocity is close to 90E. In the presence of external noise sources.12 Pa Force on microphone diaphragm is . (c) Applications include: sound power measurement. (iv) Problem 3.20 (a) See part (a) in previous question. determining the importance of flanking sound transmission paths in noise control applications. ampl ' 2u rms S ω 2 × 2.557 × 10&2 Thus the particle velocity is pωV u rms ' .0122 4 × 2 × π × 500 ' ' 1.0122 × (π / 4) ' 2.01 × 0. velocity of the diaphragm is the same as the air particle velocity.188 × 10&5 × π × 0.m. Thus .206 × 3432 × 0.022 × (π / 4) 1. the ratio of the sound pressure to particle velocity in a cavity of dimensions much smaller than a wavelength is p jρc 2 ' & uA Vω The sound pressure measured by the monitoring microphone is 65dB which corresponds to an r. From equation 9.19 × 10&5 m/ s (c) The volume displacement in the cavity corresponding to a sound pressure level of 65dB can be calculated using the same equation as used in part (b).11 × 10&12 m 3 (d) Mechanical input impedance.82 Solutions to problems Frms ' 1.s.557 × 10&2 × 2π × 500 × 0.m. Thus the volume displacement is Vol. ( S is the area of the microphone diaphragm) ρc 2S ' 3.s.displ.12 × π × (0. Zm = F/u.35 in the text.pressure of p rms ' 2 × 10&5 × 1065 / 20 ' 3.012)2 / 4 ' 1.27 × 10&4 N (b) r. 65) below the displacement of the test microphone.5 = 3% which will not affect the sound pressure sensed by the test microphone significantly. This is explained in Chapter 7 in the text. (f) Upper test frequency is limited by the onset of resonant cavity modes.Instrumentation and measurement Zm ' 1. This represents a percentage difference of 100/101.27 × 10&4 2.8 N&s / m 83 (e) The volume displacement of the monitoring microphone will be 30dB (95 . As the largest dimension is the radius. cross modes will occur before axial modes. .188 × 10&5 ' 5. 3 hours Problem 4.32) % 109.9 ) ' 100 dB(A) (b) Allowed daily exposure time in Australia is: Ta ' 8 × 2& ( 100 & 90) / 3 ' 0.86) % 10(9. .4 Solutions to problems relating to criteria Problem 4.1 (a) Using equation 4.8h = 90dB(A) be x dB(A).5 & 0.2 (a) A-weighted SPL is given by: LpA ' 10 log10( 10(9.7 & 0.3 Fan noise = 91dB(A) Saw idling noise = 88dB(A) Saw cutting noise = 93dB(A) Let required fan noise for LAeq.8 hours Allowed daily exposure time in USA is: Ta ' 8 × 2& ( 100 & 90) / 5 ' 2 hours Problem 4.42 in the text. the allowable exposure time using European criteria is Ta ' 8 × 2&(99 & 90 ) / 3 ' 1 hour (b) The allowable exposure using USA criteria is Ta ' 8 × 2&(99 & 90 ) / 5 ' 2. Assuming a precision of 0.667 log10 0.39 with the integral replaced with a sum and with with LB = 90 and L = 5.3 × [ 10 log10 (109.3 × [ 10 log10 (108.6 10 0.8h = 10Log10(1/8)[2 ×1095/10 + 6 ×1070/10] = 89.4 (a) LAeq.3 % 10x / 10) & 90 ] / 5 % 90 In solving for x.85.2dB(A) (b) USA criteria Using equation 4.8 = 11. Problem 4.0 dB(A) (b) EA.79.4 10 8 % 1. the allowed fan noise plus saw idle noise is 89.6 1093 / 10 % 10x / 10 8 85 Solving for x gives: 10x / 10 ' 9. This is because if the fan noise plus saw idle noise ) is greater than 90dB(A).8 ' 85. so we .4 = 5. x = 79. we must remember the proviso that combined fan and saw noise levels of less than 90dB(A) at any time do not contribute to the noise exposure results in a value of x as close to 90dB(A) as possible. Thus the maximum allowed fan noise is x ' 10 log10 108.3 or 4. we obtain: 90 ' 16. Thus the required fan noise reduction is 91 .9dB(A).54 Pa2 @h (c) We may assume that the 70 dB(A) does not contribute significantly.8 % 10x / 10) & 90 ] / 5 1 6.4 dB(A) .6dB(A).39 with LB = 90 and L = 3: 90 ' 10 log10 1 6.618 × 107 Thus.8 ' 32 × 10(89 & 100) / 10 ' 2.1dB(A). the overall LAeq is greater than 90dB(A).4 1088 / 10 % 10x / 10 % 1.Criteria (a) European criteria Using equation 4.99 & 108.8dB(A) The required fan noise reduction is then 91 . 5 × 1010.667 × log10 ( 1 / 8 ) 100.8h ' 10 log10 (European criteria) Using equation 4. and he will be 23 years old before he joins the old folks (assuming that he is in the 20% more sensitive part of the population).33 hours (USA criteria) ) 1 4.41 in the text with the integral replaced with a summation sign: LAeq.1 ] ' 92.5 % 100. 59.3 in the text: LAeq.42 in the text.5 ' 10 log10 10110 / 20 × T ' 55 % 10 log10 T where T is the number of years to cross the hearing loss criterion. Thus. Problem 4.52 hours (d) SPL due to machine only is: 10 log10 [ 109.5 & 109.5 8 ' 102.5 HDI ' 10 log10 10 m 0 t L p / 20 dt In this case.86 Solutions to problems need to find the allowed exposure to 95 dB(A).6 (a) Using equation 4. T .5 % 90 ' 101.6 dB(A) .5 % 1.4 dB(A) (b) European criteria Using equation 4. 3 years.5 × 109.8 dB(A) Problem 4. Ta ' 6 × 2&(102.8h ' 16.3 × (105 & 90) / 5 × 4.6 & 90 ) / 3 ' 0.3 × (95 & 90) / 5 × 1. This is given by Ta ' 8 × 2& ( 95 & 90 ) / 3 ' 2. 39 with LB = 90 and L = 3.8 dB(A) (L ) ) % 90 & 90) / 5 . we obtain: LAeq.43 & 90 ) / 5 ' 1.7 HDI ' 10 log10 j Ti × 10 i Lpi / 20 87 '10 log10 5 × 108.6 × 1088 / 10 8 % 2 × 1091 / 10 % 2 × 1096 / 10 ' 91.4(b) in the text. Problem 4.23 hours Problem 4.8 (a) USA criteria Using equation 4. LAeq ' 10 log10 1 2.3 × (96 & 90) / 5 8 ' 88. there is a 22% risk of developing a 22dB handicap.667 1 × log10 2 × 100.9 dB Daily noise dose = 2 Aeq ' 0.86 No reduction in exposure time is necessary.5 / 2 % 3 × 109 / 2 % 6 × 109.41 with LB = 90 and L = 5.3 × ( 91 & 90) / 5 % 2 × 100. (b) European criteria Using equation 4.3 or 4.6 From figure 4.8h ' 16.Criteria USA criteria Ta ' 6 × 2&(101.4 × 1085 / 10 % 1.5 / 2 % 1 × 1010 / 2 % 10 × 108 / 2 ' 58. European criteria La = 112. criteria La = 121dB.040 B-duration = 100 msec B-duration × number of impacts = BN = 2.A.S.5dB.120.29 × 106.5 in text.4 = 16. and noise dose = 2.4 BN = 1.40 × 105.304/1.S. Problem 4.3 × 106 Peak SPL = 125dB Allowable level for BN = 2.3 × 106 is obtained from fig 4. Required work day decrease European criteria Assuming that the press accounts entirely for the exposure of the employee.8 hours is required.8 & 90) / 3 Thus a reduction of 2. and noise dose = 2(125 .A.112. and noise dose = 23.9 Allowable BN for 125dB peak . European criteria U.see fig 4.8 The small differences in results obtained using the two methods (allowable level vs allowable BN) are due to difficulties in reading the figure any more accurately.5)/3 = 18 U. and noise dose = 2(125 . .29 = 1.2 hours 8 2 (91.04/1.53 ' 5.9 Number of impacts per day = 80 × 60 × 8 × 0.6 = 23. The operator is overexposed according to both criteria.5 in the text.88 Solutions to problems Daily noise dose = 2 Aeq Allowable exposure time: Ta ' (L & 90) / 3 ' 1. criteria BN = 1.5)/5 = 1. The background noise when the press is not operating does not contribute to the exposure according to USA criteria.0. Thus the allowable time of press operation is 0.6.2 × 106.29 hours of press operation.29). Thus required work day decrease = 4.7 × 0. allowable BN product = 1.10 Number of impacts per day = 40.29 × 106 /(80 × 60 × 100) = 2. In terms of operating hours. Allowable BN for 135dB peak .112)/3 = 200.4 × 106 Peak SPL = 135dB Allowable level for BN = 2. Thus required workday decrease = 4.A. and noise dose = 2(135 . European criteria BN = 1.29 = 0.000 B-duration = 60 msec B-duration × number of impacts = BN = 2.2 = 200 .6.4 × 106 is obtained from fig 4.2 × 104 and noise dose = 240/1.6 hours of press operation. In terms of hours. and noise dose = 2(135 .3.121)/5 = 7.4 × 105 /(80 × 60 × 100) = 0. This corresponds to a noise dose of (7.7 = 2.8 .2 = 4.6 in the text.4 × 105.71 hours (8 . U. USA criteria Background noise of 85dB(A) does not contribute From figure 4.20 hours.0.1 hours of press operation. [Iterating again does not affect the result significantly]. this is equal to: 1.see fig 4. the allowed BN product is 1.6 in text. Problem 4. criteria La = 121dB. this is equal to: 1. the exposure will be controlled by this for a minimum of 7.2.8 .Criteria 89 then from figure 4. European criteria La = 112dB.S.90)/3 = 0. Accounting for the background noise.69 hours of press operation.71/8) × 2(85 . Thus required decrease = 40.2 × 104 /60 = 200.5 due to this alone.7 in the text. Thus the allowable dose due to the impact noise is 0. Problem 4. that the 87dB(A) background can be considered to dominate the exposure for almost 8 hours.5. The answers are: No.000 . Problem 4. Too Loudly."shout". which corresponds to 100 impacts.11 Use figure 4. Problem 4.A. Expected level . then from figure 4.13 The one third octave band levels would have to be first combined into octave band levels by logarithmically adding together three third octave bands for .7 in text. the allowed BN product is 1.700 = 34. criteria BN = 3. However this represents such a small part of the 8-hour day.S.4 × 105 /60 = 5700. No.6.4 × 105 and noise dose = 240/34 = 7 Allowable number of impacts USA criteria The background noise of 87dB(A) contributes nothing to the daily noise dose because it is less than 90dB(A).300.90 Solutions to problems U. resulting in a noise dose of 0.12 See figure 4. No. European criteria Assuming that the press accounts entirely for the exposure of the employee."very loud voice to shout" Required level . The allowable number of impacts is then 1. Thus the allowed number of impacts = 3.2 × 104.5. 2 0. However overall dB(A) numbers are adequate for the purposes of assessing hearing damage risk and for comparing noise (either occupational or environmental) with permitted levels according to local regulations. then the 250Hz octave band level would be: Lp250 ' 10 log10 1060 / 10 % 1065 / 10 % 1063 / 10 ' 67.15 (a) A-weighted levels are calculated and tabulated below Frequency (Hz) Lp (dB re 20µPa) 63 100 125 101 250 500 97 91 1k 90 2k 88 1. and 63dB respectively.14 A-weighted overall sound levels are inadequate for noise level specification and control because they give no indication of the frequency content of the noise which is necessary for assessing annoyance and determining the type of noise control approach which may be feasible.2 -16.1 79.0 8k 81 -1. 250Hz and 315Hz one third octave band levels were 60dB.1 -8.6 -3. although often for specification purposes. For example. Problem 4.9 dB Problem 4.0 73.8 84.2 87 . NC or RC numbers are adequate as they take into account the spectral content of the noise. The three bands to add would be one with a centre frequency the same as the octave band and one band above and one below that one. if the 200Hz.Criteria 91 each octave band result.8 90 89. 65dB.9 88. NR.9 A-weighting A-weighted level -26.4 87. It is preferable to have data as octave band levels for control purposes.2 4k 86 1. 3 [ 28.49 % 108.78 % 109 % 108.3 38 33 The overall level in sones is calculated using equation 2. 1.33.5 % 35.1 dB(A) (b) Using equation 4. where it may be seen that NR = 91.42 in the text.99 ' 96. Thus: L ' 38 % 0.5 % 33 % 35.plot levels on NR curves as shown below.1 & 90 ) / 3 .5 sones .84 % 108.95 hours (c) NR level of noise .92 % 108.7 % 107.3 % 28.92 Solutions to problems The overall A-weighted level is: Lp ' 10 log10 107. the allowable number of hours is: Ta ' 8 × 2& (96.5 33 35. (d) Loudness level Frequency (Hz) Lp (dB re 20µPa) 63 100 28.3 % 38 % 33 ] ' 107.3 28.5 125 101 38 250 97 500 91 1k 90 2k 88 4k 8k 86 81 Sones 35.38 % 108. 5 phons 93 Note that the above equation is inaccurate for levels above 100 phons.1 & 1.16 (a) The levels are first plotted on NC and NCB curves 50 Octave band sound pressure level (dB re 20 Pa) 45 40 35 30 25 20 MAF 15 10 0 63 125 250 500 1k 2k 4k 8k Octave band center frequency (Hz) The result is NC = 33 and NCB = (38 + 30 + 20 + 16)/4 = 26 .5 ) / 10 ' 90.4.1 / 10 & 10( 96.32: P ' 40 % 10 log10 S log10 2 ' 107.Criteria From equation 2.8 dB(A) Problem 4. the contribution of the machine to the overall level is x dB(A) where x is defined as: x ' 10 log10 1096. (e) Following example 1. 5 63 125 250 500 1k 2k 4k 8k Octave band center frequency (Hz) The system is sufficiently quiet for churches holding less than 250 people (see table 4. . Best fit between 125Hz and 500Hz is NCB = 33. Edn. This conclusion can be checked by plotting on RC curves. (b) This NCB level is exceeded by more than 3dB in 125Hz. Although this conclusion is different to that drawn using NCB curves. Note RC = (38+30+20)/3 = 29. however for larger churches. the noise will be neutral (not rumbly or hissy -see 3rd.8 in the third edition of the textbook and table 4. text. Note that RC criteria would result in a neutral classification (not rumbly or hissy). No levels in the octave bands between 1000Hz and 8000Hz are above NCB = 33 so sound is not hissy.94 Solutions to problems 50 45 NCB Octave band sound pressure level (dB re 20 Pa) 40 35 30 25 20 15 10 10 0 0 16 31. the level should be about 5dB lower. it can be seen that the RC classification is close to rumbly. As can be seen from the following RC plot. page 159) as no levels in bands below 500 Hz exceed the RC level by more than 5 dB. 250Hz and 500Hz bands so it will sound rumbly.2 in the first edition or look up AS2107-1987). 3 48 48 0 43 43 0 38 35 0 30 33 30 20 28 27 16 23 22 12 .3 dB Desired spectrum levels of masking noise are listed in the table below. Octave band centre frequency (Hz) Sound pressure level (dB) RC-30 values Desired added levels 63 125 250 500 1000 2000 4000 8000 48 50 51.5 63 125 250 500 1k 2k 4k 8k 30 25 95 Octave band center frequency (Hz) (c) Optimum spectrum levels of added masking noise would be equal to the RC-33 levels (as this corresponds to the highest existing spectrum levels) with the existing levels logarithmically subtracted from it. the RC-33 value is 53dB and the desired added level is: LA ' 10 log10 1053 / 10 & 1048 / 10 ' 51. in the 63Hz band. For example.Criteria Octave band sound pressure level (dB re 20 Pa) 50 (RC) 40 50 45 30 40 35 20 10 16 31. 0 / 10 % 10(45 & 1.1) / 10 ' 59.6) / 10 % 10(50 & 3.2) / 10 % 10(50 % 1.2) / 10 % 10(55 & 16. Three high frequency bands exceed this curve so the noise sounds hissy.17 Solutions to problems (a) The NCB value for the noise is (37 + 33 + 33 + 32)/4 = 34 (b) The line of best fit for the NCB curve between 125 Hz and 500 Hz is 32 or 33 NCB.18 (a) A-weighted level is: LA ' 10 log10 10(60 & 26.5 63 125 250 500 1k 2k 4k 8k Octave band center frequency (Hz) Problem 4. No bands below 500 Hz exceed 34 NCB so noise is not rumbly.2) / 10 % 10(55 & 0) / 10 % 10(55 % 1.1) / 10 % 10(55 & 8.9 dB(A) . 50 45 NCB Octave band sound pressure level (dB re 20 Pa) 40 35 30 25 20 15 10 10 0 0 16 31.96 Problem 4. See following figure. 60 Octave band sound pressure level (dB re 20 Pa) 50 (RC) 40 50 45 30 40 35 20 10 16 31. 97 (b) The levels are plotted on RC curves below.Criteria NR level from following figure (NR curves) = 58. the resulting allowable noise levels and the expected public reactions are: .5 63 125 250 500 1k 2k 4k 8k 30 25 Octave band center frequency (Hz) (c) The noise level is 59.9dB(A). where it can be seen that the spectrum would sound hissy. The dB(A) adjustments to the base level of 40dB(A). (d) The noise reductions between inside and outside and the resulting inside levels are in the table below. . Octave band centre 63 125 250 500 1000 2000 4000 8000 frequency (Hz) Exterior sound pressure levels Expected noise reduction (dB) Interior sound pressure levels 60 5 55 55 5 50 55 8 47 50 10 40 55 14 41 55 16 39 50 20 30 45 21 24 The spectrum in the last line of the table is plotted in the figure below and represents an NR value of 42.98 Solutions to problems day: +20 -2 = 58dB(A) evening: +20 -5 -2 night: +20 -10 -2 (marginal public reaction) = 53dB(A) (little public reaction) = 48dB(A) (medium public reaction) See table 4.11 in the text for public reaction estimates. 25 × 1059.Criteria 99 From Table 4. the expected community response would be widespread complaints. When the windows are open.9 dB(A) .19 From table 4.11 in text. . From table 4. the NR criteria would indicate that it should still only operate during the day.10 + 15 = 45dB(A). 5dB is added to levels in all octave bands and the NR value of the interior noise becomes NR 47.11 in the text. However from the results of (d) above.12 in the text. From the results of (c) above. it can be seen that the factory could now operate in the evening as well.10 in the text. the acceptable level for nighttime operation is LAeq = 40 . (e) Factory could be built provided it only operated during the day. daytime base level = NR 30 and nighttime base level = NR 25. which gives an allowable NR = 35 Thus we would expect complaints during the evening and night but not during the day if the windows are closed. If the noise occurred only 25% of the time: LAeq ' 10 log10 0. but not at night.9 / 10 ' 53. and table 4. Problem 4. Daytime adjustments = +5 +10. This would result in a few complaints during the day and an increase in the number of nighttime complaints. which give an allowable NR = 45 nighttime adjustment = +10. 18 mWatts/m2 (b) prms ' Iρc ' 3.52 ) Watts/m2 = 3. the acoustic pressure and particle velocity may be written respectively as: A p ' j ωρ e j( ωt & kr) r and u ' A r 2 e j( ωt & kr) % jkA j( ωt & kr) e r ' ' A j( ωt & kr) 1 e % jk r r p jωρ 1 % jk r ' p j 1& ρc kr .6 and 1.1 (a) Intensity.5 Solutions to problems relating to sound sources and outdoor sound propagation Problem 5.206 × 343 × 10&3 ' 1. equation 1.18 × 1.40c may be written as: A φ ' ej( ωt & kr) r Using equations 1. (c) For outwardly travelling spherical waves in free space.15 Pa and thus the pressure amplitude = prms/2 = 1.7.62 Pa. I = W/S = 10-2 /(4π × 0. 01 × 343 π × 4002 × 1.832.12 in the text. At 10m.Sound sources and outdoor sound propagation Thus: * u * ' or.32.8945 Pa .38 × 10&3 m 3 / s ' 0.2 dB (e) Lw ' 10 log10 W % 120 ' 100 dB (f) Source strength.13(b) in the text.62 1. kr = 18. kr = 2πf/c = 2π × 100/343 = 1. Q.1(c) above.2 (a) From equation 5.06 mm/ s (d) Lp ' 20 log10 prms % 94 ' 95.s. may be calculated using equation 5.5 2 ' 4.945 Pa and the amplitude at ¯ 10m is p ' 2 × 10&5 × 2 × 1090 / 20 ' 0.206 Problem 5. Thus the sound pressure level at 10m would be 110 20log10(10/1) = 90dB. (b) We can use the result of 5. pressure2 is inversely proportional to the distance2 from the source.206 × 343 343 2 × π × 400 × 0. it can be seen that for a simple source the r. ¯ At 1m. *u* ' *p* ρc 1% c 2πfr 1% 2 101 *p* j *1 & * ρc kr ' 1. The . The acoustic pressure amplitude at 1m is p ' 2 × 10&5 × 2 × 10110 / 20 ' 8. Thus: Q ' 4πW k ρc 2 ' Wc πf 2ρ ' 2.m. 322 relative to the acoustic pressure is: β ' &tan&1 1 kr ' &tan&1 1 ' & 3.5 = 6.8945 1 1% ' 2.102 Solutions to problems result of 5. the acoustic power radiated is: W ' Q 2 k 2 ρc 6. the acoustic power radiated at 100Hz is: W ' Q 2 k 2 ρc 6.1(c) may be written as: u ' ¯ p ¯ ρc 1% 1 (kr)2 Thus at 1m.206 × 343 1.12 in the text. .832 0.6 mm/ s and the phase 1.655 m-1.206 × 343 ' ' 2.79 milli Watts 4π 4π The corresponding sound power levels are calculated using: Lw ' 10 log10 W % 120 dB Thus at 100Hz.945 1 1% ' 24. Lw = 94.206 × 343 ' ' 43.832 m-1 and at 800Hz.5dB.2 mm/ s and the phase 1. k = 14.6552 × 1. k = 2π × 100/343 = 1.012 × 0. u ' ¯ 8.282 × 10&8 × 1.4dB and at 800Hz.8322 × 1.32 Problem 5.282 × 10&8 × 14.6E 1.6 µ Watts 4π 4π and at 800Hz.8322 relative to the acoustic pressure is: β ' &tan&1 ¯ At 10m.28 × 10-4 At 100Hz.3 Source volume velocity = 4πr2urms = 4π × 0. Using equation 5.1E 18.206 × 343 18. u ' 1 kr ' &tan&1 1 ' &28. Lw = 76. 2(c). Problem 5. whereas at 10m from the source. and the amplitude of the pressure fluctuations is then: p ' ¯ Qkρc 1. This indicates that close to the source the acoustic pressure field is dominated by near field effects.79 × 10&4 Pa 4πr 4π × 10 (b) Using the equation from problem 5. β = 6.2E.916.13(a) in the text. is given by: ¯ ¯ Q ' 4 π r 2 u ' 4 π × 0.4E and at r = 10m. The amplitude of the volume velocity.257 × 10&4 m 3 / s At 50Hz.916 × 1.1 ' 1. Q .5 The radiation impedance per unit area is equivalent to the specific acoustic impedance. the near field effects will be small and the field may be approximated as a propagating plane wave. equation 1.5m. For outwardly travelling spherical waves in free space. the phase of the pressure minus the phase of the particle velocity is given by: β ' & tan&1 1 kr At r = 0.4 103 (a) The amplitude of the pressure fluctuations can be calculated using ¯ equation 5.206 × 343 ' ' 3. k = 2πf/c = 2π × 50/343 = 0.257 × 10&4 × 0.40c may be written as: A φ ' e j( ωt & kr) r . the above expression gives β = 65.012 × 0.Sound sources and outdoor sound propagation Problem 5. Z which is simply p/u. 1 to 5.6 Follow the analysis in the text described by equations 5. that is. replace the mean square volume velocity Q2 with the product of half the velocity amplitude and the surface area of the pulsating sphere.7 The wave equation is: 1 M M r2 2 Mr Mr r % M M sinθ 2 Mθ r sinθ Mθ 1 % 1 M2 r 2sin2θ Mψ2 φ & 1 M2φ c 2 Mt 2 ' 0 . setting r = a.104 Solutions to problems Using equations 1. 2 Problem 5. * U *2 Q2 ' ( 4πa 2 )2 . In equation 5. the acoustic pressure and particle velocity may be written respectively as: A p ' j ωρ e j( ωt & kr) r and u ' A r 2 e j(ωt & kr) % jkA j( ωt & kr) e r ' ' A j(ωt & kr) 1 e % jk r r p jωρ 1 % jk r ' p j 1& ρc kr j ka 1 Thus.6 and 1. p j ' ρc 1 & u ka &1 1% ' ρc 1% (ka)2 ' ρc (ka)2 % jka (ka) % 1 2 ' ρc ω2 a 2 % jωac c 2 % ω2 a 2 Problem 5. The required result is then obtained.7.12.12. 32 and 5. (b) Equations 5. This is true provided that r is very large compared to h.33 in the text may be written as: p ' ρ Mφ A cos θ j ' e j(ωt & kr) 1& Mt kr (kr) A cos θ 2 2 j j(ωt & kr) e 1& & 2 kr ρc kr (kr) ur ' & Lφ ' .Sound sources and outdoor sound propagation (a) The solution given by equation 5.25 in the text is: φ ' 2f )(ct & r) (h / r) cosθ 105 Substituting the solution into the various terms in the wave equation gives: Mφ r2 ' & 2f ))(ct & r) (h r) cosθ & 2f )(ct & r) h cosθ Mr 1 M Mφ r2 Mr r 2 Mr ' & 2f )))(ct & r) (h / r) cosθ & 2f ))(ct & r) (h / r 2) cosθ % 2f ))(ct & r) (h / r 2) cosθ sinθ Mφ ' & 2f )(ct & r) (h / r) sin2θ Mθ ' & 2f )(ct & r) (h / r 3) 2 cosθ 1 2 2 M Mφ sinθ Mθ Mθ r sinθ 1 2 M2 r sin θ Mψ2 & 1 M2φ c Mt 2 2 ' 0 ' & 2f )))(ct & r) (h / r) cosθ Adding all the above terms together gives: & 4f )(ct & r) (h / r 3) cosθ which must equal zero to satisfy the wave equation. 106 Solutions to problems Using the first of the above equations and omitting the integration constant: φ ' 1 jA cosθ j e j( ωt & kr ) p dt ' & 1& 2 ρm kr ρck r 0 T Checking the expression for u. by evaluating -Lφ. we obtain: & Mφ jA cosθ j ' & e j( ωt & kr ) 1& Mr kr ρck 2r 2 % jA cosθ ρck r jA cosθ ρck r 2 2 j kr 2 e j( ωt & kr ) j (&jk) e j(ωt & kr ) kr % 1& To verify that the expression obtained above for φ is a solution to the wave equation we substitute it into the wave equation and calculate the result term by term as follows: r2 Mφ A r cosθ 2j 2 ' & 1& e j( ωt & kr ) Mr ρck kr (kr)2 1 M Mφ r2 2 Mr Mr r ' & A cosθ r ρck 2 1& 2j kr 2 1& 2j 2 & e j( ωt & kr ) kr (kr)2 % 4 k 2r 3 e j( ωt & kr ) & A r cosθ r 2ρck A r cosθ r ρck 2 & 2j 2 & (&jk) e j( ωt & kr ) kr (kr)2 ' & A cosθ 1 2 2j & % & jk % e j( ωt & kr ) r ρck r k 2r 3 kr 2 . 33 satisfy the spherical wave equation exactly.32 and 5. Problem 5.35 in the text is: φ ' f (ct & r) r To verify that the expression obtained above for φ is a solution to the wave equation we substitute it into the wave equation and calculate the result term by term. r2 Mφ ' & f (ct & r) & r f )(ct & r) Mr .8 Equation 1.Sound sources and outdoor sound propagation sinθ 1 Mφ jA sin2θ j ' e j(ωt & kr ) 1& 2 Mθ kr ρck r ' 2jA cosθ ρck r M2 2 2 3 107 M Mφ sinθ Mθ r sinθ Mθ 2 1& j e j(ωt & kr ) kr 1 2 r sin θ Mψ2 1 M2φ c Mt 2 2 ' 0 & ' ω2 c 2 φ ' & j k 2 A cosθ ρck r 2 1& j e j( ωt & kr ) kr Adding all the above terms together gives: jA cosθ j 2j 2 2 2j j e j( ωt & kr ) & % %1& % & &1% 3 2 2 3 ρcr kr (kr) kr (kr) (kr) (kr) which is equal to zero. Thus the solutions given by equations 5. 58.584 × (0.08 / 2)2 × 2.108 1 M Mφ r2 2 Mr Mr r Solutions to problems f )(ct & r) r2 f ))(ct & r) f )(ct & r) & r r2 ' 0 ' % 1 M Mφ sinθ Mθ r 2sinθ Mθ 1 2 2 M2 r sin θ Mψ2 & 1 M2φ c 2 Mt 2 ' & ' 0 f ))(ct & r) r Adding all the above terms together gives 0.35 in the text is a solution.0224 watts . Problem 5. of the monopole may be calculated using equation 5. That is: Q ' 4πW k ρc 2 ' 4 π × 0.206 × 343 2 ' 2. The source strength. Q. may be calculated using equation 5. is equal to (2πf/c) = (2π × 250/343) = 4.205 × 343 × 4.58 × 1. WD.29 in the text to give: k 4h 2Q 2 WD ' ρc 3π ' 1. so equation 1.69 × 10&2 m 3 / s The dipole acoustic power.12 in the text.6912 × 10&4 3π ' 0. k.9 The wavenumber.5 4. Sound sources and outdoor sound propagation Sound power level = 10log10W + 120 = 103. r1 . for phase accuracy purposes. r & h cosθ where it has been assumed that h << r. or 1% q2 q1 e 2jkh cosθ0 ' 0.10 For a single source: p(r) ' jωρ q e&jkr 4πr 109 For 2 sources separated by a distance 2h. the total pressure. The above equation may be rewritten as: p(r. p = 0. is the sum of the pressures p1 and p2 from each source. θ) 1 % When θ = θ0. Thus.5dB. Substituting this into the preceding equation for p gives: . p. Problem 5. Thus: p(r) ' p1 % p2 ' jωρ 4π q1 r1 e & jkr1 % q2 r2 e & jkr2 As shown on page 179 in the text. r ' p1(r. θ) ' jωρ & j kr e q1 e& jkh cosθ % q2 e jkh cosθ 4π r q2 q1 e2jkh cosθ r2 . the following approximations are adequate: r1 . q2 q1 ' &e & 2jkh cosθ0 . Noting that for amplitude purposes. r % h cosθ and r2 . 110 p(r, θ) ' p1(r, θ) 1 & e Solutions to problems & 2jkh (cosθ0 & cosθ ) If θ0 = 90E, then p(r, θ) ' p1(r, θ) 1 & e2jkh cosθ Taking the modulus of the preceding equation gives: * p * ' 2 * p1 * 1 & cos(2kh cosθ) This function contains the directivity information and is plotted in the figure above. Problem 5.11 The wavenumber, k, is equal to (2πf/c) = (2π × 500/343) = 9.16. The source strength, Q, of each monopole making up the dipole source may be calculated using equation 5.12 in the text. That is: Q ' 4πW k 2ρc ' 4 π × 0.01 9.162 × 1.206 × 343 ' 1.903 × 10&3 m 3 / s (a) The dipole intensity at θ = 45E is given by equation 5.28 in the text and is: ID ' ρc ' k 4h 2Q 2 (2 π r) 2 cos2θ × 0.7072 1.205 × 343 × 9.164 × (0.005 / 2)2 × 1.9032 × 10&6 (2 π × 0.5)2 ' 3.34 µ Watts / m 2 (b) ¢ p 2 ¦ ' ρcI ' 1.205 × 343 × 3.33 × 10&6 ' 1.38 × 10&3 Pa 2 Sound sources and outdoor sound propagation Lp ' 10 log10 ¢p 2¦ p ref 2 111 ' 65.4 dB (c) Required driving force can be calculated by taking the mean square value calculated using equation 5.39 in the text. Thus: Frms ' 4πaA 3k 2 1% 1 (ka)2 From equation 5.35 in the text: A 2 ' ρchQk 3 2π Substituting in the previously calculated values for k and Q gives: A 2 ' 1.206 × 343 × (0.005 / 2) × 1.903 × 10&3 × 9.163 ' 0.241 2π As ka is very small, equation 5.40 in the text may be used. Thus: Frms ' A 4π 2 3k 2 ' 0.241 × 4π 3 × 9.162 ' 12.0 mN Problem 5.12 The arrangement is illustrated in the figure. As R is 1m off the floor, the distance to it is 20 % 1 ' 20.02 m 2 2 R 20m 0.5W 0.1m 1W 0.1m 0.5W h = L =0.05m (a) The strength of each source may be calculated using equation 5.12 in the text: W 4π Q2 ' M ρck 2 112 Solutions to problems In this case, WM = 0.5W and k = 2π × 125/343 = 2.290. Thus: Q2 ' 0.5 × 4 π 1.206 × 343 × 2.292 ' 2.897 × 10&3 The arrangement shown is a longitudinal quadrupole and the mean square sound pressure at any location may be calculated using equations 5.54 and 5.55 in the text. Note that the equation for the mean square pressure is multiplied by 2 in this case because the radiation is into half space. Thus: ¢p ¦ ' 2 5ρc Wlong cos4θ 4πr 2 &3 ×2 ' Q 2 ρck 3hL cos2θ πr 2 ×2 2 ' 2.897 × 10 1.206 × 343 × 2.293 × 0.052 × cos2 (30) π × 20.02 ×2 ' 1.271 × 10&4 Pa 2 The sound pressure level is then: Lp ' 10 log10 1.271 × 10&4 4 × 10&10 ' 55.0 dB (b) 125Hz random noise will make the sources act independently and the power radiated will be the arithmetic sum of the individual sources. Thus W = 1.5W and Lw ' 10 log10 (1.5 / 10&12 ) ' 121.8 dB . Using equation 5.108 in the text, the sound pressure level may be calculated using S = 2πr2 and ρc = 413.6. Thus: Lp ' 121.8 & 10 log10(2 π × 20.022) & 10 log10(400 / 413.66) ' 87.9 dB Sound sources and outdoor sound propagation Problem 5.13 (a) T h e arrangement approximates a simple dipole and is illustrated in the figure. 0.2m The angle β = sin-1 0.1/5 = 1.145E. The azimuthal angle given in the problem is irrelevant. The wavenumber, k = (2π × 250)/343 = 4.58. speaker hole 5m O 113 0.1m The sound pressure levels radiated by the hole alone (monopole) and hole + speaker (dipole) may be calculated using equations 5.13a, 5.30b and 5.29 in the text. Using these equations, the ratio of the mean square pressures (dipole/monopole) may be written as: ¢ pM ¦ 2 ¢ pD ¦ 2 ' 4(kh)2 cos2(90 & 1.145) ' 4(4.58 × 0.1)2 (0.02)2 ' 3.36 × 10&4 Thus the reduction in sound pressure level due to the presence of the speaker is: Reduction ' 10 log10 3.36 × 10&4 &1 ' 34.7 dB (b) If a speaker is placed below the hole as well, a longitudinal quadrupole is formed with L = h = 0.1. In this case, β = 0, and as can be seen from equation 5.55 in the text (where θ = 90 - β), the theoretical mean square sound pressure will be zero, implying a reduction of infinity dB. Problem 5.14 In equation 5.62, (W/b) is effectively the power per unit length of source (as W is the power of each source separated by b) and in equation 5.70, W/D is the same quantity. Thus the difference between the finite length and infinite 114 Solutions to problems length source is the quantity ( αu & αl ) / π which is the ratio of the angle subtended by the source at the observer in each case. Thus, by logical argument, equation 5.65 can be rewritten as follows for a finite coherent line source. ¢ p 2 ¦ ' ρc W 2 π r0 D 2 [αu & αl] Problem 5.15 pipe 20m 80m u l A The arrangement is as shown in the figure. αu ' & αl ' tan&1(10 / 80) ' 0.124c Finite length pipe, D = 20m, r0 = 80m and Lw = 130dB. Turbulent flow, so assume an incoherent source. Also assume incoherent addition of the direct and ground reflected waves. Taking logs of equation 5.70 in the text gives for the direct wave: Lp ' Lw & 10 log10 (4πr0D ) % 10 log10 ( αu & αl ) % 10 log10 (ρ c / 400 ) ' 130 & 10 log10 (4π × 80 × 20) % 10 log10 (0.248) % 10 log10 (1.0342) ' 81.1 dB The ground reflected wave level is then 81.1 - 3 = 78.1dB. Thus the total Sound sources and outdoor sound propagation level at the receiver is: Lp ' 10 log10 108.11 % 107.81 ' 82.9 dB Problem 5.16 115 The situation is as shown in the figure below. Equation 5.70 in the text may be used to calculate the sound pressure level. The equation must be multiplied by the directivity factor (2 in this case). pipe 50m 200m u l A Thus: ¢ p 2 ¦ ' [ Wρc / 4πr0D ] [ αu & αl ] × 2 25 ' 0.124 radians . 200 αu ' αl ' tan&1 W = 2, r0 = 200, D = 50. Thus: ¢ p 2 ¦ ' [ 2 × 1.206 × 343 / ( 4 π × 200 × 50 ) ] × 0.249 × 2 ' 3.275 × 10&3 Pa 2 Lp ' 10 log10 3.275 × 10&3 4 × 10&10 ' 69.1 dB Ag + Aa = 1. so for 200m.19.3dB Problem 5. of one vehicle by: ¢ p1 ¦ ' ρc 2 W 4πr0 2 × DF where DF is the directivity factor for the source/ground combination.1 .17 The traffic may be treated as an infinite line source.2dB(A).1 dB Thus.3/5 = 3. of one vehicle by: ¢ p2 ¦ ' ρc 2 W × DF 4br0 Thus: ¢ p1 ¦ 2 2 ¢ p2 ¦ ' br2 π× 2 r1 ' 6 × 50 ' 95. At 1m r0 < b/π and the mean square sound pressure is related to the source sound power.8dB and the sound pressure level at the receiver is: Lp = 69.10log10 ¢ p2 ¦ = 88 . W.3 in the text. Aa = 19. the sound pressure is related to the sound power.116 Solutions to problems From table 5.9dB. Thus the ground effect is given by equation 5.1. Aa = 19.8 = 67. Sound intensity loss due to ground reflection is 2dB.8 = 68. W.5 π ¢ p1 ¦ 2 2 The level at 50m = level at 1m . At 50m.175b in the text as: Ag ' &10 log10 [ 1 % 10&2 / 10 ] ' &2.3dB per 1000m. . 6 % 3.19 (a) The arrangement is shown in the figure. &3 dB) Problem 5. So the sound pressure level at the residence is: Lp ' 84. -3 dB.Sound sources and outdoor sound propagation Problem 5. For 250 m air absorption 0. air absorption ranges from 2.7 dB. The sound power is then: 2 r .118 ( 2 × 10&5 )2 ' 84.62: ¢p 2¦ ' ρc W 413 × 2 ' 0.7 dB 117 Lp ' 10 log10 Concrete ground. From table 5. Assume no obstacles blocking line of sight to the road from the residence.18 From Equation 5.6 to 2.3. meteorological influence is +6. Assume 20EC.0 ' 87 dB (% 6. We are given: I ' (¯o / 3ρcr 2) (2 % cosθ) p The sound power is obtained by integrating the intensity over an imaginary hemispherical surface centred at the centre of the speaker.7 & 0.118 Pa 2 ' 4 b r0 4 × 7 × 250 0. so ground effect is Ag = -3 dB.8 dB per 1000 m. (b) The function 2J1(x)/x vs x is plotted out in the figure below. This effectively means that an expression for the sound pressure at some distance. r. the presence of a baffle will have no influence on the power radiated. & 2 cosθ & 0.0127p0 Watts ¯ 2 (b)& (c) As the speaker only radiates into a hemispherical space. regardless of whether the source is constant volume or constant pressure.20 (a) If the piston is assumed to be made up of an infinite number of point monopole sources. due to a monopole on the piston surface must be integrated over the piston surface. . all pulsating in phase. the monopole source must be replaced with a hemispherical source which radiates twice the pressure. then the sound pressure at any location can be calculated by summing the contributions from each source.25 cos2θ 0 ' 2 % 1 / 2 ' 0. If a baffle is present. Problem 5. where x = ka sinθ.118 m S Solutions to problems 3ρc m 0 W ' I dS ' 2 p02 ¯ 2π π/2 dψ m 0 (2 % cosθ) r 2 sinθ r2 dθ p0 2 π π / 2 ¯ ' (2 sinθ % cosθ sinθ ) dθ 3ρc m 0 ' p0 2 π ¯ 3ρc p0 2 π ¯ 3ρc 2 2 π/2 . etc. 7.1.7 in the text) may be used with the x-axis crossings representing the nodal locations of each lobe and the peaks in the curve representing the relative amplitude of each lobe.8. 23E. Directivity patterns are shown for each of these cases in the figures on the next page.4. For sketching purposes. The side lobes for the 10kHz case have been expanded for clarity. At 10. ka = 2π × 2. 16. etc.000 × 0.Sound sources and outdoor sound propagation 1.91.1. 13.6.58. Nodes in the radiation pattern occur when 18.0 2J1(x) x 119 0. the figure shown at the beginning of part (b) (figure 5. That is.6 will only provide information for the first three lobes so it must be extended for the 10kHz case.8.500Hz. ka = 2π × 10. 7.1/343 = 0. Nodes in the radiation pattern occur when 4.31 sinθ = 3. 7. and 19.58 sinθ = 3. 33E. there is only one node at θ = 60E.500 × 0. That is.91 sinθ = 3.3. At 2.] .000Hz.8.5 0 5 10 15 20 x At 500Hz. 10. 10. 47E and 64E. there are 5 nodes at θ = 12E.1/343 = 4. That is. Nodes in the radiation pattern occur when 0.1/343 = 18. [Note that fig 5.31.1. ka = 2π × 500 × 0. 10. there are no nodes. 8.21 (a) Using equations 5. (b) The radiation efficiency at low frequencies is: σ ' RR ρcS ' πa 2ω2 2πc 2 ' (ka)2 / 2 which described the solid line in figure 5.9 in the text.22 (a) Piston radiating from an infinite baffle.95b and 5.98b in the text: W ' RR πa 2ρcU 2 / 2 . (c) See fig 5.120 Solutions to problems 500Hz 2. k = 2πa/λ = 2 and a = 0. Problem 5.5kHz 10kHz Problem 5.1m. the radiation resistance for a piston can be written as: RR ' ρcπa 2 × (2ka)2 / 8 ' ρc(πa 2)24ω2 / (8πc 2) ' ρcS 2ω2 / (2πc 2) where it has been assumed that ka is sufficiently small that all but the first term of equation 5. From equation 5.96 in the text.96 in the text is negligible.9 in the text for ka < 0. the on-axis intensity is: I ' ρck 2 8π r 2 2 F 2(w) where w = kasinθ = 0. RR = 1.8 in the text).0002 × 343 / 0. is given by: ¯ ¯ U ' ξω ' 2¯ / λ ' 2ξc / a ' 2 × 0. U. Thus: I ' ρch 2U 2π2a 4 8π2r 2 ' ρcU 2a 2 8r 2 ' 0.0 dB Problem 5.55 ' 7. Thus.01 2 × 22 (c) Radiation mass loading = πa2ρc[X(2ka)]. The piston velocity amplitude.372 m/ s ξc Thus the radiated power is: W ' π × 0.Sound sources and outdoor sound propagation 121 and from fig 5.206 × 343 × 1.12 × 1.3722 × 0. for ka = 2. F(0) = Uπa2.5mm.97 W / m 2 ' 1.01 × 1.84 in the text.1 ' 1.2 W (b) From equation 5.1 kg / s (d) Sound pressure level at 2m: Lp ' 10 log10 ¢p 2¦ 2 p ref ' 10 log10 ρcI 4 × 10&10 ' 120. Thus the mass loading is π × 0.3722 / 2 ' 12.206 × 343 × 0.8.23 (a) The arrangement is shown in the figure on the next page where it can be seen that h = L = 102.55 (see fig 5. .206 × 343 × 1. where X(2ka) = 0. 12 in the text and multiply by 2 to account for radiation into half space): WM ' ( QH ρck 2 / 4π) × 2 2 The power radiated by the longitudinal quadrupole may be calculated using equation 5.122 Solutions to problems 200mm speaker hole speaker 205mm (b) Power radiated by original opening (assuming a constant volume velocity source in an infinite baffle) is (see equation 5. Thus: Wlong ' [ ( 2k 3hLQL )2 ρc / 5π ] × 2 where QL = QH/2.1025)4 ' 9.94 × 10&12 f 4 5 3434 Frequency. Hz 63 125 250 500 dB reduction (-10log10(Wlong/Wm) 38 26 14 2 .54 in the text. Thus: Wlong WM ' 4k 6h 2L 2QH ρc4π 2 20πQH 2 ρck 2 ' 4 4 2 2 k h L 5 ' 4 (2π)4 f 4 (0. Again the equation in the text must be multiplied by 2. (b) The expected sound pressure level due to an omni directional source on a hard floor is: Lp ' Lw & 10 log10 [4 π r 2] ' 120 & 10 log10 [2 π × 100] ' 92 dB The actual sound pressure level is 110dB so the directivity due to the source characteristics is -18dB.55 in the text.Sound sources and outdoor sound propagation 123 (c) The ratio of the mean square pressures may be obtained using equations 5. 3π/2. frequency (Hz) 63 125 θ=0 31 19 θ = π/4 37 25 θ = π/2 4 4 When the speakers are turned on. Problem 5. . π and minima at θ = π/2.13b and 5. Thus: ¢ p long ¦ 2 2 ¢ pM ¦ ' [ 5ρcWlong cos4θ / 4πr 2 ] × 2 [ WM ρc / 4πr ] × 2 2 ' 5 cos4θ Wlong WM The reduction in sound pressure level is then: ΔLp ' & 10 log10 [5cos4θ Wlong / WM ] Values of sound pressure reduction (dB re 20µPa) are tabulated below for the required values of θ.24 (a) The directivity index due to the hard floor is 3dB. the sound field amplitude distribution has 2 lobes with maxima at θ = 0. 25 Solutions to problems 25m hole The radiated sound power is W = IS = 0.5. Thus: ¢p 2¦ ' 2ρcW &1 HL tan πHL 2 2r H % L 2 % 4r 2 2 × 1.01S Watts. the total sound pressure level at the receiver is: Lp ' 10 log10 2 × 2. With no ground reflection. H = L = 0.25 % 0.25 tan S×π 2 × 25 0.105 in the text (as we may assume that the opening behaves like an incoherent plane source).25 % 4 × 252 ' ' 2.01 × S = 0.63 × 10&4 Pa 2 Assuming a similar travel distance for the ground reflected wave. . S = HL and r = 25. The 2m arrangement is shown in the figure. Here.105.01 × S &1 0.124 Problem 5.633 × 10&4 4 × 10&10 ' 61.106 had been used instead of equation 5.206 × 343 × 0. in this case. the on-axis sound pressure level may be calculated using equation 5. the receiver is sufficiently far from the source for the source to appear as a point source and the same result would have been obtained if equation 5.2 dB Interestingly. From table 5.26 (a) Sound power level. Aa = 15. From similar triangles. so β = 1.3dB. For grass covered ground.206 × 2000 ' ' 0.011 1/2 ' 16. 125 150m hole 3m R x y 1.72 1 0.3dB. .5 × 0.4 dB Thus the effect of the ground is to increase the level at the receiver by 2. R1 = 2. x = 2y. So for a distance of 150m.Sound sources and outdoor sound propagation Problem 5. Aa = 15.011 R1 2.72E.3 in the text.5dB per 1000m (25% RH and 20EC). Lw = 10log10W + 120 = 123dB.6 From figure 5. The ground effect is then: Ag ' &10 log10 1 % 10 &A R / 10 ' & 2.20 in the text.25 × 105 (middle of range). (c) Loss due to atmospheric absorption. Thus x = 100 and y = 50. tanβ = 3/100.15 = 2.5m (b) The arrangement for calculating the ground effect is shown in the figure.4dB. AR = 1. Thus: ρf 1.25 × 105 β R1 ρf 1/2 ' 1. 852 × 10&3 4 × 10&10 ' 71.11) and we may use equation 5.852 × 10&3 Pa 2 The sound pressure level is then: Lp ' 10 log10 5. thus the sound pressure level at the receiver is equal to 71. However as we found in the previous problem. the receiver is far enough from the source for it to appear as a point source (see figure 5. Thus: ¢p 2¦ ' 1.27 10000 x 5 r1 r2 r 90 10000 . Thus the sound pressure level at the community location of (b) above would be 74. Using similar triangles: .1.126 Solutions to problems (d) The opening should be treated as an incoherent plane source and equation 5.x The arrangement is illustrated in the figure.8dB.206 × 343 × 2 2π × 1502 ' 5.8dB.7 dB Aa + Ag = -0. Problem 5.106.105 in the text used to calculate the sound pressure level. (e) Adding a second opening will add 3dB to the sound pressure levels at the receiver as it may be assumed that the sound fields from the two sources are incoherent. f.0900 m Destructive interference will occur if λ/2 = 0. Also the machine is far enough from the wall for its sound power to be unaffected by the wall.3 kHz λ 2 × 0. The reflected wave path length is 6m and the direct path length is 2m.31579)2 ' 9474.31579 m 127 r1 ' 52 % (526. the level at the operator's position may be calculated as the logarithmic sum of the direct and reflected waves.33954 r2 ' 902 % (10000 & 526. the operator is not in the hydrodynamic near field of the machine (although he/she could be in the geometric near field).111701 r ' 852 % 100002 ' 10000. As the sound is predominantly in the 500Hz to 2000Hz band.315)2 ' 526.0900 Problem 5.0900m. Thus the sound pressure level of the reflected wave is .36124 r1 % r2 & r ' 0. This occurs at a frequency.28 We would NOT expect to measure 84dB(A) at the operator's position due to reflected energy from the nearby wall. assuming that geometric near field effects are negligible assuming that the machine does not act as a barrier to the reflected sound and assuming that the loss on reflection from the wall is negligible. given by: f ' c 1500 ' ' 8.Sound sources and outdoor sound propagation x 10000 & x ' 5 90 90x ' 50000 & 5x x ' 50000 / 95 ' 526. Assuming incoherent addition of direct and reflected waves. *Rp*2. when . Also.5 dB(A) Problem 5.29 (a) Specific acoustic impedance is a complex quantity characterised by an amplitude and a phase and is the complex ratio of acoustic pressure to acoustic particle velocity at any point in an acoustic medium. cosψ = 1 and: * Rp * ' * (Zs / ρc) cosθ & 1 * * (Zs / ρc) cosθ % 1 * The maximum absorption coefficient will occur when the modulus squared of the reflection coefficient is a minimum. it is a real quantity and is equal to the specific acoustic impedance of a plane wave propagating in an acoustic medium of infinite size.45 ' 84.129 in the text may be used with θ = 0 to give: Rp ' (Zs / ρc) & 1 (Zs / ρc) % 1 If θ is not equal to 0. that is.4 % 107. For an acoustic medium where the viscous and thermal losses are small (such as air or water). including the interface between two different media. then Rp ' (Zs / ρc)cosθ & cosψ (Zs / ρc)cosθ % cosψ If it is assumed that the wavenumber in the material is much larger than that in air. Thus the total sound pressure level expected at the operator's position is: Lp ' 10 log10 108.5 dB .128 Solutions to problems 84 & 20 log10(6 / 2) ' 74. the normal specific acoustic impedance of the surface of an acoustic medium of infinite extent is the characteristic impedance of the medium. Characteristic impedance is equal to ρc. Thus equation 5. for an infinitely thick medium. as α = 1 . (b) The absorption coefficient (assuming plane incident waves) is defined in terms of the reflection coefficient. Rp. 5 % 3 & 2. so Ag = -3 dB Assume 20 EC temperature.30 Power radiated by window = 0.2774 % 1 ' 1 & 0.8.10.Sound sources and outdoor sound propagation * Rp * 2 ' (2 cosθ & 1) 2 % 9cos2θ (2 cosθ % 1) 2 % 9cos2θ 13 cos2θ & 4cosθ % 1 13 cos2θ % 4cosθ % 1 129 ' is a mimimum. Use Aa = 2. we obtain the minimum value when cosθ = 0. &6 dB ) Problem 5. so Aa ranges from 2.686α (dB / m) Important factors are air temperature and humidity.75 & Am ' 45.0769 % 4 × 0.4343 ln eα ' 8. The decibel decay rate per unit distance is 20log10 of the reciprocal of the above expression when x = 1 and is thus given by: Decay rate ' 20 log10 eα ' 20 × 0. Range due to meteorological conditions is: (+8.7.6 to 2. Thus: Lp ' Lw & 10 log10 ( 2πr 2 ) & Ag & Aa & Am'110 & 65. -6 dB) from Table 5. Assume no barriers between the source and receiver.5 dB (%8. Differentiating the above expression wrt cos2 using the chain rule.2774 or θ = 74E.71 13 × 0. .0769 & 4 × 0.7 × 0.31 The pressure amplitude at any location x is given by: p ¯ ' e& αx p0 ¯ ¯ where p0 is the amplitude at x = 0.2774 % 1 Problem 5. The corresponding maximum value of the absorption coefficient is given by: α ' 1 & * Rp * 2 ' 1 & 13 × 0. The receiver is far enough away for the window to be treated as a point source in a baffle. and setting the result equal to zero.1 W = 110 dB Concrete ground.286 ' 0. 190 in the text.32 Solutions to problems The situation is illustrated in the figure below.130 Problem 5. The distance d is then: r . the radius of curvature of the wave is thus 343 × 10 = 3430m. Using equation 5. of the wave and hence the distance. We may use the aircraft as the reference frame. r. We need to find the radius of curvature. d. The total sonic gradient is 1/10 = 0.600 r A2 600 . The wave which hits the ground at grazing incidence will be the one heard first.1s-1. Thus we assume a coordinate system moving horizontally at the speed of the aircraft and then later on calculate the distance that the aircraft travels during the time it takes for the sound to reach the ground (with an assumed aircraft speed). 1h 0 0 θ0 θ0 ' 10 × dθ 34. where h = height above the ground and c0 = 343m/s is the speed of sound at ground level. then it would travel 710m in 6.0.1 cosθ m c0 & 0. The speed of sound as a function of θ (with θ = 0 corresponding to ground level and θ = θ0 corresponding to the aircraft level) is given by cθ = c0 .4E 3430 Thus the time taken for the sound to travel from the aircraft to the ground is r dθ dθ ' t ' m (c0 / r) & 0. The value of θ0 is given by θ0 ' cos&1 3430 & 600 ' 34.1h. .4E ' 6.710m) = 1230m from the observer. Thus the aircraft emerges from ground shadow (1940m .Sound sources and outdoor sound propagation d ' r 2 & (r & 600)2 ' 34302 & 28302 ' 1940 m 131 We now need to take into account the speed of the aircraft and the distance it will travel in the time the sound wave travels to the observer.4 seconds ' 10 loge [ secθ % tanθ ] 0 m cosθ 0 θ0 If the aircraft travels at 400km/hour (not given in the question).1 % 0. Let the angle subtended at the centre of the circular arc shown in the previous figure be θ0.4 seconds. For a constant power source in the corner of the room. However. the radiated power would be concentrated over a one eighth sphere instead of a sphere and the sound pressure level in the direct field would thus be increased by 9dB. there would be no change in the reverberant field sound pressure level. its use and measurement Problem 6. (c) A good approximation would be a constant pressure source model because the noise is originally generated by a fluctuating pressure.6 Solutions to problems relating to sound power. For a constant volume velocity source in the corner of the room.1 (a) (b) This is discussed in detail on p246–247 in the text. but the reverberant field sound pressure level would be reduced by 9dB. the amplitude of which is controlled by the aerodynamics and not the acoustics of the problem. corresponding to the power increase. . For a constant pressure source in the corner of the room. the radiated power would be increased by a factor of 8 due to there being 3 reflecting surfaces and in addition the power would be concentrated over a one eighth sphere instead of a sphere and the direct field sound pressure level would thus be increased by 18dB. corresponding to a reduction of 9dB in the radiated power. However. the reverberant field sound pressure level would be increased by only 9dB. the direct field radiated sound pressure level would be unchanged. (d) The desirable quantity is usually sound power level as it indicates the amount of acoustic energy which will be added to an environment and allows the increase in sound pressure level to be calculated at . Beginning with W ' ¢ p 2 ¦ S / ρc and taking logs of both sides gives: 10log10W ' 10log10¢ p 2 ¦ % 10log10 S & 10log10(ρc) (b) Dividing both sides by 10-12 and remembering that the reference sound power level is 10-12W and the reference sound pressure level is 2 × 10-5Pa. the preceding equation may be written as: W 10log10 Wref ' 10log10 ¢p 2¦ p ref 2 % 10log10 S % 10log10(400) & 10log10(ρc) If the quantity ρc is approximated as 400.2 (a) 133 Sound power level is a measure of the rate of total energy radiated by an acoustic source while sound pressure level is a measure of the fluctuating sound pressure at a particular location.Sound sources and outdoor sound propagation Problem 6. Sound power level is a source property whereas sound pressure level depends on the measurement location as well as the strength and size of the source. the sound power level would be: Lw ' Lp % 10log10 S ' 85 % 10log10(2 π × 22) ' 85 % 14 ' 99 dB Assumptions: ρc = 400 and the sound pressure level measurements were made in the acoustic far field of the machine. then the preceding equation becomes: Lw ' Lp % 10log10 S (c) Using the above equation. r > 3λ / 2π.134 Solutions to problems any location in the far field of the source as a result of the introduction of the machine (provided the sound radiation is omnidirectional or directivity information is also given).5.250 in text).1 in the text). so γ = 2 and κ = π × 550/343 .4 (a) The anechoic room should be sufficiently large that sound pressure measurements can be made in the far field of the source. if only the noise exposure of the operator of a machine is of concern. (c) Problem 6. for the far field to be dominant. Sound power is also independent of the presence of nearby reflecting surfaces. 5. 550 Hz r = R = 1.5m is the distance at which the far field would be dominant. and therefore 2r / R ' 10 λ / (π R ) ' 10 × 343 / (π R f ) f ' 10 × 343 / (2 π r) ' 3430 / (2 × π × 1) .2 on p43 in the text) and the highest frequency is 11. then perhaps it is better to specify sound pressure level at the operator's location measured in the presence of specified reflecting surfaces than sound power level as the operator may not be in the acoustic far field of the source. p. However.3 (a) (b) This is discussed in detail on pages 249-251 in the text. γ = 15. . r > 3R. r > 3π R2 / 2λ The lowest frequency of interest is 44Hz (see table 1. 6.300Hz. Problem 6. Thus the following criteria should be satisfied (see eq. provided that these are more that a quarter of a wavelength from the acoustic centre of the machine or noise source. On the other hand.1. Thus r = 7. the frequency would be given by γ ' 10 / κ (see figure 6. Assuming that "negligible" is a factor of 10. From figure 6. the sound pressure level at a specified location is affected by the presence of reflecting surfaces and also does not necessarily allow the sound pressure level at other locations to be calculated. 2 × 6 (4.8 0.2. the following minimum interior room dimensions are needed. the required room dimensions are 14.6m.8 / 4) > ' 14.36 349 3πR2 /2λ R = 0.8 / 4) by (3.4m × 14.8m × 5.3 m As standard measurements use a hemispherical surface.5 The sound power is related to the average mean square pressure by: W ' ¢p 2¦S ¢p 2¦2πr 2 ' ρc ρc or in terms of sound power level.015 The last criterion in the above table for r represents the transition from the geometric near field to the far field.03 3R R = 1.Sound power use and measurement frequency λ (m) 3λ/2π (Hz) 44 11.1 24.5 1.8 0.5 % 0.5. Lw: . the required room volume is: 1. Thus the required dimensions are 3.7m × 9.91m.4m × 7.72 % 0.6 % 7.6 0. or x = 1.4 m × 11. (b) For measurements in octave bands. As all measurements must be taken at least λ/4 from the room walls.7 m × 6.300 7. 1.75 % 7.72 4.9 135 3.72 % 0. Thus 2x × 3x × 5x = 210.2m high. the geometric near field will suffice.2 % 7.4 R = 0. 1. Thus for high frequencies it is not practical to take measurements in the far field.3 λ3 ' 1.3 343 63 3 ' 210 m 3 Optimum dimensions are in the ratios 2:3:5.8 / 4) > by (3.22 55.8 0. Problem 6. 042 I ' ' ' 1.6 (a) r. The distance r must be larger than the following quantities for the location to be in the far field: 3 λ / 2π. λ = 1481/250 = 5. so the location is not in the far field but rather in the transition between the hydrodynamic near field and the far field. for a spherical source by equation 5.13a in the text as: (Qkρc)2 ¢p 2¦ ' (4πr)2 (d) (e) . r = 2m.6 ' 83. acoustic pressure: prms ' 2 × 10&5 × 10 (b) L p / 20 ' 2 × 103.9 dB The sound power. (c) Intensity: 2 p rms 0. 3 R.027 µW The mean square sound pressure is related to the source volume velocity.92m.1m.08 × 10&9 × 2 × π × 22 ' 0.245 mW Problem 6. 0.m.3 & 5 ' 0.s. W ' I S ' 1.04 Pa Source dimension. R = 0. Q.82.08 nano&watts / m 2 ρc 998 × 1481 Power. wavelength.39 ' 0. W. is thus: W ' 10&12 × 10 L w / 10 ' 10&12 % 8.008 respectively.3 and 0.136 Solutions to problems Lw ' Lp % 10 log10(2πr 2) % 10 log10400 / ρc ' 70 % 10 log10(8π) % 10 log10400 / 413. and 3 π R2 / 2λ Substituting values for λ and l into the preceding expressions and evaluating gives 2. 0 1.4 4.9m3 and the surface area.2 0.69 0.7 + 3.6 106.4 96. The volume velocity. is 2(3.8 7.3 . by: d ' 2Q 2πf × 4πa 2 ' 1.7 Equation 6.4 1.9 dB From problem 6.8m × 5.5 110.1 0.1 103.8 × 5.6 93.4 95. Thus the same result is obtained as for a spherical source radiating into spherical space.582 r p rms d ' ρf 2 Substituting values for the variables gives: d ' 3.8 × 9.2 99.4 6.37 0.5 5.7 0.13 in the text is: Lw ' Lp % 10 log10 V & 10 log10 T60 % 10 log10 1 % Sλ 8V & 13.9 102.44 2. V.Sound power use and measurement 137 In this case.17 0. is related to the surface displacement.086 0.6) = 225.791 Q / f Combining the above two equations gives: 3. Using the above equation.4 0. is 207.34 0. Q.7m2. S.043 10log10(1+Sλ/8V) 2. The volume. the source is a hemisphere and it is radiating into hemispherical space.4 0.6 × 10&9 m Problem 6.7m × 9.4 the room size is 3.1 8.6 + 5.1 0.582 × 2 × 0.7 × 9. d.0 Lw 87.0 3.8 λ 5.6m. the following table may be constructed.74 1.04 998 × 2502 ' 4. f 63 125 250 500 1000 2000 4000 8000 Overall Lp 85 105 100 90 95 98 90 88 10log10T60 9. 828 × 0.7 85. Ratio of area of measurement surface to machine surface = 40/8 = 5. 6.9 The data may be used to construct the following table. The machine surface area is then Sm = 1. Assuming that the machine is in a large enclosure.0 + 108.5 + 108. p.8 83.7 84. and the factory volume V = 20 × 20 × 5 = 2000m2.138 Problem 6. Thus V/S = 50 and from table 6.25 in the text which is: Lw ' Lp % 10 log10 S & Δ1 & Δ2 where S = 40m2.828m × 0. Then 5x2 = 40 and x = 2.7 Lp(av) ' 10 log10 Lpi / 10 1 j 10 N i ' 86.828m. Machine surface area is the area of a cube of dimensions 1m smaller than the test cube. Δ1 = 2.4 88. measured background 85 80 88 82 86 80 90 81 84 80 85 79 87 81 88 79 89 80 90 81 90 88 83 83 87 82 88 80 89 80 85 79 machine 83. 86.7 89.828 × 0.3 + 108.3 87 only 3 88.4 89 86.3 85.8 + 10log1040 = 100.828m high (as the machine is resting on the floor). Let the side of the test cube = x.828 = . so Δ2 = 0.7 87.4.8 dB The sound power level may be calculated using equation 6.25 in text).828 × 4 + 0. 266 in the text.5dB. Lw ' Lp % 10log10S & Δ1 & Δ2 (eq.7 + 108. Δ1 = 0 as well.8dB Area of measurement surface = 2(2 × 4 + 2 × 3) + 3 × 4 = 40m2.8 Solutions to problems Average Lp = 10 log10 {(1/5)[108.8dB.4 83. Thus Lw = 84. Thus the machine size is 0.4 89.4 81. Problem 6.6]} = 84.828m × 1. Thus radiated power. 8 % 10log10 40 & 3 . Lw is calculated using equation 6.24 in the text which is: Lw ' Lp2 & 10 log10 [ S1 & 10 log10 Thus: Lw ' 84 & 10 log10 188&1 & 428&1 % 10log10 102/10 & 1 & 10log10 ' 84 %25.7 400 &1 & S2 ] % 10 log10 [ 10 &1 ( Lp1 & Lp2 ) / 10 & 1] ρc 400 . Area ratio (surface1 to surface2) = 40/120 = 0.25 & 2.22 in the text: 4 (1 / 188) & (1 / 428) [102 / 10] ' ' 2.Sound power use and measurement 139 6.8dB.11 Machine surface area = 2(8 × 3 + 4 × 3) + 8 × 4 = 104m2. Assuming ρc = 400: Lw ' 86.3 or equation 6.8.9 and from table 6. Δ2 = 0.8 dB re 10&12W 413.3. Using equation 6. Sound power level.763 × 10&3 R 102 / 10 & 1 Thus R = 1448 m2. From figure 6.3 dB re 10&12 W Problem 6.8 % 10 log10 40 & 2.5 ' 100.6dB and from problem 6. p266 in the text.33 & 0. Δ2 = 0.3333.8 . Area of test surface 1m from machine = 2(10 × 4 + 6 × 4) + 10 × 6 = 188m2 = S1. Area of test surface 3m from machine = 2(14 × 6 + 10 × 6) + 14 × 10 = 428m2 = S2. Thus: Lw ' 86.10 Second surface average = 86. Δ1 = 2.7m2. 100 dB re 10&12W Problem 6.15 ' 106.7 = 5. Thus S/Sm = 40/6.27 in the text.2 = 84. 3 in the text. Surface 1.4 & 1 ] % 10 log10 [ 1 & 96 / 280 ] ' 0. Δ2 = 1. Thus: Lp2 ' 10 log10 1 108. Machine surface area.5 × 2) + 6 × 6 = 96m2.5 & 85.68 % 108. Thus from table 6.140 Solutions to problems However.6 / 10 & 1080 / 10 ' 85.5 & 10 log10 [ 100. S1 /Sm = 96/65 = 1.65 11 % 108. which for this case is -1 dB. First calculate the areas of the test surfaces. Sm = 2(5 × 2 × 2) + 5 × 5 = 65m2.6 % 108.72 % 108. Problem 6.4 dB (c) Correction for non-normal sound propagation is Δ2. Thus the noise level on test surface 1 due only to the machine is: Lp1 ' 10 log10 1090 / 10 & 1080 / 10 ' 89. .6 dB The background level may be subtracted from the overall averaged levels as this will give the same result as subtracting it from the individual levels and then averaging. S2 = 2(10 × 4.48.58 % 108.12 (a) The average noise level measured on the larger surface is calculated by logarithmically averaging the given values.5 dB (b) Reverberant field correction.25) in the text. So if we used equation 6.53 ' 86.5 dB and the noise on test surface 2 due only to the machine is: Lp2 ' 10 log10 1086.5 % 108.25 and also applied the correction due to ρc not equal to400.6 % 108.5 × 2) + 10 × 10 = 280m2. Thus: Δ1 ' 89.8 % 108. use equation 6.27 in the text.7 % 108. S1 = 2(6 × 2. Δ1.75 % 108. Surface 2. this excludes the near field correction term of equation (6.8 dB. the result would be Lw = 105. 5 & 0. ! Allows good results to be obtained at low frequencies.5 % 10 log10(96) & 0. Maximum allowable total reverberant sound pressure level generated by the three new machines is: Lp ' 10 log10 1085 / 10 & 1082 / 10 ' 82 dB(A) Thus the maximum allowable sound power level is 87. Disadvantages of sound intensity for sound power measurement: .4 dB(A) ρc Lw ' Lp % 10 log10(2πr 2) % 10 log10 (b) (c) Sound power level of existing machinery is 92.9 dB Problem 6.1 ' 92.4 & 10 log10 3 ' 82. (d) If all new machines emit the same sound power level and the total allowed is 87. Lw is calculated using equation 6.4dB(A).14 (a) Advantages of sound intensity for sound power measurement: ! reduces errors arising from presence of reflecting surfaces. ! reduces errors from near field effects resulting from measurements taken close to the source.87 = 87.4dB(A).Sound power use and measurement (d) 141 Sound power level.4 + 82 .13 (a) sound power level: 400 ' 75 % 17. ! reduces errors caused by background noise generated by sources other than the one under test.25 in the text which is: Lw ' Lp % 10 log10 S & Δ1 & Δ2 ' 89.4dB(A). the upper bound on the level generated by each machine is: Lw ' 87.6 dB(A) Problem 6.4 & 1 ' 107. Disadvantages of sound intensity for localisation and identification of noise sources: ! instrumentation is more expensive.142 ! ! ! Solutions to problems Usually more time consuming. Disadvantages of sound intensity for transmission loss measurement: ! Usually more time consuming.15 (a) "radiation efficiency" is a measure of the amount of sound power radiated by a vibrating surface compared to that carried by a plane wave having the same mean square acoustic velocity (as the vibrating surface) and equal area. (b) Advantages of sound intensity for transmission loss measurement: ! reduces errors arising from flanking path transmission. ! thus contamination from other nearby sources is reduced. It can be expressed as: W σ ' Sρc ¢ v 2 ¦ (b) Referring to the above equation. ! only requires a single reverberant room rather than two. instrumentation is more expensive. Problem 6. When the radiated sound field is complex. mean square velocity ¢ v 2 ¦ and radiation efficiency σ is: . sound intensity measurements can provide too much data which is time consuming to analyse and can be confusing. ! instrumentation is more expensive. (c) Advantages of sound intensity for localisation and identification of noise sources: ! more reliable than a directional microphone due to greater spatial resolution and the ability to measure very close to a source. it can be seen that the sound power radiated by a surface of area S. ! Allows good results to be obtained at low frequencies. 16 (a) Equation 6.Sound power use and measurement W ' σ Sρc ¢ v 2 ¦ 143 To identify possible paths of sound transmission between rooms. In this case there will always be some radiation angle at which the bending waves on the surface will match the trace acoustic wavelength in the surrounding medium with a resulting strong coupling between the surface vibration field and the radiated acoustic field. fc ' 0.551c 2 / (cL h) ' 0. the radiation efficiencies of all walls ceiling and floor can be used together with their measured mean square velocities and the above equation to calculate the relative contributions of each surface to the overall sound power transmitted into the room. Problem 6.003) ' 4001 Hz 10log10S = 0.003/1 = 0. . for finite size surfaces) the critical frequency of the surface.551 × 3432/(5400 × 0.31 in the text is: Lw ' 10 log10 ¢ v 2 ¦ % 10 log10 S % 10 log10 σ % 146 dB re 10&12W The critical frequency. (c) The radiation efficiency of a surface is close to one when the bending wavelength of bending waves in the surface is less than or equal to the wavelength of the radiated acoustic waves. that is. thus allowing the flanking paths to be identified and ranked.012. at frequencies equal to or above (and in practice just below. Ph/S = 4 × 0. 81 ' 55.s. Thus the following table can be constructed.0 -46.0 -54.39 % 104.8 76.0 (b) From figure 5.39 & 0.86) % 10(5.0 Lp (dB re 20µPa) 44.9 48.12 0.8 -18. Octave band centre frequency (Hz) 250 500 1000 Lw (dB re 10-12W) 72. acceleration levels (approximate). . Octave band centre frequency (Hz) 250 500 1000 f/fc 10log10σ 10 log10¢ v 2 ¦ Lw (dB re 10-12W) 0.3) % 104. the panel will appear as a point source and the sound pressure level is given by: ρc Lp ' Lw & 10 log10(2 π r 2) % 10 log10 400 The quantity 10log102πr2 = 28 and 10log10(ρc/400) = 0.43 & 0.06 0.43 % 105.m.81 ' 52.2 81.8 76.144 Solutions to problems Thus the following table may be constructed.1 The overall sound pressure level is: Lp ' 10 log10 104.0 72.2 -16.24 -19.3 53.8 dB(A) (d) r.11 in the text we can see that for r / H L ' 10 .1.3 dB re 20µPa (c) The A-weighted sound pressure level is: LpA ' 10 log10 10(4.2 81.0 -54. Sound power use and measurement a = 2πfv (see following table) f (Hz) 250 500 1000 v (mm/s) 2 5 2 a (m/s2) 3.6 145 .1 15.7 12. bandwidth from table 1.2 250Hz octave band. Reverberant field The reverberant field of a source is defined as that part of the sound field radiated by a source which has experienced at least one reflection from a boundary of the room or enclosure containing the source. is (from equation 7.3 (a) If nz = 0.7 Solutions to problems relating to sound in enclosed spaces Problem 7. Problem 7. For nz = 1 and nz = 3.24 in the text) given by: M ' (20 % 25 % 30 % 32) / 177 ' 0.176 = 177Hz. M.1 Direct field The direct field of a sound source is defined as that part of the sound field which has not suffered any reflection from any room surfaces or obstacles.6 Problem 7. For nz = 2. there will be an antinode at Lz/2 and the mode will be excited by a monopole source. (b) A dipole will be ineffective at exciting the nz = 0 and nz = 2 modes . Thus the modal overlap.2 on p43 in the text is 353 . the pressure distribution in the room is uniform and the monopole source will excite the mode. there will be a node at Lz/2 and the monopole will not excite the mode. On the other hand. hard walled duct).Sound in enclosed spaces 147 because the phase of each part of the dipole is opposite to the other and the modal response is not. (d) The required shape is shown in the figure below where the nodes are represented by lines and the relative phases of acoustic pressure are shown by plus and minus signs. κ becomes real and the mode begins to propagate down the duct without decaying in amplitude (assuming a rigid. the particle velocity is zero. Using equation 7.1 ' 343 2 1 10 2 % 1 5 2 % 1 22 ' 93. The required derivation is described on pages 278 and 279 in the text.1.17 in the text: f1.9 Hz Problem 7. for a single mode the acoustic .4 (a) Cut-on frequency is the frequency at which the wavenumber. one would expect good excitation of the nz = 3 mode because the phase on one side of a node is 180E different to that on the other side and this can be matched by the dipole. (c) At a rigid wall. (b) From the equation given in the problem. 2 ' (ω / c)2 & [ 45 π2 / Lz ] 2 Cut-on frequency is thus given by: ω ' c [ 45 π / Lz ] Thus at 1/3 of the cut-on frequency.2 ' [5 π2 / Lz ] & [ 45 π2 / Lz ] ' ± j 40π / Lz 2 2 Thus Δ = 172/Lz (dB/m).2 = c 1 & 45π2c 2 / (ω2 Lz ) 2 ω (ω2 / c 2) & 45 π2 / Lz 2 ' ' c 1 & const / ω2 c32 c fco f . n = 2.2 = ω/κ3. and Ly = Lz/3: κ3. κ3.2 is given by: κ3. where p1 is the sound pressure ¯ p1 ¯ amplitude at 1m. For m = 3.148 Solutions to problems pressure amplitude is given by p ' p0 e ¯ ¯ &j κmn x .4343 × jκmn . Thus the dB decay per unit distance is: p ¯ Δ ' 20 log10 0 ' 20 × 0. (c) Phase speed. c3. 7 Hz 2 6. The pressure is uniform across a given cross section defined by the 4.32 in the text as: ¢p 2¦ ψ ' ρc 2 The sound pressure level in a room corner is 80dB and this corresponds to the maximum sound pressure. Problem 7.2 as a function of frequency is shown in the figure.04 Pa 2 2 The sound pressure at any other location is (from equation 7.2 (b) Pressure distribution follows a half cosine wave in the 6.19 in the text): πx ¢ p 2 ¦ ' 0.6m × 6. Energy density is given by equation 7.2m direction as shown in the figure and calculated from equation 7.2m × 3. Thus: f ' c 1 343 1 ' × ' 27.5m.6m × 3. Thus: ¢ pmax ¦ ' ( 2 × 10&5 )2 ×1080 / 10 ' 0.5 (a) Room dimensions are 4.5m dimensions.19 in the text.2 2 6.04 × cos2 L Thus the energy in the room is given by: . The lowest resonance frequency is the axial mode corresponding to the longest room dimension.Sound in enclosed spaces 149 The variation of c3. Near the mode cut-on frequency it can be seen that the phase speed approaches infinity and at high frequencies it approaches the speed of sound in free space. we can write: α ¢ p 2 ¦ ' ¢ p0 ¦ e& Sc¯t / V 2 Thus: Lp &Lp0 ' 4.6 × 3..6 × 3.150 Solutions to problems L E ' m V ψ dV ' 4.343 c¯t / L α α .54 × 10&6 x Lsin(2πx / L) % 2 4π Substituting L = 6.206 × 343 2 m 0 cos2 πx L dx L 0 ' 4. (c) The analysis procedure used on p290 and 291 in the text may be used here. Alternatively.5 × 0.5 L m ρc 2 0 ¢p 2¦ dL ' 4.343 c¯tS / V ' 4.4 × 10-5 Joules.04 1.2 gives E = 1. we may start with the given equation and write: dE Sc¯ α ' & dt E V Integrating gives: dE Sc¯ α ' & dt m E m V 0 E t E0 which gives: logeE & logeE0 ' & Sc¯t α V or α E ' E0 e& Sc¯t / V As E % +p2. c = 343.250 ' ' × × 4.5 = 16. m = 1 and nz = 0) gives: f ' 343 0. Wa ' 2I¯Sw . in one direction is related to the acoustic pressure in a diffuse field by: ¢p 2¦ ' I 4ρc Substituting 0. I. the expression given in part (c) may be used to give: Wa ' dE ESc¯ α 1.206 × 343 Alternatively.9µW dt V 6. the effective intensity. As shown on p287 in the text.9 µW 4 × 1.5 and the values of the characteristic function ψ given in the problem table into the equation for f given in the problem gives for the lowest order resonance frequency (n = 0.5861 × ' 18. a = 5. is the sound intensity in the direction of one wall.25/5 = 0.6 × 3.5 .04 × 0.1 m2.407 × 10&5 × 343 × 0.04Pa2 for the maximum mean square pressure and the above equation into the equation for Wa gives for the absorbed power: Wa ' 2 × 0.343 343 × T60 T60 (d) T60 = 5 seconds. so ¯ = 0. and Sw = 4.343 c¯T60 / L α 151 Therefore: α ¯ ' 60 L 60 6.1 ' 38. I.6 (a) Substituting L = 2. α Power consumption.343 cT60 4.Sound in enclosed spaces and 60 ' 4.2 0.05.05 ' ' ' 38.2 Problem 7.05 × 16. The α quantity.7.3 Hz 2 5. 4 7.4 6.0 0.7 Hz Modes with resonances below this with nz = 1 are found from the table in the problem as those with ψ less than the value of ψ corresponding to .697 5.2 0.0 2.152 Solutions to problems (b) The lowest order mode pressure distribution is shown in the figure at right. Tangential modes have one zero and oblique modes have none.5Hz.1 2.3 4.1 4.2 6.2 3.3 5. tangential and oblique modes.7) = 63.3 7.3 3. where the nodal plane which runs the full length of the cylinder is indicated.3 5.2 7.0 4. (d) This problem may be answered by inspection of the given table and equation. These criteria can be used to list any number of axial.4 6.4 6.3 6.0 4. column 1 followed by column 2 etc.0 3. The resonance frequency of the first oblique mode is: 343 2 1 2.1 3. The sound pressure will be at a maximum along two axial lines at the surface of the cylinder furthest from the nodal plane.1 ' % ' 82.2 5. 1. The modes are listed in the table below in order of ascending frequency. The sound field is in opposite phase from one side of the nodal line to the other.3 2.1 5.3 1.1 0.1 3. (f) Axial modes have 2 zeroes in the subscripts nz.4 7. n.1. (c) The air particles are oscillating with a velocity in-quadrature with the local acoustic pressure as the mode is characterised by a standing wave generated by the interference of two acoustic waves travelling in opposite directions across the nodal plane.2 1.4 7. and m.5 2 f1.4 5.4 (e) f = 343/(2 × 2.0 2.1 1.2 4.1 0.72 1.0 1.2 2. 0. n).1.0 0.1.651 The modes with resonance frequencies below the (1. where I is the incident wave intensity. For nz = 0.7 × 2 × 5.7 (a) A 2-D space would be one where the dimensions in 2 directions were very much larger than in the third direction.2 That is.4. area of circular region πr 2 π ' ' 2 area of square region 4 4r Time for sound to travel through the circular region (length of encompassing square) = 2r/c.2. Energy in the square region of length 2r (and unit thickness) as a result of a wave travelling normally to any of the 2r sides is I .0 0. r Following the procedure for a 3-D space described on p285-287 in the text.0 1.0 0.0. Problem 7.1 0. 1.0 1.2.0 1.3. Of these.2.1 0. a 2-D square region enclosing a circle is considered as shown in the figure.1 0.1 0.4. An example would be a large factory with a low ceiling in the frequency range below the first floor/ceiling axial resonance frequency. the value of ψ must be below that given by: ψ < 82.0 1. 9 are axial and 8 are tangential modes.1.6.3. there are 17 modes with resonance frequencies below that of the first oblique mode.0 0.5 / 343 ' 2.5.0 0. The energy per unit c perimeter in the circular region as a result of an incident wave 2r wide is: . m.1.0.0 1.1 0.0.0 0.3.1) mode are listed below in the form (nz.Sound in enclosed spaces 153 n = 1 and m = 1. E = I/c. ψ. Thus for a plane wave: ψ ' 1 1 A I ¢p 2¦ ' c ρc 2 Equation 7. Thus ψ = E. is given by: ψ ' E I r 2 π2 πI ' ' 2 S c cπr ψc π Thus: I ' Consider a plane wave travelling unit distance. Thus for a diffuse 2-D field: ψ ' πI ¢p 2¦ ' c ρc 2 ¢p 2¦ πρc Rearranging gives: I ' (b) An example of a 1-D field would be the inside of a rigid tube closed at both ends at frequencies below the first higher order mode cut-on .32 in the text also gives the relation between sound pressure and energy density. Thus: E ' m 2π Δ E dx ' 2πr m 0 Δ E r dθ ' π Ir 2 I r 2 π2 dθ ' 2 c m c 0 2π Energy density.154 Solutions to problems ΔE ' π 2r I 4 c The total energy in the circular region due to waves from all directions is found by integrating the preceding expression over the perimeter of the circle. The energy is also equal to the energy density multiplied by unit area. The energy in unit area is the intensity multiplied by the time it is present which is. Thus. Wa. The total energy in the volume is also ψL. Using this relation and equation 7. The energy in the wave travelling from left to A right is IL/c. Assuming unit cross-sectional area and L a wave travelling from left to right over a distance of L. Rate of energy absorbed. I = ψc/2 and ψ = 2I/c. W0 rate absorbed. Thus: W ' S Mψ ψc ' W0 & P¯ α Mt π Introducing the dummy variable: X ' [ πW0 / Pc¯ ] & ψ α into the preceding equation.Sound in enclosed spaces 155 frequency. is L/c. we may write: dX Mψ 1 ψc ' & ' & P¯ α W0 & dt Mt S π and 1 P¯c α ψc ' P¯ α W0 & X π π &1 . Wa. of length P and α absorption coefficient ¯ is given by: Wa ' I P ¯ ' α ψc P¯ α π Rate of change of energy in the reverberant field = rate of supply. The total energy (due to left and right 1 travelling waves) is thus 2IL/c.32 in the text gives for a 1-D field: I ' ¢p 2¦ 2ρc (c) Let S = area of 2-D room of unit height as shown in the figure. L. The time for a wave to travel a distance. around the perimeter. Wa. X. then at any time t. Rate of energy absorbed. Using the preceding equation. at the ends of the tube of length L and absorption α coefficient ¯ is given by: Wa ' 2 I ¯ ' 2 α ψc α ¯ 2 L Rate of change of energy in the reverberant field = rate of supply. X = X0 = -ψ0. From the preceding discussion and the equation above we may write: α ψ ' ψ0 e& P¯ct / πS We know that ψ % +p2. W0 - . The same equation may be used to show that as W0 = 0.156 Thus: Solutions to problems 1 dX P¯c α ' & X dt πS Integrating gives: dX P¯c α ' & m X m πS 0 X t X0 Thus: logeX & logeX0 ' & P¯ct α πS and α X ' X0 e& P¯ct / πS For a decaying sound field.. X = -ψ. Thus: α ¢ p 2 ¦ ' ¢ p0 ¦ e& P¯ct / πS 2 (d) Let L be the length of the 1-D tube of unit cross sectional area. Also the definition of the dummy variable. W0 = 0 at t = 0. X = X0 when t = 0. above can be used to show that when W0 = 0. X = X0 when t = 0. then at any time t. X = -ψ. From the preceding discussion and the equation above we may write: . X = X0 = -ψ0. Thus: W ' L Mψ α ' W0 & ψ c¯ Mt 157 Introducing the dummy variable: X ' [ W0 / c¯ ] & ψ α into the preceding equation. Also the definition of the dummy variable.Sound in enclosed spaces rate absorbed. W0 = 0 at t = 0. above can be used to show that when W0 = 0. we may write: dX Mψ 1 ' & ' & W0 & ψc¯ α dt Mt L and 1 ' ¯c W0 & ψc¯ α α X &1 Thus: 1 dX α ¯c ' & X dt L Integrating gives: dX αc ¯ ' & m X m L 0 X t X0 Thus: logeX & logeX0 ' & α ¯ct L and ¯ X ' X0 e& αct / L For a decaying sound field. Wa. The same equation may be used to show that as W0 = 0. Using the preceding equation. X. in an angular direction of ψ from the x-axis as shown in the figure. we have: y Ly 0 x Lx cx ' c cosθ and cy ' c sinθ The number of reflections per unit time for a wave travelling in the θ direction is given by: c cosθ c sinθ % Nr ' Lx Ly Averaging over all directions in one quadrant (the average for the other three quadrants would be the same) we find that the average number of reflections per unit time is: Nav ' 1 2c N dθ ' π/2 m r π m 0 0 π/2 π/2 cosθ sinθ % Lx Ly 2c 1 1 % π Lx Ly ' 2c sinθ cosθ & π Lx Ly π/2 ' 0 .158 Solutions to problems ¯ ψ ' ψ0 e& αct / L We know that ψ % +p2. Thus: ¯ ¢ p 2 ¦ ' ¢ p0 ¦ e& αct / L 2 (e) Mean free path. Consider a sound wave propagating with speed c. we need to consider only one of the 4 quadrants and find an average path length between reflections for sound propagating in these directions. Due to symmetry. Sound propagation in any direction in the 2D plane is equally likely (see figure). 2-D space. Resolving into x and y components.. first mode in each direction given by: . Thus: Lw ' Lp % 10 log10 V & 10 log10 T60 % 10 log10 (1 % Sλ/8V) & 13.9 8 × 105 ' 95 % 20. In this case.0 % 0. The sound power in the reverberation room is related to the sound pressure level by equation 6. Thus: Λ ' π Lx Ly πS ' 2 Lx % Ly P 159 For a 1-D space (see figure). Wavelength of sound.Sound in enclosed spaces But Nav = c/Λ. the c sin length of the space. Lw = 97. Volume = 105m3. where Λ is the mean free path. surface area = 2(7 × 5 + 7 × 3 + 5 × 3) = 142m2. (1 + Sλ/8V) is already included in Lp. Power conversion efficiency is given by: η ' acoustic power 10&12 × 1097. which is the distance between reflections is equal to L.5) % 10log10 1 % 142 × 0.8 / 10 ' ' 6 × 10&4 ' 0.5 & 13. we could assume that the correction term.0. c cos Problem 7.13 in the text. λ = 343/500 = 0.8 (a) Room size = 7m × 5m × 3m.06% electrical power 10 (b) Fire box dimensions are: 10 × 12 × 20m.8 dB re 10&12W Alternatively.2 & 4.3dB.5 = 97.8 . Axial resonances . the mean free path.686 & 13.9 ' 95 % 10 log10(105) & 10log10(2.686m.9 ' 97. 21. Pressure at wall is the sum of the incident and reflected pressures. because on reflection the pressure amplitude is doubled.2 Hz . . Force on each wall = 1. pr ' ( 1 & ¯ ) pi . 36 Hz . (pi % pr )rms ' 1.59 × 200 = 318kN. it seems likely that it is associated with the axial mode across the width in the 12m direction. Thus. [ 1 % ( 1 & ¯ ) ] pi ' 1.160 f ' Solutions to problems c 864 864 864 ' . (iv) Power absorbed by 2 side walls. ¢ p 2 ¦ ' ¢ p0 ¦ e& c¯ t / L . But c¯ α .343 ) × L . 2L 2 × 10 2 × 12 2 × 20 ' 43. α For a 1-D sound field. Amplitude is then: p ' ¯ 2 p rms ' 2 2 × 10&5 × 10155 / 20 ' 1. where Aw = area of one wall. α So.59 kPa Side wall area = 10 × 20 = 200m2. Thus: 2 Lp0 & Lp ' 4. (iii) Amplitude of cyclic force acting on walls.125 × 103 . Acoustic pressure is 155dB. However.6 Hz (i) As the frequency of instability is 36Hz.125 × 103 α The absorbed power is controlled by the incident sound intensity which is: I ' pi 2 ρc ' ( 1.343c ¯ t / L α and T60 ' ( 60 / 4. (ii) The acoustic pressure will be largest at the two side walls normal to the 12m dimension.125 × 103 ) 2 ( 1 % 1 & ¯ ) 2 ρc α α The absorbed power is Wa ' 2¯ I Aw . 343 c ¯ t / L α Thus: 60 ' 4.2 Q 2.2 × 30 ' 1.343 × 343 × 0.05 × T60 / 5 which gives.63 seconds.1047 ) 2 × 1. (b) Again taking logs of the equation given in the problem gives: ¯ Lp0 & Lp ' 4.25 × 343 × T60 / (π × 25) which gives.105 864 × 1.1047 × ( 1. T60 = 0.833 The absorbed power is then: Wa ' 2 × 0.Sound in enclosed spaces T60 ' 2.16 × 864 ' 14 kW (v) Power conversion efficiency is: η ' 13953 800 × 103 ' 1.9 (a) Taking logs of the equation given in the problem for a 1-D field.7% Problem 7. (c) S = 2(5 × 5 + 5 × 5 + 5 × 5) = 150m2 . we have: Lp0 & Lp ' 4.1246 × 103 ) 2 × 200 (1 % 1 & 0.833 secs ' fn 36 161 Thus: α ¯ ' ( 60 / 4.343 P α c t / (π S) P = 4 × 5 = 20m and S = 5 × 5 = 25m2.343 × 20 × 0.0 seconds. T60 = 4.343 ) × 12 ' 0. Thus: 60 ' 4. 265 A ' × 30 ' 0.162 Solutions to problems V = 5 × 5 × 5 = 125m3 α ' ¯ 50 × 0. Thus.265 Pa 2 and the energy density is then: Energy density = 1.05 % 100 × 0. Thus in the situation under consideration here.73 seconds.10 In a reverberant field the sound intensity in any particular direction is equal to that in any other direction resulting in a net active intensity (averaged over all directions) of zero and a large reactive intensity. accurate measurements of the active field will not be generally feasible in a reverberant room. Problem 7.265/(413 × 343) = 8. the active intensity will only be contributed to by the direct field of the source and the reactive intensity field will dominate the active field by a large amount.9 × 10-6 J/m3 (b) Sound power incident on a wall is given by: ¢p 2¦ 1.023 W ' 103.5 ' 1.11 (a) Energy density in a reverberant field is ¢ p 2 ¦ / ρc 2 and the given SPL is 95 dB. we obtain: 60 ' 1.6 dB re 10 &12 W W ' IA ' 4ρc 4 × 413 .50 in the text.086 × 150 × 343 × 0. T60 = 0. Due to the dominance by the reactive field together with limitations on the phase accuracy between the two microphones of any measurement system. especially at large distances from the source. Problem 7.25 ' 0. ¢ p 2 ¦ ' (2 × 10 &5 )2 × 109.183 150 Using equation 7.1833 × T60 / 125 which gives. 05 × 15.51 in the text for reverberation time.206 × 343 × (1 & 0.40 to 7. the reverberant field level is α α equal to the direct field level when Dθ/(4πr2) = 4/R = 4 ( 1 & ¯ )/( S¯ ) .42 × 10&7 W / m 2 4 × 1.1 + 3.24 = 283.1m2. = 4 × 10-10 × 1060/10 = 4 × 10-4 Pa2. Assuming the acoustic centre of the source is within a quarter wavelength from the hard floor.25 × 400 ' 1. V = 3.33 in the text: I ' 4 × 10&4 ' 2.Sound in enclosed spaces Problem 7. S = 2(10 × 10 + 10 × 4 × 2) = 360m2. Using equation 7.1 × 15.05 × 6. S = 2(3.1 (b) Lp = 60dB corresponds to +p2. at which the fields are equal is: r ' S¯ α 8π ( 1 & ¯ ) α 1/2 ' 36 8π × 0.24 + 6.13 (a) Room 3.26 m Problem 7.54m3. (c) Using equation 7.9dB re 10-12W.05 × 6. Lw = 10log10(9. Dθ = 2 and the distance. r.1 × 4 × 10&4 ' 9. we may write: .1) Sound power level.8 seconds T60 ' 343 × 360 × 0.1m × 15. V = 10 × 10 × 4 = 400m3.12 163 (a) Room 10m × 10m × 4m.9 1/2 ' 1.51 in the text for reverberation time.42 in the text.67 × 10-6) + 120 = 69. Using equation 7.206 × 343 (d) From equations 7. Using equation 7.24m.05m × 6.41 in the text we obtain: W ' 360 × 0.10 × 15.67 µW 4 × 1.24) = 316. we may write: 55. 0722.42. we need to increase R by a factor of 10. expanding the above equation gives: 316.7 dB re 10&12W (b) To lower the reverberant field by 10dB.1 × 0.15 ' 81.33 seconds Sc¯ α 316.0. The old value of ¯ α α is 0.1 + 120 = 124. thus required new S¯ / (1 & ¯) = α α α α 246. the sound power output may be written as: Lw ' Lp % 10 log10 ' 74 & 10 log10 4(1 & α) ¯ S¯ α % 10 log10 ρc 400 4 × 0.25V 55.9 dB re 10-12 W (b) c ' γRT ' M 1.1¯ α α which results in a new required value of ¯ = 0. so the increase in absorption needed is: Δ S¯ = 316.51 in the α text.54 ' ' 0.25 V 55.035 .25 × 283.54 ' 0.4 × 8.5m2.314 × 1473 ' 700 m/ s 0.072 & 0.1 × 343 × 0.14 (a) Sound power level = 10 log10W + 120 = 10 log103.438 obtained above and equation 7.25 × 283.438 Problem 7.1 & 246.438. we obtain: T60 ' 55.072 ' ScT60 316.1¯ ' 246.438 . α (c) Using the value of ¯ = 0.1m2.928 316. Thus in the new situation.1(0.61m2.1 × 343 × 2 α ¯ ' Using equation 7.0722) = 115. Old S¯ / (1 & ¯) = 24.164 Solutions to problems 55. 1 m 6.395 400 .15 γP 1. 102.78 m 2 (1 & ¯) α 0.92 Adding sound absorbing treatment to the walls and ceiling results in a new mean absorption coefficient calculated as follows: αn ' ¯ 100 × 0.9 ) ' 102.1 dB(A) Lp ' Lw % 10 log10 Q 4πr 2 % 4(1 & α ) ¯ S¯ α 2 4πr 2 Thus.29 kg/ m 3 2 &α) c4(1 ¯ %10 log ρc 7002 Lp ' Lw % 10 log10 10 Sα ¯ 400 ' 124. S = 2(10 × 10 + 10 × 5 × 2) = 400m2.0358 Problem 7. S = πdL + πd2/2 = π × 4 × 6 + π × 16/2 = 32π = 100 m2 Problem 7.05 400 Lp ' 10 log10 ( 109. r ' % 3. The room constant before treatment is Rb ' S¯ α 400 × 0.95 700 × 0.28r 2 1 ' 2.09 6.28 × 0.1 & 110 ' 10 log10 and.9 % 10 log10 4 × 0.5 ' 0.8 dB 100 × 0.4 × 103 ρ ' ' ' 0.9 / 10 & Thus.08 % 300 × 0.7 % 109.16 Room 10m × 10m × 5m. 10&7.Sound in enclosed spaces 165 Surface area.29 % 10 log10 ' 120. V = 10 × 10 × 5 = 500m3.64 320 × 0.5 % 109.4 × 101.09 3.64 1 ' 320 × 0.08 ' ' 34. we can write: ¢ pR ¦ ' 2 4 × 25 × 10&3 × 1.2 Problem 7.41.605 Using equation 7.0767) ' 2.78 & 10 log10 2 4 261.0 dB % 4 34. The mean Sabine absorption coefficient may be calculated using equation 7.0767 The reverberant field sound pressure level is then: LpR ' 10 log10 (2.166 Solutions to problems Thus the new room constant after treatment is: Rn ' S¯ α 400 × 0.15 % 132 × 0.0767 180 Using equation 7.206 × 343 (1 & 0.2 m 2 (1 & ¯) α 0.42 in the text allows the difference in sound pressure level (assuming that the sound power is constant) to be calculated as follows (assuming the acoustic centre of the machine is within a quarter wavelength of the hard floor): Δ Lp ' 10 log10 Dθ 4πr 2 4π3 2 2 % 4 R & 10 log10 old Dθ 4πr 2 2 4π3 % 4 R % new ' 10 log10 ' 6.4 dB (b) Direct and reverberant fields equal (see equation 7.78 in the text to give: α ' ¯ 48 × 0. S = 2(8 × 6 + 8 × 3 + 6 × 3) = 180m2.395 ' ' 261.766 Pa 2 180 × 0.766) % 94 ' 98.05 ' 0.42 in the text) when .17 (a) Room 8m × 6m × 3m. 2 dB (b) Sabine room. Pressure reflection coefficient amplitude = 0.0767 16 π × (1 & 0. V ' 10 × 10 × 5 ' 500 m 2 The total sound pressure level (direct plus reverberant) in the room is: .7.2 dB The direct field sound pressure level is: Lp (direct) ' Lw % 10 log10 ρc 400 & 10 log10 ( 4 πr 2 ) ' 130 % 0.55 m Problem 7. S ' 2 [10 × 10 % 10 × 5 × 2 ] ' 400 m 2 and volume.Sound in enclosed spaces 167 Dθ / 4πr 2 ' 4 / R . From the figure.72 and α ' 1 & 0. r.97 ' 105. area.18 (a) This is a flat room as on pages 314-319 in the text.14 & 18. Assuming that the acoustic centre of the source is well above the floor.0767) 1/2 ' 0. the reverberant field sound pressure is given by: 10log10 ¢ p 2 ¦ ' 10log10 W ρc πa 2 &8 Thus the reverberant field sound pressure level is: Lp (reverb) ' Lw % 10 log10 ρc 400 & 10 log10 ( πa 2 ) & 8 ' 130 % 0.14 & 24. at which the two fields are equal is: r ' S¯ α 16 π (1 & α ) ¯ 1/2 ' 180 × 0.95 & 8 ' 103. Thus r/a = 1. a = 5 m and distance r = 5 m. Dθ = 1 and the distance.72 ' 0. β = 0.51 ¯ Room height. 3 % 300 × 0.25 V 55.08. S = 400. the sound level at the receiver without the enclosure at a distance of 50 + 5 m is: Lp ' Lw % 0.9 ' 62.9 m 16 π × 0.51 Thus the noise reduction = 24. r = 1. where C ' 10 log10 0.51 DSα ¯ ' 4π × 4( 1 & α ) ¯ 400 × 0. Thus: Lp ' Lw % 10 log10 Q 4πr 2 % 4(1 & ¯) α S¯ α % 0.9 dB and the SPL at 50 m is: Lp ' 87. Lw = 95dB.5 dB Problem 7.49 ' 0. α = 0.49 (e) Treat the room like an enclosure.1 dB 400 × 0.14 & 10 log10 2πr 2 ' 130 % 0.51 ¯ ¯ (c) Direct and reverberant fields equal when D / 4πr 2 ' 4(1 & α) / S α .15 .19 Q = 2.C.25 × 500 ' ' 0.39 seconds Scα ¯ 400 × 343 × 0.14 & 42.168 Solutions to problems Lp ' Lw % 10 log10 ' 130 % 10 log10 D 4πr 2 % 4(1 & α) ¯ Sα ¯ 1 4 × 0.3 % SE( 1 & α ) ¯ Si α ¯ ' 10 log10 0.4 dB The enclosure noise reduction is given by NR = TL . Thus.49 % ' 111 dB 4π × 25 400 × 0.4 & 24.8 ' 87.51 ' 2. Thus: r ' (d) T60 ' 55. Sound in enclosed spaces Lp ' 95 % 10 log10 Problem 7. The allowable contribution from each 107.15dB correction for ρc different from 400). Thus: Lp ' Lw % 10 log10(4 / R) % 0.54 % 11. Sound power level is: Lw ' Lp % 20 log10 r % 10 log10(4π) & 3 % 10 log10 ' 80 % 9.43 in the text gives: 328 α ¯ ' 71. Thus the allowable contribution from the three new line printers is 10 log10 (108 & 107.15 % 6.15 Substituting in values for the parameters gives: 10 log10 R ' 97.42 in the text (with a 0.38 & 85 % 0.15 ' 89. ¯ = 0. Using equation 7.6.42 in the text (using only the reverberant field part) as follows: line printer is then 10 log10 .02 ' 18. The allowable sound 3 power level is the obtained using equation 7.5 dB . R = 71.83 ' 73.08 169 (a) Reference source on hard asphalt (DI = 3).3 dB .01 & 0.20 2 4 (1 & 0. (c) Room 14m × 6m × 4m.55 Thus.08) % % 0.4 dB 400 ρc (b) Assuming a negligible direct field. the room constant may be calculated using equation 7.18. S = 2(14 × 6 + 14 × 4 + 6 × 4) = 328m2. (d) Existing Lp = 75dB and the allowable total = 80dB.5 dB 4 × π 400 × 0.15 ' 97.5 ) ' 78.00 & 3. Lp = 80dB at 3m.62 (1 & ¯) α α Thus. Average α ' ¯ 0. corresponding ¯ = 0.2.1 dB Thus the level in the room before addition of the new printers will now be 75 .8 dB Problem 7.5 × 84 % 0. Using equation 7.29 & 10 log10 4 116 & 0.120 in the text. An Lp of 83dB corresponds to a +p2.6 Lw ' 73.261 ' 116 m 2 (1 & 0.9dB.21 (a) Considering only the reverberant field of the machine we may use equation 7. The allowable contribution from each line printer is then 107.5.91 ' 74.42 in the text (using only the reverberant field part) as follows: 10 log10 Lw ' 74.15 ' 88.84 = 244m . Thus: 328 R ' 328 × 0.6 & 10 log10 & 0.29 ) ' 79.6 ' 2. The allowable sound pressure level of the three new line printers together is now 10 log10 (108 & 107.179. The surface area of the factory is S = 2(10 × 10 × 3) = 600m2.1 = 72. α corresponding ¯ = 0.98 × 10&2 Pa 2 .261) Using equation 7.3 dB .41.170 Solutions to problems 4 71.41 in the text.179 × 244 ' 0.15 ' 86. the required mean absorption coefficient is given by: .261 .0 dB (e) Area of ceiling = 14 × 6 = 84m2. the reduction in sound pressure level is thus: Δ Lp ' 10 log10 116 71.1 dB . of 4 × 10&10 × 1083 / 10 ' 7. The allowable sound power level is the 3 obtained using equation 7. α 2 Area of floor and walls = 328 . and we let the required absorption coefficient of the walls and ceiling be x.2dB. If we assume that the concrete floor has an absorption coefficient of 0.01. The sound power level. R.01 × 1.78 in the text to write: 600 × 0.22 (a) From Table 4. For a total Lp of 90dB we may write (assuming a directivity factor.346 × 600 = 208m2. then we can use equation 7. Problem 7.2dB in that band.10 = 45dB(A).257 ' 100 × 0.01) + 120 = 100dB. Dθ = 2 as the source is assumed close to a hard floor and other surfaces are more absorptive): 90 ' 100 % 10 log10 2 4πr 2 % 4 208 Solving the above gives r = 1. (b) The room constant.346 (1 & ¯) α 7.01 % 500x which gives x = 0. However this would be the required absorption α coefficient if the floor were lined as well.31. The existing level is 44dB so the allowed increase is 4.2dB A-weighting at this frequency.306.42 in the text) of: . If the only noise is in the 500Hz octave band then the -3.Sound in enclosed spaces α ¯ 4 × 0.981 × 10&2 × 600 171 Thus ¯ ' 0. Lw = 10log10(0. A reverberant field sound pressure level corresponds to a sound power level (see equation 7.40m as the radius around the machine within which the sound pressure level will exceed 90dB.206 × 343 ' ' 0.10 in the text. the allowable community noise level to ensure minimal risk of complaints is 40 + 15 . results in an allowable level of 48. is from part (a) 0.257 . Thus the required absorption coefficient for the walls and ceiling is 0. 4 m 2 α corresponding to ¯ = 0.132) ' 220.5 m 2 and new R = (1450 × 0.15 dB Room 25m × 20m × 5m.10log10(5) = 100.132) 1450 × 0.8 m 2 .132) / (1 & 0. S = 2(25 × 20 + 25 × 5 + 20 × 5) = 1450m2.5 % 950 × 0. V = 25 × 20 × 5 = 2500m3 T60 = 2.132 343 × 2.2 = 109.4 dB As there are 5 new machines.25 × 2500 ' 0. Thus the allowable total sound power of existing + new equipment is 105.132 Thus: Lw ' 88 & 10 log10 & 0.15 ' 105.5dB.1 secs. Thus the allowed power level for the new equipment is: Lw ' 10 log10 10109.172 Solutions to problems Lw ' 88 & 10 log10 4(1 & ¯) α S¯ α & 0. α α (b) If ceiling tile with ¯ = 0. Thus the allowed increase in sound power level for the same reverberant .259) / (1 & 0.3 dB This is the sound power level of the existing equipment.51 in the text: α ¯ ' 55. walls and ceiling Air temperature of 20EC.4dB Assumptions ! ! ! Only absorption is due to floor.3 / 10 ' 107. From equation 7.132 ' 375.1 × 1450 4(1 & 0. the allowed power level for each is Lw = 107.3 + 4.5 / 10 & 10105. The relative contribution of direct and reverberant sound energy to the community noise levels will be the same for the new machines as for the old machines.259) ' 506.259 α Old R = (1450 × 0.4 .5 were added then the new S¯ is: S¯ ' 500 × 0. 205 × 3432) ' 0.12) = 0.52 ' 0.8) & 10 log10(220. Problem 7.7W.41 in the text.85 216 × 0.15 ' 74.35 W α (c) The power generated is equal to the power absorbed by the walls.15 + 0.28 J / m 3 (b) Enclosure surface area. S = 2(0.2 × 0.15. ψ = ¢ p 2 ¦ / ρc 2 Thus ψ = (2 × 10&5 ) 2 × 10140 / 10 / (1.2 × 0.4 dB (b) Compare 4(1 & α) As the acoustic centre of the source is Sα 4πr well above the hard floor.2 = 1.035 .5) ' 3. Dθ = 1.6 = 104dB.1 / 4 ' 0.15 × 0. Problem 7.15 % 0. Power flow into walls (equation 7.Sound in enclosed spaces field sound pressure level is: Δ Lw ' 10 log10(506. ¯ = 0. so: Lp ' 84 % 10 log10 Dθ 4 × 0.144m2.6 dB 173 Assuming that the reverberant field dominates the direct field.144 × 0.23 (a) Energy density.24 (a) Reverberant field Lp from taking logs of equation 7. the new allowed power level of each machine is 100.4 + 3.12 + 0.348/0.39a) is given by Wa ' ψcS¯ / 4 ' 0.282 × 343 × 0. Thus the power required to drive the source is 0. so: 2 with Dθ 4πr 2 ' 1 4π × 1. Lp ' Lw % 10 log10 4(1 & ¯) α S¯ α %10 log10 ρc 400 α S = 2(10 × 6 + 10 × 3 + 6 × 3) = 216m2. 1dB. For the original machine: Lp ' 80.15 ' 71.105 S¯ α 216 × 0.42 in the text (and allowing for ρc not equal to 400) the sound pressure level corresponding to ¯ = 0. The total sound power output is 10 log10 108.6 dB(A) and the sound pressure level corresponding to ¯ = 0.42 in the text.5 dB(A) This corresponds to a reduction of 4.85 ' ' 0.15 is: α Lp ' 84 % 10 log10 0.9 % 108.15 ' 0. (d) From equation 7.85 216 × 0.85 r ' When the machines are running together.15 Thus the reverberant field dominates.15 % 0.15 . the reverberant field contribution will be the sum of the two reverberant fields originating from each machine.5 % 0.0354 % 4 × 0.87 m 4 × 4 × π × 0.6 ' 84 & 10 log10(4πr 2) % 0.15 ' 80.5 216 × 0.6 dB(A) Thus the required distance is that at which the direct field Lp is equal to 80. when each machine is running separately. Dθ 4(1 & α) the fields are equal when .4 ' 90.2 % 10 log10 4 × 0.15 ' 75. Thus for each machine: ' 2 Sα 4πr 216 × 0. The reverberant field sound pressure level is then: Lp ' 90.2 dB .15 % 0.0354 % 4 × 0.174 Solutions to problems 4(1 & α) 4 × 0.85 216 × 0. (c) Using equation 7.5 is: α Lp ' 84 % 10 log10 0.6dB(A). 15 Thus r = 0.5 ' 2. For the new machine. ! Both machines exhibit similar frequency spectra.0 ' 89 & 10 log10(4πr 2) % 0.81m (distance from new machine at which new machine direct field = reverberant field with both machines running). From equation 7. the fields are equal when Dθ 4(1 & α) ' .6 ' 89 & 10 log10(4πr 2) % 0. .15 Thus r = 1.5 The reverberant field sound pressure level when both machines are running is.07 m 4 × 4 × π × 0. Lp ' 80. Lp ' 73. Lp ' 73.0dB(A). For the original machine.0 dB(A) Lp ' 90.02m (distance from original machine at which original machine direct field = reverberant field with both machines running).Sound in enclosed spaces 175 Thus r = 0.5 % 0. For the new machine.5.15 ' 73. (e) Increase α to 0.42m (distance from original machine at which original machine direct field = reverberant field with both machines running). 4 × 0.0 ' 84 & 10 log10(4πr 2) % 0. 2 Sα 4πr r ' 216 × 0.5 Thus the required distance is that at which the direct field Lp is equal to 73.15 Thus r = 1. Thus for each machine. ! The frequency averaged absorption coefficient is obtained using a sound source with a frequency spectrum similar to that of the machines under test. when each machine is running separately. (f) Assumptions: ! Each machine is radiating omni-directional sound.2 % 10 log10 216 × 0.75m (distance from new machine at which new machine direct field = reverberant field with both machines running).42 in the text. 15 (87.25 Solutions to problems (a) Room 10m × 15m × 6m.25 × 900 ' 2.176 Problem 7.16) (d) When quiet machines are installed.4 secs 600 × 343 × 0.0 respectively.43 in the text: R ' 600 × 0.5 % 108.67 m 2 1 & 0.0 dB The existing reverberant field level is 75dB prior to installation of the new machines. Thus: T60 ' 55. Lw = 94 + 10log104.1 (c) For each machine. Thus the total reverberant field level after installation of quiet machines is: Lp ' 10 log10 107. Thus the required reduction due to ceiling tile if untreated machines are used is: Δ Lp ' 88.1 (b) Room constant from equation 7.16 & 85 ' 3. V = 10 × 15 × 6 = 900m3. Thus the total level after installation is: Lp ' 10 log10 107.2 dB (88.5 % 107.795 ' 79.7) % 10log10(4) % 0. Thus the reverberant field sound pressure level due to the 4 new machines is: Lp ' Lw % 10log10(4 / R) % 10log10 4 % 0. The room reverberation time can be calculated using equation 7.15 ' 88.51 in the text. the sound power and the sound pressure level contribution due to the new machines is reduced by 10dB to 94 and 78.16 dB ' 10 log10 Rf / Ri . S = 2(10 × 15 + 10 × 6 + 15 × 6) = 600m2.22 % 6.1 ' 66.80 ' 88. Lw = 94dB.95) ' 94 & 12.15 ' 94 % 10log10(4 / 66.02 % 0.7 dB (e) The design goal is 85dB. Thus for 4 new machines. the new sabine absorption coefficient would be: α ' ¯ (1450 & 500) × 0.187 .5 ' 0. Thus the new requirement is R = 138m2.9 (b) In this case.53 × 21/2 = 3.07 .67m2.6m (c) If the ceiling were covered with ceiling tiles. Thus ¯ ' 0. The initial room constant is 66. The required average absorption coefficient is then given by 600 × α ' 138 & 138 × ¯ .5 m 16 π × 0.2 m 16 π × 0. Area of ceiling = 10 × 15 = 150m2.238 ' 4.6x 600 which gives x = 105m2.26 (a) Distance at which direct and reverberant fields are equal is given by: D 4πr 2 ' 4( 1 & α ) ¯ or r ' Sα ¯ D S¯ α 16 π ( 1 & α ) ¯ The room surface area. S.238 1450 2 × 1450 × 0. Then: 0. So covering the ceiling with ceiling tile would be adequate.Sound in enclosed spaces 177 Thus Rf / Ri ' 103. D=4 instead of 2 and r = 2.1 ' 2.187 ' 0. This would cost $50 + 3 × 150 = $500 which is much less than the machine noise control and is thus the preferred option.16 / 10 ' 2.1(600 & x) % 0. Problem 7.762 So r ' . To achieve this. is given by S ' 2( 20 ×25 % 20 ×5 % 25 × 5 ) ' 1450 m2 So r ' 2 × 1450 × 0. let x ¯ α α square metres of room surface be covered by tile.1 % 500 × 0. 51 in the text and the modal density may be calculated using equation 7.02 0.6 69.5m from the machine so he/she is in the direct field.408 2. S = 2(5 × 5.5 85.5m3.238 4π × 0.03 (b) The reverberation time may be calculated using equation 7.1dB Problem 7. Equations .9 % 2 1450 × 0.01 0. Assuming that measurements of Lp are made far enough from the machine that the direct field is negligible compared to the reverberant field.5 & 10log10 2 4 × 0.5 × 3 = 82. Octave band centre frequency (Hz) 63 250 1000 4000 Overall Lp 75 85 84 70 α ¯ R 1.02 0. the sound power level is given by: Lw ' Lp & 10 log10(4 / R) & 0. V = 5 × 5.5 + 5 × 3 + 5.649 Lw 69.6 82.5 × 3) = 118m2.192 2.1 4π × 0.901 ' 0.5 ' &1. Could calculate this (but not necessary).5 + 5 + 3) = 54m. The reverberant field contribution at this distance is small so the ceiling tiles will have only a very small effect.795 % 1.762 % 2 1450 × 0.15 This allows the following table to be constructed.5m × 3m.27 (a) Room 5m × 5.408 3.4 0. L = 4(5.178 Solutions to problems (d) The operator is only 0. Perimeter.6 81. L p1 & L p2 ' 10log10 D % 4(1 & ¯1 ) α S ¯1 α 4πr 2 & 10log10 D 4πr 2 % 4(1 & α2 ) ¯ S α2 ¯ ' 10log10 2 4 × 0.21. 01 m Assuming that the acoustic centre of the source is within a quarter of a wavelength from the hard corner.04 0.575 × 10&3 2 × 2.569 × 10&5 % ( 1.75 Δf 0. The following table may be constructed.575 × 10&3f % 0.391 0.569 × 10&5f 2 % 1.23 and 7.67.03 0.391 0.569 × 10&5 ' & 31 % 547 ' 516 Hz Thus the modal overlap is greater than 3 for frequencies above 520Hz.5 2 2 % 5 2 2 % 3 2 2 ' 4.02 0.01 0.02 0. (c) The effect of acoustic tile may be calculated using equation 7.673 Thus: f ' & 1. Dθ = 8 and so .3 417.6 246.590 M 0.63 5. so we need to calculate the frequency where dN/df = 3/0.Sound in enclosed spaces 179 7.391 = 7.42.0 27.26 5.24 are then used to calculate the modal overlap.569 × 10&5 × 7. Octave band centre frequency (Hz) 63 250 1000 4000 dN df α ¯ T60 11. The quantity Lw will remain the same in each case so we need to calculate the change in the second term in the equation.673 2 × 2.195 0.3 0.22 2.0197 ' 7.79 10.575 × 10&3 )2 % 4 × 2. That is: 2.4 The reverberation time may be assumed constant between 250 and 1000Hz. Distance from room corner to centre is given by: d ' 5.63 3. 25 0.78 in the text. The following two tables (one with no 4π × 4.3 2.84 1.048 0.649 3.9 3.01 0.408 3.3 -3.45 The noise reductions in each octave band are given by the last column in the above table.37 0.39) % 10(8.27 9.4 % 107 & 10 log10 10(7.72 0.1 dB .5 % 108.0396 .20 0. 2 α ¯ Octave band centre (no tile) frequency (Hz) R Dθ 4πr 2 % 4 R 10 log10 Dθ 4πr 2 % 4 R 63 250 1000 4000 0.408 2.46) & 10(7 & 0.180 Dθ ' 8 Solutions to problems ' 0.4 -1. (d) Assuming that the overall space average level will be reduced by the same amount as the level at the centre of the room. the overall absorption coefficient is calculated using equation 7.02 0.08 0.4 1.7 1.5 2 % 4 R Improvement (dB) 3. the difference between new and old overall levels is: 10log10 107.5 & 0.5 % 108.6 In the following table (which includes the effects of acoustic tile.38) % 10(8.59 0.077 3.012 4πr tile and the other with 25m2 of acoustic tile) may be constructed.01 0.058 0.95 7.02 0.5 -2.7 4.3 0.7 1.6 4.8 & 83.3 2.02 0.03 0. Octave band centre frequency (Hz) 63 250 1000 4000 α ¯ α ¯ α ¯ (wall) (tile) (overall) R Dθ 4πr 2 % 4 R 10 log10 Dθ 4πr 1.02 0.192 2.7 ' 4.4 & 0.0 5.025 0.14 5.1 0.5 & 0.03 1.41) ' 87.15 0. 2m2.28 181 (a) Using equation 6.565m × 4. (and allowing for ρc … 400) we can write: Lp ' 95.565 + 6.72 = 179.84 × 5.565 × 4. The room reverberation time can be calculated using equation 7.72 + 5. .72) = 193.2 × 343 × α ¯ α ¯ The results for each third octave band are given in the following table. S = 2(6.84m × 5.5 dB re 10&12W (b) Using equation 7.5 % 0.150 ' ' Sc¯ α 193.51 in the text.5 % 10 log10 2 4π × 0.84 × 4.42 in the text.022) 193.5 2 % 4(1 & 0.7 0.7m3.13 in the text and substituting in the appropriate values gives: Lw ' 95 & 8.2 × 0. Thus: T60 ' 55.Sound in enclosed spaces Problem 7.022 % 0.9 ' 95.25 × 179.25V 55.3 % 22.29 (a) Room 6.72m.84 × 5.6 dB re 20µPa Problem 7.15 ' 97. V = 6.565 × 4.2 & 13. 3 1/3 ' 5. Thus: λ ' 179.8 4.050 (b) Lowest 1/3 octave band given by V = 4.031 0.018 0.4 3.393 m This corresponds to a frequency.6 1/3 ' 3.025 0.3 7.9 7.037 0.011 0.013 0.010 0.017 0.5 6.0 5.022 0.047 0.015 0.028 0.018 0.6 13.011 0.182 One third octave band centre frequency (Hz) 63 80 100 125 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000 6300 8000 Solutions to problems α ¯ T60 15.171 = 66Hz. Thus the room is suitable for measurements down to and including the 100Hz 1/3 octave band.0 3. (c) Lowest octave band given by V = 1. Thus: λ ' 179.2 3.0 0.171 m This corresponds to a frequency.020 0.017 0.044 0.7 / 1.019 0.4 4.6λ3 (see p259 in text).3λ3 (see p259 in text).393 = 101Hz.6 11.0 13.8 8.0 15.4 4.8 8.0 8.5 10.040 0. Thus the room may just be suitable for measurements down to and including the 63Hz .8 6. f = 343/3. f = 343/5.7 3.034 0.7 / 4.3 8.010 0. it can be seen that the lowest acceptable frequency for tonal noise is 430Hz.6 13.7m2.0 8.5dB.3 8. Lpr = 87dB for reference source. the lowest acceptable frequency is given by f ' T60 / V 1 / 2 (see p259 in text).0 15.5 ' 87 & 10 log10(4 / R) & 0.30 (a) Lw = 92.8 8.0 13. We need to solve for the frequency by trial and error as illustrated in the table below.272 and so R = 14.6.5 & 0. The room constant is calculated using only the reverberant part of equation 7.42 in the text and allowing for ρc = 413. Thus: Lw ' Lp & 10 log10(4 / R) % 10 log10(400 / ρc) 92. One third octave band centre frequency (Hz) 63 80 100 125 160 200 250 315 400 500 630 T60 corresponding lowest acceptable frequency 578 578 550 550 506 472 443 443 430 430 419 15. 183 (d) For pure tone noise.5 10.8 8.3 7. .15) / 10 ' 0.Sound in enclosed spaces octave band.9 From the table.15 (4 / R) ' 10(87 & 92.6 11. Problem 7. ρc = 413.3 dB(A) re 10&12W Same assumptions as for part (b).43. so allowed Lp from each new machine is: Lp ' 10 log10 108.5 & 108. Using equations 7.15 ' 85.42 and 7. 4 new machines. then calculations should be done in octave bands.77 & 10 log10(4 / 29. Allowed total Lp = 85dB(A).78 and is: .7) & 0.4) & 0.184 Solutions to problems (b) Existing Lp is 81dB(A). (c) Doubling room constant gives the allowed sound power level of Lw ' 76. If not. the noise reduction is given by: Dθ 4πr 2 10 log10 % 4(1 & α1) ¯ S ¯1 α & 10 log10 Dθ 4πr 2 % 4(1 & ¯2) α S¯2 α The mean absorption coefficient before treatment is calculated using equation 7. S = 2(15 × 5 × 2 + 15 × 15) = 750m2.8 dB(A) re 20µPa The corresponding allowed sound power level of each machine is then: Lw ' 76.6 Spectral content of noise from new machines is similar to that of existing noise.3 dB(A) re 10&12W Assumptions: ! ! ! Direct field small compared to reverberant field at measurement locations.1 & 10 log10(4) ' 76. Problem 7.15 ' 82.31 Room 15m × 15m × 5m.77 & 10 log10(4 / 14. 75secs.493) % 4 × π × 16 750 × 0.42 omitted.5dB .493 ' & 11. Thus the following table may be constructed. T60 = 0. V = 100m3.7 ' 0. the quantity S¯ is calculated using equation 7. thus the noise reduction at the specified location is: ΔLp ' 10 log10 & 10 log10 2 4(1 & 0.32 Room: Lw = 105dB. For the room and the office.493 750 Assuming that the acoustic centre of the machine is within a quarter of a wavelength of the floor.42 and 7. Office: Lw = 85dB.0 dB Problem 7. T60 = 1. Partition area: 15m2.739 0.0dB 10.1 ' 0.073 750 185 and after treatment it is: α ¯2 ' 15 × 15 × 0.51 α in the text and the reverberant sound pressure level existing in each space is calculated by combining equations 7.1718 78.43 with the direct field term of equation 7.15 Office 17. Room S¯ α α ¯ Lp ' Lw % 10 log10 4(1 & ¯) α S¯ α % 0.1 % 18.Sound in enclosed spaces α ¯1 ' 15 × 15 × 0.01 % 525 × 0.0826 100. V = 80m3. S = 100m2.073) % 4 × π × 16 750 × 0. S = 130m2.01 % 525 × 0. Dθ = 2.5secs.1 ' 7.182 0.073 2 4(1 & 0. 812 α ¯ 4 × 1. T10 = 0.1 = 28.8 ' 72. (c) New ¯ is given by: α α ' ¯ 10 × 0.9 & 107.172 (b) Using equation 7.172 Thus.094secs. Problem 7.5dB.172 ' 0. surface area.2 × 50 ' ' 4.298) ' 77.206 × 343 × (1 & 0. S = 50m2 Reverberant field mean square pressure is: ¢ p 2 ¦ ' 4 × 10&10 × 10116 / 10 ' 159.33 (a) Room volume V = 30m3.41 in the text we obtain: 1&α ¯ 159.96 Pa 2 50 × 0.51 in the text: T60 ' 55.72.206 × 343 × 1 α Thus ¯ = 0.5 .25 × 30 ' 0.8 % 40 × 0.1 dB The level in the room is 100.56 secs 343 × 50 × 0. so the noise reduction required of the wall is 100.298 .41.186 Solutions to problems Allowable level in office = 78 + 1 = 79dB.4dB which should be rounded up to 30dB for specification purposes.56/6 = 0.2 Pa 2 Using equation 7.298 50 Using equation 7. we obtain for the new mean square sound pressure: ¢p 2¦ ' 4 × 1 × 1. Allowable level due to new machine is: Lp ' 10 log10 107. Problem 7.34 porous acoustic fibrous material (fibreglass. Porous plastic and rubber (polyurethane foam etc. Applications: air handling duct mufflers. not oil. ceramic fibre. high absorption coefficient in mid to high frequency range.9 = 3.9 dB 187 which corresponds to a reduction of 116 . the increase in total sound pressure level over that corresponding to one source would be ΔLp ' 10 log10(4) ' 6 dB . water or chemical resistant.1dB. Disadvantages: fibre loss can be a health hazard so care has to be taken to properly contain the material. (d) Adding 3 more sources increases the existing number by a factor of 4. pipe lagging. steel wool): Absorption mechanisms: viscous friction losses due to difference in velocities of air particles adjacent to fibres and those in the centre of the gap between fibres. rockwool. susceptible to powdering in the presence of vibration. Absorption .): Absorption mechanisms: viscous friction losses due to difference in velocities of air particles adjacent to capillary walls and those in the centre of the capillaries.96) ' 112.Sound in enclosed spaces The corresponding sound pressure level is then: Lp ' 94 % 10 log10(77. reverberant space absorption.112. Absorption characteristics: good at mid to high frequencies but at low frequencies need a large thickness of material to be effective. Avantages: Inexpensive. Providing all sources produce the same sound pressure level. 188 characteristics: Applications: Avantages: Disadvantages: Solutions to problems good at mid to high frequencies but at low frequencies need a large thickness of material to be effective. vehicle cabins, pipe lagging, reverberant space absorption. High absorption coefficient in mid to high frequency range; no health risk due to fibres. not oil, water or chemical resistant; expensive, fire and smoke hazard; will not tolerate high temperatures. Helmholtz resonators: Absorption mechanisms: viscous friction losses around neck of resonator associated with large air particle motion in the centre of the neck and zero motion at the walls of the neck at resonance. good in a narrow band of frequencies around the design frequency. electrical transformer enclosures, vehicle mufflers, mufflers for tonal noise generated by large industrial fans. can be made to be immune to moisture, oil and chemicals. Can withstand high temperatures; no health risk. Narrow frequency range of effective absorption. Absorption characteristics: Applications: Avantages: Disadvantages: Resonant panels Absorption mechanisms: Absorption characteristics: Applications: Avantages: Disadvantages: panel and backing cavity damping losses. good in a narrow band of frequencies around the design frequency. electrical transformer enclosures, auditoria, concert halls, reverberant spaces. can be made to be immune to moisture, oil and chemicals. Can withstand high temperatures; no health risk. Narrow frequency range of effective absorption. Sound in enclosed spaces Problem 7.35 189 We require ¯ = 0.8 at 125Hz. Referring to figure 7.8 in the text (p.307) it is α clear that we need to use curve "D" which implies the use of sound absorbing material behind the panel. From figure 7.9, it can be seen that the 125Hz line crosses curve "D" when the cavity depth is 110mm and the panel mass is 2kg/m2. Thus this is the required design. Note that the guide notes in the caption of figure 7.8 should also be included as design specifications. Problem 7.36 Following equation 7.78 in the text: α ' ¯ α j S i¯ i j Si 2 × 6.84 × 5.565 × 0.02 % 2 × 5.565 × 4.72 × 0.05 %2 × 6.84 × 4.72 × 0.06 ' 2 × 6.84 5.565 % 2 × 5.565 × 4.72 % 2 × 6.84 × 4.72 ' 8.023 / 193.2 ' 0.042 Problem 7.37 Room volume = π × 72 × 2.5 = 384.85m3. The optimum reverberation times are calculated using equation 7.121 in the text with K = 5. The calculated values are listed in the following table. Octave band centre frequency (Hz) 125 250 500 1000 2000 4000 8000 T60 1.45 1.06 0.96 0.96 0.96 0.96 0.96 190 Solutions to problems (b) Using equation 7.51 in the text, the recommended S¯ is 65m2 at 500Hz α and above, 59m2 at 250Hz and 43m2 at 125Hz. (c) Area ceiling = π × 49 = 154m2. At 125Hz, the Sabine absorption coefficient of tile is 0.2. Thus if all the ceiling were covered the α maximum S¯ would be 154 × 0.2 = 30.8m2 (assuming that the floor and walls contributed a negligible amount). Alternatively, if it is assumed α that the floor and walls are of concrete with ¯ = 0.01, then the total α ¯ = 30.8 + 0.01 × (154 + 2π × 7 × 2.5) = 33.4m2. The recommended S¯ α at 125Hz is 43m2 (from part (b)), so the ceiling tile would NOT be adequate. (d) A compromise would be to use sufficient tiles to achieve as closely as possible the required absorption over the range 500 to 1000Hz and then design a panel absorber with a maximum absorption at 125Hz and no absorption at 500Hz. The required absorption in the range 500 to 4000Hz is 65m2. Thus the optimum amount of tile is 65/0.8 = 81m2, which will be OK at 500 and 2000Hz, a bit much at 1000Hz and a bit little at 4000Hz, but nevertheless a good compromise. We are then left with an area of ceiling of 154 - 81 = 73m2 for panel absorbers. The amount of absorption needed at 125Hz is 43 - 3 - 81 × 0.2 = 24m2. So the panel absorber must have an absorption coefficient of 24/73 = 0.33 at 125Hz and nothing at 500Hz. Choosing curve H from figure 7.8 in the text and allowing for the fact α that the 125Hz band includes the peak, the average ¯ for the 125Hz band is approximately 0.35. The absorption coefficient of the panel at 500 and above is likely to be 0.08 and at 250Hz it will be 0.12. To optimise the required absorption coefficients, we can vary the relative areas. The area of floor and walls is 264m2. Thus the amount of absorption due to the floor and walls is 2.6m2 in the 125, 250 and 500Hz bands and 5.3m2 in the 1000, 2000 and 4000Hz octave bands. Including this and with an area of 70m2 of tile and 65m2 of panel, the amount of absorption in the octave bands from 125Hz to 4000Hz is 41, 53, 64, 70, 67, and 63m2 which is a little low at 125Hz and 250Hz (optimum is 43 and 59 respectively) and a little high in the other bands (optimum at 500Hz and higher frequencies is 62m2). Choosing the area of panel = 75m2 and the area of tile = 65m2, gives the following amounts of absorption: 42, 51, 61, 67, 63, 60m2 which is close enough. Note that there are many other adequate solutions to this problem. Sound in enclosed spaces Problem 7.38 (a) sound power of the source is given by: Lw ' Lp % 10log10(2πr 2) & 0.15 ' Lp % 21.85 191 Thus the following table may be constructed. Octave band centre frequency (Hz) Lp (dB re 20µPa) Lw (dB re 10-12W) 63 90 125 250 500 85 78 73 95 1k 70 92 2k 65 87 112 107 100 (b) Room dimensions 5m × 3m × 2m. Using equation 7.17 in the text, we have: 343 1 × ' 34.3 Hz f1,0,0 ' 2 5 343 1 × ' 57.2 Hz f0,1,0 ' 2 3 343 1 × ' 85.8 Hz f0,0,1 ' 2 2 343 2 × ' 68.6 Hz f2,0,0 ' 2 5 343 × (1 / 5)2 % (1 / 3)2 ' 66.7 Hz f1,1,0 ' 2 Thus the 3 lowest order modes are in the 31.5Hz and 63Hz third octave bands. Room volume, V = 5 × 3 × 2 = 30m2 Area, S = 2(5 × 3 + 5 × 2 + 3 × 2) = 62m2. Perimeter, L = 4(5 + 2 + 3) = 40 The modal density is given by equation 7.21 in the text. Thus: dN 4 × π × 1252 × 30 π × 125 × 62 40 ' ' 0.264 % % df 8 × 343 3433 2 × 3432 192 Solutions to problems From table 1.2 on p43 in the text, the 125Hz third octave bandwidth is 141 - 113 = 28Hz, so the number of modes in the band is 28 × 0.264 = 7 to 8 modes. Alternatively equation 7.20 could be used to calculate the number of modes occurring below 141Hz and 113Hz and then take the difference. The number of modes below 141Hz is: N ' 4 × π × 1413 × 30 3 × 343 3 % π × 1412 × 62 4 × 343 2 % 40 × 141 ' 19 modes 8 × 343 The number of modes below 113Hz is: N ' 4 × π × 1133 × 30 3 × 343 3 % π × 1132 × 62 4 × 343 2 % 40 × 113 ' 11.4 modes 8 × 343 Thus the number in the 125Hz octave band is between 7 and 8 modes. (c) Equations 7.51, 7.43 and 7.42 (with a correction for ρc = 413.6) in the text, and the knowledge that S = 62m2, may be used to construct the following table. Octave band centre frequency (Hz) Lw (dB re 10-12W) T60 S¯ α 63 112 5.5 0.877 125 107 5 0.964 250 100 4 1.206 500 95 3 1.608 1k 92 2 2.411 2k 87 1.5 3.215 Overall α ¯ R ' S¯ α 1&α ¯ 0.0141 0.015 0.019 0.025 0.038 0.0519 6 4 9 9 0.890 118.7 -26.2 92.5 0.979 113.3 -16.1 97.2 1.230 105.3 -8.6 96.7 1.650 99.0 -3.2 95.8 2.509 94.2 0.0 94.2 3.391 87.9 1.2 89.1 102.8 Lp(reverb) A-weighting Lp (dB(A)) Sound in enclosed spaces Assumptions: ! ! ! 193 A-weighting assumed uniform across each octave band when in fact it varies continuously with frequency. Direct field contribution assumed negligible. Reflections from and absorption of surfaces other than room boundaries is excluded. (d) If 2 more generators were added, the sound pressure level would increase by 10 log10(2 % 1) ' 4.8 dB(A) . (e) Ceiling tiles added with area = 15m2. Remaining room surface area = 62 - 15 = 47m2 = Sfloor, walls. The following table may be constructed, where: αoverall ' αnew ' ¯ ¯ S αfloor walls % S αceiling ¯ ¯ Sfloor walls % Sceiling 63 125 250 . Octave band centre frequency (Hz) 500 1k 2k Overall Lw (dB re 10-12W) 112 0.66 0.15 2.25 107 0.73 0.25 3.75 100 0.91 0.55 8.25 95 1.22 0.85 12.75 92 1.83 1.0 15 87 2.44 1.0 15 0.2813 24.27 S¯floor walls α α ¯ceiling S¯ceiling α α ¯overall ' ¯new α Rnew ' S¯new α 1 & αnew ¯ 0.0469 0.0722 0.1477 0.2253 0.2715 3.05 4.83 10.75 18.03 23.10 Lp(reverb) A-weighting 113.3 -26.2 87.1 106.3 -16.1 90.2 95.9 -8.6 87.3 88.6 -3.2 85.4 84.5 0.0 84.5 79.3 1.2 80.5 114.2 Lp (dB(A)) 94.5 Similar assumptions as made in part (c). 0810 0.25V/ ScT60 Thus the following table may be constructed.111 0. The existing mean statistical absorption coefficient may be calculated using equation 7.096 There are many possible solutions to achieve the desired mean absorption coefficients.1 2.6 2. The amount of material required is then calculated on the basis of achieving the optimum reverberation time in the octave bands most important for speech. S = 2(20 × 15 + 20 × 4 + 15 × 4) = 880m2. One alternative is to look for the frequency at which the additional absorption required is the largest and choose a material which has a maximum absorption coefficient at this frequency.211 0.1040 0. with 10% increase at 250Hz.93 0.107 which would satisfy the requirements at and above 1000Hz.211 0.8 0. so a material thickness of 25mm should be chosen.3 2. let x be the fraction of room surface area to be covered with absorbing material.064 0. namely.0993 0.0 1.107 0.93 0. 500Hz to 2000Hz.39 Solutions to problems Room 20m × 15 × 4m. it would seem that the required increase in mean absorption coefficient is 0.93 0.194 Problem 7.0911 0.1149 1. In this case.145 0.040 0.86 1.40 1. V = 20 × 15 × 4 = 1200m3. the maximum increase in absorption is needed at 500Hz. we .0706 0.211 0.103 0. In this case.107 0. The desired reverberation times are calculated using equation 7.02 0. Octave Existing Existing Desired Desired Required band mean αst T60 mean increase in T60 centre αst αst frequency (Hz) 63 125 250 500 1000 2000 4000 8000 3. with a compromise of a little less than desired at 500Hz.1040 0. 329 in the text.211 0.121 on p.93 0.211 0.0 2. 50% increase at 125Hz and 100% increase at 63Hz.107 0.0 2.58.56 in the text rearranged to give: αst ' 1 & e &55.112 0.194 0.1040 0. Using equation 7.0 2.93 0. 995 × 10&3 × 0.0711. we obtain: log10 (0. T 10 level is 88dB.15) % (1 & x)log10(0. Problem 7.11.106 in the text.7 ' 1.206 × 343 The power radiated through the window is then W ' For an incoherent plane source.995 × 10&3 he reverberant sound pressure ¢ p i ¦ ' 4 × 10&10 × 1088 / 10 ' 0.5 ' 40 . Thus: ¢p 2¦ ' 1.211) ' (1 & 0.106.349 × 10&9 Pa 2 .79) ' x log10(0. Thus 2 &TL / 10 &2.40 Assume that the house is approximately in a direction along the normal axis from the window. Thus 4ρc The power radiated through the window is then W ' τ ¢ pi ¦ S 2 W ' ¢ pi ¦ S 2 4ρc where τ ' 10 = .206 × 343 × 4.Sound in enclosed spaces have: (1 & 0.896) which gives x = 0. Thus from figure 5. The power incident on the window is the intensity in the direction of the window multiplied by the area of the window.565 × 10&7 2 × π × 60 2 ' 8.104)(1 & x) 195 Taking logs of both sides. The quantity r / HL ' 60/1.252 Pa 2 1.252 × 1.457 µW 4 × 1. So the required area of 25mm thick acoustic material is 63m2. the on-axis sound pressure at the receiver is given by equation 5. it is clear that we can treat it as a hemispherically radiating point source producing a sound pressure described by equation 5.85)x × (1 & 0.5 ' 0. Choosing a segment of plate as shown in the figure we can calculate the ratio of holes to total area of the segment and set this equal to 0.2 dB re 20µPa As the ground is hard asphalt. of the text on p528.0022 ' 6. This gives a hole spacing for the parallel holes of: q ' 1000 × π × 0.7 mm 4 × 0.07 × 2 3 and for the staggered holes. we have: .7mm.07. we may add 3dB to the level to account for the effect of ground reflection.07 π × 0. Problem 7. edn. Let q be the distance between holes as shown in the figures. Thus the expected level at the house is 16dB.0022 2 × 0. Using equation E.1 mm For the purposes of this problem we will use q = 6. Could assume parallel or staggered holes as shown in the two figures to follow. q ' 1000 × ' 5.41 (a) First calculate distance between holes.7 in the 2nd.349 × 10&9) % 94 ' 13.196 Solutions to problems This corresponds to a sound pressure level of : Lp ' 10 log10(8. we would get 691 Hz.85 × 0. but the error is greater than 15% because the condition.77 in the 3rd edn.1 343 197 7 × 0.5256 as illustrated in the table below.06592 % 0.1696) × 0.63259.5263 1.9: a(X) ' 3.5000 1.06590.41 and to evaluate that we need equations C.43.5256 Can solve by trial and error.1 is not satisfied.0339 1.22 × 0. Referring to equation C.1 / 100 0.002 × (1 & 0.63259 & 3. If we used Equation 7.06592 & 0.206 × 543/10000 = 0.5254 Thus the frequency of maximum absorption is 543Hz.80849 and .16632 × 1. Referring to equation C.Sound in enclosed spaces fmax ' 2 × π × 0.00183 ' 1. Thus. T3 = 3. fL/c < 0. (b) Specific normal impedance is given by equation C.0067) Rewriting gives: 0.002 / 0. To evaluate this equation we need to use equation C. fmax(Hz) 100 500 600 550 540 543 542. choosing values of fmax until the LHS = 1.1663.0655.4. T2 = 0.1696.632592 ' & 0.632592 3.003 % 0.00183 × fmax tan fmax × 0.1467 1.1696(1.5902 1. T1 = 1.16962 × 0.15.9 LHS 0.3 and C. T4 = 0.1892 2.16632 × 1.1 343 tan fmax × 2 × π × 0. X = 1. Thus.9667 ' 2.80849 + j0.1915 % j0.25285 % j0. T3 = 3.19923 & j0.632592 ' 0.8375 ' 1.24893 )&1 Using equations C.056362 % 0.3 .6270e&j0.6616 . we obtain: Zm ' ρm κ ρc ' 1 (0.093188) 1 0.28068 ' ' 1. T4 = 0.8085 % j0.54152 ' & 0.6 as follows: κ ' ( 1 & 0.24893.4 × (& 0.24893) × (1.5415 & 3.23297 Thus τ = -0.84133×0.4188 ' 1.856 × 1.0785(1. T2 = 0.4864 & j0.84133 and b(X1) ' 1.23297 and σ = -0.05636.54152 ' 0.6471 e& j0.07852 × 0.206 × 543/10000 = 0.198 b(X) ' Solutions to problems 1.24893 X1 = 0.2489 )&1 ' ( 0.16632 × 1.165262 × 1.165262 × 1.056362 & 0. The quantities κ and ρm may be calculated using equations C.07850. Referring to equation C.165262 × 1.06592 % 0.056362 % 0.54152 3.592 + j0.0659 × 0. T1 = 1.093188 ) &1 and ρm ' ( 1 & 0.1663 3.23297 )) &1 ' ( 1.632592 × 0.16526 3.9: a(X1) ' 3.07852 × 0.7717 & j1.16962 × 0.49807 % j0.54150.1915 % j0.05636 × 0.19923 % j0.16526.07850) × 0.05605.5 and C.54152 × 0. 3891 To calculate the overall impedance. Using the previous result for km.24893 199 ' 9.7118) % e& 0.41 (assuming a rigid backing for the porous material). but first we need to evaluate R and tan(kR).1375 & j0.093188) × (0.19923 & j0. we can write: e0.9269 cos(1.4b we obtain: km ' 2πfmax c ρm κ ' 2π × 543 343 1. and setting R = 0.1915 % j0.098638 ' 9.1915 & j0.9468 (1. From equation 9. we can write: ZN ρc ' (0.7118) & jsin(1.1. we use equation C.3480 Using the previous results and equation C.9269 cos(1.09319 & j3.3480 & j0.7118) &0.7118) e0.4663e& j0.4864) 2.7127 ' 0.1097 j tan(kmR) ' ' Thus: tan(kmR) ' &0.43.9269 cos(1.0834 & j0. it will be useful to evaluate the quantity.99259 ' 19.7118) % jsin(1.25 in the text.2692 Before continuing.1205 % j1.9468 2.8934 ' 1.9269 cos(1.7118) % jsin(1.Sound in enclosed spaces and using equation C.24893) 0.4107 % j2.8316 ' 1.8300 e& j0.2074 ' 9.9468 3.1205 &0.7118 & jsin(1. the effective length of the holes in the perforated sheet is: . tan(kmR).7118 & e& 0.1178 & j9.3480 1.19923 & j093188 0.49629 ' 17.6616 % j1.2995 % j2.1205 & j 1. 9467 .0022 / 4 ' 3. A ' π × 0.1416 × 10&6 m 2 .002 ' 0. is the largest of the half plate thickness or t.003 % Thus: Solutions to problems 16 × 0. The quantity.001 / 0.43 × 0.8 × 10&5 ' 9. To evaluate this equation we need the following quantities: k ' (2 × π × 543) / 343 ' 9. The above quantities may be inserted into equation 9.29 in the text. D ' π × 0.006283 t ' 2 × 1.0067 ' 0.206 × 2 × π × 543 ε = 1.001 1 & 0.0 as radiation from a baffle.0015m.200 R ' 0.004588 3×π tan(kR) ' tan (2 × π × 543 × 0.004588 / 343 ) ' 0. Thus: h = w/2 = 0.29 to give: . h.04567 We also need to calculate the acoustic resistance of the holes using equation 9.3538 × 10&5 1. Sound in enclosed spaces Ra A ρc ' 9.9467 × 9.3538 × 10&5 × 6.283 × 10&3 × 0.003 2 × 3.1416 × 10&6 × 1 % (1.4 & 1) 5 3 × 1.4 4 × 3.1416 × 10&6 π × 0.00152 201 % 0.288 × 9.9467 × 9.3538 × 10&5 × log10 % 3.1416 × 10&6 × 9.94672 2×π ' 4.00923 × 10&3 % 6.69555 × 10&5 % 4.9468 × 10&5 ' 4.1256 × 10&3 We can now use equation C.43 to evaluate the overall impedance. The second term on the right is the impedance due to the perforated sheet and is: ZP ρc ' (100 / 7) (0.04567j % 4.1256 × 10&3 ) 100 × 1.206 × 343 × 0.04567 & 4.1256j × 10&3 1% 2 × π × 543 × 21.762 × 7 0.05894 % 0.6524j 1.00363 & 3.2836j × 10&4 ' ' 0.05894 % 0.6548j Thus the total impedance is: ZN ρc % ZP ρc ' 1.1375 & 0.3891j % 0.05894 % 0.6548j ' 1.1964 % 0.2657j ' 1.2255 e0.21854j cosβ = 0.9762; cos2β = 0.9060; sinβ = 0.2168 and ξ = 1.2255. Using the above data, the statistical absorption coefficient may be calculated using equation C.37 in the text as follows: 202 αst ' 8 × 0.9762 1.2255 Solutions to problems 1& 0.9762 1.2255 × loge ( 1 % 2 × 1.2255 × 0.9762 % 1.22552 ) % 0.9060 1.2255 × (0.2168) × tan&1 1.2255 × ( 0.2168) 1 % 1.2255 × 0.9762 ' 6.373 × (1 & 0.7966 × 1.5881 % 3.410 × tan&1 [0.12097]) ' 6.373 × (1 & 1.2651 % 0.4105) ' 0.93 Problem 7.42 The NRC is given by: 0.6 % 0.8 % 1.0 % 1.0 ' 0.85 ' 4 4 So the material is adequate for the purpose. NRC ' Problem 7.43 α250 % α500 % α1000 % α2000 ¯ ¯ ¯ ¯ Truck emits 110 dB. This is equal to: W ' 10 &12 × 10110 / 10 ' 0.1 watts . r/a = 60/6 = 10. From Fig 7.14, reverberant field pressure squared is: 10 log10 ¢ pR ¦ & 10 log10 2 2 Wρc πa 2 ' & 4 dB 0.1 × 413.6 & 4 ' & 18.4 dB π × 36 Thus, 10 log10 ¢ pR ¦ ' 10 log10 Direct field pressure: 10 log10 ¢ pD ¦ ' 10 log10 2 so reverberant field pressure is, LpR ' &18.4 % 94 ' 75.6 dB 0.1 × 413.6 4 π × 602 ' & 30.4 dB Sound in enclosed spaces So direct field pressure is, LpD ' &30.4 % 94 ' 63.6 dB Total pressure, Lp ' 10 log10 1075.4 / 10 % 1063.6 / 10 ' 75.7 dB 203 Assumptions: • Effective acoustic source location is in the centre of the cross section • Specularly reflecting surfaces • Ambient temperature of 20 EC 8 Solutions to problems relating to sound transmission loss, acoustic enclosures and barriers Problem 8.1 (a) The Transmission Loss of a partition is an inverse decibel measure (bigger TL means a smaller amount of transmitted energy) of the amount of incident energy which is transmitted to the space on the side opposite that on which the energy is incident. It is defined in terms of the transmission coefficient, τ, which is the fraction of transmitted to incident energy, as follows: TL ' &10 log10 τ It may be measured using two reverberant rooms with the panel to be measured acting as a partition between the two rooms with the space around the panel of high transmission loss construction so that all of the significant acoustic energy transmitted between the two rooms passes through the panel under test. The test is conducted by exciting the one of the reverberant rooms with 1/3 octave band noise and then measuring the difference in the space averaged sound pressure level in the two rooms. The appropriate mathematical analysis is embodied in equations 8.13 to 8.16 in the text (page 342). (b) Measurements often do not agree with theoretical calculations because the latter do not take into account the size of the panel exactly. Also the experimental determination of space average sound pressure level is often characterised by errors, especially at low frequencies when the sound fields in the two rooms are not sufficiently diffuse. Sometimes, energy is transmitted from one room to the next by way of paths not through the panel (called "flanking"), resulting in Transmission Loss measurements which are too small. Sound transmission loss, acoustic enclosures and barriers Problem 8.2 205 Mass Law Transmission Loss is obtained by combining equations 8.34 (with θ = 0E) and equation 8.35b, which gives: TL ' 10 log10 1 % πfm ρc 2 & 5.5 (dB) Substituting in values for the variables gives: 4 ' 10 log10 1 % π × f × 7800 × 0.01 988 × 1481 2 & 5.5 Rearranging gives: f ' 100.95 & 1 2.8046 × 10&8 ' 16,800 Hz The Transmission Loss in air at this frequency is much greater because the impedance of the panel compared to the characteristic impedance in the propagating medium is much larger for air than water. Problem 8.3 75 25 75 z0 25 All dimensions in mm 45 o zn 45 o 30 75 (a) First find location of neutral axis by taking moments about an axis through the centre of the angled section and shown as z0 in the figure. In the following equations bi is the length of the ith section. If the neutral axis is denoted as zn where z is the vertical coordinate on the figure, then: 206 i Solutions to problems j b izi0 z0 & zn ' j bi i ' 75 × 15 % 2 × 25 × 2.5 & 25 × 10 % 75 × 15 & 75 × 15 75 % 3 × 25 % 75 % 2 2 × 30 %75 1000 ' 2.6 mm 384.9 ' Thus the neutral axis is 12.4mm from the centre of the top of the section. The section thickness, h = 1.2mm and the horizontal length, R, before repeating itself is 250 + 60mm = 0.31m. The bending stiffness in the direction along the ribs may be calculated with E = 207GPa and ν = 0.3 using equation 8.10 in the text, which is incorrect in the first printing of the text and should be: B ' Eh (1 & ν2)R j b n zn % 2 n h 2 % bn 24 2 % h 2 & bn 24 2 cos 2θn Thus: B1 ' 207 × 109 × 0.0012 0.00122 0.15 0.01242 % 0.91 × 0.31 12 % 0.05 0.00012 % 0.0252 0.00122 % 0.025 0.01262 % 12 12 0.00122 12 0.00122 % 2 × 0.032 24 % 0.075 0.01762 % % 0.06 2 0.00262 % ' 8.805 × 108 ( 2.3082 × 10&5 % 2.6046 × 10&6 % 3.9720 × 10&6 % 2.3241 × 10&5 % 6.9427 × 10&6 ) ' 5.27 × 104 kg m 2 s &2 acoustic enclosures and barriers 207 The stiffness in the direction across the ribs may be calculated using equation 8. the first resonance frequency of the panel may be calculated using equation 8.3 % 40.6 and: .385 ' 40.22 in the text with i = n = 1.5 5.62 1/4 1/4 ' 650 m/ s cB2 ' ' 108 m/ s (c) The lower and upper critical frequencies for the panel may be calculated using equation 8.00123 0.62kg m-2 and the frequency is 1000Hz.7 × 0.91 (b) The bending wavespeed is calculated using equation 8.62 40.62 5.62 40.5 as follows: B2 ' 207 × 109 × 0. Thus: fc1 ' 3432 2π 11.7 kg m 2 s &2 × 0.3 in the text.1 in the text.Sound transmission loss.31 12 × 0. Thus the lower and upper bending wavespeeds corresponding to waves propagating parallel and perpendicular to the ribs respectively are: cB1 ' 5.27 × 104 × 4π2 × 106 11.00123 3 × 2.7 1/2 1/2 ' 278 Hz fc2 ' 3432 2π ' 10.385/0.3 % ' 7934 207 × 109 × 0. Thus: Bab ' 0.27 × 104 × 0.0012 × 0.27 × 104 11.000 Hz (d) Assuming that the enclosure wall edge condition is simply supported (a good approximation in practice for most enclosures). The surface mass of the panel is m = 7800 × 0.7 × 4π2 × 106 11.31 = 11. the curve should only be used to find 1/3 octave values and the octave band levels must then be calculated from the following equation: TLoct ' &10Log10 6 (1/3) [10 &TL1/10 % 10 &TL2/10 % 10 &TL3/10 ]> .2 ' 34.4 dB At point D (20. the TL is: TLD ' 10 log10(11.2 ' 19. the TL is: TLB ' 20 log10(278) % 10 log10(11.000Hz).2 dB At point B (278Hz).000Hz). Strictly speaking.4 Hz (e) The sound transmission loss of the panel may be calculated using figure 8.62) & 54 ' 16.27 × 104 24 40.62 % % ' 28.7 24 7934 24 1/2 f1.4 dB These points are plotted on the following graph and interpolation is used to find the octave band TL values.1 dB At point C (5.8b in the text.1 ' 2 11.62) & 10 log10(278) & 20 log10[ loge(4)] & 13.62) & 10 log10(278) & 20 log10[ loge(20000/278)] & 13. Point A is at 139Hz and the corresponding TL is given by: TLA ' 20 log10(278 × 11.208 π Solutions to problems 5.62) % 15 log10(10000) & 5 log10(278) & 17 ' 41. the TL is: TLC ' 20 log10(5000) % 10 log10(11. 48mm.4 (a) Only one half of the sine wave section needs to be considered. Following the figure. .5fc1 fc1 0. for most practical purposes. y1 = 20 sin(15π/40) = 18. acoustic enclosures and barriers 209 C A B 0. the octave band results are summarised in a table.5fc2 However. in the case of isotropic panels.Sound transmission loss. However. the results obtained that way are little different to the results obtained by reading the octave band levels directly from the figure. care should be taken to avoid errors near the dip in the curve corresponding to the critical frequency. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Transmission Loss (dB) 9 15 19 21 25 29 33 37 Problem 8. 8 + 10)/40 = 18.482 % ' 8.210 Solutions to problems b1 = 23.48 / 2)2 % 1.62 % 23.9E b2 = 10mm. The lower and upper critical frequencies are calculated using equation 8.8 × 2 (18.91 The surface mass is m = 7800 × 0.0016 × (23.8E) 24 % 10 18.9E l = 40. θ1 = 50.0kg/m2.8 + 23.04 × 10 % 9 23.82 cos(101.82 24 1. we can write the following for the bending stiffness for waves travelling parallel to the corrugations: B1 ' 207 × 109 × 0.3 as: fc1 ' 3432 2π 3432 2π 18 8.63 × 10&9 57.3 y1 1 15 10 15 Using the corrected form of equation 8.10 in the text.62 12 1. θ2 = 0 b3 = 23.500 Hz . θ3 = 50.04 × 104 kg m 2 s &2 The bending stiffness for waves travelling perpendicular to the corrugations can be calculated using equation 8.91 × 0.6 ' 111.8 kg m 2 s &2 40 12 × 0.8mm.8 1/2 1/2 ' 280 Hz fc2 ' ' 7. E = 207GPa ν = 0.0016 0.8mm.62 & 23.11 in the text as: B2 ' 207 × 109 × 1.04 × 104 18 111. 0 dB At point B (280Hz).5 ' 12.00163 34 × 3 × 2.8 × 0.Sound transmission loss.3 34 (b) The sound transmission loss of the panel may be calculated using figure 8.22 in the text.2 ' 34. Strictly speaking.0) & 10 log10(280) & 20 log10[ loge(4)] & 13.2 ' 21.0 dB At point C (3.000Hz).6 1/2 8.4 dB At point D (15.8 34 % 111.8b in the text.04 × 104 3 4 % 111.0) % 15 log10(7500) & 5 log10(280) & 17 ' 41. Point A is at 140Hz and the corresponding TL is given by: TLA ' 20 log10(280 × 18. the TL is: TLC ' 20 log10(3750) % 10 log10(18. Thus: f1.4 dB These points are plotted on the graph below and interpolation is used to find the octave band TL values. the TL is: TLD ' 10 log10(18.750Hz).1 ' π 2 18 % 0. the first resonance frequency may be calculated using equation 8. the TL is: TLB ' 20 log10(280) % 10 log10(18.0) & 54 ' 20.5 Hz 8. the curve should only .04 × 104 × 0.0) & 10 log10(280) & 20 log10[ loge(15000/280)] & 13.3 34 % 207 × 109 × 0. acoustic enclosures and barriers 211 Assuming that the enclosure wall edge condition is simply supported. for most practical purposes. in the case of isotropic panels.212 Solutions to problems be used to find 1/3 octave values and the octave band levels must then be calculated from the following equation: TLoct ' &10 log10 6 (1/3) [10 &TL1/10 % 10 &TL2/10 % 10 &TL3/10 ]> However. the results obtained that way are little different to the results obtained by reading the octave band levels directly from the figure.5fc1 fc1 0. C B A 0.5fc2 Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Transmission Loss (dB) 13 19 21 23 26 30 35 38 . care should be taken to avoid errors near the dip in the curve corresponding to the critical frequency. However. the octave band results are summarised in a table. Following the figure. m = 0.55 × 3432 / (2000 × 0.5fc (d) Second panel.6 dB At point B.8 1/2 ' 10.013 × 1000 = 13kg/m2. A 0.600 × 36) & 54 ' 57.1) & 45 ' 56.6 dB The results are plotted on the graph below from which can be read TL values as a function of 1/3 octave band centre frequency. m. acoustic enclosures and barriers 213 (c) With viscoelastic damping the panel may be treated as isotropic with the surface mass.300Hz (10. . Surface mass.Sound transmission loss. The 1 subscript is used because this critical frequency is smaller than that for the other panel.600 Hz On the isotropic panel curve.600 × 36) % 10 log10(0. fc1 ' 0.600Hz and the TL is: TL ' 20 log10(10. the frequency is 10. point A is at 5.013) ' 2500 Hz . The critical frequency is: fc ' 3432 36 2π 111. now equal to 2 × 18 = 36kg/m2.600/2) and the TL is: TLA ' 20 log10(10. 1 and the corresponding TL is: TLA ' 20 log10(36 % 13) % 20 log10(81. TLB = 80dB.01) ' 66 dB At point D. Problem 8.5 A double wall partition may perform more poorly than a single partition of the same weight at resonance frequencies corresponding to acoustic modes in the cavity and also at the critical frequencies of the individual panels (if they are lightly damped).7 dB and as there are rubber spacers. Note that the panel with the higher critical frequency (the damped corrugated panel in this case) must be the one which is point supported to obtain the high TL predicted.6) % 40 log10(10600) & 99 ' 80 dB Assuming that there is sound absorbing material in the cavity. The value of TLB2 for linepoint support is: TLB2 ' 20 log10(13 × 0. one panel may be considered to be point supported. . The TL data are plotted in the following figure from which the 1/3 octave values can be read directly. the frequency is f1 = 55/0.214 Solutions to problems Cavity resonance frequency is: f0 ' 80 36 % 13 ' 82 Hz 36 × 13 × 0. At point C.1 = 550Hz.600Hz and the TL is: TLC ' 80 % 6 % 10 log10(0.86) & 48 ' 24 dB At point B the frequency is 5300 Hz and the TL is: TLB1 ' 24 % 20 log10(2500 / 82) & 6 ' 47. the frequency is 10. 5fc2 Problem 8.6 Following the procedure on page 360 in the text.2 Hz 0.1 × 1000 × 0.5fc1 ' 1280 Hz 2100 × 0.55 × 3432 ' 2570 Hz and 0. we have at point A: 1000 × 0.4 and the corresponding Transmission Loss is: TLA ' 20 log10 (12 % 12.55 × 3432 ' 7490 Hz 5400 × 0.2) & 48 ' 20.012 fc2 ' 0.012 × 7800 × 0.Sound transmission loss.0016 .0016 f0 ' 80.0 dB The critical frequencies are: fc1 ' 0.012 % 7800 × 0.0016 ' 103.48) % 20 log10(103. acoustic enclosures and barriers 215 D B C A f0 fR 0. 01) ' 47.7 % 6 % 10 log10(0.48 2570 12 7490 & 78 At point C: TLC ' 61.7 dB 12.1 = 550Hz.6) % 30 log10(7490) % 20 log10 1 % ' 61.7 dB The frequency fR = 55/0.2) & 6 ' 42 dB and: TLB2 ' 20 log10(12) % 10 log10(0. D B C A f0 fR 0. The Transmission Loss may thus be plotted as shown in the following figure and 1/3 octave (but not octave) values may be read directly from the figure.5fc2 fc2 .216 Solutions to problems The Transmission Loss at point B is the larger of: TLB1 ' 20 % 20 log10(2568 / 102. 00238 which corresponds to a Transmission Loss of: TL ' &10log10 (2. so τ = 10-30/10 = 0.25 × 10&28 / 10 % 2 × 10&25 / 10 18 From which τw = 7.3dB (c) 25mm crack under the door.0.8 % 1.001. acoustic enclosures and barriers Problem 8. TLw = -10log10(7.8 When designing an enclosure. (b) Maximum TL possible corresponds to τw = 0.73 × 10-4) = 34.25 & 2) % 0. Thus: τ ' 0.975 × 10&2.Sound transmission loss.161 × 10-4. Thus the overall transmission coefficient is: τ ' 7. always (if possible) include sound absorptive material on the ceiling and walls.73 × 10&4 18 which corresponds to TL = 10log10(3.001 ' τw (18 & 0.5dB. as listed in the table below (along with the required noise reduction and corresponding wall TL given by TL = NR + C). then we can write: 0.16 × 10&4 × 15.161 × 10-4) = 31. and τcrack = 1.2 dB Problem 8. . Thus the required wall Transmission Loss is.25 × 10&28 / 10 % 2 × 10&25 / 10 ' 3.25 × 10&2. Effective crack height is 50mm due to reflection in the floor.38 × 10&3) ' 26.7 217 (a) TLoverall = 30dB.025 × 1 × 1 % 0. If we let the required transmission coefficient of the wall be τw. C. thus resulting in an "average" enclosure with the coefficients.75 % 0.5 18 ' 0. 1m.1 × 19 × 23.4) % 20 log10(78) & 48 ' 22 dB at 78Hz This is insufficient as we need 27dB at 63Hz.4 ' 78 Hz 0.55 × 3432 .218 Octave band centre frequency (Hz) Required NR C required wall TL Solutions to problems 63 125 250 500 1000 2000 4000 8000 14 13 27 18 11 29 25 9 34 35 7 42 50 5 55 40 4 44 40 3 43 40 3 43 100mm studs implies gap.4 TLA ' 20 log10(19 % 23. Assume that the door and window have the same TL as the walls.4 kg / m 2 f0 ' 80 19 % 23. TL63 ' TLA % 60 log10(63 / f0) ' 27 That is: 20 log10(m1 % m2) % 20 log10 f0 % 60 log1063 & 60 log10 f0 ' 27 % 48 .003 ' 23.55 × 3432 . fc(steel) ' fc(plaster) ' 0.025 m1 ' 760 × 0.003 0. 4000 Hz ' fc2 5400 × 0. d = 0. To begin. 1600 Hz ' fc1 1600 × 0. It seems that we need to lower f0 below 63Hz. to put this point on the 18dB/octave slope. try 3mm steel and 25mm plasterboard.025 ' 19 kg / m 2 and m2 ' 7800 × 0. 8 dB Assume a stud spacing of 0.1 × 38 × 39 TLA ' 20 log10(38 % 39) % 20 log10(58) & 48 ' 25 dB at 58Hz TLB1 ' 24. acoustic enclosures and barriers or 20 log10(m1 % m2) & 40 log10f0 ' & 33 219 (1) Using the equation for f0 on page 357 in the text and taking logs gives: 40 log10f0 ' 40 log1080 % 20 log10(m1 % m2) & 20 log10(m1 × m2) % 20 ' 96 % 20 log10(m1 % m2) & 20 log10(m1 × m2) Substituting the above expression into equation (1) above gives: 20 log10(m1 × m2) ' 63 dB or m1m2 ' 1410 Try 50mm plasterboard (same as gypsum board).Sound transmission loss.7) & 6 ' 41.9 % 20 log10(810 / 57. m = 760 × 0. Required steel weight = 1410/38 = 37kg/m2.05 0.0 kg / m 2 f0 ' 80 38 % 39 ' 58 Hz 0. fc1 ' 0.005 fc2 ' m1 ' 760 × 0.05 = 38kg/m2.55 × 3432 . use 50mm thick plasterboard and 5mm thick steel plate. That is. 2400 Hz 5400 × 0.7mm thick. which is 4. Thus: .6m and line-line support.55 × 3432 .005 ' 39. 810 Hz 1600 × 0.05 ' 38 kg / m 2 and m2 ' 7800 × 0. The required thickness of steel panel is such that steel weight = 1410/57 = 25kg/m2.005 (see table on p.220 Solutions to problems TLB2 ' 20 log10(38) % 10 log10(0. An easy solution to the problem is to use point supports for the steel panel but this may not be practical. where B B C C A f0 fc2 fR fc2 0. It is clear that the first critical frequency must be increased or the quantity TLB (and hence TLC) must be increased.609 in text).6) % 30 log10(2400) % 20 log10 1 % 39 × 8101 / 2 38 × 24001 / 2 & 78 ' 57 dB As the cavity is filled with sound absorbing material. calculated using Figure 8. Thus try 75mm thick plaster board (m = 760 × 0.9 in the text).5fc2 0. Then: TLC ' 57 % 6 % 10 log10(0. The TL for this construction is plotted on the following figure (where the dashed line is the required Transmission Loss and the solid line is the predicted Transmission Loss.005) ' 40 dB The frequency at point D is fR = 55/0. TLB = 57dB.1 = 550Hz.075 = 57kg/m2. Assume a loss factor for the steel of 0.5fc2 it can be seen that the design is deficient between 1000Hz and 2500Hz. which . 4 TLA ' 20 log10(66.088 ' 66.9 % 23.6) % 30 log10(4000) % 20 log10 1 % 23. This is an odd thickness. where it can be seen that the proposed construction easily meets the noise reduction requirements and is even a little too good. 460 Hz 1600 × 0. which is 2.8 % 20 log10(460 / 61) & 6 ' 38. use 88mm thick plasterboard and 3mm thick steel plate.7mm thick.088 0.088 = 66.4 ' 61 Hz 0. fc1 ' 0.4) % 20 log10(61) & 48 ' 27 dB at 61Hz TLB1 ' 26. That is.55 × 3432 .9 kg / m 2 and m2 ' 7800 × 0. so try 88mm thick plasterboard (3 × 25 + 13) m = 760 × 0. Required steel weight = 1410/66.9 × 40001 / 2 & 78 ' 65 dB As the cavity is filled with sound absorbing material. acoustic enclosures and barriers 221 is 3.4 × 4601 / 2 66.005) ' 48 dB The frequency at point D is fR = 55/0.9 × 23.2mm thick.4 kg / m 2 f0 ' 80 66.9 = 21kg/m2. Assume a loss factor for the steel of 0.003 fc2 ' m1 ' 760 × 0. TLB = 65dB.003 ' 23. .3 dB Assume a stud spacing of 0. Thus: TLB2 ' 20 log10(66. 4000 Hz 5400 × 0.6m and line-line support.005.9 % 23.Sound transmission loss.55 × 3432 .9kg/m2. Then: TLC ' 65 % 6 % 10 log10(0.1 = 550Hz. The TL for this construction is plotted on the previous figure as the grey line.1 × 66.9) % 10 log10(0. 5× 10-4 1.005 × 1 = 0.01m.73× 10-4 0. Effective gap under door = 0.11 in the text. the area of gap in subsequent calculations is determined without doubling the width.22× 10-5 47 52 2. Area under door = 0.005 1.65 and 8.36× 10-5 1.9 Solutions to problems There is a 5mm air gap under door. it can be seen that the required enclosure noise reduction will no longer be achieved in the 1000Hz octave band.19× 10-5 0. Problem 8.005m2.001 2.92× 10-4 2.5 4. Area of walls and ceiling = 2(4 × 2. ΔT = 3EC and H = 0.5 1. 8. the reflection in the floor effectively doubles the width of the gap.79× 10-5 47 63 2. Cp = 1010m2s-2C-1.206kg/m2.0938 0. However. We can construct the following table using equations 8.003 7.05× 10-4 0.12.26× 10-6 55 ¯ τ ¯¯ TL From the table it can be seen that the effect of the gap under the door is to significantly reduce the effective wall TL and on comparing the results in the above table with the table of problem 8.5× 10-4 1.63× 10-5 44 60.05 × 104W. Octave band 63 centre frequency (Hz) 125 250 500 1000 2000 4000 8000 TLwall Swallτwall Sgapτgap 27 0.3× 10-4 3.222 Problem 8.5) + 4 × 3 = 47m2. in which ρ = 1.84 in the text. For the purpose of calculating the transmission coefficient.66 in the text.18× 10-5 41 56.0016 3. τgap is calculated using figure 8. Thus: .97× 10-4 5.00149 0.38× 10-4 39 51 3.8.5 + 3 × 2.11 in the text.10 The required air flow may be calculated using equation 8.10× 10-3 27 45 0.005 2. once the transmission coefficient has been determined using figure 8.80× 10-5 47 49 5. 14 m 3 / s 3 × 1010 × 1.05 × 104 ' 0. . Calculate the noise reduction from the compressor site to the nearest community location due to atmospheric absorption. then the criterion may be more conservative. Use table 4. Check local noise regulations for allowable levels in the residential area. Choose as the design criterion for the compressor noise in the community the smallest of the regulation level and the lowest measured existing level plus 5dB(A). Use the sound power data for the compressor and the excess attenuation data together with equation 5. or if this is impractical use a sound level meter). turbulence.158 in the text to calculate the noise levels due to the compressor at the nearest community location in octave bands. 3.11 The following steps would need to be taken.206 223 The required Insertion Loss specifications for the silencer would be the same as the TL of the walls (not the noise reduction required of the enclosure as this excludes reverberant build-up in the enclosure) and this is found in problem 8.8) to specify the allowable community noise levels in octave bands. 6.Sound transmission loss. 5.8 to convert the allowable dB(A) level in the community to an NR level and then use the corresponding NR curve (figure 4. Problem 8.8. 2. Measure existing levels in dB(A) on the perimeter of the supermarket property at the closest location to the proposed compressor location over an extended period (about a week with a statistical noise analyser. 1. Thus calculate the required enclosure noise reduction and following that. ground effects and meteorological influences. such as the existing level plus 2dB(A). 4. acoustic enclosures and barriers V ' 0. or place a loudspeaker at the proposed compressor location and measure the noise reduction as a function of distance from it. If existing levels were determined using spot checks with a sound level meter. So 500Hz is in the mass law range for all choices of panel.158 in the text with DIM = AE = 0. so 500Hz is well above the first panel resonance frequency (you can calculate this using equation 8. where h is in metres .224 Solutions to problems the corresponding required wall transmission loss. the required panel TL = NR + C. (b) The wall TL may be calculated in the mass law range by using equation 8.12 (a) Compressor is 80m from perimeter. Critical frequency is: fc ' 0.36 in the text (with the assumption NOT made that fm/ρc > 1). Check compressor cooling requirements and if necessary design lined inlet and outlet ducts (with forced air ventilation) with the same Insertion Loss as the Transmission Loss of the enclosure walls. with a constant of 4 instead of 5.55 × 3432 / (5400 × h) ' 12 / h Hz.206 × 343 &4 dB . C = 7 (see table 8. so reduction required = 21dB at 500Hz. Problem 8.79 in the text. The sound level at the receiver with no enclosure (assuming hard ground.6m (stud spacing). so that: Lp ' Lw & 10 log10(2πr 2) ' 105 & 10 log10(2π × 802) ' 59 dB The required noise level = 38dB.21 to be sure).5 to account for octave band calculations. Thus the required enclosure TL = 21 + 7 = 28dB at 500Hz. and for an enclosure lined on the inside.4. zero reflection loss) may be calculated using equation 5. 7. Smallest panel of enclosure = 0. p376 in the text).6m × 0. Thus: TL ' 10 log10 1 % ' 10 log10 1 % πfm ρc 2 &4 2 π × 500 × 7800 × h 1. From equation 8. Trying h = 0.6mm.0 dB TLB1 ' 19 % 20log10(3100 / 114) & 6 ' 42 dB . (c) Should consider the effect of a door and the design of appropriate door seals. Problem 8. TL = 15 + 5 = 20dB Problem 8.1 × 9.79 in the text gives TL = NR + C.4dB which is too low. C = 5dB for an enclosure with surfaces lined with sound absorbing material.88 × 2) % 20log10(113. As we are in mass law range a doubling of panel thickness will increase the TL by 6dB which is a bit high. gives TL = 29. Thus. m1 = m2 = 760 × 0. (b) We may use equation 8. TL = 25. the need for cooling air. f0 ' 80 2 / (0.5dB which is OK. the introduction of the inlet air.0016m.Sound transmission loss.14 (a) The performance of a machinery noise enclosure should not be expressed as a single number dB(A) rating because the dB(A) performance will be dependent on the spectrum shape of the noise generated by the enclosed source.88kg/m2.8) & 48 ' 19. From Table 8.13 Equation 8. the pipe penetration (to be isolated from the enclosure wall) for the compressed air and the need for vibration isolation of the enclosure from the compressor.013 TLA ' 20 log10(9.88) ' 114 Hz fc1 ' fc2 ' 0. The TL due to the panel may be calculated using the procedure on p360 in the text.001m.1m. so choose a wall thickness of 1.013 = 9.55 × 3432 ' 3100 Hz 1600 × 0. d = 0.3.79 in the text to relate TL and noise reduction and thus define the minimum required TL. acoustic enclosures and barriers 225 When h = 0. 4 % 6 % 15log10(0.02) ' 31 dB fR ' 55 / 0. TL2. and TL3.6) % 30log10(3100) % 20log10(2) & 78 ' 50 dB Assuming that there is sound absorbing material in the wall cavity.226 For line-line support: Solutions to problems TLB2 ' 20log10(9.5fc2 fc2 . TLC ' 50. TLB = 50dB. are the (1/3) octave band data read from the figure.1 ' 550 Hz The corresponding TL curve is plotted in the figure below and the octave band data are listed in the following table.88) % 10log10(0. D B C A f0 fR 0. where octave band results are obtained by using 1/3 octave values read from the graph and calculated using: TLoct ' &10log10 6 (1/3) [10 &TL1/10 % 10 &TL2/10 % 10 &TL3/10 ]> where TL1. Sound transmission loss. Use panels of different thickness.5) + 4 × 5 = 65m2 and 10log10SE = 18dB.02 × 50000 ' 0. .84 in the text: V ' H 0. acoustic enclosures and barriers Octave Band Centre Frequency 227 63 125 250 500 1000 2000 4000 8000 TL 14 (from plot) C NR Required NR 13 1 10 21 11 10 15 35 9 26 20 41 7 34 25 47 5 42 30 43 4 39 35 36 3 33 40 52 3 49 13 It can be seen that the enclosure is deficient in the 63Hz. (c) ! ! ! ! ! Use double.70 in the text. (d) From equation 8. so it is not adequate. The power level calculations may be summarised in the following table. Use point-line or point-point support by placing rubber grommets between the panel and stud at attachment points. and 4000Hz octave bands. staggered stud wall. 125Hz.5 + 4 × 2. Thus: Lw ' Lp1 % TL % 10log10S E & C ' Lp1 % NR % 10log10S E where SE = 2(5 × 2.206 × 1010 × 5 (e) The sound power level is given by equation 8. Use thicker panels. Fix additional 13mm thick panels to existing panels with patches of silicone sealant.164 m 3 / s ' ρCp ΔT 1. Machine surface area Sm = 2(2 × 1 + 1 × 1) + 2 × 1 = 8m2.78 in the text and sound powers are converted to sound power levels and vice versa using equation 1. so measurement surface is 4m × 3m × 2m (machine assumed to be resting on the ground). Thus the mean square sound pressure in the enclosure is: ¢ p 2 ¦ ' ¢ pd ¦ % ¢ pR ¦ 2 2 .228 Octave Band Centre Frequency Lp1 NR Lw Solutions to problems 63 125 250 500 1000 2000 4000 8000 80 10 108 83 15 116 78 20 116 73 25 116 70 30 118 60 35 113 60 40 118 60 20 98 (f) Test surface 1m from machine of dimensions 2m × 1m × 1m. Sound pressure levels are converted to mean square sound pressures using equation 1. The ratio. the mean square sound pressure is equal to the sum of the direct and reverberant field contributions. S/Sm = 5.25 in the text with Δ1 = 0 because the measurements are made outdoors.80 in the text. The sound pressure is related to the sound power by equation 6. having an area of S = 2(4 × 2 + 3 × 2) + 4 × 3 = 40m2 (4 sides and 1 top). so the near field correction Δ2 = 0 and the sound pressure level 1m from the machine surface is given by: Lpd ' Lw & 10 log10S ' Lw & 16 dB The results are tabulated in the following table: Octave Band Centre Frequency Lw Lpd 63 125 250 500 1000 2000 4000 8000 108 92 116 100 116 100 116 100 118 102 113 97 118 102 98 82 (g) With the enclosure in place. 2 0. Octave Band Centre Frequency C Lw W 4/R ¢ pd ¦ 2 2 63 125 250 500 1000 2000 4000 8000 13 108 11 116 9 116 7 116 5 118 4 113 3 118 3 98 6.25 31 109 -26 126 115 -16 79 113 -9 50 111 -3 52 111 0 14 105 1 32 109 1 0.43) to give: 4 4 ' R 3.14 0.2 25.20 0. ρc = 400.40 0. acoustic enclosures and barriers The reverberant mean square pressure is: ¢ pR ¦ ' 2 229 4Wρc R where R is the room constant and W is the source sound power.10 0.31 89 -1 LpT dB(A) correct. The enclosure constant C is given in terms of R by equation 8.21 0.3× 10-2 10-3 1.3 97 2.0 102 6.5 102 82 6.40 0.3 × S E S E The results are summarised in the following table.82 in the text which can be rearranged (with the aid of equation 7.29 0.63 0.4 45.5 122 75.063 Lpd ¢ pd ¦ % ¢ pR ¦ 2 2 ¢ pR ¦ 30.2 46. dB(A) sound pressure levels 92 100 100 100 102 97 103 88 .Sound transmission loss.18 0. where for convenience.0 100 4.0 100 4.76 0.3× 0.40 0.4 11.3 0.63 6.33 10C / 10 1 & 0.10 92 0.63 100 4.47 0. so the near field correction Δ2 = 1 and the sound power level of the enclosure is given by: . Noise level on surface. Test surface 1 m from enclosure of dimensions 3m × 3m × 3m.25 in the text with Δ1 = 0 because the measurements are made outdoors.10) Thus the range of variability is 18 to 27dB. The sound pressure is related to the sound power by equation 6.3 on page 225 in the text). Ag is included in Dθ above as the asphalt is hard resulting in essentially hemispherical spreading.72 in the text (with Dθ = 2 due to the hard ground surface) as: Lp2 ' 60 % 10 log10(65) % 10 log10 ' 24 & AE (dB) 2 4 × π × 2002 & AE AE is the excess attenuation given by equation 5.15 Noise level inside enclosure = 101dB at 1000Hz. Am = +5. having an area of S = 2(5 × 4 × 2) + 5 × 5 = 105m2 (4 sides and 1 top). Noise level at 50m distance = 70dB. Lp1 = 60dB and the sound pressure level at distance r from the enclosure is given by equation 8. The ratio.25 in the text. Sm = 2(3 × 3 × 2) + 3 × 3 = 45m2.230 Solutions to problems (h) In the 2000Hz band. S/Sm = 105/45 = 2.165 in the text. Problem 8.8/5 = 2dB (see table 5.3. so measurement surface is 5m × 5m × 4m (assuming that the enclosure is resting on the ground). (a) Sound power level radiated by enclosure may be calculated using equation 6. Aa = 10. -4dB (see table 5. Machine surface area. 1m from enclosure = 91dB. Problem 8.161 with r = 50m and DIM = AE = 0 so that: Lp ' 110 & 20 log10(50) & 10 log10(2π) ' 68 dB (c) As the measured noise level is 70dB.16 Adequate internal absorption is necessary to prevent the build-up of reverberant energy which will compromise the predicted acoustic performance. acoustic enclosures and barriers Lw ' 91 % 10 log10(105) & 1 ' 110 dB 231 (b) We may use equations 5. causing the enclosure wall to vibrate and radiate noise. so the total excess attenuation is -5dB. (d) The machine is probably not well vibration isolated from the enclosure. Vibration could be transmitted by way of the floor or by direct connection of parts of the machine (or attached pipework) to parts of the enclosure. Also.158 and 5. pipework or other equipment not included in the enclosure but attached to the noisy machine could radiate noise which was not apparent prior to installation of the enclosure.Sound transmission loss. (b) One possible disadvantage associated with vibration isolation from the floor is increased difficulty in producing an adequate acoustic seal around the base of the enclosure.17 (a) The enclosure should be vibration isolated to prevent the walls from being excited to vibrate and thus radiate sound which in turn will compromise the enclosure performance. (c) Other factors which could degrade the enclosure performance are ! poor seals around doors and windows . the excess attenuation due to atmospheric absorption and meteorological influences is 68 . The excess attenuation due to the ground effect is -3dB. Problem 8.70 = -2dB. 232 ! ! ! ! Solutions to problems inadequate TL performance of doors and windows inadequate internal absorption pipe penetrations in the enclosure walls not vibration isolated from the walls or poorly sealed acoustically. pipework or other equipment not included in the enclosure but attached to the noisy machine could radiate noise which was not apparent prior to installation of the enclosure. . If the noise external to the enclosure with the loudspeakers operating is lower. then the problem is airborne flanking. pipework or other equipment not included in the enclosure but attached to the noisy machine could radiate noise which was not apparent prior to installation of the enclosure. A test to determine whether the problem was airborne or structure-borne would be to turn the machines off and use loudspeakers in the enclosure to generate the same noise levels inside the enclosure. then the problem is likely to be structure-borne vibration or radiation from equipment attached to the noisy machine but not included in the enclosure. If the exterior noise levels are then the same with the loudspeakers operating as they were with the machine. Problem 8.18 ! ! ! ! ! ! ! ! ! ! ! ! acoustic absorbing material left out of wall cavity or too rigid and touching both walls poor seals around windows and doors glass in double glazing too thin doors of insufficient acoustic performance floor vibration transmitted to enclosure walls because of inadequate vibration isolation wall stud spacing incorrect incorrect wall thickness or wall materials poor seal at base of enclosure poor bricklaying (if brick walls) leading to gaps in the mortar change from original noise sources tonal noise from the machine corresponding to a structural resonance or an acoustic resonance (wall cavity or enclosure space). 5 % 0. p118 in text) as illustrated in the following figure.2 & 0.7. acoustic enclosures and barriers Problem 8. .12) % 10(4.62) % 10(6.5.2 & 0. then the allowable level is Lp = 40 + 20 = 60dB(A) and the existing level is thus acceptable.61) % 10(6. The actual level is over 9dB(A) above the allowable level.3 & 2.5 & 0.32) % 105. A-weighted sound level is given by: LpA ' 10 log10 10(6. it will generate widespread complaints from the community. If the noise only occurs during the hours of 7am and 6pm.10 = 50dB(A). The NR value is 52. a situation which is not acceptable and according to table 4.86) % 10(5.Sound transmission loss.1) % 10(4.7 & 1.10.19 233 (a) The Noise Rating (NR) is obtained by plotting the un-weighted octave band data on a set of NR curves (see fig 4.11.11) ' 59.2 % 10(5 % 0. p167 in the text and is Lp = 40 + 20 .2 dB(A) (b) The acceptable noise level is obtained using table 4. ρ = 1. With a barrier.206 × 500 ' ' 0. there are two propagation paths.d1 ! ! around each end with no ground reflections around each end with a ground reflection (e) Attenuation of ground reflected wave (see figure) Using similar triangles: 0.5 10 d1 ' 10 m ' 2. so the noise will sound neutral.2 on p209.206.862E From table 5.234 Solutions to problems (c) The octave band levels essentially follow the shape of the NR curve (except at 63Hz. (d) With no barrier.5m d1 40 . the direct path and the ground reflected path.5m 1.5 ' .5 1.003 R1 2 × 105 . R1 = 2 × 10-5 . there are 8 paths as listed below: ! over the top with no ground reflections ! over the top with a ground reflection on the source side ! over the top with a ground reflection on the receiver side ! over the top with ground reflections on both sides R S 0. where the level is much lower). d1 40 & d1 β ' tan&1 0. Thus at 500Hz: ρf 1. d1 ' 2.20. β ' 181E and from figure 5. First calculate the reflection loss for each path which involves a ground reflection.5m 3m 0.5m d2 0. ! Over the top. source side reflection R 20m 20m 1.20.5 2.Sound transmission loss.5m 20m 20m Using similar triangles: .5 β ' tan&1 ' 20 & d1 3 .9dB ! over the top.857 m R1 ρf 1/2 0.5m S d1 Using similar triangles: d1 0. AR = 7.9dB. receiver side reflection R S 3m 1. (f) Attenuation due to barrier.857 ' 9. p231 in the text.93E . acoustic enclosures and barriers 1/2 235 β R1 ρf ' 52E From figure 5. the reflection loss on ground reflection is thus 3. d2 ' 6.5m S / d3 20m 20m Using similar triangles: d3 0.5m A d1 20m 20m .5 ' 40 & d3 1.9dB Now calculate Fresnel numbers for all paths over and around the barrier. ' 231E and from figure 5.667 β R1 ρf 1/2 1.20.5m 0.7E . so AR = 3. AR = 7. d3 ' 10 m which is the same as if there were no barrier.0dB ! around each end of the barrier R S 3m 1.5 . Over the top (see figure) B A B B R 1.5 β ' tan&1 Solutions to problems 20 & d2 3 ' . p231 in the text.5 6.667 ' 12.5m 3m ! A S 0.236 d2 1. Sound transmission loss. ! Around the ends (see figure) Top view S A d B R .5 .8 + 0.92 ' 2. between source and receiver depends on the path which is being considered. Δb2 ' 12. Ab2 = 12.156 % 6.05 m A ' [ 202 % 2.834 % 13. B ' [ 202 % 1.01) × 2.156 m.88 are: Ab1 = 11.9.156 % 20.05) × 2.663 m d ) ' [ 22 % 402 ] 1 / 2 ' 40. acoustic enclosures and barriers 237 The distance d.92 ' 0.404 % 20.05) × 2.52 % 6. Ab3 = 15.2. is that from the image source to the receiver.58 N2 ' (2.834 % 13.8572 ]1 / 2 ' 2.3332 ]1 / 2 ' 13. the corrections of equation 8.404 % 6.52 ]1 / 2 ' 20. B )) ' [ 32 % 13.1 = 15.5.8 .8 .900 % 17.5 + 0.056 & 40.8 + 0.91 N3 ' (20.9.92 ' 1. Δb3 ' 15. Δb4 ' 17.14.01) × 2. Ab4 = 17.31 From figure 8. p389 in the text: Δb1 ' 11.52 ]1 / 2 ' 20.834 m.2 = 17. the value of d (denoted d') below.9004 m. for waves reflected from the ground on the source side of the barrier only.92 ' 0.663 & 40. A )) ' [ 32 % 17.1432 ]1 / 2 ' 17.900 % 17.663 & 40.01 m N1 ' (20.0 dB It can be shown that in this case.76 N4 ' (2. etc.056 m d ' [ 402 % 12 ] 1 / 2 ' 40.6672 ]1 / 2 ' 6.056 & 40.52 % 2. A ) ' [ 0.1 = 12.0 = 11.404 m B ) ' [ 1. For example.0 + 0. 3 ' 19.01 m.0 dB NR due to barrier is then calculated using equation 1.9 % 2 × 10&(1.8 ' 18.5m S A 10m A 0.70) % 10&(1.5 % 9.0 dB Reflected waves (one each side) R 1.622 m.52 % 2. d ' 40.7 dB .622 × 2 & 40.3 ' 19.97 in the text as: NR ' 10 log10 10&0 / 10 % 10& 3. 6 ' 18.52 ]1 / 2 ' 10.32 m B ) ' [ 202 % 12 % 52 ] ' 20.90 % 0.05) × 2.59 % 0.7 dB Thus Ab7.92' 3.6 ' (20.5m S 0. 8 ' (10.238 Solutions to problems Direct waves.29 % 0.14 in the text: N7.2 ' 10.64 & 40.05 m From figure 8. From figure 8.9 / 10 & 10 log10 10&1. no reflection (one each side): A ' B ' [ 202 % 52 % (1.5)2 / 4 ]1 / 2 ' 20.49) % 2 × 10& 1.7 dB Thus Ab5. Δb5. Δb7.64 m.72 % 1.6 .15 % 10&(1.01) × 2. 8 ' 18.6 ' 18.39) ' 1.92 ' 3.5 & 0.7 % 0.7 % 0.14 in the text: N5.5m B 3m 20m 20m A ) ' A )) ' [ 102 % 0.6 .79) % 10&(1. d ) ' [ 402 % 22 ] 1 / 2 ' 40.32 × 2 % 20. It is left to the reader to verify this. .8.Sound transmission loss. acoustic enclosures and barriers Problem 8. Octave band centre frequency (Hz) Existing noise NR-50 Required reduction 63 125 250 500 100 200 400 8000 0 0 0 77 67 10 65 60 5 67 54 13 63 50 13 58 48 10 45 46 0 40 44 0 68 76 0 Noise reduction is a function of Fresnel number which is directly proportional to frequency. p155 in the text and the following table may be generated.20 239 A d 1.5m 2m 48m NR-50 octave band values may be read from figure 4.5m B ?? 1. If this is satisfied. Total of 8 paths (4 over top and 4 around sides) to consider to get an overall reduction of 10dB at 125Hz. the reduction at 500Hz will be OK. The important frequency is then 125Hz. Consider first the paths around the barrier edges as the attenuation around these paths is independent of barrier height. 6m Path e = ( 252 % 1.0 % 0.38m .52 % 2.6 ' 17. b/5 = 25/48.240 Solutions to problems Around edge (elevation) S 1.744m and N = (2/ λ ) × 3.6 dB .180 m By similar triangles. λ = 343/125 = 2.88 in the text (assuming an omnidirectional source) is 20 log10(53.64 / 50) ' 0.5m h c f g j e 1.260 & 50 ' 3.0dB.14 on p389 in the text. f = 1. Path with ground reflection From the figure.385m Path j = (482 + 52)1/2 = 48.260m A % B & d ' 5.65. The correction term given by equation 8.6042 ) 1 / 2 ' 25.6 dB . Path with no ground reflection Path h = (22 + 52)1/2 = 5. f/1.5 = 23/25.64 = 2. Δb = 17.385 % 48. so Ab ' 17. b = 2.5m ?? 2m 23m 25m Plan 2m S 5m 23m b 25m R As the source and receiver are at the same height. the location of the ground reflection will be mid-way (in a horizontal direction only) between the source and receiver.64 m At 125Hz. and using similar triangles. From figure 8. A4 1.5m A1. Δb = 17. B2 h-1.64 m Thus.382 % 2.387 m A % B & d ' 5. B4 h 1. the subscript "2" implies a reflection on the source side.88 in the text (assuming an omnidirectional source) is added to the barrier attenuation Δb to give Ab. from figure 8.65. A3 2m 48m A2. so Ab ' 17. We will begin the trial with a barrier height of 3.5m 2m 48m The required barrier height can be found by trial and error. .6 dB . B1.0m.166 m Path c = ( 0.6 ' 17.5 h 1.5m d 1. The correction term given by equation 8. Losses due to ground reflection are zero.6 dB .64 = 2. acoustic enclosures and barriers 241 Path g = ( 232 % 1. The distances "A" refer to the source side of the barrier (source to barrier top along the particular path specified by the associated subscript) and the distances "B" refer to distances on the receiver side.387 % 23.180 & [ 502 % 32 ] 1 / 2 ' 3. where the subscript "1" refers to waves travelling over the barrier with no reflections. N = (2/ λ ) × 3.0dB. the subscript "3" implies a reflection on the receiver side and the subscript "4" implies a reflection on both sides.09) ' 0.122 % 22 % 52 ) 1 / 2 ' 5.166 % 25.0 % 0.14 on p389 in the text.5m B3.88 in the text (assuming an omnidirectional source) is 20 log10(53. The correction term given by equation 8.3962 ) 1 / 2 ' 23. The results are summarised in the following table.74 / 50.Sound transmission loss. 828 5.9 13.0m 3. .7 10.042 48.1 18 12.3 13.2 16.6 17.4 4.02 48.6 9.923 2.43 0.66 11.5 17 10.02 48.5 12.92 2.385 2.92 48.314 48.04 3.5m 2.828 5.21 0.0 17.385 48.036 12.065 48.79 1.0 Thus the wall height should be about 4.8 17.202 5.1 12.0m.260 0.97 in the text which can be expanded to give: NR ' 10 log10 100 % 100 & 10 log10 4 × 10& 1.21 The layout is illustrated in the following figure.202 5.314 0.50 4.852 3.07 0.50 4.260 48.3 3.21 48.63 2.5 8.45 2.38 2.242 Solutions to problems The barrier overall noise reduction is given by equation 1.76 % 10 & Ab1 / 10 % 10 & Ab2 / 10 % 10 & Ab3 / 10 % 10 & Ab4 / 10 Barrier height A1 (m) A2 (m) A3 (m) A4 (m) B1 (m) B2 (m) B3 (m) B4 (m) N1 N2 N3 N4 Δb1 Δb2 Δb3 Δb4 Ab1 (dB) Ab2 (dB) Ab3 (dB) Ab4 (dB) NR (dB) 3.9 17 12. Problem 8.28 10.1 10.3 18.1 16.6 17.0 17.0m 2.2 17.6 10.852 48.73 2.065 48.042 48. 686 Path 2. so d2 = 3. so d1 = 20.52 ' 3.1496 m 6.0100 m B ' 102 % 1.3332 % 1.6550 m A ' 502 % 12 ' 50.0m 3 1.686m A ) ' 202 % 22 ' 20.1119 m There are 4 paths which will contribute to the sound level at the receiver so we need to calculate the Fresnel Number corresponding to each. A Y B N1 ' 2 50.35 0.5 .52 ' 10.52 )1 / 2 ' 0. A' Y A'' Y B N2 ' 2 20.52 )1 / 2 ' 0.1496 % 10.333m 3 λ = 343/500 = 0.Sound transmission loss.5m d2 From similar triangles.3109 m 3.0998 m A )) ' B) ' B )) ' 302 % 32 ' 30.1119 & (602 % 0.1119 & (602 % 3. acoustic enclosures and barriers A 2m A' A'' d1 50m 10m B' B 243 3m B'' 1.76 0. d2 10 & d2 ' d1 50 & d1 ' 2 .686 .0100 % 10.6672 % 32 ' 7. Path 1.0998 % 30. the attenuations corresponding to N1 to N4 are ΔN1 = 9. source side.82 m . 9.5.9.3106 % 3. tanβ1 ' 25 & x x over top. receiver side.5 / 10 ' 1.0100 % 7. 20dB.6553 & (602 % 0.76 % 8. 15dB.52 )1 / 2 ' 2. ΔN2 = 12. Adding 3dB to all ground reflections results in the following noise reductions corresponding to the 4 paths: Path 1.5 0.5dB.2E y ' 6. tanβ2 ' 25 & y y x ' 2.9 / 10 % 10&15 / 10 % 10&20 / 10 % 10&24.97 in the text.5 0.78 m .31 ' 10.52 )1 / 2 ' 3.9dB. A Y B' Y B'' N3 ' Solutions to problems 2 50. Path 3.88 in the text is greatest for path 4 and is equal to 0.1 dB Problem 8. we have for the reflection angles: over top. 4 0. 4 1.14. From equation 1. ΔN3 = 17.1496 % 7. no sound is diffracted around its edges.686 Path 4. Path 2. A' Y A'' Y B` Y B'' N4 ' 2 20. β2 ' 12.244 Path 3.3106 % 3. the noise reduction due to the enclosure is: NR ' 10 log10 10&0 / 10 % 10&3 / 10 & 10 log10 10&9.5 ' .0998 % 30.22 Referring to the following figures. Path 4. From figure 8.686 As the wall completely surrounds the factory.5 ' . ΔN4 = 18. β1 ' 10. 24. The correction term given by equation 8.4E . so the correction will be ignored.6553 & (602 % 3.09dB which is negligible. 5m Around edge (elevation) R 1.Sound transmission loss.5m S 25m 7.5m 25m R .5m 4m Plan 7.5m S 0. acoustic enclosures and barriers 245 Over top (elevation) R S 0.5m 25m 25m 7.5m x 25m 25m 1 2 y 1.5m S z 25m 1m h1 4m 3 25m Plan 7. 2 0.598 and A + B .20 in the text and the results are tabulated in the table below.0×10-5 4.010m.487m Path 3 .244.0 6.52)1/2 + (42 + 22.52)1/2 = 26.402.0 1.3 0.106m and A + B .359m Path 2 .010m.over top of barrier with ground reflection on both sides d = 50.6×10-4 3.222)1/2 = 25.3 3.5 ' .246 Solutions to problems 1.0×10-5 8.d = 0.3 3.d = 0. B = 25.0 AR3 and AR4 1.201m .0×10-5 2.782 + 0.182)1/2 = 25.990m Paths 5&6 .125 A = (2. z ' 12.0 1.5 0.over top of barrier with no ground reflections d = 50. B = 25. B = 25.around edges with no reflection B = A = (0.2 0.5 0.206kg/m3.5×10-6 5.5 5.5 2.5 2. A = 25.040m.5 2.402m and A + B .d = 0.5 0.0×10-5 1. A = 25.5 3.010m. β3 ' 2.0×10-6 1.2×10-4 630 447 316 223 158 112 79 56 Path 1 .3E. The reflection loss. AR is calculated using figure 5.0 2.244 B = (6.52 + 252 + 7. around edge. β4 = β3 = 2.5 m .822 + 1.125 and A + B .0 AR2 0.3 0.52)1/2 + (42 + 18. tanβ3 ' Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 ρf R1 R1 ρf 1/2 AR1 0.over top of barrier with ground reflection on the receiver side d = 50.0 6.3E 50 & z z Flow resistivity of ground = R1 = 3 × 107 MKS rayls/m and ρ = 1.5 2.802m Path 4 .0 7.d = 0. A = 25.7 1. With no barrier.7 1.d = 2.over top of barrier with ground reflection on the source side d = 50.598m and A + B . 4 12.18 0.3 103 N7&8 0.61 3. The numbers in brackets indicate the sum of the ground reflection losses and barrier diffraction loss.52 + 12.80 1.52)1/2 = 26.60 3.30 0.120m B = 2(12 + 252 + 7.5m.44 2. The correction term given by equation 8.84 5.14.42 2. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 N1 0.13 0.09 4. Values of N for each path are 343 tabulated below.71 1. N ' (A % B & d) .5 = 0.52 1.5/37.5 × 12.72 1.120m and A + B .21 6.05 2. p389 in the text and are tabulated below.5/2)2)1/2 = 26.19 8.36 0.Sound transmission loss.7 N3 0. .34 4. acoustic enclosures and barriers 247 Paths 7&8 .37 16.8 26 51 103 The noise reductions (NR or Δb) corresponding to the above Fresnel Numbers are calculated using figure 8.52 + (7.80 1.1dB overall and will be ignored here.42 12.77 11.68 9.8 25.7 37.d = 2.7 N2 0.26 0.5 23.7 51. All quantities are in dB.4 N4 0.59 1.68 11.1 46.89 5.4 22.20m 2f Fresnel Number.21 6. A = 2(0.36 0.around edges with reflection in ground Intersection height of diffracted wave with barrier edge = 1.2 N5&6 0.17 2.35 18.88 in the text (assuming an omnidirectional source) will be less than 0. 8 13.1 (10.0 8.2 24 24 24 24 12.8) 23.6 1.8 (14.1 21.9 (22.7 (22.5 10.0 (12.7) 23.0 (17.1) 13.9 (18.5) 24 (26.9 1. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 10 log10 j 10 & NR Ai / 10 j 10 & 10 log10 & NR Bi / 10 NR SPL at receiver 2.5 13. Thus the noise reduction due to the barrier is 9dB(A).4) 12.8 (12. The A-weighted level without the barrier is calculated using the octave band levels given in the problem and is 67dB(A).5 44.1 5.3 15.8) 24 (28.3 (13.5 (15.0 (10.2 65.7 (19.1) 10.3 2.248 Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Solutions to problems NR1 NR2 NR3 NR4 NR5&6 NR7&8 8.8 7. .3 4.1 (20.8 (15.8 (26.1) 24 (27) 9.3 15. All quantities are in dB.3) 17.1 49.7 40.0 (12.3) 24 (27) 10.2 22.5 18.5) The barrier noise reduction is given by equation 1.7 6.2 1.9 (10.1 14.0) 18.2dB(A).8 3.5) 14.1) 11.97 in the text and the results of the calculations are summarised in the table below.9 (9.0 (24.7) 19.8) 15.8 0.8 17.6) 21.5 33. The A-weighted level with the barrier is calculated using the numbers in the last column of the preceding table and is 58.2 9. The subscript "A" refers to the condition with no barrier and the subscript "B" refers to the condition with barrier.7) 24 (29) 24 (30) 24 (31) 24 (30.9 12.6) 14.3 24 8.5) 17.7 17.2 (24.1 19.1) 20.4) 12.0 0.5) 17.1 1.9) 20.8 The overall A-weighted level is calculated using the octave band levels as described on pages 101 and 102 in the text.9 (26.9 10.0 (17.2 32.0 16.6) 24 (30) 12.0 18.4) 23.9 69.8 16.2 64.8 (20.9 1. 2 2.6.4 4.17 in the text and trigonometry is used to calculate the angles. θ and φ .0 1.8 6. p394 of the text which may be rewritten as: Δ C ' K log10( 2πf b / 343) The values of K are calculated using figure 8. the additional noise reduction may be calculated using equation 8.7 5.7&8 1.5 2. Path 1 θ (degrees) φ (degrees) Path 2 100 96 1.5 1.1 8.8 15.23 249 If the barrier were a building.7 4.1 13.6 3.2 2.2 1.6 4.98. .Sound transmission loss.0 5.0 3.3 Path 4 100 102 2.3 2.8 1.4 7. acoustic enclosures and barriers Problem 8.3 ΔC2 ΔC3 ΔC4 ΔC5.7 6.9 2.5 The noise reductions are calculated as before with the attenuation due to the barrier thickness added to each path.9 3.8 3.4 1.6 98 96 1.9 2.8 10.7 Paths 5&6 117 117 5.2 5. The results (in dB) are summarised in the table below. The results are summarised in the two following tables.2 3.0 5.4 1.6 Paths 7&8 117 117 5.8 2.7 7.4 12. Values for ΔC are in dB and the subscript on C refers to the path number.2 1.3 5.5 Path 3 98 102 2.2 K Octave band centre ΔC1 frequency (Hz) 63 125 250 500 1000 2000 4000 8000 0.6 3.4 3. 4 42.2 1. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 10 log10 j 10 & NR Ai / 10 j 10 & 10 log10 & NR Bi / 10 NR SPL at receiver 64.6 35.3 26.6dB(A) to a total of approximately 11dB(A).97 in the text and the results of the calculations are summarised in the table below.6 12. Thus the effect of thickening the barrier is to increase the noise reduction by 2.3 14.4 36.6 2.3 17.3 27.7 32.5 22.1 1.6 16.9 8.3 27.7 25. The A-weighted level with the thick building as the barrier is calculated using the numbers in the last column of the preceding table and is 55.1 35.7 26.1 44.6(A) which is 2.2 29.8 31.6 14.8 14.8 39.8 25.3 2.9 3.0 33.9 20.0 19.8 33.5 46.1 37.5 39.6 1.4 37.1 66.7 19.6dB(A) less than for the thin barrier.9 18.9 61.1 11.4 16.4 41. .2 11.5 5.4 17.4 27.0 15.3 10.6 29.0 34.8 46.1 10.9 23.250 Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Solutions to problems NR1 NR2 NR3 NR4 NR5&6 NR7&8 9.4 25.2 21.3 21.6 6.3 23.9 1.4 The overall A-weighted level is calculated using the octave band levels as described on pages 101 and 102 in the text.8 0.9 24.0 0.4 12.8 21. The subscript "A" refers to the condition with no barrier and the subscript "B" refers to the condition with barrier.7 25.7 12.0 8.5 25.3 22.2 15.0 The barrier noise reduction is given by equation 1.5 17.0 10.5 13.2 30.2 18.5 33. 24 A S 0. Problem 8.5m 251 3m 25m 25m If the barrier is moved closer to the refrigeration unit.Sound transmission loss.348 × 10&3 α . equation 8.01 m A ' (252 % 2.52) 1 / 2 ' 25. As the barrier is indoors. acoustic enclosures and barriers Problem 8. ¯0 = 0.52) 1 / 2 ' 25. 4 / S0¯0 ' 4. For an omnidirectional source Dθ ' 6. and 4πr 2 r ' d ' [ 502 % 12 ]1/2 ' 50.366 × 10&5 .045 m A + B .500m2.109 on p402 in the text is appropriate.25 An elevation view of the situation is shown in the figure.160m.08.5m d B R 1. Thus: IL ' 10 log10 Dθ 4πr 2 % 4 S0¯0 α & 10 log10 Dθ F 4πr 2 % 4K1K2 S(1 & K1 K2) ' term1 & term2 Source and receiver are each 25m from barrier. S0 = 2(100 × 5 + 50 × 5 + 100 × 50) = 11.125 m B ' (252 % 1. on a hard floor Dθ = 2. α Thus.d = 0. the noise reduction will increase because the Fresnel numbers will increase due to the increased path length from source to receiver over the top of the barrier and around the edges. S = 2 × 50 = 100m2 = area of gap between barrier and ceiling. 900m2.5 5.117 0.34×10-6 5. Thus term 2 can be written as: term2 ' 10 log10 6.5 Thus: S1α1 ' S2α2 ' 482.1 -29.348 × 10&3 ' & 23.240 0.1 5.46 0.85 and figure 8.29 × 10&3 The calculations which are a function of frequency are summarised in the following table.233 0. where the barrier IL is given by IL = term 1 .1 -29.172 100 % 482.term 2.16×10-6 2.131 0.15 ' 0.93×10-6 1.366 × 10&5 F % 1.5 5.188 0.466 0.28 0.933 1. α1 ' α2 ' ¯ ¯ 5750 × 0.1722 100(1 & 0.252 Solutions to problems S1 = S2 = S0/2 + 3 × 50 = 5.5 5.732 7. calculated using equation 8.865 3.081 0.08 % 150 × 0.5 5.4 5.013 1.1 -29.059 0.13 on page 388 in the text.59×10-6 8.025 0.12×10-5 8. where N is the Fresnel number for the path.082 5900 100 ' 0.1 -29.046 0.1722) ' 10 log10 6. so F = (3 + 10N)-1.4 5. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 N= 2 (A % B & d) λ F Dθ F 4πr 2 term 2 IL 0.5 m 2 and K1 ' K2 ' ¯ ¯ Referring to the first equation: term1 ' 10 log10 6.78×10-5 1.53×10-5 1.6 dB There is only one path over the top of the barrier.366 × 10&5 % 4.20×10-7 -29 -29 -29.1 -29.5 .5 5.366 × 10&5 F % 4 × 0. R S We are to ignore any paths contributing less than 0.160 in the text with DIM and AE = 0.8 ¯ (2 % 1.1 α (1 % 1. the correct way to do this problem is to use equation 7.26 253 (a) The most dominant five paths from C source to receiver are shown in the figure. Of course.8)2 % 42 42 The fourth path (reflection from the floor then ceiling) will be attenuated by: .2dB to the total. here we are happy with an approximate solution.2)2 % 42 42 The third path (reflection from the ceiling) will be attenuated by: & 10 log10 (1 & ¯c ) % 10 log10 α ' & 10 log10(1 & αc ) % 2. The direct path contribution is given by equations 5. acoustic enclosures and barriers Problem 8. we will ignore paths which contribute 10dB or more lower than the direct path. That is. Thus: Lpd ' Lw & 10 log10(4 π × 16) ' Lw & 23 dB The second path (reflection from the floor) will be attenuated by: ¯ & 10 log10(1 & αf ) % 10 log10 ' & 10 log10(1 & ¯f ) % 1.110 in the text for a flat room with a diffusely reflecting floor (as there is furniture present) and specularly reflecting ceiling.Sound transmission loss.158 and 5. However. 254 Solutions to problems & 10 log10(1 & αf ) (1 & αc ) % 10 log10 ¯ ¯ ' & 10 log10(1 & ¯f ) (1 & ¯c ) % 4.2 21.2 % 3 % 2)2 % 42 42 The contributions of each path to the final sound pressure level is obtained by subtracting the attenuations from the direct wave sound pressure level (arithmetically). Perhaps you could prove this by trial and error. The total sound pressure level is then obtained by logarithmically combining the contributions from each path (assuming incoherent combination).7 21.8)2 % 42 42 The fifth path (reflection from the ceiling then floor) will be attenuated by: & 10 log10(1 & αf ) (1 & αc ) % 10 log10 ¯ ¯ ' & 10 log10(1 & ¯f ) (1 & ¯c ) % 5.2dB to the total and are thus not included.9 . Note that paths involving more reflections will result in a contribution of less than 0.5 55.9 48.2 52.2 38.5 29.9 α α (1 % 3 % 1.2 32.1 2000 75 52 45.4 31.3 α α (1. frequency (Hz) Lw Lpd Lp2 Lp3 Lp4 Lp5 Lpt 500 70 47 42.8 38.8 1000 77 54 47. The results are tabulated in the table below.2 31 30. as all other paths are blocked by the barrier (do a sketch of the arrangement to prove this to yourself). we can write the following for the Fresnel number: 2×f N ' 1. the only contributions to the sound level on the other side will be the wave reflected from the ceiling (Lp3) and the wave diffracted over the top.0 10 1000 54 33. Lpb of the wave diffracted over the top of the barrier to the sound field at the receiver is calculated using: Lpb ' Lpd & Ab The results are summarised in the following table. For the wave diffracted over the top.65 16 31 39.3 20 .6 22 30 33.282 % 2.3 19 35 39.14 in the text (point source) and corrected using equation 8.31 × 10&3f c The attenuation.4 6.88 to give: Ab ' Δb % 20 log10 2. Δb is read from figure 8.282 4 ' Δb % 1.22 % 22 % 12 % 22 & 4 ' 3. frequency (Hz) Lpd Lp3 N Ab Lpb Lpt reduction due to barrier 500 47 39.2 The contribution.9 3.9 1. acoustic enclosures and barriers 255 (b) When the barrier is present.Sound transmission loss.9 15 2000 52 24. Problem 8. Thus: IL ' 10 log10 Dθ 4πr 2 % 4 S0¯0 α & 10 log10 Dθ F 4πr 2 % 4K1K2 S(1 & K1 K2) ' term1 & term2 First we calculate the Fresnel Numbers for diffraction over the top and around the edges using equation 8.256 Assumptions: A S 1m 5m Solutions to problems B d 5m 50m a R 1.109 on p402 in the text is appropriate. In addition the barrier Fresnel number will be increased. ! ! (c) The noise reduction could be increased by moving the barrier sufficiently close to either the noise source or receiver that the wave reflected from the ceiling will also be blocked.13 on page 388 in the text. As the barrier is indoors.5m 3m 2m 3m 5m R 20m 2m 10m 5m S b ! The barrier is sufficiently wide that the contribution at the receiver position due to refraction around the edges is negligible. . equation 8. further reducing the energy diffracted over the top of the barrier. All waves at the receiver combine together incoherently.85 and figure 8.27 (a) The situation is illustrated in the figure. The directivity of the source is uniform in all directions. 5)2 ] 1 / 2 ' 5. h = 3m YR = 2 × 5 / (5 % 5) ' 1 d ' [ ( 5 % 5)2 % (1 % 1)2 % (1.ZR = 3m YR = 0.d) = 3.25 d ' [ ( 5 % 5)2 % 4 × 0.5 & 1)2 ] 1 / 2 ' 10.d) = 3.575 × 10-2f Diffraction around edge "b" XR = XS = 5m.16 and Dθ/4πr2 = 2. Na = (2f/343)(A + B .252 % 52 ] 1 / 2 ' 7.5 × 5 / (5 % 5) ' 0.252 % 4] 1 / 2 ' 10.ZS = 2m.836 m Fresnel Number.ZS = 5m. ZR = 1.606 m Fresnel Number.5m.210 m A ' [ 52 % 12 % (3 & 1)2 ] 1 / 2 ' 5.Sound transmission loss.252 % 52 ] 1 / 2 ' 7.08 ' 216. h .ZR = 7m YR = 0.5m.5 × 5 / (5 % 5) ' 0.0 m 2 α S ' 2 × 5 × 3 % 20 × 2 ' 70 m 2 S1 ' S2 ' 30 % 1350 ' 1380 m 2 α1 ' α2 ' (216 / 2 % 30 αb) / (1350 % 30) ¯ ¯ ¯ 257 r = d = 10.210m.51 × 10-3 . acoustic enclosures and barriers Diffraction over the top XR = XS = 5m.393 × 10-3f Diffraction around edge "a" XR = XS = 5m.210 m A ' [ 52 % 0.477 m B ' [ 52 % 12 % (3 & 1.ZS = 2m.210 m A ' [ 52 % 0.190 × 10-2f S0¯0 ' 2(5 × 20 % 5 × 50 % 20 × 50) × 0. ZR .315 m Fresnel Number. h . Y = 0.075 m B ' [ 52 % 0.ZS = 5m. h .252 % 72 ] 1 / 2 ' 8. ZS = 1m. DI = 5.25 d ' [ ( 5 % 5)2 % 4 × 0. Y = 0. h .5m. thus Dθ = 3. Y = 2m.252 % 32 ] 1 / 2 ' 5.252 % 4] 1 / 2 ' 10. Nb = (2f/343)(A + B . N = (2f/343)(A + B .d) = 1. ZR .075 m B ' [ 52 % 0. 8 dB a. They are frequency dependent quantities.337 1000 2000 3.42 1.123 0.95 0.337 % 7.98 500 1.02× 10-3 10-3 10-2 0.99 2.75 31.01 125 0.08 0.57 63.08 0.0 0.337 0.068 1.79 4000 13.10 0.345 1.0 0.85 3.08 0.10 (table 7.10 0.10 1.39× 7.5 4.036 0. K2 Dθ F 4πr 2 0.337 0.9 63.2 4.6 15.46× 7.1) α ¯ ¯1 .205 0.21 0.315 0.377 0.09× 9.258 Solutions to problems Dθ 4πr 2 term1 ' 10 log10 and F ' j % 4 S0α0 ¯ ' & 16.94 7. c 1 3 % 10Ni The quantities K1 and K2 are calculated using equation 8.010 1.6 .5 31. Octave band centre frequency (Hz) N (Top) Na Nb F α ¯b 63 0. but all other variables needed for their calculation have been evaluated above.8 3.99 250 0.0 0. b.88 15.0 127.10 1. α2 K1.2 4.5 4.25 0.019 0. Results for the second term as well as the overall Insertion Loss of the barrier are tabulated in the following table.97 3.111 in the text.0 0.39 6.32× 10-3 10-3 10-3 10-3 4K1K2 S(1 & K1K2) Insertion Loss (dB) 2.98× 8.5 4. We now proceed to evaluate the second term of equation 8.109 as a function of octave band centre frequency.83 0.8 0.34× 7.388 0.70 7. The quantity.22mm and the diameter of the jacket. The thickness of the jacket is h = 6/2700 = 2. The cost effectiveness of this action can only be assessed by completing the calculations as was done in part (a) of the problem and fully evaluating the benefit of the increased noise reduction vs the inconvenience of restricted passage.4022m.323 Hz cL h 5476 × 0. This effect should be slightly larger than the effect in part (b) above but would still be restricted to a few dB.6 / [2700 ( 1 & 0. Thus we can expect an increase in the barrier insertion loss.00222 . 8.55 c 2 0.20m corrected in equations 8.119. This effect is expected to be only 1 or 2dB but calculations as were done in part (a) are necessary to verify this.1 + 0. The value of K1 will be reduced and the term involving K1K2 will be reduced slightly.4022 The critical frequency is obtained using the equation on p337 as: fc ' 0.0022 = 0. The longitudinal wavespeed is: cL ' E / [ρm ( 1 & ν 2 ) ] ' 71. d = 0.55 × 3432 ' ' 5.342 ) ] ' 5476 m/s cL πd The ring frequency is: fr ' ' 5476 ' 4.334 Hz π × 0. Problem 8. 1000(m/d)1/2 = 3862.116.2 + 2 × 0.28 Jacket This is a pipe lagging problem so we follow the procedure on pages 404 0.Sound transmission loss.40m pipe and 405 in the text with the errata 0.117 and 8. (c) Extending the barrier length will increase the barrier Insertion Loss because it will remove the contribution from waves diffracted around the barrier ends and will also reduce the reverberant field contribution because of a smaller gap area and also because of an increase in the effective room absorption. acoustic enclosures and barriers 259 (b) Moving the barrier closer to the source will increase the value of N which will reduce the value of F and thus the direct field contribution. 2 26.0288 0.7305 3.112 on p404 for octave bands of 4000Hz and below and equation 8.9220 15.2 11.115 0.0145 0.8 The Insertion Loss results are somewhat different between the two methods of calculation and as we shall see in the next problem.5 1307 1731 2120 2479 2717 1991 4289 4.9 Insertion Loss eqn.8 7.00587) which is valid for frequencies defined by f $ 120 / 6 × 0.4326 0.2163 0. it is supposed that the true results lie somewhere between the two.7 34.119 -18.8.115 for the 8000Hz octave band. . Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 f/fr or f/fc Cr or Cc Xr or Xc Insertion Loss eqns.1 ' 155 Hz . neither prediction scheme is particularly good.1090 0.3 41.120. However.849 789. the Insertion Loss is given by: IL ' 40 f 6 × 0.4 22.112 . In addition.4610 6. if three 1/3 octave bands are averaged. Of course.7 19.29 (a) The jacket cross section is shown schematically in the following figure. With equation 8.12 / 0.503 0.923 1.461 0.2 132 ' 25 log10 ( f × 0.0577 0.8653 1.5 29. then the result would be slightly different.6 17. Problem 8. equation 8.120 has also been used to generate an alternative set of Insertion Loss predictions.9 25. The results of the calculations are summarised in the table below at octave band centre frequencies.231 0.6 15.4 19. 8. we use equation 8.8.260 Solutions to problems For the Insertion Loss calculations.120 0.1 log10 1 % 0. 308 & 0.598)2 12 1 & 0.384 / 2 ' 0.342 16.6 % 16.616 = 0.3842 % 12 × (0.616mm = thickness of Aluminium.5 × 0.Sound transmission loss. acoustic enclosures and barriers hPb 1mm Lead y Aluminum neutral axis 261 hAl Let x mm be the thickness of the aluminium part of the jacket.6162 % 12 × (0.0.616 / 2) × 71. Thickness of lead = 1 .6 % 16.598 & 0.616 0.442 % ' 2.384mm Effective surface mass is given by: meff ' ρ1 h1 % ρ2 h2 ' 6.6 × 0.392 ' 3.384 % 0.5 ' The bending stiffness is given by: EAl h Al h Al % (h Pb % h Al / 2 & y)2 12 2 2 Beff ' 1 & νAl 2 % EPb h pb h Pb % (y & h Pb / 2)2 12 1 & νPb 2 ' 71. Then: x × 10&3 × 2700 % (1 & x) × 10&3 × 11300 ' 6 Thus x = 0.00 kg / m 2 The neutral axis location is given by: y ' (h Pb % h Al / 2)EAl % EPb h Pb / 2 EAl % EPb (0.017 % 1.598 mm 71.5 × 0.41 kg m 2 s &2 .192)2 12 1 & 0.384 % 0.384 0. 05m.112 to 8.05 = 0.05 ' 220 Hz .120 are only valid for frequencies defined by f $ 120 / 6 × 0. Note that the calculations using equation 8. d = 0.409 ' 2611 m/ s 6 cL ' ' The critical frequency is given by equation 8.15 + 2 × 0.3 in the text as: c2 2π m 3432 ' B 2π 6 ' 24. .25m and D = 0.262 Solutions to problems Assume that the effective Poisson's ratio is that of Aluminium and equal to 0. The results of the calculations at the 1/3 octave band centre frequencies are summarised in the following table. We will use both methods here as a comparison.251 The jacket insertion loss may be calculated using either equation 8.409 fc ' and the ring frequency is given on p404 in the text as: fr ' cL πd ' 2611 ' 3311 Hz π × 0.001 3.84 kHz 3.119 in the text.120 or equations 8. The quantities used in the equations are m = 6kg/m2.34.15m. Thus the longitudinal wave speed is: 12 h Beff meff 12 × 0. R = 0. 5 33.5 20.8.4 37.1805 0.119 1279 1500 1690 1869 2056 2220 2380 2543 2708 2859 3010 3157 3280 3378 3435 3403 3193 2258 2714 3356 4134 5038 5893 0.9 31.942 3.6096 4.2256 0.1 15.2560 2.6852 1. second edition) are shown for comparison.0 The data in the table are plotted in the following figure where the solid black line represents the theory embodied in equations 8.8877 3.112 .112-8.736 4. 8.Sound transmission loss.1 21.0 9.0 Eq.335 2.3 28.9 7.5 31.1444 0.119 and the dashed black line represents the theory embodied in equation 8.2 31.4439 1.6 24.9024 1.18.7219 0.1 8.120.6 13.5 12.7 15.5120 5.670 -2.868 2.7 30.1 29.3610 0. In addition.4 22. some experimental data (solid grey line) from the text book (figure 8.9 18.120 0.2888 0.1 24.5685 0.9 26.1370 1. .7 18.8 36.0 29.3 20.4 30. 8. Unfortunately there are no better theories available at present. acoustic enclosures and barriers 263 Octave band centre frequency (Hz) 63 80 100 125 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000 6300 8000 10000 Insertion Loss (dB) Xr or Xm Cr or Cc Eqs.8048 2.7 30.35 2.3 11.8 27.3 31.3 23.1 30. It can be seen that neither theory provides a very good prediction of the measured data.6 21.6 4.4512 0.1137 0.4 4.4 24. it will turn to powder.264 Solutions to problems (b) It is clear from the equations used for either calculation method that increasing the mass of the liner will increase the low frequency Insertion Loss. For the first prediction scheme. (c) One advantage of porous acoustic foam over rockwool is that the porous foam will support the weight of the jacket indefinitely whereas rockwool will gradually compress and in a high vibration environment. . Another disadvantage is that the foam is much more expensive than rockwool. reducing cL will also result in increased values for the low frequency Insertion Loss. A disadvantage of foam is that it is not fireproof and if ignited it emits toxic gas. The percent open area is given by: P ' 100 N πd 2 . . The total backing volume is then LB2 and the volume associated with one hole is LB2 /N.9 Solutions to problems relating to muffling devices Problem 9. N. ' P P 4 B where A is the neck cross-sectional area. B2 4 N holes B Thus the effective resonator volume is: V ' 100 L πd 2 100L A .1 L V R t Referring to the figure. Let L be the depth of the backing cavity and lets consider a section of sheet of dimensions B × B with a number of holes. we can imagine that each hole in the perforated sheet represents a neck of a Helmholtz resonator with the volume associated with each neck being equal to the total volume behind the perforated sheet divided by the number of holes. pr ' B e j ( ωt % kx % θ ) pT ' A e j ( ωt & kx ) % B e j ( ωt % kx % θ ) uT ' pT uT 1 A e j ( ωt & kx ) & B e j ( ωt % kx % θ ) ρc ' ρc A % B e j ( 2kx % θ ) A & B e j ( 2kx % θ ) Z ' ' ρc A e & j kx % B e j ( kx % θ ) Ae & jkx & Be j ( kx % θ ) ' ρc A / B % e j ( 2kx % θ ) A / B & e j ( 2kx % θ ) Minimum pressure occurs when θ ' & 2kx % π . x ' 1.2E A 108 / 20 % 1 3. Thus at 200 Hz only plane waves propagate.85d(1 & 0. gives θ = 2.266 Solutions to problems From equation 9.5119 10 &1 At x = 2.656 and 2kx + θ = 4. θ ' & 2 × 3.048c.4 ' 500 Hz .38 in the text: f0 ' c 2π A c ' LV 2π P 100 L R where R = 2R0 + t is the effective length of the neck and 2R0 is the total effective end correction for the hole. 0.2 (a) Diameter.586c / 0. so higher order mode cut on frequency is f ' 0.22d / a) The Helmholtz model is appropriate because the system is effectively a small mass of air vibrating against a stiffness represented by the backing volume.5119 ' ' 2. Problem 9. k ' 2πf / c ' 2π × 200 / 343 ' 3.4m.8 % π = -10.77 in the text gives the effective end correction of the perforate as: 2R0 ' 0.664 × 1. Adding 4π. Thus: .518c = 144. pi ' A e j ( ωt & kx ) .323 ' 8 / 20 B 1.608c = 264 degrees. Comparing the above equation with equation 7.8 m At 200 Hz. 2kx = 14.664 . j278 (b) Re{ZA} ' Re { Z / Aorifice} The hole impedance is in parallel with the rigid plate containing it.995) ' 413 2. If the rigid plate impedance is effectively infinity. Z = 264 . the combined specific acoustic impedance is that of the hole and this is what is measured by the standing wave. because of the large particle motions near the edges of the entrance.219 & j0. .427 % j0.25 × 0.608) & jsin(4.608) % jsin(4.219 & j0.995) × (2.3 A quarter wave tuning stub works by changing the wall impedance of a duct on which it is mounted so that downstream propagating waves are reflected back upstream. The acoustic impedance of the hole is the specific acoustic impedance divided by the area of the hole only.608) 2.05 m Thus total end correction is: 8 × 0.Muffling devices Z ' 413 267 2.995 (2. Hole radius = 0.M)2.639 )2 ' 0.25) × (1 & 0.323 % cos(4.05 R0 ' 0.427 & j0.623 6.995) ' 413 2.395 & j4.219 % j(& 0.0078 m 3π 4.995 (2.323 & cos(4. M ' Re{ZA} Aorifice / ρc Thus. 263.61 × 0.995) Multiplying numerator and denominator by complex conjugate of the denominator gives: Z ' 413 2.995) × (2.64 M ' Re{Z} / ρc ' 413 (c) End correction = no flow correction × (1 . If cross flow dominates resistance.427 % j0. The stub can also act by changing the radiation impedance (and thus the amount of power radiated) of the source producing the noise.608) 2. The viscous losses at the entrance to the stub also account for some of the energy loss.427 & j(& 0.995) ' 413 Thus.8 ' 0.05 % 1 & (1.88 Problem 9.427 & j0. 268 Problem 9. then Z1 = 0 and .2 (a) The impedance looking into the stub must be zero as we are ignoring the resistive component. The equivalent electrical circuit is shown in the figure below: Zb Zd Z3 Zf Z1 Za Z2 Zc Ze Zg Impedance looking in at location 3 = Z3 ' Ze(Zf % Zg) Ze % Zf % Zg Zc(Zd % Z3) Zc % Zd % Z3 Za(Zb % Z2) Za % Zb % Z2 Impedance looking in at location 2 = Z2 ' Impedance looking in at location 1 = Z1 ' If we neglect resistive impedance.4 Solutions to problems Z1 a b c d e f g 0. 25 3π 0. Equation (1) can now be solved by computer for d. although the bandwidth of significant attenuation will increase.5 × λ/4 = c/8f = 343/800 = 0. Using equation 9. 1 & 1. Zb ' Zd ' Zf ' j and the effective length of the hole A is given by: R ' 2 R0 ' 8d d . The volumes between the orifices are given by: πD 2 L ' 0. If we now substitute the expression for Z3 into the preceding equation.4288 × 2 × π × 100 Za = Zg = 2Zc = -j1. 0.01047 × 0.01 × 105.206 × 3432 ' & j 5. we obtain: Zc ' Ze ' &j 1. ρ = 1. The result is d = 44mm.Muffling devices Zb ' & Z2 ' & Zc(Zd % Z3) Zc % Zd % Z3 269 . (b) The device with baffles would have a much larger resistive impedance than the one without baffles because of all the cross-sectional changes at which there will be viscous losses.5Ve ' 0.206. The larger resistive impedance will lower the quality factor and thus the peak attenuation.429m.03 × 104 .2 The density.005236 L . .35. A = πd2/4 and ω = 200π. we obtain: Zb Zc % Zd % Ze(Zf % Zg) Ze % Zf % Zg % Zc Zd % Ze(Zf % Zg) Ze % Zf % Zg ' 0 (1) We now need to substitute in physical dimensions in place of the ρωR impedances. L = 0.01047L and Ve ' Vc ' 4 3 Va ' Vg ' 0. Required length. 180 × 10&4 log10 0. the reactive part of the impedance Zd will theoretically be infinite (based on equation 9.29. t = h = 2 × 1. Problem 9. (b) Equation 9.0442/4 = 0. although the bandwidth of significant attenuation will increase.180 × 10&4)2 ' 146 MKS Rayls (b) At the design frequency (100Hz).00152 0.270 Problem 9. Thus the quality factor and therefore the peak attenuation for open ended tubes will be smaller. ρc = 343 × 1. equation 9. ε ' 0 and M = 0.206 × 2π × 100 ) ' 2. A = π × 0.46 is given by: .832 × 2.288 × 1.6 (a) Closed end side branches have less viscous losses than open ended tubes because of one less cross-sectional change and corresponding edge. in practice.832. w = 0. This results in the Insertion Loss being finite and usually limited to 25 to 30dB. First we will evaluate the variables used in the equation.46 in the text to give: IL ' 20 log10 1 % Zd Rs If the side branch is mounted an odd number of quarter wavelengths from the end of the duct. k = 2π × 100/343 = 1. Thus the ratio Zd/Zs in equation 9.00152 m2. Substituting these values into equation 9.5 Solutions to problems (a) To calculate the resistive impedance. The downstream duct impedance for an infinite duct is simply ρc/A. However. the quarter wave tube is of finite cross section and Zd cannot therefore be too large.8 × 10&5 / ( 1.14) and the side branch Insertion Loss will also be infinite as indicated by the above equation. the Insertion Loss is found by substituting Rs for Zs in equation 9.29 on p417 in the text may be used.00152 π × (2.206 = 414.180 × 10&4 . we obtain: Rs ' 414 4 × 0.46 in the text may be used here. 7 The equation is simply a differential equation of the vibration of a mass acting against a spring. Zs is zero. K. The quantity. V = π × (0. Volume of cylinder.0742 × 6. 8 × 0.61 × 0.03 % ( 1 & 1. M. the side branch reactive impedance.6 rad / sec . Problem 9.03)2 = 2. 100mm radius.83 × 10&3 0.3 ) ' 0. the quantity. Thus: ω2 ' 104 and ω ' 100 rad / s The Insertion Loss is then: IL ' 20 log10 * 1 % Zd Zs * ' 20 log10 * 1 % 104 * ' 40 dB ω Problem 9.8 (a) Helmholtz resonator .38 in the text: ω0 ' c A / RV ' 343 2. From equation 9.0342 m 3π so R = 0. represents the vibrating mass of air in the resonator neck.83 × 10-3 m2. 200mm high. Length of neck.1)2 × 0.2 = 6.28 × 10&3 1/2 ' 844. represents the viscous damping losses K at the entrance and exit of the neck caused by motion of the air particles relative to the edges of the neck and the quantity.0742m.cylindrical cavity. represents the M stiffness of the volume of air in the cavity of the C resonator as shown in the figure.03 end correction ' 0. A = π × (0.25 × 0. C.Muffling devices Zd Zs 1 j(ω & 10 / ω & jω × 10&4 ) 4 271 ' When the Insertion Loss is at its peak. R = 0.28 × 10-3 m3 Area of neck. Neck 40mm long with a radius of 30mm.04m + end corrections. 6 × 40 × 10&6 / 0.04 ' 40 µWatts 1000 ¢p 2¦A ' 40 × 10&6 (e) Power in a plane wave = IA = ρc Thus area of plane wave.28 × 10&6 1/2 ' 26.414 = 2.314 × 268 / 0. S¯ = 0.4136 m 2 Thus the power dissipated = (f) The sabine absorption of the resonator.4 resonators per square metre.4 × 8. f0 = 844.0742 R / AV ' Rs 1000 2. (d) Incident plane wave of 80dB Lp and 135Hz. (g) From equation 1.031 m 2 .22 in the text as Δf ' f / Q ' 135 / 26. Thus: Q ' ρc 413.7 (c) If we define effectiveness as the frequency range within 3dB of the maximum.4 in the text. Thus the crosssectional area of the resonator volume is 13 times smaller than its Sabine absorption area. an ambient temperature variation from 5EC to 45EC corresponds to a speed of sound change from 1.029 to 1.4 × 8.04 ' 0.0 Hz . (b) The quality factor may be calculated using equation 9.04 Pa 2 0.83 × 6.029 . we would need 1/0. Thus we can expect the device to be effective between 132 and 137Hz.414m2 α 2 Cross-sectional area of resonator = π × 0.314 × 318 / 0. The problem could be overcome by . which is a range from 328 to 357m/s.66 0. Power dissipated = ¢p 2¦ ¢p 2¦ = at resonance.272 Solutions to problems Thus. Thus to make the wall look anechoic. A = 413.1 ' 0.40 in the text. * Zs * Rs The mean square sound pressure is: ¢ p 2 ¦ ' 4 × 10&10 × 108 ' 0. From equation 9.6/2π = 134Hz.38.7 ' 5. it can be seen that the resonance frequency of the resonator will vary from 808 to 880 rad/sec which corresponds to a range from 129 to 140Hz which is outside the 3dB range discussed in part (c). then the bandwidth of effectiveness is obtained using equation 7. the acoustic particle velocity may be written as: 1 ut ' ui % ur ' A ej(ωt % kx) & Ar ej(ωt & kx) ρc i At x = 0. Problem 9. pt Su t ' ZL .9 (a) The total acoustic pressure Zi ZL anywhere along the tube is the sum of the incident and end reflected pressures and may be L written in terms of two 0 x complex constants. representing the complex amplitudes of the incident and reflected (from x = 0) waves respectively.6.Muffling devices 273 using two additional types of resonator with resonance frequencies of 131. and 1.7. Thus: A % Ar ρc Ai % Ar ' ZL ' Zc i S Ai & Ar Ai & Ar At x = L: pt S ut ' Zi ' Zc Ai e jkL % Ar e &jkL Ai e jkL & Ar e &jkL Expanding the exponents gives: .5Hz respectively at 20EC. Ai and Ar. Thus: p t ' p i % p r ' Aiej(ωt % kx) % Arej(ωt & kx) Using equations 1.5 and 137. 5 Thus.5 ' 3 1 & 0.29 / 343) 31050 ' 31000 MKS Rayls 10350 . ZL = 3Zc = 31.000 MKS Rayls (iii) Zi ' 10350 ' 10350 31050 % j 10350 tan(200π × 10.274 Solutions to problems Ai cos(kL) % jAi sin(kL) % Ar cos(kL) & jAr sin(kL) Ai cos(kL) % jAi sin(kL) & Ar cos(kL) % jAr sin(kL) Ai % jAi tan(kL) % Ar & jAr tan(kL) Ai % jAi tan(kL) & Ar % jAr tan(kL) Zi ' Zc ' Zc ' Zc Ai % Ar % jtan(kL)(Ai & Ar) Ai & Ar % jtan(kL)(Ai % Ar) Rearranging gives: Ai % Ar Zi ' Ai & Ar 1 % % jtan(kL) ' Zc jtan(kL) ZL Zc 1 % % jtan(kL) ZL Zc jtan(kL) Ai % Ar Ai & Ar Rearranging gives: Zi ' Zc ZL % jZc tan (kl) Zc % jZL tan (kl) (b) (i) (ii) Zc ' ρc 414 ' ' 10. thus Ar/Ai = 0. pr/pi = 0.400 MKS Rayls S 0.29 / 343) 10350 % j 31050 tan(200π × 10.2 At x = 0.2 × 0.5.5 and Ai % Ar Ai & Ar ' 1 % 0. where ω = 2πf.2 = 3614/Ly rad/sec.Muffling devices Problem 9.ny (b) Drop in Lp over a distance of Ly/4 for 2.2 mode. cn c0 fnx. Thus: cn ' ω c ω2 2 1/2 2 2 & n xπ Lx & n yπ Ly This equation is shown sketched in the figure below. The wavenumber is then: .10 (a) The wavenumber is: κn ' (ω/c)2 & [(n xπ /Lx)2 % (n y π / Ly)2 ] 2 275 (i) Cut-on when κ2 = 0.75ω2. That is when: (ω/c)2 ' [ (n xπ /Lx)2 % (n y π / Ly)2 ] (ii) Phase speed is defined as cn ' ω / κn . Cut-on frequency is given by: ω2.2 ' 343 ' 4π Ly 2 f % 2π Ly 1/2 2 1/2 343 16π2 % 4π2 Ly ' 4819 Ly Excitation frequency = 0. 5Re{Zc}* v2 *2 ( ( A similar expression can be derived for the incident power.2 1 .2 ' ± j 9.2L y / 4 ' 20 log10 1 e&9.5Re{Zc}* v * 2 ' * v2 v *2 The characteristic impedance is defined as Zc = ρc/A. we can write the following ' &jκ x p(x2) e 2.5Re{p2v2 } ' 0.294 / 4 ' 20.2 dB Problem 9.11 v Zc Zs v1 Zc v2 (a) The transmitted power is given by: Wt ' 0.276 2 Solutions to problems 3614 343Ly 2 κn ' & 16π2 % 4π2 Ly 2 Thus.293 at the excitation frequency. Thus the transmission coefficient is given by: τ ' Wt Wi ' 0.2 2 for the reduction in sound pressure level as the wave travels from x1 to x2.5Re{Zc}* v2 *2 0.5Re{Zcv2v2 } ' 0. Setting x1 = 0 and x2 = Ly/4. Thus: Z R % jX ' s Zc The circuit equations are v ' v1 % v2 and v1Zs ' v2Zc from which: . Δ Lp ' 20 log10 1 e &jκ2. Thus &jκ x p(x1) e 2. κ2. The acoustic pressure Ly &j κ x is related to the distance along the duct as p(x) % e n . Muffling devices Zc Zs v2 v 1 277 v ' v2 1 % and * *2 ' *1 % 1 *2 Rs % jXs The transmission coefficient can then be written as: τ ' * v2 v *2 ' *1 % 1 1 *2 Rs % jXs ' * Rs % jXs *2 * Rs % jXs % 1 *2 ' Rs % Xs 2 2 2 (Rs % 1 )2 % Xs (b) The power reflection coefficient, * Rp *2 , is related to the transmission coefficient, τ, by * Rp *2 ' 1 & τ . Thus: * Rp * ' 1 & 2 Rs % Xs 2 2 2 ( R s % 1 )2 % X s ' (Rs % 1)2 % Xs & Rs & Xs (Rs % 1)2 % Xs 2 2 2 2 Rearranging gives: * Rp *2 ' 2Rs % 1 (Rs % 1)2 % Xs 2 Problem 9.12 (a) The Insertion Loss is calculated using equation 9.54 as it is a constant volume velocity source. Assuming that the damping term of Equation (7.33) is negligible, the resonance frequency of the muffler is defined by: IL ' 10 log10 2π × 20 1& ω0 2 2 ' 10 . Thus ω0 = 61.6rad/sec. To be a little conservative, use ω0 = 60rad/sec. The required chamber volume is then found using equation 9.38 in the 278 Solutions to problems text. The cross-sectional area of the inlet pipe is A ' π × 0.152 / 4 ' 0.01767 m 2 . Thus the required chamber volume is: V ' 3432 × 0.01767 0.3 × 652 ' 1.92 m 3 (b) The attenuating device could be made smaller by using a low pass filter as described on pp.433-438 in the text. Problem 9.13 Low pass acoustic filter (see figure 9.11 in text). Head loss . 4 velocity heads due to tube inlets and exits, so total pressure drop, Δp = 2ρU2. We now must calculate the flow speed, U. Assume that the choke tube diameter is the same as the inlet and exit pipes and equal to 0.1m. Flow rate at STP = 250,000m3/day = 2.894m3/sec. The mass flow rate can be calculated from the Universal Gas Law. Thus: m 0 0 RT PV ' M and: 0 MPV 0.029 × 101.4 × 103 × 2.894 ' ' 3.55 kg / s RT 8.314 × 288 m ' 0 At operating conditions, T = 623EK and P = 12 × 106 Pa. Thus: mRT 0 3.55 × 8.314 × 623 0 ' V ' ' 0.0529 m 3 / s MP 0.029 × 12 × 106 The velocity is thus: U ' 0.0529 × 4 π × 0.12 ' 6.735 m/ s The gas density is: ρ ' m 0 3.54 ' 66.9 kg / m 3 ' 0.0529 0 V Thus the pressure drop, Δp is: Muffling devices Δ p ' 2 × 67.2 × 6.7352 ' 6.1 kPa 279 which is much less than 0.5% of 12MPa, so it is OK. Speed of sound in the gas is: cg ' γRT / M ' (1.3 × 8.314 × 623 / 0.029) 1 / 2 ' 482 m/ s Following the design procedure on pages 437 and 438 of the text, we have 1. 2. 4. 7. f0 = 0.6 × 10 = 6Hz Try V1 = V2 = 1m3 Choke tube diameter = 0.1m and length = 1.8m (assuming the chambers are cylinders 1m long and 0.64m diameter). Resonance frequency is: f0 ' cg Ac 2π Rc 1 1 % V1 V2 1/2 ' 482 π × 0.12 1%1 2π 4 × 1.8 1/2 ' 7.2 Hz which is close enough to 6Hz for now. Choke tube x-sectional area = π × 0.01/4 = 0.007854m2 Equation 9.71 in the text may be used to calculate the Insertion Loss for the muffler. Substituting the values into this equation gives IL = 30dB which is too much. Thus try changing the volumes to 0.75m3 and the choke tube length to 1.6m. This gives an Insertion Loss of 17.6dB which is too small. Try changing the volumes to 0.9m3 and the choke tube length to 1.7m. This gives an Insertion Loss of 25.9dB which is too large. Try changing the choke tube length to 1.4m. This gives an Insertion Loss of 20.8dB which is OK. Thus, the final design is for 2 volumes, each of 0.9m3 with a 0.1m diameter choke tube, 1.4m long connecting them. In practice, conservatism would usually dictate sticking with the 30dB design if it is practical. 280 Problem 9.14 As the tubes are short, they may be treated as lumped elements and the design procedure outlined on pages 437 and 438 in the text may be used. The impedance of the three tubes is three times that for a single tube. Solutions to problems The design frequency is 40Hz. Thus the required resonance is 0.65 × 40 = 26Hz. To simplify matters, use the largest allowable chamber volumes and smallest allowable choke tube diameter. Using equation 9.80, we obtain: 26 ' 343 2×π π ×0.052 (0.03&1 % 0.03&1 ) 3 × 4 × Rc 1/2 Thus 26Rc 1/2 ' 11.403 and Rc = 192mm. The filter thus consists of 2 volumes, each 0.03m3, connected by three 0.05m diameter tubes which are approximately 0.2m long. Problem 9.15 4 3 L 1 2 compressor Muffling devices 281 Z1 v2 Z3 Z4 v3 ZL vL v Z2 (a) The equivalent acoustical circuit is shown above. (b) The Insertion Loss of the muffler is given by equation 9.64 in the text. The circuit equations are: v ' v2 % v3 % v L v2Z2 ' v3(Z3 % Z4) ' VLZL v3 ' v L v2 ' v L ZL Z3 % Z4 ZL Z2 ZL Z3 % Z4 % ZL Z2 v ' vL 1 % Thus the Insertion Loss is given by: IL ' 20 log10 * 1 % ZL Z3 % Z4 % ZL Z2 * (c) For no reflections from the pipe exit: ZL ' ρc 413.6 × 4 ' ' 1.317 × 106 A π × 0.022 The volume impedances are: Z2 ' Z4 ' & j ρc 2 1.206 × 3432 ' &j ' & j 1.129 × 104 Vω 0.2 × 2π × 10 282 Solutions to problems and the tube impedances are: Z1 ' Z3 ' j ρc 2πL tan A λ ' j 413.6 × 4 π × 0.02 2 tan 6π 343 ' j7.24 × 104 The Insertion Loss is then: IL ' 20 log10 * 1 % 1.317 × 106 j(7.24 × 104 & 1.129 × 104 ) % 1.317 × 106 & j1.129 × 104 * ' 20 log10 * 1 & j21.54 % j116.7 * . 39.6 dB Problem 9.16 a f c d b e g (a) Equivalent circuit diagram. Zf v Za Zb/2 v1 Zc Zd Zb/2 v2 Ze v3 Zg v4 (b) System equations: v ' v1 % v2 % v3 % v4 v1Za ' (v2 % v3 % v4) (Zb / 2) % (v3 % v4) (Zb / 2) % v4Zg v2(Zc % Zd) ' (v3 % v4) (Zb / 2) % v4Zg v3Ze ' v4Zg Zf and Zg.17 10mm 0. Capacitative impedances are Za.5m Z2 0.5m Z3 Zi Z3 Z2 Zi Z1 (a) Impedance looking into the tube is: Zi ' Z3 % Z1Z2 1 ' Z3 % (1 / Z2) % (1 / Z1) Z1 % Z2 Z1 % Z2 4jρc πd3 2 ' Z3Z1 % Z3Z2 % Z1Z2 (b) Z3 ' tan (kL3).5m 0.Muffling devices (c) Inductive (with resistive part) impedances are Zb.4m 10mm 283 Z1 0. Problem 9. Zc. Zd and Ze. Z2 ' & 4jρc 2 πd2 L2ω 2 . 429 (& tan(ω / 686) % cot(ω / 686)) ' 0 Solving by trial and error: .01 and d2 = 0.5. d1 = d3 = 0. L1 = L2 = L3 = 0. or Zi = 0. we obtain: & ω × 8 × 10&2 cot(ω / 686) tan(ω / 686) & 343 × 10&4 tan(ω / 686) % 343 × 10&4 cot(ω / 686) ' 0 Simplifying gives: & ω % 0. we obtain: & cot(kL1) tan(kL3) d1 d3 2 2 & c tan(kL3) d2 d3 L2ω 2 2 % c cot(kL1) d1 d2 L2ω 2 2 ' 0 Substituting for k = ω/343.4. Eliminating (4jρc/π)2 from previous expression for Zi and setting the result = 0.284 and: Solutions to problems Z1 ' & 4jρc πd1 2 cot(kL1) Thus: & (4jρc)2 2 2 π2d1 d3 cot(kL1)tan(kL3) & % (4jρc)2c π2d1 d2 L2ω 2 2 (4jρc)2c π2d2 d3 L2ω 2 2 tan(kL3) cot(kL1) Zi ' & 4jρc cot(kL1) c % 2 2 π d1 d2 L2ω (c) Resonance occurs when inductive impedance = capacitative impedance. Thus v ' v1 % v2 % v3 .15 17.7Hz. Problem 9. The Insertion Loss is given by: .28 0.1586 Value of expression 0.000045 285 Thus the resonance frequency = 17.18 Z1 Z2 Z3 Z4 Z5 ZL reciprocating compressor (a) Z1 v v1 reciprocating compressor Z3 Z4 Z2 v2 Z5 ZL v3 (b) Constant volume velocity source.3 -0.3 17.3 -1.6 ω 17 17. there will be a pressure maximum at the closed end and a maximum particle velocity at the open end.017 -0.Muffling devices ω 1 10 20 18 Value of expression 293 19.4 -5. (d) At resonance.1586/2π = 2. 286 Solutions to problems IL ' 20 log10 * v * v3 Examining pressure drops. thus v2 ' Substituting for v2 in the expression for v1 gives: v1 ' v3 Z3ZL Z2(Z4 % Z5) % Z3 Z2 % Z4ZL Z2(Z4 % Z5) % Z5ZL Z2(Z4 % Z5) The Insertion Loss is then: IL ' 20 log10 * (v1 % v2 % v3) v3 * ZL Z3ZL Z ' 20 log10 /0 1 % % % 3 00 Z4 % Z5 Z2(Z4 % Z5) Z2 0 % Z4ZL Z2(Z4 % Z5) % /0 Z2(Z4 % Z5) 000 Z5ZL (c) The result of (b) is no longer valid if the dimensions of the chambers represented by impedances Z2 and Z5 exceed one quarter of a wavelength of sound. . we may write: Z2v1 ' Z3(v2 % v3) % v2(Z4 % Z5) Thus: v1 ' Z3v2 Z2 % Z3v3 Z2 % Z4v2 Z2 % Z5v2 Z2 v3ZL Z4 % Z5 Also: v2(Z4 % Z5) ' v3ZL. 5a 3π 4V 1/3 . If the cylinder effective length of the hole is then R ' 3π diameter is equal to its length.5 a π 4V 1/3 .16.5a / D) and the total the side of the hole in the enclosure is R0 ' 3π 8a (2 & 2. Thus: 1 Vω πa 2 ' j & j Zs ρωR ρc 2 ' j Vω ρc 2 & j 3π2a 8ρω 2 & 2.Muffling devices Problem 9.19 287 Za Zv Volume. then the end correction for 8a (1 & 2. From equation 9. A.5a / D) . V v Zv Za cross-sectional area. the end correction for the side of the hole in free space is 8a/3π. Zs given by: 1 1 1 ' % Zs Zv Za where: Zv ' &j ρc 2 ρωR and Za ' j Vω A and R is the effective length of the orifice. then D ' (4V / π)1 / 3 and R ' 8a π 2 & 2. of hole (a) Load seen by speaker (ignoring external air load). If we assume that the enclosure is cylindrical of diameter D. which is when: 16Vω2 (1 & 1.153aV & 1 / 3) & 3π2ac 2 (b) At low frequencies. As the particle displacement also leads the velocity by 90E. the particle displacement at the orifice and the acoustic pressure will be in phase. the second term in the denominator will be larger than the first and so the phase of the impedance will be +j which means that the acoustic pressure leads the acoustic particle velocity by 90E. the second term in the denominator will become smaller than the first and the phase of the impedance will be -j. The crossover frequency is thus when the two terms in the denominator are equal. thus the out flow from the orifice will be in the same direction as the cone motion. Thus in this case the out flow from the orifice will be in the opposite direction to the cone motion and will thus reinforce the out flow at the cone as shown in the figure. At higher frequencies. Rearranging the above equation and substituting values for variables.153aV &1 / 3) 16Vω2 (1 & 1.153aV & 1 / 3) (c) We need to solve the above equation for V. given that ω0 = 2π × 100.288 Thus: Zs ' & j Solutions to problems 16ρωc 2 (1 & 1. we obtain: . a 180E shift from the lower frequency case. As the box is small compared to a wavelength. the acoustic pressure will be in phase with the displacement of the cone.153aV & 1 / 3) ' 3π2ac 2 or: ω0 ' 3π2ac 2 16V (1 & 1. Problem 9.1m long 0.01 / π)1 / 2V &1 / 3) 0.153 × (0.2m Equivalent acoustical circuit Za vb p Zb Zd Zf ZL Zc vd Ze vf vL Referring to the figures above and using equation 9.55 in the text for a constant pressure source.2m dia 0.0311 1 & 0.0651V &1 / 3 2 289 ' ' Solving by trial and error gives V = 0.153aV & 1 / 3) 3π2 × (0.Muffling devices V ' 3π2ac 2 16ω0 (1 & 1.2m dia 0.1m long e 1.2m dia 10m long 0.039m3.01 / π)1 / 2 × 3432 16 × 4 × π2 × 104(1 & 1.20 0.5m3 a 1m3 b d 2m3 L f c 0. we may write the following Insertion Loss and acoustical circuit equations: . 290 Solutions to problems IL ' 20 log10 * p * vL ZL (1) (2) (3) (4) (5) p ' (vb % vd % vf % vL )Za % vb Zb vb Zb ' vd Zd % Zc (vd % vf % vL ) vd Zd ' vf Zf % Ze(vf % vL ) Using eq. (2). (7) and (8): vb ' vL Zd Zb % Zc Zb ZL Zd % Ze ZL Zd Zf % Ze Zd % Zc ZL Zb Zf % Zc Zb (9) Using eqs. (5). (3): vb ' vd Zd Zb % Zc Zb % vf Zc Zb % vL Zc Zb (8) Using eqs. (7) and (9): Z % Zb p ' a ZL vL ZL % Za ZL Zd Zd Zb % Zc Zb % Ze Zd ZL Zd % Ze ZL Zd Zf %1 Ze Zd Zc ZL Zb Zf Zc Zb % % ZL Zf % % % Ze ZL Zd Zf The Insertion Loss is then: . we can write: vd ' vf vf Zf ' vL ZL Zf Zd Ze Zd vL Ze Zd % % (6) Using eqs. (5). we can write: vd ' vL ZL Zd % Ze ZL Zd Zf % Ze Zd (7) Using eq. (4). (5) and (6). 206 × 343 = 414.22/4 = 0. R0 = (8 × 0. Ae = Ac = Aa = π × 0. Pipe end corrections. p.2/1)/(3 × π) = 0. Ze ' j tan(k Re ) % Re .00314m2. ( µ ' 1. ha = largest of t or tube inlet radius.29.1. ZL ' Ac Ae AL From equation 9. we = wc = 0.25 × 0.288 4Aa ω d log10 2 c πh a ω2 A a % M c 2 2π Rc and Re are defined similarly to Ra. except that subscripts c and e are substituted for subscript a respectively. γ = 1.8 × 10&5 kg m &1 s &1) hc = he = largest of t or 0.05.Muffling devices IL ' 20 log10 /0 00 0 Za % Zb ZL % Za Zd Zb ZL Zd % Zc Zb % ZL Zd Ze Zd Ze ZL Zd Zf ZL Zf Ze Zd Zc ZL Zb Zf Zc Zb 291 % % % % % Ze ZL Zd Zf % % 1 /0 00 0 where the impedances are defined as: jρc ρc 2 ρc 2 Za ' tan(k Ra) % Ra .4. Zd ' &j Aa Vb ω Vd ω Zc ' j ρc ρc ρc tan(k Rc ) % Rc . Vd = 1.00550 / ω . ρc = 1.1)(1 .5. εa ' εc ' εe ' 0 .064m t ' 2µ / (ρω) ' 0. Zb ' & j . Vb = 1.1m. wa = 10m.417: Ra ' ρc ω tDawa 1 % (γ & 1) Aa c 2Aa % ε 5 3γ % 0. Vf = 2m3 Assume that M is small enough to neglect. Z1. Re{Z0} = 0. (c) The resistive component consists of viscous friction losses due to air particles vibrating back and forth against the edge of the inlet.2 of a wavelength. . and so W0 = 0 (b) For Z2. The expressions are valid over the frequency range from zero up to where the neck diameter or plenum dimensions approach 0.14 for the imaginary part and equation 9. (d) Varying the fan position along the duct will have the effect of adding an inductive impedance between the fan and plenum and will also change impedance. use the analysis leading up to equation 9.292 Problem 9. Equating 0 circuit pressure drops gives: v1(Z1 % Z0) ' v2 Z2 ' Δ p v1 ' Δp Z1 % Z0 W0 ' * v1 * 2 2 Re 6 Z0 > ' * Δ p * 2 Re 6 Z0 > 2 * Z1 % Z0 * 2 At low frequencies.35 in the text. Thus the radiated sound power will vary. p Z1 p Z1 v1 Z2 v2 Z0 Z2 v2 v1 Z0 Power flow through Z0 is the radiated sound power and is equal to: *v * 2 1 1 ( ( Re Δ p Z v1 ' Re Z0 v1 v1 ' 1 Re Z0 0 2 2 2 v1 is an rms quantity and Δ pZ is the pressure drop across Z0.21 Solutions to problems (a) Fan and plenum equivalent circuit diagram is shown in the figure below. For Z1 use the analysis leading to equation 9.29 for the real part. 432 in the text. this will not satisfy the minimum requirement for A1. ' 72.5V.Muffling devices Problem 9. Thus try V = 0.52 A 1.3 × 10&4 m 2 Vmin ' 3 × 5. Thus increase A and V by a factor of 2. Thus: Amin ' 3 × 4.3 314.707 × 106 A 1.769 & 0. which is a bit large .73 × 10&3 ' 0.734 × 10&8 0. ω0 ' ω 18. the design equation is: 1 18 × 0.60 in the text.2 V A 2 × 6502 10 π × 8 × 10&3 × 0.5 V > 1 & V A2 To begin.6 × 10-4m2 (d = 35mm).5 rad / s Using equation 9.04m3 and A = 9.77 However.41 ' 2.78 ' 2π × 50 18.01 Rewriting gives: 0.78 2 .1m/s.73 × 10&3 4.03 1/2 293 ' 650 m/ s Following the design procedure on p.77 1/2 ' 4. Design equation becomes: [ 0.77 & 5.3 × 8.051 ] 4. The speed of sound in the gas is: cg ' γRT ' M 1.5 V & & > 1 1.734 × 10&8 3.1 2 × 0. U0 = 0.022 m 3 0.22 The speed of flow.143 & 0.77/3). the desired resonance frequency is given by: 25 ' 10 log10 1 & ( ω / ω0 )2 Thus.01 72.53 .314 × (273 % 900) 0. set each term in brackets = (0. is7.143 & 0.6m From Figure 9.3m = 2h Liner thickness.15 in the text.095 ] 2. As M = -0.15m Thus R/h = 1.3)2 = 4). top right figure. Thus the required length of liner = 0.42 m Design summary (to achieve 25dB at 50Hz) Volume = 40litres Tail pipe diameter = 30mm tail pipe length = 1.7dB per length of duct equal to the duct radius.3 +7.3 × 500 ' ' 0.22 in the text. the figure with the highest attenuation is the top right figure corresponding to R1R/ρc = 2.437. A = 7. use an attenuation rate of 0.1 dB per length of duct equal to the duct radius.06 × 10&4 0.05.1. Problem 9.42m Could use a smaller tail pipe length and larger tail pipe diameter if a larger volume were chosen.17.15 × 10.7 = 0.04 650 72.8/6. R/h = 1.294 Solutions to problems Try reducing d to 30mm. R = 0. for M = 0.24m . For a square duct lined on 4 sides (equivalent to a circular duct).06 × 10-4 and the design equation is [ 0.1) = 6.6m.769 & 0.3m 0.6/0. the attenuation for M = -0.0 and 2h/λ = 0.48 .05 At 500Hz.5 2 ' 1. From Figure 9.0 M = -17/343 = -0.78 ' 1.437 λ 343 0.5(6. is: R ' 7. R. (expansion ratio = (0. From figure 9. the required liner attenuation for an overall attenuation of 15dB is 10.23 Use maximum allowable OD = 0.8dB. the attenuation would be 6. 2h 0. Inside diameter = 0. which is OK The required tail pipe length.3dB per length of lined duct equal to the duct radius. 9 19. where: S ' λ 0.39 343 From the figure. if the duct were venting a machine enclosure) can be found from figure 9. and for the small dimension.0.6 2.464.875 1.1 24.21.21.9 short 0. numbers in brackets) ! liner attenuation (figure 9.449) Mach no.875 The total loss may now be calculated with the aid of the following table. p. the additional attenuation is 7dB.5 24. p.2 29.6 30. The following table may now be generated. p.5.24 75mm 450mm 300mm The total attenuation is made up of ! entrance losses (figure 9. Problem 9. . σ = 0.. For the large dimension. M = 0.15.459 in the text.2 31.9 24.219 0. Octave band centre frequency (Hz) 500 1000 2000 4000 2h1 / λ 2h2 / λ attenuation (dB) .109 0.459) ! exit losses (table 9. 2h = 0. p.749 3.32 π× 4 1/2 × 500 ' 0.15 which corresponds to M = 0.0225.0 26.5 4.075 and the duct length is equal to 12h.4 0. To maximise the attenuation choose the curve corresponding to R/h = 4 and R1R/ρc = 4 in figure 9.Muffling devices 295 Additional noise reduction if direction of sound is distributed equally in all directions at the duct inlet (for example. 2h = 0.3 and duct length is equal to 3h. Area of open duct = 75 × 300 × 10-6 = 0.0.lining only short sides long sides all sides 7.437 0.499 long 0.437 0. 25 m2 &3 and S / λ ' f × S / c ' 1.0 4.91 If we use the largest outer cross section allowed. outlet and lined section (no expansion loss as it is mounted on an enclosure).0 = .6 10.1 42.296 Solutions to problems Octave band Entrance Exit loss (dB) lining total S/λ centre loss loss loss (S=0.6 30. needed 9 15 15 Liner atten needed 1.4 dB 2000 Hz 1. we need a length of duct equal to 5 × 0. Thus.46 2.9 33. Problem 9.1 34. S = 0. (b) The best attenuation possible at 500Hz is 34dB.25 m length of duct lined on all 4 sides.5 5 5 0.2 dB 1000Hz 3.0.458 × 10 f and 2h / λ ' 0.3 9.219 0.16.2 1. Using curve 3 in Figure 9.0 dB It is clear that the critical frequency is 2000 Hz and to satisfy the requirement of 5 dB at this frequency.169 (dB) (dB) frequency (dB) D' π (Hz) 500 1000 2000 4000 0.5 0 0 total atten.9 (a) It is clearly possible to achieve 30dB or more in each of the octave bands.3 attenuations: inlet.458 × 10 &3 f frequency (Hz) 125 1000 2000 S 2h ' λ λ inlet loss 2 10 10 outlet loss 5.2 31. the ratio of liner thickness to half airway width is 1.1. In fact a thinner liner would most likely be adequate.6 0 26. we obtain the following attenuations for 0.1 7.3×0. and using R1R/ρc = 8. 1 2 & 4kHz.437 0. assuming a flow speed of M=0.9 24.9 39.25 Dissipative muffler . Try beginning with the smallest allowed cross section of duct.749 3.875 1.5 f / c ' 1.075) 4A ' 0.18 1.25/1. 125 Hz 1.6 0. . R/h is 100/200 = 0. The effective duct 45o length.5 in 300mm the text and the inlet loss is from figure 9. a b 297 b 100mm 200mm a (c) h = 2R = 200//2 = 141. The lined duct loss is from figure 9. λ = 343/f.4mm. (b) For the ratio R/h = 0.2//2) × 0.Muffling devices 1. Problem 9.700 MKS Rayls. an acceptable value of R1R/ρc is 2. The results are summarised in the following table.5 (as one side only of the airway is lined). Duct cross-sectional area = (0. L = 300/2 = 424mm = 3h. top right figure).21 in the text (assuming diffuse field input). Wavelength.0566m2.15 in the text (curve 2. Note that many other solutions would be acceptable as well.25 m. It may also be assumed that the flow speed is small enough to ignore. it can be seen that the ratio.6/0.0707 = 11. using similar triangles.4 = 0.5. The exit loss is obtained from Table 9. Thus the required flow resistance of the liner is R1 = 2 × 413.26 (a) From the figure to the right. Power transmitted with muffler = τ.2 4.3 8.298 Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 Solutions to problems lined duct loss (dB) length. (dB) 14 9 9 14 18 13 11 10 0.69 1.2 1. (b) r Power transmitted down duct with no muffler = 1. Transmission Loss is the difference in sound pressure level measured at the inlet and outlet of the silencer.39 2. Power transmitted by muffler = τ Thus Transmission Loss.103 0.30 6.087 0.206 0.2 0.1 6.9 0.27 (a) Insertion Loss is the difference in sound level at the end of the duct with and without the silencer in place.1 8. 3h 0. Reactive mufflers change the radiation impedance "seen" by the sound .052 0.35 0.8 10 10 10 Exit Loss (dB) 12.412 0.2 3.0 0.1 0.17 0.60 0.6 2. TL = -10log10τ = IL (c) Dissipative attenuators absorb energy and contain surfaces lined with sound absorbing material.3 2h/λ /S/λ Inlet Loss (dB) 0 0 2 6. They are cost effective for high frequency noise.1 0 0 0 Total Atten.6 8. Thus IL = -10log10τ Power incident on muffler = 1.044 0.55 Problem 9.78 5.65 3.825 1. Doors at each end are 2.32m long.06 × 0.95m high.95) × 0. They can also dissipate energy through viscous losses at entrances and exits of small air passage ways. These mufflers are cost effective for low frequency tonal noise problems and in many cases they are preferred to reactive mufflers because of their relatively small size and installation convenience.32m 2. 2.82m (a) Room 3.1 / 0. sound reflection or absorption is provided by a sound source such as a loudspeaker.627m2.82 × 2.79 = 1.32 × 2. Reactive mufflers are cost effective for low frequency broadband noise.106 m 2 Thus: . Active mufflers act in a similar way to reactive mufflers but the impedance change.95 % 2.32m R ' S¯ α (1 & α) ¯ ' 2(3. r = 3.9 ' 6.82m wide and 2.28 3.Muffling devices 299 source (tonal noise or system resonances) and reflect energy back to the source. Can treat it like a plenum chamber.32 × 2. They are also preferred in dirty environments where reactive mufflers can become clogged.06m high.79m wide. They can also be tuned to attenuate tonal noise at one or more frequencies.1 and the sound power attenuation or transmission loss is: TL ' & 10 log10 A A cosθ % R πr 2 A = 2.82 % 3. α ¯ ' 0. 0. Problem 9. 244 × 10&4 W 4ρc 4 × 413. Using equations 7.627 1.627 % ' 11.0 dB re 10&12 W Thus the sound power incident on the second doorway is: 81. TL ' & 10 log10 That is.5 × ' 54.5. TL ' & 10 log10 1.106 × 0.6 The sound power level is: Lw ' 10 log10 W % 120 ' 81.95 m 2 0.0dB. there is little difference in the first case where the reverberant field contribution dominates.95 π × 3. Thus for case (a) above: TL ' & 10 log10 1.0 dB 6.95 and for case (b) above.627 A ' ' 1.627 ' 5. then the direct field term contribution will be zero. we obtain: W ' ¢p 2¦ 4 × 10&10 × 108.2 dB 54.5 Thus. this corresponds to a sound pressure level in the doorway given by equation 6.5 gives: α R ' 6.5 × 1.75 in the text.322 TL ' & 10 log10 (b) Increasing ¯ to 0.0 = 76. a 6dB improvement.627 ' 15.627 % ' 5.322 That is. (c) If the direct line of sight were prevented.7 dB 6.2 in the text as (where .106 π × 3.1 0.33 and 1. Assuming no reflection from the doorway.106 1.0 .9 0.300 Solutions to problems 1. (d) Sound power leaving the doorway of the first room is given by the sound intensity directed towards the door multiplied by the door opening area.627 1.3 dB 54. DIM = -9.Muffling devices the correction for ρc … 400 has been included): Lp ' Lw & 10 log10 A % 0.161 in text) AE = Aa + Ag + Am + Ab + Af (equation 5. The Strouhal number is: Ns ' fd / c ' 500 × 1 / 343 ' 1.3dB (table 5.2 . -1)dB (table 5.627) % 0.158 in the text which may be written as: Lp ' Lw & K %DIM & AE (dB re 20µPa) Lw = 135dB K = 10log10(2πr2) = 10log10(2π × 1002) = 48dB (equation 5.15 ' 74. the angular orientation from the stack axis of the line joining the stack to the observer is approximately 90E.243 in text) Ab = Af = 0 . p. 225 in text) Ag = -3dB (attenuation of ground reflected wave = attenuation of direct wave) Am = (+3.0 dB 301 Problem 9.165 in text) Aa = 0. The sound pressure level at the receiver is related to the sound power radiated by the stack using equation 5.0.458 Thus from figure 9.3.29 90 2m o 1 2 100m R At a distance of 50m.0 & 10 log10(1.27 in the text.10. The directivity index may be obtained using figure 9.27 in the text.15 ' 76. p.6dB. 6 & 0. .302 Thus: Solutions to problems Lp ' 135 & 48 & 9. Lp = 80dB re 20µPa.2 % 3 & (%3. &1) ' 77 & 81 dB re 20µPa If meteorological influences are ignored. 10 Solutions to problems in vibration isolation Problem 10. Thus improved vibration isolation will only help the low frequency noise problem and after allowing for the A-weighting correction of table 3. the problem is dominated by structure-borne noise whereas at higher frequencies.2 The single degree of freedom model gives inaccurate estimations of vibration transmission in the audio frequency range because it treats the spring and supported mass as lumped elements and does not include the effects of wave transmission along the spring. it can be seen that reducing the 63Hz problem will not significantly reduce the A-weighted noise level in the apartment. airborne noise dominates.1 (a) The difference in levels between the plant room and apartment for the two cases of the plant operating and the test source operation are obtained by subtracting appropriate rows in the table given in the problem and are given in the table below Octave band centre frequency (Hz) Equipment operating Test source operating 63 25 34 125 36 37 250 38 38 500 40 39 Inspection of the above table indicates that in the 63Hz band. The wave transmission effects are negligible at sub-audio frequencies but are often important mechanisms of vibration transmission in the audio frequency range.1. Problem 10. . then x(t) ' ω A sin( ωt ) and: Tmax ' m 1 m % s (ωA)2 2 3 The maximum potential energy stored in the spring is: V ' 1 2 1 k xmax ' k A 2 2 2 where k is the spring stiffness. a distance of y from the fixed base will have a dy displacement of (y/L)x(t). An intermediate point on the spring. m s L Ts ' 1 y 2 1 ρS 2 L 3 ρS x dy ' 0 x 0 m2 L 2 L2 3 0 1 x2 0 ( ρSL ) 2 3 1 m x2 0 2 s ' ' The total kinetic energy of the spring and mass is: Ttot ' Ts % Tm ' 1 2 m m 1 1 x 2 % mx 2 ' 0 0 m % s x2 0 3 2 2 3 0 If x(t) = Asin(ωt). Equating Tmax with Vmax gives: m 1 1 m % s (ωA)2 ' k A 2 2 2 3 .304 Problem 10. The y total kinetic energy of the spring is: L m k.3 Solutions to problems x 0 (a) Assuming that the displacement of the end of the spring attached to the mass is described by x(t). The solution for this equation is the same as for the acoustic case. E.25 SρL k ms (c) The wave motion in the spring satisfies the one dimensional wave equation given by: M2ξ Mx 2 ' 1 M2ξ 2 2 cL Mt where ξ is the spring longitudinal displacement as a function of axial location x. That is: ξ ' Ae j (ωt & ( ω / c L ) x ) % Be j (ωt % ( ω / c L ) x % θ) .Vibration isolation Thus: ωn ' k m% ms 3 305 and the resonance frequency f0 is thus: f0 ' ωn 2π ' 1 2π k m % ms / 3 (b) Effective Young's modulus. is given by: E ' force / area kx / S kL ' ' spring extension / spring length x/L S Longitudinal wave speed is thus given by: cL ' E ' ρ kL ' fs λs ' 4fs L Sρ Thus the surge frequency is given by: fs ' 1 4 k ' 0. Substituting the wave equation solution into this gives: & m ω2 e ' jkL j (ωt & ( ω / c L ) L ) & e j (ωt % ( ω / c L ) L ) j (ωt % ( ω / c L ) L ) ω j (ωt & ( ω / c L ) L ) e % e cL which can be rewritten as: m ω2 e ' jkL j ( ω / cL ) L ) & e & j ( ω / cL ) L ) j ( ω / cL ) L ) ω & j ( ω / cL ) L ) e % e cL As shown in part (b). where ω is the solution of the above transcendental equation. Substituting this into the solution to the wave equation gives B ejθ ' & A . Using this relation and rearranging the above equation gives: ωL ωL tan cL cL ' m ρSL 1 ' s ' m N m The surge frequency is 2π × ω. the upper frequency bound will be 0. ξ = 0. That is: M2ξ Mξ m ' &kL 2 Mx Mx where k is the spring stiffness. . Thus the wave equation solution becomes: ξ ' Ae j (ωt & ( ω / c L ) x ) & e j (ωt % ( ω / c L ) x ) The second boundary condition is that the inertia force of the mass at x = L is equal to the spring force.9 × this value. As stated in the problem. kL = cL2ρS.306 Solutions to problems One boundary condition is that at x = 0. 5 (a) The equivalent mobility electrical circuits are shown in the two figures below.Vibration isolation Problem 10. The effective stiffness. ke is then given by: k1 k2 ke ' k1 % k2 Thus the resonance frequency is given by: ω1 ' k1 k2 (k1 %k2)m k1 307 x 0 m k2 With a rigid frame.7) is: ω1 ω2 k2 k1 % k2 1 1 % k1 / k2 ' ' Problem 10. let the frame stiffness be represented by k2 and the isolator stiffness by k1.4 As shown in the figure. fin fin Mi Mm fm Mf ff Mm fm Mf ff without isolator with isolator . the resonance frequency is given by: ω2 ' k1 m The ratio of the two (which is plotted in figure 10. which is: T2 ' ff fin ' Mm Mm % Mf % Mi The force transmissibility with the isolator compared to that without the isolator is then T2/T1 and is: TF ' T2 T1 ' Mm % Mf Mm % Mf % Mi which is the same as equation 10.308 Solutions to problems (b) For the circuit without the isolator: fm Mm ' ff Mf and: fin ' fm % ff ' ff 1 % Mf Mm ' ff Mm %Mf Mm The force transmissibility is the ratio of ff /fin.31. . which is: T1 ' ff fin ' Mm Mm % Mf For the circuit with the isolator: fm ' ff and: fin ' fm % ff ' ff 1 % Mi % Mf Mm ' ff Mm % Mi % Mf Mm Mi % Mf Mm The force transmissibility is the ratio of ff /fin. 19. Ωa = 0. Thus: fa ' 0.5 Mz ' a / δx ' 0.281 From figure 10.80 Hz 0.847 Hz 50 309 Now the rocking modes will be calculated.35 / 0.2/2)2 ] / 3 ' 0.0 m. Ωa = 0.33 and Ωb = 1.446 Wx ' (δz / b) kx / ky ' 0.3873 1 / 4 ' 0. machine depth =1.3122 ' 1.6/2)2 % (1. first in the x-y plane (about the zaxis) and then in the z-y plane (about the x-axis): 0.18 in text): δx ' [ (0.847 × 0.38 × 2.09.35 Mx ' a / δz ' 0.35 / 0.3873 0.7 m.4-h) = 0.Vibration isolation Problem 10.3122 1 / 4 ' 0.18 in text): δy ' [ (0.21 Hz and fb ' 1.6m k = 4000 N/m Radius of gyration about vertical y-axis (eq.h) = 0.4/0.6m vertical dimension of mass.2m 2h = 0.4330 m Radius of gyration about horizontal x-axis (eq.2/2)2 ] / 3 ' 0. Thus: .2 m a = 0.6 2b = 0.38 and Ωb = 1.19 × 2.h = 0. machine height = 2×(0.3873 m Radius of gyration about horizontal z-axis (eq. 10.033 Wz ' (δx / e) kz / ky ' From figure 10.9/2)2 % (1.3122 ' 3.18 in text): δz ' [ (0. 2d = 2(a . 10.5 in the text. machine width = 0.9m 2e = 1.6/2)2 % (0.4 m.847 × 0.3122 m f0 ' 1 2π k 1 ' m 2π 4000 × 4 ' 2.9/2)2 ] / 3 ' 0.4/0. 10.3873 ' 1.5 in the text.5 .3122 ' 1. 1 % 0.310 Solutions to problems fc ' 0.11 in the text and the discussion on pages 496. Problem 10.352 × 4000 % 0.52 × 4000 50 × 0.7 From equation 10.2 ' 0. Mf = 0.2 % 1 which corresponds to a reduction in force transmission by a factor of 4.847 × 0.09 × 2.3873 ' 1.21 Hz and fd ' 1. the increase in force transmission is: TF ' Mm % Mf Mm % Mf % Mi From the question.2Mi = 2Mm.1 % 0.31.01 Hz The resonance frequency of the rotational mode is calculated using equation 10.8 Referring to figure 10. the optimal absorber is characterised by: k2 k1 ' m1 m2 ( m1 % m2 ) 2 k2 m2 2 π f2 ' ζ2 ' 3 ( m2 / m1 ) 8 ( 1 % m 2 / m 1 )3 The excitation frequency is 3000/60 = 50Hz and this should be equal to the .3873 ' 4.5 / 0.231 0.4332 fy ' ' 4.847 × 0.01 Hz Problem 10.17 in the text and is: 1 π 0.33 × 2.3.5 / 0. Thus: TF ' 0. The text indicates that the damping mass should be as large as possible. and the given values for k1 and m1. Problem 10. then it is likely that the sound radiation will be dominated by vibration modes which are being forced to vibrate at frequencies well above their resonances.47 as: ζ ' 3 × 200 / 1000 8 ( 1 % 200 / 1000 )3 ' 0. Thus (100 π) 2 ' k2 / m2 or k2 ' 9. thus it seems impractical to try to satisfy the second of the above three design equations. However. if the surface is excited by an acoustic wave on the side opposite that which is causing the radiation problem.7 MN / m The required damping may be obtained from equation 10. This is generally the case when the surface is excited mechanically.9 Damping a vibrating surface will only reduce the resonant response. Thus the required stiffness is given by: k2 ' (100π)2 × m2 ' 19.10 (a) 100dB below 1 volt corresponds to a voltage of 1 × 10-100/20 Volts = .Vibration isolation resonance frequency of the absorber mass and spring system. Thus let the design mass m2 =20% of m1 = 200kg.21 Problem 10. In this case damping the surface will not reduce the sound radiation directly but may reduce it a little because adding mass to the vibrating structure will decrease its mobility for excitation by the incoming sound field. See pages 504-506 in the text. we obtain: 9.87 × 104 × m2 311 Using the first equation and substituting the above for k2. Thus damping will only result in a reduction in sound radiation if the resonant modes are contribution most to the radiated sound field.87 × 104 × m2 107 1000 m2 (1000 % m2 ) 2 ' which results in a negative mass m2. Thus the condition mfu < 15. then the lightest accelerometer which can be used is 1 gram. Thus at any point the pressure and acoustic particle velocity are related by p = ρcu.s. sound will be radiated as plane waves with no near field.01g. the velocity of the plate is given by: u ' p 2 × 10&5 × 1080 / 20 ' ' 4.s. (b) The required relation can be determined by considering the mass loading effect of the accelerometer. As the acoustic particle velocity adjacent to the plate is equal to the plate velocity. (c) From part (a).954.6 The r.836 × 10&4 m/ s ρc 413. For steel. Solutions to problems Accelerometer mass. is approximately equal to sensitivity in mV/g. acceleration is thus given by: u rms ' 2π f u ' 2 π × 1000 × 4. m (grams). Smallest detectable acceleration (in g) is the smallest detected voltage in milli-volts divided by m. Problem 10. ρ = 7800 and cL = 5150/0.580 h 2 must be satisfied.836 × 10&4 ' 3.m.312 10µVolts. Substituting m = 1 and h = 1 in the relation of (b) above gives fu = 15.7 × 10&4(ρ cL h 2 / f ) .04 m/ s 2 0 (b) The r.m.078 µm 2π f 2π × 1000 . To obtain results within 3dB of the correct level.836 × 10&4 ' ' 0.58kHz.01 divided by the accelerometer weight in grams. Thus smallest detectable acceleration is 10-2/m "g" which in metres/sec is 0.11 (a) For a large plate vibrating as a piston. displacement is given by: d rms ' u 4. the accelerometer mass must satisfy the requirement that m < 3. if the smallest acceleration to be detected is 0. .12 (a) Adding damping will reduce the sound radiated by a vibrating surface if the surface vibration modes which are excited are resonant as is usually the case if the structure or surface is excited mechanically (but not acoustically). If the surface is excited at resonance by a tonal excitation source. (c) Adding mass to a vibrating surface will reduce the sound radiation and increase the panel transmission loss at frequencies above the first resonance frequency of the surface and below the surface critical frequency. provided that the accelerometer did not significantly mass load the plate (see equation 10. Problem 10.53 in the text). (b) Adding stiffness to a vibrating surface will only decrease its sound radiation or increase its transmission loss at frequencies below the first resonance frequency of the surface.Vibration isolation 313 (c) At high frequencies accelerations are generally large compared to displacements. then adding stiffness will increase its resonance frequencies and if the carefully done will result in a reduction in radiated sound. a measurement of the sound pressure close to the plate may be the best way of determining the plate response.13 Vibration velocities measured in octave bands on a diesel engine are listed in the following table. whereas the opposite is true at very low frequencies. An accelerometer is thus the best means of measuring the acceleration at 1kHz. Problem 10. A full discussion of this concept may be found on pages 504-506 of the text. For thin plates at high frequencies. (b) Longitudinal waves would cause the accelerometer to vibrate normal to its axis and the cross-axis sensitivity of the accelerometer will result in a signal due to the longitudinal wave.4 mm/s (b) Overall velocity in dB re 10-6 mm/s = 20 log10(12. Problem 10.1Hz d .85 12. we obtain f0 = 11.5 3. For a properly selected and aligned accelerometer the effect can be very small but for any other accelerometer the effect could be significant depending on the relative levels of the bending and longitudinal displacements and the actual cross axis sensitivity of the accelerometer.14 (a) I would mount the accelerometer so that its axis was normal to the beam surface and thus parallel to the beam displacement.29/10-6) = 142dB re 10-6m/s2 (d) Estimate of the overall displacement level in dB re 10-6µm = 20 log10(18.85 3.08 (a) Overall rms velocity = (52 + 102 + 52 + 22 + 0.15 (a) Using the relation.64 0.52)1/2 = 12.18 2 6.22/10-6) = 145dB re 10-6µm Problem 10.7 5 7.98 12.14 0.6 10 7.4/10-6) = 142dB re 10-6 mm/s (c) Estimate of the overall acceleration level in dB re 10-6 m/s2 = 20 log10(13.28 0.314 Octave band centre frequency (Hz) rms vibration velocity (mm/s) rms acceleration estimate (v2πf) (m/s2) rms displacement estimate (v/2πf) (µm) Solutions to problems 63 125 250 500 1k 5 1. f0 ' 1 2π g . 81/0.492 )2 % (2 × 0.14) in the text. 3000/(60×11. First we must calculate the isolator mobility.05 × 4.002 = 4.183 × 10&6 & 2 × 10&5 % 6.057) = 25dB (c) We may use equation (10.31) in the text.905MN/m.Vibration isolation (b) We may use equation (10. equal. The isolator mobility is then jω calculated using Mi ' to give: ki Mi ' j 2 π × 50 ' j6.405 × 10&5 0 /0 ' 0. Problem 10.567 00 Thus the reduction in transmitted vertical force is now -20 log10(0. thus there is an increase of 20dB.49)2 (1 & 4. Thus: 1 % (2×0.5 .905e%6 The mobility of the supported mass is calculated using Mm = 1/j2πfm = -j3.567) = 5dB. we obtain: &3. The result is k = 1000×9. Using equation (10.16 The critical damping ratio is given by: 4πζ 1 & ζ2 ' 0.49.05×4.057 Thus the reduction in transmitted vertical force is -20 log10(0.405 × 10&5 m/s/N 4.2) and (10.31).49)2 315 The value of X is TF ' ' 0.3) in the text.183 × 10&6 & 2 × 10&5 TF ' / 00 &3. The overall spring stiffness is found by setting equations (10.183 × 10-6 m/s/N.1) = 4. 316 Thus: Solutions to problems ζ ' and η = 0.25 16 π2 % 0. .04 This loss factor is about 10 to 20 times greater than would be expected from a sheet of steel.25 ' 0.02. so one might conclude that the product would be effective. 0. One application would be for lining of parts bins. Mechanism would be reflection and absorption of primary energy. (c) Not feasible if global control is needed. Possible to establish zones of local noise reduction.2m downstream of the reference sensor (depending on the controller time delay) and the error sensor would be placed 0. Mechanism involved is suppression of primary source by changing its radiation impedance. A reference signal could be obtained from a microphone (with a turbulence filter) placed upstream. control source should be downstream of primary source and remote from turbulence generating parts of the duct system.1 Acoustic mechanisms associated with active noise control include: 1. then active control may be feasible. (b) Not feasible if the source of noise is the grille. reference sensor would be tacho signal. Absorption of energy by the control source 4. If the source of noise is upstream of the grille. . Reflection of energy as a result of causing an impedance mismatch at the control source 3. This is because it would be difficult to obtain a causal reference signal. and it may even work better if placed on the room side of the grille. Local cancellation at the expense of increased levels elsewhere Applications: (a) Feasible. Suppression of the primary source by changing its input impedance with a control source 2. The control source would be placed at least 1.5 to 1m downstream of the control source.11 Solutions to problems in active noise control Problem 11. and error sensor should be downstream of control source (out of source near field). control sources could be shakers on the tank or loudspeakers surrounding the transformer and very close to it. Reference signal would be derived from electricity mains signal. Error sensors would need to surround the transformer and be located further away than the control sources. Mechanism involved is suppression of primary source by changing its radiation impedance for small room and local cancellation for large room.318 Active noise control (d) Not feasible due to difficulty in obtaining a causal reference signal. control sources would be placed in cabin ceiling or in seat headrests. (i) Feasible. Microphones would be best located in passenger head rests. Feedback system is needed. . although local cancellation may dominate in some cases (g) Not generally feasible due to complexity of radiated sound field. and error sensors should be as close as possible to the passengers. otherwise they should be near the locations where noise reduction is needed. reference sensor would be tacho signal on rotating shaft related to noise producing machine. (h) Feasible. control sources should be in factory corner if room is small. Control sources could be actuators on the fuselage skin or loudspeakers in the passenger headrests. (f) Feasible. (e) Feasible. Mechanism is suppression of primary noise by changing the radiation impedance of the transformer tank. Mechanism involved is suppression of primary source by changing its radiation impedance. reference sensor would be tacho signal from aircraft engine. (1. . DF DF EW 2 R ρw 2 % ν t ρ In Eq. 1995" First line under Eq. change "332" to "331" Change Eq. p16.36" to "1. (1. p27. remove the subscript. p16. (1. p45. p41.10.40a to φ ' f (k ( ct ± r )) r 3 lines above Eq. change DP = 1.65" to "1. In line 3.90). (1.12 Errata in the 3rd edition of Engineering Noise Control p xi.. p34.3) to DC ' 1% p18.67). p29.099E Change Eq. 1. (1.50).41" Change the reference just above Eq. replace "pet" with "per" 2 lines under Eq. Change “Noise Reduction Index (NRI)” to “Noise Reduction Coefficient (NRC)” p xv. (1.89) and in Eq. change "1.5). change "1. change the equation to (1 / h f) E / ρ > 2 line 10.64" 4 lines above Section 1. " t " from pt. change “Noise Reduction Index” to “Noise Reduction Coefficient” p16. change “FWHA” to “FHWA” p xviii In line 19.346E to DP = 1. p35. (1.69) to "Fahy. 42) with: "If the number of hours of exposure is different to 8. 3rd equation down should be: Ta ' 8 × 2&(91. p72.2 & 90.6 with the more accurate figure below. the text should read.37) to (4. 3rd line. p87. change "1252" to "61252".1 hours p147. replace H with H) p142.3.0) / 3 ' 8 / 20. line 5. add "is the" after the word.01" and "0. line 4. "ordinate" Line 13. the "8" in Equation (4. then to find the actual allowed exposure time to the given noise environment. p51. That is: DND ' 2 (LAeq. replace "Zd " with "ZA " Heading 1." p144.3" should be replaced by "3.10(b) is an alternative representation of Figure 2.39 ' 6. . "Figure 2.301" respectively in Equations (4. replace "U " with "u" Table 1. Replace the sentence following equation (4.8h & 90 ) /L ) p143. line 3.2.41) inclusive p143. p134. change "sound" to "sounds" 2 lines above Example 2.3. is defined as equal to 8 hours divided by the allowed exposure time.320 p51. Replace Figure 4. or "noise exposure".1.10(a)" p111. Ta with LB set equal to 90.43 and the 2 lines preceding it with: The daily noise dose (DND).12. The number "3" and "0. Errata for third edition text book Table 1. p51. Replace equation 4.42) is replaced by the actual number of hours of exposure. replace "Z " with "Zs" line immediately below the figure. p76. 9.6. On y-axis.9 caption.1 1 2 5 321 impulse 5 dB / doubling steady state impact and steady state (equal energy) 8-hour dB(A) equivalent 5 10 5 102 5 10 3 5 10 4 105 10 6 5 10 7 5 108 5 109 B-duration x number of impulses (ms) p147.6 to Figure 4. First paragraph in section 4. add "each of which has a radius of ai" immediately after "sources" . p150. p157.y)" with "O" and label the observer as O in Figure 5.2. change "r" to "a" p179.6". In Eq (5.6). (5. 5th and 6th lines from the top.6). 13 lines from the bottom. replace "1995" with "1995. change "645" to "60645" in four places.7).Errata for third edition text book Peak sound pressure level (dB re 20 Pa) 180 170 160 150 140 130 120 110 100 90 0.2 p192. 2 lines above Eq.6.71). 2 lines under figure 5. (5. 4 lines under Figure 4. change "0. In Eq. (5.06" to "0. First line after the headings in Table 4. p149. change "r" to "a" p177. p153.7. add "MAF" = minimum audible field. change Figure 4. p176. 1999". change "r" to "r = a" p176. replace "(x. Line above Eq. p160. change "1414" to "1474". change label from "dB re 20 mPa" to "dB re 20 µPa" p165. Fig 4. 5 should be replaced with -3. Interchange the 63 Hz and 2000 Hz labels on the curves in Fig. Thus section 5.322 Errata for third edition text book p192. change "ka" to "kai" p192. 5.9" p236.4.71 and below. -3.171). In Eq.19. change "-0. In Figure 6.0<<<+0. change "a" to "ai" in 5 places ¯ p192 Eq. (5. In Eq. p229.12.72).5 p244. Eq.9.should be deleted and replaced with the paragraph above.09" to "-0. change Q to Q in 4 places p192 last line add "amplitude" immediately after "velocity" ¯ p193 Eq 5.either they should be included in the barrier calculations or calculated separately but not both.11. Thus the only correction term (Equation (5. Table 5. p232. in the centre on the right hand side replace γ ' 1 / κ with γ ' κ . p226.3 caption.0" p241. (5.73 and below change Q to Q in 2 places p225. (5.72). 2 lines above Eq.0<<<-0. Paragraph beginning "Note that ISO" only applies to overall AWeighted calculations and should be deleted here. change "Sutherland et al.. Line above Eq.1. 5. p251.3" to "10. 13 lines above Eq. change "2613" to "9613". change "a" to "ai" p192. ISO 9613-2 procedures for calculating ground effects and shielding effects are based on an assumption of downwind propagation from the sound source to the receiver. (5. The paragraph following this one should also be deleted as the meteorological effects should not be taken into account in two separate places .193)) that is offered by ISO for meteorological effects is a term to reduce the A-weighted calculated sound pressure level for long time averages of several months to a year. (5.181. 1979" p226.188) change "10. In Table 5. 5.72). 1974" to "Sutherland and Bass. 24" p264. Eq.20. 7. change ¢ pk (t) ¦ to ¢ pk (0) ¦ 2 2 p293. 7. add "energy" before "reflection" p295.16V with S S2 p295.52. Eq.52.16V 0. 2 lines beneath Eq. 2 lines above Eq. lines 2 and 3.11" p264. (7.12" should be numbered "6. The equation numbered "6. change "2001" to "2000" p296. change "NRI" to "NRC" in three places and change "Noise Reduction Index" to "Noise Reduction Coefficient" in two places. The lower curve should be labelled "5" p292. 6. The first equation should be numbered "6. Sx and Sx " to . replace p295.55. 6.52. change "Sx . 2 lines above Equation (7. 2 lines above section 6. The equation numbered "6. 3 lines above Eq. In each of the top two lines of the table. (7. p259.64). (7.64).Errata for third edition text book p253.6. change (2000) to (2001) 0. Sy and Sz " p301.76 to: . replace S2 with 1/S2 p267. replace S1 with 1/S1 p264.2. there are two curves labelled "4". there should be a minus sign before loge p294. Section 7. change ¢ pk (t) ¦ to ¢ pk (0) ¦ 2 2 p292.20. 3 lines above Eq. 7. Also change Eq. In Eq. multiply each of the three terms in brackets by -1 p294.25" should be numbered "6. 2 lines below Eq. "Sx . change pk to pk(0) p293 6 lines from the bottom.59). 7. 7.26" 323 p267. In Fig 6. 5 lines from the bottom. add "(m2)" after Sα p303. change ¢ pk (t) ¦ to ¢ pk (0) ¦ and add "at time t=0" after "mode k" 2 2 p292. change "1989" to "1995".62). 3 lines below Eq.3.7. 2. add the following: "Note that for square. replace T60 u with th 1 T60 u p330.37" with "8. 10 line. Caption of Figure 7. change x-axis label to f (Hz) (log scale)" p354. 3.99. Equation 7.38" p355. change "2000" to "2001" p339.36). 8. End of second full paragraph.25 and 2.122).4 and the last word in the line above that with "contour value at 2000 Hz is increased by 1 dB.2. change "ASTM E90-66T" to "ASTM E41387" p347. 2 lines from bottom. 2nd line after Eq. 8 and 10 the factors are 1.44. change "20 mm" to "20 µm" p304." and add "Note that IIC. 12th line from the bottom.88).324 Errata for third edition text book α250 % α500 % α1000 % α2000 ¯ ¯ ¯ ¯ 4 NRC ' (7. line 1. 5 lines above the figure. 2nd and 3rd lines from the bottom.11." p310. replace "8.6. change "porous surface" to "rigidly backed porous material" and in the last line.5. p353. 2." p352. change "below" to "above". clamped-edge panels. Eq. change "1973" to "1988" p343.76) p303. Rw and STC values are all reported as integers. change "L" to R p310. 2. 6.26 respectively. the fundamental resonance frequency is 1.83 times that calculated using Equation (8. 2. Immediately following Equation (7.23. 3 lines under Equation (8. For panels with aspect ratios of 1. replace the line immediately above section 8. (7. change "Elbert" to "Elfert" p329.21).85 should be: ξc ' f fc 1/2 p311. replace fc2 /2 with fc1 /2 .89. 1. 5 fc1" 325 p360. change f l to fR p361. Under "Point B". first line of item (b) in the caption. last line. replace 10 log10 m1 with 20 log10 m1 p360.0 dB p363.5 fc2" to "0. add the term. In Eq. replace "40log10 fc2" with "20log10 fc1 + 20log10 fc2" p360.. change "61" to "60" and "52" to "51" . change to "LineBpoint support ( fc2 is the critical frequency of the point supported panel)" p360. replace "8. items (b) and (c). 4 lines from the bottom of the page. Eq (a) under "Point C". 3rd line.9 × fc1 h ' 1& f fc1 2 1& f fc2 2 2 p363. last Eqn. Under "Point B". change the equation to: 20 % 20 log10 (2500/100) & 6 ' 42. 8.37" with "8. item (a). change "77" to "78" and "61" to "60" in 2 places p363. 6 lines from the bottom of the page.55 with: 2 h D ' π fc1 8 f η1 η2 fc2 f 2 if f < 0. change x-axis label to "frequency (Hz) (log scale)" p360. replace "30log10 fc2" with "20log10 fc1 + 10log10 fc2" p360. 8.50. "20 log10 (fc2 / fc1)" to the RHS of the equation p360.Errata for third edition text book p355. on the x-axis of the figure.9 × fc1 if f > 0. change "0. replace Eq.38" p359. 525 0. Equation (8.013 0. p395.5/30" should be "30.98. replace the sentence beginning with "Alternatively" with the following: "This mechanism can be considered to approximately double the loss factor of the base panels. Alternatively. p391. the panels could be connected together with a layer of visco-elastic material to give a loss factor of about 0. after the words "(0. In the 500 Hz column.05 0." 3 lines following Eq.76) should be (8.326 Errata for third edition text book p365.04 α f from Table 7.06 0.2 30 36 37 40 46 54 α w from Table 7. 8.1 0.02 0. p371. replace S1" with "51" p379. replace the four equations for Ab with the following in the same order p381. add the following sentences.36 67 67 59 45. in section 3.6).015 0. the straight line distance.8" p380. 4th Eq. replace the example table with the following table. should be (8. "or connected with a layer of visco-elastic material or even nailed together". d.2. 9 lines down.8 SE / Si α i 10log10( SE / Si α i ) 18 18 18 17 15 14 NR (dB) 12 18 19 23 31 40 p381.66) and 8 lines down. Equation (8.02 0.02 1." p365.04 15. .2.85) is the distance between the image source and the receiver. used in Equation (8.75) should be (8.5/31" At the end of the paragraph above the figure.5 22.03 0.01 0.68 1. 7th number from the bottom. "When paths involving the ground reflected wave on the source side are considered.03 2. Section 8.2.7" to "Example 8. p394.2.01 0. 3rd line down.2.05 1. "30.1 0.6.65).02 Si α (m) 0. 6 lines down Equation (8. Octave band centre frequency (Hz) 63 125 250 500 1000 2000 4000 8000 57 0. Section 8.6. 5 lines down.2 12 47 TL from Table 8.01 0.463 0.67 18.6 29.6 13 44 59 0. The same reasoning applies to paths involving ground reflections on the receiver side.4". change "Example 8.463 0.65). replace "barier" with "barrier".013 0.6 m)". add the words.01 0.3 to 0. 2 lines above "Example 8. 19. Item 3.0 % 20 log10[4.9 dB dB.9 dB p396.20". 24.9 dB p396.Errata for third edition text book 327 Ab ' 15. last line. lines 2 and 3.5 dB p395.5/4.8 % 20 log10[5. replace "4. AR ' 5 dB.3 dB.0 dB Item 3. replace the two equations for Ab with the following in the same order. Ab ' 19. line 4.2 / 4] ' 24. Ab % AR ' 27.3 dB. change "10 dB" to "12 dB" Figure 8.5 Ab ' 19.20".18" to "5.7" Solution.6" with "4. p396. Item 3.9 dB. 27.2/4] ' 24.5] ' 18.9 dB. Ab ' 12. AR ' 2.3 dB Ab % AR ' 19.19. 6 lines from the bottom. p399.8 % 20 log10[7.3 dB.8 % 20 log10[7.9 dB.5/4] ' 13. change "5. Ab % AR ' 28.6 dB.5] ' 23.0 dB. 13 dB. item 1. p396. p395. line 4. replace r with R .3 dB Ab ' 19. 28.9 dB. 28.18" to "5.8/4.9 dB and 24.5 % 20log10[7. change the numbers to 19. change "5. Ab % AR ' 19. AR ' 1. p400. 6 kg/m2.107) should be: 1/2 1/2 2 2 2 XS % (hb & ZS )2 % XR % (hb & ZR )2 %b λ Figure 8. Errata for third edition text book Replace Eq.12" to "Figure 8.100) with: Rs ' R θ cos α hs ' Hb & R θ sin α α ' 1 (π & θ) & β 2 ) ) β ' cos&1 (Hb / A) θ ' ± cos&1 [1 & (A 2 / 2R 2 ) ].328 p399. (8. . 150 mm and 360 mm). (8. The I symbols represent variations in measured values for three pipe diameters (75 mm. density 70-90 kg/m3. st *R* > A / 2 1 paragraph. p401. covered with a lead / aluminium jacket of surface density. change "Figure 8.21 Typical pipe lagging insertion loss for 50 mm glassfibre. N ' ± p404.21 is missing (see following figure) 2 % Y2 1/2 &d 50 Octave band insertion loss 40 30 20 10 0 63 125 250 500 1000 2000 4000 8000 Octave band centre frequency (Hz) Figure 8.14" Eq. p432. replace µ with fm In Table 9. Item 5. (9.25) and (9. and which only applies for flow speeds such that uτ / (ω d ) > 0. 329 Replace Equations (8. change "6" to "5" 4 lines below the equation in the middle of the page.52) up one line. lines 6 and 7 under Eq 9.117) and (8.Errata for third edition text book p405.03 . "19" should be "-19" curve no 1 2 3 4 p453.232ξc R / h Xm ' [ 226(m / h)1 / 2 ξc (1 & ξc ) ] & [258h / (Rξc )] 2 (8.2.5" to "7" .52)" line following Equation (9.21. (8.16. for grazing flow across the holes. Replace "Equation (8. Move Equation (9. In this case. Figure 9. change "S" to "A" and in the caption add "open" immediately before "duct". p439. replace with.116).117) (8. 454.119) p415.5 1 p459. change "5.1 0. In the equation in the centre of the page.116) (8.25). For a" p417.81). Replace the legend in the figures with h 0. line 1. p461.48)" with "Equation (9.6( m / h )1/2 ξc ( 1 & 1 / ξc )&1/4 ] & [258h / R ξc )] Cc ' 0. x-axis label. 2001)" p429. p444. is (Dickey and Selamet. "the end correction. replace the text between Eqs. p461.26) with: "An alternative expression for the effective length. which may give slightly better results than Equation (9. > = 0.119) with the following: Xc ' [41.01 0. 27.18) and 2 lines above it. change "1. p462. and Eq. p462. change "12. replace "e" with "q" to avoid confusion with the distance. (9. caption. e. change "1979" to "1978" Figure 10. replace "D" with "d" Eq. p471.115). replace the lowest y-axis label (currently 0) with 0.02 In Equation (10. p464. 63 18(20) 16(18) 14(16) 13(14) 10(12) 9(10) 8(9) 7(8) 6(7) 5(6) 4(5) 2(3) 125 13(14) 11(12) 9(11) 8(9) 6(7) 5(6) 4(5) 3(4) 2(3) 2(3) 1(2) 1(1) 250 8(9) 6(7) 5(6) 4(5) 2(3) 2(2) 1(2) 1(1) 1(1) 1(1) 0(1) 0(0) 500 4(5) 2(3) 2(2) 1(2) 1(1) 1(1) 0(1) 0(0) 0(0) 0(0) 0(0) 0(0) 1000 1(2) 1(1) 1(1) 0(1) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 2000 0(1) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) 0(0) Figure 9.2.5 with the following: Octave band centre frequency (Hz) Duct diameter (mm) 150 200 250 300 400 510 610 710 810 910 1220 1830 p470.2" to "1. replace "D" with "d" 3rd and 6th line of the first paragraph. not .28. 9. the force should be shown as acting on mass m2. In Figure 10.23 caption. between spring supports. p478.0" Figure 9. p483.6. p484. (9.117) and 2 lines below Fig. change "1992" to "1987" Replace Table 9. change "1979" to "1978" line above Equation (10.116) and (9. p476. p479. Errata for third edition text book 8 lines below the equation in the middle of the page.5" to "13" line 1.14). last line.330 p461. 47). last Equation. replace "d" with |F|/ k1 Equation (10. p485 p487. line above Equation (10.10). the numerator on the RHS should be 3(m2 /m1)3 8 lines from the top of the page. 32. P496. change "8. 27. 72 74 61 55 77 79 66 60 80 82 69 63 81 83 70 64 80 82 69 63 76 78 65 59 69 71 58 52 63 65 52 46 p513. 3rd line in 2000 Hz column should be "25" Table 11. 15.31).5 log10 kW last line. p517. Equation (10. p514. the 8000 Hz column should be replaced with 13. p498.2.2) Example 11. 42 and 44 respectively and the BFI column for the two tubeaxial entries should be " 7 " Remove the paragraph containing Equation (11.1 table. (10.b). the left hand side should be squared. change "1986" to "1988" 331 Equation (10. p513. p515.2) and remove "(11.8" to "8.Errata for third edition text book mass m1. 35.2. p496. (dB re 10&12 W) p526. 34. 35. change "1979" to "1978" Table 11.25a. 26. p495. change to: Lw ' 72 % 13. p528. p513.3" replace the values in the table with the following. 0 60 120 180 . 32.2)" in the second to bottom line. In Eqs.48). 18.42). replace "30" with "36" Equation (11. label (11. remove the symbol "d" from the right hand side. 11. Following Eq. 11.29 with the following: "The road surface or condition correction is taken as zero for either sealed roads at speeds above 75 km/hr or gravel roads. the correction is -1 dB.0. insert the statement.37".65) and (11. Ccond = 4 . Errata for third edition text book 4 lines above Eq. change "534" to "60534". p536. "If the second term in brackets of Equation (11.332 p535.67)." 1 line and 4 lines above Eq. add the following: "Note that the final spectrum levels must all be adjusted by adding or subtracting a constant decibel number so that when A-weighted and added together. the correction is.. the correction is -3. p542.73. (11. change "534" to "60534".(11. it is set equal to 0.64. p541. (11...33). Equation 11. not "53". p558..34). For pervious road surfaces. p544.73) and (11.3".76) with octave band sound power levels used in Equation (11. p552. Immediately before Equation (11. line 3. change "534" to "60534". so it reads "17.70).. the result is identical to the A-weighted overall levels form Equations (11.66)." Replace the nine lines following Eq. . "The octave band external sound pressure levels may be calculated using Equations (11. and 2 lines after Eq.89) should be "55". p544. p542.102 with the following: "Low barriers such as twin beam metal crash barriers can have less p543.. 4 lines from the bottom. For concrete roads with deep random grooves greater than 5 mm in width.76) instead of overall sound power levels.)" Replace the last paragraph with. change "534" to "60534". Replace the paragraph following Table 11.03P where P is the percentage of heavy vehicles. p559.. For speeds below 75 km/hr on impervious sealed roads..64) exceeds 0.37(.27" replaced with "17..3. second term on the right should have the "log10" removed and "17." The constant in Equation (11.5 dB.. and 3 lines under Equation (11. change "FWHA" to "FHWA" 6 lines from the bottom. replace "1995" with "U. excluding the barrier correction. So if these are used with any proportion. In addition the track correction. Pd . "Note that the two values for $ must add up to 180E " In the heading and first line. DOT.. change SEL to SELref. 1995a". For any specific train type consisting of N identical units.0). p565. p563. 2 axles" should actually be "Freight vehicles. the quantity SELref is calculated by adding 10log10N to SELv." before "1998". The second entry of "Freight vehicles." before "1998". their effect should be calculated by looking at the lower noise level (or the most negative correction) resulting from the following two calculations: $ $ Soft ground correction (0 < Pd < 1. replace "1995" with "U. 4 lines under Equation (11.111). C2 from Table 11. et al. and hard-ground correction (Pd = 0) plus the barrier correction. . 4 axles" Lines 1 and 3.K.Errata for third edition text book 333 effect than soft ground. tread braked.” p560.108). add "Menge. Remove the sentence beginning 12 lines from the bottom of the page. of soft ground. p563.32 must also be added so that: SELref ' SELv % 10 log10 N % C2 p564.K. add "Menge. p562..109). 3 lines under Equation (11. DOT. p563.b". 1995a. Replace the last two lines of page 563 and the top three lines of page 564 with the following: "Note that different vehicle types must be considered as separate trains. p564. et al. p561. 5th line in first paragraph. disc braked. p561. (1993).27.1).015 line in table for concrete. p609.32. Auditorium acoustics and architectural design . p609. 10 lines above Equation (12. loss factor = 0. Errata for third edition text book table 11.C2. p617. 15-23.6" to "C. line in the table for Nylon. and Champoux. p622. E / ρ = 1.320. line in the table for iron. 1.30).5" and "C. p609. p621. Barron. E&FN Spon: London. add . after "Correction" in the column 2 label.02 the last column of numbers is the density and the 2nd last column is Young’s modulus.29). Young’s Modulus = 206.0005 and < = 0. 11.0. Missing references. M. loss factor = 0. p623. p609. move the "6. replace ZN with ZN /Dc In Equation (C. remove the minus sign Add equation numbers. J. p567. In Equation (11. line 2 in the table for fresh water.122 and 11.6". p645. change "C. line in table for lead. materials. Noise Control Engineering Journal 32. density =1. p609.600. (1989). E / ρ = 4910.121). Change number of Eq.5" to "C.6" next to "nylon" and Young’s Modulus = 2. In figure captions. In Equation (C.334 p565.123 to the equations at the top of the page.005 . density =7. .F.24. change "988" to "998". 0 = 0. p568. Allard. p580.140. In situ two-microphone technique for the measurement of acoustic surface impedance of . Y. p610.36 to C. replace 2 with $ in three places. change "1985" to "1986". A. (1995). Cazzolato. D. In Proceedings of Internoise '98. p648.H. (1983). (1980). Davy.C: Institute of Noise Control Engineering. Revised edition. p647. J.Paper #44.G. Problems in the theoretical prediction of sound insulation. J. (1999).L.Errata for third edition text book p646. R.L. B.. Comparison of aircraft noise-contour prediction programs.L. PhD thesis.H. A criterion for predicting the annoyance due to lower level low frequency noise. Lindvall. (1993). S. Chapkis. B. (ed.. Concert and Opera Halls. T. (1981). Acoustical Society of America: New York. Journal of Aircraft.pp. 975B978. (1998). 3732B3735. and Hansen. L. C.L. Stockholm: Stockholm University and Karolinska Institute. (1996). N. Broner. Missing references. Davy. Missing references. South Australia. Missing references. Beranek. Lindvall. B. pp. Chapkis. Journal of the Acoustical Society of America. 18. and Marsh. . 160B168. Jan.. R.H.H. L. In Proceedings of Internoise '93. Journal of Low Frequency Noise and Vibration 2. (1999). Impact of technical differences between methods of INM and NOISEMAP. Adelaide University. Bragg. Community Noise. Geneva: World Health Organization. Blankenship. How They Sound. H. (1963). Guidelines for Community Noise.S. B.) (1988).. Eds. L. Noise and Vibration Control. 106. and Schwela. In Proceedings of Internoise '80. G.S.L. D. Washington D. 12B16. The sound transmission of cavity walls due to studs. Sensing systems for active control of sound transmission into cavities. p649. Cazzolato.L. 335 Missing references. Structural radiation mode sensing for active control of sound radiation into enclosed spaces. Berglund. Beranek. and Leventhall. Combustion noise.L. T. 831B834. Journal of the Institute of Fuel. 926 B 933. . Berglund. and Schwela. (1999). In Proceedings of Acoustics 2000. and Hansen. Jr. G.Q. 1028B1041. F. 48. .L. (1976).A. Edge. Rapoza. Relation of acoustical parameters with and without audiences in concert halls and a simple method for simulating the occupied state. Washington. (2000).336 Errata for third edition text book Davy. Selected methods for quantification of community exposure to aircraft noiseNASA TN ..E.M. Harland. London: Academic Press. Missing references. . p650. 155-160. C.G. Highway Traffic Noise Analysis and Abatement Guide.. and Walker. and Gulding.H. (1959). (1998).. 16B24.M.L. Noise Control Engineering Journal 48. Fleming. Hood. Applied Acoustics. J. U. D. Reverberation formula which seems to be more accurate with nonuniform distribution of absorption. E-M. (1976).. C. Howard. J. J. N. FHWA (1995). (2001). Western Australia. 893-97. Journal of the Acoustical Society of America. Vigran. pp. M. R. Burstein. Dept. D. Nishihara.. Missing references..109. Exhaust stack silencer design using finite element analysis. Ground effects in FAA's integrated noise model. Noise Control Engineering Journal. 113-120. Cazzolato.E. (2000).G.. The regulation of sound insulation in Australia. D. 305-25.. D. and Cawthorn. Australian Acoustical Society Conference. London: E&FN Spon. Add Wosal. G.G. . Fahy. and Scholes.S. B. Fitzroy. and Kristiansen. Delaney. Journal of Sound and Vibration 48. F.M. Fahy. L. Dutilleaux.R.C. of Transportation. T. A. to the authors of the Hodgson (2002) paper. P. 555-572. (2000).J. November 15-17. Federal Highway Administration. D-7977.A. Journal of the Acoustical Society of America. J. Foundations of Engineering Acoustics. Fundamentals of Noise and Vibration. J.31. and Beranek. (2001). Senzig. U. W.E. The prediction of noise levels L10 due to road traffic. p652.J. Hidaka.S.S. T. 62. An in situ transfer function technique for the assessment of acoustic absorption of materials in buildings. (2001). . (1971). The present and future of aircraft noise models: a user's perspective. E. J. Response of ribbed panels to reverberant acoustic fields.S. and Lifsltitz.M. C.. Maidanik. Journal of the Acoustical Society of America. 2-6 July. Li. 809B826.M. Journal of the Acoustical Society of America. Active noise control systems . U. (2001). (1994).M. Applied Acoustics.J. (1993). (1996). (1998). S. In Proceedings of the 8th International Congress on Sound and Vibration. p655. 2929B2936. A high frequency approximation of sound propagation in a stratified moving atmosphere above a porous ground surface.H.D. K. B. pp969 B 974. L. Kuo. J-W.W. and Lifshitz. and Morgan. Kurze. In Proceedings of the 8th International Congress on Sound and Vibration. L. p654.S. Li. E. 2-6 July. Cazzolato.D.R. S. Ph. On the validity of the heuristic rayBtraceBbased modification to the WeylBVan der Pol formula.M.. and Anderson. 1727B1735.Errata for third edition text book p653. K. 337 Missing reference Jean.W." should be "Landau. K. Hong Kong." Missing references. (2001). In Proceedingsof Noise-Con '94. the reference. 2077B2083. 34. Active vibration control to reduce the low frequency vibration transmission through an existing passive isolation system. G. (1994). Journal of the Acoustical Society of America. pp. Rondeau. K. Lee. X. Li. Hansen. Taherzadeh. Larson. New York: John Wiley. 4. "Landau. D. Numerical models for noise prediction near airports. and Attenborough. Hong Kong. 1840B1852. 35B53. Sound attenuation by barriers. D. G. (1962). An improved rayBtracing algorithm for predicting sound propagation outdoors.. p654.M.104. Missing references..93. K. Journal of the Acoustical Society of America. 95. and van Maercke. and Li.-F. 0. (1968). U. R.. Missing references. In Proceedings of the Sixth International Congress on Sound and Vibration. 61B70. pp. Hearing Loss Due to Exposure to Steady State Broadband Noise. Anderson. Transportation. p657.C. C. Missing references. Version 1. p656. 154B158. (2001). (1969). 36. Report No. pp. 685B694.338 Errata for third edition text book Menge. PhD thesis.O. .J. Nilsson. Hong Kong. D.W. Active minimization of energy density in a threeBdimensional enclosure. (2000).S. J. p658. W.O. In Proceedings of the 8th International Congress on Sound and Vibration. Denmark. (1999). Parkins.. Institute for Public Health Eng. Technical Manual. In Proceedings of the Eighth International Congress on Sound and Vibration. Garmisch-Partenkirchen. Copenhagen. Washington. Hearing Levels of Non-Noise Exposed Subjects and of Subjects Exposed to Constant Noise During Working Hours. R. Hong Kong. Existing reverberation time formulae .a comparison with computer simulated reverberation times. July. Passchier-Vermeer.. M. Germany. Journal of the Acoustical Society of America.F. Estimation of reverberation times in nonrectangular rooms with non-uniformly distributed absorption using a modified Fitzroy equation. W.S. A. July. Pennsylvania State University.J. Passchier-Vermeer. Dept. 805-812. (1998). (1998). Neubauer. C. Sound transmission through double panels using Statistical energy analysis. Report B367. Neubauer. 1709B1716. USA. Outdoor sound propagation over complex ground. (1977). The Netherlands. Wave propagation and sound transmission in sandwich composite plates. July. The Netherlands.W. and Crocker. G. FHWA Traffic Noise Model.J.47. A. Missing references. B. Plovsing. Price. and Bajdek. Research Institute for Environmental Hygiene. 7th International Congress on Sound and Vibration. C. (2001). Rossano. Optimal design of linear and nonlinear vibration absorbers for damped systems. and Yamamoto.A.E. ..M. J. Raspet.Errata for third edition text book 339 Raney. A. Aircraft noise. p659. Yokohama. The effect of realistic ground impedance on the accuracy of ray tracing.. Acoustical Society of America. Applied Acoustics. Experimental evaluation. VIC.. K.M. 683B693. (1998). A. Computational Acoustics in Architecture.E. Leach. Vermont South. change "1985" to "1986" Missing references. and Mateus. J. DoE Traffic Noise Prediction Method. Australia.B. Missing references.R. change "1979" to "1978". Calculation methods for road traffic noise propagation proposed by ASJ. Applied Acoustics. (2001). Saunders.J. edited by C. R. A critical review of some traffic noise prediction models. A. and Cawthorn. and Lee. K. Southampton: WIT Press. Sutton. and Hall.M. D. Sendra. In Proceedings of Internoise ‘94. Micrometeorology. 122. L'Esperance. Acoustics. New York: MGraw-Hill. Soom. M. 3rd edn..P. New York. Samuels. p660. O.97. 307B325. S. pp.G. A. (1995).K. Steele. In the Shepherd reference. . 62. (1953). Australian Road Research Board. Japan. J. (1999). Journal of Vibration. double and triple glazing. C. p660. U. (2001). International Institute of Noise Control Engineering. Noise Emissions of Road Vehicles: Effect of Regulations. Harris. 271-287. (1983). Sound transmission through single. p659. Tadeu. (2001). An Evaluation of the U. R. Takagi.J. R. Journal of the Acoustical Society of America. 112B119. Sandberg. reprint. 62. Research Report ARR No. Chapter 47 in Handbook of Acoustical Measurements and Noise Control. Stress and Reliability in Design 105. G.289B294. Tse reference. (1983). Final Report 01-1 of the I-INCE Working Party on Noise Emissions of Road Vehicles. and Daigle. (1998). (1955). Journal of the Acoustical Society of America. Journal of the Acoustical Society of America. Specification for Audiometers. 535-548. L. Wiener. 724. DOT (1995a). "ANSI S3. and Keast.. London: HMSO. S. V. In the Zinoviev reference.I. DOT (1995b). Acoustical behavior of some engine test cell structures. D. Department of Transport. Zaporozhets. p667. London: HMSO. DOT (1988). Evaluation and prediction of airport noise in Japan.K. (1959). London: HMSO. add. .K. Experimental study of the propagation of sound over ground. B.L. Watters. 341B344. Yoshioka.G.M. 99B127. p661. U. Supplement 1. 55. Journal of the Acoustical Society of Japan (E). p715. 31.I. Errata for third edition text book Missing references. 21. Aircraft noise modelling for environmental assessment around airports. p663. 449B456.D. and Tokarev. Department of Transport. F. H.6 B 1997. Calculation of Railway Noise. Labate.340 p661. (2000). U." line 1. Change "Noise Reduction Index" to "Noise Reduction Coefficient". O. replace "In print" with " 269. and Beranek. U. Department of Transport.K. replace "E90-99" with "E90-02".27. Calculation of Railway Noise. Calculation of Road Traffic Noise. Applied Acoustics." after last line.