Noise Control: From Concept to Application [2 ed.] 1138369012, 9781138369016

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Noise Control

Noise Control

From Concept to Application Second Edition

Colin H Hansen Kristy L Hansen

Second edition published 2021 by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN and by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 © 2021 Colin H Hansen and Kristy L Hansen First edition published by CRC Press 2005 CRC Press is an imprint of Informa UK Limited The right of Colin H Hansen and Kristy L Hansen to be identified as authors of this work has been asserted by them in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data Names: Hansen, Colin H., 1951- author. | Hansen, Kristy L., author. Title: Noise control : from concept to application / Colin H Hansen, Kristy Hansen. Description: Second edition. | Boca Raton : CRC Press, 2021. | Includes bibliographical references and index. Identifiers: LCCN 2021002295 (print) | LCCN 2021002296 (ebook) | ISBN 9781138369016 (hbk) | ISBN 9781138369023 (pbk) | ISBN 9780429428876 (ebk) Subjects: LCSH: Noise control--Textbooks. | Acoustical engineering--Mathematics--Textbooks. Classification: LCC TD892 .H359 2021 (print) | LCC TD892 (ebook) | DDC 620.2/3--dc23 LC record available at https://lccn.loc.gov/2021002295 LC ebook record available at https://lccn.loc.gov/2021002296

ISBN: 978-1-138-36901-6 (hbk) ISBN: 978-1-138-36902-3 (pbk) ISBN: 978-0-429-42887-6 (ebk) Typeset in Latin Modern font by KnowledgeWorks Global Ltd.

Dedication This book is dedicated to Susan, to Laura, to Branko, to Vladimir and to Dimitrij.

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Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 1.2

1.3 1.4

1.5

1.6 1.7 1.8 1.9 1.10 1.11 1.12

1.13

1.14

2

Introduction . . . . . . . . . . . . . . . . . . . . Noise-Control Strategies . . . . . . . . . . . . . 1.2.1 Sound Source Modification . . . . . . . 1.2.2 Control of the Transmission Path . . . 1.2.3 Modification of the Receiver . . . . . . 1.2.4 Existing Facilities . . . . . . . . . . . . 1.2.5 Facilities in the Design Stage . . . . . . Acoustical Standards and Software . . . . . . . Acoustic Field Variables . . . . . . . . . . . . . 1.4.1 Variables . . . . . . . . . . . . . . . . . 1.4.2 Magnitudes . . . . . . . . . . . . . . . 1.4.3 The Speed of Sound . . . . . . . . . . . 1.4.4 Acoustic Potential Function and the Wave 1.4.5 Complex Number Formulations . . . . Plane, Cylindrical and Spherical Waves . . . . 1.5.1 Plane Wave Propagation . . . . . . . . 1.5.2 Cylindrical Wave Propagation . . . . . 1.5.3 Spherical Wave Propagation . . . . . . 1.5.4 Wave Summation . . . . . . . . . . . . 1.5.5 Plane Standing Waves . . . . . . . . . Mean Square Quantities and Amplitudes . . . . Energy Density . . . . . . . . . . . . . . . . . . Sound Intensity . . . . . . . . . . . . . . . . . . Sound Power . . . . . . . . . . . . . . . . . . . Decibels . . . . . . . . . . . . . . . . . . . . . . Spectra . . . . . . . . . . . . . . . . . . . . . . 1.11.1 Frequency Analysis . . . . . . . . . . . Combining Sound Pressures . . . . . . . . . . . 1.12.1 Coherent Sounds . . . . . . . . . . . . 1.12.2 Incoherent Sounds . . . . . . . . . . . . 1.12.3 Subtraction of Sound Pressure Levels . 1.12.4 Combining Level Reductions . . . . . . Impedance . . . . . . . . . . . . . . . . . . . . 1.13.1 Mechanical Impedance, Zm . . . . . . . 1.13.2 Specific Acoustic Impedance, Zs . . . . 1.13.3 Acoustic Impedance, ZA . . . . . . . . Additional Problems . . . . . . . . . . . . . . .

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Loudness, Descriptors of Noise, Noise Criteria and Instrumentation 2.1 2.2

Introduction . . . . . . . . . . . . . . . . . Loudness . . . . . . . . . . . . . . . . . . 2.2.1 Comparative Loudness and the Phon 2.2.2 Low-Frequency Loudness . . . . . 2.2.3 Relative Loudness and the Sone . 2.2.4 Weighting Networks . . . . . . . .

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xiii 1 1 2 3 4 4 4 5 6 7 7 8 8 11 11 11 12 17 17 21 22 26 27 27 31 37 43 44 45 45 50 52 54 56 56 57 57 58 61 61 62 62 63 63 68 vii

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Contents 2.3

2.4 2.5

2.6 2.7 2.8

2.9

2.10 2.11

2.12

2.13

Descriptors of Noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Equivalent Continuous Sound Pressure Level, Leq . . . . . . . 2.3.2 A-Weighted Equivalent Continuous Sound Pressure Level, LAeq 2.3.3 Noise Exposure Level, LEX,8h or Lex or Lep0 d . . . . . . . . . 2.3.4 A-Weighted Sound Exposure, EA,T . . . . . . . . . . . . . . . 2.3.5 A-Weighted Sound Exposure Level, LAE or SEL . . . . . . . . 2.3.6 Day-Night Average Sound Level, Ldn or DNL . . . . . . . . . 2.3.7 Community Noise Equivalent Level, Lden or CNEL . . . . . . 2.3.8 Statistical Descriptors . . . . . . . . . . . . . . . . . . . . . . . 2.3.9 Other Descriptors, Lmax , Lpeak , LImp . . . . . . . . . . . . . . Hearing Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hearing Damage Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Requirements for Speech Recognition . . . . . . . . . . . . . . 2.5.2 Quantifying Hearing Damage Risk . . . . . . . . . . . . . . . . 2.5.3 United States Standard Formulation . . . . . . . . . . . . . . 2.5.4 Occupational Noise Exposure Assessment . . . . . . . . . . . . 2.5.5 Impulse and Impact Noise . . . . . . . . . . . . . . . . . . . . Implementing a Hearing Conservation Programme . . . . . . . . . . . Speech Interference Criteria . . . . . . . . . . . . . . . . . . . . . . . . Psychological Effects of Noise . . . . . . . . . . . . . . . . . . . . . . . 2.8.1 Noise as a Cause of Stress . . . . . . . . . . . . . . . . . . . . 2.8.2 Effect on Behaviour and Work Efficiency . . . . . . . . . . . . Ambient Sound Pressure Level Specification . . . . . . . . . . . . . . . 2.9.1 Noise Weighting Curves . . . . . . . . . . . . . . . . . . . . . . 2.9.1.1 NR Curves . . . . . . . . . . . . . . . . . . . . . . . . 2.9.1.2 NC Curves . . . . . . . . . . . . . . . . . . . . . . . . 2.9.1.3 NCB Curves . . . . . . . . . . . . . . . . . . . . . . . 2.9.1.4 RC, Mark II Curves . . . . . . . . . . . . . . . . . . 2.9.2 Comparison of Noise Weighting Curves with dBA Specifications 2.9.3 Speech Privacy . . . . . . . . . . . . . . . . . . . . . . . . . . Environmental Noise Criteria . . . . . . . . . . . . . . . . . . . . . . . 2.10.1 A-Weighting Criteria . . . . . . . . . . . . . . . . . . . . . . . Environmental Noise Surveys . . . . . . . . . . . . . . . . . . . . . . . 2.11.1 Measurement Locations . . . . . . . . . . . . . . . . . . . . . . 2.11.2 Duration of the Measurement Survey . . . . . . . . . . . . . . 2.11.3 Measurement Parameters . . . . . . . . . . . . . . . . . . . . . 2.11.4 Measurement Uncertainty . . . . . . . . . . . . . . . . . . . . 2.11.5 Noise Impact . . . . . . . . . . . . . . . . . . . . . . . . . . . . Measuring Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . 2.12.1 Microphones . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12.1.1 Microphone Sensitivity . . . . . . . . . . . . . . . . . 2.12.2 Sound Level Meters . . . . . . . . . . . . . . . . . . . . . . . . 2.12.2.1 Calibration . . . . . . . . . . . . . . . . . . . . . . . 2.12.2.2 Measurement Accuracy . . . . . . . . . . . . . . . . . 2.12.3 Statistical Analysers . . . . . . . . . . . . . . . . . . . . . . . 2.12.4 Personal Exposure Meter . . . . . . . . . . . . . . . . . . . . . 2.12.5 Data Acquisition and Recording . . . . . . . . . . . . . . . . . 2.12.6 Spectrum Analysers . . . . . . . . . . . . . . . . . . . . . . . . 2.12.7 Sound Intensity Meters . . . . . . . . . . . . . . . . . . . . . . 2.12.8 Acoustic Cameras . . . . . . . . . . . . . . . . . . . . . . . . . Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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70 70 70 71 71 72 73 73 74 74 74 75 75 76 77 78 83 84 85 87 87 87 88 88 90 92 93 93 98 99 100 100 104 105 105 105 106 108 108 108 109 109 110 110 111 111 111 112 114 115 116

Contents

3

Sound Sources and Sound Power Determination from Measurement

. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simple Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dipole Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Quadrupole Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Lateral Quadrupole . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Longitudinal Quadrupole . . . . . . . . . . . . . . . . . . . . . . . 3.5 Line Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Infinite Line Source . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Finite Line Source . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Piston in an Infinite Baffle . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Incoherent Plane Radiator . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Single Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.2 Several Walls of a Building or Enclosure . . . . . . . . . . . . . . 3.8 Radiation Field of a Sound Source . . . . . . . . . . . . . . . . . . . . . . 3.9 Directivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10 Reflection Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.1 Simple Source Near a Reflecting Surface . . . . . . . . . . . . . . 3.10.2 Observer Near a Reflecting Surface . . . . . . . . . . . . . . . . . 3.10.3 Observer and Source Both Close to a Reflecting Surface . . . . . 3.11 Determination of Sound Power . . . . . . . . . . . . . . . . . . . . . . . . 3.11.1 Measurement in Free or Semi-Free Field . . . . . . . . . . . . . . 3.11.2 Measurement in a Diffuse Field . . . . . . . . . . . . . . . . . . . 3.11.2.1 Substitution Method . . . . . . . . . . . . . . . . . . . . 3.11.2.2 Absolute Method . . . . . . . . . . . . . . . . . . . . . . 3.11.3 Field Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11.3.1 Semi-Reverberant Field Measurements Using a Reference Source to Determine Room Absorption . . . . . . . . . . 3.11.3.2 Semi-Reverberant Field Measurements Using a Reference Source Substitution . . . . . . . . . . . . . . . . . . . . . 3.11.3.3 Semi-Reverberant Field Measurements Using Two Test Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11.3.4 Near–Field Measurements . . . . . . . . . . . . . . . . . 3.11.4 Uncertainty in Sound Power Measurements . . . . . . . . . . . . . 3.12 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sound Propagation Outdoors . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Geometric Divergence, Adiv . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Atmospheric Absorption, Aatm . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Ground Effects, Agr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Excess Attenuation Using Simply Hard or Soft Ground . . . . . . 4.5.2 Excess Attenuation Using the Plane Wave Method . . . . . . . . 4.6 Meteorological Effects, Amet . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.1 Attenuation in the Shadow Zone (Negative Sonic Gradient) . . . 4.7 CONCAWE Propagation Model . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 Geometrical Divergence, K1 . . . . . . . . . . . . . . . . . . . . . 4.7.2 Atmospheric Absorption, K2 . . . . . . . . . . . . . . . . . . . . . 4.7.3 Ground Effects, K3 . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.4 Meteorological Effects, K4 . . . . . . . . . . . . . . . . . . . . . . 4.7.5 Source Height Effects, K5 . . . . . . . . . . . . . . . . . . . . . .

3.1 3.2 3.3 3.4

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6

Contents 4.7.6 Barrier Attenuation, K6 . . . . . . . . . . . . . . . . . . . . . . . 4.7.7 In-Plant Screening, K7 . . . . . . . . . . . . . . . . . . . . . . . . 4.7.8 Vegetation Screening, Kv . . . . . . . . . . . . . . . . . . . . . . . 4.7.9 Limitations of the CONCAWE Model . . . . . . . . . . . . . . . . 4.8 ISO 9613-2 (1996) Noise Propagation Model . . . . . . . . . . . . . . . . . 4.8.1 Ground Effects, Agr . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.2 Meteorological Effects, Amet . . . . . . . . . . . . . . . . . . . . . 4.8.3 Source Height Effects . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.4 Barrier Attenuation, Abar . . . . . . . . . . . . . . . . . . . . . . 4.8.5 In-Plant Screening, Asite . . . . . . . . . . . . . . . . . . . . . . . 4.8.6 Housing Screening, Ahous . . . . . . . . . . . . . . . . . . . . . . . 4.8.7 Vegetation Screening, Afol . . . . . . . . . . . . . . . . . . . . . . 4.8.8 Effect of Reflections Other Than Ground Reflections . . . . . . . 4.8.9 Limitations of the ISO9613-2 Model . . . . . . . . . . . . . . . . . 4.9 Propagation Model Prediction Uncertainty . . . . . . . . . . . . . . . . . 4.9.1 Type A Standard Uncertainty . . . . . . . . . . . . . . . . . . . . 4.9.2 Type B Standard Uncertainty . . . . . . . . . . . . . . . . . . . . 4.9.3 Combining Standard Uncertainties . . . . . . . . . . . . . . . . . 4.9.4 Expanded Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . 4.10 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sound-Absorbing Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Flow Resistance and Resistivity . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Sound Propagation in Porous Media . . . . . . . . . . . . . . . . . . . . . 5.4 Measurement of Absorption Coefficients of Porous Materials . . . . . . . . 5.4.1 Measurement Using the Moving Microphone Method . . . . . . . 5.4.2 Measurement Using the Two-Microphone Method . . . . . . . . . 5.5 Calculation of Statistical Absorption Coefficients of Some Porous Material Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 Porous Liner with a Backing Cavity . . . . . . . . . . . . . . . . . 5.5.2 Porous Liner Covered with a Limp Impervious Layer . . . . . . . 5.5.3 Porous Liner Covered with a Perforated Sheet . . . . . . . . . . . 5.5.4 Porous Liner with a Limp Impervious Layer and a Perforated Sheet 5.6 Measurements of the Sabine Absorption Coefficient and Room Constant . 5.6.1 Reference Sound Source Method . . . . . . . . . . . . . . . . . . . 5.6.2 Reverberation Time Method . . . . . . . . . . . . . . . . . . . . . 5.6.3 Measurement of α ¯ for a Particular Material . . . . . . . . . . . . 5.7 Panel Sound Absorbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Noise Reduction Coefficient (NRC) . . . . . . . . . . . . . . . . . . . . . . 5.9 Sound Absorption Coefficients of Materials in Combination . . . . . . . . 5.10 Reverberation Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sound in Rooms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Low Frequency Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Bound between Low-Frequency and High-Frequency Behaviour . . . . . . 6.3.1 Modal Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Modal Damping and Bandwidth . . . . . . . . . . . . . . . . . . . 6.3.3 Modal Overlap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.4 Cross-Over Frequency . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 High-Frequency Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . .

188 189 190 190 190 191 194 195 195 197 198 198 199 200 206 206 207 207 208 210 213 213 214 214 216 216 233 234 234 235 236 241 241 241 242 243 244 246 247 247 248 251 251 252 260 260 261 261 261 262

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Contents 6.4.1

7

8

Relation between Source Sound Power and Room Sound Pressure Level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Relation between Room Absorption and Reverberation Time . . . 6.5 Flat Room with Diffusely Reflecting Surfaces . . . . . . . . . . . . . . . . 6.6 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Partitions, Enclosures and Barriers . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Sound Transmission through Partitions . . . . . . . . . . . . . . . . . . . 7.2.1 Bending Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 Transmission Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3 Single-Leaf Panel Transmission Loss Calculation . . . . . . . . . . 7.2.4 Double Wall Transmission Loss . . . . . . . . . . . . . . . . . . . 7.2.4.1 Staggered Studs . . . . . . . . . . . . . . . . . . . . . . . 7.2.4.2 Panel Damping . . . . . . . . . . . . . . . . . . . . . . . 7.2.5 Triple Wall Sound Transmission Loss . . . . . . . . . . . . . . . . 7.2.6 Sound-Absorptive Linings . . . . . . . . . . . . . . . . . . . . . . 7.2.7 Common Building Materials . . . . . . . . . . . . . . . . . . . . . 7.3 Composite Transmission Loss . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Enclosures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Enclosure Leakages (Large Enclosures) . . . . . . . . . . . . . . . 7.4.2 Enclosure Access and Ventilation . . . . . . . . . . . . . . . . . . 7.4.3 Enclosure Vibration Isolation . . . . . . . . . . . . . . . . . . . . 7.5 Barriers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Diffraction at the Edge of a Thin Sheet . . . . . . . . . . . . . . . 7.5.2 Outdoor Barriers . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.2.1 Thick Barriers . . . . . . . . . . . . . . . . . . . . . . . . 7.5.2.2 Shielding by Terrain . . . . . . . . . . . . . . . . . . . . 7.5.2.3 ISO 9613-2 Approach to Barrier Insertion Loss Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.3 Indoor Barriers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Muffling Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Measures of Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Design for a Required Performance . . . . . . . . . . . . . . . . . . . . . . 8.4 Diffusers as Muffling Devices . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Classification of Muffling Devices . . . . . . . . . . . . . . . . . . . . . . . 8.6 Acoustic Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Impedances of Reactive Muffler Components . . . . . . . . . . . . . . . . 8.7.1 Impedance of an Orifice or Short, Narrow Tube . . . . . . . . . . 8.7.1.1 End Correction . . . . . . . . . . . . . . . . . . . . . . . 8.7.1.2 Acoustic Resistance . . . . . . . . . . . . . . . . . . . . . 8.7.2 Impedance of a Volume . . . . . . . . . . . . . . . . . . . . . . . . 8.8 Reactive Mufflers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8.1 Acoustical Analogues of Kirchhoff’s Laws . . . . . . . . . . . . . . 8.8.2 Side Branch Resonator . . . . . . . . . . . . . . . . . . . . . . . . 8.8.2.1 End Corrections . . . . . . . . . . . . . . . . . . . . . . . 8.8.2.2 Quality Factor . . . . . . . . . . . . . . . . . . . . . . . . 8.8.2.3 Power Dissipation . . . . . . . . . . . . . . . . . . . . . . 8.8.2.4 Insertion Loss Due to a Side Branch . . . . . . . . . . . 8.8.2.5 Transmission Loss Due to a Side Branch . . . . . . . . .

263 269 273 274 279 279 279 279 283 286 295 298 298 301 301 302 302 303 309 314 315 317 317 318 322 334 334 335 341 343 343 344 345 346 347 347 348 348 349 351 353 360 360 361 363 364 365 365 367

xii

Contents

Expansion Chamber . . . . . . . . . . . . . . . . . . . . . 8.8.3.1 Insertion Loss . . . . . . . . . . . . . . . . . . . 8.8.3.2 Transmission Loss . . . . . . . . . . . . . . . . . 8.8.4 Lowpass Filter . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Dissipative Mufflers . . . . . . . . . . . . . . . . . . . . . . . . . 8.9.1 Liner Specification . . . . . . . . . . . . . . . . . . . . . 8.9.2 Lined Duct Design . . . . . . . . . . . . . . . . . . . . . 8.9.2.1 Temperature Effects . . . . . . . . . . . . . . . 8.9.2.2 Higher Order Mode Propagation . . . . . . . . 8.9.3 Inlet Attenuation . . . . . . . . . . . . . . . . . . . . . . 8.9.4 Cross-Sectional Discontinuities . . . . . . . . . . . . . . . 8.9.5 Splitter Mufflers . . . . . . . . . . . . . . . . . . . . . . . 8.10 Insertion Loss of Duct Bends or Elbows . . . . . . . . . . . . . . 8.11 Insertion Loss of Unlined Ducts . . . . . . . . . . . . . . . . . . . 8.12 Effect of Duct End Reflections . . . . . . . . . . . . . . . . . . . 8.13 Pressure Loss Calculations for Muffling Devices . . . . . . . . . . 8.13.1 Pressure Losses Due to Friction . . . . . . . . . . . . . . 8.13.2 Dynamic Pressure Losses . . . . . . . . . . . . . . . . . . 8.13.3 Splitter Muffler Pressure Loss . . . . . . . . . . . . . . . 8.13.4 Circular Muffler Pressure Loss . . . . . . . . . . . . . . . 8.13.5 Staggered Splitter Pressure Loss . . . . . . . . . . . . . . 8.14 Flow-Generated Noise . . . . . . . . . . . . . . . . . . . . . . . . 8.14.1 Straight, Unlined Air Duct Noise Generation . . . . . . . 8.14.2 Mitred Bend Noise Generation . . . . . . . . . . . . . . . 8.14.3 Splitter Muffler Self-Noise Generation . . . . . . . . . . . 8.14.4 Exhaust Stack Pin Noise . . . . . . . . . . . . . . . . . . 8.14.5 Self-Noise Generation of Air Conditioning System Elements 8.15 Duct Break-Out Noise . . . . . . . . . . . . . . . . . . . . . . . . 8.15.1 Break-Out Sound Transmission . . . . . . . . . . . . . . 8.15.2 Break-In Sound Transmission . . . . . . . . . . . . . . . 8.16 Lined Plenum Attenuator . . . . . . . . . . . . . . . . . . . . . . 8.16.1 Wells’ Method . . . . . . . . . . . . . . . . . . . . . . . . 8.16.2 ASHRAE (2015) Method . . . . . . . . . . . . . . . . . . 8.17 Directivity of Exhaust Ducts . . . . . . . . . . . . . . . . . . . . 8.18 Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . A Properties of Materials . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8.3

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375 375 377 386 395 396 397 401 402 403 403 404 407 407 408 413 413 414 414 418 419 419 419 420 421 423 424 424 424 426 426 427 428 431 435 439 443 457

Preface to the Second Edition This is a textbook intended for undergraduate or graduate students wanting to learn how to solve problems in industrial noise control. Occupational Health and Safety engineers as well as consultants in noise control should also find the book useful. The book covers relevant material by first introducing the physical principles and then by illustrating these principles with many examples that are worked out in detail. There are very few derivations of equations, but reference is made to texts where they are derived. Complex analyses are not included and neither are some of the advanced measurement and control techniques (such as active control) that are discussed in other texts by the first author. Additional problems are provided at the end of each chapter and corresponding solutions are available on causalsystems.com, where we will also post any errata that we or our readers find. The book is an excellent learning tool for those who wish to apply their noise-control knowledge to solving real problems. This second edition has been thoroughly revised and updated with a considerable amount of new material, many new example problems and many more references to procedures in acoustic standards. Colin H. Hansen Kristy L. Hansen December, 2020

xiii

1 Fundamentals

LEARNING OBJECTIVES In this chapter, the reader is introduced to: • • • • • • • • •

1.1

fundamentals and basic terminology of noise control; noise-control strategies for new and existing facilities; the speed of sound and the wave equation; plane and spherical waves; energy density; sound intensity and sound power; decibels and sound pressure level; frequency analysis and sound spectra; adding and subtracting sound pressure levels and combining level reductions; and three kinds of impedance.

Introduction

The World Health Organisation (Goelzer et al., 2000) has recognised noise as the most significant health hazard to the working population in terms of the number of people affected. In many industries there is much that can be done to alleviate harmful noise. Perhaps the most insidious aspect of noise-induced hearing loss is that in most cases, damage accumulates over time and is only recognised as a problem when it is too late to do anything about it. Noise can also affect our daily living away from the work place. This type of problem is called environmental noise pollution and it is unhealthy for us in terms of our psychological well-being, which in turn can affect our physical health. In this context, noise is defined as sound that is unwanted by one or more individuals, even though it may be wanted by someone else. When considering noise control, it is well-known that the most cost-effective solution to a problem is often to control the noise-generating mechanism right at its source. This often has the added benefit of making the process more efficient in addition to being less noisy. However, it is often (although not always) only manufacturers of equipment who can alter the noise-generating processes in their equipment, which leaves the engineer who is responsible for an existing item of noisy machinery to find some other way of reducing the noise that reaches places or people where it is either a hazard or unwanted. Often this treatment takes the form of enclosures, barriers, mufflers or vibration isolation and is referred to as “add-on” noise-control technology. Poorly designed “add-on” measures can prove cumbersome in use and are likely to be sabotaged by employees who see them as adversely affecting their efficiency and productivity. 1

2

Noise Control: From Concept to Application

The purpose of this text is to address the causes and methods of control of both occupational and environmental noise with many example problems to illustrate the principles. Much of the background to the discussion presented here is provided in the more comprehensive text by Bies et al. (2018), on which this one is based. The example problems in this book are intended to illustrate and further expand the material. It is not intended that readers should be able to complete all parts of each problem without inspecting the solution. The solutions are provided as a different type of learning tool, as an alternative to spelling out all the principles in the absence of any problem to which to apply them. Problems without solutions, which can be used to test the understanding of the reader, are provided at the end of each chapter. Solutions to these additional problems are available to teaching staff by request to either author. The treatment here is restricted to passive noise control as opposed to active noise control where “anti-noise” sources are introduced to “cancel” unwanted noise. Active noise control is very difficult to apply successfully in an industrial environment; development costs are generally high and on-going maintenance is currently an issue. The principles of active noise control are discussed in detail in another text (Hansen, 2001). The discussion in this chapter begins with an outline of noise-control strategies and a discussion of the fundamental principles of acoustics, followed by an explanation of how noise is quantified, with the intention of providing a basis for understanding the applications discussed in the remainder of the book. In Chapter 2, noise criteria are discussed and instrumentation for measuring noise is described. Chapter 3 describes the sound fields produced by different sources and various means for measuring the sound they produce. Chapter 4 is concerned with sound propagation outdoors and in rooms. Chapter 5 is a detailed treatment of sound absorbing materials and in particular, their properties and measurement and Chapter 7 is a comprehensive treatment of sound in rooms. Chapter 7 includes methods for calculating the sound transmission loss of partitions and the design of enclosures, while Chapter 8 is concerned with the design of dissipative and reactive mufflers. Properties of materials relevant to calculations described in the textbook are provided in Appendix A.

1.2

Noise-Control Strategies

Possible strategies for noise control are always more numerous for new facilities and products than for existing facilities and products. Consequently, it is always more cost-effective to implement noise control at the design stage than to wait for complaints about a finished facility or product. In existing facilities, controls may be required in response to specific complaints from within the work place or from the surrounding community, and excessive noise levels may be quantified by suitable measurements. In proposed new facilities, possible complaints must be anticipated, and expected excessive noise levels must be estimated by some procedure. Often it is not possible to eliminate unwanted noise entirely and more often to do so is very expensive; thus minimum acceptable sound pressure (or noise) levels of noise must be formulated, and these sound pressure levels constitute the criteria for acceptability. Criteria for acceptability are generally established with reference to appropriate regulations for the work place and community. In addition, for community noise it is advisable that at worst, any facility should not increase background (or ambient) sound pressure levels in a community by more than 5 dBA over existing levels without the facility, irrespective of what local regulations may allow. Note that this 5 dBA increase applies to broadband noise and that clearly distinguishable tones (single frequencies) are less acceptable. In addition, noise that is amplitude modulated (varying in level regularly with a constant time interval between maximum levels) or heavily weighted towards the low-frequency part of the spectrum is more annoying than its time-averaged A-weighted sound pressure level would suggest. The A-weighting that results in an A-weighted or dBA sound pressure level is discussed in more detail in Section 2.2.4.

Fundamentals

3

When dealing with community complaints (predicted or observed) it is wise to be conservative; that is, to aim for adequate control for the worst case, noting that community noise levels (or more accurately, sound pressure levels) may vary greatly (±10 dB) about the mean as a result of atmospheric conditions (wind and temperature gradients and atmospheric turbulence). It is worth careful note that complainants tend to be more conscious of a noise after making a complaint and thus subconsciously tend to listen for it. Thus, even after considerable noise reduction may have been achieved and regulations satisfied, complaints may continue. Clearly, it is better to avoid complaints in the first place and thus yet another argument supporting the assertion of cost-effectiveness in the design stage is provided. For both existing and proposed new facilities and products, an important part of the process will be to identify noise sources and to rank order them in terms of contributions to excessive noise. When the requirements for noise control have been quantified, and sources identified and ranked, it is possible to consider various options for control and finally to determine the costeffectiveness of the various options. Community sound pressure level predictions and calculations of the effects of noise control are generally carried out in octave frequency bands. Current models for prediction are not sufficiently accurate to allow finer frequency resolution and less fine frequency resolution does not allow proper account of frequency-dependent effects. Generally, octave band analysis provides a satisfactory compromise between too much and too little detail. Where greater spectrum detail is required, 1/3-octave band analysis is often sufficient. If complaints arise from the work place, then regulations should be satisfied, but to minimise hearing damage compensation claims, the goal of any noise-control programme should be to reach a sound pressure level of no more than 85 dBA. Criteria for other situations in the work place are discussed in Chapter 3. Measurements and calculations are generally carried out in standardised octave or 1/3-octave bands, but particular care must be given to the identification of any tones that may be present, as these must be treated separately. Any noise problem may be described in terms of a sound source, a transmission path and a receiver, and noise control may take the form of altering any one or all of these elements. When faced with an industrial noise problem, reducing its hazard can be achieved in a number of ways and these are listed below in order of effectiveness. 1. Eliminate the hazard, which means physically removing it (modification of the source). 2. Substitute the noisy process with a quieter one (modification of the source). 3. Reduce the hazard by good design (modification of the source). 4. Isolate personnel from the hazard via physical barriers or mufflers (modification of the transmission path). 5. Change the way people work by rotating them out of noisy areas or by introducing quieter ways of doing things (modification of the receiver). 6. Provide earplugs and earmuffs (modification of the receiver). When considered in terms of cost-effectiveness and acceptability, experience puts modification of the source well ahead of either modification of the transmission path or the receiver. On the other hand, in existing facilities the last two may be the only feasible options.

1.2.1

Sound Source Modification

Modification of the energy source to reduce the noise generated often provides the best means of noise control. For example, where impacts are involved, as in punch presses, any reduction of the peak impact force (even at the expense of the force acting over a longer time period) will dramatically reduce the noise generated. Generally, when a choice between various mechanical processes is possible to accomplish a given task, the best choice, from the point of view of minimum noise, will be the process that minimises the time rate of change of force or jerk (time

4

Noise Control: From Concept to Application

rate of change of acceleration). Alternatively, when the process is aerodynamic a similar principle applies; that is, the process that minimises pressure gradients will produce minimum noise. In general, whether a process is mechanical or fluid mechanical, minimum rate of change of force is associated with minimum noise. Mechanical shock between solids should be minimised; for example, impact noise may be generated by parts falling into metal bins. This problem could be reduced by minimising the height that parts fall and treating the container that they fall in to. Guidelines for the design of low-noise machinery and equipment are provided in ISO/TR 11688-1 (1995) and ISO/TR 11688-2 (1998).

1.2.2

Control of the Transmission Path

In considering control of the noise path from the source to the receiver some or all of the following treatments need to be considered: barriers (single walls), partial enclosures or full equipment enclosures, local enclosures for noisy components on a machine, reactive or dissipative mufflers (the former for low frequency noise or small exhausts, the latter for high frequencies or large diameter exhaust outlets), lined ducts or lined plenum chambers for air-handling systems, vibration isolation of machines from noise-radiating structures, vibration absorbers and dampers, active noise control and the addition of sound-absorbing material to reverberant spaces to reduce reflected noise fields.

1.2.3

Modification of the Receiver

In some cases, when all else fails, it may be necessary to apply noise control to the receiver of the excessive noise. This type of control may involve use of ear-muffs, ear-plugs or other forms of hearing protection; the enclosure of personnel if this is practical; moving personnel further from the noise sources; rotating personnel to reduce noise exposure time; and education and emphasis on public relations for both in-plant and community noise problems.

1.2.4

Existing Facilities

In existing facilities or products, quantification of the noise problem requires identification of the noise source or sources, determination of the transmission paths from the sources to the receivers, rank ordering of the various contributors to the problem and finally determination of acceptable solutions. To begin, sound pressure levels must be determined at potentially sensitive locations or at locations from which the complaints arise. For community noise, these measurements may not be straightforward, as such noise may be strongly affected by variable weather conditions and measurements over a representative time period may be required. This is usually done using remote data logging equipment in addition to periodic manual measurements to check the validity of the logged data. Guidelines for taking community noise measurements are provided in ASTM E1780-12 (2012) and ASTM E1503-14 (2014). The next step is to apply acceptable noise level criteria to each location and thus determine the required noise reductions, generally as a function of octave or 1/3-octave frequency bands. Noise level criteria are usually set by regulations and appropriate standards. Next, the transmission paths by which the noise reaches the place of complaint are determined. For some cases, this step is often obvious. However, cases may occasionally arise when this step may present some difficulty, but it may be very important in helping to identify the source of a complaint. For example, the noise may propagate as structure-borne vibration, which then radiates as noise when it reaches an efficient acoustic radiator such as a wall or large panel.

Fundamentals

5

Having identified the possible transmission paths, the next step is to identify (understand) the noise generation mechanism or mechanisms, as noise control at the source always gives the best solution. In existing facilities and products, altering the noise-generating mechanism may or may not be too expensive, but should always be considered as a means for possible control. Having identified the noise sources and determined their radiated sound power levels, the next task is to determine the relative contribution of each noise source to the level at each location where the measured sound pressure levels are considered to be excessive. For a facility involving just a few noise sources this is a relatively straightforward task. However, for a facility involving tens or hundreds of noise sources, the task of rank ordering can be intimidating, especially when the locations of complaint are in the surrounding community. In the latter case, the effect of the ground terrain and surface, air absorption and the influence of atmospheric conditions must also be taken into account, as well as the decrease in sound pressure level with distance due to the “spreading out” of the sound waves (see Chapter 5). Commercial computer software is available to assist with these calculations if necessary. Measured sound pressure levels can be compared with predicted levels to validate the calculations. Once the analytical model is validated, it is then a simple matter to investigate various options for control and their cost-effectiveness. In summary, a noise-control programme for an existing facility includes (see ISO 11690-1 (1996); Bakker et al. (2009) for more details): • undertaking an assessment of the current environment where there appears to be a problem, including the preparation of both worst case and average sound pressure level contours where required; • establishment of the noise-control objectives or criteria to be met; • identification of noise transmission paths and generation mechanisms; • rank ordering noise sources contributing to any excessive levels; • formulating a noise-control program and implementation schedule; • carrying out the program; and • verifying the achievement of the objectives of the program.

1.2.5

Facilities in the Design Stage

In new facilities the problems are the same as for existing facilities and products; they are identification of the source or sources, determination of the transmission paths of the noise from the sources to the receivers, rank ordering of the various contributors to the problem and finally determination of acceptable solutions. Most importantly, at the design stage the options for noise control are generally many and may include rejection of the proposed design. Consideration of the possible need for noise control in the design stage has the very great advantage that an opportunity is provided to choose a process or processes that may avoid or greatly reduce the need for noise control. Experience suggests that processes chosen because they make less noise, often have the additional advantage of being a more efficient process than that which was originally considered. The first step for new facilities is to determine the noise criteria for sensitive locations, which may typically include areas of the surrounding residential community that will be closest to the planned facility, locations along the boundary of the land owned by the industrial company responsible for the new facility, and within the facility at locations of operators of noisy machinery. Again, care must be taken to be conservative where surrounding communities are concerned so that initial complaints are avoided. In consideration of possible community noise problems following establishment of acceptable noise criteria at sensitive locations, the next step may be to develop a computer model or to

6

Noise Control: From Concept to Application

use an existing commercial software package to estimate expected sound pressure levels (in octave frequency bands) at the sensitive locations, based on machinery sound power level and directivity information (the latter may not always be available), and outdoor sound propagation prediction procedures as described in Chapter 6 and ISO 9613-2 (1996). Previous experience or the local weather bureau can provide expected ranges in atmospheric weather conditions (wind and temperature gradients and turbulence levels) so that a likely range and worst case sound pressure levels can be predicted for each community location. When directivity information is not available, it is generally assumed that the source radiates uniformly in all directions. If the estimated sound pressure levels at any sensitive location exceed the established criteria, then the equipment contributing most to the excess levels should be targeted for noise control, which could take the form of: • specifying lower equipment sound power levels, or sound pressure levels at the operator position, to the equipment manufacturer; • including noise-control fixtures (mufflers, barriers, enclosures, or factory walls with a higher sound transmission loss) in the factory design; or • rearrangement and careful planning of buildings and equipment within them. Sufficient noise control should be specified to leave no doubt that the noise criteria will be met at every sensitive location. Saving money at this stage is not cost-effective. If predicting equipment sound power levels with sufficient accuracy proves difficult, it may be helpful to make measurements on a similar existing facility or product. More detail on noise-control strategies and noise prediction for facilities at the design stage can be found in ISO 11690-1 (1996), ISO 11690-2 (1996) and ISO 11690-3 (1997). Example 1.1 You are responsible for a large factory containing many items of noisy equipment. You have been informed that some of your employees are suffering from severe hearing loss and you have also received threats of legal action from members of the surrounding community because of excessive noise made by your facility. List the steps (in order) that you would take to quantify and rectify the problem. Solution 1.1 • undertake an assessment of the current environment where there appears to be a problem, including the preparation of sound pressure 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 programme and implementation schedule; • carry out the programme; and • verify the achievement of the objectives of the programme.

1.3

Acoustical Standards and Software

Acoustical standards are available that describe standardised methods for undertaking various calculations in acoustics and noise control. International standards are published by the International Standards Organisation (ISO) and the International Electrotechnical Commission (IEC). Relevant American standards are published by the American National Standards Institute (ANSI) and the American Society of Mechanical Engineers (ASME). Relevant standards are referenced throughout this text book but readers should also search the relevant standards

7

Fundamentals

organisation database for a possible standard to describe any measurement or calculation that is to be undertaken. Where guidelines in American standards (ANSI and ASTM) differ from those in international standards (ISO and IEC), most countries (except USA and Canada) base local standards on international guidelines. Software that is able to do the calculations outlined in this book and according to the various standards is also available for purchase. Examples of packages for outdoor sound propagation and prediction of sound pressure levels at various locations due to various sound sources include SoundPlan and CadnaA. An example of a package to perform room acoustics calculations is Odeon and a package that can perform all of the calculations in this book (and more) is ENC (causalsystems.com). These packages are widely used by consultants in acoustics and noise control, but the packages that provide sound pressure level contours are usually quite expensive. However, the purpose of this book is for readers to gain an understanding of the fundamental principles that underpin these software packages, such that the input data required by the various programs will be better understood in terms of which data needs to be accurate and which can be estimated. In addition, the knowledge gained from this book and the example problems will be of assistance in understanding and interpreting the results obtained from the various software packages.

1.4 1.4.1

Acoustic Field Variables Variables

Sound is the sensation produced at the ear by very small pressure fluctuations in the air. The fluctuations in the surrounding air constitute a sound field. A sound field is described as a perturbation (or fluctuation) of steady-state variables, which describe a medium through which sound is transmitted. For a fluid, expressions for the pressure, Ptot , velocity, Utot , temperature, Ttot and density, ρtot , may be written in terms of the steady-state (mean values), shown as Ps , U , T and ρ and the variable (perturbation) values, p, u, τ and σ, as follows: Pressure : Velocity : Temperature : Density :

Ptot = Ps + p(r, t) Utot = U + u(r, t) Ttot = T + τ (r, t) ρtot = ρ + σ(r, t)

(Pa) (m/s) (◦ C) 3 (kg/m )

where r is the position vector, t is time and the variables in bold font are vector quantities. Pressure, temperature and density are familiar scalar quantities that do not require discussion. However, an explanation is required for the particle velocity u(r, t) and the vector equation above that involves it. The notion of particle velocity is based on the assumption of a continuous rather than a molecular medium. The term, “particle”, refers to a small part of the assumed continuous medium and not to the molecules of the medium. Thus, even though the actual motion associated with the passage of an acoustic disturbance through the conducting medium, such as air at high frequencies, may be of the order of the molecular motion, the particle velocity describes a macroscopic average motion superimposed upon the inherent Brownian motion of the medium. In the case of a convected medium moving with a mean velocity, U , which itself may be a function of the position vector, r, and time, t, the perturbating particle velocity, u(r, t), associated with the passage of an acoustic disturbance may be thought of as adding to the mean velocity to give the total velocity. Any variable could be chosen for the description of a sound field, but it is easiest to measure pressure in a fluid, so this is the variable usually used. It has the additional advantage of providing a scalar description of the sound field from which all other variables may be derived. For example,

8

Noise Control: From Concept to Application

the particle velocity is important for the determination of sound intensity, but it is a vector quantity and would require three measurements as opposed to one for pressure. Additionally, the particle velocity is often too small to be measured directly by any practical means, although it can be inferred from acoustic pressure measurements.

1.4.2

Magnitudes

The minimum acoustic pressure audible to the young human ear judged to be in good health, and unsullied by too much exposure to excessively loud music, is approximately 20 ×10−6 Pa, or 2 ×10−10 atmospheres (since one atmosphere equals 101.3 ×103 Pa). The minimum audible level occurs between 3000 and 4000 Hz and is a physical limit; lower sound pressure levels would be swamped by thermal noise due to molecular motion in air. For the normal human ear, pain is experienced at sound pressures of the order of 60 Pa or 6 ×10−4 atmospheres. Evidently, acoustic pressures ordinarily are quite small fluctuations about the mean atmospheric pressure.

1.4.3

The Speed of Sound

Sound is conducted to the ear by longitudinal waves travelling through the surrounding medium. Consequently, it is customary to refer to longitudinal waves as sound waves for which the associated particle motion in the transmitting medium is parallel to the direction of wave propagation. In general, the medium surrounding the ear will be air and sometimes water but sound energy may be transported by all gases, fluids and solids. Consequently, the speed of travel of sound waves has been generalised to mean the speed of travel of longitudinal waves propagating through any medium. For gases, the speed of sound is: c=

p

γPs /ρ =

p

γRT /M ≈ 331 + 0.6(T − 273.15)

(m/s)

(1.1)

where γ is the ratio of specific heats (1.40 for air), T is the temperature in Kelvin (K), R is the universal gas constant which has the value 8.314 J mol−1 K−1 and M is the molecular weight, which for air is 0.029 kg/mol. For sound propagating in free space, Ps is the atmospheric pressure. Equation (1.1) is derived in many standard texts: for example, Morse (1948), Pierce (1981), and Kinsler et al. (1999). For calculations in this text, unless otherwise stated, a temperature of 20◦ C for air will be assumed, resulting in a speed of sound of 343 m/s and an air density of 1.206 kg/m3 at sea level, thus giving ρc = 413.7. Some representative speeds of sound are given in Appendix A. Example 1.2 (a) Verify from fundamental principles that the speed of sound in Helium is 1/0.34 times that in air. Helium has a molecular weight of 4 g/mole and it is a monatomic gas for which the average number of excited degrees of freedom is 3. Thus the ratio of specific heats γ = (3 + 2)/3. Air has a molecular weight of 29 g/mole. (b) Explain why taking a mouthful of helium from a balloon makes you speak with a high pitched voice. Solution 1.2 (a) Speed of sound is given by Equation (1.1). The only variables that are different for the two gases are γ and M . Thus:

9

Fundamentals cair = cHe



MHe γair × γHe Mair

1/2

=



4 5/3

1/2 

1.4 29

1/2

= 0.34.

(b) The wavelength of sound emitted from one’s mouth is a function of the vocal cord properties which remain unchanged by the presence of helium. As f λ = c, and λ is fixed (as the vocal cord properties are fixed) and c is faster in helium, the sound emanating from your mouth will be higher in pitch. Example 1.3 Given the first-order approximation that cwet = (1 + 0.16h)cdry , calculate the speed, c, of sound in air at 30◦ C with a relative humidity of 95%. The quantity, h, is the fraction of total molecules which are H2 O. Vapour pressure of water at 30◦ C is 4240 Pa and h = (vapour pressure/total pressure) × (%relative humidity/100). Assume the total pressure is atmospheric (101.4 kPa). Solution 1.3 Using the relation for h given in the question and knowing that atmospheric pressure is 101.4 kPa, we can write, h =

95 4240 = 0.0397. × 101.4 × 103 100

The speed of sound in dry air at 30◦ C is given by Equation (1.1) as: cdry =

γRT /M =

p

p

1.4 × 8.314 × 303.2/0.029 = 348.8 m/s.

Using the equation given in the question, the speed of sound in wet air is then: cwet = 348.8(1 + 0.0397 × 0.16) = 351.0 m/s. Example 1.4 A reciprocating compressor installation is suffering piping joint failures due to excessive fluid pulsations at the compressor discharge. Prior to designing pulsation dampeners (see Chapter 8), it is necessary to calculate the speed of sound in the compressed gas and this must include the gas flow speed. Assume a gas discharge pressure of 8 MPa, a temperature of 120◦ C, a pipe diameter of 0.1 m, ratio of specific heats of the gas of 1.4 and a molecular weight of 29 grams/mole. The volume of gas flowing in the system is 250,000 m3 per day, measured at 15◦ C and standard atmospheric pressure. This volume flow rate would be different inside the pipe where the pressure is much greater than atmospheric pressure. Calculate: (a) gas mass flow rate (kg/s) using the universal gas law, Ps V = (m/M )RT , where Ps is the gas absolute pressure, V is the gas volume, m is the mass of gas, M is the molecular weight of the gas, T is its temperature in Kelvin (K) and R is the universal gas constant (Joules mol−1 K−1 ); (b) density of gas in the discharge pipe; (c) gas flow speed in the discharge pipe (m/s); and (d) speed of sound in the gas relative to the pipe and in the direction of gas flow. Solution 1.4 Gas volume flow rate = 250000 m3 per day at STP. Gas Law also applies to a moving fluid, so Equation (1.1) gives: Ps V˙ =

m ˙ RT . M

10

Noise Control: From Concept to Application

Ps and T are the absolute pressure (Pa) and temperature (K), respectively, and V˙ = 2.894 m3 /s.

250000 = 24 × 3600

(a) The mass flow rate (which is the same no matter where it is measured according to the law of conservation of mass) is given by: m ˙ =

Ps V˙ M 101, 400 × 0.029 × 2.8935 = = 3.550 kg/s. RT 8.314 × 288.2

(b) The density of gas in the pipe is equal to the mass flow rate divided by the volume flow rate. Thus: Ps M 8 × 106 × 0.029 m ˙ 3 = = = 71.0 kg/m . RT 8.314 × 393.2 V˙

ρ=

(c) The gas flow speed in the discharge pipe is equal to the volume flow rate divided by the pipe cross-sectional area. Thus: U=

4 m 4 3.551 ˙ 4 ˙ V = = × = 6.37 m/s. πd2 πd2 ρ π × 0.01 71.0

(d) The speed of sound (relative to the fluid) is given by Equation (1.1) as: c=



γPs ρ

1/2

=

p 1.4 × 8 × 106 /71.0 = 397 m/s.

Speed of sound relative to the pipe is thus: 397.2 + 6.4 = 404 m/s. Example 1.5 What is the speed of sound in a gasoline engine cylinder just after combustion when the pressure is 200 times atmospheric pressure and the temperature is 1000◦ C? The ratio of specific heats of the gas mixture is 1.35 and the gas density is 1.4kg/m3 at 0◦ C and atmospheric pressure. Solution 1.5 The speed of sound is given by Equation (1.1). Thus, c = γRT /M and for any gas, (R/M ) is fixed. We can find (R/M ) by using the properties at 0◦ C and the expression from Equation (1.1):

p

Ps V =

m RT M

or

R Ps V = . M mT

R 101400 As (m/V ) = 1.4 kg/m , = and thus c = M 1.4 × 273 3



1.35 × 101400 × 1273 1.4 × 273

1/2

= 675 m/s.

Example 1.6 Show, using the Universal Gas Law, that the value of ρc of air is equal to 400 at sea level and at a temperature of 40◦ C. Solution 1.6 Using Equation (1.1), it can be shown that:

p

γPs γRT /M γPs M c γPs 1.4 × 101400 ρc = = = p = p = 400.4. γRT γRT /M γRT /M 1.4 × 8.314 × 313.2/0.029

11

Fundamentals

1.4.4

Acoustic Potential Function and the Wave Equation

The hydrodynamic equations, from which the equations governing acoustic phenomena derive, are, in general, quite complex, and well beyond solution in closed form. Fortunately, acoustic phenomena are associated with very small perturbations. Thus, it is possible to greatly simplify the governing equations to obtain the relatively simple equations of acoustics. The equations are said to have been linearised, and the phenomena described are referred to as linear acoustics, which is adequate for the cases discussed in this text and for sound pressure levels less than 130 dB re 20 µPa. The potential function, φ, provides a means for determining both the acoustic pressure and particle velocity by simple differentiation, as will be shown, and is thus a very useful function. The potential function, φ, is defined such that the particle velocity is equal to its negative gradient: u = −∇φ

(1.2)

The acoustic potential function may also be used to determine the acoustic pressure, which for small, negligible or absent convection velocity, U , is given by the following expression: p=ρ

∂φ ∂t

(1.3)

The acoustic potential function satisfies the well-known linearised wave equation as follows: ∇2 φ = (1/c2 )∂ 2 φ/∂t2

(1.4)

The same equation applies if the acoustic pressure variable, p, is used to replace φ in Equation (1.4). However, the wave equation for the acoustic particle velocity is more complicated. Derivations of the wave equation in terms of acoustical particle velocity and in the presence of a mean flow are given in Chapter 1 of Hansen (2018). Other useful books containing derivations of the wave equation are Fahy and Thompson (2015) and Fahy (2001).

1.4.5

Complex Number Formulations

In the remainder of the book, sound waves will often be represented using complex numbers with real and imaginary parts, so that, for example, the time-varying sound pressure, p, at a particular location of an acoustic wave of frequency, ω (Hz), may be represented as, p = (a + jb)e jωt√= pˆe jωt = |ˆ p|e j(ωt+β) , where pˆ is the complex pressure amplitude of the sound wave and |ˆ p| = a2 + b2 . The angle, β (radians), is included in pˆ and is the phase of p relative to some reference which could be its particle velocity, the sound pressure at another location or the sound pressure of another sound wave of the same frequency. The purpose in using complex numbers to represent a relative phase is to enable representation of the phase and amplitude of a sound wave in a form that is easily amenable to mathematical manipulation. In the remainder of this book, for any sinusoidally time-varying complex quantity, x, its complex amplitude will be denoted, x ˆ, and its scalar amplitude will be denoted |ˆ x|.

1.5

Plane, Cylindrical and Spherical Waves

In general, sound wave propagation is quite complicated and not amenable to simple analysis. However, sound wave propagation can often be described in terms of the propagation properties of plane, cylindrical and spherical waves.

12

1.5.1

Noise Control: From Concept to Application

Plane Wave Propagation

For the case of plane wave propagation, only one spatial dimension, x, the direction of propagation, is required to describe the acoustic field. An example of plane wave propagation is sound propagating along the centre line of a tube with rigid walls. In this case, Equation (1.4) reduces to: ∂ 2 φ/∂x2 = (1/c2 )∂ 2 φ/∂t2 (1.5) The same equation applies for acoustic pressure by replacing φ with p in Equation (1.5). A solution for Equation (1.5) which may be verified by direct substitution, is: φ = f(c t ± x)

(1.6)

Using Equations (1.2) and (1.3), the following equations may be obtained: u = ∓f 0 (c t ± x)

(1.7)

p = ρcf 0 (c t ± x)

(1.8)

where the prime denotes differentiation with respect to the argument of f. Division of Equation (1.8) by Equation (1.7) gives: p/u = ±ρc (1.9) which is a very important result – the characteristic impedance, ρc, of a plane wave. The characteristic impedance is one of three kinds of impedance used in acoustics. It provides a very useful relationship between acoustic pressure and particle velocity in a plane wave. It also has the property that a duct terminated in its characteristic impedance will respond as an infinite duct, as no wave will be reflected at its termination. In Equation (1.9), the positive sign is taken for waves travelling in the positive x-direction, while the negative sign is taken for waves travelling in the negative x-direction. Fourier analysis enables the representation of any function, f(c t ± x), as a sum or integral of harmonic functions. Thus, it will be useful for consideration of the wave equation to investigate the special properties of harmonic solutions. Consideration will begin with the following harmonic solution for the acoustic potential function: φ = A cos(k(c t ± x) + β)

(1.10)

where k is a constant, which will be investigated, and β is an arbitrary constant representing an arbitrary relative phase, which is only relevant when there is more than one wave. As β is arbitrary in Equation (1.10), for fixed time, t, β may be chosen so that: kc t + β = 0

(1.11)

In this case, Equation (1.6) reduces to the following representation of the spatial distribution: φ = A cos(kx) = A cos(2πx/λ)

(1.12)

From Equation (1.12) it may be concluded that the unit of length, λ, defined as the wavelength of the propagating wave and the constant, k, defined as the wavenumber are related as: 2π/λ = k

(1.13)

An example of harmonic (single frequency) plane wave propagation in a tube is illustrated in Figure 1.1. The type of wave generated is longitudinal, as shown in Figure 1.1(a) and the corresponding pressure fluctuations as a function of time are shown in Figure 1.1(b).

13

Fundamentals

(a)

Wavelength Acoustic pressure pmax +

+

(b) patm

FIGURE 1.1 Representation of a sound wave: (a) compressions and rarefactions of a sound wave in space at a fixed instance in time; (b) graphical representation of sound pressure variation.

The distribution in space has been considered and now the distribution in time for a fixed point in space will be considered. The arbitrary phase constant, β, of Equation (1.10) will be chosen so that, for fixed position, x: β ± kx = 0 (1.14) Equation (1.10) then reduces to the following representation for the temporal distribution: φ = A cos(kc t) = A cos

2π t Tp

(1.15)

The period, Tp , of the propagating wave is given by: 2π/kc = Tp

(s)

(1.16)

Its reciprocal is the more familiar frequency, f . Since the angular frequency, ω = 2πf (radians/s), is quite often used as well, the following relation should be noted: 2π/Tp = 2πf = ω

(rad/s)

(1.17)

and from Equations (1.16) and (1.17): k = ω/c

(rad/m)

(1.18)

(m/s)

(1.19)

and from Equations (1.13), (1.17), and (1.18): fλ = c

The wavenumber, k, may be thought of as a spatial frequency, where k is the analogue of frequency, f , and wavelength, λ, is the analogue of the period, Tp . The relationship between wavelength and frequency, for sound propagating in air, is illustrated in Figure 1.2. The wavelength of generally audible sound varies by a factor of about one thousand. The shortest audible wavelength is 17 mm (corresponding to 20000 Hz) and the longest is 17 m (corresponding to 20 Hz), although humans can detect sound via their vestibular system (which

14

Noise Control: From Concept to Application Wavelength (m) 20 20

10

5 50

2 100

200

1

0.5 500

0.2

1000 2000

0.1

0.05

0.02

5000 10000 20000

Audible frequency (Hz) FIGURE 1.2 Wavelength in air versus frequency under normal conditions.

the ear is part of) at much lower frequencies if it is sufficiently loud. Letting A = B/ρω in Equation (1.10) and use of Equation (1.18) and either (1.2) or (1.3) gives the following useful expressions for the acoustic pressure and the particle velocity, respectively, for a plane wave: p = Bsin(ωt ∓ kx + β) u=±

B sin(ωt ∓ kx + β) ρc

(Pa) (m/s)

(1.20) (1.21)

Acoustic velocity potential and acoustic pressure harmonic solutions to the wave equation are complex and may be written as follows in terms of either exponential or trigonometric functions: φ = Ae j(ωt±kx+β) = A cos(ωt ± kx + β) + jA sin(ωt ± kx + β)

(1.22)

p = Be j(ωt±kx+β) = B cos(ωt ± kx + β) + jB sin(ωt ± kx + β)

(1.23)

√

where j = −1 and B = A/(jρω) = Ae−jπ/2 /(ρω), indicating that the pressure phase lags the velocity potential phase by 90◦ . In Equations (1.22) and (1.23), the negative sign represents a wave travelling in the positive x-direction, while the positive sign represents a wave travelling in the negative x-direction. The real parts of Equation (1.22) are just the solutions given by Equation (1.10). The imaginary parts of Equation (1.22) are also solutions, but in quadrature (90◦ out of phase) with the former solutions. It can also be seen that the 90◦ phase difference between the real and imaginary solutions represents a fixed time shift so the two solutions may be considered the same if the start time of one is adjusted by the time equivalent to a 90◦ phase shift. By convention, the complex notation is defined so that what is measured with an instrument corresponds to the real part; the imaginary part is then inferred from the real part. The complex exponential form of the harmonic solution to the wave equation is used as a mathematical convenience, as it greatly simplifies mathematical manipulations, allows waves with different phases to be added together easily and allows graphical representation of the solution as a rotating vector in the complex plane. Setting β = 0 and x = 0, allows Equation (1.22) to be rewritten as: Ae jωt = A exp {jωt} = A(cos ωt + j sin ωt) (1.24) Equation (1.24) represents harmonic motion that may be represented at any time, t, as a rotating vector of constant magnitude, A, and constant angular velocity, ω, as illustrated in Figure 1.3. Referring to the figure, the projection of the rotating vector on the abscissa, x-axis, is given by the real term on the right hand side of Equation (1.24) and the projection of the rotating vector on the ordinate, y-axis, is given by the imaginary term. Use of the complex form of the solution makes integration and differentiation particularly simple. Also, impedances are conveniently described using this notation. For these reasons, the complex notation will be used throughout this book. However, care must be taken in the use of the complex notation when multiplying one function by another. In the calculation of products of actual quantities expressed in complex notation it is important to remember that for these actual quantities, the product implies that only like quantities are multiplied, so the product is not a full complex product. In general, the real parts of the quantities are multiplied, rather than

15

Fundamentals

y(t) X

Im y w

y(t) = X sin wt

X

r 0

p

2p

3p

wt 4p

x

wt 0

Re

-X

p 0

X

x(t)

x(t) = X cos wt

2p

3p

wt 4p

-X

FIGURE 1.3 Harmonic motions represented as a rotating vector.

the imaginary parts. This is important, for example, in the calculation of intensity associated with single frequency sound fields expressed in complex notation. Plane wave propagation in a duct that may be either circular or rectangular in cross section is a common phenomenon. Plane waves propagate at all frequencies below the duct cut-on frequency, which is given by: fco

( c/(2Ly ) for rectangular − section ducts = 0.586c/d for circular − section ducts

(1.25)

where Ly is the largest duct cross-sectional dimension for rectangular ducts and d is the duct diameter for circular-section ducts. Above the duct cut-on frequency, higher-order modes propagate and these are characterised by waves reflecting back and forth from the interior duct walls as they propagate down the duct. Example 1.7 The complex pressure amplitude of the time-varying sinusoidal signal produced by a plane wave travelling in a semi-infinite, hard-walled, lossless tube having e jωt time dependence can be described by pˆ(x) = e−jkx where B is a real constant, p(x, t) = pˆ(x)e jωt and x is the distance from the sound source located at the end of the tube. (a) Derive an expression for the complex acoustic particle velocity. (b) Explain the difference between particle velocity and sound speed.

16

Noise Control: From Concept to Application

(c) Derive an expression for the specific acoustic impedance at any location, x, in the tube. (d) Derive an expression for the specific acoustic impedance corresponding to the tube being terminated rigidly instead of infinitely. Solution 1.7 (a) The acoustic pressure may be written using Equation (1.23) as p(x, t) = B e j(ωt−kx) , where B is an unknown coefficient. Note that the phase angle, β of Equation (1.23) can be set equal to zero as it is arbitrary when only considering one parameter. Using Equations (1.2) and (1.3), the acoustic particle velocity may be written as: u(x, t) =

B j(ωt−kx) k B j(ωt−kx) e = e . 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 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: Z=

p(x, t) B e j(ωt−kx) = ρc = ρc. u(x, t) Be j(ωt−kx)

Z = pT /uT

0

-L

x

FIGURE 1.4 Arrangement for Example 1.7.

(d) To simplify the algebra, set the origin of the coordinate system at the rigid end of the tube as shown in Figure 1.4. 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 0◦ 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 add an additional quantity, kx1 (where x1 is the distance from the origin to the point of reflection) to the exponent representing 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 using the result of part (a) above and Equation (1.23), as: pt = B e j(ωt−kx) + e j(ωt+kx) and ut =





B j(ωt−kx) e − e j(ωt+kx) . ρc

The specific acoustic impedance is then: e−jkx + e jkx Z cos(kx) = −jkx = = j cot(kx). ρc (e − e jkx ) −j sin(kx)




17

Fundamentals

1.5.2

Cylindrical Wave Propagation

A second important case is that of cylindrical wave propagation; an example is the propagation of sound waves from a pipeline at low frequencies such that the pipeline wall is vibrating in phase along the length of the pipeline (which is many wavelengths long) and the pipe radius is very small compared to a wavelength. In this case, sound radiates from the pipe in the form of cylindrical waves which satisfy Equation (1.26). 1 ∂ 2 (φ) ∂ 2 (φ) 1 ∂φ + = ∂r2 r ∂r c2 ∂t2

(1.26)

The solution to Equation (1.26), which is the wave equation in cylindrical coordinates is complicated, involving Bessel and Neumann functions, so will not be discussed further here. More detail is provided in (Skudrzyk, 1971). An important property of cylindrical waves is that at distances greater than a wavelength from the source, the sound pressure level decays at a rate of 3 dB for each doubling of distance from the sound source whereas for spherical waves (discussed in the next section), the decay rate is 6 dB for each doubling of distance from the source. Note that plane waves (such as found in a duct at low frequencies) do not decay as the distance from the source increases.

1.5.3

Spherical Wave Propagation

A third important case is that of spherical wave propagation; an example is the propagation of sound waves from a small source in free space with no boundaries nearby. In this case, the wave Equation (1.15) may be written in spherical coordinates in terms of a radial term only, since no angular dependence is implied. Thus Equation (1.15) becomes (Morse and Ingard, 1968, p. 309): ∂ 2 (rφ) 1 ∂ 2 (rφ) = ∂r2 c2 ∂t2

(1.27)

The difference between, and similarity of, Equations (1.5) and (1.27) should be noted. Evidently, rφ = f(ct ∓ r) is a solution of Equation (1.27) where the source is located at the origin. Thus: φ=

f(ct ∓ r) r

(1.28)

For harmonic waves, the solution given by Equation (1.28) can also be written as: φ=

f(k(ct ± r)) f(ωt ± kr) A = = e j(ωt±kr) r r r

(1.29)

which is the same as the solution for acoustic pressure (but not particle velocity), except that the coefficient, A, would be different (but related to the coefficient, A, in Equation (1.29) by a factor of jρω). This factor is obtained by substituting Equation (1.29) into Equation (1.3), resulting in the following expression for the acoustic pressure for outwardly travelling waves (corresponding to the negative sign in Equation (1.29)). p=

jkρcA j(ωt−kr) jωAρ j(ωt−kr) e = e r r

(1.30)

Substitution of Equation (1.29) into Equation (1.2) gives an expression for the acoustic particle velocity, as: jkA j(ωt−kr) A e (1.31) u = 2 e j(ωt−kr) + r r

18

Noise Control: From Concept to Application

Dividing Equation (1.30) by Equation (1.31) gives the following equation, which holds for a harmonic wave characterised by a wavenumber k, and also for a narrow band of noise characterised by a narrow range of wavenumbers around k: p jkr = ρc u 1 + jkr

(1.32)

For inward-travelling waves, the signs of k are negative. Note that when the quantity, kr, is sufficiently large, Equation (1.32) reduces to the outward travelling plane wave form, Equation (1.9). Example 1.8 (a) Show that the particle velocity amplitude at distance, r, from a point monopole in a j pˆ 1− for a single frequency, ω = kc. (Begin with the spherical free field is, u ˆ = ρc kr wave solution to the wave equation.) (b) At what kr value is the velocity amplitude twice pˆ/ρc? Solution 1.8 (a) From Equation (1.29), with φ replaced with p, we can write the acoustic pressure B j(ωt−kr) solution to the spherical wave equation as p = e , where B is a different r constant to the one in Equation (1.29). Using Equation (1.3), the velocity potential B j(ωt−kr) is φ = e . jωρr It would be equally valid to start with the acoustic velocity potential function solution given by Equation (1.29), but in this case, the pressure solution is more convenient. Using the one-dimensional form of Equation (1.2), the particle velocity is: jkB j(ωt−kr) B ∂φ = e + 2 e j(ωt−kr) ∂r jrωρ jr ωρ     B j(ωt−kr) j p j = e 1− = 1− . rρc kr ρc kr

u=−

Thus u ˆ =

pˆ j 1− . ρc kr





(b) If the particle velocity amplitude is twice pˆ /(ρc), this implies that |1 − j/ kr| = 2 or √ Thus kr = 1/ 3 = 0.58.

p

1 + (kr)−2 = 2.

Example 1.9 The acoustic pressure at distance, r, from a small source, radius r0 , and surface velocity ampliˆ e jωt , is of the form: p = B e jω(t−r/c) = B e j(ωt−kr) . tude, U = U r r (a) Find an expression for the particle velocity at any arbitrary distance from the source. ˆ. (b) Show that the constant, B, is given by B = jωr02 ρU

19

Fundamentals Solution 1.9 B jω(t−r/c) B j(ωt−kr) (a) At any location, p = e = e . From Equation (1.3), the velocity r r R B j(ωt−kr) 1 p dt = e , and from Equation (1.2), the particle potential is, φ = ρ jρrω ∂φ Bk j(ωt−kr) B velocity is, u = − = e + e j(ωt−kr) . ∂r ρrω jρr2 ω (b) When r = r0 , u = U , and the particle velocity may be written as: B e j(ωt−kr0 ) U= ρr0 ω



1 k+ . jr0

ˆ ejωt , so U ˆ= But U = U Thus, B = 



B ρr0 ω



k+

1 jr0



e−jkr0 .

ˆ ρr0 ω U  e jkr0 . 1 k+ jr0

However, e jkr0 = cos kr0 + j sin kr0 ≈ 1 + jkr0 Thus, B = 

for small kr0 .

ˆ ρr0 ω U  (1 + jkr0 ) = jρr02 ω Uˆ . 1 k+ jr0

Example 1.10 The acoustic pressure of a harmonic spherical wave may be written as: p(r, t) =

B j(ωt−kr) e . r

(a) Derive an expression for the radial acoustic particle velocity. (b) Derive an expression for the specific acoustic impedance at a distance r from a monopole source. (c) At what distance (in wavelengths) from the sound source is the modulus of the specific acoustic impedance half that for a plane wave. The specific acoustic impedance is defined in Table 1.5. Solution 1.10 (a) The acoustic pressure solution to the spherical wave equation is p = Using Equation (1.3), the velocity potential is φ =

B j(ωt−kr) e . r

B j(ωt−kr) e . jωρr

Using the one dimensional form of Equation (1.2), the particle velocity is: ∂φ jkB j(ωt−kr) B B j(ωt−kr) j u=− = e + 2 e j(ωt−kr) = e 1− ∂r jrωρ jr ωρ rρc kr



p = ρc

j 1− kr





 .

−1

j (b) Specific acoustic impedance, Zs = p/u. Thus, Z = ρc 1 − . kr (c) The modulus of the impedance of the spherical wave is half that of a plane wave (ρc) when



20

Noise Control: From Concept to Application

ρc 1 + j/(kr) ρc(1 + j/kr) ρckr(kr + j) = = |Z| = × 1 − j/(kr) 1 + j/(kr) 1 + 1/(kr)2 (kr)2 + 1 √ √ ρckr 1 + k2 r2 1 + k2 r2 = ρckr = ρckr = √ = ρc/2. 1 + k2 r2 1 + k2 r2 1 + k2 r2 √ Thus, 4k 2 r2 = 1 + k 2 r2 or k 2 r2 = 0.3333 and r = (λ/2π) 0.3333 = 0.092 λ. Example 1.11 A bubble in the ocean acts as a small resonator having a characteristic resonance frequency determined by its size and local hydrostatic pressure. (a) Derive a simple relationship for the bubble acoustic pressure in terms of the bubble radius, r, and the surface acceleration, a. [Hint: Use the spherical wave specific acoustic impedance and assume kr α, β, δ, the function, F = 1, and the corresponding term in Equation (4.8) is zero.

4.4

Atmospheric Absorption, Aa

Atmospheric absorption, Aatm , is dependent upon temperature and relative humidity. Calculated values of the absorption rate, m (dB/kilometre) (ISO 9613-1, 1993; ANSI/ASA S1.26, 2014) for nominal octave band centre frequencies are listed in Table 4.1 for representative values of temperature and relative humidity. These values are based on the model developed by Sutherland et al. (1974), as summarised in Bies et al. (2018), and can be used in both propagation models described later (CONCAWE and ISO 9613-2 (1996)). Values for other temperatures, atmospheric pressures and relative humidities may be calculated using ENC software available on www.causalsystems.com.

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Sound Propagation Outdoors

The standards, ISO 9613-1 (1993) and ANSI/ASA S1.26 (2014) both provide values of m corresponding to pure tones at the exact 1/3-octave band centre frequencies, which are slightly different to the nominal frequencies (see Table 1.2). ANSI/ASA S1.26 (2014) also provides equations for calculating the absorption for bands of noise with varying bandwidths. However, it is sufficiently accurate to use the 1/3-octave or octave band centre frequency, tonal absorption value as the value for a 1/3-octave or octave band of noise (ISO 9613-1, 1993). For propagation over distance X (in kilometres), the absorption Aatm is: Aatm = mX TABLE 4.1

Attenuation due to atmospheric absorption

Relative humidity %

Temperature ◦ C

63

25

15 20 25 30 35 15 20 25 30 35 15 20 25 30 35

0.2 0.2 0.2 0.2 0.2 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1

50

75

(4.10)

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000 m (dB per 1000 m) 0.6 0.7 0.7 0.6 0.6 0.5 0.4 0.4 0.3 0.3 0.4 0.3 0.3 0.2 0.2

1.2 1.4 1.6 1.8 1.8 1.2 1.3 1.3 1.3 1.2 1.1 1.1 1.0 0.9 0.8

2.4 2.5 2.9 3.5 4.2 2.2 2.7 3.2 3.6 3.7 2.4 2.8 3.0 3.0 2.9

6.4 5.5 5.4 6.0 7.1 4.2 4.7 5.7 7.0 8.4 4.1 5.1 6.3 7.4 8.2

21.9 16.9 14.0 12.7 12.8 10.8 9.9 10.2 11.6 14.1 8.5 9.0 10.5 13.0 16.1

74.7 59.4 46.9 38.4 33.3 36.5 29.7 25.8 24.6 25.6 25.1 22.2 21.5 23.0 26.8

For temperatures and relative humidities not covered in Table 4.1, see Bies et al. (2018), ISO 9613-1 (1993) or ANSI/ASA S1.26 (2014). Example 4.1 A single frequency sound wave propagates in the x direction with a decaying pressure amplitude proportional to e−αx . Find the decibel decay rate (dB per unit distance) in terms of α. Describe the factors influencing α in the audio frequency range. Solution 4.1 As the constant of proportionality of the acoustic pressure is the same at all locations, the proportionality constant will cancel in an expression for the ratio of the acoustic pressures at any two locations. Thus, the pressure amplitude at any location, x, is given by: pˆ/ˆ p0 = e−α(x−x0 ) , where pˆ0 is the amplitude at x0 = 0. The decibel decay rate per unit distance is 20 log10 of the reciprocal of the above expression when x = 1 and is thus given by: Decay rate = 20 log10 eα = 20 × 0.4343 loge eα = 8.686α (dB/m), 1 logb b and = 0.4343 and where where the change of logarithm base formula, loga b = logb a loge 10 a = 10 and b = e. Important factors are air temperature and humidity.

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Example 4.2 A pipe of length 50 m is radiating uncorrelated noise to a community at a distance of 200 m in a direction normal to the centre of the pipe. The sound field generated by the pipe has a directivity of two in the direction of the community. If the sound power radiated by the pipe in the 2 kHz octave band is 2 W, calculate the sound pressure level (in dB re 20 µPa) in the community for the 2 kHz octave band. Include the effects of air absorption but ignore other atmospheric effects. Assume that the sound intensity loss due to ground reflection is 2 dB, and that the pipe is 2 m above the ground. Atmospheric temperature is 15◦ C and relative humidity is 25%. Assume no obstacles exist between the pipe and the community and assume incoherent combination of the direct and ground-reflected waves. Solution 4.2 The situation is as shown Figure 4.1. Equation (3.25) may be used to calculate the sound pressure level. The equation must be multiplied by the directivity factor (2 in this case).

pipe

50 m

200 m

au aR

A

FIGURE 4.1 Arrangement for Example 4.2.

Thus, hp2 i = [W ρc/4πr0 D][αu − α` ] × 2 and αu = α` = tan−1

25 = 0.124 radians. 200

Given, W = 2, r0 = 200, D = 50, then, hp2 i = [2 × 1.206 × 343/(4π × 200 × 50)] × 0.249 × 2 = 3.275 × 10−3 Pa2 , 3.275 × 10−3 and Lp = 10 log10 = 69.1 dB (re 20 µPa). 4 × 10−10 From Table 4.1, Aatm = 19.3 dB per 1000 m, so for 200 m, Aatm = 19.3/5 = 3.9 dB. Sound intensity loss (or sound pressure level reduction) on ground reflection is 2 dB. Thus the ground effect (or the overall attenuation at the observer due to the ground) can be derived using the first term on the RHS of Equation (1.75) with the attenuation due to the direct path set equal to 0. Thus, Agr = −10 log10 [1 + 10−2/10 ] = −2.1 dB, Agr + Aatm = 1.8 dB and the sound pressure level at the receiver is calculated using Equation (4.2) as: Lp = 69.1 − 1.8 = 67.3 dB (re 20 µPa).

4.5

Ground Effects, Agr

Progressively more complicated procedures for taking account of ground effects will be considered in this section. The discussion will begin with the simplest procedures based upon experimental data and will conclude with a reasonably complex calculation procedure. The most complex

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Sound Propagation Outdoors

procedure, which requires extensive information about weather conditions, the local geography and the flow resistance of the earth is beyond the scope of this text and is discussed in detail in Bies et al. (2018).

4.5.1

Excess Attenuation Using Simply Hard or Soft Ground

As a rough first approximation, it is often assumed that the effect of hard ground such as concrete, water or asphalt is to increase sound pressure levels by 3 dB and that soft ground (such as grass, snow, etc.) has no effect. Thus, Agr = −3 or 0 dB depending on the ground surface. In this case, the assumption is implicit that reflection in the ground plane is incoherent. Example 4.3 A line of traffic is radiating sound into the community. The nearest residence is 250 m away. The average sound power of each vehicle is 2 W in the 500 Hz octave band and the average vehicle spacing is 7 m. Assuming that only noise in the 500 Hz octave band is of interest, calculate the sound pressure level at the nearest residence if the ground surface between the road and the residence is concrete. Assume a community location 1.5 m above the ground and a vehicle acoustic centre 0.5 m above the ground. Meteorological effects can be ignored for this example. State any other assumptions that you make. Solution 4.3 Assume that the situation can be modelled as a semi-infinite row of incoherent sources above a hard ground. From Equation (3.20), the mean square pressure, excluding the effect of the hard ground is: h p2 i = ρc

413.7 × 2 W = 0.118 Pa2 , = 4 b r0 4 × 7 × 250

Lp = 10 log10

and

0.118 hp2 i = 10 log10 = 84.7 dB re 20 µPa. ( 2 × 10−5 )2 p2ref

The ground is concrete, so the ground effect is Agr = −3 dB. Assume an air temperature of 20◦ C. Then the air absorption effect ranges from 2.6 to 2.8 dB per 1000 m. For a 250 m distance, the air absorption effect, Aatm = 0.7 dB. Assume that there are no obstacles blocking the line of sight to the road from the residence. So the sound pressure level at the residence can be calculated using Equation (4.2) as: Lp = 84.7 − 0.7 + 3.0 = 87.0 dB.

4.5.2

Excess Attenuation Using the Plane Wave Method

A better approximation than use of the simple method just described is obtained by calculating the ground amplitude reflection coefficient, which can be done with varying degrees of sophistication and accuracy. The simplest procedure is to assume plane wave propagation and that the effect of turbulence is sufficient that the waves combine incoherently; that is, on an energy basis such that squared pressures add. The plane wave reflection loss in decibels is given by Arf = −20 log10 |Rp |, where |Rp | is the modulus of the complex plane wave reflection coefficient of the ground, which is given by Bies et al. (2018) as: Zm cos θ − ρc cos ψ (4.11) Rp = Zm cos θ + ρc cos ψ where Z m is the impedance looking into the ground surface (see Chapter 5) and θ is defined in Figure 4.2.

176

Noise Control: From Concept to Application receiver, R

q

r=dSR source, S rR

rS air

hS

q

b0

hR

b0

O ground

d image source

FIGURE 4.2 Geometry illustrating reflection above a ground plane.

Equation (4.11) only applies to infinitely extending media or media that extend for a sufficient distance that waves reflected from any termination back towards the interface have negligible amplitude on arrival at the interface (such as the ground). The quantity ψ, in Equation (4.11) is defined as:

s cos ψ =

1−



k km

2

sin2 θ

(4.12)

where km is the complex wavenumber describing propagation in the ground (see Chapter 5). Equation (4.36) is derived in Bies et al. (2018) using Snell’s law of refraction for sound travelling from medium 1 to medium 2, which may be stated in terms of the speeds of sound, c1 and c2 , in media 1 and 2, respectively, as: sin ψ c2 (4.13) =n= c1 sin θ where θ is the angle between the incident wave in medium 1 and the normal to the interface plane between the two media and ψ is the angle between the normal to the interface plane between the two media and the refracted wave in medium 2. Reference to Equation (4.36) shows that when km  k, the angle, ψ, tends to zero and Equation (4.11) reduces to the following form: Rp =

Zm cos θ − ρc Zm cos θ + ρc

(4.14)

which is the equation for a locally reactive surface (one where the response at any given point is independent of the response at any other point). The excess attenuation, Agr , is then calculated as:     (4.15) Agr = −10 log10 1 + |Rp |2 = −10 log10 1 + 10−Arf /10 The quantity, Arf , is plotted in Figure 4.3 for various values of the dimensionless parameter ρf /R1 . Here, f is the tonal frequency, or the centre frequency of the measurement band, ρ is the density of air and R1 is the flow resistivity of the ground. Values of R1 for various ground surfaces are given in Tables 4.2 and 4.8. As can be seen from Table 4.2, there are great variations in measured flow resistivity data from different sources. Figure 4.3 is used to determine the decrease in energy, Arf (dB) of the reflected sound wave on reflection from the ground. The quantity Agr will vary between 0 and −3 dB, depending on the value of Arf (which is always positive). Alternatively, if the condition, km  k, is not satisfied (that is, local reaction cannot be assumed for the ground surface), then Rp should be calculated using Equation (4.11)).

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Sound Propagation Outdoors 0

Reflection loss Ar f (dB)

less than 0.001 0.01 0.02 0.05 10 0.1

0.2 rf /R1=1 20

1

10

100

1000

10,000

1/2

Reflection parameter, b (R1 /r f ) (degrees)

FIGURE 4.3 Reflection loss, Arf (dB), as a function of the reflection parameter, which consists of the reflection angle β (where β is measured from the horizontal and β = 90 − θ degrees), the surface flow resistivity, R1 , the air density, ρ and the frequency, f . Curves are truncated when β reaches 90◦ or the reflection loss exceeds 20 dB. The surface is assumed to be locally reactive.

4.6

Meteorological Effects, Amet

The two principal meteorological variables are vertical wind and temperature gradients. When the temperature increases with height and the temperature gradient is thus positive, the condition is termed an inversion. When the temperature decreases with height and the gradient is thus negative, the condition is termed a lapse. If the weather conditions are not accurately known, then it is generally assumed that meteorological effects will result in a sound pressure level variation about the predicted level at the receiver, as shown in Table 4.3. The range of values in Table 4.3 are primarily due to variations in vertical wind shear and temperature gradients (discussed in the following two subsections), which cause variations in the vertical sound speed gradient. This, in turn, causes the sound rays to bend towards or away from the ground, depending on whether the resulting vertical sound speed gradient is positive or negative, and this causes significant variations in the sound pressure level arriving at the receiver. Wind shear is the common name for a vertical wind speed profile in which the wind speed changes with height above the ground. Due to surface friction effects, the wind is almost always characterised by an increasing speed as the distance from the ground increases. Generally, wind shear is greater at lower ground level wind speeds and the spread in values of wind shear is usually greater at night. In the direction upwind from the source (that is, the direction into the wind from the source), sound rays are diffracted upwards away from the ground and may result in a shadow zone. In this zone, the only sound in the mid- and high-frequency ranges that reaches a receiver is what is scattered due to atmospheric turbulence or due to other obstacles impacted by the sound ray, although this may not be the case for low-frequency sound and infrasound. Thus in the upwind direction, the sound experienced at a receiver is usually less than would be expected in the absence of wind. In the direction downwind from the source, sound rays are refracted towards the ground so that in this direction the sound pressure level at large distances from the source is likely to be

178

Noise Control: From Concept to Application TABLE 4.2

Flow resistivities measured for some common ground surfaces

Ground surface type Dry snow, newly fallen 0.1 m over about 0.4 m older snow Sugar snow Soft forest floor with blueberry greens and moss Forest floor covered by weeds Pine or hemlock forest floor Soft forest floor covered with pine needles Sandy forest floor Dense shrubbery, 20 cm high Soil and bark, sparse vegetation Peat or turf area, homogeneous organic material Soil covered with leaves and twigs Soil mixed with sawdust Relatively dense soil sparsely covered by grass and other low greens Short grass, green moss and blueberry greens Rough grassland and pasture Grass, soccer field Lawn, moderately stepped on Lawn, seldom stepped on Lawn Agricultural field Hard soil Soil, exposed and rain packed Wet, sandy loam Moistened sand Bare sandy plain Dry sand Sandy silt, hard packed by vehicles Quarry dust, hard packed by vehicles Mixed paving stones and grass Old gravel field with sparse vegetation Gravel road, stones and dust Gravel parking lot Asphalt sealed by dust and light use Concrete

Flow resistivity, R1 (kPa s/m2 ) 10–30 25–50 40 63–100 20–80 160 630–2000 100 100 100 160–250 250 630 40 100–300 630–2000 160–250 250 250–400 160–250 400–2000 4000–8000 1500 500 250–500 60–140 800–2500 5000–20000 630–2000 2000 2000 630–2000 30000 20000

greater than would be expected in the absence of wind. At a sufficient distance from the source, there will be multiple rays striking the ground, which have all been reflected from the ground one or more times. At distances greater than the shortest distance from the source that multiple reflections occur, there will be less attenuation of sound with distance than might otherwise be expected and this is the main reason that most noise propagation models are not considered reliable at distances greater than 1000 m from the source. In the early hours of the morning when there is a clear, cloud-free sky, the atmospheric temperature usually increases with altitude close to the ground, instead of the more normal condition where it decreases with altitude. The layer of air in which the temperature is increasing is known as the inversion layer and this is usually just a few hundred metres thick. The effect of an atmospheric temperature inversion is to cause sound rays to bend towards the ground in much the same way as they do when propagating downwind. Conversely, the more normal (especially

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Sound Propagation Outdoors

during the daytime) non-inverted atmospheric temperature profile results in sound rays that are refracted upwards, away from the ground, thus resulting in reduced sound pressure levels at the receiver. However, only a light downwind condition needs to exist to counteract the effect of the usual atmospheric temperature gradient in non-inverted conditions. This can be verified using Equation (4.19), which gives for a ground temperature of 20◦ C and an atmospheric temperature gradient of −3◦ C per 300 m, a sonic gradient of −5.85 m/sec per 1000 m, which is equivalent to the gradient associated with a light surface wind (less than 1 m/s at 10 m – see Equation (4.22)). When predicting outdoor sound, it is usual to include downwind or temperature inversion meteorological conditions, corresponding to CONCAWE category 5 or 6 (see Section 4.7.4). The downwind condition implies that the wind direction makes an angle of less than 45◦ to the line joining the source to the receiver. It has been established (Rudnick, 1947) that the radius of curvature of a sound ray propagating in the atmosphere is dependent on the vertical gradient of the speed of sound, which can be caused either by a wind gradient or by a temperature gradient or by both. It has also been shown (Piercy et al., 1977) that refraction of a sound ray due to either a vertical wind gradient or a vertical temperature gradient produces equivalent acoustic effects, which are essentially additive. A sound ray travelling at an angle, ψS , above (or below) the line parallel to the ground plane will have a curved path with a radius of curvature, Rc , given by the following equation (De Jong and Stusnik, 1976). When Rc is positive the sound rays are curved downward and when Rc is negative the sound rays are curved upward: Rc = 

TABLE 4.3

c

(4.16)

dc cos ψS dh



Variability in sound pressure level predictions due to meteorological influences

Octave band centre frequency (Hz)

100

200

63

+1

+4, −2

+7, −2

+8, −2

125

+1

+4, −2

+6, −4

+7, −4

250

+3, −1

+5, −3

+6, −5

+7, −6

500

+3, −1

+6, −3

+7, −5

+9, −7

1000

+7, −1

+11, −3

+12, −5

+12, −5

2000

+2, −3

+5, −4

+7, −5

+7, −5

4000

+2, −1

+6, −4

+8, −6

+9, −7

8000

+2, −1

+6, −4

+8, −6

+9, −7

Distance from the source (m) 500

1000

Amet (dB)

A non-zero sonic gradient, dc /dh, implies that the speed of sound varies with altitude, which means that the radius of curvature of a sound ray will vary as the ray height varies. The sonic gradient, dc /dh, is a direct result of vertical atmospheric wind and temperature gradients. In the following analysis, it will be assumed that the wind contribution to the sound speed gradient adds to the temperature gradient contribution. The total vertical sonic gradient can thus be expressed as:   dc dU ∂c = + (4.17) dh dh ∂h T

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Noise Control: From Concept to Application

∂c , on the right-hand side is evaluated by assuming a linear vertical temwhere the term, ∂h T perature profile. The sonic gradient due to the vertical atmospheric temperature profile can be calculated from a knowledge of the vertical atmospheric temperature profile and the use of Equation (1.1). The latter equation may be rewritten more conveniently for the present purpose as:





c = c00

p

(4.18)

T /273

where T is the temperature in Kelvin and c00 is the speed of sound at sea level, one atmosphere pressure and 0◦ C (331.3 m/s). The vertical sound speed gradient resulting from the atmospheric temperature gradient is found by differentiating Equation (4.18) with respect to h to give:



∂c ∂h

 T

=

dT ∂c dT −1/2 c00 dT −1/2 √ = T [T0 + 273] = Am = 10.025 dh ∂T dh dh 2 273

(4.19)

In Equation (4.19), dT /dh is the vertical temperature gradient (in ◦ C/m) and T0 is the ambient temperature in ◦ C at 1 m height. Typical values of Am range from −0.1 to +0.1 s−1 . To make the problem of finding the ray path length from the source to the receiver tractable, an equivalent linear sound speed profile is determined and used, enabling the radius of curvature to be approximated as a constant value, independent of the altitude of the sound ray. This results in a circular ray path. The component of the sonic gradient due to the wind, dU /dh, is added to the sonic gradient due to the atmospheric temperature gradient of Equation (4.19) and the total sonic gradient, dc/dh, is then used with Equation (4.16) to obtain the radius of curvature of the sound ray travelling in a circular arc, which is a very rough approximation of the true ray path. The wind gradient is a vector. Hence U is the velocity component of the wind in the direction from the source to the receiver and measured at height h above the ground. The velocity component, U , is positive when the wind is blowing in the direction from the source to the receiver and negative for the opposite direction. In a downward refracting atmosphere (positive radius of curvature), at a sufficient source– receiver separation distance, one or more ground-reflected rays will arrive at the receiver in addition to the direct ray with no ground reflection. Calculation of the contribution of the wind to the sonic gradient begins with the following equation, which calculates the wind speed as a function of height, h, above the ground at low altitudes. U (h) = U0



h h0

ξ

(4.20)

In Equation (4.20), ξ is the wind shear coefficient, which is a function of surface roughness and atmospheric stability (see Figure 4.4), although the value for neutral atmospheric conditions is often used without justification for all atmospheric conditions. U (h) is the wind speed at height h and U0 is the wind speed at some reference height, h0 , normally taken as 10 m, although a height of 5 m is sometimes used. The surface roughness, z0 , can be obtained from Table 4.4 or it can be estimated using measurements of the wind speed at two different heights in a stable atmosphere (IEC 61400-11 Ed.3.0, 2012): U (h) loge h0 − U0 loge h z0 = exp U (h) − U0





(4.21)

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Sound Propagation Outdoors

0.6

Wind shear coefficient, x

F

0.5

0.4 E

0.3 D

0.2 B

C A

0.1 0 0.01

0.1

1

Surface roughness, z0 (m) FIGURE 4.4 Estimates of the wind shear coefficient as a function of surface roughness, based on averaged measurements between 10 m and 100 m above the ground (Irwin, 1979). The exponent is a function of Pasquill stability category, A–F, (see Table 4.7) and surface roughness.

To obtain the wind gradient (and hence the sonic gradient due to the wind), Equation (4.20) is differentiated with respect to height, giving the expected wind gradient at height, h as: U (h) dU =ξ (4.22) dh h Equations (4.22) and (4.19) are substituted into Equation (4.17) to obtain the total sonic gradient. The height, h, is fixed to a height midway between the source and receiver heights, resulting in a linear sonic gradient. Equation (4.22) is then used with Equations (4.16), (4.19) and (4.17) to calculate the radius of curvature due to the combined effects of atmospheric wind and temperature gradients. TABLE 4.4

Surface roughness estimates for various ground surface types (IEC 61400-11 Ed.3.0, 2012)

Surface type

Surface roughness, z0 (m)

Water, snow or sand Open flat land, mown grass, bare soil Farmland with some vegetation Suburbs, towns, forests, many trees & bushes

0.0001 0.01 0.05 0.3

The angle, ψS , at which the ray actually leaves the source is determined iteratively, starting with ψS = 0, calculating the corresponding value of Rc using Equations (4.16), (4.19), (4.17) and (4.22) and then calculating the distance, dc , from the source at which the ray is at a height equal to the receiver height. Referring to Figure 4.5, we may write the following for dc : dc = 2Rc sin(ψS + ϕ) cos ϕ

(4.23)

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Noise Control: From Concept to Application

yS

S

dSR

j hS

yR

yS

R

j Rc

hR

yS

yR Rc

j

dc=d

FIGURE 4.5 Geometry for a sound ray originating at source, S, and arriving at receiver, R.

where ϕ = arctan

(hS − hR ) dc

(4.24)

As dc appears on the left-hand side of Equation (4.23) and also in the equation for ϕ, and we eventually require that dc ≈ d, it is more efficient to replace dc with d in Equation (4.24). The above equations are also valid for the receiver being higher than the source, in which case, ϕ is negative. The value of ψS is incremented (or decremented) by small amounts until dc ≈ d to the required accuracy (usually a few percent). The radius of curvature is used together with the barrier analysis in Chapter 7 to determine the effect of wind and temperature gradients on barrier attenuation.

4.6.1

Attenuation in the Shadow Zone (Negative Sonic Gradient)

A shadow zone is defined as that region where direct sound cannot penetrate due to upwards diffraction. Of course, a small amount of sound will always transmit to the shadow zone as a result of scattering from objects so that one might expect an increase in attenuation in the shadow zone to be up to 30 dB. To create a shadow zone, a negative sonic gradient must exist, as this causes the sound rays emitted by the source to have a radius of curvature, Rc , that is finite. By convention, in a negative sonic gradient, the radius of curvature is negative (sound rays curved upwards), whereas in a positive sonic gradient, the radius of curvature is positive (sound rays curved downwards). In the case of a negative sonic gradient and corresponding negative radius of curvature, Rc , the distance, x, between the source (height, hS ) and receiver (height, hR ) beyond which the receiver will be in the shadow zone is: x=

p

−2Rc

hp

hS +

p

hR

i

(4.25)

Note that in the presence of no wind, the shadow zone around a source will only exist when there is no temperature inversion close to the ground and will be symmetrical around the source. In the presence of wind, the distance to the shadow zone will vary with direction from the source, as the sonic gradient in a given direction is dependent on the component of the wind velocity in that direction. Thus, it is likely that in most cases, a shadow zone will exist in the upwind direction but not in the downwind direction.

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Sound Propagation Outdoors

The angle subtended from a line drawn from the source towards the oncoming wind beyond which there will be no shadow zone is called the critical angle, βc , and is given by: βc = cos

−1



(∂c/∂h)T dU/dh



(4.26)

This is illustrated in Figure 4.6, where r is the distance of the receiver from the source, r1 is the distance from the source in the direction of the receiver at which a sound ray emitted by the source grazes the ground and x is the distance of the source from the receiver at which the receiver will be on the border of the shadow zone.

Sound source Ground

Sound ray

hS

r1

Receiver on shadow border hR

Receiver well into shadow

x d

bc

u r1 b

Sound source

Shadow zone x Receiver

bc -b

FIGURE 4.6 Illustration of the shadow zone and the limiting angle βc , beyond which the shadow zone does not exist. β is the angle subtended between the wind direction and the source–receiver line.

The actual excess attenuation due to the shadow zone increases as the receiver moves away from the source further into the zone. It is also dependent on the difference between the angle β, between the receiver/source line and the line from the source towards the oncoming wind (see Figure 4.6) and the critical angle βc . The actual excess attenuation, as a function of β and βc , may be determined using Figures 4.7 and 4.8 for wind speeds of 5 to 30 km/hr, for octave bands from 250 Hz to 4000 Hz, for ground cover heights less than 0.3 m, for sound source heights of 3 to 5 m and a receiver height of 2 m (Wiener and Keast, 1959). Note that for wind speeds greater than 30 km/hr, noise due to wind blowing over obstacles and rustling leaves in trees usually dominates other environmental noise, unless the latter is particularly severe. The attenuation for a particular value of β is calculated using Figure 4.7. Note that this is only the excess attenuation due to a negative vertical sonic gradient (upwind propagation and/or

184

Noise Control: From Concept to Application b#65E

Excess attenuation (dB)

30

b=75E

b=80E

20

b=90E

10

0 0.5

1

2

4

8

Normalised distance from sound source (d /x)

FIGURE 4.7 Excess attenuation in the shadow zone due to the negative vertical sonic gradient for various angles, β, between the wind direction and the line joining the source and receiver.

Limiting value of excess attenuation (dB)

30

20

10

0 -20

0

20

40

60 80 100 ( bC - b) degrees

120

140

160

FIGURE 4.8 Limiting value of excess attenuation due to a negative sonic gradient only in the shadow zone.

no temperature inversion close to the ground). The attenuation cannot exceed the value given by Figure 4.8, which is a function of βc − β. Example 4.4 An aircraft in level flight, 600 m above ground level, travelling at 400 km/hr, approaches an observer on the ground. The atmosphere is still, and the sound speed falls off by 1 m/s per 10 m height. At what range does the aircraft emerge from the ground shadow? You may assume that to a first approximation, the radius of curvature of the sound ray from the aircraft to the ground is independent of altitude and may be based on an average speed of sound (for sound travelling between the aircraft and the ground) of 343 m/s. Solution 4.4 The situation is illustrated in Figure 4.9. We need to find the radius of curvature, rc , 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

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Sound Propagation Outdoors

(with an assumed aircraft speed). The wave which hits the ground at grazing incidence will be the one heard first. q0

r - 600

r

d A2

q0 600

FIGURE 4.9 Layout for Example 4.4.

The total sonic gradient is −1/10 = −0.1 s−1 . Using Equation (4.16), the radius of curvature of the wave, rc = −343 × 10/ cos(θ0 ) = −3430/ cos(θ0 ) m. The negative sign simply means that the centre of the ray arc is above the source so the rays curve upwards. In Figure 4.9, we set r = −rc :     3430/ cos θ0 − 600 3430 − 600 × cos θ0 −1 −1 θ0 = cos m. = cos 3430/ cos θ0 3430 By trial and error, the above equation gives θ0 = 31.67◦ . Thus, r = 3430/ cos(31.67◦ ) = 4030 m. The distance, d, is then: p √ d = r2 − (r − 600)2 = 40302 − 34302 = 2116 m. 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. The time taken for the sound to travel from the aircraft to the ground is the integral of distance travelled divided by the speed of sound. It is necessary to integrate as the speed of sound varies with altitude. Thus: t=

Rθ0 0

Rθ0 Rθ0 dθ dθ rdθ 31.67◦ = = 10 × = 10 loge [sec θ + tan θ]0 c0 − 0.1h (c /r) − 0.1 + 0.1 cos θ cos θ 0 0 0

= 5.83 seconds (as c0 /r ≈ 0.1). If the aircraft travels at 400 km/hour, then it would travel 648 m in 5.83 seconds. Thus the aircraft emerges from the ground shadow at a horizontal distance of (2116 m − 648 m) = 1468 m from the observer.

4.7

CONCAWE Propagation Model

CONCAWE is a noise propagation model developed for environmental noise estimations for noise radiated by petroleum and petrochemical complexes to surrounding communities (Manning, 1981a). Since that time, the original version or its modified form (Marsh, 1982) has been used in almost all commercially available software for calculating the level of noise radiated into surrounding communities by any sound source.

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The equation used in the CONCAWE model is a derivative of Equation (4.2) and may be written for the ith source producing a sound pressure level, Lpik , at the kth community location as: Lpik = LW i + DIik − AEik (dB re 20 µPa) (4.27) where Lpik is the octave band sound pressure level at community location, k, due to the ith source, and LW i (dB re 10−12 W) is the sound power radiated by the ith source. DIik is the directivity index of source, i, in the direction of community location, k. DIik is usually assumed to be 0 dB, unless specific source directivity information is available. Calculations are done in octave bands from 63 Hz to 8 kHz and the overall A-weighted sound pressure level is calculated by applying the A-weighting correction (see Table 2.4) to each octave band level and then summing the levels logarithmically (see Section 1.12.2). C-weighted sound pressure levels can be calculated in a similar way. The coefficient, AEik , is the excess attenuation experienced by a sound pressure disturbance travelling from source, i, to community location, k, and is given by: AEik = (K1 + K2 + K3 + K4 + K5 + K6 + K7 + Kv )ik

(4.28)

Each of these attenuation factors are discussed in the following paragraphs.

4.7.1

Geometrical Divergence, K 1

Sound sources are treated as point sources (see Section 4.3) and the attenuation (K1 = Adiv ) as a result of the sound waves spreading out as they travel away from the sound source is given by Equation (4.5).

4.7.2

Atmospheric Absorption, K 2

This is discussed in detail in Section 4.4 and is the same for all propagation models. Note that K2 = Aatm , and is calculated using Equation (4.10).

4.7.3

Ground Effects, K 3

For a hard surface such as asphalt, concrete or water, K3 = Agr = −3 dB. For all other surfaces a set of empirical curves is used (see Figure 4.10). However, these curves were developed for noise sources close to the ground and will over-predict the actual ground effect for high sources such as wind turbines, with the error becoming larger as the distance from the source increases. A conservative approach would be to use the hard ground value for K3 for all ground surfaces.

4.7.4

Meteorological Effects, K 4

In the CONCAWE model (Manning, 1981b), meteorological effects (K4 = Amet ) have been graded into six categories based on a combined vertical wind and temperature gradient. In Table 4.5, incoming solar radiation is defined for use in Table 4.6. In Table 4.6, the temperature gradient is coded in terms of Pasquill stability category A–G. Category A represents a strong lapse condition (large temperature decrease with height). Categories E, F and G, on the other hand, represent a weak, moderate and strong temperature inversion, respectively, with the strong inversion being that which may be observed early on a clear morning. Thus, category G represents very stable atmospheric conditions while category A represents very unstable conditions and category D represents neutral atmospheric conditions. The wind speed in this table is a non-vector quantity and is included for the effect it has on the temperature gradient. Wind gradient effects are included in Table 4.7.

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Sound Propagation Outdoors

15

Excess attenuation, K3 (dB)

250 10

500 125 1000

5 2000 4000

0

63 -5 100

500 1000 Distance from source (m)

2000

FIGURE 4.10 CONCAWE ground attenuation curves.

TABLE 4.5 etc.)

Daytime incoming solar radiation (full cloud cover is 8 octas, half cloud cover is 4 octas,

Latitude of sun

Cloud cover (octas)

Incoming solar radiation

< 25 25◦ –45◦ > 45◦ > 45◦

0–7 3.0

— — v < −3.0 −3.0< v < −0.5 −0.5< v < +0.5 +0.5< v < +3.0

The Pasquill stability category is combined with the magnitude of the wind vector using Table 4.7 to determine one of the six meteorological categories (CAT) for which attenuations, Amet = K4 , were experimentally determined from surveys of three European process plants (Manning, 1981a). Attenuations are shown in Figures 4.11(a)–(g) for the octave bands from 63 Hz to 4 kHz. The wind speed used in Tables 4.6 and 4.7 is measured at a height of 10 m above the ground and is the proportion of the wind vector pointing from the source to the receiver.

4.7.5

Source Height Effects, K 5

A higher than normal noise source may act to reduce the attenuation (that is, increase the level) of sound arriving at the receiver. The CONCAWE approach only applies a correction if Figure 4.10 is used to calculate the ground effect. In this case, for source heights greater than 2 m the additional correction, K5 must be added to the excess attenuation term, AE . The correction, K5 , is given by:

( K5 =

(K3 + K4 + 3) × (γ − 1) dB

if (K3 + K4 ) > −3 dB

0 dB

if (K3 + K4 ) ≤ −3 dB

(4.29)

where: γ = 1.08 − 0.478(90 − θ) + 0.068(90 − θ)2 − 0.0029(90 − θ)3 and γmax = 1

(4.30)

Note that K5 is always negative, which means that it acts to reduce the excess attenuation. The angle, θ, is defined in Figure 4.2. If propagation is to a receiver located on a hillside, or across a valley floor, the value of K5 should be reduced (made more negative) by up to 3 dB to account for multiple reflections from the hillside.

4.7.6

Barrier Attenuation, K 6

Barriers are any obstacles that represent interruptions of the line-of-site from the sound source to the community location. The attenuation, K6 = Abar , due to barriers is taken into account in the CONCAWE propagation model by modelling them as thin screens. The corresponding attenuation is calculated using the procedure outlined in Chapter 7, Section 7.5. In practice, allowance is also made for the bending of sound over the barrier as a result of atmospheric wind and temperature gradients (Bies et al., 2018).

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Sound Propagation Outdoors 15

Attenuation (dB)

10

(b)

(a) CAT 1

CAT 1

5

CAT 2 CAT 3 CAT 5 CAT 6

0

CAT 2 CAT 3 CAT 5 CAT 6

-5

-10

(c)

Attenuation (dB)

10

(d) CAT 1 CAT 2

CAT 1 CAT 2 CAT 3

5

CAT 3

0 CAT 5 CAT 6

-5

CAT 5 CAT 6

-10 15 (e)

Attenuation (dB)

10 5

(f)

CAT 2 CAT 1 CAT 3

CAT 2 CAT 1

CAT 3

0 CAT 5

-5

CAT 5 CAT 6

CAT 6

-10 -15 100

500 1000 2000 100 Distance from source (m) 15

1000 500 Distance from source (m)

2000

Attenuation (dB)

(g)

10

CAT 1 CAT 2

5

CAT 3

0 -5

CAT 5 CAT 6

-10 -15 100

1000 500 Distance from source (m)

2000

FIGURE 4.11 CONCAWE meteorological curves for various octave bands. (a) 63 Hz, (b) 125 Hz, (c) 250 Hz, (d) 500 Hz, (e) 1000 Hz, (f) 2000 Hz, (g) 4000 Hz.

4.7.7

In-Plant Screening, K 7

Manning found that in-plant screening was only significant for large petrochemical plants and for these, a value for K7 was difficult to estimate due to the limited amount of data available. However, a conservative estimate would be to set K7 = 0. CONCAWE also ignores the screening effect of housing.

190

4.7.8

Noise Control: From Concept to Application

Vegetation Screening, Kv

Marsh (1982) suggested that the following equation could be used to estimate the excess attenuation of sound (in an octave band of centre frequency, f ) as a result of travelling a distance, r, through vegetation such as a forest. Marsh did point out that this equation was likely to underestimate the attenuation in European forests. Kv = 0.01rf 1/3

(4.31)

For sound propagating a distance, r, through long grass or shrubs, the attenuation may be calculated using: Kv = (0.18 log10 f − 0.31)r (4.32)

4.7.9

Limitations of the CONCAWE Model

The limitations of the CONCAWE model are: • The model has been validated for distances from the source to the receiver between 100 m and 2000 m. • The model was developed using empirical data from petrochemical plants with noise sources less than 20 m in height. When used with the hard ground option, it has been shown to overpredict wind farm noise (Evans and Cooper, 2011). • The model is only valid for wind speeds less than 7 m/s. • The model is only valid for octave band analysis for octave band centre frequencies ranging from 63 Hz to 4000 Hz. Of course, it is also valid for calculating overall A-weighted sound pressure levels over the frequency range covered by the 63 Hz to 4000 Hz octave bands. • The attenuation due to the ground is based on a very simplified model derived from experimental data taken at two airfields. Its accuracy is questionable for propagation over other types of ground surface. • The model used to calculate the excess attenuation due to vegetation is not reliable. • The calculation of the attenuation due to shielding by terrain is oversimplified and may overestimate the terrain shielding effect. • Most software packages that implement the CONCAWE model only implement its standard form, in which case, the following aspects discussed in this section are excluded. 1. Vegetation effects, and if they are included, they are limited to forests. 2. Ground-reflected rays are not included when considering barrier attenuation; only sound rays travelling over the top of the barrier with no ground reflection are included. 3. The reduced attenuation of barriers in the presence of a downward refracting atmosphere (downwind and/or temperature inversion conditions) is not taken into account. 4. Terrain shielding is not included.

4.8

ISO 9613-2 (1996) Noise Propagation Model

The ISO 9613-2 (1996) standard is a noise propagation model that recommends calculations be done in octave bands with centre frequencies between 63 Hz and 8 kHz, and that for each octave

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Sound Propagation Outdoors

band, the same basic equation as used by the CONCAWE model (Equation (4.27)) be used. The ISO 9613-2 (1996) model claims to be calculating results for worst case (downwind) weather conditions, yet there is no explicit correction that would indicate this meteorological condition. In fact the meteorological correction that is suggested for the overall A-weighted level reduces the calculated sound pressure levels by a factor related to the percentage of time that worst case downwind weather conditions exist. Thus, it seems that the ISO 9613-2 (1996) method may not produce sound pressure levels corresponding to worst case weather conditions when there are no barriers present. The equation used to calculate the sound pressure level, Lpik , at location, k, due to the sound power, LW i (dB re 10−12 W), radiated by sound source, i, with a directivity index, DIik , in the direction of the receiver location, is: Lpik = LW i + DIik − AEik

(dB re 20 µPa)

(4.33)

The overall A-weighted sound pressure level is calculated by applying the A-weighting correction (see Table 2.4) to each octave band level and then summing the levels logarithmically (see Section 1.12.2). The overall C-weighted sound pressure level can be calculated by following a similar procedure. In the standard, the excess attenuation AEik is given by: AEik = [Adiv + Aatm + Agr + Abar + Amisc ]ik

(4.34)

The geometrical divergence attenuation component, Adiv , is calculated assuming a point source using Equation (4.5) as described in Section 4.3 and the component, Aatm , is calculated using Equation (4.10), as described in Section 4.4. Both of these excess attenuation effects are applied to the final total calculated sound pressure level at the receiver for each sound source. The attenuation components, Agr and Abar , are equivalent to the CONCAWE attenuation components, K3 and K6 , respectively, but are calculated differently. The ISO model only includes a correction to the long-time average A-weighted sound pressure level due to meteorological effects for downwind sound propagation only and for distances from the source greater than 10 times the sum of the source and receiver heights. The attenuation components, Agr and Abar , are discussed separately in Sections 4.8.1 and 4.8.4, respectively. The attenuation component, Amisc , is made up of the following subcomponents. • Attenuation, Asite , due to propagation of sound through an industrial site on its way from the source to the receiver (identical to the quantity, K7 , in the CONCAWE model). • Attenuation, Ahous , due to propagation of the sound through a housing estate on its way from the source to the receiver. • Attenuation due to propagation of sound through foliage, (Afol ). This the same as the quantity, Kv , which is estimated in Section 4.7.8. However, the calculation of Afol relies on a different procedure, which is outlined in Section 4.8.7 below.

4.8.1

Ground Effects, Agr

The method recommended in ISO 9613-2 (1996) for calculating the ground effect is moderately complex and yields results of moderate accuracy. In this method, the space between the source and receiver is divided into three zones: source, middle and receiver zones. The source zone extends a distance of 30hS from the source towards the receiver and the receiver zone extends 30hR from the receiver towards the source. The middle zone includes the remainder of the path between the source and receiver and will not exist if the source–receiver separation (projected onto the ground plane) is less than d = 30hS + 30hR , where hS and hR are defined in Figure 4.2.

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Noise Control: From Concept to Application

The acoustic properties of each zone are quantified using the parameter, Gg . According to ISO 9613-2 (1996), this parameter has a value of 0.0 for hard ground, a value of 1.0 for soft (or porous) ground and for a mixture of hard and soft ground it is equal to the fraction of ground that is soft. However, a more recent document from the European Commission (Kephalopoulos et al., 2012) lists a more refined gradation for values of Gg , as shown in Table 4.8. It is assumed that for downwind propagation, most of the ground effect is produced by the ground in the vicinity of the source and receiver so the middle part does not contribute much. TABLE 4.8 Flow resistivities for some common ground surface types (Values of Gg in brackets are the original values recommended in ISO 9613-2 (1996))

Representative flow resistivity R1 (kPa s/m2 )

Ground surface class

Value of ISO9613-2 parameter, Gg

A

1 (1)

12.5

Very soft (snow or moss)

Ground surface description

B

1 (1)

31.5

Soft forest floor

C

1 (1)

80

Uncompacted, loose ground

D

1 (1)

200

Normal uncompacted ground (pastures, forest floors)

E

0.7 (0)

500

F

0.3 (0)

2000

G

0 (0)

20000

H

0 (0)

200000

Compacted fields, lawns and gravel Compacted dense ground (gravel road, parking lot)

Asphalt, concrete Water

The total excess attenuation due to the ground is the sum of the excess attenuations for each of the three zones. That is: Agr = AS + Amid + ARec

(4.35)

Values for each of the three quantities on the right-hand side of Equation (4.35) may be calculated using Table 4.9. Note that if the source–receiver separation distance is much larger than their heights above the ground, then d ≈ dSR . See Figure 4.2 for a definition of the quantities, d and dSR . TABLE 4.9 (1996))

Octave band ground attenuation contributions, AS , ARec and Amid (after ISO 9613-2

63 AS (dB) ARec (dB) Amid (dB)

−1.5 −1.5 −3q 1000

AS (dB) ARec (dB) Amid (dB)

−1.5 + GS dS −1.5 + GR dR −3q(1 − Gm )

Octave band centre frequency (Hz) 125 250 −1.5 + GS aS −1.5 + GR aR −3q(1 − Gm )

−1.5 + GS bS −1.5 + GR bR −3q(1 − Gm )

Octave band centre frequency (Hz) 2000 4000 −1.5(1 − GS ) −1.5(1 − GR ) −3q(1 − Gm )

−1.5(1 − GS ) −1.5(1 − GR ) −3q(1 − Gm )

500 −1.5 + GS cS −1.5 + GR cR −3q(1 − Gm ) 8000 −1.5(1 − GS ) −1.5(1 − GR ) −3q(1 − Gm )

In Table 4.9, GS , GR and Gm are the values of Gg corresponding to the source zone, the receiver zone and the middle zone, respectively. The quantity, Amid , is zero for source–receiver

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Sound Propagation Outdoors

separations of less than dSR = 30hS + 30hR , and for greater separation distances it is calculated using the lines labelled Amid in Table 4.9 with: q =1−

30(hS + hR ) d

(4.36)

The coefficients aS , bS , cS and dS and the coefficients aR , bR , cR and dR of Table 4.9 may be calculated using the following equations. In each equation, hS,R is replaced with hS for calculations of aS and by hR for calculations of aR . 2

2



aS , aR = 1.5 + 3.0e−0.12(hS,R −5) (1 − e−d/50 ) + 5.7e−0.09hS,R 1 − e−2.8×10 2

bS , bR = 1.5 + 8.6e−0.09hS,R (1 − e−d/50 ) cS , cR = 1.5 + 14.0e

−0.46h2S,R

dS , dR = 1.5 + 5.0e

−0.9h2S,R

(1 − e

−d/50

)

(1 − e−d/50 )

−6

×d2



(4.37) (4.38) (4.39) (4.40)

There is yet another more complex and maybe more accurate method that involves the use of the spherical wave reflection coefficient and this is discussed in detail in Section 5.2.3 of Bies et al. (2018). Example 4.5 A square opening (3 m × 3 m) in the side of a building is leaking noise into the surrounding community. The internal room surfaces are sufficiently hard that the sound field incident at the opening may be considered diffuse. The centre of the opening is 3 m above ground level. The closest community location is at a distance of 150 m from the opening in a direction normal to the plane of the opening. The sound power radiated through the opening in the 2000 Hz octave band is 2 W. The ground between the opening and the community is grass covered. The following questions refer only to the 2000 Hz octave band. (a) Calculate the sound power level LW radiated through the opening. (b) Calculate the excess attenuation Agr due to ground reflection for sound travelling from the opening to the nearest community location (1.5 m above ground) (use Figure 4.3). (c) Calculate the loss due to atmospheric absorption (in dB). Assume RH = 25%, and temperature = 20◦ C. (d) Ignoring all other losses not mentioned above, calculate the sound pressure level at the community location of (b) above. (e) If a second opening of the same size (and radiating the same power), with its centre horizontally 5 m from the centre of the existing opening, were introduced, what would be the total sound pressure level at the community location of (b) above? Solution 4.5 (a) From equation (1.65), the sound power level, LW = 10 log10 W + 120 = 123 dB. (b) The arrangement for calculating the ground effect is shown in Figure 4.12. From similar triangles, x = 2y. Thus x = 100 and y = 50. tan β = 3/100, so β = 1.72◦ . For grass covered ground, R1 = 2.25×105 (middle of the range – see Table 4.2). Thus: ρf 1.206 × 2000 = = 0.011, and R1 2.25 × 105



R1 β ρf

1/2

1 = 1.72 0.011



1/2

= 16.6.

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Noise Control: From Concept to Application

150 m hole R

3m

b

x

b

1.5 m

y

FIGURE 4.12 Arrangement for Example 4.5.

From Figure 4.3, Arf = 1.3 dB. The ground effect is then: Agr = −10 log10 1 + 10−Arf /10 = −2.4 dB.





Thus the effect of the ground is to increase the level at the receiver by 2.4 dB. (c) Loss due to atmospheric absorption. From Table 4.1, Aatm = 15.5 dB per 1000 m (25% RH and 20◦ C). So for a distance of 150 m, Aatm = 15.5 × 0.15 = 2.3 dB. (d) The opening should be treated as an incoherent plane source (as the interior sound field in the room may be treated as diffuse, due to all of the room surfaces being √ hard) and Equation (3.38) used to calculate the sound pressure level. However, r/ HL = 150/3 = 50, so from Figure 3.18, it can be seen that the receiver is far enough from the source for it to appear as a point source and we may use Equation (3.39). Thus: hp2 i =

1.206 × 343 × 2 = 5.852 × 10−3 Pa2 . 2π × 1502

From Equation, (1.62), the sound pressure level is: Lp = 10 log10

5.852 × 10−3 = 71.7 dB re 20 µPa. 4 × 10−10

Aatm + Agr = −0.1, thus the sound pressure level at the receiver is equal to 71.8 dB. (e) Adding a second opening will add 3 dB 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 location of (b) above would be 74.8 dB. The different position of the second source will have a negligible effect.

4.8.2

Meteorological Effects, Amet

The ISO model provides a meteorological correction term for the overall long-time average Aweighted sound pressure level. where the average is over several months to a year. Essentially, the correction is to allow for the fact that downwind propagation does not occur 100% of the time. The corrections are only to be subtracted from the overall A-weighted level calculations (not individual octave band values) and they are not to be included for locations closer to the source than ten times the sum of the source and receiver heights. The correction to account for downwind propagation not occurring 100% of the time is: Amet = A0 [1 − 10(hS + hR )/d]

(dB)

(4.41)

where hS and hR are the source and receiver heights, respectively, and d is the horizontal distance between the source and receiver. The quantity, A0 , depends on local meteorological statistics and varies between 0 and 5 dB with values over 2 dB very rare.

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Sound Propagation Outdoors

4.8.3

Source Height Effects

The ISO9613 model is only valid for source heights less than 30 m. No correction is provided for higher sources.

4.8.4

Barrier Attenuation, Abar

For an obstacle to be classified as a barrier for which the calculations in this section are valid, it must satisfy the following conditions. 2

• Obstacle mass must be greater than 10 kg/m . • No cracks or gaps that would allow sound to travel through. • The obstacle length normal to the line between the sound source and receiver should be greater than a wavelength at the frequency corresponding to the lower frequency limit of the octave band centre frequency of interest. Obstacles that fulfil the above conditions are replaced for the purposes of the calculation with an equivalent flat rectangular panel with height above the ground equal to the average height of the obstacle. For any path, zi, i = 1 − 8, over or around the barrier, the octave band excess attenuation, Abar , is given by:

h

Abar = −10 log10 10−(Dz1 −Agr )/10 + 10−(Dz2 +Arf 2 −Agr )/10 + 10−(Dz3 +Arf 3 −Agr )/10 + 10−(Dz4 +Arf 4 −Agr )/10 + 10−Dz5 /10 + 10−Dz6 /10 + 10−(Dz7 +Arf 7 )/10 + 10−(Dz8 +Arf 8 )/10

i

(4.42) where Abar is set = 0 if Abar < 0. Dzi is calculated for path zi using Equation (4.43). The use of Equation (4.42) is a more straightforward alternative way to that described in ISO 9613-2 (1996) for finding the excess attenuation due to a barrier. ISO 9613-2 (1996) proceeds by calculating the sound pressure level at the receiver as a result of each path, combining the results to obtain a total sound pressure level and then comparing the result to the total sound pressure level without a barrier to find the excess attenuation due to the barrier. The (ISO 9613-2, 1996) method is efficient if only one path is considered but becomes tedious when more than one path is considered. In Equation (4.42), Dz1 refers to the non-reflected path over the top, Dz2 , Dz3 and Dz4 refer to the three paths over the top that have a ground reflection loss of Arf 2 , Arf 3 and Arf 4 dB, respectively, Dz5 and Dz6 refer to the two non-reflected paths around the ends of the barrier, and Dz7 and Dz8 refer to the two ground-reflected paths around the ends of the barrier (one at each end) that have a ground reflection loss of Arf 7 and Arf 8 dB, respectively. Any of the terms may be omitted from Equation (4.42) if it is not desired to consider a particular path. In particular, if C2 = 20 is used in Equation (4.43), then the terms corresponding to i values of 2, 3, 4, 7 and 8 must be excluded from Equation (4.42). If diffraction around the ends of the barrier is to be ignored, then paths 5–8 are excluded from Equation (4.42). The paths around the barrier and the calculation of their lengths are described in Section 7.5.2. Usually, C2 = 20 is chosen in Equation (4.43), so that the only terms to be used in Equation (4.42) are those corresponding to i = 1, 5 and 6. In Equation (4.42), the term, Agr , which is the ground attenuation in the absence of the barrier (usually negative), is only included for paths that go over the top of the barrier. However, the Agr term is not included in Equation (4.42) for paths over the top of the barrier when the barrier is shielding multiple sources and is a building more than 10 m high, or if the sound

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Noise Control: From Concept to Application

sources are high (ISO 9613-2, 1996). It is important to note that the Agr term is still included in Equation (4.34), which is used to calculate the total excess attenuation due to all effects. The excess attenuation, Dzi , i = 1 − 8, for any propagation path over and around the ends of the barrier is given by: Dzi = 10 log10 [3 + (C2 /λ)C3 Kmet ∆zi ]

(dB)

(4.43)

where λ is the wavelength of sound at the octave band centre frequency, C2 = 20 if groundreflected paths over and around the barrier are included and C2 = 40 if ground-reflected paths are calculated separately. The coefficient, C3 = 1 for single edge diffraction (Figure 4.13). For double edge diffraction (Figure 4.14), C3 is: 1 + (5λ/e)2 (4.44) C3 = (1/3) + (5λ/e)2 The quantity, ∆zi is the difference between the line-of-sight distance between source and receiver and the path length over the top or around the side of the barrier, for path, i.

Source at (XS ,YS ,hS )

Barrier of height, hb

A1

B1

dSR1

hS

Receiver at (XR ,YR ,hR )

hb YS

YR

Y

hR

XR

XS X

FIGURE 4.13 Geometry for determining the additional path length between source and receiver as a result of sound propagation over a single-edge barrier.

Double-edge barrier b Source at (XS ,YS ,ZS )

A1

ZS = hS

e B1

dSR1

Receiver at (XR ,YR ,ZR )

hb1 hb2

YS a YR

Y XS

ZR = hR

XR X

FIGURE 4.14 Geometry for double edge diffraction (single, thick barrier or two thin, parallel barriers).

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Sound Propagation Outdoors

The upper allowed limit of the attenuation, Dz1 , for the diffracted wave over the barrier top is 20 dB for single-edge diffraction and 25 dB for double-edge diffraction. If ground-reflected rays are treated separately, the upper diffraction limit is not applied to those rays; it is only applied to the non-reflected ray over the barrier top, prior to combining the attenuations of the various paths using Equation (4.42). No upper limit is applied to diffracted rays around the barrier ends. The path length difference, ∆z1 for use in Equation (4.43) for diffraction over the top of the barrier may be calculated using Figure 4.13 for single-edge diffraction (a thin obstacle) or Figure 4.14 for double-edge diffraction (a wide obstacle). If the barrier does not interrupt the line of sight between the source and receiver, ∆zi is negative. Otherwise, ∆zi is positive. From Figure 4.13 for single diffraction, the path difference, ∆z1 , is given by: ∆z1 = ±(A1 + B1 − dSR1 ) where

dSR1 = X 2 + Y 2 + (hR − hS )2 A1 = XS2 + YS2 + (hb − hS )2 2 B 1 = XR + YR2 + (hb − hR )

and

(4.45)

YR = Y XR /X

and


1/2


1/2

(4.46)


2 1/2

YS = XS YR /XR

(4.47)

From Figure 4.14 for double edge diffraction, the path difference, ∆z1 , for the two edges of heights, hb1 and hb2 , respectively, is given by: ∆z1 = ±(A1 + B1 + e − dSR1 )

(4.48)

where the quantities, A1 , B1 and dSR1 are calculated using the same equations as for the single diffraction barrier (Equation (4.46)) with the additional equation for edges of the same height: e = (a2 + b2 )1/2

(4.49)

and where in this case, X = XR + XS + b and Y = YR + YS + a. For edges of different heights, hb1 and hb2 : e = a2 + b2 + (hb1 − hb2 )2


1/2

(4.50)

The term, Kmet , in Equation (4.43) is a meteorological correction factor to the barrier attenuation for downwind propagation and is given by: 1 = exp − 2000



Kmet

r

A1 B1 dSR1 2(A1 + B1 + e − dSR1 )



(4.51)

where the dimension, e, is zero for single edge diffraction, all dimensions are in metres and exp(x)=ex . For diffraction around the vertical edge of a screen, or if ∆zi < 0, Kmet = 1.

4.8.5

In-Plant Screening, Asite

In large industrial facilities, significant additional attenuation can arise from shielding due to other equipment blocking the line of sight from the source to the receiver. Usually it is best to measure this, but if this cannot be done, it may be estimated using the values in line 1 of Table 4.10 as described in ISO 9613-2 (1996). The distance, r1 , to be used for the calculation is only that part of the curved sound ray (close to the source) that travels through the process equipment and the maximum attenuation that is expected is 10 dB. The distance is equivalent to distance r1 in Figure 4.15 with the foliage replaced by process equipment. However, in many cases, it is not considered worthwhile taking Asite into account, resulting in receiver sound pressure level predictions that may be slightly conservative.

198

4.8.6

Noise Control: From Concept to Application

Housing Screening, Ahous

The excess attenuation due to screening by housing (Ahous ) may be calculated using (ISO 9613-2, 1996): Ahous = 0.1Brb − 10 log10 [1 − (P/100)]

(dB)

(4.52)

where B is the density of buildings along the path (total plan area of buildings divided by the total ground area (including that occupied by the houses), rb is the distance that the curved sound ray travels through the houses and P is the percentage (≤ 90%) of the length of housing facades relative to the total length of a road or railway in the vicinity. The second term in Equation (4.52) is only used if there are well defined rows of houses perpendicular to the direction of sound propagation. The second term may also not exceed the calculated Insertion Loss for a continuous barrier the same height as the building facades (see Chapter 7). The quantity, rb is calculated in the same way as rf (= r1 + r2 ) in Figure 4.15 for travel through foliage, except that the foliage is replaced by houses. Note that if Ahous of Equation (4.52) is non zero, the ground attenuation through the built up area is set equal to zero, unless the ground attenuation with the buildings removed is greater than the first term in Equation (4.52) for Ahous . In that case, the ground attenuation for the ground without buildings is substituted for the first term in Equation (4.52).

4.8.7

Vegetation Screening, Afol

ISO 9613-2 (1996) gives the attenuation values in Table 4.10 for sound propagation through dense foliage. For distances less than 20 m, the values given are absolute dB. For distances between 20 and 200 m, the values given are dB/m and for distances greater than 200 m, the value for 200 m is used. TABLE 4.10 9613-2 (1996))

Octave band attenuation, Afol , due to dense foliage and process equipment(after ISO

63 Asite (dB/m) Afol (dB) for 10 m ≤ rf ≤20 m Afol (dB/m) for 20 m ≤ rf ≤200 m

0 0 0.02

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000 0.015 0 0.03

0.025 1 0.04

0.025 1 0.05

0.02 1 0.06

0.02 1 0.08

0.015 2 0.09

8000 0.015 3 0.12

The distance of travel through the foliage is not equal to the extent of the foliage between the source and receiver. It depends on the height of the source and receiver and the radius of curvature of the propagating ray as a result of wind and temperature gradients. ISO 9613-2 (1996) recommends that a radius of 5 km be used for downwind propagation. The centre (always below the ground plane) of the circular arc, representing the sound ray path from the source to the receiver, is easily found, using a scaled drawing, as the intersection of two lines of length equal to 5 km, with one line intersecting the source location and the other intersecting the receiver location. The distance rf = r1 + r2 , where r1 and r2 are defined in Figure 4.15. r1

Source

r2

Receiver

FIGURE 4.15 Path lengths for sound propagation through foliage.

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Sound Propagation Outdoors

4.8.8

Effect of Reflections Other Than Ground Reflections

If the receiver is close to the wall of a building, the expected sound pressure level will be increased as a result of reflection from the wall. ISO 9613-2 (1996) provides a means for calculating this effect and the procedure is summarised below. As can be seen from inspection of Figure 4.2, if a reflecting plane is present, sound arrives at the receiver by a direct path and also by a reflected path. If the reflecting plane is not the ground, but is a reflecting surface such as a tank or wall of a house, then provided the reflecting surface satisfies the conditions below, it can be taken into account using the procedure outlined here. Where the reflecting surface is a cylinder, the reflection point is deemed to be the centre of the cylinder cross section. The reflecting plane is represented by the line passing through the cylinder centre, which makes equal angles with the line from the cylinder centre to the source and the line from the cylinder centre to the receiver. • The magnitude of the surface absorption coefficient, αr , is less than 0.8. • The surface is sufficiently large in extent such that the following equation is satisfied: 2 1 > λ (Lmin cos θ)2





rS rR rS + rR



(4.53)

In equation (4.53), λ is the wavelength of the incident sound (at the 1/3-octave or octave band centre frequency of interest), Lmin is the minimum dimension (width or height) of the reflecting surface and rS , rR and θ are defined in Figure 4.2, where the ground is replaced by the reflecting surface. The sound pressure level at the receiver location is then determined by calculating the sound pressure levels due to the direct and reflected waves separately (using all the excess attenuation parameters for each of the direct and reflected waves) and then adding the two results logarithmically (see Section 1.12.2). The sound power used for the reflected wave, LW r , is derived from the sound power level of the source, LW , which is the sound power level used for the direct wave. Thus: LW r = LW + 10 log10 (1 − αr ) + DIr (dB re 10−12 W) (4.54) where DIr is the source directivity index in the direction of the reflecting surface and αr is the absorption coefficient of the surface. If measured absorption coefficient data are unavailable, then ISO recommends using the values in Table 4.11. TABLE 4.11

Estimates of sound absorption coefficient (derived from ISO 9613-2 (1996))

Object

αr

Flat, hard walls with no openings Walls of building with windows and small additions or bay Walls of building with 50% openings, installation or pipes

0.0 0.2 0.6 D sin(90 − θ) 1− 2rS where D=cylinder diameter (m) rS is the distance from the source to the cylinder centre and the cylinder centre is at location O in Figure 4.2



Cylinders with hard surfaces (tanks, silos)a

a



Applies only if the cylinder is much closer to the source than it is to the receiver.

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Noise Control: From Concept to Application

A number of adjustments were suggested during investigations of the most appropriate EU noise model (European Commission, 2010). The most appropriate ones (in addition to the refinement of the G values shown in Table 4.8) are listed below. • Section 7.3.2, which is an alternative method for calculating A-weighted sound pressure levels at the receiver, is to be entirely disregarded. • Under Section 7.4: ◦ The value of Dz calculated using Equation (14) in ISO 9613-2 (1996) (Dzi in Equation (4.43) above) cannot be less than 0. ◦ The quantity, C2 , in Equation (4.43) above is always equal to 20.

4.8.9

Limitations of the ISO9613-2 Model

The ISO9613-2 model, when applied to the prediction of noise in surrounding communities, has the following uncertainties and issues associated with it. • It has only been validated for distances between source and receiver of less than 1 km. • The model is only valid for octave band analysis for octave band centre frequencies ranging from 63 Hz to 4000 Hz. Of course, it is also valid for calculating overall A-weighted sound pressure levels over the frequency range covered by the 63 Hz to 4000 Hz octave bands. • Downwind propagation is assumed by the ISO model, but only wind speeds between 1 m/s and 5 m/s (measured between 3 m and 11 m above the ground) are valid. • Significant deviations from the ISO model may be expected for wind speeds above the 5 m/s limit (Kalapinski and Pellerin, 2009). • The ISO model has only been validated for source heights of less than 30 m above the ground. • No allowance is made for the attenuation effects of scattering due to atmospheric turbulence. • The ISO standard states that the expected accuracy of sound pressure level predictions is ±3 dB for source heights less than 30 m and for distances between the source and receiver of less than 1000 m. It is expected that the errors will increase as the source–receiver distance increases. Also as the source height increases above 30 m the calculation of the excess attenuation due to the ground becomes less accurate. Accuracy and prediction uncertainty are discussed in more detail in Section 4.9. • The ISO model assumes that sound propagation is from point sources, which makes the calculated sound pressure levels close to large sound sources subject to greater errors than those estimated above. Example 4.6 A new compressor is to be located in an industrial processing facility. The most sensitive community location is at a distance of 910 m from the edge of the industrial facility and the compressor is to be located another 90 m inside the facility (in the exact opposite direction from the facility boundary to the sensitive community location), with a 20 m band of process equipment of average height 6 m, located between 20 m and 40 m from the compressor. The existing sound pressure level due to other equipment in the facility at the community location is 40.0 dBA at night during moderate (2–3 m/s) downwind conditions, when the temperature is 20◦ C and the relative humidity is 50%. Other noise sources contribute insignificantly to the noise at this community location during downwind conditions. The ground between the facility and the community

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Sound Propagation Outdoors

location is grass covered and the ground from the compressor to the edge of the industrial facility is concrete. The acoustic centre of the compressor noise source is 3 m above the ground and the community location to be evaluated is at 1.5 m above the ground. The ground between the compressor and community location is flat. In the following calculations, state any assumptions that are made. The arrangement is illustrated in Figure 4.16.

20 m 20 m S

hmax R

hS hR

x

a

b

Rc

Rc

d - d0

d0 d

FIGURE 4.16 Arrangement for Example 4.6.

(a) For the purposes of this problem, we will assume that the compressor only radiates noise in the 250 and 500 Hz octave bands, and the sound power level in the 250 Hz band is 3 dB greater than in the 500 Hz band. If the allowable sound pressure level at the community location is 45 dBA for the atmospheric conditions stated above, what would be the maximum allowable sound power level (according to ISO 9613-2 (1996)) in the 500 Hz octave band of the new compressor so the allowable level will not be exceeded. (b) Repeat the calculations using the CONCAWE noise model. (c) It is proposed to place a wall between the compressor and the community location. If the wall is 5 m high and 10 m long, and is placed 2 m from the compressor, what will be the expected noise reduction according to ISO 9613-2 (1996) in the 250 and 500 Hz octave bands? What will be the allowable compressor sound power in the 500 Hz octave band in this case, according to the ISO 9613-2 (1996) model? (d) If the compressor radiates easily detectable tones, what would be the allowable sound power level in the 500 Hz band in that case (without the presence of the wall) according to the ISO 9613-2 (1996) model and according to the CONCAWE model if the relevant regulations stipulate a 5 dBA penalty for tonal noise. Solution 4.6 (a) As explained in Section 1.41, the total sound pressure level allowed in the 250 and 500 Hz octave bands is given by:
LpA (tot) = 10 log10 1045/10 − 1040/10 = 43.35 (dBA re 20 µPa). Rearranging Equation (4.2) gives LW i = Lpi + Adiv,i + AEi (dB re 10−12 W). √ The distance, dSR , between source and receiver is dSR = 10002 + 1.52 = 1000.0 m,

202

Noise Control: From Concept to Application which is practically the same as the horizontal distance, d. From Equation (4.5), Adiv = 10 log10 (4π × 10002 ) = 71.0 dB. The excess attenuation, AEi = [Aatm + Agr + Abar + Amet + Amisc ]i . The excess attenuation factors that make up AEi are calculated as follows. From Table 4.1 Aatm (250) = 1.3 dB Aatm (500) = 2.7 dB Ground effect calculations are undertaken in accordance with Section 4.8.1. Agr = AS + Amid + ARec , where GS = 1.0, Gmid = GRec = 0. In both 250 Hz and 500 Hz bands, AS = −1.5 + GS bS = −1.5. In both 250 Hz and 500 Hz bands,
= −3 × 0.865 = −2.6 dB. Amid = −3q(1 − Gmid ) = −3 1 − 30×4.5 1000 In the 250 Hz band,
2 ARec = −1.5 + bR = −1.5 + 1.5 + 8.6e−0.09×1.5 1 − e−1000/50 = 7.0 dB. In the 500 Hz band,
2 ARec = −1.5 + cR = −1.5 + 1.5 + 14e−0.46×1.5 1 − e−1000/50 = 5.0 dB. Thus Agr (250) = −1.5 − 2.6 + 7.0 = 2.9 dB, and Agr (500) = −1.5 − 2.6 + 5.0 = 0.9 dB. For meteorological effects, the final A-weighted level is corrected which means the same correction is applied to all octave bands. Assuming A0 = 1.5 dB, Amet = 1.5(1 − 45/1000) = 1.4 dB in both octave bands. Amisc = Asite , as Ahouse and Afol are zero due to there being no housing nor foliage between the source and receiver. Calculating Asite to account for the 20 m of process equipment is complicated so is often omitted to provide a conservative result. However, it will be done here to illustrate the process. The ISO 9613-2 (1996) model assumes a radius of curvature of the downwind propagating wave, Rc = 5000 m. We need to calculate the distance that this curved ray path travels through the process equipment. Referring to Figure 4.16, we can write the following equations. d20 + (x + hS )2 = Rc2 and p (d − d0 )2 + (x + hR )2 = Rc2 , which can be re-written as: d0 = d − Rc2 − (x + hR )2 . p From the first equation, x = Rc2 − d20 − hS . Solving the above two equations iteratively gives d0 = 492.54 m and x = 4972.68 m. Therefore hmax = 5000 − 4972.7 = 27.3 m. Referring p to Figure 4.16 again, we can see that: a = Rc2 − (d0 − 20)2 = 4977.62 m and p b = Rc2 − (d0 − 40)2 = 4979.48 m. Comparing the value of b to (hS + x), the difference in height of the sound ray at the source compared to the far side of the process equipment is 4979.48−4975.68 = 3.8 m. As the process equipment is only 3 m higher than the source, the sound ray only travels partway through the process equipment. The distance from the source at which the sound ray is equal in height to the average of the process equipment is found, by trial and error adjustment of the “20” value in the above equation for a, to be approximately 31 m. Thus the sound ray travels 11 m through the process equipment. From table 4.10, this results in an excess attenuation Asite (250) = Asite (500) = 0.3 dB. The total excess attenuation in the 250 Hz octave band is 1.3+2.9+1.4+0.3=5.9 dB. The total excess attenuation in the 500 Hz octave band is 2.7+0.9+1.4+0.3=5.3 dB. The sound power level in the 250 Hz octave band is 3 dB greater than in the 500 Hz

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Sound Propagation Outdoors octave band. Let the sound power level in the 500 Hz octave band be x dB. The A-weighting for the 250 Hz band is -8.6 dB and for the 500 Hz band it is -3.2 dB (see Table 2.4). The overall allowed A-weighted sound pressure level at the receiver is then given, in terms of x, by: Lp = 10 log10 10(x−3.2−5.3−71.0)/10 + 10(x+3−8.6−5.9−71.0)/10 = 43.35 dBA. 1043.35/10 = 1.284 × 1012 . Thus, 10x/10 = (−3.2−5.3−71.0)/10 10 + 10(3−8.6−5.9−71.0)/10 So x = 10 log10 1.284 × 1012 = 121.1 dB (re 10−12 W) = maximum allowable sound power level in the 500 Hz octave band. (b) Repeating with the CONCAWE model, the only changes are with the excess attenuation associated with the ground effect, meteorological effects and process equipment screening.




Note that K1 is the attenuation due to spherical spreading and is 71.0 dB. From Table 4.1, K2 (250) = 1.3 dB and K2 (500) = 2.7 dB. From Figure 4.10, K3 (250) = 13.0 dB. From Figure 4.10, K3 (500) = 8.6 dB. The meteorological effects may be calculated using Tables 4.6 and 4.7, where it can be seen that we have meteorological category 6. Thus, from Figure 4.11: K4 (250) = −6.1 dB and K4 (500) = −6.8 dB. Source height effects must be considered as K3 + K4 > −3. From Figure 4.2, the angle θ is given by θ = tan−1 [d/(hS + hR )] = tan−1 [1000/4.5] = 89.742 degrees. From Equation (4.30), γ = 1.08 − 0.478 × 0.258 + 0.068 × 0.2582 − 0.0029 × 0.2583 = 0.961. Thus, K5 (250) = (K3 + K4 + 3) × (−0.039) = (13.0 − 6.1 + 3) × (−0.039) = −0.4 dB. K5 (500) = (8.6 − 6.8 + 3) × (−0.039) = −0.2 dB. K6 = K7 = Kv = 0. The total excess attenuation in the 250 Hz octave band is: 1.3 + 13.0 − 6.1 − 0.4 = 7.8 dB. The total excess attenuation in the 500 Hz octave band is: 2.7 + 8.6 − 6.8 − 0.2 = 4.3 dB. Let the sound power level in the 500 Hz octave band be x dB (re 10−12 W). The overall A-weighted sound pressure level at the receiver, in terms of x, is: Lp = 10 log10 10(x−3.2−4.3−71.0)/10 + 10(x+3−8.6−7.8−71.0)/10 = 43.35 dB (re 20 µPa).




1043.35/10 = 1.218 × 1012 . + 103−8.6−7.8−71.0/10 ) So x = 10 log10 1.218 × 1012 = 120.9 dB (re 10−12 W) = maximum allowable sound power level in the 500 Hz octave band, which is 0.2 dB less than calculated using the ISO model. (c) According to ISO 9613-2 (1996), the excess attenuation as a result of diffraction over the top of a barrier or around each side is given by Equation (4.43). The quantity, Kmet , in Equation (4.43) is defined in Equation (4.51), where the variables are illustrated in Figure 4.17. For diffraction over the top, we should take into account that the effective source height will be slightly higher than the actual source height due to curvature of the Thus, 10x/10 =

(10(−3.2−4.3−71.0/10

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Noise Control: From Concept to Application

998 m

2m A

B

S Hb

hS

dSR

R

Elevation view

hR

d = 1000 m

B A

5m

Plan view

S

R

5m

FIGURE 4.17 Wall arrangement for Example 4.6(c).

sound ray. This effect is not included as a requirement in ISO 9613-2 (1996), but is included here for completeness and for illustrating how ISO 9613-2 (1996) may be modified to include ray curvature effects on barrier attenuation. Referring to Figure 4.18, as the ray path is now curved, the new distance, A0 , from the source to the top of the barrier is longer than the actual path, A, as shown in Figure 4.18 and is equal to Rc θ. The extent of the diffraction that will be experienced by the ray that is incident on the barrier top is a function of the angle between the barrier face and the incident ray. This angle changes from β to (90 − α), as illustrated in Figure 4.18. The angle, α, is found by drawing a tangent to the ray path at the top of the barrier and extending this line on the source side of the barrier a distance equal to the actual path length, which is the curved ray path length, A0 , shown in Figure 4.18. The end of this line is then the effective source position. With this new source position, the barrier calculations can proceed as they would for a situation of no ray curvature, but with the source moved to its new “effective” source position. All that is left is to find the variables that define the new position, `0S (horizontal distance from the barrier face to the “effective” source position) and h0S (increase in height of the “effective” source position compared to the original source position). By definition, A0 is equal in length to the curved ray path. Thus the variables `0S , and h0S may be calculated given the radius of curvature, Rc , using simple trigonometry to obtain the following equations: A0 = Rc θ `0S = Rc θ cos α h0S = Hb − hS − Rc θ sin α α = 21 (π − θ) − β β = cos−1 [(Hb − hS )/A] The cosine rule for θ is : A2 = Rc2 + Rc2 + 2Rc Rc cos θ As θ must be between 0 and 90 degrees, θ = cos−1 [1 − (A2 /2Rc2 )];

√ Rc > A/ 2

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Sound Propagation Outdoors

Wind

Curved sound ray trace

Effective source position

a A'

Top of barrier

b Hb-hS

h'S

A Source

RS R'S

hS Rc

a

q

FIGURE 4.18 Wall arrangement including sound ray curvature for Example 4.6(c).

where Hb −hS is the actual difference between source and barrier heights, and A is the distance from the actual source to the barrier top. Substituting values into the above −1 2 2 2 −4 equations gives √ θ = cos [1 − ((5 − 3) + 2 )/(2 × 5000 )] = 5.657 × 10 radians. −1 2 2 β = cos [2/ 2 + 2 ] = 0.78540 radians α = 12 (π − 5.657 × 10−4 ) − 0.78540 = 0.78511 radians h0S = 5 − 2 − 5000 × 5.657 × 10−4 sin(0.78511) = 5.66 × 10−4 m `0S = 5000 × 5.657 × 10−4 cos(0.78511) = 2.00057 m As h0S is very small and `0S is close to `S (2 m), we can ignore the ray curvature effect and use the arrangement shown in Figure 4.17. The ∆ztop = A + B − d ppath length difference, √ = (2)2 + (2)2 + 9982 + 3.52 − 1000 = 2.828 + 998.006 − 1000 = 0.834 m. C2 = 20, C3 = 1. From r Equation (4.51), with e= 0:  1 2.828 × 998.006 × 1000 = 0.522. Kmet = exp − 2000 2 × 0.834 Wavelength, λ250 = 343/250 = 1.372 m and λ500 = 343/500 = 0.686 m. Using Equation (4.43): Dtop (250) = 10 log10 [3 + (20/1.372) × 0.522 × 0.834] = 9.7 dB. Dtop (500) = 10 log10 [3 + (20/0.686) × 0.522 × 0.834] = 12.0 dB. For√diffraction√around the sides, the path length difference, ∆zside = A + B − d = 22 + 52 + 9982 + 52 − 1000 = 5.385 + 998.002 − 1000 = 3.387 m. C2 = 20, C3 = 1. From r Equation (4.51):   1 5.385 × 998.002 × 1000 Kmet = exp − = 0.641. 2000 2 × 3.387 Dside (250) = Abar (250) = 10 log10 [3 + (20/1.372) × 0.641 × 3.387] = 15.4 dB. Dside (500) = Abar (500) = 10 log10 [3 + (20/0.686) × 0.641 × 3.387] = 18.2 dB. To obtain the overall excess attenuation (or noise reduction) due to the barrier, we use Equation (4.42). Thus,   Abar (250) = −10 log10 10−9.7/10 + 10−(15.4+2.9)/10 − 2.9 = 6.2 dB  −12.0/10  Abar (500) = −10 log10 10 + 10−(18.2+0.9)/10 − 0.9 = 10.3 dB

206

Noise Control: From Concept to Application Adding the wall excess attenuations to the other excess attenuations, we can use the same previous equations to calculate the allowed sound power level of the compressor.

Thus, Lp = 10 log10 10(x−3.2−5.3−10.3−71.0)/10 + 10(x+3−8.6−5.9−6.2−71.0)/10 = 43.35 dB (re 20 µPa) 1043.35/10 and 10x/10 = = 2.065 × 1013 (−3.2−5.3−10.3−71.0)/10 (10 + 10(3−8.6−5.9−6.2−71.0)/10 ) So x = 10 log10 2.065 × 1013 = 133.2 dB (re 20 µPa) = maximum allowable sound power level in the 500 Hz octave band. (d) If a 5 dBA penalty is imposed due to the presence of tonal noise, the maximum allowable level will be 40 dBA re 20 µPa, so the compressor will be able to increase the existing level insignificantly (or by less than 0.5 dBA), which means its contribution is only allowed to be 30 (dBA re 20 µPa) (10 dBA less than 40, which when added to 40 logarithmically represents an increase of between 0.4 and 0.5 dBA). So in the equations in parts (a) and (b), we can replace the “43.35” quantity with “30”. This results in an allowable sound power level in the 500 Hz band of 107.7 dB (re 10−12 W) for the ISO model and 107.5 dB (re 10−12 W) for the CONCAWE model.




4.9

Propagation Model Prediction Uncertainty

The accuracy of noise propagation modelling is often the subject of discussions in court, especially when predicted sound pressure levels are close to allowable sound pressure levels. When testing a prediction model with measured data, uncertainty in the comparison is a function of the uncertainty in the prediction model results and the uncertainty in the measured results. The uncertainty in the predicted results is a function of the accuracy of the propagation model in representing all of the physical parameters that influence the sound pressure level at the receiver. This type of uncertainty is often expressed as an accuracy and is an estimate based on experience. It is referred to in the statistics literature as a type B standard uncertainty. The uncertainty in the measured results is discussed in detail in Section 2.11.4, and as it is based on the statistics of measured data, it is referred to as a Type A standard uncertainty. Uncertainty is most conveniently expressed as an expanded uncertainty, ue . These uncertainty types are discussed in the following sections.

4.9.1

Type A Standard Uncertainty

This type of uncertainty estimate is usually associated with repeated measurements to obtain a number of measurement sets, each of which is characterised by a mean and standard deviation. The means of all of the data sets also follow a normal distribution, characterised by its own mean and standard deviation. However, the standard deviation, us , of the means of all data sets can be calculated from the standard deviation, σ, of a single data set using Equation (4.55), where N is the number of measurements in the data set. To avoid confusion, us is referred to as the standard uncertainty. σ us = √ (4.55) N where, for Nm individual measurements, Lpi, i=1,...Nm , for which the mean is Lpm , the standard deviation is defined as: s PNm Lpi /10 − 10Lpm /10 )2 i=1 (10 σ= (4.56) (Nm − 1)

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Sound Propagation Outdoors

1 PNm Lpi /10
10 dB. For outdoor sound pressure level measurements, Nm i=1 Lp,i is usually an Leq measurement averaged over 10 minutes. The terms, “accuracy” and “range” do not apply to this type of uncertainty. This type of uncertainty could be applied to a series of sound pressure level measurements at a single location to determine the uncertainty in the estimate of the mean sound pressure level (see Section 2.11.4). where Lpm = 10 log10

4.9.2

Type B Standard Uncertainty

Type B uncertainties are not based on statistics but usually on judgement and experience. This type of uncertainty usually assumes a rectangular distribution of values between an upper and a lower limit (range or accuracy) or in terms of an accuracy. The terms, “accuracy” and “range”, are used interchangeably and are expressed as ±x dB. In many cases, the expected occurrence of all values between the maximum and minimum limits is considered equally likely, so the distribution is rectangular rather than √ normal. In this case, the standard uncertainty, us , is related to the accuracy, x, by us = x/ 3 and this is what is used in assessing the 95% confidence limits for a sound propagation model. In cases where the occurrence of values increases in likelihood near the centre of the range, a normal distribution may be more appropriate and in this case, the standard uncertainty, us , is related to the accuracy or range, x, by us = x/2, where x (dB) is half the difference between the maximum and minimum values of the sound pressure level.

4.9.3

Combining Standard Uncertainties

Type A and Type B standard uncertainties can be combined by summing the squares of each standard uncertainty and then taking the square root of the result, provided that all quantities contribute equally to the quantity for which the overall standard uncertainty is to be determined. If the quantities that are represented by the various uncertainties contribute in different amounts to the quantity for which the overall uncertainty is to be determined, the individual uncertainties must be weighted by their relative contribution. For standard uncertainty, us,i , this weighting is referred to in ISO 1996-2 (2017) as a fractional quantity, ci . For N sound sources contributing different amounts to the total sound pressure level at a receiver, where the contribution from source, i, is represented by a standard uncertainty, ui (dB), the overall uncertainty, us,tot , is given by:

q PN us,tot =

us,i × 10Li /10

i=1

PN

i=1


2

10Li /10

(dB)

(4.57)

where Li is the sound pressure level contribution at the receiver from source, i. Equation (4.57) shows that as the number of sources contributing significantly to the sound pressure levels at a receiver increases, the standard uncertainty of the sound pressure level prediction at a specific receiver will decrease. The individual standard uncertainties in Equation (4.57), for the case of sound source, i, are themselves made up of two standard uncertainties: that due to the uncertainty in the sound power level data, us,source , (see Section 3.11.4), and that due to the uncertainty associated with the propagation model, us,prop . Thus: us,i =

q

u2s,source + u2s,prop

(4.58)

The uncertainty associated with the propagation model can itself be made up of a number of terms, with each term contributing to one of the attenuation effects that make up the propagation model. There would be a term for each of the following parameters.

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Noise Control: From Concept to Application

These uncertainties would be combined as indicated by Equation (2.39), with sensitivity coefficients set equal to 1 in the absence of any other information. Usually propagation model uncertainty is expressed as an overall uncertainty (or accuracy) based on experience, as it is simpler and avoids the necessity of estimating the individual uncertainties associated with the items in the above list.

4.9.4

Expanded Uncertainty

The expanded uncertainty, ue , is calculated by multiplying the standard uncertainty, us , by a factor to account for the uncertainty confidence limits. For a confidence limit of 95% (that is, a 95% chance that the true sound pressure level will be in this range), the factor is 1.96 if a twosided normal distribution of the standard error is assumed. However, if the standard uncertainty is related to a one-sided distribution, so that it reflects the chance that a predicted value will be greater than the actual value (or alternatively, less than the actual value), then the expanded uncertainty for a 95% confidence limit is calculated by multiplying the standard uncertainty by 1.65 (instead of 1.96), which makes the expanded uncertainty very close numerically to the accuracy or range of possible values. The expanded uncertainty for a two-sided distribution is the quantity that should be used in assessing the uncertainty associated with acoustic predictions and measurements. The expanded uncertainty for a 95% confidence limit may be interpreted as there being a 95% chance that the difference between the predicted (or measured) sound pressure level and the actual sound pressure level at any time for which the model meteorological conditions (usually downwind or temperature inversion) are satisfied, will be ± ue . Generally, most noise models allow noise predictions to be made for the worst-case meteorological conditions that are expected to occur on a regular basis. However, the best accuracy (or expanded uncertainty) that can realistically be expected for sound pressure level predictions at distances of 100 m or more from a noise source is ± 3 dB. However, an expanded uncertainty of ± 4 dB is probably more realistic when uncertainties in the source sound power levels are taken into account. Some practitioners claim an accuracy of better than ± 2 dB, but there are insufficient data available to confirm that. The difficulty in obtaining accurate predictions is mainly associated with the variability of the atmospheric wind and temperature profiles over time and geographic location. There is another type of uncertainty called model bias, which is the difference between predictions and measured data averaged over many different locations. This result is also expressed as a 95th percentile, but it is inappropriate for use in a noise level prediction report as it does not indicate the error that could exist at a single location. Rather it tells us that, although in some locations the predictions may be high, in others they will be low, so that the average is usually less than the uncertainty for a single location. The most extensive study of propagation model prediction uncertainty was done by Marsh (1982) for the CONCAWE model for source–receiver separations spanning distances of 200 to 2000 m. His results are summarised in Table 4.12 for the various meteorological categories discussed in Section 4.7.4. For downwind propagation only, the expanded uncertainty (for 95% confidence limits) for the CONCAWE model is approximately ± 4.5 dBA. The ISO 9613-2 (1996) model includes an estimate of the accuracy of the predicted overall Aweighted sound pressure level. For a mean source and receiver height less than 5 m ([hS +hR ]/2 < 5 m), the estimated accuracy is ±3 dB. For a mean source and receiver height between 5 m and 30 m, the estimated accuracy is ±1 dB for source–receiver distances of less than 100 m, and ±3 dB for source–receiver distances between 100 m and 1000 m. It is expected that for higher sources such as wind turbines, the upper distance limits mentioned above may be extended further, which implies that the propagation models are more accurate for higher sound sources, as the distance from the source at which multiple ground reflections begin to occur increases with increasing difference between the source and receiver height. Perhaps the greatest uncertainty

209

Sound Propagation Outdoors

TABLE 4.12 95% confidence limits of the CONCAWE model, which is representative of the expected reliability of the predicted value at a single residence (after Marsh (1982)). The average of the measured value minus the predicted value for many receiver locations is shown in brackets, indicating the extent of bias (or error in the model in terms of predicting average values)

Meteorological category 2 3 4 5 6 Meteorological category 2 3 4 5 6

dBA 6.8 6.9 5.7 4.7 4.5

(0.5) (0.6) (0.5) (0.0) (0.5)

500 9.4 10.1 9.8 8.1 9.3

Octave band centre frequency (Hz) 63 125 250 5.4 5.0 4.8 3.9 5.2

( 0.1) ( 0.0) ( 0.3) (−0.1) (−0.8)

5.4 6.2 6.5 5.4 6.1

( 0.1) ( 0.5) ( 0.8) ( 0.0) (−0.3)

9.1 9.4 8.7 8.4 6.7

( 2.0) ( 1.6) (−1.2) (−2.3) (−1.7)

Octave band centre frequency (Hz) 1000 2000 4000

( 2.2) ( 0.4) (−0.2) ( 0.4) ( 1.2)

7.8 8.5 6.6 5.2 4.9

( 2.2) ( 0.8) ( 0.1) (−0.6) (−0.2)

9.8 8.5 5.6 5.6 5.5

(−0.2) ( 0.8) ( 1.4) ( 0.9) ( 0.1)

12.4 9.4 6.7 6.7 8.2

( 0.4) ( 0.4) ( 0.2) (−0.9) (−0.9)

lies in the input data used in the prediction models, especially meteorological data, including atmospheric wind and temperature variation as a function of height above the ground. In summary, for most A-weighted environmental noise predictions, it would be wise to suggest that the variation between prediction and measurement for any particular location in a downward refracting atmosphere is of the order of ± 4 − 5 dBA. Example 4.7 (a) For the compressor in Example 4.6, calculate the 95% confidence limits, representing the range of differences between the measured and predicted sound pressure levels that would include 95% of the data, for the ISO 9613-2 (1996) propagation model. Assume that all quantities follow a normal distribution and that the accuracy of the sound power measurement is ±2 dB. (b) What would the 95% confidence limits be if a second compressor were introduced adjacent to the first one and had a sound power level of 3 dB less than the first one in all octave bands and the same uncertainty in sound power level data. Solution 4.7 (a) The ISO standard suggests that for a distance of 1000 m, the expected accuracy of the propagation part of the sound pressure level calculation is ±3 dB. Assuming a normal distribution, the corresponding standard uncertainty is 3/2 = 1.5 dB. The standard uncertainty for the sound power measurement (assuming a normal distribution) is 2/2 = 1 dB. We √ can use Equation (4.58) to calculate the combined standard uncertainty, us = 1.52 + 12 = 1.80 dB. The 95% confidence limits are then 1.80 × 2 = ±3.6 dB. (b) The combined standard uncertainty for the second compressor will be the same as the first; that is, us = 1.80 dB. The overall standard uncertainty is given by Equation (4.57) as:

p us,tot =

1.82 + (1.8 × 10−3/10 )2 = 1.34 dB 1 + 10−3/10

210

Noise Control: From Concept to Application where Lpi = 0 for the first compressor has been used for convenience. Any value of Lpi for the first compressor would give the same result, as it is the dB difference between sound source contributions that is associated with the final result. Thus the 95% confidence limits are ±1.34 × 2 = ±2.7 dB.

4.10

Additional Problems

1. (a) A sound wave in air with a frequency of 500 Hz and a pressure level of 60 dB re 20 µPa is normally incident on a boundary between air and a second medium having a characteristic impedance of ρc = 830 MKS rayls. Calculate the RMS pressure amplitudes of the incident and reflected waves. (b) For what angle of incidence will the sound be totally reflected for the conditions specified in part (a)? 2. The side of a factory contains a square opening (3m × 3m) which is radiating noise to a community 500 m away. The centre of the opening is 3 m above the ground and the community location of interest is 1.5 m above the ground. The average sound pressure level measured just outside the opening was 100 dB re 20 µPa in the 500 Hz octave band and the level at the nearest residence varies between 46 and 58 dB re 20 µPa, depending on the weather conditions and time of measurement. (a) Determine the sound pressure level at the community location in the 500 Hz octave band that is due only to the noise radiated from the square opening. You may use the method encapsulated in Figure 4.3 for calculating the ground effect. You may assume the following: (i) The ground between the source and receiver is classed as “normal, uncompacted ground” (see Table 4.8). (ii) There are no trees or other barriers in the way. (iii) The measurement at the opening only includes sound propagating to the outside from the opening. (iv) All of the sound energy is radiated into half space as if the dimensions of the factory wall are infinite. (v) The relative humidity varies between 25% and 75%. (vi) The temperature varies between 15 and 25◦ C. (vii) All noise at the residence is originating from the factory. (b) What approximate noise reduction at the nearest residence would you expect in the absence of the opening? 3. The plane wave reflection coefficient for normally incident waves reflected from a plane surface of specific acoustic impedance, Zs is given by Rp = (Zs − ρc)/(Zs + ρc). (a) Show that the absorption coefficient (fraction of incident energy absorbed and equal to α = 1 − |Rp |2 ) of the surface is given by α = 4ρc Re{Zs }/|Zs + ρc|2 [Hint: express |Zs + ρc|2 as |Re{Zs } + ρc|2 + [Im{Zs }]2 .] (b) What value of Zs gives the highest absorption coefficient? 4. A window of area 1 m2 in the side of a building has its centre 2 m above the ground and is radiating sound into the community. The nearest residence is 750 m away. The average sound intensity over the outside of the window is 0.1 W/m2 in the 500 Hz octave band. Assuming that only noise in the 500 Hz octave band is of interest, calculate the sound pressure level at the nearest residence if the ground surface between

211

Sound Propagation Outdoors the window and the residence is concrete. Assume a community location 1.5 m above the ground. State any other assumptions that you make. 5. It is desired to calculate the sound pressure level at a community location at a normal distance of 100 m from the line of traffic on a freeway, assuming that vehicles are spaced an average of 10 m apart and on average, each radiates 0.1 W of acoustic power in the 500 Hz 1/3-octave band from a height of 0.5 m above the road surface. Assume that the community location is 1.5 m above the ground and that the terrain between this location and the freeway is flat and that the air temperature is 20◦ C. (a) Calculate the sound pressure level in the 500 Hz 1/3-octave band at the community location mentioned above, ignoring all excess attenuation effects. (b) Calculate the excess attenuation due to the ground effect, the excess attenuation due to atmospheric absorption and the resulting sound pressure level at the community location in the 500 Hz 1/3-octave band if all other excess attenuation effects are ignored. Assume that the ground between the freeway and the community location is covered with rough thick pasture 10 cm high and is represented by a flow resistivity of 2 ×105 MKS Rayls/m. In this instance, the ground reflection coefficient may be calculated by assuming plane wave propagation. (c) Calculate the strength of the wind (at a height of 10 m) required in the direction from the freeway to the community location for the community measurement point to be in the shadow zone in the absence of any barrier. You may assume that no temperature gradient exists and that the wind shear coefficient, ξ = 0.15. 6. An industrial facility is already the major contributor of noise in the surrounding community and during moderate downwind (3–4.5 m/s) atmospheric conditions, the sound pressure level measured at the most sensitive community location is 42 dBA re 20 µPa (wind blowing from the facility to the sensitive community location). It is planned to introduce some noisy process equipment into the facility at a location that is 500 m from the community location. The sound power level of the equipment is only significant in the 500 Hz, 1000 Hz and 2000 Hz octave bands. In these bands the sound power level is 121 dB, 119 dB and 117 dB re 10−12 W respectively. The acoustic centre of the introduced process equipment is 4 m above the ground and the community location may be assumed to be 1.5 m above the ground. The ground between the facility and community location is flat and the ground cover types between the facility and community are shown in Figure 4.19. You may assume a temperature of 20◦ C, a relative humidity of 50% and 4 octas nighttime cloud cover.

Elevation view

S

R 4m

concrete 50 m

normal uncompacted ground 450 m

1.5 m

500 m FIGURE 4.19 Arrangement for Problem 6.

(a) What is the total sound pressure level expected at the sensitive community location during moderate downwind conditions, according to ISO 9613-2 (1996)? (b) What is the total sound pressure level expected at the sensitive community location during moderate downwind conditions, according to CONCAWE?

212

Noise Control: From Concept to Application (c) What noise reduction would be required to meet an allowed 45 dBA re 20 µPa at the most sensitive community location? (d) What would be the expected noise reduction according to ISO 9613-2 (1996), if foliage of height 5 m and width 50 m existed between the facility and the sensitive community location with its boundary finishing 10 m from the community location? For this calculation, assume that the existing noise before the addition of the new process equipment originates from sources at the same distance from the community location as the new process equipment. Thus, the same noise reduction due to the foliage will occur for all noise sources contributing significantly to the noise at the community location. (e) According to ISO 9613-2 (1996), how high would a wall, located 2 m from the process equipment, have to be in order for the allowable sound pressure level to be achieved at the sensitive community location during moderate downwind conditions? Include the effect of the foliage in part (d) above. You may assume that the noise emitted by the original facility at frequencies below 500 Hz is much lower than the 500 Hz level of the new process equipment. You may also assume that the wall is sufficiently wide that diffraction around the ends is negligible.

7. Three items of process equipment in a facility are responsible for all significant noise emission. The accuracy of measurement of the sound power levels is ±2 dB for each item in each octave band. The sound pressure level contributions at the closest sensitive community location (at a distance of 500 m from the equipment) are calculated using the ISO 9613-2 (1996) model to be 38 dBA, 40 dBA and 42 dBA re 20 µPa, respectively for the three items of equipment. (a) Calculate the total A-weighted sound pressure level at the community location. (b) What are the 95% confidence limits within which 95% of overall A-weighted sound pressure level measurements will be found, for the same downwind conditions as assumed in the ISO 9613-2 (1996) model? 8. Sound being generated by an item of process equipment is resulting in octave band sound pressure levels at the nearest sensitive community location shown in Table 4.13. The source height is 2 m and the receiver height is 1.5 m and they are separated horizontally by 40 m. TABLE 4.13

63 Octave band sound pressure level (dB re 20 µPa)

75.0

Data for Problem 8

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000 64.0

55.0

50

48.0

46.0

42.0

(a) Calculate the overall A-weighted noise reduction achieved with a 3 m high × 6 m long brick wall built 2.5 m from the process equipment and 37.5 m from the sensitive community location, using the ISO 9613 method encapsulated in Equation (4.43). For this exercise, you may ignore the excess attenuation due to the ground. (b) More noise reduction is required, and one suggestion is to build a second brick wall of the same height on the receiver side of the first wall and separated from the first wall by 2 m. A second suggestion is to increase the height of the first wall by 1 m. Which suggestion would provide the greatest noise reduction?

5 Sound Absorbing Materials: Properties and their Measurement

LEARNING OBJECTIVES In this chapter, the reader is introduced to: • concepts of flow resistance and how it relates to sound absorption; • statistical sound absorption coefficient and its calculation and measurement for porous materials; • Sabine absorption coefficient and its measurement; • use of porous materials in sound absorption; and • panel absorbers.

5.1

Introduction

The ability of materials to absorb sound can be quantified in terms of their absorption coefficient, which can vary from 0.001 for highly reflective materials to 1 for very absorptive materials. There are two types of absorption coefficients that are commonly used: the statistical absorption coefficient, αst and the Sabine absorption coefficient, α ¯ . The main difference in these two absorption coefficients lies in the way that they are measured. Statistical absorption coefficients are determined from measurements of the sound field in a tube terminated at one end by a sample of the material to be tested. On the other hand, Sabine absorption coefficients are measured in a reverberation room using a large sample of material mounted on the floor. The statistical absorption coefficient has a maximum theoretical value of approximately 0.95 and it is always less than the Sabine absorption coefficient for the same material, which can exceed 1 (meaning that in principal, there is more energy absorbed than is incident on the test material). This peculiar phenomenon arises from the sample of absorbing material distorting the sound field so that more energy falls on its surface than fell on the floor in the absence of the material. The difference between the two absorption coefficients increases as the magnitude of the absorption coefficient increases. Below a value of about 0.4, the difference is negligible. In this chapter the measurement and characterisation of both types of absorption coefficient are discussed in depth, beginning with a discussion of flow resistance, from which the normal specific acoustic impedance and hence statistical absorption coefficient of a porous acoustic material may be determined. This is followed by a discussion of the direct measurement of normal incidence absorption coefficients and normal specific acoustic impedance and their application 213

214

Noise Control: From Concept to Application

to the calculation of statistical absorption coefficients of materials with backing cavities or covered by perforated sheets or thin impervious material. The measurement of Sabine absorption coefficients is discussed next and this is followed by a discussion of the design of absorbers made from impervious flexible panels.

5.2

Flow Resistance and Resistivity

Porous materials are often used for the purpose of absorbing sound. It is the porous nature of many surfaces, such as grass-covered ground, that determines their sound reflecting properties. The property of a porous material which determines its usefulness for acoustical purposes, is the resistance of the material to induced flow through it, as a result of a pressure gradient. Flow resistance, an important parameter, which is a measure of this property, is defined according to the following simple experiment. A uniform layer of porous material of thickness, `, and area, S, is subjected to an induced mean volume flow, V0 (m3 /s), through the material and the pressure drop, ∆P , across the layer is measured. Very low pressures and mean volume velocities are assumed. The flow resistance of the material, Rf , is defined as the induced pressure drop across the layer of material divided by the resulting mean volume velocity per unit area of the material: Rf = ∆P S/V0

(5.1)

The units of flow resistance are the same as for specific acoustic impedance, ρc; thus it is sometimes convenient to specify flow resistance in dimensionless form in terms of numbers of ρc units. The flow resistance of unit thickness of material is defined as the flow resistivity R1 which has the units Pa s m−2 , often referred to as MKS rayls per metre. Experimental investigation shows that porous materials of generally uniform composition may be characterised by a unique flow resistivity. Thus, for such materials, the flow resistance is proportional to the material thickness, `, as follows: Rf = R1 ` (5.2) Flow resistance and flow resistivity of porous materials are discussed in depth in the literature (Bies and Hansen, 1979, 1980). Its measurement is discussed in detail in Bies et al. (2018).

5.3

Sound Propagation in Porous Media

For the purpose of the analysis, the porous, gas-filled medium is replaced by an effective medium, which is characterised in dimensionless variables by a complex density, ρm , and complex compressibility, κ. In terms of these quantities, a complex characteristic impedance and propagation coefficient are defined, analogous to the similar quantities for the contained gas in the medium. Implicit in the formulation of the following expressions is the assumption that time dependence has the positive form, e jωt , consistent with the practice adopted throughout the text. The characteristic impedance of porous material may be written in terms of the gas density, ρ, the gas speed of sound c, the gas compressibility, κ0 , the porous material complex density, ρm , and the porous material complex compressibility, κ, as: Zm = ρc

r

ρm κ ρ κ0

(5.3)

Similarly, a propagation coefficient, km , may be defined as: km =

2π (1 − jαm ) = (ω/c) λ

r

ρm κ0 =k ρκ

r

ρm κ0 ρκ

where ω = 2πf is the angular frequency (rad/s) of the sound wave.

(5.4)

215

Sound-Absorbing Materials

The quantities, ρm /ρ and κ/κ0 , may be calculated using the following procedure (Bies, 1981). This procedure gives results for fibrous porous materials within 4% of the mean of published data (Delany and Bazley, 1969, 1970), and unlike the Delaney and Bazley model, it tends to the correct limits at both high and low values of the dimensionless frequency, ρf /R1 . However, this model and the Delaney and Bazley model have only been verified for fibreglass and rockwool materials with a small amount of binder and having short fibres smaller than 15 µm in diameter, which excludes such materials as polyester and acoustic foam. The normalised compressibility and normalised density of a porous material can be calculated from a knowledge of the material flow resistivity, R1 , using the following Equations (5.5) to (5.15). κ/κ0 = [1 + (1 − γ)τ ]−1

(5.5)

ρm /ρ = [1 + σ]−1

(5.6)

where γ is the ratio of specific heats for the gas (=1.40 for air), ρ is the density of gas (=1.205 kg/m3 for air at 20◦ C), f is the frequency (Hz), R1 is the flow resistivity of the porous material (MKS rayls/m) and: τ = 0.592a(X1 ) + jb(X1 )

(5.7)

σ = a(X) + jb(X)

(5.8)

a(X) =

T3 (T1 − T3 )T22 − T42 T12 T32 T22 + T42 T12

(5.9)

T12 T2 T4 2 T3 T22 + T42 T12

(5.10)

b(X) =

T1 = 1 + 9.66X

(5.11)

T2 = X(1 + 0.0966X)

(5.12)

T3 = 2.537 + 9.66X

(5.13)

T4 = 0.159(1 + 0.7024X)

(5.14)

X = ρf /R1

(5.15)

The quantities, a(X1 ) and b(X1 ), are calculated by substituting X1 = 0.856X for the quantity, X, in Equations (5.9)–(5.14). The impedance and wavenumber relationships that have been generally accepted in the past (Delany and Bazley, 1969, 1970), and which are accurate in the flow resistivity range R1 = 103 to 5 × 104 MKS rayls/m, are: Zm /(ρc) = [1 + C1 X −C2 − jC3 X −C4 ]

(5.16)

km /k = [1 + C5 X −C6 − jC7 X −C8 ]

(5.17)

The quantities X, Zm , km , c and ρ have all been defined previously. Values of the coefficients C1 − C8 are given in Table 5.1 for various materials from various references.

216

Noise Control: From Concept to Application TABLE 5.1

Material type reference

Values of the coefficients C1 –C8 for various materials

C1

C2

C3

C4

C5

C6

C7

C8

Rockwool/fibreglass Delany and Bazley (1970)

0.0571

0.754

0.087

0.732

0.0978

0.700

0.189

0.595

Polyester Garai and Pompoli (2005)

0.078

0.623

0.074

0.660

0.159

0.571

0.121

0.530

Polyurethane foam of low flow resistivity Dunn and Davern (1986)

0.114

0.369

0.0985

0.758

0.168

0.715

0.136

0.491

Porous plastic foams of medium flow resistivity Wu (1988)

0.212

0.455

0.105

0.607

0.163

0.592

0.188

0.544

5.4 5.4.1

Measurement of Absorption Coefficients of Porous Materials Using an Impedance Tube Measurement Using the Moving Microphone Method

Statistical absorption coefficients may be determined using impedance tube measurements of the normal-incidence, specific acoustic impedance, as described in the following paragraphs. The normal-incidence absorption coefficient of a porous material may be determined directly from the sound pressure field in the impedance tube whereas the statistical absorption coefficient is determined from the normal-incidence specific impedance of the material, which, in turn, is determined from measurements of the sound pressure field in the impedance tube. When a tonal (single frequency) sound field is set up in a tube terminated in a specific acoustic impedance, Zs , a pattern of regularly spaced maxima and minima along the tube will result, which is uniquely determined by the driving frequency and the terminating impedance. The absorption coefficients are related to the terminating impedance and the characteristic impedance, ρc, of air. An impedance tube is relatively easily constructed and therein lies its appeal. Any heavy walled tube may be used for its construction. A source of sound should be placed at one end of the tube and the material to be tested should be mounted at the other end. Means must be provided for probing the standing wave within the tube. An example of a possible configuration is shown in Figure 5.1. The older and simpler method by which the sound field in the impedance tube is explored, using a moveable microphone which traverses the length of the tube, will be described first (see ASTM C384-04 (2016)). In this case, the impedance of the test sample is determined from measurements of the sound field in the tube. This method is slow but easily implemented. A much quicker method, which makes use of two fixed microphones and a digital frequency analysis system, is described in Section 5.4.2 (see ASTM E1050-12 (2012)). It is also possible to measure the specific acoustic surface impedance and absorption coefficient of materials in situ, without using an impedance tube, by using two microphones close to the surface of the material (Dutilleaux et al., 2001). However, the procedure is quite complex.

217

Sound-Absorbing Materials

sound level meter and band pass filter

frequency counter

variable frequency oscillator

heavy metal plug

amplifier

speaker

microphone location if a probe tube is used

test sample

moveable microphone

ruler

FIGURE 5.1 Equipment for impedance tube measurement using the older and simpler method.

Implicit in the use of the impedance tube is the assumption that only plane waves propagate back and forth in the tube. This assumption puts upper and lower frequency bounds on the use of the impedance tube. Let d be the tube diameter if it is circular in cross section, or the larger dimension if it is rectangular in cross section. Then the upper frequency limit (or cut-on frequency) is: ( 0.586c/d; for circular section ducts fu = (5.18) 0.5c/d; for rectangular section ducts Here, c is the speed of sound, and the frequency limit is given in Hz. The required length, `, of tube is a function of the lowest frequency, f` , to be tested and is given by: `=d+

3c 4f`

(5.19)

In general, the frequency response of the apparatus will be very much dependent on the material under test. To reduce this dependence and to ensure a more uniform response, some sound absorptive material may be placed permanently in the tube at the source end. Furthermore, as energy losses due to sound propagation along the length of the tube are undesirable, the following equations may be used as a guide during design to estimate and minimise such losses: Lmin2 − Lmin1 = aλ/2 (dB) (5.20)

p

a = 0.19137

f /cd

(dB/m)

(5.21)

In the above equations, a is the loss (in dB per metre of tube length) due to propagation of the sound wave down the tube, and Lmin1 and Lmin2 are the sound pressure levels at the first and second minima relative to the surface of the test sample (see Figure 5.2). The other quantities are the frequency, f (Hz), the corresponding wavelength, λ (m), and the speed of sound, c (m/s). For tubes of any cross section, d (m) is defined as d = 4S/PD , where S is the cross-sectional area and PD is the cross-sectional perimeter. The propagation loss in dB/m is related to the attenuation coefficient, ζ, in nepers/m for a propagating wave, expressed as follows for the sound pressure as a function of distance, x, along

218

Noise Control: From Concept to Application Sound pressure level in tube l /2

D1

Lp (dB) L max2 L min2

Lmax- Lmin

L max1 L min1

Z = pt /ut

sample 0

x = -L

x

FIGURE 5.2 Schematic of an impedance tube.

a tube from the sound source:

p(x) = p0 e−ζx

(5.22)

where the sound wave has an RMS sound pressure of p0 at location x0 and an RMS sound pressure of p(x) at a location that is a distance, x + x0 , from the sound source. One neper/m = 20/(loge 10) = 8.686 dB/m. The sound field within the impedance tube may be explored either with a small microphone placed at the end of a probe and immersed in the sound field, as illustrated in Figure 5.1, or with a probe tube attached to a microphone placed externally to the field. Equation (5.21) may be useful in selecting a suitable tube for exploring the field; the smaller the probe tube diameter, the greater will be the energy loss suffered by a sound wave travelling down it, but the smaller will be the disturbance to the acoustic field being sampled. An external linear scale should be provided for locating the probe. As the sound field will be distorted slightly by the presence of the probe, it is recommended that the actual location of the sound field point sampled be determined by replacing the specimen with a heavy solid metal face located at the specimen surface. The first minimum will then be λ/4 away from the solid metal face. Subsequent minima will always be spaced at intervals of λ/2. Thus, this experiment will allow the determination of how far in front of the end of the probe tube that the sound field is effectively being sampled. It will be found that the minima will be subject to contamination by acoustic and electronic noise; thus, it is recommended that a narrow band filter, for example, one which is an octave or 1/3-octave wide, be used in the detection system. When the material to be tested is in place and the sound source is excited with a single frequency, a series of maximum and minimum sound pressures in the impedance tube will be observed. The maxima will be effectively constant in level but the minima will increase in level, according to Equation (5.20), as one moves away from the surface of the test material. For best results, it is recommended that losses in the tube be taken into account by extrapolating the minima back to the surface of the sample by drawing a straight line joining the first two minima to the location corresponding to the surface of the sample on the plot of sound pressure level in dB vs distance along the tube (see Figure 5.2). The standing wave ratio, L0 , is then determined as the difference between the maximum level, Lmax , and the extrapolated minimum level, Lmin . It is of interest to derive an expression for the normal incidence absorption coefficient of a sample of acoustic material in an impedance tube, as shown in Figure 5.2. For the following analysis, the impedance tube contains the material sample at the right end and the loudspeaker sound source at the left end. For simplicity, it is assumed that there are no losses due to dissipation in the tube; that is, the quantity, a, in Equation (5.21) is assumed to be zero. However, losses can be included approximately by replacing the minimum sound pressure level closest to the sample by the extrapolated minimum sound pressure level obtained from the two minima

219

Sound-Absorbing Materials

closest to the sample, as described in the previous paragraphs. The loudspeaker is not shown in the figure and the origin is set at the right end of the tube at the face of the test sample. Reference should be made to Section 1.5.4, where it is shown that multiple waves travelling in a single direction may be summed together to give a single wave travelling in the same direction. For the case considered here, the multiple waves travelling in each direction are a result of multiple reflections from each end of the tube. As the sample is at the right end of the tube, the resultant incident wave will be travelling in the positive x-direction. Assuming a phase shift between the incident and reflected waves of β at x = 0, the incident wave and reflected wave pressures may be written as: pi = Ae j(ωt−kx) and pr = Be j(ωt+kx+β)

(5.23)

where A and B are real amplitudes. The total sound pressure is thus: pt = Ae j(ωt−kx) + Be j(ωt+kx+β)

(5.24)

The first maximum pressure (closest to the sample) will occur when: β = −2kx

(5.25)

β = −2kx + π

(5.26)

and the first minimum will occur when:

Thus, the maximum and minimum sound pressure amplitudes in the standing wave are, respectively: pˆmax = |e−jkx | (A + B) = A + B

and

pˆmin = |e−jkx | (A − B) = A − B

(5.27)

and the ratio of maximum to minimum pressures is (A + B)/(A − B). The standing wave ratio, L0 , is the difference in decibels between the maximum and minimum sound pressures in the standing wave and is defined as: A+B (5.28) 10L0 /20 = A−B Thus, the ratio (B/A) is:  L /20  10 0 −1 B = (5.29) A 10L0 /20 + 1 The amplitude of the pressure reflection coefficient squared is defined as |Rp |2 = (B/A)2 , which can be written in terms of L0 as: 10L0 /20 − 1 |Rp | = 10L0 /20 + 1 2



2

(5.30)

The normal incidence absorption coefficient is defined as: αn = 1 − |Rp |2

(5.31)

It is also of interest to continue the analysis to determine the specific acoustic normal impedance of the surface of the sample. This can then be used to determine the statistical absorption coefficient of the sample, which is the absorption coefficient averaged over all possible angles of an incident wave. The total particle velocity can be calculated using Equations (1.2), (1.3) and (5.23) to give: ut =

1 (pi − pr ) ρc

(5.32)

220

Noise Control: From Concept to Application

Thus: ut =


1 Ae j(ωt−kx) − Be j(ωt+kx+β) ρc

(5.33)

The specific acoustic impedance (or characteristic impedance) at any point in the tube may be written as: Ae−jkx + Be jkx+jβ A + Be j(2kx+β) pt = ρc −jkx = ρc (5.34) Zs = jkx+jβ ut Ae − Be A − Be j(2kx+β) At x = 0, the impedance is the normal specific acoustic impedance, ZN , of the surface of the sample. Thus: pt A + Bejβ ZN = = (5.35) ρc ρcut A − Be jβ The above impedance equation may be expanded to give: ZN A/B + cos β + j sin β (A/B)2 − 1 + (2A/B)j sin β = = ρc A/B − cos β − j sin β (A/B)2 + 1 − (2A/B) cos β

(5.36)

In practice, the phase angle, β, is evaluated by measuring the distance, D1 , of the first sound pressure minimum in the impedance tube from the sample surface, and the tube that corresponds to Figure 5.2 is known as an impedance tube. Referring to Equation (5.26) and Figure 5.2, the phase angle, β, may be expressed in terms of D1 (which is a positive number) as: β = 2kD1 + π = 2π



2D1 1 + λ 2



(5.37)

Equation (5.36) may be rewritten in terms of a real and imaginary components as: ZN (A/B)2 − 1 (2A/B) sin β = R + jX = +j 2 2 ρc (A/B) + 1 − (2A/B) cos β (A/B) + 1 − (2A/B) cos β

(5.38)

where β is defined by Equation (5.37) and the ratio, A/B, is defined by the reciprocal of Equation (5.29) where L0 is the difference in dB between the maximum and minimum sound pressure levels in the tube. In terms of an amplitude and phase, the normal specific acoustic impedance is: ZN /(ρc) = ξe jψ

(5.39)

where: ξ=

p

R2 + X 2

(5.40)

ψ = tan−1 (X/R)

(5.41)

and:

The statistical absorption coefficient, αst , may be calculated as:

αst

1 = π

Z2π 0

dϕ

Zπ/2

α(θ) cos θ sin θ dθ

0

where the angles, θ and ϕ are defined in Figure 5.3.

(5.42)

221

Sound-Absorbing Materials Incoming sound wave

z

q Panel j y

x

FIGURE 5.3 Definition of the incidence angles for an incoming sound wave of Equation (5.44).

Rewriting the absorption coefficient in terms of the reflection coefficient using: |R(θ)|2 = 1 − α(θ)

(5.43)

Equation (5.42) can then be written as: αst = 1 − 2

Z

π/2

|R(θ)|2 cos θ sin θ dθ

(5.44)

0

For bulk reacting materials, Equations (4.11) and (4.36) may be used to calculate the reflection coefficient, R(θ), and for locally reacting materials, Equation (4.37) may be used. Note that use of these equations requires a knowledge of the normal specific acoustic impedance, ZN , of the material. This can be measured using an impedance tube in which a sample of the material, in the same configuration as to be used in practice, in terms of how it is backed, is tested. Alternatively, the normal specific acoustic impedance for any configuration may be calculated using the methods described in Section 5.5. Using Equations (5.44) and (4.37), Morse and Bolt (1944) derive the following expression for the statistical absorption coefficient for a locally reactive surface of normal specific acoustic impedance, ZN /(ρc), given by Equation (5.39): αst =



8 cos ψ ξ



1−



cos ψ loge (1 + 2ξ cos ψ + ξ 2 ) ξ



cos(2ψ) ξ sin ψ + tan−1 ξ sin ψ 1 + ξ cos ψ









(5.45)

Some results of such calculations for a simple porous blanket mounted on a rigid wall are shown in Figure 5.4. Implicit in the calculations is the assumption that sound propagation within the porous material is predominantly normal to the surface. This condition is sufficiently well satisfied if the porous material is fairly dense. Alternatively, the porous material could be contained between solid partitions arranged to prevent lateral propagation. The calculated statistical absorption coefficient is optimum when the total flow resistance, Rf = R1 `, through the material is between 2ρc and 5ρc, where ` is the material thickness. This is shown in Figure 5.4, where it can be seen that a porous liner as little as one-tenth of a wavelength thick will give a statistical absorption coefficient of about 0.92. This is close to the maximum that is theoretically possible.

222

Noise Control: From Concept to Application 1.0 0.9 0.8 0.7 0.6

ast 0.5 0.4 0.3

R1R /rc = 3 4 5 10 20 50 100 200 500 1000 10000 100000

0.2 0.1 0 -3 10

-2

-1

10

10

0

10

1

10

f R /c

FIGURE 5.4 Calculated statistical absorption coefficient for a rigidly backed porous material in a reverberant field, as a function of a frequency parameter, for various indicated values of flow resistance. The quantity R1 is the flow resistivity of the material (MKS rayls/m), ρ is the density of air (kg/m3 ), c is the speed of sound in air (m/s), f is the frequency (Hz) of the incident sound and ` is the porous material thickness (m).

The performance of a porous blanket material can be improved by mounting it so that an air gap between the material and the hard backing wall is provided. This is discussed in Section 5.5.1. Calculations show that the optimum air gap depth is equal to the thickness of the porous liner, and that the absorption will be the same as that which would characterise a porous layer equal in thickness to the air gap and liner. To achieve this optimum, it would be necessary to place partitions in the air gap to prevent lateral sound propagation in the gap parallel to the wall. Example 5.1 (a) Write expressions for the plane wave acoustic pressure and particle velocity at any location in an impedance tube shown in Figure 5.5, in terms of the complex pressure amplitude, pˆ0 , and complex particle velocity amplitude, u ˆ0 , at the surface of the acoustic material. The acoustic material is mounted at one end of the tube. Also show that the specific acoustic impedance at any point in the tube is given by: ρc jρcˆ u0 sin(kx) − pˆ0 cos(kx) = S ρcˆ u0 cos(kx) − jˆ p0 sin(kx)



Zs



where x is the distance along the tube from the material sample which is mounted at the left end of the tube at x = 0 and S is the tube cross-sectional area. (b) How would the expression in (a) above for the impedance differ if the tube were open at the left end and the sample were placed at the closed right end at x = 0, with the coordinate system arranged such that x = 0 at the right-hand end and the right-hand direction is positive. Show that the two expressions are equivalent when evaluated at the open end of the tube (assuming a tube length of L.)

223

Sound-Absorbing Materials Solution 5.1

ZA = -pT /SuT L 0

x

FIGURE 5.5 Arrangement for Example 5.1, part(a)

(a) As the sample 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 β at x = 0, the incident wave and reflected wave pressures may be written in terms of arbitrary real constants, A and B, using Equation (1.23), as: pi = A e j(ωt+kx) and pr = B e j(ωt−kx+β) . The total pressure is thus, pt = A e j(ωt+kx) + B e j(ωt−kx+β) .

(i)

The total particle velocity can be calculated using Equations (1.2) and (1.3) as: 1 ut = (pr − pi ). ρc
1 B e j(ωt−kx+β) − A e j(ωt+kx) . (ii) Thus, ut = ρc At x = 0, Equations (i) and (ii) above become: pt = pˆ0 e jωt and ut = u ˆ0 e jωt . Thus: pˆ0 = A + B e jβ and ρcˆ u0 = B e jβ − A. Thus, A = 0.5(ˆ p0 − ρcˆ u0 ) and Be jβ = 0.5 (ˆ p0 + ρcˆ u0 ), and the total acoustic pressure and particle velocity may be written as: pt = 0.5(ˆ p0 − ρcˆ u0 ) e j(ωt+kx) + 0.5 (ˆ p0 + ρcˆ u0 ) e j(ωt−kx)
0.5 (ˆ p0 + ρcˆ u0 ) e j(ωt−kx) − (ˆ p0 − ρcˆ u0 ) e j(ωt+kx) . and ut = ρc 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 Figure 5.5 and Table 1.5). Thus: ZA = −

pt ρc (ˆ p0 − ρcˆ u0 ) e jkx + (ˆ p0 + ρcˆ u0 ) e−jkx =− . Sut S (ˆ p0 + ρcˆ u0 ) e−jkx − (ˆ p0 − ρcˆ u0 ) e jkx

As ejkx = cos(kx) + j sin(kx) and e−jkx = cos(kx) − j sin(kx), the acoustic impedance may be written as: ρc jρcˆ u0 sin(kx) − pˆ0 cos(kx) ZA = . S ρcˆ u0 cos(kx) − jˆ p0 sin(kx) 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 as shown ρc pˆ0 cos(kx) − jρcˆ u0 sin(kx) in Figure 5.6: ZA = . S ρcˆ u0 cos(kx) − jˆ p0 sin(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, the quantity, −ˆ p0 /ˆ u0 , is equal to the specific acoustic termination impedance, Z0 in part (a), while in part (b), the quantity, pˆ0 /ˆ u0 , is equal to the specific acoustic termination impedance, Z0 . Making the appropriate substitutions,

224

Noise Control: From Concept to Application both expressions give for the acoustic impedance at the open end of the tube: ZA =

ρc Z0 cos(kL) + jρc sin(kL) . S ρc cos(kL) + jZ0 sin(kL) ZA = pT/SuT

0

-L

x

FIGURE 5.6 Arrangement for Example 5.1, part(b)

Example 5.2 As illustrated in Figure 5.7, a loudspeaker is placed at one end (x = 0) of a tube of cross-sectional area S = 10 cm2 , and at the other end (x = 0.3 m) a second loudspeaker is placed. The first loudspeaker is driven to produce a single frequency wave and the resulting sound field in the tube is sampled with a microphone. The sound field has a pressure maximum of 100 dB re 20 µPa at x = 0.03 m, 0.15 m and 0.27 m. The sound field has a pressure minimum of 96.5 dB re 20 µPa at x = 0.09 m and 0.21 m. It may be assumed that propagation losses in the tube are negligible. Find: (a) the frequency of the sound wave in the tube; (b) the volume velocity of the first loudspeaker; and (c) the mechanical impedance of the second loudspeaker. [Hint: Write an expression for the total sound pressure in the tube in terms of a left and right travelling wave. Include a phase angle, β in the wave which is reflected from the second speaker. The mechanical impedance is Sp/u. Let A be the amplitude of the right travelling incident wave (on the second loudspeaker) and let B be the amplitude of the left travelling reflected wave.] Solution 5.2

speaker 2

speaker 1

0

x

FIGURE 5.7 Arrangement for Example 5.2

(a) From Equation (1.23), the incident wave and reflected wave pressures may be written as, pi = A e j(ωt−kx) and pr = B e j(ωt+kx+β) , respectively. The total pressure is thus, pt = A e j(ωt−kx) + B e j(ωt+kx+β) . The maximum pressure amplitude occurs when the left and right going waves are in-phase which is at location, x, such that the phase, β = −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

225

Sound-Absorbing Materials β = −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, 20 log10 [(A + B)/(A − B)]. The location of the minimum closest to the end where x = 0 (left end) is xmin = 0.09 m. Thus, β = −2kxmin + π = −0.18k + π. As β is a constant, π − 2kxmin = −2kxmax . λ c π = = and thus: Therefore, xmin − xmax = 2k 4 4f f=

c 343 = = 1430 Hz. 4(xmin − xmax ) 4(0.09 − 0.03)

(b) The total particle velocity can be calculated using Equations (1.2) and (1.3) as: ut =

1 (pi − pr ). ρc

Thus, ut =


1 A e j(ωt−kx) − B e j(ωt+kx+β) . ρc

(i)

The complex particle velocity amplitude at x = 0 is then: u ˆt =


1 A − Be jβ . ρc

(ii)

The phase angle, β, is given by part (a) as: β = −2kxmax = −0.06 k = −0.06



π 2(xmin − xmax )



= −π/2.

Using Equations (i) and (ii), the particle velocity amplitude can be written as:


1/2 1 A2 + B 2 . 413.7 From part (a), the standing wave ratio is given by:

|u ˆt | =

pmax pmin

A+B = A−B

= 10(100−96.5 )/20 = 1.496.

Thus, A = 5.03B. (iii) However, the maximum pressure amplitude is A + B. Therefore, A + B = 2 × 10−5 × 1.414 × 10100/20 = 2.828. (iv) From Equations (iii) and (iv), we have A = 2.36 and B = 0.469. Thus the velocity amplitude at x = 0 is: u ˆt =

1 413.7

2.362 + 0.4692


1/2

= 5.8 mm/s.

Thus, the volume velocity amplitude is 5.8 × 10−6 m3 /s (0.001m2 area) which is equivalent to an RMS volume velocity of 4.1 × 10−6 m3 /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 (see Table 1.5). Thus Zm = pS/u. Using the relationships derived in parts (a) and (b), we have: Zm = ρcS

A + Be j(2kL+β) . A − Be j(2kL+β)

We previously found that β = −π/2 and k = −β/0.06. Also L = 0.3 and 2kL = (2π/0.12) × 0.3 = 5π. Thus Zm may be written as:

226

Noise Control: From Concept to Application Zm = 413.7 × 0.001 ×

5.032 + e jπ/2 5.032 + j = 0.4137 × jπ/2 5.032 − j 5.032 − e

= 0.01571 × (5.032 + j)2 = 0.382 + 0.158j. Example 5.3 As illustrated in Figure 5.8, a loudspeaker backed by a small enclosure, radiating low-frequency sound into a circular cross section tube terminated in a specific acoustic impedance, Z0 = R0 + jX0 , produces a constant local particle velocity amplitude, UL , at its surface. (a) Derive an expression for the sound power transmitted down the tube in terms of R0 , X0 and UL . (b) If R0 = 0, what is the transmitted power? (c) If all losses, including propagation losses and losses at the termination are negligible, what is the impedance presented to the loudspeaker? What does this suggest about the power generated? Note that if all losses are zero, then the resistive impedance is zero. (d) In case (a), what is the phase between the pressure and particle velocity at the pressure maxima in the resulting acoustic pressure field in the tube? (e) Show that when R0 = ρc, the magnitude of the amplitude reflection coefficient is given by |Rp | = X0 [4ρ2 c2 + X02 ]−1/2 Solution 5.3

Zs = -pT /uT 0

L x

FIGURE 5.8 Arrangement for Example 5.3

(a) Assume a horizontal tube with the left end containing the termination specific acoustic impedance, Z0 at x = 0. As the loudspeaker is at the right-hand 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 can be related to the specific acoustic impedance, Zs , it “sees” using Equations (1.57) and (1.61), and Table 1.5 as follows: S Su ˆ2t W = IS = Re { pˆt u ˆ∗t } = Re {Zs }, 2 2 where the hat represents the complex amplitude (or modulus) and S is the tube cross-sectional area. Referring to Example 5.1, the specific acoustic impedance looking towards the left in the negative x-direction may be written as: (ˆ p0 − ρcˆ u0 ) e jkx + (ˆ p0 + ρcˆ u0 ) e−jkx . Zs = −ρc (ˆ p0 + ρcˆ u0 ) e−jkx − (ˆ p0 − ρcˆ u0 ) e jkx As e jkx = cos(kx) + j sin(kx) and e−jkx = cos(kx) − j sin(kx), the impedance may be written as: jρcˆ u0 sin(kx) − pˆ0 cos(kx) Zs = ρc . ρcˆ u0 cos(kx) − jˆ p0 sin(kx)

227

Sound-Absorbing Materials Dividing through by ρcˆ u0 cos(kx), replacing x with L and replacing −ˆ p0 /ˆ u0 with Z0 gives:     (R0 + jX0 )/ρc + j tan kL Z0 /ρc + j tan kL = ρc . Zs = ρc 1 + j(Z0 /ρc) tan kL 1 + j((R0 + jX0 )/ρc) tan kL Thus:   (R0 /ρc) ( 1 + tan2 kL) Re{Zs } = ρc (1 − X0 /ρc)2 tan2 kL + (R0 /ρc)2 tan2 kL and the power is then: Sρc u ˆ2L W = 2



(R0 /ρc) ( 1 + tan2 kL) . (1 − X0 /ρc)2 tan2 kL + (R0 /ρc)2 tan2 kL



(b) It can be seen from the equation derived in part (a) that when R0 = 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 Zs with R0 = 0. In this case: Zs = ρc



jX0 /ρc + j tan kL 1 − ( X0 /ρc) tan kL



= jρc



X0 /ρc + tan kL , 1 − ( X0 /ρc) tan kL



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 90◦ out of phase. (e) The pressure amplitude reflection coefficient is given by, Rp = (B/A)e jβ . From Example 5.1, A = 0.5(ˆ p0 − ρcˆ u0 ) and Be jβ = 0.5 (ˆ p0 + ρcˆ u0 ). Thus: B e jβ (ˆ p0 + ρcˆ u0 ) = . A (ˆ p0 − ρcˆ u0 ) Dividing the numerator and denominator by u ˆ0 and putting Z0 = ρc + jX0 = −ˆ p0 /ˆ u0 (negative sign as we are looking at the impedance in the negative x-direction), we obtain: −ρc − jX0 + ρc jX0 Rp = = . −ρc − jX0 − ρc 2 ρc + jX0 Rp =

Thus:

−ρc − jX0 + ρc |jX0 | = = X0 [ 4ρ2 c2 + X02 ]−1/2 . | Rp | = −ρc − jX0 − ρc |2ρc + jX0 | Example 5.4 A loudspeaker mounted in one end of a uniform, 150 mm diameter tube, open at the opposite end, is modelled as a rigid piston of infinite internal impedance, driven at a fixed amplitude at 250 Hz. An infinite impedance source is one for which the vibration causing the radiated sound power is unaffected by the surrounding acoustic medium. The radiation impedance of the open end of the tube (mounted in a large baffle) of diameter, 2a, may be approximated as: Zr = (ρcπa2 )[(ka)2 /2 + j0.8ka] Qualitatively describe why the sound power output of the loudspeaker depends on the tube length and find the length of tube which will maximise the power output. Solution 5.4 Using the result derived in the answer to Example 5.3, the specific acoustic impedance seen by the loudspeaker may be written as:

228 Zs = ρc

Noise Control: From Concept to Application



Zr /ρc + j tan kL . 1 + j(Zr /ρc) tan kL



which on substituting Zr = Rr + jXr may be rewritten as: (Rr /ρc + j (Xr /ρc + tan (kL) ) ) ( 1 − (Xr /ρc) tan(kL) − j(Rr /ρc) tan(kL) ) Zs = . 2 2 ρc (1 − (Xr /ρc) tan(kL)) + ((Rr /ρc) tan(kL))

(i)

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 with 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 Equation (i) is zero. However, there is no value of kL for which this occurs, so we must find the maximum value of the entire expression. Of course the real power output is only dependent on the resistive impedance while the imaginary power output is dependent on the reactive impedance. Taking the real part of the RHS of Equation (i) gives the specific acoustic resistive impedance as: Rr /ρc 1 + tan2 (kL) Rs = 2 2 ρc (1 − (Xr /ρc) tan(kL)) + ((Rr /ρc) tan(kL))




(ii).

At 250 Hz, k = 2πf /c = 2π × 250/343 = 4.58. As a = 0.075, ka = 4.58 × 0.075 = 0.343 and πa2 = 0.0177. Thus, Rr /ρc = 0.0177 × 0.3432 /2 = 1.04 × 10−3 . and Xr /ρc = 0.0177 × 0.343 × 0.8 = 0.00486. Substituting in Equation (ii), we obtain: 1.04 × 10−3 1 + tan2 (kL) Rs = 2 2 ρc (1 − 0.00486 tan(kL)) + (1.04 × 10−3 tan(kL))




(iii).

By trial and error it can be shown that the maximum value of Equation (iii) occurs when tan(kL) = 205.9, or kL = 1.566. Thus, the optimum L for maximum power out = 1.566/4.58 = 342 mm.

Example 5.5 A loudspeaker introduces sound into one end of a small diameter (compared to a wavelength) tube of length, L, and diameter, d. The loudspeaker is at axial coordinate location, x = 0, and the other end of the tube is rigid and at x = −L. (a) Write expressions for the acoustic velocity potential, acoustic pressure and acoustic particle velocity in terms of one real unknown constant, A, the tube length, L, location, x, along the tube and angular excitation frequency, ω. ˆ0 , of the speaker (b) Evaluate the constant, A, in terms of the velocity amplitude, U diaphragm, assuming it to be a rigid piston. (c) Rewrite the expressions of part (a) for acoustic pressure and particle velocity in terms ˆ0 and use the results to write an expression for the real and imaginary parts of of U the acoustic intensity as a function of axial location, x, along the tube. (d) Give a physical interpretation of the results obtained in (c) above. Solution 5.5 The coordinate system is as shown in Figure 5.9.

229

Sound-Absorbing Materials

Rigid termination

Speaker

Z s = -pT /uT

0

-L

x

FIGURE 5.9 Arrangement for Example 5.5

(a) For a rigid termination, the phase shift on reflection is 0◦ 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, using Equation (1.23), the total acoustic pressure at any point in the
tube may be written as, pt = A e j(ωt−kx) + e j(ωt+kx+2kL) . (i) The velocity potential and acoustic particle velocity may be derived using Equations (1.2), (1.3) and (i) above, as:

A A e j(ωt−kx) + e j(ωt+kx+2kL) and ut = e j(ωt−kx) − e j(ωt+kx+2kL) . (ii) φt = jρω ρc ˆ0 ρc
U A ˆ0 e jωt . Thus, U ˆ0 = (b) At x = 0, ut = U 1 − e j2kL , so A = . ρc (1 − e j2kL ) ˆ0 , we obtain: (c) Rewriting Equations (i) and (ii) in terms of U ˆ0 ρc
U e j(ωt−kx) + e j(ωt+kx+2kL) j2kL (1 − e ) ˆ
U0 and ut = e j(ωt−kx) − e j(ωt+kx+2kL) . j2kL (1 − e ) From Equation (1.55), the real part of the acoustic intensity (where the hat denotes the complex amplitude) is:

pt =

1 ˆ 2 Re I = Re {ˆ pt u ˆ∗t } = ρcU 0 2 ˆ 2 Re = ρcU 0 =

(

(

e−jkx + e j(kx+2kL) × e jkx − e−j(kx+2kL) 2 ( 1 − e j2kL ) × ( 1 − e−j2kL )




e−jkx + e j(kx+2kL) × e jkx − e−j(kx+2kL) 2 ( 2 − e j2kL − e−2jkL )





)


)

ˆ2  ρcU 0 Re 1 − 1 + e 2jk(x+L) − e−2jk(x+L) = 0. 2 (2 − 2 cos(2kL))

(iii)

The amplitude of the imaginary part of the acoustic intensity can be derived using Equation (1.56), and from the last line in the Equation (iii), it is: ˆ2 1 ρcU 0 Q = Im { pˆt u ˆ∗t } = sin[2k(x + L)]. 2 [2 − 2 cos(2kL)] (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. Example 5.6 The plane wave reflection coefficient for normally incident waves reflected from a plane surface of specific acoustic impedance, Zs , is given by, Rp = (Zs − ρc)/(Zs + ρc).

230

Noise Control: From Concept to Application

(a) Show that the absorption coefficient (fraction of incident energy absorbed and equal 4ρcRe{Zs } . to α = 1 − |Rp |2 ) of the surface is given by, α = |Zs + ρc|2 [Hint: express |Zs + ρc|2 as [Re{Zs } + ρc]2 + [Im{Zs }]2 ]. (b) What value of Zs gives the highest absorption coefficient? Solution 5.6 (a) α = 1 − | Rp2 | = 1 −

| 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. Example 5.7 (a) For the tube shown in Figure 5.10, calculate the specific acoustic impedance looking into the left end of the tube at 100 Hz if the reflection coefficient of the surface of the sample at the right end of the tube is 0.5 + 0.5j. The higher order mode cut-on frequency of the tube is 200 Hz and the temperature of the air in the tube is 20◦ C. In your analysis, let the surface of the sample be the origin of the coordinate system. (b) Calculate the normal incidence sound absorption coefficient of the sample.

2m

FIGURE 5.10 Arrangement for Example 5.7

Solution 5.7 (a) As the frequency of interest is well below the first higher order mode cut-on frequency, we can assume that only plane waves can propagate in the tube. Set the origin, x = 0, at the surface of the sample. Then the entrance of the tube is at x = −2.0. As the sample is at the right-hand end of the tube, the incident wave will be travelling in the positive x-direction. Assuming a phase shift between the incident and reflected waves

Sound-Absorbing Materials

231

of β at x = 0, Equation (1.23) may be used to obtain expressions for the incident and reflected wave pressures as, pi = A e j(ωt−kx) and pr = B e j(ωt+kx+β) . The total pressure is thus, pt = A e j(ωt−kx) + B e j(ωt+kx+β) . (i) The total particle velocity can be calculated using Equations (i), (1.2) and (1.3) as:
1 A e j(ωt−kx) − B e j(ωt+kx+β) . ut = ρc The pressure reflection coefficient, Rp , is defined as pr /pi . Thus: Rp = (B/A) ejβ = 0.5 + 0.5j. The reflection coefficient amplitude is B/A and the phase is β. Thus, B/A = 0.707 and β = 45◦ = 0.7854 radians. Thus the specific acoustic impedance at any point in the tube may be written as (see table 1.5): A e−jkx + B e jkx+jβ A + Bej(2kx+β) pt = ρc = ρc Zs = −jkx jkx+jβ ut Ae −Be A − Bej(2kx+β) = ρc

A/B + cos(2kx + β) + j sin(2kx + β) . A/B − cos(2kx + β) − j sin(2kx + β)

k = ω/c = 2 π × 100/343 = 1.832 and x = −2. Thus, 2kx + β = −6.542. cos(2kx + β) = 0.9667, sin(2kx + β) = −0.25587 and A/B = 1.414. Thus, Zs = 413.7 × (2.3809 − j0.25587)/(0.4475 + j0.25587) = 1558(1 − j0.7237) = 1558 − j1128. (b) From Equation (5.31), the normal incidence absorption coefficient is defined as: αn = 1 − | Rp | 2 , so αn = 0.5. Example 5.8 The maximum and minimum sound pressure levels measured in an impedance tube at 250 Hz are 95 dB and 80 dB re 20 µPa, respectively. The distance from the face of the sample to the nearest pressure minimum is 0.2 m. Calculate the following: (a) (b) (c) (d) (e)

the the the the the

normal incidence absorption coefficient of the sample at 250 Hz; normal specific acoustic impedance of the sample at 250 Hz; random incidence (statistical) absorption coefficient of the sample at 250 Hz; sound intensity of the wave travelling towards the sample in the tube; magnitude and direction of the total sound intensity in the tube.

Solution 5.8 (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 Figure 5.11. As the sample 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 β at x = 0 and using Equation (1.36), the incident wave and reflected wave pressures may be written as, pi = A e j(ωt+kx) and pr = B e j(ωt−kx+β) . The total pressure is thus, pt = A e j(ωt+kx) + B e j(ωt−kx+β) . The maximum pressure will occur when β = 2kx, and the minimum will occur when β = 2kx + π. Thus, pˆmax = e jkx (A + B) and pˆmin = e jkx (A − B) and the ratio of maximum to minimum pressure amplitudes is (A + B)/(A − B). A+B From equation (5.28), the standing wave ratio, L0 , is defined as, 10L0 /20 = . A−B

232

Noise Control: From Concept to Application

sample speaker Z = -pT /uT

0

L

x

FIGURE 5.11 Arrangement for Example 5.8

10L0 /20 − 1 B = . A 10L0 /20 + 1 The amplitude of the pressure reflection coefficient squared is: |pr /pi |2 = |Rp |2 = (B/A)2 , which can be written in terms of L0 (= 95 − 80) as:



Thus, the ratio, (B/A), is,

|Rp |2 =



10L0 /20 − 1 10L0 /20 + 1

2

=



1015/20 − 1 1015/20 + 1



2

(i)

.

The absorption coefficient is defined as α = 1 − | Rp | 2 , so: 1015/20 − 1 α=1− 1015/20 + 1



2

= 0.51.

(b) The total particle velocity can be calculated using the equation in part (a) for the 1 (pr − pi ). acoustic pressure and Equations (1.2) and (1.3) as, ut = ρc
1 Thus, ut = B e j(ωt−kx+β) − A e j(ωt+kx) . ρc The specific acoustic impedance at any point in the tube may then be written as (see Table 1.5): pt A e jkx + B e−jkx+jβ A + Be j(−2kx+β) Zs = − = −ρc . = −ρc ut B e−jkx+jβ − A e jkx Be j(−2kx+β) − A At x = 0, the specific acoustic impedance is the normal impedance, Zs , of the surface of the sample. Thus: pt Zs A + B e jβ =− = . (ii) ρc ρcut A − B e jβ Equation (ii) may be expanded to give: Zs A/B + cos β + j sin β (A/B)2 − 1 + (2A/B) j sin β = = . ρc A/B − cos β − j sin β (A/B)2 + 1 − (2A/B) cos β The modulus of the impedance is then:

q Zs ρc

=

2

[(A/B)2 − 1] + (2A/B)2 sin2 β

, (A/B)2 + 1 − (2A/B) cos β and the phase is given by:     Im{Zs } 2(A/B) sin β ψ = tan−1 = tan−1 . Re{Zs } (A/B)2 − 1 A 10L0 /20 + 1 = L /20 = 1.433. B 10 0 −1 At the pressure minimum, β = 2kx − π, where k = 2π/λ. x = 0.2 m and f = 250 Hz; thus, k = 2πf /c = 4.58. β = 2 × 4.58 × 0.2 − π = −1.31 radians, cos β = 0.258 and sin β = −0.966. Using the previous analysis,

(iii)

(iv)

233

Sound-Absorbing Materials Substituting these values into Equations (iii) and (iv) gives: (1.4332 − 1) 2 + (2.8662 × 0.9662 ) | Zs | jψ = |ξ| = = ξe = 1.28, ρc 1.4332 + 1 − 2.866 × 0.258 and the phase is:   2 × 1.433 × (−0.966) −1 ψ = tan = −69.2◦ . 1.4332 − 1

p

(c) The statistical absorption coefficient is given by Equation (5.45) as: αst =



8 cos ψ ξ

 

1−





cos ψ ξ



loge (1 + 2ξ cos ψ + ξ 2 ) +

cos(2ψ) ξ sin ψ



tan−1



ξ sin ψ 1 + ξ cos ψ

 .

Substituting in the modulus and phase of the impedance, we obtain: αst =



8 × cos(−69.17) 1.28



1−



cos(−69.17) 1.28

 ×

× loge 1 + 2 × 1.28 × cos(−69.17) + 1.282 +



cos(−138.4) 1.28 sin(−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. (d) The sound intensity of the wave travelling towards the sample at the end of the tube is given by Equation (1.57) as, I = hp2i i/(ρc), where hp2i i is the RMS sound pressure of the wave incident on the sound sample. In terms of the peak sound pressure, pˆ, the sound intensity of the incident wave is, Ii = pˆ2i /(2ρc). From Equation (5.23), the amplitude of the incident wave is pˆi = A and the amplitude of the wave reflected from the sample is pˆr = B. From Equation√(5.27), A = 0.5(ˆ pmax + pˆmin ) and B = 0.5(ˆ pmax − pˆmin ), where, p√ ˆmax = 2 × 2 × 10−5 × 1095/20 = 1.591 Pa and pˆmin = 2 × 2 × 10−5 × 1080/20 = 0.283 Pa. Thus, pˆi = A = (1.591 + 0.283)/2, and the intensity of the incident wave is: Ii = A2 /(2ρc) = (1.591 + 0.283)2 /(8 × 1.206 × 343) = 1.06 mW. (e) The intensity of the reflected wave is: Ii = B 2 /(2ρc) = (1.591 − 0.283)2 /(8 × 1.206 × 343) = 0.52 mW. So the total sound intensity in the direction of the sample is 0.54 mW.

5.4.2

Measurement Using the Two-Microphone Method

The advantage of the two-microphone method over the moving microphone method for determining material absorption coefficients and normal acoustic impedance (discussed in the previous subsection) is that it considerably reduces the time required to determine the normal specific acoustic impedance and normal incidence absorption coefficient of a sample, allowing the sample to be evaluated over the frequency range of interest in one or two measurements. The upper frequency limit for valid data is related to the diameter of the tube used for the measurement in the same way as for the moving microphone method. The lower frequency limit is a function of the spacing, `s , of the two microphones (which should exceed 1% of the wavelength at the lowest frequency of interest) and the accuracy of the analysis system. However, the microphone spacing must not be larger than 40% of the wavelength at the highest frequency of interest. In

234

Noise Control: From Concept to Application

some cases, it may be necessary to repeat measurements with two different microphone spacings and two tubes with different diameters. The microphones are mounted through the wall of the tube and flush with its inside surface, as illustrated in Figure 5.12. The microphone closest to the sample surface should be located at least one tube diameter from the surface for a rough surface and half a tube diameter for a smooth surface. White noise generator

Equaliser

FFT analyser or computer Ch1

Ch2 Amplifier

Speaker Mic 2

Mic 1

L Rs Test sample Heavy metal plug

Sound absorbing material

FIGURE 5.12 Arrangement for measuring the normal incidence complex reflection coefficient, normal specific acoustic impedance and absorption coefficient of a sample of acoustic material using the two-microphone method.

The sample is mounted in a holder attached to the end of the impedance tube as illustrated in Figure 5.12 and described in ASTM E1050-12 (2012). An FFT analyser is used to determine the transfer function between the two microphones as a function of frequency. The loudspeaker at the end of the tube is excited with white noise (see Section 1.11) and the resulting transfer function (or frequency response function) between the two microphones is measured using an FFT analyser (see Section 2.12.6). Details of this method are provided in ASTM E1050-12 (2012) and Bies et al. (2018).

5.5 5.5.1

Calculation of Statistical Absorption Coefficients of Some Porous Material Configurations Porous Liner with a Backing Cavity

For porous acoustic materials, such as rockwool or fibreglass, the normal specific impedance of Equation (5.39) may also be calculated from the material characteristic impedance and propagation coefficient of Equations (5.3) and (5.4). For a material of infinite depth (or sufficiently thick that waves transmitted through the material from one face and reflected from the opposite face are of insignificant amplitude by the time they arrive back at the first face), the normal specific acoustic impedance is equal to the characteristic impedance of Equation (5.3). For a porous blanket of thickness, `, backed by a cavity of any depth, L (including L = 0), with a rigid back, the normal specific acoustic impedance (in the absence of flow past the cavity) may be calculated using an electrical transmission line analogy (Magnusson, 1965) and is given by: ZN = Zm

ZL + jZm tan(km `) Zm + jZL tan(km `)

(5.46)

235

Sound-Absorbing Materials

The quantities, Zm and km , in Equation (5.46) are defined in Equations (5.3) and (5.4). The normal specific acoustic impedance, ZL , of a rigidly terminated, partitioned backing cavity is given by: ZL = −jρc/ tan(2πf L/c) (5.47) and for a rigidly terminated, non-partitioned backing cavity, the impedance, ZL , for a wave incident at angle, β, is: ZL = −jρc cos β/ tan(2πf L/c) (5.48) where β is the angle of incidence of the sound wave measured from the normal to the surface. A partitioned cavity is one that is divided into compartments by partitions that permit propagation normal to the surface, while inhibiting propagation parallel to the surface of the liner. The depth of each compartment is the same and is equal to the overall cavity depth. If the porous material is rigidly backed so that L = 0 or, equivalently, L is an integer multiple of half wavelengths, Equation (5.46) reduces to: ZN = −jZm / tan(km `)

5.5.2

(5.49)

Porous Liner Covered with a Limp Impervious Layer

In practice, it is often necessary to protect a porous liner from contamination by moisture, dust, oil, chemicals, etc. which would render it useless. For this reason the material is often wrapped in a thin, limp, impervious blanket made of polyurethane, polyester, aluminium or PVC. The effect of the introduction of a wrapping material is generally to improve the low-frequency absorption at the expense of the high-frequency absorption. However, if the wrapping material is sufficiently thin (for example, 20 mm thick polyurethane), then the effect is not measurable. The effect can be calculated by calculating the combined normal impedance due to the membrane and the sound-absorbing material (see Equation (5.50)) and substituting the result into Equation (5.45). If the porous material is protected by covering or enclosing it in an impervious blanket of thickness, h, and mass per unit area, σ 0 , the effective normal specific acoustic impedance, ZNB , at the outer surface of the blanket, which can be used together with Equations (5.39) and (5.45) to find the statistical absorption coefficient of the construction, is: ZNB = ZN + j2πf σ 0

(5.50)

where f is the frequency of the incident tonal sound or tone, or, alternatively, the centre frequency of a narrow band of noise. Typical values for σ 0 and cL are included in Table 5.2 for commonly used covering materials. TABLE 5.2 Properties of commonly used limp impervious wrappings for environmental protection of porous materials

Material Polyethylene (LD) Polyurethane Aluminium PVC Melinex (polyester) Metalised polyester a

Density (kg/m3 )

Typical thickness (microns = 10−6 m)

σ0 (kg/m2 )a

cL (approx.) (m/s)

930 900 2700 1400 1390 1400

6–35 6–35 2–12 4–28 15–30 12

0.0055–0.033 0.005–0.033 0.0055–0.033 0.005–0.033 0.021–0.042 0.017

460 1330 5150 1310 1600 1600

σ 0 and cL are, respectively, the surface density and speed of sound in the wrapping material.

Guidelines for the selection of suitable protective coverings are given by Andersson (1981).

236

5.5.3

Noise Control: From Concept to Application

Porous Liner Covered with a Perforated Sheet

When mechanical protection is needed for a porous liner, it may be covered using perforated wood, plastic or metal panels. If the open area provided by the perforations is greater than about 20%, the expected absorption is entirely controlled by the properties of the porous liner, and the panel has no effect. Alternatively, if the facing panel has an open area of less than 20%, the statistical absorption coefficient of the panel plus sound-absorbing material can be determined by calculating the normal specific impedance of the combined construction using Equation (5.51). According to (Bolt, 1947) the normal specific impedance is:

ZNP

  100  jρc tan k`e (1 − M ) + RA Sh P = ZN +   100  1+ jρc tan k`e (1 − M ) + RA Sh jωmP

(5.51)

where ZN is the normal specific acoustic impedance of the porous acoustic material with or without a cavity backing (and in the absence of flow), ω is the radian frequency, P is the % open area of the holes, RA is the acoustic resistance of each hole (see Equation (8.21)), Sh is the area of each hole, M is the Mach number (where Mach number = speed of flow/speed of sound = U/c) of the flow past the holes and m is the mass per unit area of the perforated sheet, all in consistent SI units. The effective length, `e , of each of the holes in the perforated sheet is: `e = `w +



  16a (1 − 0.43a/q) 1 − M 2 3π 

(5.52)

where M is the Mach number of the flow, `w is the thickness of the perforated sheet, a is the hole radius and q > 2a is the distance between hole centres. If the perforated panel has an open area of less than 20%, then the frequency at which absorption is maximum may be calculated by treating the facing as an array of Helmholtz resonators in a manner similar to that used in Section 8.8.2. This procedure leads to the following approximate equation: fmax

c = 2π



P/100 Ltot [`w + 0.85d(1 − 0.22d/q)]

1/2

(5.53)

In Equation (5.53), P is the percentage open area of the panel, Ltot is the depth of the backing cavity including the porous material layer, `w is the panel thickness, d is the diameter of the perforations and q is the spacing between hole centres. If the porous material fills the entire cavity so that the thickness of the porous material is also Ltot , then the speed of sound c should be replaced with 0.85c to account for isothermal rather than adiabatic propagation of sound in the porous material at low frequencies. The condition fmax Ltot /c < 0.1 must be satisfied for Equation (5.53) to give results with less than 15% error. If this condition is not met, then it is better to use the following transcendental equation and solve by trial and error. (2πfmax Ltot /c) tan (2πfmax Ltot /c) =

P Ltot /100 `w + 0.85d(1 − 0.22d/q)

(5.54)

Measured data showing the effect of a perforated sheet on the performance of a soundabsorbing material for a panel of 10% open area are presented in Figure 5.13 (see Davern (1977) for additional data).

237

Sound-Absorbing Materials

Statistical absorption coefficient ast

1.0

>25% open area

0.8 10% open area

0.6 0.4

Panel absorber 0% open area

0.2 0

0.01

0.02

0.04 0.063 0.1

0.2

0.4

0.63

1.0

f L/c

FIGURE 5.13 Effect of perforations on the sound absorption of a panel backed by a porous liner. The panel surface weight is 2.5 kg/m2 and its thickness is 3 mm. The porous liner is 50 mm thick with a flow resistance of about 5ρc.

Example 5.9 A porous material of thickness 100 mm and a flow resistivity of 104 MKS rayls/m, is bonded to a rigid wall and covered with a perforated steel facing. The perforated facing is 3 mm thick, of 7% open area with uniformly spaced holes of 2 mm diameter. The wall is part of a low temperature curing room filled with air with an ambient temperature of 132.5◦ C. Calculate the following quantities. (a) (b) (c) (d)

Adiabatic and isothermal speeds of sound. Frequency of maximum absorption. Specific normal impedance. Statistical absorption coefficient at the frequency of maximum absorption.

Solution 5.9 (a) The adiabatic speed of sound is given by Equation (1.1).

r

1.40 × 8.314 × (273.15 + 132.5) = 403.5 m/s. 0.029 cisothermal = 0.85 × cadiabatic = 403.5 × 0.85 = 343 m/s. (b) First, calculate the distance between holes. It is possible to assume either a parallel or a staggered hole configuration as shown in Figure 5.14. Let q be the distance between holes as shown Figure 5.14. 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: Thus: cadiabatic =

s q=

π × d2h = 4 × (P/100)

r

π × 0.0022 = 0.0067 m. 4 × 0.07

where dh is the hole diameter and P is the % open area of the panel. For the staggered holes, it is a bit more complicated, as the row spacing of the holes is different to the hole spacing in an individual row as can be seen in Figure 5.14. In this case, we can assume a perforated sheet of dimensions, x × y and an individual hole area of πd2 /4, where d is the hole diameter. The number of holes in one row across is then, x/q and the number of holes in the vertical direction in Figure 5.14 is

238

Noise Control: From Concept to Application

q/2 q/2 q

q

q q

staggered holes

parallel holes

FIGURE 5.14 Parallel and staggered hole configurations for a perforated sheet.

√ √ 2y/(q 3). The total number of holes in the sheet is then 2xy/(q 2 3) and these have √ a total area of πd2 xy/(2q 2 3). The total area of the√panel is xy and the ratio of the total hole area to panel area is 0.07. Thus, πd2 /(2q 2 3) = 0.07.

s Therefore, q =

π × d2h √ = 2 × (P/100) 3

r

π × 0.0022 √ = 0.0072 m. 2 × 0.07 3

For the purposes of this problem, we will use q = 0.0067 m. Using equation (5.53) and assuming isothermal propagation (as the material fills the entire space behind the perforated sheet), we have:  1/2 343 0.07 fmax = = 674 Hz. 2π 0.1[0.003 + 0.85 × 0.002(1 − 0.22 × 0.002/0.0067)] However, this frequency does not satisfy the required condition of fmax Ltot /c < 0.1 in order for the error to be less than 15%. Thus we should use the transcendental equation, (5.54) to give:     2π × 0.1 7 × 0.1/100 2π × 0.1 fmax tan fmax = . 343 343 0.003 + 0.85 × 0.002 × (1 − 0.22 × 0.002/0.0067) Rewriting gives, 0.00183 × fmax tan (fmax × 0.00183) = 1.5256, which we can solve by trial and error, choosing values of fmax until the LHS ≈ 1.5256, giving fmax = 543 Hz. We will use this value in the following calculations. (c) The normal specific acoustic impedance is given by Equation (5.51). To evaluate this equation we need to use Equation (5.49) and to evaluate that we need Equations (5.3) and (5.4). Referring to Equation (5.15), X = 1.206 × 343/10000 = 0.0655. Thus, T1 = 1.63259, T2 = 0.06590, T3 = 3.1696, T4 = 0.1663. Referring to Equation (5.9): 3.1696(1.63259 − 3.1696) × 0.06592 − 0.16632 × 1.632592 = −0.80849, 3.16962 × 0.06592 + 0.16632 × 1.632592

a(X) =

and referring to Equation (5.10): b(X) =

1.632592 × 0.0659 × 0.1663 = 0.24893. 3.16962 × 0.06592 + 0.16632 × 1.632592

X1 = 0.856 × 1.206 × 543/10000 = 0.05605. Thus: T1 = 1.54150, T2 = 0.05636, T3 = 3.07850, T4 = 0.16526. Referring to Equation (5.9): a(X1 ) =

3.0785(1.5415 − 3.07850) × 0.056362 − 0.165262 × 1.54152 = −0.84133 3.07852 × 0.056362 + 0.165262 × 1.54152

and, b(X1 ) =

1.54152 × 0.05636 × 0.16526 = 0.23297. 3.07852 × 0.056362 + 0.165262 × 1.54152

239

Sound-Absorbing Materials Thus τ = −0.84133 × 0.592 + j0.23297 and σ = −0.80849 + j0.24893. The quantities, κ and ρm , may be calculated using Equations (5.5) and (5.6) as follows: κ/κ0 = (1 − 0.4 × (−0.49807 + j0.23297))−1 = (1.19923 + j0.093188)−1 , and ρm /ρ = (1 − 0.8085 + j0.2489)−1 = (0.1915 + j0.24893)−1 . Using Equation (5.3), we obtain: Zm = ρc

r

ρm κ = ρκ0

=

r

1 (0.1915 + j0.24893) × (1.19923 − j0.093188)

√ √ 1 = 1.7717 − j1.9667 = 2.6471e−j0.8375 0.25285 + j0.28068

r

= 1.6270e−j0.4188 = 1.4864 − j0.6616, and using Equation (5.4), we obtain: km

2πfmax = c

r

ρm κ0 2π × 543 = ρ0 κ 343

r

1.19923 − j093188 0.1915 + j0.24893

r

(1.19923 − j0.093188) × (0.1915 − j0.24893) 0.098638 √ √ = 9.9468 2.09319 − j3.2074 = 9.9468 3.8300e−j0.99259

= 9.9468

= 19.4663e−j0.49629 = 17.1178 − j9.2702. Before continuing, it will be useful to evaluate the quantity, tan(km `), where the porous material thickness, ` = Ltot . Using the previous result for km , setting ` = 0.1 and using the identity, j tan z = (e jz − e−jz )/(e jz + e−jz ), we can write: j tan(km `) = =

e0.9269 (cos(1.7118) + j sin(1.7118)) − e−0.9269 (cos(1.7118 − j sin(1.7118)) e0.9269 (cos(1.7118) + j sin(1.7118)) + e−0.9269 (cos(1.7118) − j sin(1.7118)) −0.2995 + j2.8934 = 1.3480 − j0.1205. −0.4107 + j2.1097

Thus, tan(km `) = −0.1205 − j1.3480. Using the previous results and Equation (5.49), assuming a rigid backing for the porous material, we can write: ZN (0.6616 + j1.4864) 2.0834 − j0.7127 = = = 1.1375 − j0.3891. ρc 0.1205 + j1.3480 1.8316 To calculate the overall impedance, we use Equation (5.51), but first we need to evaluate the effective length of the holes in the perforated sheet, `e , and tan(k`e ). From Equation (5.52), assuming no flow so that M = 0, the effective length of the holes in the perforated sheet is: `e = 0.003 +

16 × 0.001 (1 − 0.43 × 0.001/0.0067) = 0.004588. 3×π

Thus, tan(k`e ) = tan(2 × π × 543 × 0.004588/343) = 0.04567. We also need to calculate the acoustic resistance of the holes (see Section 8.7.1.2, which contains the following equation). RA =





ρc ktD`w 1 + (γ − 1) S 2S

r

5 3γ



+ 0.288kt log10



4S Sk 2 + + 0.7M , 2 πh 2π





where the variables are defined in Section 8.7.1.2. To evaluate this equation we need

240

Noise Control: From Concept to Application the following quantities: k = (2 × π × 543)/343 = 9.9469, S = π × 0.0022 /4 = 3.1416 × 10−6 m2 , D = π × 0.002 = 0.006283,

r

2 × 1.8 × 10−5 = 9.3538 × 10−5 , 1.206 × 2 × π × 543 `w = 0.003, γ = 1.4, M = 0 and  = 1.0, as the situation is radiation from a baffle. t=

The quantity, h, is the largest of the half plate thickness or t. Thus: h = `w /2 = 0.0015 m. The above quantities may be inserted into Equation (8.21) to give: RA S 9.9467 × 9.3538 × 10−5 × 6.283 × 10−3 × 0.003 1 + (1.4 − 1) = ρc 2 × 3.1416 × 10−6



+ 0.288 × 9.9467 × 9.3538 × 10−5 × log10 +



4 × 3.1416 × 10−6 π × 0.00152

r

5 3 × 1.4





3.1416 × 10−6 × 9.94672 = 4.00935 × 10−3 + 6.69555 × 10−5 + 4.9470 × 10−5 2×π

= 4.1258 × 10−3 . We can now use Equation (5.51) to evaluate the overall impedance. The second term on the right is the impedance due to the perforated sheet and is: (100/7)(0.04567j + 4.1256 × 10−3 ) 100 × 1.206 × 343 1+ × [0.04567 − 4.1256j × 10−3 ] 2 × π × 543 × 21.902 × 7 0.05894 + 0.6524j = = 0.05852 + 0.6502j, 1.00363 − 3.2836j × 10−4

ZP = ρc

where the value of m = 21.902 used in the equation is obtained from: m = ρp `w (100 − P )/100 = 7850 × 0.003 × 0.93 (`w is the perforated panel thickness and ρp is its material density = 7850 kg/m3 for mild steel). The total impedance is: ZP ZN + = 1.1375 − 0.3891j + 0.05894 + 0.6524j ρc ρc = 1.1960 + 0.2610j = 1.2242e0.2149j , ZN ZP + = ξe jψ , where ρc ρc cos ψ = 0.9770; cos 2ψ = 0.9091; sin ψ = 0.2132 and ξ = 1.2242. Using the above data, the statistical absorption coefficient may be calculated using Equation (5.45) as follows: or

αst =



8 × 0.9770 1.2242



0.9770 1− × loge (1 + 2 × 1.2242 × 0.9770 + 1.22422 ) 1.2242





0.9091 1.2242 × (0.2132) + × tan−1 1.2242 × (0.2132) 1 + 1.2242 × 0.9770









= 6.385 × (1 − 0.7981 × 1.5873 + 3.483 × tan−1 [0.1188]) = 6.385 × (1 − 1.2668 + 0.4119) = 0.93.

241

Sound-Absorbing Materials

5.5.4

Porous Liner Covered with a Limp Impervious Layer and a Perforated Sheet

In this case, the impedance of the perforated sheet and impervious layer of thickness, h, and mass per unit area, σ 0 , are both added to the normal specific acoustic impedance of the porous acoustic material, so that:

ZNBP

  100  jρc tan k`e (1 − M ) + RA Sh 0 P = ZN +   + j2πf σ 100  1+ jρc tan k`e (1 − M ) + RA Sh j ωmP

(5.55)

It is important that the impervious layer and the perforated sheet are separated, using something like a mesh spacer (with a grid size of at least 2 cm); otherwise, the performance of the construction as an absorber will be severely degraded, as the impervious layer will no longer be acting as a limp blanket.

5.6

Measurements of the Sabine Absorption Coefficient and Room Constant

Measurements of the room constant, R = S α ¯ /(1 − α ¯ ), discussed in Section 6.4.1, or the related Sabine absorption coefficient averaged over all room surfaces, α ¯ , may be made using either a reference sound source or by measuring the reverberation time of the room in the frequency bands of interest. These methods are described in the following two subsections. Alternatively, yet another method is offered in Section 3.11.3.1. If the Sabine absorption coefficient of a particular material is required, then measurements of the material must be done in a reverberant test chamber as described in Section 5.6.3.

5.6.1

Reference Sound Source Method

The reference sound source is placed at a number of positions chosen at random in the room to be investigated, and sound pressure levels are measured at a number of positions in the room for each source position. In each case, the measurement positions are chosen to be remote from the source, where the reverberant field of the room dominates the direct field of the source. The number of measurement positions for each source position and the total number of source positions used are usually dependent on the irregularity of the measurements obtained. Generally, four or five source positions with four or five measurement positions for each source position are sufficient, giving a total number of measurements between 16 and 25. The room constant, R, for the room is then calculated using Equation (6.13) rearranged as: R = 4 × 10(LW −Lp )/10

(5.56)

In writing Equation (5.56), the direct field of the source has been neglected, following the measurement procedure proposed above and it has been assumed that ρc = 400. In Equation (5.56), Lp is the average of all the sound pressure level measurements and is calculated using the following equation:

" Lp = 10 log10

N 1 X (Lpi /10) 10 N

# (dB re 20 µPa)

(5.57)

i=1

The quantity LW is the sound power level (dB re 10−12 W) of the reference sound source, and N is the total number of measurements.

242

5.6.2

Noise Control: From Concept to Application

Reverberation Time Method

The second method is based on a measurement of the room reverberation time, which is the time (seconds) for the sound in the room to decay by 60 dB after the sound source is turned off. The sound source is usually a loudspeaker driven by a random noise generator in series with a bandpass filter. When the sound is turned off, the room rate of decay can be measured simply by using a sound level meter attached to a level recorder (or data acquisition system and computer) as illustrated in Figure 5.15. Alternatively, there are many acoustic instruments such as spectrum analysers that can calculate the reverberation time internally for all 1/3-octave bands simultaneously. In this case, the “data acquisition system” box, “bandpass filter” box and the “sound level meter” box are replaced with a “sound analyser” box. However, it is important to ensure that the signal level in each band is at least 45 dB above any background noise. If it is less, the reverberation time results will be less accurate. Noise generator

Bandpass filter

Power amplifier

Loudspeaker

Microphone Data acquisition system

Bandpass filter

Sound level meter

Reverberant enclosure

FIGURE 5.15 Equipment for reverberation time (T60 ) measurement.

The reverberation time, T60 , in each frequency band is determined as the reciprocal sound pressure level decay rate obtained using the level recorder or the spectrum analyser. As shown in Figure 5.16, the recorded level in decibels should decay linearly with time. The slope, generally measured as the best straight line fit to the recorded decay between 5 dB and 35 dB down from the initial steady-state level, is used to determine the decay rate. The time for the sound to decay 30 dB (from −5 dB to −35 dB below the level at the time the sound source was switched off) is estimated from the slope of the straight line fitted to the trace of sound pressure level vs time. The reverberation time, T60 , for a 60 dB decay, is found by multiplying the 30 dB decay result by 2. For the example shown in Figure 5.16, the reverberation time, T60 = (1.55 − 0.15) × 2 = 2.8 seconds. The 35 dB down level should be at least 10 dB above the background noise level for any 1/3-octave or octave band measurement. It is usual to employ several different microphone positions in the room and to determine the reverberation times at each position. The value of T60 used in the subsequent calculations is then the average of all the values obtained for a given frequency band. Once found, the reverberation time, T60 , is used in Equation (6.15), rearranged as follows, to calculate the room absorption. Thus: Sα ¯ = (55.25V )/(cT60 ) (m2 ) (5.58) At this point, a word of caution is in order. When processing the data, determine average decay rates not decay times, even though ISO 354 (2003) suggests that similar results are obtained by arithmetically averaging reverberation times! Thus, what is required is: N 1 1 X 1 = T60 N T60i i=1

(5.59)

243

Sound-Absorbing Materials

Sound pressure level (dB)

90 80

Fitted straight line between -5 dB and -35 dB down points

70 Measured decay curve

60 50 0

1.0

2.0

3.0

Time (seconds) FIGURE 5.16 Measurement of reverberation time.

When observing reverberation decay curves (average sound pressure level versus time), it will be noted that for almost any room, two different slopes will be apparent. The steeper slope occurs for the initial 7 to 10 dB of decay, the exact number of dB being dependent on the physical characteristics of the room and contents. When this initial slope is extrapolated to a decay level of 60 dB, the corresponding time is referred to as the early decay time (EDT). The slope of the remainder of the decay curve, when extrapolated to 60 dB, corresponds to what is commonly referred to as the reverberation time (RT). The ratio of EDT to RT as well as the absolute values of the two quantities are widely used in the design of architectural spaces. For more information see Beranek (1962), Mackenzie (1979), Cremer and Müller (1982), Bies and Hansen (2009) and Egan (1987).

5.6.3

Measurement of α ¯ for a Particular Material

The Sabine absorption coefficient of a sample of material can be measured in a specially designed test chamber called a reverberation room, which has long reverberation times. Procedures and test chamber specifications are described in various standards (ASTM C423-17, 2017; ISO 354, 2003; AS 1045, 1988). The material to be tested is placed in a reverberant room and the rever0 beration time, T60 , is measured. The test material is removed and the reverberation time, T60 , of the room containing no test material is measured next. Provided that the absorption of the reverberation room in the absence of the test material is dominated by the absorption of the walls, floor and ceiling, the reverberation times are related to the test material absorption, S α ¯, by the following equation (derived directly from Equation (5.58)): 55.3V Sα ¯= c



1 (S 0 − S) − 0 T60 S 0 T60



(m2 )

(5.60)

The quantity, S 0 , is the total area of all surfaces in the room including the area covered by the material under test. Equation (5.60) is written with the implicit assumption that the surface area, S, of the test material is large enough to measurably affect the reverberation time, but not so large as to seriously affect the diffusivity of the sound field, which is basic to the measurement procedure. The standards recommend that S should be between 10 and 12 m2 with a length-tobreadth ratio between 0.7 and 1.0. The tested material should also be more than 1 m from the nearest room edge and the edge of the test specimen should not be parallel to the nearest room edge. Absorption coefficients for materials other than flat sheets can be determined in a similar

244

Noise Control: From Concept to Application

way to flat sheets, but there should a sufficient number of test specimens to achieve a value of Sα between 1 and 12 m2 . In many cases, the absorption of a reverberation room is dominated by things other than the room walls, such as loudspeakers at low frequencies, stationary and rotating diffuser surfaces at low- and mid-frequencies and air absorption at high frequencies. For this reason, the contribution of the room to the total absorption is often considered to be the same with and without the presence of the sample. In this case, the additional absorption due to the sample may be written as:   1 1 55.3V − 0 (m2 ) (5.61) Sα ¯= c T60 T60 Equation (5.61) is what appears in most current standards. Reverberation times used in Equations (5.60) and (5.61) are averages over many samples (10 to 20). Averages are obtained using reciprocal reverberation times as indicated by equation (5.59). The measured value of the Sabine absorption coefficient is dependent on the sample size, sample distribution and the properties of the room in which it is measured. Because standards specify the room characteristics and sample size and distribution for measurement, similar results can be expected for the same material measured in different laboratories (although even under these conditions significant variations have been reported). However, these laboratory-measured values are used to calculate reverberation times and reverberant sound pressure levels in auditoria and factories that have quite different characteristics, which implies that in these cases, values of reverberation time, T60 , and reverberant field sound pressure level, Lp , calculated from measured Sabine absorption coefficients (using Equations (5.58) and (6.12), respectively) are approximate only. This is why many practitioners prefer to do their calculations of reverberation time in building spaces that are not very reverberant using statistical absorption coefficients and Equation (6.16), rather than using Sabine absorption coefficients and Equation (6.15). Values of Sabine absorption coefficient for various materials and wall constructions are given in Table 6.2 of Bies et al. (2018). The approximate nature of the available data makes it desirable to either use manufacturer’s data or take measurements (if possible).

5.7

Panel Sound Absorbers

Figure 5.4 indicates that porous liners are not very effective in achieving low-frequency absorption. When low-frequency absorption is required, panel absorbers are generally used. These consist of flexible panels mounted over an air space; for example, on battens (or studs) attached to a solid wall. Alternatively, such panels may be mounted on a suspended ceiling. In any case, to be effective the panels must couple with and be driven by the sound field. Acoustic energy is then dissipated by flexure of the panel. Additionally, if the backing air space is filled with a porous material, energy is also dissipated in the porous material. Maximum absorption occurs at the first resonance of the coupled panel-cavity system. It is customary to assign a Sabine absorption coefficient to a resonant panel absorber, although the basis for such assignment is clearly violated by the mode of response of the panel absorber; that is, the absorption is not dependent on local properties of the panel but is dependent on the response of the panel as a whole. Furthermore, as the panel absorber depends on strong coupling with the sound field to be effective, the energy dissipated is very much dependent on the sound field and thus on the rest of the room in which the panel absorber is used. This latter fact makes the prediction of the absorptive properties of panel absorbers difficult. A typical example of absorption coefficients for a panel absorber is shown in Figure 5.13. An empirical prediction scheme (Hardwood Plywood Manufacturers’ Association, 1962) for flexible panel absorbers that has been found useful in auditoria and concert halls will be outlined. The essence of the prediction scheme is contained in Figures 5.17 and 5.18. First of all, the type

245

Sound-Absorbing Materials

of Sabine absorption curve desired is selected from curves A to J in Figure 5.17. The solid curves are for configurations involving a blanket (25 mm thick and flow resistance between 2ρc and 5ρc) in the air gap behind the panel, while the dashed curves are for no blanket. Next, the frequency f0 , which is the fundamental panel resonance frequency and the frequency at which maximum absorption is required, is determined and Figure 5.18 is entered for the chosen curve (A to J) and the desired frequency f0 . The required air gap (mm) behind the panel and the required panel surface density (kg/m2 ) are read directly from the figure. 1.2 A

1.0

B 0.8

C D

a 0.6 E 0.4

F G H I J

0.2 0.0 0.15 0.2

0.5

1.0 f /f0

2

4

6

8 10

FIGURE 5.17 Sabine absorption coefficients for resonant plywood panels. The panel configurations corresponding to the curves labelled A–J may be identified using Figure 5.18. Dashed curves (G–J) represent configurations with no absorptive material in the cavity behind the panel. Configurations A–F require a sound-absorbing blanket between the panel and backing wall. The blanket must not contact the panel and should be between 10 and 50 mm thick and consist of glass or mineral fibre with a flow resistance between 1000 and 2000 MKS rayls. Panel supports should be at least 0.4 m apart.

The resonance frequency used in the preceding procedure is calculated using: 1 f0 = 2π

r

ρc2 mL

(Hz)

(5.62)

which does not take into account the panel rigidity or geometry. A more accurate equation for a plywood panel is (Sendra, 1999): 1 f0 = 2π

s

ρc2

√ mL + 0.6L ab

(Hz)

(5.63)

where m is the mass per unit area of the panel (kg/m2 ), L is the depth of the backing cavity and a, b are the panel dimensions. Thus, it is recommended that before using Figure 5.18, the frequency, f0 of maximum desired absorption be multiplied by the ratio:

r

m √ m + 0.6 ab

(5.64)

246

Noise Control: From Concept to Application

10.0

J

E

50

5 12

C

G

2.0

85 0 10

D

70

H

60

3.0

B

0 15 5 17 00 2

1.0

0

25

0.5 0.3 25 30

0 30 0 35

40

A

Surface density, m (kg m-2 )

I

F

z)

H f 0( 40

5.0

0

50

100

200

500

1000 1500

Cavity depth, L (mm) FIGURE 5.18 Design curves for resonant plywood panels, to be used in conjunction with Figure 5.17. The quantity, f0 , is the frequency at which maximum sound absorption is required.

Example 5.10 Design a panel absorber to have a maximum Sabine absorption coefficient of 0.8 at 125 Hz. Solution 5.10 We require α ¯ = 0.8 at 125 Hz. Referring to Figure 5.17, it is clear that we need to use curve “D” which implies the use of sound-absorbing material behind the panel. From Figure 5.18, it can be seen that the 125 Hz line crosses curve “D” when the cavity depth is 110 mm and the panel mass is 2.1 kg/m2 . Thus this is the required design. Note that the guide notes in the caption of Figure 5.17 should also be included as design specifications.

5.8

Noise Reduction Coefficient (NRC)

Sometimes it is useful to use a single number to describe the absorption characteristics of a material. This is particularly useful when a comparison of the relative benefit of a number of different materials has to be made quickly. For this purpose, the frequency-averaged Noise Reduction Coefficient (NRC) has been introduced. It is defined as: NRC = (¯ α250 + α ¯ 500 + α ¯ 1000 + α ¯ 2000 ) /4

(5.65)

where α ¯ may be replaced by αst , as the equation applies to both types of absorption coefficient. Example 5.11 Sound-absorbing material with a noise reduction coefficient of 0.8 has been specified for use in hanging sound absorbers in a noisy factory. You are offered rockwool material characterised by the sound absorption coefficients shown in the table below. Does the material satisfy an NRC of 0.8?

247

Sound-Absorbing Materials Octave band centre frequency (Hz) 125 250 500 1000 2000 Sabine absorption coefficient

0.4

0.6

0.8

1.0

1.0

Solution 5.11 From Equation (5.65), the NRC is given by: α ¯ 250 + α ¯ 500 + α ¯ 1000 + α ¯ 2000 0.6 + 0.8 + 1.0 + 1.0 = = 0.85 4 4 So the material is adequate for the purpose.

NRC =

5.9

Sound Absorption Coefficients of Materials in Combination

When a surface is characterised by several different materials having different values of the absorption coefficient, it is necessary to determine an average Sabine absorption. For example, for a room characterised by q different materials, the average absorption coefficient will be the area-weighted average absorption of the q different materials calculated as: q P

α ¯=

Si α ¯i

i=1 q P

(5.66) Si

i=1

where Si are the areas and α ¯ i are the Sabine absorption coefficients of the q materials.

5.10

Reverberation Control

If the reverberant sound field dominates the direct field, then the sound pressure level will decrease if absorption is added to the room or factory. The decrease in reverberant sound pressure level, ∆Lp , to be expected in a Sabine room for a particular increase in sound absorption, expressed in terms of the room constant, R (see Equation (6.14)), may be calculated by using Equation (6.13) with the direct field term set equal to zero. The following equation is thus obtained, where Ro is the original room constant and Rf is the room constant after the addition of sound-absorbing material.   Rf (5.67) ∆Lp = 10 log10 Ro As discussed in Section 6.4.1, the room constant, R = S α ¯ /(1 − α ¯ ). It can be seen from Equation (5.67) that if the original room constant, Ro , is large then the amount of additional absorption to be added must be very large so that Rf  Ro and ∆Lp is significant and worth the expense of the additional absorbent. Clearly, it is more beneficial to treat hard surfaces such as concrete floors, which have small Sabine absorption coefficients, because this will have the greatest effect on the room constant. To affect as many room modes as possible, it is better to distribute any sound-absorbing material throughout the room, rather than having it only on one surface. However, if the room is very large compared to the wavelength of sound considered, distribution of the sound-absorbing material is not so critical, because there will be many more oblique modes than axial or tangential modes in a particular frequency band. As each mode may be assumed to contain approximately

248

Noise Control: From Concept to Application

the same amount of sound energy, then the larger percentage of sound energy will be contained in oblique modes, as there are more of them. Oblique modes consist of waves reflected from all bounding surfaces in the room and thus treatment anywhere will have a significant effect, although distribution of the sound-absorbing material equally between the room walls, floor and ceiling will be more effective than having it only on one of those surfaces. Example 5.12 A machine to be operated in a factory produces 0.01 W of acoustic power. The building’s internal dimensions are 10 × 10 × 3 m.The reverberant field sound pressure level is 83 dB and the mean Sabine absorption coefficient of the room is 0.39. Estimate the required room constant and the mean absorption coefficient of the room surfaces if the reverberant sound pressure level must be reduced to 80 dB re 20 µPa. Solution 5.12 Using Equation (5.67), the room constant needed to achieve a reverberant level of 80 dB can be calculated from the room constant corresponding to a reverberant level of 83 dB. The room surface area, S = 2(10 × 10 + 2 × 10 × 3) = 320 m2 . The original room constant corresponding to a reverberant field level of 83 dB (mean absorption coefficient of 0.39) is: Ro = 320 × 0.39/(1 − 0.39) = 204.6. In this case, ∆Lp = 3, and the final room constant is: Rf = Ro × 10∆Lp /10 = 204.6 × 100.3 = 408 m2 . Rearranging the equation for the room constant, the corresponding mean absorption coefficient is, α ¯ = Rf /(Rf +S) = 408.2/(408.2+320) = 0.56.

5.11

Additional Problems

1. (a) For the tube shown in Figure 5.19, calculate the specific acoustic impedance looking into the left end of the tube at 400 Hz if the reflection coefficient of the surface of the sample at the right end of the tube is 0.75 + 0.3j. The higher order mode cut-on frequency of the tube is 1200 Hz and the temperature of the air in the tube is 20◦ C. In your analysis, let the surface of the sample be the origin of the coordinate system. (b) Calculate the normal incidence sound absorption coefficient of the sample. (c) Calculate the statistical absorption coefficient for the sample. 2m

FIGURE 5.19 Arrangement for Problem 1.

2. The maximum sound pressure level measured in a 2.5 cm diameter impedance tube containing a test sample at the opposite end to a loudspeaker, generating sound at 500 Hz, is 90 dB and the two minima closest to the test sample have values of 82 dB and 83 dB, respectively. The distance from the face of the sample to the nearest pressure minimum is 0.1 m.

249

Sound-Absorbing Materials (a) Derive an expression for the ratio, (B/A), of the scalar amplitude of the reflected wave to that of the wave incident on the sample as a function of the difference between the maximum and extrapolated minimum sound pressure levels in the tube. (b) Calculate the normal incidence absorption coefficient of the sample at 500 Hz. (c) Calculate the losses in the tube in dB/m at 500 Hz. 3. A small diameter tube with a loudspeaker mounted at one end may be used to measure the normal incidence absorption coefficient, αn , of a sample of material mounted at the other end. The quantity, αn , is defined as the ratio of the energy absorbed by the sample to that incident upon it. The energy reflection coefficient, |Rp |2 , is simply 1 − αn . For single frequency sound, use the solution to the wave equation to derive an expression for the total sound pressure amplitude as a function of axial location, x, in the tube and the pressure amplitude reflection coefficient, Rp , of the sample. Let the sample surface be at x = 0 and the loudspeaker be at x = L. Assume a phase shift of β between the waves incident and reflected from the sample. 4. Sound-absorbing material with a noise reduction coefficient of 0.8 has been specified for use in hanging sound absorbers in a noisy factory. You are offered rockwool material characterised by the sound absorption coefficients shown in Table 5.3. What is the NRC value? TABLE 5.3

125 Sabine absorption coefficient

0.40

Data for Problem 4

Octave band centre frequency (Hz) 250 500 1000 0.70

0.80

1.0

2000 1.0

5. Find the mean Sabine absorption coefficient for a room of dimensions 6.84 × 5.565 × 4.72 m high, if the floor and ceiling have a mean absorption coefficient of 0.02, the two smaller walls a coefficient of 0.05 and the large walls a coefficient of 0.06. 6. A loudspeaker is used to introduce sound into an impedance tube, 2.5 m long and 10 cm diameter, to determine the absorption coefficient (at 315 Hz) of a sample placed at one end. The maximum measured sound pressure level in the tube is 99.0 dB and the levels of the two minima closest to the sample are 90.0 and 91.5 dB. The distance between the surface of the sample and the first minimum in the tube is 0.15 m. (a) Calculate the normal incidence sound absorption coefficient of the sample. (b) Calculate the sound intensity (W/m2 ) of the wave propagating towards the sample in the tube. (c) Calculate the sound intensity (W/m2 ) of the wave propagating away from the sample (at the sample surface) in the tube. (d) Calculate the acoustic power (W) dissipated by the sample. (e) Calculate the acoustic power (W) dissipated in the tube. (f) What will be the minimum continuous electrical power rating (W) required of the loudspeaker? State any assumptions that you make. 7. A perforated sheet is used to cover a 100 mm thick blanket of rockwool inside a brick enclosure around a large item of equipment emitting a 500 Hz tone. The aim is to use the perforated sheet to maximise sound absorption at 500 Hz. Determine the optimum percentage of open area of a steel perforated sheet, 3 mm thick with holes 3 mm in diameter.

250

Noise Control: From Concept to Application

8. In a rectangular shaped auditorium of dimensions 30 m × 20 m × 3 m high, the existing reverberation times at 20◦ C and 50% relative humidity are listed in Table 5.4. Using a combination of panel absorbers and 25 mm thick rockwool (covered with a thin layer of felt), determine how many square metres of the wall and ceiling surface you must cover with each type of absorption to achieve the reverberation times in the second line of the table. At higher frequencies, air absorption is likely to be a contributor and should be included. The absorption coefficients of the rockwool are listed in line 3 of the table. TABLE 5.4

Octave band centre frequency (Hz) 250 500 1000 2000

125 Existing reverberation times (s) Desired reverberation times (s) Rockwool Sabine absorption coefficient

Data for Problem 8

4000

3.0

1.6

1.2

0.9

0.8

0.8

1.0

0.8

0.5

0.4

0.4

0.4

0.08

0.2

0.7

0.9

1.0

1.0

9. A machine in a small factory produces the octave band sound power levels listed in line 1 of Table 5.5. The factory dimensions are 10 m × 8 m × 4 m high. (a) Calculate the octave band reverberant field sound pressure levels at 20◦ C and 50% relative humidity if the average Sabine absorption coefficients of the room surfaces are as in line 2 of the table. These coefficients do not include air absorption (which may be calculated using the procedure outlined in Section 6.4.1). (b) Calculate the overall reverberant field sound pressure level in dBA re 20 µPa. (c) Determine how much of the ceiling surface would need to be covered with 50 mm thick rockwool (sprayed with polyurethane and protected with a 25% open perforated panel) to reduce the overall reverberant sound pressure level by 5 dBA. Absorption coefficients for this material are listed in line 3 of the table.

TABLE 5.5

125 Octave band sound power level (dB re 10−12 W) Existing Sabine absorption coefficients Rockwool Sabine absorption coefficient

Data for Problem 9

Octave band centre frequency (Hz) 250 500 1000 2000

4000

90.0

84.0

82.0

80.0

78.0

76.0

0.08

0.1

0.15

0.2

0.3

0.4

0.1

0.6

1.0

1.0

1.0

1.0

6 Sound in Rooms

LEARNING OBJECTIVES In this chapter, the reader is introduced to: • low and high-frequency descriptions of sound in rooms; • transient response of Sabine rooms and reverberation decay; and • reverberation and absorption in rooms.

6.1

Introduction

Sound in an enclosed space is strongly affected by the reflective properties of the enclosing surfaces and to the extent that the enclosing surfaces are reflective, the shape of the enclosure also affects the sound field. When an enclosed space is bounded by generally reflective surfaces, multiple reflections will occur, and a reverberant field will be established in addition to the direct field from the source. Thus, at any point in such an enclosure, the overall sound pressure level is a function of the energy contained in the direct and reverberant fields. In general, the energy distribution and variation with frequency of a sound field in an enclosure with reflective walls is difficult to determine with precision. Fortunately, average quantities are often sufficient and procedures have been developed for determining these quantities. Accepted procedures divide the problem of describing a sound field in a reverberant space into low- and high-frequency ranges, loosely determined by the ratio of a characteristic dimension of the enclosure to the wavelength of the sound considered. In enclosures with lightweight walls (e.g. passenger spaces in cars), the coupling of the sound field with the enclosing wall dynamics has a significant effect on the internal sound field. This type of problem will not be considered here. Rather, consideration will be limited to building enclosures, where in general the coupling of the interior sound field with the wall dynamics may be ignored. This is discussed in more detail in Chapter 6 of Bies et al. (2018). When the reflective surfaces of an enclosure are not too distant from one another and none of the dimensions is so large that air absorption becomes important, the time-averaged sound energy density (Equation (1.47)) (but not the sound pressure) of a reverberant field will tend to uniformity throughout the enclosure. Generally, reflective surfaces will not be too distant, as intended here, if no enclosure dimension exceeds any other dimension by more than a factor of about three. Such an enclosure is often referred to as a “Sabine” type enclosure, after the person who first investigated the properties of enclosed sound fields. As the distance from the sound source increases in this type of enclosure, the relative contribution of the reverberant field to the overall sound field will increase until it dominates the direct field. 251

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Noise Control: From Concept to Application

In the high-frequency region, it is possible to characterise the sound field using space-average quantities, whereas at low frequencies, the sound field needs to be described as a superposition of modes, each of which is characterised by an acoustic resonance and a mode shape. The mode shape describes the distribution of time-averaged sound pressure corresponding to the particular mode. At any one frequency the sound field is made up of the superposition of a number of modal pressure distributions corresponding to modes, which may or may not be resonant. The number of acoustic resonances in an enclosure increases very rapidly as the frequency of excitation increases. Consequently, in the high-frequency range, the possible resonances become so numerous that they cannot be distinguished from one another. Thus, one observes a generally uniform time-averaged sound field in the regions of the reverberant field removed from the source. In this frequency range, the resulting sound field is essentially diffuse and may be described in statistical terms or in terms of space-average properties. Enclosed spaces are occasionally encountered in which some of the bounding surfaces may be relatively remote or highly absorptive, and such spaces also are of importance. For example, lateral surfaces may be considered remote when the ratio of enclosure width-to-height or widthto-length exceeds a value of about three. Among such possibilities are flat rooms, characteristic of many industrial sites in which the side walls are remote or simply open, and long rooms such as corridors or tunnels. These two types of enclosure, which have been recognised and have received attention in the technical literature, are discussed in detail by Bies et al. (2018). Only flat rooms are discussed in this chapter (Section 6.5). Example 6.1 Define what is meant by “direct field” and “reverberant field”. Solution 6.1 Direct field The direct field is defined as that part of the sound field radiated by a source, which has not suffered any reflection from any room surfaces or obstacles. Reverberant field The reverberant field is defined as that part of the sound field radiated by a source which has experienced at least one reflection from the boundary of the enclosure containing the source.

6.2

Low Frequency Behaviour

In the low-frequency range, an enclosure sound field is dominated by standing waves at certain characteristic frequencies called resonance frequencies. A resonance frequency is a frequency at which a sound wave reflected from a wall is in phase with the one reflected immediately previously and the one reflected next, so that a standing wave is generated where the average sound pressure as a function of location in the room varies substantially. At the resonance frequencies, the sound field in the enclosure is said to be dominated by resonant or modal response. At these frequencies, the standing wave is characterised by locations of maximum and minimum sound pressures, referred to as antinodes and nodes, respectively. For a lossless room (no surface or air absorption), the nodal locations represent zero sound pressure. However, in practice there is always some loss, which influences the relative phase of the direct and reflected waves, so that the nodal sound pressure is never zero. In a hard-walled enclosure, the pressure at the walls is always a maximum, due to the nonzero phase shift when the sound wave is reflected. On the other hand, the particle velocity is always zero at a hard wall that is not vibrating, and at pressure nodes in a lossless enclosure, the particle velocity is always maximum, so the particle velocity mode shapes are similar to the

253

Sound in Rooms

sound pressure mode shapes, except that the nodal and antinodal locations are interchanged. The sound pressure distribution for 1-D, 2-D and 3-D enclosures is shown in Figure 6.1 for some of the low-order resonant modes. A 1-D room could be modelled as a tube closed at each end containing sound at frequencies below the cut-on frequency of higher order modes (see Equation (1.25)). A 2-D room is a room where one of the dimensions is considerably smaller than half a wavelength of the sound of interest. A lossless enclosure is one for which there is no air absorption and the reflection coefficient of all room surfaces is 1.

(1,0,0) (1)

(1,0)

(2)

(1,1)

(3)

(2,0)

(4)

(2,2)

(5)

(3,1)

(6) (a)

(3,2) (b)

(1,1,1)

(2,0,0)

(2,1,1)

(3,1,1)

(2,2,2) (c)

FIGURE 6.1 Sound pressure mode shapes for 1-D, 2-D and 3-D rooms, where the mode numbers correspond to the number of nodal points, nodal lines or nodal planes respectively (theoretical zero sound pressure). (a) 1-D room, modes (1), (2), (3), (4), (5) and (6); (b) 2-D room, modes (1,0), (1,1), (2,0), (2,2), (3,1) and (3,2); 3-D room modes (1,0,0), (1,1,1), (2,0,0), (2,1,1), (3,1,1), (2,2,2).

If the enclosure is excited at frequencies between resonance frequencies, the spatial variation in the sound field will not be nearly as large as it is likely that more than one mode will be contributing to the total sound field and each mode will have nodes and antinodes in different

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Noise Control: From Concept to Application

locations. If the enclosure is excited at frequencies below the first resonance frequency, where half of a wavelength is greater than any room dimension, there will be no nodes or antinodes in the sound field. The sound field will become closer to uniform throughout the room as the frequency is decreased. Deviations from uniformity are a result of sound absorption by the room surfaces, the air and any objects within the room. When a source of sound in an enclosure is turned on, the resulting sound waves spread out in all directions from the source. When the advancing sound waves reach the walls of the enclosure they are reflected, generally with a small loss of energy, eventually resulting in waves travelling around the enclosure in all directions. If each path that a wave takes is traced around the enclosure, there will be certain paths of travel that repeat upon themselves to form normal modes of vibration, and at certain frequencies, waves travelling around such paths will arrive back at any point along the path in phase. Amplification of the wave disturbance will result and the normal mode will be resonant. When the frequency of the source equals one of the resonance frequencies of a normal mode, resonance occurs and the interior space of the enclosure responds strongly, being only limited by the absorption present in the enclosure. A normal mode has been associated with paths of travel that repeat upon themselves. Evidently, waves may travel in either direction along such paths so that, in general, normal modes are characterised by waves travelling in opposite directions along any repeating path. As waves travelling along the same path but in opposite directions produce standing waves, a normal mode may be characterised as a system of standing waves, which in turn is characterised by nodes and anti-nodes. Where the oppositely travelling waves arrive, for example, in pressure anti-phase, pressure cancellation will occur, resulting in a pressure minimum called a node. Similarly, where the oppositely travelling waves arrive in pressure phase, pressure amplification will occur, resulting in a pressure maximum called an anti-node. In an enclosure at low frequencies, the number of resonance frequencies within a specified frequency range will be small. Thus, at low frequencies, the response of a room as a function of frequency and location will be quite irregular; that is, the spatial distribution in the reverberant field will be characterised by pressure nodes and anti-nodes. If a sound source in a rectangular room produces a single frequency which is slowly increased, the sound pressure level at any location (other than at a node in the room for that frequency) will at first rapidly increase, momentarily reach a maximum at resonance, then rapidly decrease as the driving frequency approaches and then exceeds a resonance frequency of the room. The process repeats with each room resonance. The measured frequency response of a 180 m3 rectangular reverberation room is shown in Figure 6.2 for illustration. The sound pressure was measured in a corner of the room (where there are no pressure nodes) while the frequency of the source (placed at an opposite corner) was very slowly swept upwards. Consideration of a rectangular room provides a convenient model for understanding modal response and will be used here. However, modal response is by no means peculiar to rectangular or even regular-shaped rooms. Modal response characterises enclosures of all shapes. Splayed, irregular or odd numbers of walls will not prevent resonances and accompanying pressure nodes and antinodes in an enclosure constructed of reasonably reflective walls; nor will such peculiar construction necessarily result in a more uniform distribution in frequency of the resonances of an enclosure than may a rectangular room of appropriate dimensions. However, it is simpler to calculate the resonance frequencies and mode shapes for rectangular rooms. Using the wave equation, it can be shown that the resonance frequencies in a rectangular room of dimensions, Lx , Ly and Lz are given by (Bies et al., 2018): c fn = 2

s

nx Lx

2

+



ny Ly

2

+



nz Lz

2

(Hz)

(6.1)

255

Sound in Rooms 90

(a)

80 70

20

40

210 121 201 002 030

120 200

111

125

101 020

100

011

31.5

110

25

001

100

50

010

Lp (dB re 20 mPa)

60

50

63

80

160 200 Frequency (Hz)

250

315

(b) 90 80 70 60 80

FIGURE 6.2 Measured frequency response of an 180 m3 rectangular room. (a) In the frequency range below about 80 Hz, room resonances are identified by mode numbers. (b) In the frequency range above about 80 Hz, peaks in the room response cannot be associated with room resonances identified by mode numbers.

In Equation (6.1), the subscript n on the frequency variable f indicates that the particular solutions or “eigen” frequencies of the equation are functions of the particular mode numbers, nx , ny , and nz . The acoustic pressure in the room at any location, (x, y, z), and at any time, t, corresponding to a particular mode can be shown to be (Bies et al., 2018): p = pˆ cos













πny y πnx x πnz z jωt cos cos e Lx Ly Lz

(6.2)

The mode numbers, nx , ny , and nz take on all positive integer values including zero. For the placement of a speaker to drive a room, it is of interest to note that every mode of vibration has a pressure anti-node at the corners of a room, so placement of a sound source in any corner will maximise the room response. There are three types of normal modes of vibration in a rectangular room, which have their analogues in enclosures of other shapes. They may readily be understood as follows: 1. axial modes for which only one mode number is not zero; 2. tangential modes for which one mode number is zero; and 3. oblique modes for which no mode number is zero. The three modal types are illustrated in Figure 6.3. The relative amplitudes of the different modes is given by pˆ in Equation (6.2) and pˆ is, in turn, affected by the sound absorption, which is a function of frequency, the surface absorption coefficients and the number of reflections. Axial modes correspond to wave travel back and forth parallel to an axis of the room. For example, the (nx , 0, 0) mode in the rectangular room of Figure 6.3 corresponds to a wave travelling back and forth parallel to the x-axis. Such a system of waves forms a standing wave having nx nodal planes normal to the x-axis and parallel to the end walls. Tangential modes correspond to waves travelling parallel to two opposite walls of an enclosure while successively reflecting from the other four walls. For example, the (nx , ny , 0) mode of the rectangular enclosure of Figure 6.3

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Noise Control: From Concept to Application

Tangential 2-D

Axial 1-D

Oblique 3-D

FIGURE 6.3 Illustration of axial, tangential and oblique modes in a rectangular room.

corresponds to a wave travelling around the room parallel to the floor and ceiling. Oblique modes correspond to wave travel oblique to all room surfaces. For example, the (nx , ny , nz ) mode in the rectangular room of Figure 6.3 would successively impinge on all six walls of the enclosure. Consequently, absorptive treatment on the floor, ceiling or any wall would be equally effective in attenuating an oblique mode. As the frequency increases, the density (modes per Hz) of oblique modes increases much faster than tangential or axial mode densities. Thus, in most rooms in the audio frequency range, the response is dominated by oblique modes. The approximate number of modes, N , which may be excited in a rectangular room in the frequency range from zero up to f Hz, is given by the following expression (Morse and Bolt, 1944): N=

πf 2 S fL 4πf 3 V + + 3 3c 4c2 8c

(6.3)

where the first term corresponds to oblique modes, the second to tangential modes and the third to axial modes. The quantities, V , S and L correspond to the room volume, surface area and perimeter (sum of lengths of all 12 wall, ceiling and floor junctions) and c is the speed of sound. It has been shown that Equation (6.3) has wider application than for rectangular rooms; to a good approximation it describes the number of modes in rooms of any shape, with the approximation improving as the irregularity of the room shape increases. It should be remembered that Equation (6.3) is an approximation only and the actual number of modes fluctuates above and below the prediction of this equation as the frequency gradually increases or decreases. Example 6.2 A simple point monopole source is located in a rectangular room at (0i, 0j, (Lz /2)k) where i, j, and k are unit vectors in the x, y and z directions, respectively and Lz is the room dimension in the z-direction. (a) For the four modes defined by mode numbers nx = ny = 0 and nz = 0, 1, 2, and 3, state whether the mode will or will not be excited by the source. (b) Undertake the same procedure for a point dipole source placed at the same point as the monopole and with its axis parallel to the z-axis. (c) For an enclosure of dimensions 10 m × 5 m × 2 m high, calculate the natural frequency of the 1,1,1 mode and sketch its shape. Solution 6.2 (a) Referring to Equation (6.2), it can be seen that if nx = ny = nz = 0, the pressure distribution in the room is independent of location (uniform) and the monopole source will excite the mode. For nz = 1 and nz = 3, there will be a node at Lz /2 and the monopole will not excite the mode. For nz = 2, there will be an antinode at Lz /2 and the mode will be excited by a monopole source.

257

Sound in Rooms (b) A dipole will be ineffective at exciting the nz = 0 and nz = 2 modes because the phase of each of the two parts of the dipole is opposite to the other and the mode will only be excited if the dipole is on a nodal plane, so that it excites each side of the nodal plane in opposite phase to the other side. On the other hand, one would expect good excitation of the nz = 1 and nz = 3 modes because the phase on one side of a nodal plane is 180◦ different to that on the other side and this can be matched by the dipole. (c) The required shape is shown in Figure ??, where the nodes are represented by planes and the relative phases of acoustic pressure are shown by plus and minus signs. The figure is constructed by inserting 3 orthogonal planes at the three middle locations in the room (as it is the (1,1,1) mode), The signs on each side of a nodal plane are opposite so as to represent the relative phase of ther acoustic pressure on each side of 1 1 1 343 + 2 + 2 = 93.9 Hz. the nodal plane. Using Equation (6.1), f1,1,1 = 2 102 5 2

FIGURE 6.4 (1,1,1) mode shape in a rectangular room.

Example 6.3 The sound pressure distribution for sound propagating in an infinite rigid walled duct (of crosssectional dimensions Ly and Lz ) in the positive axial direction (x-axis) is obtained by summing the contributions from all of the modes, as: P∞ mπz nπy j(ωt−κmn x) p = m,n Amn cos cos e , where Am,n is the amplitude of mode, (m, n) and Lz Ly p the wavenumber, κmn is defined as, κmn = k 2 − [(mπ/Lz )2 + (nπ/Ly )2 ]. Plane wave propagation corresponds to m = n = 0. (a) Explain the meaning of “cut-on frequency”. (b) For a duct of dimensions Ly = Lz /3, derive an expression as a function of Lz for the acoustic pressure drop in dB per unit length of duct for the m = 3, n = 2 mode for an excitation frequency equal to one third of the cut-on frequency. (c) Derive an expression for the phase speed of the (3,2) mode and sketch its variation as a function of frequency, giving a qualitative explanation of the dependence, with particular reference to frequencies near the mode cut-on frequency. Solution 6.3 (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).

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Noise Control: From Concept to Application

(b) From the equation given in the problem, for a single mode the acoustic pressure amplitude is given by, pˆ = pˆ0 e−jκmn x . Thus, the dB decay per unit distance is: ∆ = 20 log10 (ˆ p0 /ˆ p1 ) = 20 × jκmn × log10 (e) = 20 × jκmn × 0.4343. where pˆ1 is the sound pressure amplitude at p 1 m. For m = 3, n = 2, and Ly = Lz /3, κ3,2 = (ω/c)2 − [45π 2 /L2z ]. √ Cut-on frequency is thus given by, ω = c[ 45π/Lz ]. Thus, at 1/3 of the cut-on frequency, κ3,2 is given by: p √ κ3,2 = [5π 2 /L2z ] − [45π 2 /L2z ] = ±jπ 40/Lz . √ Thus, ∆ = 20 × (j)(−j)π 40/Lz × 0.4343 = 173/Lz (dB/m). ω (c) Phase speed, c3,2 = ω/κ3,2 = p (ω 2 /c2 ) − 45π 2 /L2z c c = p = p . 2 2 2 2 1 − 45π c /(ω Lz ) 1 − const/ω 2 The variation of c3,2 as a function of frequency is shown in Figure 6.5. 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.

c32

c

fco

f

FIGURE 6.5 Variation of c3,2 as a function of frequency.

Example 6.4 Sound in a rectangular room has been considered and the response has been shown to be modal. A rectangular room was chosen for study for mathematical convenience: rooms of any shape will respond modally but generally the modes will be more complicated than for rectangular rooms. A relatively simple example of modes in a non-rectangular room is furnished by a cylindrical room in which the resonance frequencies are given as: c f (nz , m, n) = 2

s  n 2 z

L

+



ψmn a

2

.

(a)

where c = 343 m/sec, a is the radius and L is the height of the room. The characteristic values ψmn are functions of the mode numbers m, n, where m is the number of diametral pressure nodes and n is the number of circumferential pressure nodes. Values of ψmn are given in Table 6.1. The quantity nz is the number of nodal planes normal to the axis of the cylinder. (a) If L = 2.7 m and a = 5.5 m, what is the lowest order resonance frequency? (b) Describe with the aid of a diagram the lowest order mode pressure distribution. (c) Recall that in a standing wave the acoustic pressure and particle velocity are always in quadrature (i.e. 90◦ out of phase). How is the air in the room behaving in the lowest order mode?

259

Sound in Rooms TABLE 6.1

Values of ψmn for Example 6.4

m

0

1

0 1 2 3 4 5 6 7

0.0000 0.5861 0.9722 1.3373 1.6926 2.0421 2.3877 2.7034

1.2197 1.6970 2.1346 2.5513 2.9547 3.3486 3.7353 4.1165

Value of n 2 2.2331 2.7140 3.1734 3.6115 4.0368 4.4523 4.8600 5.2615

3

4

3.2383 3.7261 4.1923 4.6428 5.0815 5.5108 5.9325 6.3477

4.2411 4.7312 5.2036 5.6624 6.1103 6.5494 6.9811 7.4065

(d) Construct a list of modes shown in the table (for nz = 0) in order of ascending frequency, identifying each mode by its mode number. (e) What is the resonance frequency of the nz = 1, n = m = 0 mode? (f) Determine axial, tangential and oblique modes, as for a rectangular room. At frequencies below the first oblique mode, how many axial and how many tangential modes are resonant? Solution 6.4 (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 nz = 0): f=

343 0.5861 × = 18.3 Hz. 2 5.5

(b) The lowest order mode pressure distribution is shown in Figure 6.6, 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 plane 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.

FIGURE 6.6 Lowest order mode pressure distribution.

(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

260

Noise Control: From Concept to Application

of two acoustic waves travelling in opposite directions across the nodal plane. (d) This problem may be answered by inspection of Table 6.1 and Equation (a). The modes are listed in Table 6.2 in order of ascending frequency, row 1 followed by row 2. TABLE 6.2

1,0 3,2 4,4

4,0 7,1 7,4

0,2 3,3 3,0

1,2 2,4 2,1

5,1 6,3 7,0

4,2 6,4 0,3

Data for Example 6.4(d)

5,2 0,1 6,1

4,3 5,0 0,4

3,4 3,1 6,2

5,4 2,2 5,3

2,0 1,3 7,3

1,1 2,3

6,0 1,4

4,1 7,2

(e) From the Equation given in the example, f = 343/(2 × 2.7) = 63.5 Hz. (f) Axial modes have 2 zeroes in the subscripts nz , 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. From the Equation given in the example, the resonance frequency of the first oblique mode is: s f1,1,1 =

343 2

1 + 2.72



1.697 5.5

2

= 82.7 Hz.

Modes with nz = 1, having resonance frequencies below f1,1,1 , are found from Table 6.1 as those with a value of ψm,n less than ψ1,1 . For nz = 0, the value of ψ must be below that given by Equation (a): ψ < f1,1,1 × 2 × a/c = 82.7 × 2 × 5.5/343 = 2.651. The modes with resonance frequencies below f1,1,1 are listed in Table 6.3 in the form (nz , m, n), beginning with the . 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. TABLE 6.3

0,1,0 0,2,1

6.3

0,2,0 0,0,2

0,0,1 1,2,0

0,3,0 1,0,1

Data for Example 6.4(f)

0,4,0 0,6,0

0,1,1 1,3,0

1,0,0 0,3,1

0,5,0 1,4,0

1,1,0

Bound between Low-Frequency and High-Frequency Behaviour

Although it is difficult to see an obvious bound in Figure 6.2, it is possible to derive an analytical bound above which modal analysis becomes increasingly more difficult and less accurate and statistical analysis achieves acceptable accuracy. This frequency bound will be referred to here as “cross-over frequency” and it will be defined in terms of a quantity known as modal overlap. Modal overlap is a measure of uniformity of the room response as a function of frequency. However, prior to discussing modal overlap in detail, two quantities needed to define it will be discussed.

6.3.1

Modal Density

For the purpose of estimating the number of modes that, on average, may be excited in a narrow frequency band, the derivative of Equation (6.3), called the modal density, is useful. The

261

Sound in Rooms expression for the modal density is as follows: πf S L dN 4πf 2 V + + = df c3 2c2 8c

6.3.2

(6.4)

Modal Damping and Bandwidth

Referring to Figure 6.2, it may be observed that the recorded frequency response peaks in the low-frequency range have finite widths, which may be associated with the response of the room that was investigated. A bandwidth, ∆f , may be defined and associated with each mode, being the frequency range about resonance over which the sound pressure squared is greater than or equal to half the same quantity at resonance. The lower and upper frequencies bounding a resonance and defined in this way are commonly known as the half-power points. The corresponding response at the half-power points is down 3 dB from the peak response. The bandwidth, ∆f , is dependent upon the damping of the mode; the greater the modal damping, the larger will be the bandwidth. For acoustical spaces such as sound in rooms the modal damping is commonly expressed in terms of the damping factor (similar to the critical damping ratio, ζ), which is a function of the viscous nature of the medium and thus is proportional to particle velocity. On the other hand, for structures, modal damping is commonly expressed in terms of a modal loss factor, η, which is based on the hysteresis properties of a material, and so is proportional to displacement. These damping quantities may be related to each other and to the energy-based quality factor, Q, of the resonant mode, or the logarithmic decrement, δ, by the following relations: 2ζ

∆f /f = 1/Q = η = p

1−

ζ2

=

δ π

(6.5)

At low frequencies, individual modal bandwidths can be identified and measured directly. At high frequencies, where individual modes cannot be identified, the average bandwidth may be calculated from a measurement of the time for an introduced sound field to decay by 60 dB using Equation (6.6) (Embleton, 1988): ∆f = 2.20/T60

6.3.3

(6.6)

Modal Overlap

Modal overlap, M , is calculated as the product of the average bandwidth given by either Equation (6.5) or (6.6) and the modal density given by Equation (6.4). The expression for modal overlap is: M = ∆f dN/df (6.7) The modal overlap is a measure of the extent to which the resonances of a reverberant field cover the range of all possible frequencies within a specified frequency range. The concept is illustrated for a hypothetical case of a low modal overlap of 0.6 in Figure 6.7. In the figure, three resonant modes, their respective bandwidths and the frequency range of the specified frequency band are indicated.

6.3.4

Cross-Over Frequency

There are two criteria commonly used for determining the cross-over frequency. The criterion that is chosen will depend upon whether room excitation with bands of noise or with pure tones is of interest.

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Noise Control: From Concept to Application

Room response, Lp (dB)

Bandwidth 3 dB

Measurement band

Frequency (Hz)

FIGURE 6.7 Three modes in a specified frequency range with a modal overlap of 0.6.

If room excitation with 1/3-octave, or wider bands of noise is to be considered, then the criterion for statistical (high-frequency) analysis is that there should be a minimum of between 3 and 6 modes resonant in the frequency band. The exact number required is dependent upon the modal damping and the desired accuracy of the results. More modes are necessary for low values of modal damping or if high accuracy is required. If room excitation with a pure tone or a very narrow band of noise is of concern, then the criterion for reliable statistical analysis is that the modal overlap should be greater than or equal to 3. Example 6.5 Calculate the modal overlap in the 250 Hz octave band for a room if there are four resonant modes, in the 250 Hz band, with bandwidths (3 dB below the peak) of 20 Hz, 25 Hz, 30 Hz and 32 Hz respectively. Solution 6.5 250 Hz octave band, bandwidth from Table 1.2 is 353 − 176 = 177 Hz. Thus the modal overlap, M , is (from Equation (6.7)): M = [(20 + 25 + 30 + 32)/4] × [4/177] = 0.6.

6.4

High-Frequency Behaviour

At high frequencies the sound field in a reverberant space may be thought of as composed of plane waves travelling in all directions with equal probability. If the mean square pressure amplitude of any travelling plane wave is on average the same, independent of direction, then the sound field is said to be diffuse. In a reverberant space, the sound field is diffuse when the modal overlap is three or greater, in which case, the sound field steady-state frequency response is essentially a random phenomenon. The concept of diffuse field implies that the net power transmission along any coordinate axis is negligibly small; that is, any power transmission is essentially the same in any direction. The reverberant sound field may be considered as of constant mean energy density throughout the room. However, the concept does not imply a sound field where the sound pressure level is the

263

Sound in Rooms

same throughout. Even in a perfectly diffuse sound field, the sound pressure level will fluctuate over time at any given location in the room and the long-time-averaged sound pressure level will also fluctuate from point to point within the room. Although the net intensity in a diffuse sound field is zero, the effective intensity in any one direction in a diffuse sound field is shown by Bies et al. (2018) to be: I = hp2 i/(4ρc)

(6.8)

Example 6.6 Explain why a sound intensity meter may cause difficulties when used in a reverberant room to determine source sound power, especially if measurements are made at some distance from the source. Solution 6.6 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. Example 6.7 The net sound intensity at any location in an ideal reverberant field is zero. Clearly this cannot be the case adjacent to a wall. Consider a rectangular reverberant room in which the reverberant field sound pressure level is 95 dB re 20 µPa. (a) Calculate the energy density of the field. (b) Calculate the sound power incident on one of the walls of area 30 m2 . Solution 6.7 (a) Energy density, ψ, in a reverberant field (plane waves) is, ψ = hp2 i/ρc2 (see Equation (1.47)) and the given Lp is 95 dB. From Equation (1.62), hp2 i = (2 × 10−5 )2 × 109.5 = 1.265 Pa2 . Thus, ψ = 1.265/(413.7 × 343) = 8.9 × 10−6 J/m3 . (b) Using Equations (1.61) and (6.8), the sound power incident on a wall is: 1.265 hp2 i W = IS = S= × 30 = 0.023 W = 103.6 dB re 10−12 W. 4ρc 4 × 413.7

6.4.1

Relation between Source Sound Power and Room Sound Pressure Level

The reverberant sound pressure level in a room due to a fixed power sound source is dependent on the sound-absorbing capabilities of the enclosure boundaries. The Sabine absorption coefficient, α ¯ , is the average fraction of incident energy absorbed by the enclosure boundaries. In addition to energy absorption on reflection, some energy is absorbed during propagation between reflections. Generally, propagation loss due to air absorption is negligible, but at high frequencies above 500 Hz, especially in large enclosures, it makes a significant contribution to

264

Noise Control: From Concept to Application

the overall loss. Bies et al. (2018) show that the overall mean Sabine absorption coefficient for a room of volume, V and surface area, S, including air absorption is: α ¯=α ¯ wcf + 9.21 × 10−4 mV /S

(6.9)

where the first term on the RHS is the average absorption coefficient of all room boundaries and the second term is the air absorption. The quantity, m, is the coefficient listed in Table 4.1. More detailed tables are provided in ANSI/ASA S1.26 (2014) and ISO 9613-1 (1993). Equations (6.8) and (6.9) may be used to write for the power, Wa , or rate of energy absorbed: Wa = hp2 iS α ¯ /(4ρc)

(6.10)

At any point in a room, the total sound energy is the sum of the sound energy due to the direct field and that due to the reverberant field. Using Equation (3.3) and introducing the directivity factor, Dθ,φ (see Chapter 3), the sound pressure squared due to the direct field at a point in the room at a distance r and in a direction (θ, φ) from the source may be written as: hp2 iD = W ρcDθ,φ /4πr2

(6.11)

The quantity Dθ,φ is the directivity factor of the source in direction (θ, φ), ρ is the density of air (kg/m3 ), c is the speed of sound in air (m/s) and W is the sound power radiated by the source (W). In writing Equation (6.11), it is assumed that the source is sufficiently small or r is sufficiently large for the measurement point to be in the far field of the source. Consider that the direct field must be once reflected to enter the reverberant field. The fraction of energy incident at the walls, which is reflected into the reverberant field is (1 − α ¯ ). Using Equation (6.10) and setting the power absorbed equal to the power introduced, W , the sound pressure squared due to the reverberant field may be written as: hp2 iR = 4W ρc(1 − α ¯ )/(S α ¯)

(6.12)

The sound pressure level at any point due to the combined effect of the direct and reverberant sound fields is obtained by adding together Equations (6.11) and (6.12). Thus, using Equations (1.62) and (1.64):   h ρc i 4 Dθ + 10 log + (6.13) Lp = LW + 10 log10 10 4πr2 R 400 At 20◦ C, where ρc = 413.7 (SI units), there would be an error of approximately 0.1 dB if the last term in Equation (6.13) is omitted. In common industrial spaces, which have lateral dimensions much greater than their height, Equation (6.13) underpredicts reverberant field sound pressure levels close to the noise source and overpredicts levels far away from the source (Hodgson, 1994). Equation (6.13) has been written in terms of the room constant R, the measurement of which is discussed in Chapter 5, Section 5.6. The room constant is defined as: R=

Sα ¯ 1−α ¯

(6.14)

Example 6.8 Returning to Example 1.15 in Chapter 1, consider the following: (a) If the average sound pressure level measured in the furnace at distances further from the wall than half a wavelength is 120 dB re 20 µPa in the 500 Hz octave band, what is the sound power level in watts generated by the burner in this frequency band?

265

Sound in Rooms Assume that the furnace is a cylinder of 4 m diameter and assume an average Sabine sound absorption coefficient (see Chapter 5) for the internal surfaces of the furnace of 0.02 in the 500 Hz band. State any assumptions that you make. (b) If a second burner with a sound power of twice the original burner were added to the furnace (and the original burner remained as well), what would be the resulting reverberant field sound pressure level in the furnace and away from the furnace walls in the 500 Hz octave band. Solution 6.8 (a) Surface area, S = πdL + 2πr2 = π × 4 × 9.86 + 2π × 4 = 149 m2 . LW = Lp − 10 log10 = 120 − 10 log10



h ρc i 4(1 − α ˆ) − 10 log10 Sα ˆ 400



4 × 0.98 789 × 0.23 − 10 log10 = 122.3 dB re 10−12 W. 149 × 0.02 400









Thus, W = 10−12 × 10122.3/10 = 1.7 W. We have assumed that the contribution of the direct field is small compared to the reverberant field and that the Waterhouse correction term in Equation (3.54) is negligible (as the temperature is high and the frequency of interest is very low, so the wavelength is long). (b) Second burner has twice the sound power, so the total new sound power is 3W , where W is the sound power of the original burner. Thus, the increase in sound power level in dB will be 10 log10 (3W/W ) = 10 log10 (3) (dB), which is an increase of 4.8 dB. From Equation (6.13), the sound pressure level will increase by the same amount. So the new sound pressure level is Lp = 124.8 dB re 20 µPa. Example 6.9 Calculate the electrical to acoustic power conversion efficiency for a speaker in the 500 Hz octave band if when driven with 10 W of electrical power it produces 95 dB re 20 µPa space-averaged sound pressure level in a reverberant room 7 m long, 5 m wide and 3 m high, which has a reverberation time in the 500 Hz octave band of 2.5 seconds. Solution 6.9 Room size = 7 m × 5 m × 3 m. Volume = 105 m3 , surface area = 2(7×5+7×3+5×3) = 142 m2 . Wavelength of sound, λ = 343/500 = 0.686 m. The sound power in the reverberation room is related to the space-averaged sound pressure level by Equation (3.54), which assumes that the average is taken from measurements at least one-half of a wavelength from the nearest room boundary. The effect of not including data closer to the room boundary is taken into account in the following equation (Equation (3.54)) by the second to last term on the RHS. Thus: LW = Lp + 10 log10 V − 10 log10 T60 + 10 log10 (1 + Sλ/8V ) − 13.9 = 95 + 10 log10 (105) − 10 log10 (2.5) + 10 log10



142 × 0.686 1+ 8 × 105



− 13.9

= 95 + 20.2 − 4.0 + 0.5 − 13.9 = 97.8 dB (re 10−12 W). Alternatively, we could assume that the correction term, (1 + Sλ/8V ) is already included in the Lp measurement. In this case, LW = 97.8 − 0.5 = 97.3 dB (re 20 µPa), which is the same result as would be obtained using Equations (6.13), (6.14) and (6.16). The power conversion efficiency is given by:

266 η=

Noise Control: From Concept to Application acoustic power 10−12 × 1097.8/10 = = 6 × 10−4 = 0.06%. electrical power 10

Example 6.10 The possibility of an acoustic interpretation for combustion instability observed in a power station furnace is investigated in this problem. The fire box is a large spacious enclosure with rigid walls and approximate dimensions as follows: 10 m length, 12 m width, 20 m height. The gas within is hot and the speed of sound is estimated to be 864 m/sec. When combustion instability occurs, the furnace shakes violently at a frequency estimated to be about 36 Hz. Furthermore, the acoustic pressure on the wall is estimated to be of the order of 155 dB re 20 µPa during the onset of instability. (a) What acoustic mode is likely to be excited by the instability and what kind of mode is it? (b) Where will the acoustic pressure be largest? (c) What is the amplitude of the cyclic force on the walls of the furnace? (d) The quality factor, Q, for the resonant mode is estimated to be 30 and the density of the hot gas in the furnace is estimated to be 1.16 kg/m3 . Estimate the power, W , which must be injected to drive the mode in steady state. Use the following equations: (i) Q = fn T60 /2.2, where fn is the resonance frequency of the acoustic mode and T60 is the reverberation time for that mode. (ii) The mean square sound pressure, hp2 i, for a 1-D diffuse sound field, at any time, ¯ t, after the sound source is turned off is, hp2 i = hp20 ie−cαt/L , where α ¯ is the fraction of energy absorbed by each wall, L is the distance between opposite walls (length of a 1-D room) and hp20 i is the steady state mean square sound pressure before the sound is turned off. [Hint: the sound pressure at the wall is made up of an incident and reflected wave, the amplitudes of which have been added together arithmetically as it has been assumed that the walls are rigid. Thus, the amplitude √ of the reflected wave is equal to the ¯ ). The power input is equal to amplitude of the incident wave multiplied by ( 1 − α the power absorbed at the walls – incident wave only]. (e) If the flame in the furnace is modulated by acoustic feed back, what is the energy conversion efficiency required to produce the observed instability if 800 kW of power is introduced into the furnace by the combustion process? Solution 6.10 Fire box dimensions are: 10 m × 12 m × 20 m. From Equation (6.1), the resonance frequency of the first axial mode in each direction is: c 864 864 864 f= = , , = 43.2 Hz, 36 Hz, 21.6 Hz. 2L 2 × 10 2 × 12 2 × 20 (a) As the frequency of instability is 36 Hz, it seems likely that it is associated with the axial mode across the width in the 12 m direction. (b) The acoustic pressure will be largest at the two side walls normal to the 12 m dimension, because on reflection the pressure amplitude is doubled. (c) Amplitude of cyclic force acting on walls. Acoustic pressure is 155 dB re 20 µPa. Amplitude is then: √ √ pˆ = 2pRMS = 2 2 × 10−5 × 10155/20 = 1.59 kPa

267

Sound in Rooms Side wall area = 10 × 20 = 200 m2 . Force on each wall = 1.59 × 200 = 318 kN. (d) Power absorbed by 2 side walls. Pressure at wall is the sum of the incident and p reflected pressures. Thus, ¯ )pi . (pi + pr )RMS = 1.125 × 103 . However, pr = (1 − α p 3 ¯ )]pi = 1.125 × 10 . So, [1 + (1 − α The absorbed power is controlled by the incident sound intensity which is: p2 (1.125 × 103 )2 √ I= i = . ρc (1 + 1 − α ¯ )2 ρc As there are two walls, each of area, Aw , and power = intensity × area, the total absorbed power is Wa = 2¯ αIAw . ¯ For a 1-D sound field, hp2 i = hp20 ie−cαt/L . Thus Lp0 − Lp = 4.343c¯ αt/L; thus, (60/4.343) × L the time to decay 60 dB is, T60 = . c¯ α 2.2Q (60/4.343) × 12 2.2 × 30 But T60 = = 1.833 seconds, thus α ¯= = 0.105. = fn 36 864 × 1.833 The absorbed power is then: 2 × 0.1047 × (1.1246 × 103 )2 × 200 √ Wa = = 14 kW. (1 + 1 − 0.1047)2 × 1.16 × 864 13953 (e) Power conversion efficiency is, η = = 1.7%. 800 × 103

Example 6.11 It is proposed to add sound-absorbing material to the walls and ceiling of the room to reduce the interior sound pressure levels produced by a machine mounted on the floor in the centre of the room. Assume that there are no other significant sound sources. If the room size is 10 m × 10 m × 5 m and the Sabine sound absorption coefficient for all surfaces in the 250 Hz octave band is 0.08 before addition of the absorbing material and will be 0.5 on the surfaces covered after addition of the sound-absorbing materials, what is the expected noise reduction (in dB) 3 m from the machine in the 250 Hz octave band. Assume that the floor is concrete and that the machine radiates noise omnidirectionally (same in all directions). Solution 6.11 Room dimensions: 10 m × 10 m × 5 m, S = 2(10 × 10 + 10 × 5 × 2) = 400 m2 , V = 10 × 10 × 5 = 500 m3 . The room constant before treatment is: Rb =

Sα ¯ 400 × 0.08 = = 34.78 m2 . (1 − α ¯) 0.92

Adding sound-absorbing treatment to the walls and ceiling results in a new mean absorption coefficient calculated as, α ¯ n = (100 × 0.08 + 300 × 0.5)/400 = 0.395. Thus, the new room constant after treatment is: Sα ¯ 400 × 0.395 Rn = = = 261.2m2 . Using Equation (6.13) allows the difference in sound (1 − α ¯) 0.605 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): ∆Lp = 10 log10



Dθ 4 + 4πr2 R



− 10 log10 old



Dθ 4 + 4πr2 R

 new

268

Noise Control: From Concept to Application = 10 log10



4 2 + 4π32 34.78



− 10 log10



4 2 + 4π32 261.2



= 6.0 dB. Example 6.12 An ultrasonic sound source, in a reverberant enclosure of dimensions 200 × 150 × 120 mm, is required to produce 140 dB re 20 µPa at 43.1 kHz. The effective Sabine absorption coefficient (including air absorption) of the enclosure walls is 0.1 and the speed of sound is 343 m/s. (a) What is the energy density in the enclosure at the required sound pressure level? (b) What is the power absorbed by the air and enclosure walls? (c) If the source is 20% efficient how much power is required to drive it?

Solution 6.12 (a) From Equation (1.45), the energy density, ψ = hp2 i/ρc2 . 3

Thus, ψ = (2 × 10−5 )2 × 10140/10 /(1.205 × 3432 ) = 0.28J/m . (b) Enclosure surface area, S = 2(0.2 × 0.15 + 0.2 × 0.12 + 0.15 × 0.12) = 0.144m2 . Power flow into the walls and air between the walls (Equation (6.10)) is given by: Wa = ψcS α ¯ /4 = 0.282 × 343 × 0.144 × 0.1/4 = 0.35 W, where the coefficient, α ¯, includes both wall and air absorption. (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 = 1.7 W. Example 6.13 A reference sound source having a sound power level of 92.5 dBA re 10−12 W produces a reverberant field of 87 dBA re 20 µPa in an equipment room. Existing equipment in the room produces a reverberant field level of 81 dBA re 20 µPa. (a) Calculate the room constant. (b) If 4 new machines, all producing the same sound power level, are to be introduced, what is the allowable maximum sound power level of each machine so that the sound pressure level in the room does not exceed 85 dBA re 20 µPa? State any assumptions that have to be made to answer this question. (c) What would be the allowed sound power level of each machine if the room constant were doubled by adding sound-absorbing material to the walls? Again state any assumptions you make. Solution 6.13 (a) LW = 92.5 dB, Lpr = 87 dB for the reference source. The room constant is calculated using only the reverberant part of Equation (6.13) and allowing for ρc = 413.7. Thus: LW = Lp − 10 log10 (4/R) + 10 log10 (400/ρc), 92.5 = 87 − 10 log10 (4/R) − 0.15, and (4/R) = 10(87−92.5−0.15)/10 = 0.272. Therefore, R = 14.7m2 .

269

Sound in Rooms (b) Existing Lp is 81 dBA. Allowed total Lp = 85 dBA. For 4 new machines, the allowed Lp from each new machine is: Lp = 10 log10 108.5 − 108.1 − 10 log10 (4) = 76.8 dBA re 20 µPa.




The corresponding allowed sound power level of each machine is then: LW = 76.77 − 10 log10 (4/14.7) − 0.15 = 82.3 dBA re 10−12 W. Assumptions: • Direct field small compared to reverberant field at measurement locations. • ρc = 413.7 • 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 the room constant gives the allowed sound power level of: LW = 76.77 − 10 log10 (4/29.4) − 0.15 = 85.3 dBA re 10−12 W. Same assumptions as for part (b).

6.4.2

Relation between Room Absorption and Reverberation Time

If sound is introduced into a room, the reverberant field level will increase until the rate of sound energy introduction is just equal to the rate of sound energy absorption. If the sound source is abruptly shut off, the reverberant field will decay at a rate determined by the rate of sound energy absorption. The time required for the reverberant field to decay by 60 dB, called the reverberation time, is the single most important parameter characterising a room for its acoustical properties. For example, a long reverberation time may make the understanding of speech difficult but may be desirable for organ recitals. As the reverberation time is directly related to the energy dissipation in a room, its measurement provides a means for the determination of the energy absorption properties of a room. Bies et al. (2018) derive the following expression linking the room reverberation time to its sound absorption properties for a Sabine room as follows: T60 =

55.25V Sc¯ α

(6.15)

where α ¯ is the average Sabine absorption coefficient for the room, including air absorption. It is possible to use a different approach to the analysis and obtain the following expression for the room reverberation time (Bies et al., 2018). This expression is known as the Norris-Eyring equation in recognition of the people who derived it. T60 = −

55.25V Sc loge (1 − α ¯ st )

(6.16)

where α ¯ st is the statistical absorption coefficient averaged over the room and is derived from measurements of the normal absorption coefficient as described in Chapter 5. Note that air absorption must be included in α ¯ st in a similar way as it is included in α ¯ (Equation (6.9)). Equation (6.16) is often preferred to the Sabine equation (6.15) by many who work in the field of architectural acoustics as some authors claim that it gives results that are closer to measured data (Neubauer, 2001). There are a number of other equations for relating reverberation time to room absorption and these are discussed in Bies et al. (2018). Perhaps the most commonly used of these is the Fitzroy (1959) equation, reproduced as follows. T60

0.16V = S2



−Sy −Sz −Sx + + loge (1 − α ¯ xst ) loge (1 − α ¯ yst ) loge (1 − α ¯ zst )



(6.17)

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Noise Control: From Concept to Application

For non-Sabine rooms (such as large factories or open plan offices), Kuttruff (1994) has proposed that Equation (6.15) be used except that α ¯ should be replaced with α defined as follows: α = − loge (1 − α ¯ st ) 1 + 0.5γ 2 loge (1 − α ¯ st ) +





Pn

i=1

βi (βi − 1 + α ¯ st )Si2 2 2 S (1 − α ¯ st )

(6.18)

In Equation (6.18), n is the number of room surfaces (or part room surfaces if whole surfaces are subdivided), α ¯ st is the statistical absorption coefficient, area averaged over all room surfaces and βi is the statistical energy reflection coefficient of surface i of area Si . The first term in Equation (6.18) accounts for room dimensions which exceed the Sabine room criterion. The quantity γ 2 is the variance of the distribution of path lengths between reflections divided by the square of the mean free path length. It has a value of about 0.4 provided that the room shape is not extreme. The second term in Equation (6.18) accounts for non-uniform placement of sound absorption. Example 6.14 An electric motor produces a steady state reverberant sound pressure level of 74 dB re 20 µPa in a room 3.05 m × 6.10 m × 15.24 m. The measured reverberation time of the room is 2 seconds. (a) What is the acoustic power output of the motor in dB re 10−12 W? (b) How much additional Sabine absorption (in m2 ) must be added to the room to lower the reverberant field by 10 dB? (c) What will be the new reverberation time of the room? Solution 6.14 (a) Room 3.05 m × 6.1 m × 15.24 m, S = 2(3.05 × 6.1 + 3.05 × 15.24 + 6.1 × 15.24) = 316.1 m2 , and V = 3.05 × 6.10 × 15.24 = 283.54 m3 . Using Equation (6.15) for reverberation time, 55.25V 55.25 × 283.54 = = 0.072. we may write, α ¯= ScT60 316.1 × 343 × 2 Using Equation (6.13), the sound power output may be written as: LW = Lp − 10 log10 = 74 − 10 log10



4(1 − α ¯) Sα ¯



4 × 0.928 316.1 × 0.072



− 10 log10



 ρc  400

− 0.15 = 81.7 dB re 10−12 W.

(b) To lower the reverberant field by 10 dB, we need to increase the room constant, R = Sα ¯ /(1 − α ¯ ) by a factor of 10 (see Equation (6.14)). Old S α ¯ /(1 − α ¯ ) = 24.61 m2 ; therefore, the required new S α ¯ /(1 − α ¯ ) = 246.1m2 . Thus, in the new situation, expanding the above equation gives, 316.1¯ α = 246.1 − 246.1¯ α, which results in a new required value of α ¯ = 0.438. The old value of α ¯ is 0.0722, so the increase in absorption needed is ∆(S α ¯ ) = 316.1(0.438 − 0.0722) = 115.5 m2 . (c) Using the value of α ¯ = 0.438 obtained above and Equation (6.15), we obtain: T60 =

55.25 × 283.54 55.25V = = 0.33 seconds. Sc¯ α 316.1 × 343 × 0.438

Example 6.15 A cubic enclosure has sides of length 5 m and an effective absorption coefficient of 0.05 for the floor and ceiling and 0.25 for the walls. What are the reverberation times of the enclosure for the following wave types (ignore air absorption)?

271

Sound in Rooms (a) Axial waves between floor and ceiling. (b) Tangential waves reflected from all four walls. (c) 3-D (diffuse) (all waves combined). [Hint: The expressions for the mean square pressure as a function of time in a decaying sound field are: ¯ hp2 i = hp2o ie−S αct/4V for a 3 − D field ¯ hp2 i = hp2o ie−P αct/πS for a 2 − D field ¯ hp2 i = hp2o ie−cαt/L

for a 1 − D field

where S is the surface area of the 2-D and 3-D fields, P is the perimeter of the 2-D field and L is the length of the 1-D field.] Solution 6.15 (a) Taking logs of the equation given in the problem for a 1-D field, we have: Lp0 − Lp = 4.343c¯ αt/L. Thus, 60 = 4.343 × 343 × 0.05 × T60 /5, which gives T60 = 4.0 seconds. (b) Again taking logs of the equation given in the problem gives: Lp0 − Lp = 4.343P α ¯ ct/(πS), where P = 4 × 5 = 20 m and S = 5 × 5 = 25 m2 . Thus, 60 = 4.343 × 20 × 0.25 × 343 × T60 /(π × 25), which gives, T60 = 0.63 seconds. (c) Again taking logs of the equation given in the problem gives: Lp0 − Lp = 4.343S α ¯ ct/(4V ) where, S = 2(5 × 5 + 5 × 5 + 5 × 5) = 150 m2 and V = 5 × 5 × 5 = 125 m3 . 50 × 0.05 + 100 × 0.25 Thus, α ¯= = 0.183. 150 Using the above equation we obtain, 60 = 4.343 × 150 × 343 × 0.1833 × T60 /(4 × 125), which gives T60 = 0.73 seconds. Example 6.16 A room 10 m × 10 m × 4 m has an average Sabine absorption coefficient, α ¯ = 0.1. (a) Calculate the room reverberation time (seconds). (b) The steady state reverberant field pressure level is 60 dB re 20 µPa. What is the acoustic power output level (dB re 10−12 W) of the noise source producing this pressure level? (c) At what rate (in W/m2 ) is the sound energy incident on the walls of the room? (d) At what distance from the noise source is the reverberant field pressure level equal to the direct field pressure level? (Assume that the noise source is on the floor in the centre of the room). Solution 6.16 (a) Room 10 m × 10 m × 4 m, S = 2(10 × 10 + 10 × 4 × 2) = 360 m2 , V = 10 × 10 × 4 = 400 m3 . Using Equation (6.15) for reverberation time, the following may be written: 55.25 × 400 = 1.8 seconds. T60 = 343 × 360 × 0.1

272

Noise Control: From Concept to Application

(b) LpR = 60 dB re 20 µPa corresponds to hp2R i = 4 × 10−10 × 1060/10 = 4 × 10−4 Pa2 . 360 × 0.1 × 4 × 10−4 = 9.67 µW. 4 × 1.206 × 343 × (1 − 0.1) Sound power level, LW = 10 log10 (9.67 × 10−6 ) + 120 = 69.9 dB re 10−12 W. We could also have used Equation (6.13) and kept everything in dB. 4 × 10−4 (c) Using Equation (6.8), I = = 2.42 × 10−7 W/m2 . 4 × 1.206 × 343 (d) From Equations (6.11) to (6.13), the reverberant field level is equal to the direct field level when Dθ /(4πr2 ) = 4/R = 4(1 − α ¯ )/(S α ¯ ). Assuming the acoustic centre of the source is within a quarter wavelength from the hard floor, Dθ = 2 and the distance, r, at which the fields are equal is: Using Equation (6.12) we obtain, W =

r=



Sα ¯ 8π(1 − α ¯)

1/2

=



36 8π × 0.9

1/2

= 1.26 m.

Example 6.17 You have been given the task of installing 4 new machines in an enclosure housing other machinery and personnel, making sure that the overall reverberant sound pressure level does not exceed 85 dB re 20 µPa when all of the new machines are operating. The enclosure is a large room of dimensions 10 m × 15 m × 6 m high and the existing reverberant field sound pressure level is 75 dB re 20 µPa. Reverberation decay measurements indicate that the average wall, floor and ceiling Sabine absorption coefficient is 0.1. Assume for the purposes of this problem that the machine noise and room absorption measurements are confined to a single octave band. (a) What is the room reverberation time? (b) What is the room constant? (c) The sound power level of each new machine is 94 dB re 10−12 W without noise control. What would be the reverberant field sound pressure level if all four machines without noise control are operating? (d) The sound power level of each new machine is 84 dB re 10−12 W with noise control. What would be the reverberant field sound pressure level if all four machines have been provided with noise control treatment and all are operating? (e) Given that the cost of noise control is $1,600 per machine and the cost of acoustic tile with an average absorption coefficient of 0.6 is $50 + $3 per square metre, what is the least expensive way of meeting the design goal. Solution 6.17 (a) Room 10 m × 15 m × 6 m, S = 2(10 × 15 + 10 × 6 + 15 × 6) = 600 m2 . V = 10 × 15 × 6 = 900 m3 . The room reverberation time can be calculated using 55.25 × 900 Equation (6.15). Thus, T60 = = 2.4 seconds. 600 × 343 × 0.1 600 × 0.1 (b) The room constant from Equation (6.14) is, R = = 66.67m2 . 1 − 0.1 (c) For each machine, LW = 94 dB re 10−12 W. Thus for 4 new machines: LW = 94+ 10 log10 4. From Equation (6.13), the reverberant field sound pressure level due to the 4 new machines is: Lp = LW + 10 log10 (4/R) + 10 log10 4 + 0.15 = 94 + 10 log10 (4/66.7) + 10 log10 (4) + 0.15 = 94 − 12.22 + 6.02 + 0.15 = 88.0 dB re 20 µPa.

273

Sound in Rooms The existing reverberant field level is 75 dB re 20 µPa prior to installation of the new machines. Thus, from Equation (1.74), the total level after installation is: Lp = 10 log10 107.5 + 108.80 = 88.2 (dB re 20 µPa). (d) When quiet machines are installed, the sound power and the sound pressure level contribution due to the new machines is reduced by 10 dB to 84 dB and 78.0 dB respectively. Thus the total reverberant field level after installation of quiet machines
is, Lp = 10 log10 107.5 + 107.795 = 79.7 dB. (e) The design goal is 85 dB re 20 µPa. From Equation (5.67), the required reduction due to ceiling tile if untreated machines are used is: ∆Lp = 88.16 − 85 = 3.16 dB = 10 log10 (Rf /Ri ). Thus, Rf /Ri = 103.16/10 = 2.07. The initial room constant, Ri = 66.67 m2 . Thus the new requirement is Rf = 138 m2 . Using Equation (6.14), and noting that the total room surface area is 600 m2 , the required average absorption coefficient is then given by 600 × α ¯ = 138(1 − α ¯ ). Thus α ¯ = 0.187. To achieve this, let x square metres of room surface be covered by tile. From Equation (5.66), 0.1(600 − x) + 0.6x , which gives x = 105 m2 . 0.187 = 600 Area of ceiling = 10 × 15 = 150 m2 . 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.




6.5

Flat Room with Diffusely Reflecting Surfaces

Many factories do not satisfy the dimensional requirements of a Sabine room and may be described as “flat” as the height is small in comparison with the lateral dimensions. One type of flat room will be considered here and that is one for which the floor and ceiling surfaces may be considered as diffusely reflecting. In practice, this means that there are irregularities or objects of the order of half a wavelength or greater in size on the two surfaces. Other surface conditions are treated in Bies et al. (2018), where a treatment of sound in tunnels will also be found. This work is based on that of Kuttruff (1985, 1989). Reflections at the boundaries of either flat rooms or long rooms will produce a reverberant field in addition to the direct field of the source but, whereas in the Sabine-type rooms discussed earlier, the reverberant field could be considered as of constant mean energy density (level) throughout the room, in the case of the non-Sabine-type rooms considered here the reverberant field will always decay away from the source; there will be no constant mean level reverberant field. However, as in the case of Sabine-type rooms, it will be useful to separately identify the direct and reverberant fields, because the methods of their control will differ. For example, where the direct field is dominant, the addition of sound absorption will be of little value. The case of a flat room with diffusely reflecting surfaces is quite complicated and only the case where the reflection coefficients, β, of the floor and ceiling are equal will be considered here. In this case, the reverberant field sound pressure level is given by the following closed form approximate solution (Bies et al., 2018): W ρcβ hp2 (r)iR ≈ πa2

"

r2 1+ 2 a

−3/2

β(Γ + 1) r2 + (Γ + 1)2 + 2 1−β a



−3/2 #

(6.19)

where a is the room height (m), r is the distance of the observer from the source, W is the acoustic power output of the source and ρc is the characteristic impedance of air. The quantity,

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Noise Control: From Concept to Application

Γ, is defined as: Γ = loge



1 − 0.6019β (1 − β)0.6019



(6.20)

The direct field contribution can be calculated using Equation (3.3). The direct and reverberant contributions to the total time-averaged sound pressure squared, normalised with respect to the sound power of the source and the room height is shown in Figure 6.8 for various values of the floor and ceiling energy reflection coefficient, β. 10

10log10+ p 2(r), - 10log10 [Wr c/(pa 2)]

b = 0.9 0

0.7 0.5 0.3 0.1

-10

-20

-30 0.1

1

10

100

r/a FIGURE 6.8 Direct and reverberant sound fields in a flat room of height, a, with diffusely reflecting floor and ceiling as a function of the normalised distance from source to receiver. The reverberant field contribution is shown as a function of the energy reflection coefficient, β, assumed the same for floor and ceiling. The direct field is indicated by the dashed diagonal straight line.

6.6

Additional Problems

1. For measurements in the 200 Hz 1/3-octave band of sound emitted by a very small source in a typical factory at 20◦ C, a microphone is placed at a distance of 0.5 m from the source. The room surface area is 300 m2 and the average Sabine absorption coefficient is 0.25. (a) If the source radiates uniformly in all directions and is placed on a concrete floor, at what distance from the source would the direct and reverberant fields be equal? (b) Describe the microphone location in the source sound field in terms of direct, near, far, transition, reverberant. 2. In a computer print room with hard floor and walls, an average sound pressure level of 87.2 dBA re 20 µPa is produced by a reference sound source having a sound power

Sound in Rooms level of 92.0 dBA re 10−12 W. The average sound pressure level is measured well away from any walls or floor or ceiling and the reference source so it can be assumed to be a reverberant field measurement. An existing printer in the room produces a reverberant field level of 76.0 dBA re 20 µPa. It may be assumed that the spectral content of the sound radiated by the new printers is the same as that radiated by the old one (and the same as that produced by the reference source) and that the room temperature is 20◦ C. (a) If 5 new printers, all producing the same sound power level as one another, are to be introduced, what is the allowable maximum sound pressure level contribution of each printer to the reverberant field if the allowable total level is 85 dBA re 20 µPa? (b) What is the corresponding allowed sound power level of each new machine? 3. Calculate the average Sabine absorption coefficient (in the 500 Hz octave band) of the floor, walls and ceiling in a factory of dimensions 16 m × 16 m × 6 m, based on the following information. • There is one item of uniformly-sound-radiating equipment in the centre of a hard floor. • The total sound power of the equipment is 113 dB re 10−12 W in the 500 Hz octave band. • The total sound pressure level measured in the 500 Hz octave band is 102 dB re 20 µPa at a distance of 2 m from the acoustic centre of the source. 4. Name the 5 parts of the sound field radiated by a sound source in a room and describe each with a sentence. 5. The reverberant sound pressure level in a production room containing 15 identical machines is 90 dB re 20 µPa in the 1000 Hz octave band. (a) What is the minimum number of machines that would have to be removed to reduce the level to less than 85 dB re 20 µPa. (b) Alternatively, if all 15 machines remained, and the existing space-averaged Sabine absorption coefficient in the 1000 Hz 1/3-octave band were 0.05, what would the new Sabine absorption coefficient need to be to reduce the reverberant field level to 85 dB re 20 µPa? 6. The sound power level of a machine on a concrete floor in a factory of dimensions 10 m × 10 m × 5 m is 95 dB re 10−12 W in the 500 Hz octave band. The average sound pressure level measured at 2 m from the centre of the machine is 85 dB re 20 µPa. What is the average Sabine absorption coefficient in the room if the temperature is 20◦ C? 7. A burner in a furnace rated at one mega-watt emits 1% of its power as broadband sound and this is distributed equally among the octave bands from and including 63 Hz up to and including 8 kHz. The combustion chamber is 6 m long with a square cross section of 3 m × 3 m. The average sound absorption coefficient of the combustion chamber walls may be assumed to be 0.03 in the 500 Hz octave band. (a) If the temperature in the furnace is 600◦ C, calculate the speed of sound and density of the air in it, assuming standard atmospheric pressure. (b) If the burner noise generating mechanism occurs well away from any wall of the combustion chamber, calculate the expected sound pressure level at a distance of 2 m in front of the burner in the 500 Hz octave band. [Hint: for this part, you will need to use Equation (6.13) with Dθ = 1.]

275

276

Noise Control: From Concept to Application (c) What is the lowest frequency that a resonance instability could be initiated?

8. You are provided with the octave band sound pressure levels in Table 6.4. TABLE 6.4

Data for Problem 8

Octave band centre frequency (Hz) 125 250 500 Sound pressure level with machine on (dB re 20 µPa) Sound pressure level with machine off (dB re 20 µPa)

91

84

80

89

80

75

(a) Calculate the overall A-weighted sound pressure level due to the machine alone. (b) Calculate the overall sound power level of the machine alone if the measurements above were taken at a distance of 10 m from the acoustic centre of the machine and the machine is mounted on a hard surface out of doors away from any other reflecting surfaces. (c) State any assumptions implicit in your answer to part (b). (d) If a brick wall were built between the machine and the measurement point, calculate the overall resulting noise reduction at the measurement point for the 250 Hz octave band only. With no wall, you may assume that there exists a direct and ground-reflected wave with the ground-reflected wave experiencing no loss at all on reflection. When the wall is in place you may assume that 6 new paths exist. For each path, the contribution to the total sound at the measurement point is less than the direct path contribution in the absence of the wall by 6, 10, 12, 14, 15 and 16 dB respectively. 9. A burner is fixed to the centre of one end of a furnace and radiates sound uniformly in all directions. The furnace is 6.0 m long and 4.0 m diameter. The average absorption coefficient of the walls is 0.050 for the 125 Hz octave band. If the sound power radiated by the burner is 3.10 W: (a) Calculate the radiated sound power level (dB re 10−12 W) (b) Calculate the sound pressure level at 3.00 m from the burner for the 125 Hz octave band if the furnace temperature is 1200◦ C, the pressure of the gas in the furnace is atmospheric, the molecular weight of the gas in the furnace is 0.035 kg/mole and the ratio of specific heats is 1.40. 10. A noisy production facility has an inside reverberant field sound pressure level of 88 dB re 20 µPa in the 500 Hz octave band and the corresponding average sound pressure level measured at the nearest residence 220 m away is 44 dB re 20 µPa. (a) The facility operates 24 hrs/day and 5 new machines, which are all identical, are about to be purchased. What maximum sound power level should you specify for each machine to ensure that only negligible or marginal public reaction would occur. The neighbourhood may be classified as suburban with some commerce or industry and the noise contains no tones or impulses. For the purposes of this problem assume that the only significant noise occurs in the 500 Hz octave band. The factory dimensions are 25 m × 20 m × 8 m high and the measured reverberation time in the 500 Hz octave band is 2.1 seconds. State clearly any additional assumptions you make in obtaining your answer.

Sound in Rooms (b) What would be the allowable sound power level for each of the five machines if a suspended ceiling having a Sabine absorption coefficient of 0.5 at 500 Hz were added to the factory and it is assumed that noise leaking through the roof was not a significant contributor to community sound pressure levels? Again, state any assumptions you make. 11. A uniformly radiating source of size 0.5 m × 0.5 m × 0.5 m, suspended from the ceiling in the centre of a reverberant room of dimensions 10 m × 8 m × 6 m high, is radiating 1 W of broadband acoustic power in the 500 Hz 1/3-octave band. The room reverberation time in this frequency band is 2.5 seconds and the temperature is 20◦ C. (a) Would a measurement location 2.5 m from the centre of the source be in the far field of the source? Explain you answer. (b) Calculate the sound pressure level at a distance of 2.5 m from the source centre. (c) Calculate the sound intensity level and the energy density at a distance of 2.5 m from the centre of the source. Give your answers in correct units and state any assumptions that you make. (d) By how much would the sound pressure level at the 2.5 m measurement location decrease if the floor of the room were covered with sound-absorbing material with a Sabine absorption coefficient of 0.9 in the 500 Hz 1/3-octave band. 12. A rectangular room of dimensions 5m × 5m × 3m has an average Sabine absorption coefficient of 0.15 in the 125 Hz octave band. (a) Calculate the number of modes resonant in the 125 Hz octave band. (b) Calculate the modal overlap at 125 Hz. (c) Can the room be used for sound power measurements in octave bands including 125 Hz and above. (d) Can the room be used for tonal sound power measurements at 125 Hz? 13. A sound source in an enclosure of volume 30 m3 radiates 1 W of sound power with a wavelength much smaller than the enclosure dimensions, resulting in a space-averaged diffuse field sound pressure level of 116 dB re 20 µPa. (a) If the surface area of the enclosure is 50 m2 , calculate the average absorption coefficient of the enclosure surfaces. (b) If the source is suddenly switched off, how long does it take for the sound pressure level to decay by 10 dB? (c) If 10 m2 of sound-absorbing material having a Sabine absorption coefficient of 0.8 is placed on the walls, what would be the reduction in reverberant field sound pressure level? 14. A single storey factory has a single window of area 1.5 m2 located in an otherwise blank concrete block wall: the window overlooks a house at a distance of 60 m across an asphalted loading yard. The space-average reverberant sound pressure level in the factory is 88 dB re 20 µPa in the 1000 Hz octave band, and the diffuse field transmission loss of the window is 27 dB. Estimate the sound pressure level at the wall of the house due to transmission through the window in this frequency band. [Hint: see Section 7.2.2 for a definition of Transmission Loss.] 15. Calculate the sound power level in dB incident on a window in the centre of one wall of a reverberant enclosure if the energy density in the centre of the enclosure is 1.5 milli-Joules/m3 and the dimensions of the window are 1 m × 1.5 m and the wall is 4 m × 4 m. State any assumptions that you need to make to solve this problem.

277

278

Noise Control: From Concept to Application

16. A machine in a factory of dimensions 50 m × 50 m × 5 m emits a sound power level of 130 dB re 10−12 W in the 500 Hz octave band. The machine is located in the centre of the factory mid-way between the floor and ceiling. (a) Calculate the direct and reverberant sound pressure levels 5 m from the acoustic centre of the machine. Assume the machine radiates uniformly in all directions, the room has a specular reflecting floor and ceiling and no other machines or reflecting surfaces are in the room. Assume that the reflection coefficient amplitude is 0.7 for both the floor and ceiling. (b) If the factory dimensions were 10 m × 10 m × 5 m, and the reflection coefficient amplitude for all surfaces was 0.7, what would be the total sound pressure level 5 m from the acoustic centre of the machine. (c) For the case in part (b), at what distance from the machine would the direct and reverberant fields be equal. (d) What would be the reverberation time in the room of part (b)? (e) If the factory walls had a Transmission Loss of 25 dB in the 500 Hz octave band, what would be the sound pressure level at a distance of 50 m from the outside of the wall across an asphalt car park? You will need to use Equations (7.29) and (7.30) in Chapter 7. 17. A room of dimensions 8 m × 6 m × 3 m high has an average surface absorption coefficient of 0.05, apart from the ceiling which is covered with acoustic tiles having an absorption coefficient of 0.15 (random incidence values, for the octave band centred at 125 Hz). (a) Estimate the average reverberant sound pressure level due to a broadband source in the room which radiates 25 mW of acoustic power in the 125 Hz octave band. (b) At what distance from the source do you expect the direct and reverberant sound pressure levels to be equal, for the room described above? (Assume the source is non-directional.) 18. Consider a reverberant room of dimensions 6.84 × 5.565 × 4.72 m. The average surface Sabine absorption coefficient for the 1000 Hz one third octave band is 0.022. (a) Calculate the sound power (mW) radiated by a source into the room in the 1000 Hz octave band if the space-averaged reverberant sound pressure level was measured as 95 dB re 20 µPa. (b) What would the sound pressure level be at 0.5 m from the sound source, assuming it to be mounted on the floor in the centre of the room, and assuming it to radiate omnidirectionally? 19. A machine to be operated in a factory produces 0.01 W of acoustic power. The building’s internal dimensions are 10 × 10 × 3 m. All surfaces except the concrete floor can be lined with acoustic material. (a) Specify the absorption coefficient for the lining material so that the sound pressure level in the reverberant field of the factory is about 83 dB re 20 µPa. (b) Specify the radius of an area around the machine in which the sound pressure level will exceed 90 dB re 20 µPa.

7 Partitions, Enclosures and Barriers

LEARNING OBJECTIVES In this chapter, the reader is introduced to: • sound transmission through partitions and the importance of bending waves; • Transmission Loss and its calculation for single (isotropic and orthotropic) and double panels; • enclosures for keeping sound in and out; and • barriers for the control of sound outdoors and indoors.

7.1

Introduction

In many situations, it is necessary to place an enclosure around a noise source or place a barrier between the noise source and the observer. To understand how to design an enclosure, it is necessary to understand sound transmission through enclosure walls and this is where this chapter will begin. Following the discussion of sound transmission through single and double walls, enclosure and barrier design is considered. Pipe lagging, which involves wrapping a pipe with a layer of sound-absorbing material such as acoustic foam or rockwool covered by a limp impervious membrane such as lead-loaded vinyl or a lead-aluminium composite, is discussed in Bies and Hansen (2003) and will not be considered here.

7.2 7.2.1

Sound Transmission through Partitions Bending Waves

Although longitudinal waves can propagate through panels and walls, the transmission of sound from one side of a panel to the other is a result of bending waves. In the discussion to follow, both isotropic and orthotropic panels will be considered. Isotropic panels are characterised by uniform stiffness and material properties, whereas orthotropic panels are usually characterised by a stiffness that varies with the direction of bending wave travel (for example, a corrugated or ribbed steel panel). Bending waves in thin panels, as the name implies, take the form of waves of flexure propagating parallel to the surface, resulting in normal displacement of the surface. The speed of propagation of bending waves increases as the ratio of the bending wavelength to solid material 279

280

Noise Control: From Concept to Application

thickness decreases. That is, a panel’s stiffness to bending, B, increases with decreasing wavelength or increasing excitation frequency. The speed of bending wave propagation, cB , for an isotropic panel is given by the following expression: cB = (Bω 2 /m)1/4

(m/s)

(7.1)

The bending stiffness per unit width, B, is defined as: B = EI 0 /(1 − ν 2 ) = Eh3 /[12(1 − ν 2 )]

kg m2 s−2

(7.2)

In the preceding equations, ω is the angular frequency (rad/s), h is the panel thickness (m), ρm is the material density, m = ρm h is the surface density (kg/m2 ), E is Young’s modulus of elasticity (Pa), ν is Poisson’s ratio and I 0 = h3 /12 is the cross-sectional second moment of area per unit width (m3 ), computed for the panel cross section about the panel neutral axis. As shown by Equation (7.1), the speed of propagation of bending waves increases with the square root of the excitation frequency; thus there exists, for any panel capable of sustaining shear stress, a critical frequency (sometimes called the coincidence frequency) at which the speed of bending wave propagation is equal to the speed of acoustic wave propagation in the surrounding medium. The frequency for which airborne and solid-borne wave speeds are equal, the critical frequency, is given by the following equation and illustrated in Figure 7.1: fc =

c2 2π

r

m B

(Hz)

(7.3)

where c is the speed of sound in air.

Wave speed

Panel bending wave speed

Speed of sound in air

Critical frequency

Frequency

FIGURE 7.1 Illustration of critical frequency.

Substituting Equation (7.2) into (7.3), the following equation is obtained for the critical frequency: fc = 0.55c2 /(cL h) (7.4) where the longitudinal wave speed, cL , for thin plates is given by: cL =

p

E/[ρm (1 − ν 2 )]

(7.5)

Representative values of E/ρm , the longitudinal wave speed in thin rods, are given in Appendix A. Using Equations (7.2) and (7.5), the longitudinal wave speed may be written as: √ r 12 B cL = (7.6) h m

p

281

Partitions, Enclosures and Barriers

At the critical frequency, the panel bending wavelength corresponds to the trace wavelength of an acoustic wave at grazing incidence, as shown in Figure 7.2(a). Grazing incidence corresponds to a wave incident at an angle that is 90◦ to the normal to the panel. A sound wave incident from any direction at grazing incidence, and of frequency equal to the critical frequency, will strongly drive a corresponding bending wave in the panel. Also, a panel mechanically excited in flexure at the critical frequency will strongly radiate a corresponding acoustic wave. As the angle of incidence or radiation between the direction of the acoustic wave and the normal to the panel becomes smaller, as shown in Figure 7.2(b), the trace wavelength of the acoustic wave on the panel surface becomes longer. Thus, for any given angle of radiation smaller than grazing, there will exist a frequency (which will be higher than the critical frequency) at which the bending wavelength in the panel will match the acoustic trace wavelength on the panel surface. This frequency is referred to as a coincidence frequency and must be associated with a particular angle of incidence or radiation of the acoustic wave. Thus, in a diffuse field, in the frequency range about and above the critical frequency, a panel will be strongly driven by an incident sound field and will radiate sound well. However, the response is a resonance phenomenon, being strongest in the frequency range about the critical frequency and strongly dependent upon the damping in the system. This phenomenon is called coincidence, and it is of great importance in the consideration of Transmission Loss.

+

+ +

+ +

+ +

+

+

+ +

+

+

+ +

+

+

+ +

+ +

+ +

+

+

+ +

+ +

+ (a)

(b)

(c)

FIGURE 7.2 Coupling of the acoustic field and the panel flexural wave for sound radiation. The horizontal lines represent the maxima of radiated longitudinal acoustic wavefronts, which are separated by a wavelength and the arrows show the direction of wave travel. The edge view of the panel shows panel bending waves. (a) At the critical frequency, the radiated grazing acoustic wave has the same wavelength as the panel bending wave and the panel radiates. (b) Above the critical frequency, for every frequency, there is a radiation angle at which a radiated acoustic wave has the same trace wavelength on the panel as the panel bending wave and the panel radiates sound. (c) At frequencies less than the critical frequency, the disturbance is local and the panel does not radiate except at the edges (shaded area on the figure). The + and − signs indicate areas vibrating in opposite phase and the lines represent vibration nodes on the panel.

At frequencies less than the critical frequency (lowest critical frequency for orthotropic panels), the structure-borne wavelength is shorter than the airborne acoustic wavelength and wave coupling is not possible (Cremer et al., 2005). In this case, an infinite panel is essentially decoupled from an incident sound field. As illustrated in Figure 7.2(c), local disturbances are produced, which tend to cancel each other and decay very rapidly away from the panel. In finite panels,

282

Noise Control: From Concept to Application

radiation coupling occurs at the edges and at stiffeners, where the disturbance is not matched by a compensating disturbance of opposite sign. At these places of coupling, the panel radiates sound or, alternatively, it is driven by an incident sound field. As sound transmission is dependent on panel stiffness, it is expected that orthotropic panels (such as ribbed or corrugated constructions) will transmit sound differently to isotropic panels. The consequence of orthotropic panels being stiffer across the corrugations than along them is that orthotropic panels are characterised by a range of bending wave speeds (dependent upon the direction of wave propagation across the plate) due to the two different values of the crosssectional second moment of area per unit width, I 0 . By contrast, isotropic panels are characterised by a single bending wave speed given by Equation (7.1). For wave propagation parallel to the ribs or corrugations the bending stiffness per unit width may be calculated by referring to Figure 7.3, and using the following equation. N

X h2 + b2i h2 − b2i Eh 2 b z + + cos 2θi B= i i (1 − ν 2 )` 24 24 



(kg m2 s−2 )

(7.7)

i=1

The summation is taken over all segments in width ` (total, N ) and the distance zi is the distance of the centre of segment, i, from the neutral axis of the entire section. The location of the neutral axis is found by selecting any convenient reference axis, such as one that passes through the centre of the upper segment (see Figure 7.3). Then the neutral axis location from the reference axis location is given by: N P

zn =

zri bi hi

i=1 N P

(7.8) bi hi

i=1

where zri is the distance of the ith segment from the reference axis. For locations below the reference axis, zri is positive, although locations below the reference axis could alternatively be classified as negative in which case, zn would be negative for the section shown in Figure 7.3. R b1 zn

Reference axis z r2 = z r 4

q2

z1

zr3 z2

z4 z3

b2

h q4

q2

q4

b3

b4

Neutral axis

FIGURE 7.3 A typical cross section of a ribbed panel. Equation (7.7) corresponds to bending waves propagating into and out of the page, while Equation (7.9) corresponds to bending waves propagating from left to right of the figure (or right to left).

For wave propagation in a stiffened panel, across the corrugations, the bending stiffness per unit width will be similar to that for an isotropic panel; that is: B=

Eh3 12(1 − ν 2 )

(7.9)

Note that Equation (7.9) follows from Equation (7.7) if in the latter equation zi and θi are set equal to zero (see Figure 7.3).

283

Partitions, Enclosures and Barriers

The derivation of Equation (7.7) is predicated on the assumption that the wavelength of any flexural wave will be long compared to any panel dimension. If this were not the case, as may occur at high frequencies, where a flexural wavelength may be of the order of a characteristic dimension of the panel structure (for example, bi , i = 1, 4 in Figure 7.3), the bending stiffness will approach that for an isotropic panel, as given by Equation (7.7). Although for an isotropic panel there exists just one critical frequency, for orthotropic panels the critical frequency is dependent upon the direction of the incident acoustic wave. However, as shown by Equation (7.3), the range of critical frequencies is bounded at the lower end by the critical frequency, fc` , corresponding to a wave travelling in the panel stiffest direction (e.g. along the ribs for a corrugated panel) and at the upper end by the critical frequency, fcu , corresponding to a wave propagating in the least stiff direction (e.g. across the ribs of a corrugated panel). For the case of an orthotropic panel, characterised by an upper and lower bound of the bending stiffness, B, per unit width, a range of critical frequencies will exist. The response will be strong over this frequency range, which effectively results in poor resistance to sound transmission over a much more extended frequency range than for the case of the isotropic panel. Another mechanism that reduces the Transmission Loss of ribbed or corrugated panels at some specific high frequencies is the resonance behaviour of the panel sections between the ribs. At the resonance frequencies of these panels, the Transmission Loss is markedly reduced.

7.2.2

Transmission Loss

When sound is incident upon a wall or partition some of it will be reflected and some will be transmitted through the wall. The fraction of incident energy that is transmitted is called the transmission coefficient τ . The Transmission Loss, TL (sometimes referred to as the sound reduction index, Rw ), is in turn defined in terms of the transmission coefficient, as follows: TL = −10 log10 τ

(dB)

(7.10)

In general, the transmission coefficient and thus the Transmission Loss will depend upon the angle of incidence of the incident sound. Normal incidence, diffuse field (random) incidence and field incidence Transmission Loss (denoted TLN , TLd and TL, respectively) and corresponding transmission coefficients (denoted τN , τd and τF , respectively) are terms commonly used; these terms and their meanings will be described in the following section. Field incidence Transmission Loss, TL, is the Transmission Loss commonly observed in testing laboratories and in the field, and reported in tables. The Transmission Loss of a partition is usually measured in a laboratory by placing the partition in an opening between two adjacent reverberant rooms designed for such tests. Noise is introduced into one of the rooms, referred to as the source room, and part of the sound energy is transmitted through the test partition into the second room, referred to as the receiver room. The resulting mean space-average sound pressure levels (well away from the sound source) in the source and receiver rooms are measured and the difference in levels, called the noise reduction, NR, is determined. The receiver room constant is determined either by use of a standard sound power source or by measurements of the reverberation decay, as discussed in Chapter 6. The Sabine absorption in the room, including loss back through the test partition, is thus determined. An expression for the field incidence Transmission Loss in terms of these measured quantities can then be derived as follows. The power transmitted through the wall is given by the effective intensity in a diffuse field (see Chapter 6) multiplied by the area, A, of the panel and the fraction of energy transmitted τ ; thus, using Equation (6.8) we may write for the power transmitted: Wt =

hp2i iAτ 4ρc

(7.11)

284

Noise Control: From Concept to Application

The sound pressure level in the receiver room (from Equation (6.12)) is: hp2r i =

4Wt ρc(1 − α ¯) hp2 iAτ (1 − α ¯) = i Sα ¯ Sα ¯

(7.12)

and the noise reduction is thus given by: NR = 10log10

A(1 − α ¯) hp2i i = TL − 10 log10 hp2r i Sα ¯

(7.13)

In reverberant test chambers used for Transmission Loss measurement, α ¯ is always less than 0.1 and thus S α ¯ /(1 − α ¯ ) may be approximated as S α ¯ . Equation (7.13) may then be rearranged to give the following expression, which is commonly used for the laboratory measurement of sound Transmission Loss: TL = NR + 10 log10 (A/S α ¯)

(dB)

(7.14)

In Equation, (7.14), S α ¯ is the Sabine absorption of the receiving room, including losses through the test partition, and A is the area of the test partition. S and α ¯ are, respectively, the receiving room total surface area, including that of the test partition, and the mean Sabine absorption coefficient (including the test partition). When conducting a Transmission Loss test, great care must be taken to ensure that all other acoustic transmission paths are negligible; that is, “flanking paths” must contribute an insignificant amount to the total energy transmitted. The test procedure is described in relevant standards publications (ISO 10140-2, 2010; ISO 16283-1, 2014; ASTM E336-19, 2019). Care should be taken in the interpretation of measured TL data for which the TL is less than 10 dB (Bies and Davies, 1977), due to the absorption in the receiving room being influenced by coupling with the source room. In practice, it is desirable to characterise the Transmission Loss of a partition with a single number descriptor to facilitate comparison of the performance of different partitions. For this reason, a single number rating scheme (see ASTM E413-87) called STC (or sound transmission class) has been introduced. To determine the STC for a particular partition, a curve fitting technique is used to fit the measured or calculated one third octave Transmission Loss (TL) data for the partition. Figure 7.4 shows a typical STC contour. STC contours consist of a horizontal segment from 1250 to 4000 Hz, a middle segment increasing by 5 dB from 400 to 1250 Hz and a low-frequency segment increasing by 15 dB from 125 to 400 Hz. The STC rating of a partition is determined by plotting the 1/3-octave band TL of the partition and comparing it with the STC contours. The STC contour is shifted vertically downwards from a large value until the following criteria are met. Note that if the STC curve is reduced by 1 dB, all frequencies are reduced by the same 1 dB amount. 1. The TL curve is never more than 8 dB below the STC contour in any 1/3-octave band. 2. The sum of the deficiencies of the TL curve below the STC contour over the 16 1/3-octave bands does not exceed 32 dB. When the STC contour is shifted to meet these criteria, the STC rating is given by the value of the contour at 500 Hz. Example 7.1 Determine the STC value for the 1/3-octave band Transmission Loss spectrum shown in column 2 of Table 7.1.

285

Partitions, Enclosures and Barriers

STC 53

55

R = TL (dB)

50

45

40

35

30 125

250

500 1k 2k 1/3 octave band centre frequency

4k

FIGURE 7.4 Example STC contour showing data from Example 7.1. The solid line is the 1/3octave band STC53 curve. The dashed line represents the data for Example 7.1 and the dotted curve represents the octave band Rw curve. The 1/3-octave band curve for Rw = 53 is the same as the STC53 curve except that it extends down to 100 Hz and does not go beyond 3150 Hz. TABLE 7.1

1/3-octave band centre frequency 125 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 Sum of deficiencies

Data and STC rating for Example 7.1

Measured TL

STC53 value

Deficiency for STC53 (dB)

Deficiency for STC54 (dB)

31 38 42 48 53 56 58 59 58 55 54 53 52 54 58 60

37 40 43 46 49 52 53 54 55 56 57 57 57 57 57 57

6 2 1 0 0 0 0 0 0 1 3 4 5 3 0 0 26

7 3 2 0 0 0 0 0 0 2 4 5 6 4 0 0 33

286

Noise Control: From Concept to Application

Solution 7.1 An STC value is estimated by plotting the data and fitting an STC curve by eye, trying to make it a line of best fit. Then criterion 1 above is checked to make sure no data point is more than 8 dB below the STC curve. Next criterion 2 is checked. If the sum of deficiencies is less than 32 dB, the STC curve is moved up in 1 dB steps until the sum of deficiencies exceeds 32 dB. Then 1 dB is subtracted from the last STC value to give the preliminary STC rating. All data points are then checked to make sure that none lie more than 8 dB below the preliminary STC rating curve. If any data points lie more than 8 dB below the preliminary STC curve, the STC curve is moved down in 1 dB steps until no data points lie more than 8 dB below the curve and the resulting STC value is the final STC value. The final result from going through this process is an STC rating of 53 (see column 3 of Table 7.1), for which the total deficiency of the data below the curve is 25 dB. If the curve is moved up 1 dB, then the deficiency becomes 33 dB which is too much. Also, it can be seen that no data point is more than 8 dB below the corresponding STC value. The ISO method for determining a single number to describe the Sound Transmission Loss characteristics of a construction is outlined in ISO 717-1 (2013). Different terminology is used, otherwise the methods are very similar. The ISO standard uses Sound Reduction Index (R) instead of Sound Transmission Loss and Weighted Sound Reduction Index (Rw ) instead of Sound Transmission Class (STC). The shape of the contour for 1/3-octave band data is identical to that shown in Figure 7.4, except that the straight line at the low-frequency end continues down to 100 Hz and at the upper-frequency end, the line terminates at 3150 Hz. In addition, there is no requirement to satisfy criterion number 1 listed above (the 8 dB criterion). The ISO standard also allows for measurements to be made in octave bands. In this case, the octave band contour is derived from the 1/3-octave band contour by connecting the values at the octave band centre frequencies. The value of 32 dB in the second criterion listed above is replaced with 10 dB for the octave band data. In addition, the ISO method includes the use of correction factors to take into account the character of the sound. These are discussed in ISO 717-1 (2013) and in Bies et al. (2018). The ability of a floor or ceiling to insulate impact sound is important, but will not be considered here. See Bies et al. (2018) for a detailed treatment of this case.

7.2.3

Single-Leaf Panel Transmission Loss Calculation

An illustration of typical panel Transmission Loss behaviour is shown in Figure 7.5(a), in which various characteristic frequency ranges are indicated. At low frequencies, the Transmission Loss is controlled by the stiffness of the panel. At the frequency of the first panel resonance, the transmission of sound is high and, consequently, the Transmission Loss passes through a minimum determined in part by the damping in the system. Subsequently, at frequencies above the first panel resonance, a generally broad frequency range is encountered, in which Transmission Loss is controlled by the surface density of the panel. In this frequency range (referred to as the mass law range, due to the approximately linear dependence of the Transmission Loss on the mass of the panel), the Transmission Loss increases with frequency at the rate of 6 dB per octave. Ultimately, however, at still higher frequencies in the region of the critical frequency, coincidence is encountered. Finally, at very high frequencies, the Transmission Loss again rises, being damping controlled, and gradually approaches an extension of the original mass law portion of the curve. The rise in this region is of the order of 9 dB per octave. The Transmission Loss of orthotropic panels is strongly affected by the existence of a critical frequency range. In this case, the coincidence region may extend over two decades for common corrugated or ribbed panels. Figure 7.5(b) shows a typical Transmission Loss characteristic of orthotropic panels. This type of panel should be avoided where noise control is important,

287

Partitions, Enclosures and Barriers (a) Isotropic

e av ct o r pe dB Damping 9 controlled

Transmission Loss (dB)

Mass law Stiffness controlled

ve octa per B 6d Coincidence region

Frequency of first panel resonance Frequency (Hz)

B

Mass law (6 dB per octave)

Stiffness controlled

Transmission Loss (dB)

pe ro cta ve

(b) Orthotropic

9

d

Damping controlled

Coincidence region First panel resonance Frequency (Hz)

FIGURE 7.5 Typical single panel Transmission Loss as a function of frequency: (a) isotropic panel characterised by a single critical frequency; (b) orthotropic panel characterised by a critical frequency range.

although it can be shown that damping can improve the performance of the panel slightly, especially at high frequencies. The resonance frequencies of a simply supported rectangular isotropic panel of width, a, length, b, and bending stiffness, B, per unit width may be calculated using the following equation: fi,n

π = 2

r

B i2 n2 + 2 2 m a b





(Hz);

i, n = 1, 2, 3, ....

(7.15)

The lowest order (or fundamental) frequency corresponds to i = n = 1. For an isotropic panel, Equation (7.6) can be substituted into Equation (7.15) to give the following. n2 i2 = 0.453cL h 2 + 2 a b



fi,n



(7.16)

The resonance frequencies of a simply supported rectangular orthotropic panel of width a and length b are (Hearmon, 1959): fi,n =

π 2m1/2



Bb n 4 Bab i2 n2 Ba i 4 + 4 + 4 a b a2 b2

1/2

;

i, n = 1, 2, 3, ....

(7.17)

where: Bab = 0.5(Ba ν + Bb ν + Gh3 /3)

(7.18)

288

Noise Control: From Concept to Application

In the above equations, G = E/[2(1 + ν)] is the material modulus of rigidity, E is Young’s modulus of elasticity, ν is Poisson’s ratio and Ba and Bb are the bending stiffnesses per unit width in directions a and b, respectively, calculated according to Equations (7.7) or (7.9). Bies et al. (2018) provide detailed analyses and calculation procedures for estimating the sound Transmission Loss of single panels. Here a single most commonly used calculation procedure due to Sharp (Sharp, 1973, 1978) is summarised. The field incidence Transmission Loss (Transmission Loss averaged over the most likely angles of incidence of incident sound – see Bies et al. (2018)) in the mass-law frequency range below fc /2 for isotropic panels or fc` /2 for orthotropic panels excited by 1/3-octave bands of noise of centre frequency, f , is: TL = 10 log10



πf m 1+ ρc

2 !

− 5.6

(dB)

(7.19)

where m is the mass per unit area of the panel. Many authors (including Sharp (1978)) do not include the “1” in Equation (7.19) as it is much smaller than the second term in brackets for most practical constructions when f > 200 Hz. However, it is included in the equations in this book for completeness so that the equations are applicable to all single panel structures. Equation (7.19) is not valid for frequencies below 1.5 times the first panel resonance frequency (see Bies et al. (2018) for estimations of TL at low frequencies), but above this frequency, it agrees reasonably well with measurements taken in 1/3-octave bands. For octave band predictions, the 5.5 should be replaced with 4.0. Alternatively, better results are usually obtained for the octave band Transmission Loss, TLoct , by averaging logarithmically the predictions, TL1 , TL2 and TL3 for the three 1/3-octave bands included in each octave band as: TLoct = −10 log10

 1  −TL1 /10 10 + 10−TL2 /10 + 10−TL3 /10 3

(dB)

(7.20)

For frequencies equal to or higher than the critical frequency, Sharp gives the following equation for an isotropic panel:

" TL = 10 log10 1 +



πf m ρc

2 #

+ 10 log10 [2ηf /(πfc )]

(dB)

(7.21)

Equation (7.21) is only used until the frequency is reached at which the calculated TL is equal to that calculated using the mass law expression given by Equation (7.19) (see Figure 7.6(a)). Values for the panel loss factor, η, which is a measure of the panel’s ability to damp out vibrations, and which appears in the above equation, are listed in Appendix A. Note that the loss factors listed in Appendix A are not solely for the material but include the effects of typical support conditions found in wall structures. The Transmission Loss between 0.5fc and fc is approximated by connecting with a straight line the points corresponding to 0.5fc and fc on a graph of TL versus log10 (frequency). The preceding prediction scheme is summarised in Figure 7.6(a), where a method for estimating the Transmission Loss for single isotropic panels is illustrated. A scheme for estimating the Transmission Loss of orthotropic panels is included in Figure 7.6(b). A full derivation of both of these schemes is provided in Bies et al. (2018). There are two important points worth noting when using the Figure 7.6 prediction scheme for orthotropic panels. 1. Particularly for small panels, the Transmission Loss below about 0.7fc` is underestimated, the error becoming larger as the frequency becomes lower or the panel becomes smaller.

289

Partitions, Enclosures and Barriers

oc

ta v e

TL (dB)

(a)

B 6d

9d

B

pe r

A

e tav oc r pe

B

0.5

1.0

pe

ro

cta

ve

f /fc (log scale)

6d

B

(b) TL (dB)

C

D

A

6d

B

pe

ro

cta

ve

B

0.5fc R

fcR

f (log scale)

0.5fcu 2fcu

FIGURE 7.6 Design charts for estimating the Transmission Loss of a single panel. See Appendix A for values of ρ(= m/h) and η for typical materials. See the text for definitions of fc , fc` and fcu (see the discussion in the two paragraphs preceding Section 7.2.2). Points A and B in parts (a) and (b) and points C and D in part (b) are connected with straight lines on a plot of TL vs log10 (frequency) (a) A design chart for an isotropic panel. The points on the chart are calculated as follows: point A: TL = 20 log10 fc m − 54 (dB) (Equation (7.19) with f /fc = 0.5 and the “+1” left out, as it makes less than 0.1 dB difference in practical constructions). point B: TL = 20 log10 fc m + 10 log10 η − 44 (dB) (Equation (7.21) with the “+1” left out and with f /fc = 1.0) (b) A design chart for an orthotropic (or ribbed) panel, with critical frequencies fc` and fcu , and small damping. For a well damped panel, see the discussion in the last paragraph of Section ??. The points in the chart are calculated as follows: point A: TL = 20 log10 fc` m−54 (dB) (Equation (7.19) with the “+1” left out and with f /fc` = 0.5) Between and including points B and C: TL = 20 log10 f + 10 log10 m − 10 log10 fc` − 20 log10 [loge (4f /fc` )] − 13.2 Point D: TL = 10 log10 m + 15 log10 fcu − 5 log10 fc` − 17

(dB)

(dB)

290

Noise Control: From Concept to Application

2. For common corrugated panels, there is nearly always a frequency between 2000 and 4000 Hz where there is a dip of up to 5 dB in the measured Transmission Loss curve, which is not predicted by theory. This frequency corresponds either to an air resonance between the corrugations or a mechanical resonance of the flat panel between the ribs. The frequency corresponding to the air resonance (there may be more than one frequency for panels with more than one corrugation spacing) may be calculated using: fr =

c 2`cs

(7.22)

where `cs is the average corrugation cavity width (in m) equal to (` − b1 + b3 )/2 in Figure 7.3, and c is the speed of sound in air (343 m/s). The Transmission Loss for a single orthotropic panel may be calculated using Figure 7.6(b). If an orthotropic panel is heavily damped, then the Transmission Loss will be slightly greater (by about 1 to 4 dB) at higher frequencies than indicated in Figure 7.6(b), beginning with 1 dB at 500 Hz and increasing to 4 dB at 4000 Hz for a typical corrugated building panel. Example 7.2 At what frequency does the mass law Transmission Loss for a 10 mm thick steel panel immersed in fresh water equal 4 dB? Explain in physical terms why the Transmission Loss in air at this frequency is much greater. Solution 7.2 Mass Law Transmission Loss is given by Equation !(7.19). Substituting in values for the variables gives, 4 = 10 log10

1+



π × f × 7850 × 0.01 998 × 1497

r Rearranging gives, f =

2

− 5.5.

100.95 − 1 = 17.0 kHz. 2.7248 × 10−8

The Transmission Loss in air at this frequency is much greater because the difference in the impedance of the panel and the characteristic impedance in the propagating medium is much larger for air than water. Example 7.3 Consider the steel panel with the cross section shown in Figure 7.7, having E = 207 GPa and Poisson’s ratio ν = 0.3. 75

25

75 zr

zn

o

o

45

30

45

25 All dimensions in mm 75

FIGURE 7.7 Arrangement for Example 7.3.

(a) Calculate the bending stiffness in two directions: along the ribs and across the ribs. The panel thickness is 1.2 mm. (b) Calculate the bending wave speed in both directions for the panel at a frequency of 1000 Hz.

291

Partitions, Enclosures and Barriers (c) Calculate the range of critical frequencies for the panel. (d) If the panel is one wall of an enclosure and has dimensions 2 m × 2 m, calculate its lowest resonance frequency. (e) Calculate the Transmission Loss for the panel in octave bands from 63 Hz to 8000 Hz. Solution 7.3 (a) We first find the location of the neutral axis by taking moments about a reference axis which we choose to lie through the centre of the angled parts. The reference axis is located a distance of zr from the centres of the top segments of the section, as shown in Figure 7.7. The centre of the entire section is chosen as the reference axis, as it makes it easier to calculate contributions from the segments that are neither horizontal or vertical. However, the top of the entire section could also be used. In the following equations, bi is the length of the ith segment and zri is the distance of the ith segment from the chosen reference axis. The distance of the neutral axis from the reference axis is denoted as zn , as shown in Figure 7.7. For this problem, distances above the neutral axis are defined as positive. As all hi are equal they cancel in Equation (7.8) and the equation becomes:

P 75 × 15 + 2 × 25 × 2.5 − 25 × 10 + 75 × 15 − 75 × 15 i bi zri √ = zn = P 75 + 3 × 25 + 75 + 2 2 × 30 + 75

i bi

1000 = = 2.6 mm. 384.9 Thus the neutral axis is 2.6 mm above the reference axis or 12.4 mm from the centre of the top section of the panel. The section thickness, h = 1.2 mm and the horizontal length, `, before repeating itself is 250 + 60 mm = 0.31 m. The bending stiffness in the direction along the ribs may be calculated with E = 207 GPa and ν = 0.3, using Equation (7.7). Thus: 207 × 109 × 0.0012 0.00122 B1 = 2 × 0.075 0.01242 + 0.91 × 0.31 12







+ (2 × 0.025) 0.00012 +



+ 0.075 0.01762 +

0.0252 12

0.00122 12



 

+ 0.025 0.01262 +







0.00122 + 2 × 0.032 + (2 × 0.03) 2 0.0026 + 24 √

0.00122 12

2





= 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 m2 s−2 . The stiffness in the direction across the ribs may be calculated using Equation (7.9): B2 =

207 × 109 × 0.00123 = 32.8 kgm2 s−2 12 × 0.91

(b) The bending wave speed is calculated using Equation (7.1). The surface mass of the panel is m = 7850 × 0.0012 × 0.385/0.31 = 11.70kg m−2 and the frequency is 1000 Hz. Thus the lower and upper bending wave speeds corresponding to waves propagating parallel and perpendicular to the ribs, respectively, are:

292

Noise Control: From Concept to Application

cB1 =



5.27 × 104 × 4π 2 × 106 11.70

1/4

= 650 m/s,

1/4

32.8 × 4π 2 × 106 and cB2 = = 103 m/s. 11.70 (c) The lower and upper critical frequencies for the panel may be calculated using Equation (7.3). Thus:



fc` =

3432 2π



11.70 5.27 × 104

1/2

= 279 Hz and fcu =

3432 2π



11.70 32.8

1/2

= 11, 200 Hz.

(d) We can assume that the enclosure wall edge condition is simply supported, which means that it cannot translate (move in any axial direction) but is free to rotate. The simply supported edge assumption is a good approximation in practice for most enclosures. The first resonance frequency of the panel may be calculated using Equations (7.17) and (7.18) with i = n = 1. Thus: 207 × 109 × 0.00123 = 0.5 5.27 × 10 × 0.3 + 32.8 × 0.3 + 3 × 2(1 + 0.3)



Bab

4

π and f1,1 = √ 2 11.70



5.27 × 104 32.8 7910 + 4 + 4 24 2 2

1/2



= 7910 kgm2 s−2 ,

= 28.3 Hz.

(e) The sound Transmission Loss of the panel may be calculated using the Equations in the caption of Figure 7.6(b). Point A is at 139 Hz and the TL at point A is: TLA = 20 log10 (279 × 11.70) − 54 = 16.3 dB. At point B (279 Hz), the TL is: TLB = 20 log10 (279) + 10 log10 (11.70) − 10 log10 (279) − 20 log10 [loge (4)] − 13.2 = 19.1 dB. At point C (5,600 Hz), the TL is: TLC = 20 log10 (5600) + 10 log10 (11.70) − 10 log10 (279) − 20 log10 [loge (22400/279)] − 13.2 = 35.1 dB. At point D (22,400 Hz), the TL is: TLD = 10 log10 (11.70) + 15 log10 (11200) − 5 log10 (279) − 17 = 42.2 dB. These points are plotted in Figure 7.8 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: TLoct = −10 log10 (1/3)[10−TL1 /10 + 10−TL2 /10 + 10−TL3 /10 ] . 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. Nevertheless, for the case of isotropic panels, care should be taken to avoid errors near the dip in the curve corresponding to the critical frequency. The octave band results are summarised in Table 7.2.





TABLE 7.2

Transmission Loss (dB)

Results for Example 7.3

63

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000

9

15

19

21

25

29

33

8000 37

293

Partitions, Enclosures and Barriers

Transmission Loss (dB)

40

C

30

A

20

B

10 0 63

125

250

0.5fcR

500

1k

fcR

2k

4k

8k

0.5fcu

Frequency (Hz) FIGURE 7.8 Results for Example 7.3.

Example 7.4 (a) Calculate the fundamental resonance frequency and the upper and lower critical frequencies for a corrugated steel panel of dimensions 3 m × 3 m and thickness 1.6 mm. The corrugations may be described by y=20sinπ(x/40) where y is the corrugation height (or depth) and x is the distance in mm across the width of the panel. You may approximate the sinusoidal shape of half of a full sine wave by three straight lines as illustrated in Figure 7.9.

b2 b1 y1 q1 15

b3 q3

10

15

FIGURE 7.9 Panel profile for Example 7.4.

(b) Calculate the Transmission Loss as a function of frequency for the panel in part (a). Express your answer graphically. Also tabulate the Transmission Loss at the octaveband centre frequencies. Assume that the panel loss factor is 0.001. Solution 7.4 Final answers are given to 3 significant figures as that is more than enough to reflect the accuracy of the prediction procedure. (a) As the panel cross section is symmetrical about the neutral axis, which is the horizontal line in Figure 7.9, the stiffness of the lower half of the sine wave is the same as the upper half. Thus, only one half of the full sine wave section needs to be considered. That is: y1 = 20 sin(15π/40) = 18.48 mm, b1 = 23.8 mm, θ1 = 50.9◦ , θ2 = 0, b2 = 10 mm, b3 = 23.8 mm, θ3 = 50.9◦ , ` = 40, E = 207GPa, ν = 0.3.

294

Noise Control: From Concept to Application Using Equation (7.7), we can write the following for the bending stiffness for waves travelling parallel to the corrugations:

(

207 × 109 × 0.0016 1.62 + 23.82 B1 = 23.8 × 2 (18.48/2)2 + 9 0.91 × 0.04 × 10 24



1.62 − 23.82 1.62 + cos(101.8◦ ) + 10 18.482 + 24 12





)

= 8.04 × 104 kgm2 s−2 .

The bending stiffness for waves travelling perpendicular to the corrugations can be calculated using Equation (7.9) as: 207 × 109 × 1.63 × 10−9 B2 = = 77.6 kgm2 s−2 . 12 × 0.91 The surface mass is m = ρh = 7850 × 0.0016 × (23.8 + 23.8 + 10)/40 = 18.1 kg/m2 . The lower and upper critical frequencies are calculated using Equation (7.3) as: fc`

3432 = 2π



18.1 8.04 × 104

1/2

= 281 Hz

and

fcu

3432 = 2π



18.1 77.6

1/2

= 9, 040 Hz.

Assuming that the enclosure wall edge condition is simply supported, the first resonance frequency may be calculated using Equation (7.17). Thus: 8.04 × 104 77.6 π + 4 = √ 4 3 3 2 18.1



f1,1

+ 0.5



8.04 × 104 × 0.3 77.6 × 0.3 207 × 109 × 0.00163 + + 4 34 34 3 × 3 × 2(1 + 0.3)

 1/2

= 12.5 Hz. (b) The sound Transmission Loss of the panel may be calculated using the equations in the caption of Figure 7.6(b). Point A is at 140 Hz and the corresponding TL is given by: TLA = 20 log10 (281 × 18.1) − 54 = 20.1 dB. At point B (281 Hz), the TL is: TLB = 20 log10 (281) + 10 log10 (18.1) − 10 log10 (281) − 20 log10 [loge (4)] − 13.2 = 21.0 dB. At point C (4,520 Hz), the TL is: TLC = 20 log10 (4520) + 10 log10 (18.1) − 10 log10 (281) − 20 log10 [loge (18080/281)] − 13.2 = 35.6 dB. At point D (18,080 Hz), the TL is: TLD = 10 log10 (18.1) + 15 log10 (9040) − 5 log10 (281) − 17 = 42.7 dB. These points are plotted in Figure 7.10 and interpolation is used to find the octave band TL values which are listed in Table 7.3. Strictly speaking, Figure 7.10 should only be used to find 1/3-octave values and the octave band levels must then be calculated as discussed in the previous example. TABLE 7.3

Transmission Loss (dB)

Results for Example 7.4, part (b)

63

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000

13

19

21

23

26

30

35

8000 38

295

Partitions, Enclosures and Barriers

Transmission Loss (dB)

40

C 30

B

A

20 10 0 63

125

250

0.5fcR

500

1k

4k

2k

8k

0.5fcu

fcR Frequency (Hz)

FIGURE 7.10 Results for Example 7.4, part (b).

7.2.4

Double Wall Transmission Loss

When a high Transmission Loss structure is required, a double wall or triple wall is less heavy and more cost-effective than a single wall. Design procedures have been developed for both types of wall. However, the present discussion will be focussed on double wall constructions. A single stud construction is illustrated in Figure 7.11(a) and a staggered stud construction is illustrated in Figure 7.11(b). For a more thorough discussion of Transmission Loss, consideration of triple wall constructions and for some experimental data for wood stud walls, the reader is referred to the published literature (Sharp, 1973, 1978; Brekke, 1981; Bradley and Birta, 2001; Davy, 2010; Davy et al., 2012). Each stud only fixed to one panel

b Wall panel

Wall panel e (for point attachment)

Wall panel

Studs (a)

(b)

FIGURE 7.11 Double wall constructions. (a) Single row of studs separated by a distance, b. This figure also shows the distance, e, which is part of the discussion later in this section. (b) Staggered stud wall with each stud only attached to one of the two wall panels.

For best results, the two panels of the double wall construction must be both mechanically and acoustically isolated from one another as much as possible. Mechanical isolation may be accomplished by mounting the panels on separate staggered studs or by resiliently mounting the panels on common studs. Resilient mounting can be achieved using rubber strips or flexible steel sections for studs. Acoustic isolation is generally accomplished by providing as wide a gap between the panels as possible and by filling the gap with sound-absorbing material, while

296

Noise Control: From Concept to Application

ensuring that the material does not form a mechanical bridge between the panels. For best results, the panels should be isotropic. The Institute for Research in Construction within the Canadian National Research Council has supported an extensive assessment program of measurements of various types of double wall construction, using both wooden and steel studs to separate and support the two wall leaves or panels. Some of these results have been reported by Bradley and Birta (2000); Quirt et al. (2008); Quirt and Nightingale (2008); Warnock (2008). The latter report contains equations for estimating the STC value for three different double wall constructions containing type TC steel studs (which are channel sections with the panels attached to the two flanges). These are listed below. Note that the flexibility of steel studs reduces the extent of structure-borne sound transmission through a wall construction resulting in higher TL values for walls with steel studs compared to walls with wooden studs (except in the low-frequency range below a few hundred Hz). • Non-load-bearing steel studs or load-bearing steel studs with resilient metal channels and sound-absorbing material almost filling the entire cavity. STC = 13.5 + 11.4(log10 m1 + log10 m2 ) + 82.6d + 8.5b

(dB)

where m1 and m2 are the masses per unit area of the two panels attached to the studs (kg/m2 ), d is the cavity depth (separation distance between the two panels) in metres and b is the stud spacing (in metres). • Non-load-bearing steel studs or load-bearing steel studs with resilient metal channels and no sound-absorbing material in the cavity. STC = −18.8 + 17.55(log10 m1 + log10 m2 ) + 165.0d + 10.0b

(dB)

• Staggered steel studs with sound-absorbing material in the cavity. STC = 17.7 + 14.54(log10 m1 + log10 m2 ) + 23.0d + 27.0t

(dB)

where t is the insulation thickness in metres. In Section 7.2.3, it was shown that the Transmission Loss of a single isotropic panel is determined by two frequencies, namely the lowest order panel resonance frequency, f1,1 , and the coincidence frequency, fc . The double wall construction introduces three new important frequencies. The first is the lowest order acoustic resonance, the second is the lowest order structural resonance and the third is a limiting frequency related to the gap between the panels. The lowest order acoustic resonance, f2 , replaces the lowest order panel resonance of the single panel construction (below which the following procedure cannot be used) and may be calculated as: f2 = c/2L (7.23) where c is the speed of sound in air and L is the longest cavity dimension. The lowest order structural resonance may be approximated by assuming that the two panels are limp masses connected by a massless compliance, which is provided by the air in the gap between the panels. In practice, it is necessary to introduce an empirical factor of 1.8 into the equation defining the frequency corresponding to this structural resonance, in order to give better agreement with existing data for ordinary wall constructions (Sharp, 1973). Introducing the empirical constant, 1.8, the following expression (Fahy, 1985) is obtained for the mass-airmass resonance frequency, f0 , for panels that are large compared to the width of the gap between them:  1/2  1/2 m1 + m2 1 1.8ρc2 (m1 + m2 ) ≈ 80.4 (Hz) (7.24) f0 ≈ 2π dm1 m2 dm1 m2 where m1 and m2 are, respectively, the surface densities (kg/m2 ) of the two panels and d is the gap width (m). The empirical constant, 1.8, has been introduced by Sharp (1973) to account for

297

Partitions, Enclosures and Barriers

the “effective mass” of the panels being less than their actual mass. Part of the 1.8 factor (a factor of 1.4) may be attributed to propagation of sound through porous material being an isothermal rather than adiabatic process. In this case, the isothermal speed of sound for propagation in air √ is γ = 1.18 times (or 18%) less than the adiabatic speed of sound. For cavity walls with no sound-absorbing material, the 1.8 factor should be reduced to 1.8/1.4 = 1.3. Davy (2018) has reported experimental results showing that if steel studs are used, the 1.8 factor in Equation (7.24) should be replaced with 1, and for wooden studs, the 1.8 factor varies between 1.7 and 3.6. So for the calculations in this book, for steel studs the 1.8 factor becomes 1, for wooden studs with no sound-absorbing material in the cavity it becomes 1.3 and for wooden studs with sound-absorbing material in the cavity, it will be left as 1.8. Finally, a limiting frequency, f` , which is related to the gap width d (m) between the panels, is defined as: f` = c/2πd ' 55/d

(Hz)

(7.25)

The frequencies, f2 , f0 and f` , given by Equations (7.23) to (7.25) for the two-panel assembly, are important for determining the transmission behaviour of the double wall. Note that f` is equal to the lowest cavity resonance frequency (corresponding to the largest cavity dimension), for wave propagation in the cavity normal to the plane of the panels, divided by π. The frequencies, fc1 and fc2 , calculated using Equation (7.3) for each panel, are also important. As the nature of the attachment of a panel to its supporting studs determines the efficiency of conduction of structure-borne sound from the panel to the stud and vice versa, it is necessary to distinguish between two possible means of attachment (line and point) and, in the double panel wall under consideration, four possible combinations of such attachment types. A panel attached directly to a supporting stud generally will make contact along the length of the stud. Such support is called line support and the spacing between studs, b, is assumed regular. Alternatively, the support of a panel on small spacers mounted on the studs is called point support; the spacing, e, between point supports is assumed to form a regular square grid. The dimensions b and e are important in determining Transmission Loss. In the following discussion it is assumed that the two panels are numbered, so that the critical frequency of panel 1 is always less than or at most equal to the critical frequency of panel 2. With this understanding, four combinations of panel attachment are possible as follows: line– line, line–point, point–line and point–point. Of these four possible combinations of panel support, point–line will be excluded from further consideration, as the Transmission Loss associated with it is always inferior to that obtained with line–point support. In other words, for best results the panel with the higher critical frequency should be point supported if point support of one of the panels is considered. A method for estimating Transmission Loss for a double panel wall is outlined in Figure 7.12. In the figure, consideration has not been given explicitly to the lowest order acoustic resonance, f2 , of Equation (7.23). At this frequency it can be expected that somewhat less than the predicted mass-law Transmission Loss will be observed, dependent upon the cavity damping that has been provided. In addition, below the lowest order acoustic resonance, the Transmission Loss will again increase, as shown by the stiffness controlled portion of the curve in Figure 7.5. The procedure outlined in Figure 7.12 explicitly assumes that the inequality, M f > 2ρc, is satisfied. The estimation of Figure 7.12 is based on the assumption that the studs connecting the two walls of the construction are infinitely stiff. This is an acceptable assumption if wooden studs are used but not if metal studs (typically thin-walled channel sections with the partition leaves attached to the two opposite flanges) are used. More details on how the compliance of metal studs may be taken into account are provided in Bies et al. (2018). Damped corrugated panels act as isotropic panels of the same thickness, allowing the prediction procedures to be used for double walls consisting of at least one wall that is a damped

298

Noise Control: From Concept to Application

corrugated panel. However, the consideration of double walls that include an undamped corrugated panel is beyond the scope of this book. An important point regarding stud walls with gypsum board leaves is that a stud spacing of between 300 and 400 mm has been shown (Rindel and Hoffmeyer, 1991) to severely degrade the performance of the double wall in the 160 and 200 Hz 1/3-octave bands by up to 13 dB. Other stud spacings (even 100 and 200 mm) do not result in the same performance degradation, although smaller stud spacings improve the low-frequency performance (below 200 Hz) at the expense of a few dB loss at all frequencies between 250 and 2000 Hz. It is also important not to use walls of the same thickness (and material) as this greatly accentuates the dip in the TL curve at the critical frequency. This is also important for double glazing constructions, although it is not commonly implemented. As an aside, one problem with double glazing is that it can suffer from condensation, so if used, drainage holes are essential. 7.2.4.1

Staggered Studs

A staggered stud arrangement is commonly used to obtain high Transmission Loss. In this arrangement, studs of a common wall are alternately displaced. Panels fastened to them on opposite sides are then supported on alternate studs. The only common support between opposite panels is at the perimeter of the common wall, for example, at the base and top. For the purpose of calculating expected Transmission Loss, the staggered stud construction could be modelled as a perimeter-supported double wall similar to that of Example 7.5. However, the introduction of studs improves the structure-borne coupling and degrades the Transmission Loss which is obtained. For example, when staggered studs are introduced on 0.61 m centres, although they provide no direct coupling, they degrade the measured Transmission Loss of the perimeter mounted configuration of Example 7.5. On the other hand, if care is taken to ensure that at least one of the panels is very well damped, even higher Transmission Loss may be obtained with staggered studs than for a perimeter-mounted double panel arrangement with no panel damping. Thus, if neither of the two panels making up the double wall construction is well damped, the expected Transmission Loss for a double wall on staggered studs will lie between that of perimeter mounting and line–line support, the latter calculated using Figure 7.12. Alternatively, if at least one panel is very well damped then the double wall may be modelled as perimeter supported, and slightly higher Transmission Loss can be expected than predicted. 7.2.4.2

Panel Damping

A simple means for achieving the very high panel damping alluded to above is to construct a thick panel of two thin panels glued together at spots in a regular widely spaced grid. Slight motion of the panels results in shear between the panels in the spaces between attachments, resulting in very effective panel damping due to the shearing action, which dissipates energy in the form of heat. 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. When two panels are glued together at spots in a regular widely spaced grid (0.3 m to 0.6 m) or connected with a layer of visco-elastic material or even nailed together, the critical frequency of the construction would equal the lowest critical frequency of the two panels, not the critical frequency based on the thickness of the complete construction. However, if the panels were bonded very firmly together, then the critical frequency may be calculated on the basis of the total thickness of the construction. When glass is the material used for the wall or for a window, damping can be increased by using laminated glass, which is a sandwich of two layers of glass separated by a plastic sheet. Sound-absorbing material may also be placed around the perimeter of the cavity between two glass walls to increase acoustic absorption without affecting the transparency of the glass.

299

Partitions, Enclosures and Barriers

o c t.

B

6

C

ct. dB/o

18

d B/

TL (dB)

15

D

dB /oc t.

t. /oc B d 12 12 dB/oct.

/oct.

6 dB

A f0

fR

pfR 0.5 fc1 fc 2 Frequency (Hz) (log scale)

FIGURE 7.12 A design chart (solid line) for estimating the Transmission Loss of a double panel wall, based on Sharp (1973). The upper dashed line is used when point B is higher than shown in the figure. The lower dashed line indicates the mass-law TL, but is not used in these TL calculations. The panels are assumed to be numbered so that the critical frequency, fc1 , of panel 1 is always less than or equal to the critical frequency, fc2 , of panel 2, i.e. fc1 ≤ fc2 ; m1 and m2 (kg m−2 ) are the respective panel surface densities; d (m) is the spacing between panels; b (m) is the spacing between line supports; and e (m) is the spacing of an assumed square grid between point supports. c and cL (m/s) are, respectively, the speed of sound in air and in the panel material, h is the panel thickness and η1 and η2 are the loss factors for panels 1 and 2, respectively. Calculate the points, A, B, C and D in the chart and then construct a line with a slope of −6 dB/octave from the left side of B to where it intersects the line from A to D. Point A: f0 = 80.4

p

(m1 + m2 )/(dm1 m2 )

(Hz);

TLA = 20 log10 (m1 + m2 ) + 20 log10 f0 − 48

(dB)

2

Point B: fc1 = 0.55c /(cL1 h1 ) (Hz) The Transmission Loss, TLB , at point B is equal to TLB1 if no sound-absorptive material is placed in the cavity between the two panels; otherwise, TLB is equal to TLB2 , provided sufficient absorption is achieved: TLB1 = TLA + 20 log10 (fc1 /f0 ) + 20 log10 (fc1 /f` ) − 22 (dB) : fc1 ≥ 2πf` TLB1 = TLA + 20 log10 (0.5fc1 /f0 ) (dB) : fc1 < 2πf`

  (a) Line–line support 1/2 m2 fc1 − 78 TLB2 = 20 log10 m1 + 10 log10 b + 20 log10 fc1 + 10 log10 fc2 + 20 log10 1 + 1/2 m1 fc2 (b) Line–point support (fc2 is the critical frequency of the point supported panel): TLB2 = 20 log10 m1 e + 20 log10 fc1 + 20 log10 fc2 − 99 (dB)   (c) Point–point support: 1/2 m2 fc1 TLB2 = 20 log10 m1 e + 20 log10 fc1 + 20 log10 fc2 + 20 log10 1 + − 105 1/2 m1 fc2 Point C:   fc2 (a) fc2 6= fc1 ; TLC = TLB + 6 + 10 log10 η2 + 20 log10 (dB) fc1 (b) fc2 = fc1 ;

TLC = TLB + 6 + 10 log10 η2 + 5 log10 η1

Point D: f` = 55/d

(dB)

(dB)

(dB)

(Hz)

The final TL curve for sound-absorbing material in the cavity is the solid line in the figure. The dotted line deviation between f0 and fc1 replaces the solid line in this range when there is no sound-absorbing material in the cavity. The lower dashed line represents the mass-law TL that would be achieved at frequencies below fc /2 with no cavity between the panels, where fc is the critical frequency corresponding to the total panel thickness.

300

Noise Control: From Concept to Application

Example 7.5 A double gypsum board wall is mounted at its perimeter in an opening of dimensions 3.0 × 2.44 m in a test facility. The spacing between the panels is 0.1 m. The surface densities and critical frequencies of the panels are, respectively, 12.67 kg/m2 and 2500 Hz. Calculate the expected Transmission Loss. The space between the walls is well damped with a 50 mm thick layer of sound-absorbing material. However, the panels themselves have not been treated with damping material. Solution 7.5 Reference is made to the equations in the caption of Figure 7.12. Calculate the co-ordinates of point A: f0 = 80.4

p

2 × 12.67/0.1 × 12.672 = 101 Hz.

TLA = 20 log10 (2 × 12.67) + 20 log10 101 − 48 = 20 dB. Calculate the co-ordinates of point B. Since the panel is supported at the edge, the area associated with each support is less than half of that assumed in the theory, as the theory is based on there being equal panel areas on either side of the stud. As the sound transmitted is proportional to the panel area, 4 dB is added to the calculated Transmission Loss. This addition represents slightly more than a factor of 2 (10 log10 2 = 3 dB). There is sound-absorbing material in the cavity so we only calculate TLB2 . As the panel is supported along its edge, we use the expression for line-line support and to be conservative, we use the smallest panel dimension (2.44 m). Thus, TLB2 = 20 log10 12.67 + 10 log10 2.44 + 30 log10 2500 + 6 − 78 + 4 = 60 dB. Calculate the co-ordinates of point C. According to Appendix A, a loss factor value of η = 0.05 should be used for each panel. Thus, TLC = 60 + 6 − 13 − 7 = 46 dB. Construct the estimated Transmission Loss curve shown in Figure 7.13. For comparison, experimentally determined points and the results for a loss factor of 0.1 (dashed line originating at B) instead of 0.05 are also shown, where it can be seen that 0.1 may be a better estimate of the panel loss factor. 70

t. t.

B

dB 15

50 dB /oc

t.

C

40

18

Transmission loss (dB)

B 6d

/oc

/oc

60

30

20

10

measured

ct. B/o d 6

calculated (h = 0.05) calculated (h = 0.1)

A

63

125

250

500

1000

2000

4000

One-third octave band centre frequencies (Hz) FIGURE 7.13 Results for Example 7.5.

8000

301

Partitions, Enclosures and Barriers

Example 7.6 Explain why a double wall partition may perform more poorly at some frequencies than a single wall partition of the same total weight. Solution 7.6 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).

7.2.5

Triple Wall Sound Transmission Loss

Very little work has been done in this area, but research reported by Tadeu and Mateus (2001) indicates that for double and triple glazed windows with the same total weight of glazing and total air gap, nothing is gained in using triple glazing over double glazing. However, this is because the cut-on frequency above which 3-D reflections occur in the cavity, is above the frequency range of interest for typical panel separations used in windows (d = 30 mm to 50 mm). The cut-on frequency is given by the following equation: fco = c/2d

(7.26)

Note that the poorest performance is achieved with panes of glass separated by 10 mm to 30 mm (Tadeu and Mateus, 2001). Above the cut-on frequency, it is possible to achieve a marked improvement with a triple panel wall (Brekke, 1981). Sharp (1973) reported that for constructions of the same total mass and total thickness, the double wall construction has better performance for frequencies below 4f0 , whereas the triple wall construction performs better at frequencies above 4f0 , where f0 is the double panel resonance frequency defined by Equation (7.24). Below f0 , the two constructions will have the same Transmission Loss and this will be the same as for a single wall of the same total mass. This is because at low frequencies, the double and triple wall leaves vibrate as a single entity and the resulting Transmission Loss is then a function of the total mass of the construction.

7.2.6

Sound-Absorptive Linings

When an enclosure is to be constructed, some advantage will accrue by lining the walls with a porous material. The lining will prevent reverberant sound build-up, which would lessen the effectiveness of the enclosure for noise reduction, and at high frequencies it will increase the Transmission Loss of the walls. Calculated Transmission Loss values for a typical blanket of porous material are given in Table 7.4. TABLE 7.4 Calculated Transmission Loss (TL) values (dB) for a typical blanket of porous acoustic material (medium density rockwool, 50 mm thick)

Frequency (Hz)

TL (dB)

1000 2000 4000 8000

0.5 1.5 4 12

302

7.2.7

Noise Control: From Concept to Application

Common Building Materials

Results of Transmission Loss (field incidence) tests on conventional building materials and structures have been published both by manufacturers and testing laboratories. Some examples are listed in Table 7.6 of Bies et al. (2018).

7.3

Composite Transmission Loss

The wall of an enclosure may consist of several elements, each of which may be characterised by a different transmission loss coefficient. For example, the wall may be constructed of panels of different materials, it may include permanent openings for passing materials or cooling air in and out of the enclosure, and it may include windows for inspection and doors for access. Each such element must be considered in turn in the design of an enclosure wall, and the Transmission Loss of the wall determined as an overall area weighted average of all of the elements. For this calculation, Equation (7.27) is used: q P

τ=

Si τi

i=1 q P

(7.27) Si

i=1

In Equation (7.27), Si is the surface area (one side only), and τi is the transmission coefficient of the ith element. The transmission coefficient of any element may be determined, given its Transmission Loss, TL, from the following equation: τ = 10(−TL/10)

(7.28)

The overall transmission coefficient is then calculated using Equation (7.27) and, finally, the Transmission Loss is calculated using Equation (7.10). It can be seen from Equation (7.27) that low Transmission Loss elements within an otherwise very high Transmission Loss wall can seriously degrade the performance of the wall; the Transmission Loss of any penetrations must be commensurate with the Transmission Loss of the whole structure. In practice, this generally means that the Transmission Loss of such things as doors, windows, and access and ventilation openings, should be kept as high as possible, and their surface areas small. A list of the Transmission Losses of various panels, doors and windows is included in Table 7.6 of Bies et al. (2018). More comprehensive lists have been published in various handbooks, and manufacturers of composite panels generally supply data for their products. Where possible, manufacturer’s data should be used; otherwise the methods outlined in this chapter may be used. Example 7.7 Calculate the overall Transmission Loss at 125 Hz of a wall of total area 10 m2 constructed from a material that has a Transmission Loss of 30 dB, if the wall contains a panel of area 3 m2 , constructed of a material having a Transmission Loss of 10 dB. Solution 7.7 For the main wall, the transmission coefficient is (from Equation (7.28)), τ1 = 1/[10(30/10) ] = 0.001, while for the panel, τ2 = 1/[10(10/10) ] = 0.100. Hence τ = [(0.001 × 7) + (0.100 × 3)]/10 = 0.0307. The overall Transmission Loss is therefore: TLoverall = 10 log10 (1/0.0307) = 15 dB.

303

Partitions, Enclosures and Barriers

7.4

Enclosures

An enclosure placed around a noisy item of equipment is often used for the purpose of noise control. Such an enclosure will produce a reverberant sound field within it, in addition to the existing direct sound field of the source. Both the reverberant and direct fields will contribute to the sound radiated by the enclosure walls, as well as to the sound field within the enclosure. As shown by Bies et al. (2018), the noise reduction (or Insertion Loss) of an enclosure around a noisy item of equipment, either indoors or out-of-doors is: NR = TL − C

(dB)

(7.29)

where: C = 10 log10 [0.3 + SE (1 − α ¯ i )/(Si α ¯ i )]

(dB)

(7.30)

α ¯ i is the mean Sabine absorption coefficient of the interior of the enclosure, Si is the interior surface area of the enclosure and SE is the area of the external surface of the enclosure. Empirical values for C are given in Table 7.5. These apply to both machine and personnel enclosures. However, the latter enclosures would usually satisfy the “dead” criterion. TABLE 7.5 Values of coefficient, C (dB), to account for enclosure internal acoustic conditions. The following criteria are used to determine the appropriate acoustic conditions inside the enclosure

Enclosure internal acoustic conditions

63

125

Live Fairly live Average Dead

18 13 13 11

16 12 11 9

Octave band centre frequency (Hz) 250 500 1000 2000 4000 15 11 9 7

14 12 7 6

12 12 5 5

12 12 4 4

12 12 3 3

8000 12 12 3 3

Live: All enclosure surfaces and machine surfaces hard and rigid. Fairly live: All surfaces generally hard but some panel construction (sheet metal or wood). Average: Enclosure internal surfaces covered with sound-absorptive material, and machine surfaces hard and rigid. Dead: As for “Average”, but machine surfaces mainly of panels.

Placing an enclosure around a source will generate a reverberant field in addition to the existing direct field and both fields will contribute to sound transmission through the enclosure. The effect of inadequate absorption in enclosures is very noticeable. Table 7.6 shows the reduction in performance of an ideal enclosure with varying degrees of internal sound absorption. The sound power of the source is assumed constant and unaffected by the enclosure. “Percent” refers to the fraction of internal surface area that is treated. For best results, the internal surfaces of an enclosure are usually lined with glass or mineral fibre or open-cell polyurethane foam blanket. Typical values of absorption coefficients are given in Table 6.2 of Bies et al. (2018). Since the absorption coefficient of absorbent lining is generally highest at high frequencies, the high-frequency components of any noise will be subject to the highest attenuation. Some improvement in low-frequency absorption can be achieved by using a thick layer of lining. It is interesting that Equation (7.29) also applies to personnel enclosures, where the intention is to reduce the noise levels for people inside the enclosure. Close fitting enclosures, where the air gap between machine and enclosure wall is small compared to a wavelength, behave somewhat differently due to coupling of air gap resonances

304

Noise Control: From Concept to Application

TABLE 7.6 Enclosure noise reduction as a function of percentage of internal surface covered with soundabsorptive material

Percent sound-absorbing material Noise reduction (dB)

10 −10

20 −7

30 −5

50 −3

70 −1.5

with the enclosure walls. This topic, together with enclosure resonances in general is discussed in Nefske and Sung (2006). The noise reduction achieved by close-fitting enclosures, may be degraded by resonance effects that have not yet been discussed. At frequencies where the average air spacing between a vibrating machine surface and the enclosure wall is an integral multiple of half wavelengths of sound, strong coupling will occur between the vibrating machine surface and the enclosure wall, resulting in a marked decrease in the enclosure wall Transmission Loss. Small, stiff enclosures are often designed where a lightweight enclosure is required to reduce low-frequency noise generated by a small item of equipment. In this case, the enclosure is designed to operate at frequencies below the first panel resonance frequency illustrated on the curve of Figure 7.5. The walls are made to be as stiff and light as possible and this is often achieved using a honeycomb structure with aluminium or steel skins on each side. An example of a material used for this purpose is “Nomex”, which has a Young’s modulus of 0.047 GPa and a density of 48 kg/m3 . The effective bending stiffness of a honeycomb panel with flat plates bonded on the top and bottom is given by: hs (hs + hc )2 h3 h3 + Ec c (7.31) B = Es s + Es 6 2 12 where hs and hc are the thicknesses of the cover plates and honeycomb core, respectively and Es and Ec are the Young’s moduli for the cover plates and honeycomb core, respectively. The first resonance frequency of the panel is given by Equation (7.15), where the mass per unit area is the sum of that for the cover plates and Nomex core. For a small air-tight enclosure of volume V0 , the noise reduction at frequencies below the first wall panel resonance frequency is given by Nefske and Sung (2006) as:



Cv NR = 20 log10 1 + Pn i=1 Cwi



(dB)

(7.32)

where the compliance of the gas in the enclosure and the compliance of the ith enclosure wall panel are given, respectively, as: Cv =

V0 (m5 /N) ρc2

and

Cwi =

10−3 Si3 Fi (α) (m5 /N) Bi

(7.33)

where Fi (α) (see Table 7.7) depends on both the support condition and the aspect ratio of the ith wall panel. TABLE 7.7

Simply supported Clamped

Values of Fi (α)

Panel aspect ratio 4 5 6

1

2

3

1.7

1.2

0.7

0.43

0.30

0.37

0.18

0.083

0.048

0.030

7

8

9

10

0.21

0.16

0.12

0.10

0.081

0.02

0.014

0.011

0.009

0.007

In Equation (7.32), Si is the area of the ith wall panel, Bi is the bending stiffness per unit width of the ith wall panel (Equation (7.9)). For curved panels or cylindrical enclosures, the

305

Partitions, Enclosures and Barriers

noise reduction will be much greater due to the much greater wall stiffness (see Nefske and Sung (2006)). For enclosure walls made of flat panels, as Fi (α) becomes smaller, the panels become stiffer and the enclosure noise reduction thus becomes greater. If the enclosure has any air leak at all, the performance predicted by the preceding procedure will not be achieved. A calculation procedure that includes the effects of leaks for small enclosures is outlined by Nefske and Sung (2006). Example 7.8 (a) Explain why the performance of a machinery noise enclosure should not be expressed as a single dBA rating. (b) Given the required noise reductions in Table 7.8 for an enclosure around an item of equipment, determine whether a standard, single-stud (0.6 m centres and line-line support), double-gypsum-board wall (100 mm wide cavity, each panel 13 mm thick) would be adequate. Assume that the machine and floor surfaces are hard, the enclosure is lined with sound-absorbing material, the panel loss factor is 0.02 and sound-absorbing material is placed in the enclosure wall cavity.

TABLE 7.8

Required noise reduction Desired SPL immediately outside enclosure

Data for Example 7.8

63

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000

10

15

20

25

30

35

40

20

80

83

78

73

70

60

60

60

8000

(c) If the above gypsum board wall is inadequate, what would you suggest to improve the noise reduction at the required frequencies? Give three possible alternatives, but retain the double wall and use gypsum board panels. Solution 7.8 (a) The performance of a machinery noise enclosure should not be expressed as a single number dBA rating because the dBA performance will be dependent on the spectrum shape of the noise generated by the enclosed source. (b) We may use Equation (7.29) 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 outlined in the caption of Figure 7.12, using the following input data. 2 9 3 d = 0.1 m, pm1 = m2 = 760 × 0.013 = 9.88 p kg/m , E = 2.1 × 10 , ρ = 760 kg/m . f0 = 80.4 (m1 + m2 )/(dm1 m2 ) = 80.4 (9.88 + 9.88)/(0.1 × 9.88 × 9.88) = 114 p Hz. p cL = E/[ρ(1 − ν 2 )] = 2.1 × 109 /[760(1 − 0.242 )] = 1712 m/s. fc1 = fc2 = (0.55 × 3432 )/(1712 × 0.013) = 2907 Hz. TLA = 20 log10 (9.88 × 2) + 20 log10 (113.8) − 48 = 19.0 dB.

306

Noise Control: From Concept to Application For line-line support: TLB2 = 20 log10 (9.88) + 10 log10 (0.6) + 30 log10 (2907) + 20 log10 (2) − 78 = 49.6 dB. TLC = 50.4 + 6 + 15 log10 (0.02) = 30 dB. f` = 55/0.1 = 550 Hz. The corresponding TL curve is plotted in Figure 7.14 and the octave band data are listed in Table 7.9, where octave band results are obtained by using 1/3-octave values read from the graph and calculated using Equation (7.20): TLoct = −10 log10 (1/3) [10−TL1 /10 + 10−TL2 /10 + 10−TL3 /10 ] ,





where TL1 , TL2 , and TL3 , are the (1/3)-octave band data read from the figure. 70

Transmission Loss (dB)

60

D B

50 40 30

C A

20 10 63

125

250

500

f0

fR

1k

4k

2k

0.5fc1

8k

fc2

Frequency (Hz)

FIGURE 7.14 Results for Example 7.8.

TABLE 7.9

TL (from plot) C NR Required NR

Results for Example 7.8

63

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000

14 13 1 10

21 11 10 15

34.5 9 25.5 20

40.5 7 33.5 25

46.5 5 41.5 30

40.5 4 36.5 35

37.5 3 34.5 40

8000 53 3 50 20

(c) It can be seen that the enclosure is deficient in the 63 Hz, 125 Hz, and 4000 Hz octave bands, so it is not adequate. Possibilities for improvement are listed below. (i) Use double, staggered stud wall. (ii) Use point-line or point-point support by placing rubber grommets between the panel and stud at attachment points. (iii) Use thicker panels. (iv) Fix additional 13 mm thick panels to existing panels with patches of silicone sealant. (v) Use panels of different thickness.

307

Partitions, Enclosures and Barriers

Example 7.9 Explain why it is important to design acoustic enclosures with adequate internal absorption of acoustic energy. Solution 7.9 Adequate internal absorption is necessary to prevent the build-up of reverberant energy which will compromise the predicted acoustic performance. Example 7.10 A noisy compressor measuring 2 m × 0.5 m × 0.5 m, generates a sound power level of 105 dB re 10−12 W in the 500 Hz octave band and is located on a concrete pad. The Local Authority require that sound pressure levels at the perimeter of the neighbouring residential premises (80 m from the compressor) should not exceed 38 dB re 20 µPa in the 500 Hz octave band. Design a suitable enclosure constructed with a single thickness, isotropic steel wall lined on the inside with 50 mm thick mineral wool using the following steps. (a) Calculate the required TL of the enclosure wall, assuming a maximum allowed thickness of 3 mm. (b) Select the necessary thickness of mild steel for the wall, assuming thicknesses are available ranging from 1.0 mm to 3.0 mm in 0.5 mm steps. (c) List anything else you should consider in the design. As the distance to the residential perimeter is so small, we may ignore atmospheric sound absorption and meteorological influences. We may also assume that the ground is hard with a reflection coefficient of 1.0. Solution 7.10 (a) The compressor is 80 m from the perimeter. The sound pressure level at the receiver with no enclosure (assuming hard ground, zero reflection loss) may be calculated using Equation (4.1) with DIM i = AEi = 0, so that: Lp = LW − 10 log10 (2πr2 ) = 105 − 10 log10 (2π × 802 ) = 59 dB re 20 µPa. The required sound pressure level = 38 dB re 20 µPa, so the reduction required = 21 dB at 500 Hz. Assume that the smallest panel of the enclosure is a standard size of 0.6 m × 0.6 m (stud spacing), so 500 Hz is well above the first panel resonance frequency (you can calculate this using Equation (7.16) to be sure). The critical frequency is calculated using Equation (7.4) as, fc = 0.55 × 3432 /(5378 × h) = 12/h (Hz), where h is the thickness of the mild steel enclosure walls and roof in metres. So 500 Hz is in the mass law range for all choices of panel. From equation (7.29), the required panel TL = NR + C, and for an enclosure lined on the inside, C = 7 (see Table 7.5). Thus the required enclosure TL = 21 + 7 = 28 dB at 500 Hz. (b) The wall TL may be calculated in the mass law range by using Equation (7.19) with a constant of 4 instead of 5.5 to account for octave band calculations. Thus:

" TL = 10 log10 1 +

" = 10 log10 1 +

2 #



πf m ρc



π × 500 × 7850 × h 1.206 × 343

−4

2 #

−4

(dB).

308

Noise Control: From Concept to Application

When h = 0.001 m, TL = 25.5 dB, which is too low. As we are in mass law range, a doubling of panel thickness will increase the TL by 6 dB, which is a bit high. Trying h = 0.0015 m, gives TL = 29.0 dB which is OK, so choose a wall thickness of 1.5 mm. (c) We 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. We should also check that there is enough clearance between the enclosure wall and machine to allow access, that the enclosure fits within its allocated space. Acoustic resonance effects are not important since the enclosure is lined. Example 7.11 A small personnel enclosure of nominal dimensions 2 m wide, 3 m long and 2.5 m high is to be constructed of single leaf brick 125 mm thick, plastered on both sides. The floor will be of concrete but the ceiling will be of similar construction to the walls (not bricks but plastered, similar weight, etc.). Determine the expected noise reduction (NR) for the basic hard wall design. Solution 7.11 1. As measured data are available for this construction in Tables 6.2 and 7.6 in Bies et al. (2018), we will use those data (and extrapolate to obtain approximate data at 63 Hz and 8 kHz). We could have calculated the TL values but that would not have provided values as reliable as the tabulated values. Enter the values from Bies et al. (2018) in Table 7.10. 2. Calculate, Si α ¯ i = [2(2 × 2.5 + 3 × 2.5) + 2 × 3]¯ αw + [2 × 3]¯ αf = 31¯ αw + 6α ¯f . Enter these values in Table 7.10. Also calculate 10 log10 (SE /Si α ¯ i ) and enter it in Table 7.10. 3. Calculate the external surface area excluding the floor (as the enclosure sits on the floor so its floor surface does not radiate sound externally): SE = 31 m2 . 4. Calculate noise reduction, NR, using Equation (7.14) rather than Equations (7.29) and (7.30), as α ¯ is small. Note that, in this case, the former test partition area, A, becomes the external area SE exposed to the external sound field: NR = TL − 10 log10



 P

SE . ¯i i Si α

The results are entered in Table 7.10. TABLE 7.10

63 TL (Table 7.6, B&H) α ¯ w (Table 6.2, B&H) α ¯ f (Table 6.2, B&H) Si α ¯i P (m2 ) SE ¯i i Si α 10 log10 (SE /Si α ¯i ) NR (dB)

30 0.013 0.01 0.463 67 18 12

Data and results for Example 7.11

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000 36 0.013 0.01 0.463 67 18 18

37 0.015 0.01 0.525 59 18 19

40 0.02 0.01 0.68 45.6 17 23

46 0.03 0.02 1.05 29.5 15 31

54 0.04 0.02 1.36 22.8 14 40

57 0.05 0.02 1.67 18.6 13 44

8000 59 0.06 0.03 2.04 15.2 12 47

309

Partitions, Enclosures and Barriers Example 7.12

Suppose that an opening of 0.5 m2 is required for ventilation in the enclosure wall of Example 7.11. What Transmission Loss (TL) must any muffling provide for the ventilation opening if the noise reduction (NR) of the enclosure must not be less than 15 dB in the 250 Hz octave band? Solution 7.12 1. Assume Si α ¯ i is not significantly changed by the small penetration through the wall, then from Example 7.11 at 250 Hz, 10 log10 Si α ¯ i = 10 log10 (0.525) = −2.8 and 10 log10 SE = 10 log10 (31) = 14.9 dB. 2. In Equation (7.14), replace A with SE ; then, NR = TL +10 log10 Si α ¯ i − 10 log10 [SE (1 − αi )]. As NR = 15, TL = 15 + 2.8 + 14.9 ≈ 32.7 dB. Thus, the net TL required is 32.7 dB. 3. Calculate transmission coefficient, τ , of the ventilation opening. Equation (7.27) gives, 31τ = (31 − 0.5)τw + 0.5τv . From the table of Example 7.11 and use of Equation (7.28): τw = 10−37/10 = 100.3 × 10−4 = 0.0002. Putting the above in Equation (7.27) gives: 0.5 30.5 × 0.0002 + τv TL = −10 log10 τ = −10 log10 31 31 From the above, the following is obtained:





= 32.7 dB.

30.5 0.5 × 0.0002 + τv = 10−3.27 = 0.000537. 31 31   31 30.5 Rearranging gives, τv = 0.000537 − × 0.0002 = 0.0211. 0.5 31 4. Calculate the required TL of muffling for the vent using Equation (7.28): TLv = −10 log10 τv = −10 log10 0.0211 = 17.0 dB.

7.4.1

Enclosure Leakages (Large Enclosures)

The effectiveness of an enclosure can be very much reduced by the presence of air gaps. Air gaps usually occur around removable panels or where services enter an enclosure. The effect of cracks around doors or around the base of a cover can be calculated with the help of Figure 7.15, which gives the transmission coefficient of a crack as a function of frequency and width. Note that if the crack is between one plane surface and another plane surface normal to it, the effective crack width must be doubled (because of reflection – for example, a crack under a door) before using Figure 7.15. However, note that the effective area of the crack is not doubled when overall TL values are calculated using Equation (7.27). Once the transmission coefficient, τ , has been determined for a particular frequency, the procedure embodied in Equation (7.27) is used for estimating the average value of τ for the enclosure wall or cover. An alternative, simple expression for the transmission coefficient of a long narrow slit (k`w λ/2 must be satisfied. Otherwise the barrier may be assumed to be thin. 180

f (degrees)

9.8

q f

9.4

150

9 8 7

120 2 90 1

90

3

4

6 K=5

Source Receiver

120 150 q (degrees)

180

FIGURE 7.20 Finite width barrier correction factor, K (Equation (7.52)).

Similar results are obtained for earth mounds, with the effective barrier width being the width of the top of the earth mound. Any trees planted on top of the earth mound are not considered to contribute significantly to the barrier attenuation and can be ignored. Example 7.20 A small source of low-frequency, broadband sound at 1 m above the ground introduces unwanted noise at a receiver, also 1 m above the ground at 4 m distance. The ground surface is grass, and the source may be treated as a point source. What is the effect in the 250 Hz octave band on the receiver, of a barrier centrally located, 2 m high and 6 m wide? Proceed as follows. (a) Calculate the reflection loss of the ground, assuming plane wave reflection. (b) Calculate the sound pressure level at the receiver due to each diffracted path. (c) Combine the attenuations of all paths before the barrier was inserted and again with the barrier inserted to obtain the overall attenuation due to the barrier. Solution 7.20 (a) The flow resistivity, R1 , for grass-covered ground is approximately 2 × 105 MKS rayls/m (see Table 4.2). This results in a value of flow resistance parameter,

323

Partitions, Enclosures and Barriers ρf /R1 = 1.5 × 10−3 . Referring to Figure 7.21, the angle of incidence from the horizontal for the ground reflection of the wave diffracted over the top of the barrier is arctan(3/2) which is 56◦ . For waves diffracted around the ends the calculation is more complicated. However, in this particular case, symmetry results in the reflection occurring at the bottom edge of the barrier so the reflection angle is the arctan of the source height divided by the distance of the point on the ground immediately √ under the source to the junction of the barrier edge with the ground, or arctan(1/ 32 + 22 ) = 15.5◦ . The x-axis parameter in Figure 4.3 corresponding to these angles of inp cidence (56◦ and 15.5◦ ) are, respectively, 56 2 × 105 /(1.206 × 250) = 1442 and p 15.5 2 × 105 /(1.206 × 250) = 399. These values can be used on the x-axis in Figure 4.3 to obtain the respective reflection, Arf,2 = Arf,3 = 1.2 dB and Arf,7 = Arf,8 = 4.8 dB.

Elevation S 1m 1m S

56o

26.6o(no barrier) R 2m 15.5o(3-D) 1 m

22+32 2m

2m

S

R 3m

22+32

6m Plan

FIGURE 7.21 Configuration for Example 7.20.

(b) For each diffracted path, i, i = 1, ....8, calculate the required path lengths (Ai , Bi and dSR,i , see Figure 4.13), Fresnel number, Ni (Equation (7.38)), and attenuation, Abi (Figure 7.17 and Equation (7.39)). Add the reflection loss, Arf,i , (where appropriate) to obtain the total attenuation for each path, Abi + Arf,i . (i) Non-reflected waves (three paths): Source–receiver path over the √ barrier top dSR1 = 4 m; A1 + B1 = 2 5 = 4.5 m. 2 2f 500 N1 = (A1 + B1 − dSR1 ) = (A1 + B1 − dSR1 ) = (4.5 − 4.0) = 0.7. λ c 343 From Figure 7.17, assuming an omni-directional source, ∆b1 = 12.0 dB; thus, Ab1 = 12.0 + 20 log10 [4.5/4] = 13.0 dB. Source–receiver path around the barrier ends (two paths): √ dSR5 = dSR6 = dSR1 = 4 m; A5 + B5 = A6 + B6 = 2 32 + 22 = 7.2 m ; N5 = N6 = 4.7. From Figure 7.17, ∆b5 = ∆b6 = 19.8 dB; thus, Ab5 = Ab6 = 19.8 + 20 log10 [7.2/4] = 24.9 dB. (ii) Reflected waves over the top (three paths): Image source–receiver path (reflection on the source side): √ √ √ dSR2 = 22 + 42 = 4.5 m; A2 + B2 = 22 + 32 + 22 + 12 = 5.8 m. 2 2f 500 N2 = (A2 + B2 − dSR2 ) = (A2 + B2 − dSR2 ) = (5.8 − 4.5) = 1.9. λ c 343 From Figure 7.17, assuming an omnidirectional source, ∆b2 = 15.8 dB. Using

324

Noise Control: From Concept to Application Equation (7.39): Ab2 = 15.8+20 log10 [5.8/4.5] = 18.0 dB; Arf,2 = 1.2 dB; Ab2 +Arf,2 = 19.2 dB. Source–image receiver path (reflection on the receiver side with the same attenuation as above): Ab3 + Arf,3 = 19.2 dB. Image source–image receiver path√(reflection on source and receiver sides): dSR4 = dSR1 = 4 m; A4 + B4 = 2 22 + 32 = 7.2 m; N4 = 4.7, and ∆b4 = 19.8 dB. Using Equation (7.39), Ab4 = 19.8 + 20 log10 [7.2/4] = 24.9 dB. Arf,4 = Arf,2 + Arf,3 = 2 × 1.2 = 2.4 dB; Ab4 + Arf,4 = 27.3 dB. (iii) Reflected waves around the barrier ends (two paths): Image source–receiver path: Due to symmetry, the reflection points are at the base of the barrier at each end. √ √ dSR7 = dSR8 = 42 + 22 = 4.47 m; A7 + B7 = A8 + B8 = 2 22 + 32 = 7.21 m; N7 = N8 = 4.0; ∆b7 = ∆b8 = 17.1 dB. Ab7 = Ab8 = 17.1 + 20 log10 [7.5/4.5] = 21.5 dB; Arf,7 = Arf,8 = 4.5 dB; Ab7 + Arf,7 = Ab8 + Arf,8 = 26.0 dB.

(c) As octave bands are considered, waves reflected from the ground are combined incoherently with non-reflected waves. In fact, all sound waves arriving at the receiver by different paths are combined incoherently. The overall attenuation due to the barrier is obtained using Equation (7.50). In this equation, the Abi for the eight paths with the barrier are 13 dB, 24.9 dB and 24.9 dB, 19.2 dB, 19.2 dB, 27.3 dB, 26 dB and 26 dB. The Arf,w term is the ground reflection loss in the absence of the barrier and the “1” term accounts for the direct wave in the absence of the barrier. The attenuation of the ground-reflected wave in the absence of the barrier is obtained by assuming plane wave reflection, allowing the use of Figure 4.3. From Figure 7.21, β0 = arctan[2/(2 + 2)] = 26.6◦ . Thus, in Figure 4.3, the parameter, β[R1 /(ρf )]1/2 = 685, which corresponds to an attenuation of 2.8 dB. From Equation (7.50), the overall attenuation (or noise reduction due to the barrier) is then: Ab = 10 log10 1 + 10−2.8/10





−10 log10 10−13/10 + 2 × 10−24.9/10 + 2 × 10−19.2/10 + 10−27.3/10 + 2 × 10−26/10





= 1.8 + 10.7 = 12.5 ≈ 13 dB. This is the same noise reduction that would have been obtained if all ground reflections were ignored (including the one prior to installation of the barrier), and only diffraction over the top of the barrier were considered. In many cases, where the width of the barrier is large in comparison with the height, results of acceptable accuracy are often obtained by considering only diffraction over the top of the barrier and ignoring ground-reflected paths and paths around the ends of the barrier. Note that the final result of the calculations is given to the nearest 1 dB because this is the best accuracy which can be expected in practice and the accuracy with which Figure 7.17 may be read is in accord with this observation. Example 7.21 A barrier 3 m high and 10 m wide is inserted mid-way between a noisy pump and a residence located 40 m away. With no barrier, sound pressure levels at the residence in octave bands are

325

Partitions, Enclosures and Barriers

as listed in Table 7.13. For the purposes of the following calculations assume the acoustic centre TABLE 7.13

Sound Pressure Level at Residence (dB re 20 µPa)

Data for Example 7.21

63

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000

63

67

62

55

52

50

45

8000 42

of the pump is 0.5 m above the ground and the point of interest at the residence is 1.5 m above the ground. The ground surface is grass. Assume sound waves travelling from the pump to the residence along different propagation paths combine incoherently at the residence. (a) Calculate the Noise Rating (NR) level and dBA level of the noise at the residence with no barrier in place. (b) Is the noise at the residence with no barrier in place acceptable, if the area is zoned as residences bordering industrial areas and the noise occurs 24 hours/day? Assume the noise has no detectable tonal or impulsive components. Is the noise acceptable if it only occurs during the hours of 7.00 a.m. and 6.00 p.m.? Explain your answer. (c) Would the noise at the residence with no barrier in place sound rumbly, hissy or neutral? Why? (d) List the possible sound propagation paths from the noise source to the residence without the barrier in place and then with the barrier in place. Assume no obstacles other than the barrier exist. (e) Estimate the attenuation of the ground-reflected wave with no barrier in place for the 500 Hz octave band. (f) Calculate the overall attenuation in the 500 Hz octave band due to the barrier for sound travelling from the pump to the residence. Solution 7.21 (a) The Noise Rating (NR) is obtained by plotting the un-weighted octave band data on a set of NR curves (see Figure 2.8) as illustrated in Figure 7.22. The NR value is 52.5, as this is the highest rating curve that just envelopes the data. The A-weighted sound pressure level is:



LpA = 10 log10 10(6.3−2.62) + 10(6.7−1.61) + 10(6.2−0.86) + 10(5.5−0.32) + 105.2 + 10(5+0.12) + 10(4.5+0.1) + 10(4.2−0.11)



= 59.2 (dBA re 20 µPa). (b) The acceptable sound pressure level is obtained using Table 2.21 and is Lp = 40 + 20 − 10 = 50 dBA re 20 µPa. The actual level is over 9 dBA above the allowable level, a situation which is not acceptable and according to Table 2.22, it will generate widespread complaints from the community. If the noise only occurs during the hours of 7 am and 6 pm, then the allowable level is Lp = 40 + 20 = 60 dBA re 20 µPa and the existing level is thus acceptable. (c) The octave band levels essentially follow the shape of the NR curve (except at 63 Hz, 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 groundreflected path. With a barrier, there are 8 paths as listed below:

326

Octave band sound pressure level (dB re 20 mPa)

Noise Control: From Concept to Application

70 65 60 55 50 45 40 35 30 25 20 0

5

15

10 31.5

63

125 250 500 1k 2k Octave band centre frequency (Hz)

4k

8k

FIGURE 7.22 Results for part (a) of Example 7.21.

(i) one over the top with no ground reflections; (ii) one around each end with no ground reflections (total two); (iii) one over the top with a ground reflection on the source side; (iv) one over the top with a ground reflection on the receiver side; (v) one over the top with ground reflections on both sides; and (vi) one around each end with a ground reflection (total two). (e) Attenuation of the ground-reflected wave with no barrier (see Figure 7.23) R dSR S

1.5m

0.5m

b0

d1

40 - d1

FIGURE 7.23 Geometry for part (e) of Example 7.21.

Using similar triangles:

0.5 1.5 = ; d1 40 − d1

d1 = 10 m.

0.5 = 2.862◦ . 10 From Table 4.2, R1 = 2 × 105 MKS Rayls. The air density, ρ = 1.206 kg/m3 .  1/2 ρf 1.206 × 500 R1 Thus at 500 Hz, = = 0.0030 and β0 = 52◦ . R1 2 × 105 ρf From Figure 4.3, the reflection loss on ground reflection is thus 3.8 dB. Thus, β0 = tan−1





327

Partitions, Enclosures and Barriers (f) Attenuation due to the barrier. Non-reflected waves (three paths): Over the top with no ground reflection (see Figure 7.24) p p √ dSR1 = 402 + 12 = 40.01 m; A1 + B1 = (20)2 + 2.52 + (20)2 + 1.52 = 40.21 m. B1

A1

R

dSR1

S

3m

1.5 m

0.5 m

20 m

20 m

FIGURE 7.24 Arrangement for diffraction over the top for Example 7.21(f).

2 2f 1000 (A1 + B1 − dSR1 ) = (A1 + B1 − dSR1 ) = (40.21 − 40.01) = 0.58. λ c 343 From Figure 7.17, ∆b1 = 11.4. Thus, Ab1 = 11.4 + 20 log10 [40.21/40.01] = 11.4 dB.

N1 =

Around each end of the barrier (two paths with no ground reflection): see Figure 7.25 √ dSR5 = dSR6 = 40.01 m; A5 + B5 = A6 + B6 = 2 202 + 0.52 + 52 = 41.24 m; R

Elevation dSR1

0.5 m 0.5 m

0.5 m

S

0.5 m

5m

Plan S

20 m

dSR1

A5

20 m 5m

R

B5

FIGURE 7.25 Arrangement for diffraction around each barrier end with no ground reflection for Example 7.21(f).

N5 = N6 = (1000/343) × 1.23 = 3.6. ∆b5 = ∆b6 = 18.7 dB; Ab5 = Ab6 = 18.7 + 20 log10 [41.24/40.01] = 19.0 dB. Alternatively, we can use Equation (7.44) to calculate h0b and then Equations (7.45) and (7.46) to calculate A5 and B5 . Thus: √ √ 0.5 202 + 52 + 1.5 202 + 52 0 √ √ hb = = 1 m. 202 + 52 + 202 + 52 and A5 + B5 =

p

(−0.5)2 + 202 + (−5)2 +

(−5)2 + 202 + (−0.52 ) = 41.24 m,

p

which is the same as obtained with the previous method above.

328

Noise Control: From Concept to Application Reflected waves over the top (three paths): Image source–receiver path (reflection on the source side) (see Figure 7.26)

R S 0.5 m

3m

1.5 m

b2

S d1 20 m

20 m

FIGURE 7.26 Arrangement for diffraction over the top with a source side reflection for Example 7.21(f).

d1 20 − d1 = ; d1 = 2.857 m; 0.5 3    1/2 R1 0.5 = 9.93◦ ; β2 β2 = tan−1 = 181◦ and from Figure 4.3, Arf,2 = 7.8 dB. 2.857 ρf √ Image source - receiver distance, dSR2 = 22 + 402 = 40.05 m.

Using similar triangles:

A2 + B2 = 202 + (3 + 0.5)2 + 202 + (3 − 1.5)2 = 20.304 + 20.056 = 40.36 m. 1000 2f (A2 + B2 − dSR2 ) = (40.36 − 40.05) = 0.90; ∆b = 12.8 dB. N2 = c 343 Using Equation (7.39), Ab2 = 12.8 + 20 log10 [40.36/40.05] = 12.9 dB. Arf,2 = 7.8 dB; Ab2 + Arf,2 = 20.7 dB.

p

p

Source–image receiver path (reflection on the receiver side) (see Figure 7.27) d2 20 − d2 Using similar triangles: = ; d2 = 6.667 m. 1.5 3 R S b3

0.5 m

3m 1.5 m

d2 20 m

20 m

R

FIGURE 7.27 Arrangement for diffraction over the top, with a receiver side reflection for Example 7.21(f).

1.5 β3 = tan = 12.7◦ ; β3 6.667 From Figure 4.3, Arf,3 = 6.8 dB. −1







R1 ρf

Source – image receiver distance, dSR3 = 202

√

= 231◦

22 + 402 = 40.05 m.

A3 + B3 = + (3 − + + (3 + 1.5)2 = 20.156 + 20.500 = 40.656 m. 2f 1000 N3 = (A3 + B3 − dSR3 ) = (40.66 − 40.05) = 1.77; ∆b3 = 15.3 dB. c 343 Ab3 = 15.3 + 20 log10 [40.66/40.05] = 15.4 dB; Arf,3 = 6.8 dB. Ab3 + Arf,3 = 22.2 dB.

p

0.5)2

1/2

202

p

Image source–image receiver path (reflection on both the source and receiver sides): From previous calculations (see above), dSR4 = dSR1 = 40.01; A4 + B4 = 20.304 + 20.500 = 40.804.

329

Partitions, Enclosures and Barriers 2f 1000 (A4 + B4 − dSR4 ) = (40.80 − 40.01) = 2.30; ∆b4 = 16.8 dB. c 343 Ab4 = 16.8 + 20 log10 [40.8/40.0] = 17.0 dB; Arf,4 (2 reflections) = 7.8 + 6.8 = 14.6 dB. Ab4 + Arf,4 = 31.6 dB. N4 =

Ground reflection around each end of the barrier (see Figure 7.28)

R S

b7

A7

0.5 m S

3m

B7 hb = 0.5 m

1.5 m

?m 20 m

20 m

FIGURE 7.28 Arrangement for diffraction around each barrier end with a ground reflection for Example 7.21(f).

We can assume that the point of reflection is on the source side of the barrier and thus use Equation (7.42) to calculate h00b . If the result is negative, then we know that the reflection point is on the receiver side and we should use Equation (7.43) (which is Equation (7.42) multiplied by −1). Thus: √ √ 1.5 202 + 52 − 0.5 202 + 52 00 √ √ hb = = 0.5 m, 202 + 52 + 202 + 52 which is positive, thus proving that our initial assumption of ground reflection on the source side of the barrier is correct. We can then use Equations (7.47) and (7.48) to calculate A7 and B7 . Note that as the point of reflection is on the source side of the barrier, the subscripts, S and R must be interchanged in those equations. Thus: √ √ A7 + B7 = A8 + B8 = 1 + 202 + 52 + 1 + 202 + 52 = 41.28 m. The distance, dSR7 = dSR8 , from the image source to the receiver was calculated previously and was 40.05 m. 2f 1000 Thus, N7 = N8 = (A7 + B7 − dSR7 ) = (41.28 − 40.05) = 3.58; c 343 ∆b7 = ∆b8 = 18.5 dB and Ab7 = Ab8 = 18.5 + 20 log10 [41.28/40.05] = 18.8 dB. Next we find the angle, β7 , in Figure 7.28. As h00b is equal to h S , the reflection  point 1 is midway between the source and the barrier, so β7 = arctan √ = 2.78◦ . 2 + 52 20  1/2 p R1 Thus, β7 = 2.78× 1/0.00302 = 50.6◦ and from Figure 4.3, Arf,1 = 3.9 dB. ρf Thus, Ab7 + Arf,7 = Ab8 + Arf,8 = 18.8 + 3.9 = 22.7 dB. Using Equation (7.50), the overall attenuation, Ab , due to the barrier is: Ab = 10 log10 10−0/10 + 10−3.8/10
−10 log10 10−1.14 + 2 × 10−(1.90) + 10−(2.07) + 10−(2.22) + 10−3.16 + 2 × 10−(2.27) = 1.5 + 9.1 = 10.6 dB.




Example 7.22 Approximately how high would a barrier need to be between a compressor and the property line of the owner so that an NR curve of 50 is not exceeded at the property line (see Figure

330

Noise Control: From Concept to Application

7.29)? The distance from the compressor to the property line is 50 m and the distance from the compressor to the barrier is 2 m. The compressor noise source is 1.5 m above the ground. Assume that the receiver location at the property line is also 1.5 m above the ground, that the ground in the vicinity of the ground reflection locations is grass covered and that losses due to meteorological effects and atmospheric absorption are negligible. The barrier is 10 m wide and the 63 Hz to 8000 Hz octave band sound pressure levels due to the compressor at the property line are, respectively: 68, 77, 65, 67, 63, 58, 45, and 40 dB re 20 µPa prior to installation of the barrier. B1

A1 dSR1

?? 1.5m

1.5m

2m

48m

FIGURE 7.29 Arrangement for Example 7.22.

Solution 7.22 The NR-50 octave band values may be read from Figure 2.8 and Table 7.14 may be generated. TABLE 7.14

Existing noise NR-50 Required reduction

NR-50 values and required reduction for Example 7.22

63

Octave band centre frequency (Hz) 125 250 500 1000 2000 4000

8000

68 75 0

76.5 65.5 11

40 43.5 0

65 58.5 6.5

67 53.5 13.5

63 50 13

58 47.2 10.8

45 45.2 0

Noise reduction is a function of Fresnel number which is directly proportional to frequency. The important frequency is then 125 Hz. If this is satisfied, the reduction at 500 Hz will be OK. It is left to the reader to verify this. There are a total of 8 paths (4 over top and 4 around sides) to consider to get an overall reduction of 11 dB at 125 Hz. Consider first the paths around the barrier edges (see Figure 7.30), as the attenuation around these paths is independent of barrier height. Paths around the barrier ends (see Figure 7.30) Path with no ground reflection (one at each end of the barrier) As the source and receiver are at the same height and the barrier is closer to the receiver, the ground reflection point will be on the receiver side of the barrier. As the source and receiver are the same height, h0b = 1.5 √ m. √ From Equations (7.45) and (7.46), A5 + B5 − dSR5 = 52 + 22 + 52 + 482 − 50 = 3.64 m. At 125 Hz, λ = 343/125 = 2.744 m and N5 = (2/λ) × 3.64 = 2.66. From figure 7.17, ∆b5 = 17.0 dB. The correction term given by Equation (7.39) (assuming an omnidirectional source) is 20 log10 (53.64/50) = 0.6 dB, so Ab5 = Ab6 = 17.0 + 0.6 = 17.6 dB. Path with ground reflection According to Equation (7.43),

h00b

√ √ 1.5 482 + 52 − 1.5 22 + 52 √ √ = = 1.2 m. 22 + 52 + 482 + 52

331

Partitions, Enclosures and Barriers

Around edge (Elevation)

S

A7 hb

1.5 m

2m

R

B5

A5 hb b7

?? 1.5 m

B7

?m

?m

Plan 48 m

2m S

R A5

5m A7

B5 B7

FIGURE 7.30 Paths around the barrier end for Example 7.22.

A7 =

(1.5 − 1.2)2 + 22 + 52 = 5.39 m; B7 =

p

dSR7 =

p

(1.5 + 1.2)2 + 482 + 52 = 48.34 m.

(1.5 + 1.5)2 + 502 = 50.09 m.

p

N7 = (250/343) × (5.39 + 48.34 − 50.09) = 2.65, so ∆b7 is the same as for the non-reflected ray, which is 17.0 dB. The correction term given by Equation (7.39) (assuming an omnidirectional source) is 20 log10 (53.74/50.09) = 0.6 dB, so Ab7 = Ab8 = 17.0 + 0.6 = 17.6 dB. To find the loss due to ground reflection, we need to find the angle, β7 , that the reflected ray makes with the ground, which is given by Equation (7.49) as: β7 = arcsin[(hR + h00b )/B7 ] = arcsin[(1.2 + 1.5)/48.34] = 3.2◦ . From Table 4.2, R1 = 2 × 10−5 . The air density, ρ = 1.206. ρf 1.206 × 125 At 125 Hz, = = 0.000754. R1 2 × 105



1/2

R1 Thus, β7 = 117◦ and from Figure 4.3, Arf,7 = 7.3 dB. ρf The reflected paths around the barrier ends have total attenuations of 17.6 + 7.3 = 24.9 dB. The non-reflected paths around the barrier ends have total attenuations of 17.6 dB. Without the barrier in place, the angle, β0 , that the reflected wave makes with the ground can be calculated using Equation (7.40) and is, β0 = arctan(3/50) = 3.4◦ .

1/2

R1 ρf 1.206 × 125 = = 0.000754. Thus, β0 = 124◦ . At 125 Hz, R1 2 × 105 ρf From Figure 4.3, Arf,1 = 7.4 dB. The angles, β2 and β3 made by the ground-reflected waves can be calculated using Equation (7.41), so that β2 = arctan[(1.5 + hb )/2] and β3 = arctan[(1.5 + hb )/48]. The correction term given by Equation (7.39) (assuming an omnidirectional source) is a function of which path is under consideration. It is added to the barrier attenuation, ∆b to give Ab .



332

Noise Control: From Concept to Application

The barrier overall attenuation is given by Equation (7.50) as:







Ab = 10 log10 100 + 10−7.4/10 − 10 log10 2 × 10−1.76 + 2 × 10−2.49



+ 10−Ab1 /10 + 10−(Ab2 +Arf,2 )/10 + 10−(Ab3 +Arf,3 )/10 + 10−(Ab4 +Arf,2 +Arf,3 )/10 , where, Arf,2 is the loss due to ground reflection on the source side of the barrier, Arf,3 is the loss due to ground reflection on the receiver side of the barrier, Ab1 is the barrier attenuation for the direct wave over the barrier top with no ground reflections, Ab2 is the barrier attenuation for the wave over the barrier top with a ground reflection on the source side of the barrier, Ab3 is the barrier attenuation for the wave over the barrier top with a ground reflection on the receiver side of the barrier and Ab4 is the barrier attenuation for the wave over the barrier top with a ground reflection on both sides of the barrier. The required barrier height can be found by trial and error. √ Referring to figure 7.31, the direct path lengths are, dSR1 = dSR4 = 50 m and dSR2 = dSR3 = 32 + 502 = 50.090 m. B1, B2

A1, A3

h -1.5

R

S dSR1=dSR4

1.5 m

2m

h 1.5 m

48 m

dx

A2 , A4 R

S B3 , B4

1.5 m b2

b3

2m

48 m

h 1.5 m

FIGURE 7.31 Paths over the barrier for Example 7.22.

The refracted path lengths are: A1 A2 B1 B3

= A3 = A4 = B2 = B4

= 22 + (hb − 1.5)2 ; p = 22 + (1.5 + hb )2 ; p = 482 + (hb − 1.5)2 ; and p = 482 + (1.5 + hb )2 .

p

The Fresnel Numbers at 125 Hz, corresponding to each of the 4 paths over the barrier top are: N1 N2 N3 N4

= (250/343)(A1 + B1 − dSR1 ); = (250/343)(A2 + B2 − dSR2 ); = (250/343)(A3 + B3 − dSR3 ); and = (250/343)(A4 + B4 − dSR4 ).

The correction term to be added to ∆b to obtain Ab is calculated using Figure 7.17 and is given by (see Equation (7.39)), Corr = 20 log10 [(Ai + Bi )/dSRi ], .....i = 1, 4. The results for 4 barrier heights are summarised in Table 7.15.

333

Partitions, Enclosures and Barriers

In Table 7.15, 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. All variables mentioned above are shown in Figure 7.31. It can be seen from Table 7.15 that the barrier height should be about 5.0 m. Example 7.23 If the excessive community noise emanating from your workplace existed only at residences adjacent to your property line, what would be the noise reduction obtained in the 500 Hz octave band by building a 3 m high brick wall around your facility, 10 m from the residences? Assume that the distance from your facility to the wall is 50 m, that the height of the noise sources is 2 m, that the height of the complainant is 1.5 m and that the noise reduction due to any ground reflection is 3 dB. State any assumptions you make. Solution 7.23

TABLE 7.15

Noise reductions for various barrier heights (Example 7.22)

Wall height

3.0 m

4.0 m

4.5 m

5.0

A1 (m) A2 (m) A3 (m) A4 (m) B1 (m) B2 (m) B3 (m) B4 (m) dSR1 (m) dSR2 (m) dSR3 (m) dSR4 (m) N1 N2 N3 N4 ∆b1 ∆b2 ∆b3 ∆b4 Ab1 (dB) Ab2 (dB) Ab3 (dB) Ab4 (dB) Arf,2 (dB) Arf,3 (dB) NR = Ab (dB)

2.500 4.924 2.500 4.924 48.023 48.023 48.210 48.210 50 50.09 50.09 50 0.381 2.083 0.452 2.285 10.2 16.1 10.8 16.8 10.3 16.6 10.9 17.3 0.8 7.3 8.4

3.202 5.852 3.202 5.852 48.065 48.065 48.314 48.314 50 50.09 50.09 50 0.923 2.790 1.039 3.037 12.7 17.5 13.0 17.9 12.9 18.1 13.2 18.6 0.8 6.8 10.0

3.606 6.325 3.606 6.325 48.094 48.094 48.374 48.374 50 50.09 50.09 50 1.239 3.155 1.377 3.424 14.0 18.2 14.5 18.5 14.3 18.9 14.8 19.3 0.8 6.3 10.8

4.031 6.801 4.031 6.800 48.127 48.127 48.438 48.438 50 50.09 50.09 50 1.573 3.526 1.734 3.818 15.0 18.7 15.5 18.8 15.4 19.5 15.9 19.7 0.8 6.0 11.3

334

Noise Control: From Concept to Application

The layout is illustrated in Figure 7.32. As the wall completely surrounds the factory, no sound is diffracted around its edges. A1, A3

S

2m

B1, B2

A2, A4 b2 50 m

b3

B3, B4 R 1.5 m

3m

10 m

FIGURE 7.32 Paths over the barrier for Example 7.23.

√ 2 2 Referring to Figure √ 7.32, dSR1 = dSR4 = 60 + 0.5 = 60.002 m and 2 2 dSR2 = dSR3p = 60 + 3.5 = 60.102 m. A1 = A3 = 502 + (3 − 2)2 = 50.010 m. A2 = A4 =

p 502 + (3 + 2)2 = 50.249 m. p B1 = B2 = 102 + (3 − 1.5)2 = 10.112 m. p

B3 = B4 = 102 + (3 + 1.5)2 = 10.966 m. λ = 343/500 = 0.686 m; thus, the Fresnel numbers and barrier noise reductions corresponding to each of the 4 paths are: N1 = (2/0.686)(50.010 + 10.112 − 60.002) = 0.35, ∆b1 = 10.0; Ab1 = 10.0 + 20 log10 [(50.010 + 10.112)/60.002] = 10.0 dB; N2 = (2/0.686)(50.249 + 10.112 − 60.102) = 0.76, ∆b2 = 12.2; Ab2 = 12.2 + 20 log10 [(50.249 + 10.112)/60.102] = 12.2 dB; N3 = (2/0.686)(50.010 + 10.966 − 60.102) = 2.55, ∆b3 = 17.0; Ab3 = 17.0 + 20 log10 [(50.010 + 10.966)/60.102] = 17.1 dB; N4 = (2/0.686)(50.249 + 10.966 − 60.002) = 3.54, ∆b4 = 18.5; and Ab4 = 18.5 + 20 log10 [(50.249 + 10.966)/60.002] = 18.7 dB. Noting that the ground reflection loss is 3 dB for all ground reflections and using Equation (7.50), the noise reduction due to the barrier is: Ab = 10 log10 10−0/10 + 10−3/10




− 10 log10 10−10.0/10 + 10−(12.2+3)/10 + 10−(17.1+3)/10 + 10−(18.7+3+3)/10




= 1.76 + 8.45 = 10.2 dB. 7.5.2.2

Shielding by Terrain

For outdoor sound propagation over undulating or mountainous terrain, the equivalent barrier effect due to the terrain is calculated using the geometry shown in Figure 7.33. The equivalent Fresnel number is: 2 N = ± [A + B + C − dSR ] λ

(7.53)

where the negative sign is used if there is direct line-of-sight between the source and receiver and the positive sign is used if there is no direct line-of-sight. 7.5.2.3

ISO 9613-2 Approach to Barrier Insertion Loss Calculations

This is discussed in detail in Section 4.8.4. It should also be noted that barrier attenuation is adversely affected by reflections from vertical surfaces such as building facades. In this case, the reflection must be considered as giving rise to an image source, which must be treated separately

335

Partitions, Enclosures and Barriers N>0 A

B

C dSR

Source

Receiver

dSR

Source

Receiver

B

A N f0 :

8.16

TLin = TLout − 3

(dB)

(8.125) (8.126)

Lined Plenum Attenuator

A lined plenum chamber is often used in air conditioning systems as a device to smooth fluctuations in the air flow. It may also serve as a sound attenuation device. As shown in Table 8.1, such a device has dimensions that are large compared to a wavelength. The plenum thus acts like a small room and, as such, an absorptive liner, which provides a high random incidence sound absorption coefficient, is of great benefit. In general, the liner construction does not appear to be

427

Muffling Devices Out Area, A

H

Area, A

In

q

L (a)

(b)

R

FIGURE 8.48 Lined plenum chamber: (a) physical acoustic system; (b) definition of angle, θ.

critical to the performance of a plenum, although some form of liner is essential. The geometry of such a device is shown in Figure 8.48(a) and, for later reference, Figure 8.48(b) shows the essential elements of the source field. The latter figure will be used in the following discussion. In all cases discussed in this section, the values obtained for TL will be the same as those for IL, provided that the plenum chamber is lined with sound-absorbing material and providing that the TL is greater than 5 dB.

8.16.1

Wells’ Method

There are a number of analytical models that have been developed in the past by various authors for estimating the sound attenuation performance of a plenum chamber. The oldest known model is the Wells model (Wells, 1958) and this will be discussed first of all. The acoustic power, Wo , which leaves the exit consists of two parts for the purpose of this analysis, a direct field, WD , and a reverberant field, WR . The acoustic power in the reverberant field is related to the input power Wi as: WR = Wi Aexit /R

(8.127)

R = Sα ¯ /(1 − α ¯)

(8.128)

where In the preceding equations, Aexit (m2 ) is the cross-sectional area of the plenum exit hole, R (m ) is the plenum room constant, S (m2 ) is the total wall area of the plenum and α ¯ is the mean Sabine wall absorption coefficient. Referring to Figure 8.48(b), the power flow in the direct field is (Wells, 1958): 2

WD = Wi (Aexit /2πr2 )cos θ

(8.129)

where θ is the angular direction, and r is the line of sight distance from the plenum chamber entrance to the exit. It is recommended (ASHRAE, 1987) that for an inlet opening nearer to the edge than the centre of the plenum chamber wall, the factor of 2 in the preceding equation be deleted. If this is done, the TL (see Section 8.2) of the plenum is: TL = −10 log10

Wo Aexit Aexit cosθ = −10 log10 + Wi πr2 R





(8.130)

Equation (8.130) is only valid at frequencies for which the plenum chamber dimensions are large compared to a wavelength of sound and also only for frequencies above the cut-on frequency for higher order mode propagation in the inlet duct (see Section 8.9.2.2).

428

Noise Control: From Concept to Application

If the room constant, R, is made large, then the effectiveness of the plenum may be further increased by preventing direct line of sight with the use of suitable internal baffles. When internal baffles are used, the second term in Equation (8.130), which represents the direct field contribution, should be discarded. However, a better alternative is to estimate the IL of the baffle by using the procedure outlined in Chapter 7, Section 7.5.3 and adding the result arithmetically to the TL of the plenum. The assumption implied in adding the IL of the baffle to the TL of the chamber is that the baffle does not affect the sound power generated by the upstream noise source and this is a reasonable approximation, provided the downstream duct is sufficiently long so that the amplitudes of the waves reflected back upstream from the duct exit are negligible. Equation (8.130) agrees with measurement for high frequencies and for small values of TL, but predicts values lower than the observed TL by 5 to 10 dB at low frequencies, which is attributed to neglect of reflection at the plenum entrance and exit and the modal behaviour of the sound field in the plenum.

8.16.2

ASHRAE (2015) Method

Mouratidis and Becker (2004) published a modified version of Wells’ method and showed that their version approximated their measured data more accurately. Their analysis also includes equations describing the high-frequency performance. The Mouratidis and Becker approach has been adopted by ASHRAE (2015). Their expression for the TL at frequencies above the first mode cut-on frequency, fco (see Equation (8.95)), in the inlet duct is:



Aexit Aexit + TL = b πr2 R

n

(8.131)

which is a similar form to Equation (8.130). If the inlet is closer to the centre of the wall than the corner, then a factor of 2 is included in the denominator of the first term in the preceding equation. The values of the constants, b and n, are 3.505 and −0.359, respectively. The preceding equation only applies to the case where the plenum inlet is directly in line with the outlet. When the value of θ in Figure 8.48(b) is non-zero, corrections must be added to the TL calculated using the preceding equation and these are listed in Table 8.18 as the numbers not in brackets. For frequencies below the duct cut-on frequency (see Section 8.9.2.2), Mouratidis and Becker (2004) give the following expression for estimating the plenum TL as: TL = Af S + We

(8.132)

where the coefficients, Af and We , are given in Table 8.19. Again, when the value of θ in Figure 8.48(b) is non-zero, corrections must be added to the TL calculated using the preceding equation and these are listed in Table 8.18 as the numbers in brackets. For an end in, side out plenum configuration (ASHRAE, 2015) corrections listed in Table 8.20 must be added to Equations (8.131) and (8.132) in addition to the corrections in Table 8.18. Example 8.26 A small room of length 3.32 m, width 2.82 m and height 2.95 m serves, by means of doors at opposite ends of the room, as a passage way connecting two larger rooms. Each door is 2.06 m high and 0.79 m wide. (a) If the average absorption coefficient of the small room is 0.1, how much is the sound power in at one door reduced on leaving the second door if the doors are opposite each other in direct line of sight? (b) If the average absorption could be increased to 0.5 what would be the effect?

429

Muffling Devices

TABLE 8.18 Corrections (dB) to be added to the TL calculated using Equation (8.131) or Equation (8.132) for various angles θ defined by Figure 8.48(b). The numbers not in brackets correspond to frequencies below the inlet duct cut-on frequency and the numbers in brackets correspond to frequencies above the duct cut-on frequency. The absence of numbers for some frequencies indicates that no data are available for these cases. Adapted from Mouratidis and Becker (2003)

1/3-Octave band centre frequency (Hz)

15

22.5

30

37.5

45

80 100 125 160 200 250 315 400 500 630 800 1000 1250 1600 2000 2500 3150 4000 5000

0 1 1 0 0(1) 1(2) 4(1) 2(1) 1(0) (1) (1) (1) (0) (0) (1) (1) (0) (0) (0)

−1 0 0 −1 −1(4) 2(4) 6(2) 4(2) 3(1) (2) (2) (2) (2) (1) (2) (2) (2) (2) (3)

−3 −2 −2 −2 −2(9) 3(8) 8(3) 6(3) 6(2) (3) (2) (4) (4) (1) (4) (3) (4) (5) (6)

−4 −3 −4 −3 −3(14) 5(13) 10(4) 9(4) 10(4) (5) (3) (6) (6) (2) (7) (5) (6) (8) (10)

−6 −6 −6 −4 −5(20) 7(19) 14(5) 13(6) 15(5) (7) (3) (9) (9) (3) (10) (8) (9) (12) (15)

Angle, θ (degrees)

(c) Would there be any worthwhile gain in Transmission Loss of sound power if direct line of sight between doors were prevented as by a screen? (d) If the sound pressure level in the 500 Hz octave band were 85 dB re 20 µPa in the first room what would be the level in the doorway of the second room? (For the case of α ¯ = 0.1, and no screen). Assume for this calculation that the second room is essentially anechoic. Use the Wells’ method for plenum chambers. Solution 8.26 (a) The room is 3.32 m long, 2.82 m wide and 2.95 m high and is illustrated in Figure 8.49. Doors at each end are 2.06 m high, 0.79 m wide. Can treat it like a plenum chamber. α ¯ = 0.1 and the sound power attenuation or Transmission Loss is calculated using Equation (8.130): TL = −10 log10



Aexit Aexit cos θ + . R πr2



Aexit = 2.06 × 0.79 = 1.627 m2 and r = 3.32 m. From Equation (8.128): Sα ¯ R= = 2(3.32 × 2.82 + 3.32 × 2.95 + 2.82 × 2.95) × 0.1/0.9 = 6.106 m2 . (1 − α ¯)

430

Noise Control: From Concept to Application

TABLE 8.19 Values of the coefficients in Equation (8.132). The numerical values given in the headings for We represent the thickness (25 mm to 200 mm) of sound-absorbing material between the facing and plenum wall. The material normally used is fibreglass or rockwool with an approximate density of 40 kg/m3 . Adapted from ASHRAE (2015) and Mouratidis and Becker (2003).

Af 1/3-Octave band centre frequency (Hz)

< 1.4m2

50 63 80 100 125 160 200 250 315 400 500

1.4 1.0 1.1 2.3 2.4 2.0 1.0 2.2 0.7 0.7 1.1

> 1.4m2

25 mm fabric facing

50 mm fabric facing

We 100 mm perforated facing

200 mm perforated facing

25 mm solid metal facing

0.3 0.3 0.3 0.3 0.4 0.4 0.3 0.4 0.3 0.2 0.2

1 1 2 2 2 3 4 5 6 8 9

1 2 2 2 3 4 10 9 12 13 13

0 3 3 4 6 11 16 13 14 13 12

1 7 9 12 12 11 15 12 14 14 13

0 3 7 6 4 2 3 1 2 1 0

Plenum volume

TABLE 8.20 Corrections to be added to the TL calculated using Equations (8.131) and (8.132) for plenum configurations with an inlet on one end and an outlet on one side. The numbers not in brackets correspond to frequencies below the inlet duct cut-on frequency and the numbers in brackets correspond to frequencies above the duct cut-on frequency. The absence of numbers for some frequencies indicates that no data are available. Adapted from Mouratidis and Becker (2003).

1/3-Octave band centre frequency (Hz)

Elbow effect correction

1/3-Octave band centre frequency (Hz)

Elbow effect correction

50 63 80 100 125 160 200 250 315 400 500

2 3 6 5 3 0 −2(3) −3(6) −1(3) 0(3) 0(2)

630 800 1000 1250 1600 2000 2500 3150 4000 5000

(3) (3) (2) (2) (2) (2) (2) (2) (2) (1)

Thus, TL = −10 log10



1.627 1.627 + 6.106 π × 3.322



= 5.0 dB.

0.9 0.5 × = 54.95 m2 . 0.1  0.5  1.627 1.627 = 11.2 dB. Thus, TL = −10 log10 + 54.95 π × 3.322 That is, a 6 dB improvement. (c) If the direct line of sight were prevented, then the directfield term contribution will 1.627 be zero. Thus for case (a) above, TL = −10 log10 = 5.7 dB, 6.106

(b) Increasing α ¯ to 0.5 gives, R = 6.106 ×

431

Muffling Devices 3.32 m

2.82 m

FIGURE 8.49 Arrangement for Example 8.26.

1.627 and for case (b) above, TL = −10 log10 = 15.3 dB. 54.95 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 (6.8) and (1.61), we obtain: 4 × 10−10 × 108.5 × 1.627 hp2 i Aexit = = 1.244 × 10−4 W. W = 4ρc 4 × 413.7 The sound power level is, LW = 10 log10 W + 120 = 81.0 dB re 10−12 W.





Thus, the sound power incident on the second doorway is 81.0 − 5.0 = 76.0 dB re 10−12 W. Assuming no reflection from the doorway, this corresponds to a sound pressure level in the doorway given by Example 3.20(b) (where the correction for ρc 6= 400 has been included below) or by Equations (1.57) and (1.61) as: Lp = LW − 10 log10 Aexit + 0.15 = 76.0 − 10 log10 (1.627) + 0.15 = 74.0 dB re 20 µPa.

8.17

Directivity of Exhaust Ducts

The sound radiation from the exit of a duct may be quite directional, as shown in Figure 8.50. The latter figure is based on model studies (Sutton, 1990) in the laboratory and includes ducts of round, square and rectangular cross section. The figure is applicable to circular ducts with a radius of a. It can also be used for vertical rectangular section duct exhausts if a is replaced with half of the duct cross-sectional dimension in the direction of the observer, multiplied by 4/π. Also, as k is the wavenumber of sound at the duct exit, the duct exhaust temperature and gas properties must be used to calculate the speed of sound which is used to calculate k. However, it should be noted that the directivity of a hot exhaust is very likely to be substantially different to that of an ambient temperature exhaust due to the strong temperature gradient near the hot exhaust exit. This temperature gradient causes sound rays to bend towards the ground resulting in higher values of directivity in directions at 90◦ from the axis of a vertical exhaust stack. These directivity values will also be a function of the distance of the receiver from the exhaust stack. A downwind cross-flow across the exit of a vertical stack will have a similar effect. Figure 8.50 only applies to unlined ducts. Ducts lined with sound-absorbing material radiate more directionally so that higher on-axis sound pressure levels are produced. As shown in Figure 8.50, a distinct advantage accrues from pointing an exhaust duct upwards. The figures show the importance of placing any vertical exhaust duct, such as a cooling tower, so that line of sight from the exhaust discharge to any nearby building is greater than about 30◦ .

432

Noise Control: From Concept to Application 20 q = 0E 15E 30E

Directivity Index (dB)

10

45E

0

60E 75E

-10

90E

-20

120E

150 E-

-30

-40 0.2

1.0

180 E

10

100

ka

FIGURE 8.50 Exhaust stack directivity index measured in the laboratory vs ka where a is the stack inside radius. Curves fitted to data reported by Sutton (1990) and Dewhirst (2002).

Data measured on large ducts in the field are presented in Chapter 8 of Bies et al. (2018). These data are similar to the curves shown in Figure 8.50, except for large angles from the duct axis. This is a result of sound leaking out of the duct wall in the field measurements on typical ducts. Curves calculated theoretically (Davy, 2008b,a) are very similar to those in Figure 8.50 for all angles from the duct axis. The directivity index is given by the intensity in any one direction divided by the intensity averaged over all possible directions which in this case is a sphere. Thus: DI(θ) = 10 log10 (I(θ)/Iav )

(8.133)

As Equation (8.133) is a ratio, absolute values of the above quantities are not needed. To calculate the average intensity, Iav , a radial distance of unity is assumed and the intensity, I(θ), will be evaluated at the same location for an arbitrary source strength. Thus, the normalised average intensity is given by:

Iav

1 = 4π

Z2π 0

dφ

Zπ 0

1 I(θ) sin(θ)dθ = 2

Zπ

I(θ) sin(θ)dθ

(8.134)

0

The directivity index, DI(θ), and the intensity, I(θ), also depend on the azimuthal angle for non-circular cross-section ducts. This dependence will be ignored when calculating the directivity index here. It is difficult to obtain an accurate estimate of the radiated sound power from sound pressure level measurements averaged over the plane of the duct exit as not all of the energy propagation is normal to the plane of the duct cross section at the exit. For values of ka greater than about 1.5, higher order mode propagation probably occurs (see Section 8.9.2.2), and consequently, sound energy propagation in the duct is not entirely axial. At values of ka greater than about 6, cross-mode propagation can be expected to be dominant and an effective intensity is required to describe axial sound energy propagation in the duct. In this frequency range, the following expression for duct axial sound intensity, assuming no reflected energy from the end of the duct,

433

Muffling Devices may be used: IA =

< p2 > ; 2ρc

ka > 2

(8.135)

The introduction of the 2 in the denominator of the expression for intensity for ka values greater than 6.0 (reverberant field as a result of cross-mode domination) is proposed while it is omitted for ka values less than 1.5 (plane-wave propagation dominant). Linear interpolation on a graph of attenuation in dB vs ka is recommended for ka values in between. Thus if the average sound pressure squared hp2D i (or sound pressure level, LpD ) is measured over the plane of a duct outlet and just outside of the duct, the sound power radiated may be estimated by multiplying Equation (8.135) by the cross-sectional area, S, of the duct outlet and the sound power level is then given by:

LW =

 400  ; LpD + 10 log10 S + 10 log10   ρc 

ka < 1.5

   LpD + 10 log10 S + 10 log10 400 − 3 ;

ka > 6

ρc

(8.136)

Linear interpolation is recommended to calculate values of LW corresponding to 0.5 < ka < 2 . Just inside the duct exit, the sound pressure level would be larger than outside the duct due to reflection of waves from the end of the duct, particularly at low frequencies. If the sound pressure level, Lp , is measured at some distance, r, and some angle, θ, from the duct outlet (see Figure 8.51), the sound power level, LW , radiated by the duct outlet may be calculated as: 400 LW = Lp − DIθ + 10 log10 4πr2 + 10 log10 + AE (8.137) ρc where the directivity index, DIθ , may be obtained from Figure 8.50 and the excess attenuation, AE , may be calculated as described in Chapter 4. If the ISO 9613-2 (1996) model is used, then AE already includes the 10 log10 4πr2 term so it should be removed from Equation (8.137).

q

2a or 2d

FIGURE 8.51 Exhaust stack directivity definition. The quantity, a, is the circular duct radius and 2d is the rectangular duct dimension in the direction of the observer.

If it is desired to add an exhaust stack to the duct outlet, the resulting noise reduction may be calculated from a knowledge of the Insertion Loss, ILs , of the stack due to sound propagation

434

Noise Control: From Concept to Application

through it, and both the angular direction and distance from the stack axis to the receiver location (if different from the values for the original duct outlet). The excess attenuation, AEs , must also be taken into account if it is different for propagation from the duct outlet without the stack. Thus: NR = ILs + AEs − AE + 20 log10 (rs /r) + DIθ − DIs (8.138) where the subscript, s, refers to quantities with the stack in place. The sound pressure level, Lps , at location rs with the stack in place is given by: Lps = LW − ILs + DIs − 10 log10 4πrs2 − 10 log10

400 − AEs ρc

(8.139)

where the directivity, DIs , of the exhaust stack in the direction of the receiver is obtained using Figure 8.50. Without the stack in place, the sound pressure level may be determined using Equation (8.137). Example 8.27 The sound power of noise in the 500 Hz octave band, emerging from a 1 m diameter circular exhaust stack, shown in Figure 8.52, is 135 dB re 10−12 W. Calculate the sound pressure level at a distance of 100 m in a horizontal direction from the top of the stack using the ISO 9613-2 (1996) propagation model. Assume that the attenuation of the ground-reflected wave is the same as the direct wave and that the two waves combine incoherently at the receiver. o

90

1 2m

2

R 100 m

FIGURE 8.52 Arrangement for Example 8.27.

Solution 8.27 At a distance of 50 m, the angular orientation from the stack axis of the line joining the stack to the observer is approximately 90◦ . The directivity index may be obtained using Figure 8.50. The ka value is, ka = 2πf a/c = 2π × 500 × 0.5/343 = 4.58. Thus, from Figure 8.50, DIθ = −7.5 dB. The sound pressure level at the receiver is given by Equation (8.137) (with the 10 log10 4πr2 term removed), where: • • • • • •

LW = 135 dB re 10−12 W, AE = Adiv + Aatm + Agr + Abar + Amisc (Equation (4.34)), Adiv = 10 log10 (4πr2 ) = 10 log10 (4π × 1002 ) = 51 dB, Aatm = 0.2 to 0.3 dB (Table 4.1), Agr = −3 dB (attenuation of ground-reflected wave = attenuation of direct wave), Abar = Amisc = 0.

Thus: Lp = 135 − 51 − 7.5 − 0.2 + 3 = 79 dB re 20 µPa.

435

Muffling Devices

8.18

Additional Problems

1. What is the main difference between a reactive and dissipative silencer? 2. We wish to calculate the extent of speech privacy (using Table 2.10) between two offices that share a common partition of area 10 m2 . Each office has a suspended ceiling and the partition separating the offices continues above the suspended ceiling to the floor above. There is an unlined air conditioning duct that extends a distance of 5 m above each office and penetrates the partition above the suspended ceiling. The penetration may be considered to be well sealed. The duct wall thickness is 0.6 mm and the cross-sectional dimensions are 2 m wide × 1 m high. There is one air conditioning outlet of dimensions 0.2 m × 0.2 m in each office. The configuration is illustrated in Figure 8.53. Partition, ceiling and air conditioning outlet TL data are listed in Table 8.21. In the table, the TL data for the air conditioning outlets, represent the combined reduction for sound power entering the outlet in the source room to sound power radiated from the outlet in the receiver room. Follow the calculation steps listed below the assumptions to find the speech privacy condition. What would be the best way to improve the speech privacy if it is inadequate? 5m

5m

0.6 mm thick 1.0 m 2.0 m

10 m2 Observer Source

FIGURE 8.53 Arrangement for Problem 2.

TABLE 8.21

TL data for Problem 2

Frequency (Hz)

TLwall

TLceiling

TLoutlet

Frequency (Hz)

TLwall

TLceiling

TLoutlet

100 125 160 200 250 315 400 500

24 27 31 35 39 42 44 46

2 2 2 2 2 2 2 3

15 15 15 15 16 16 17 17

630 800 1000 1250 1600 2000 2500 3150

45 44 43 44 45 47 49 51

3 3 3 3 3 3 3 3

18 19 19 20 20 20 20 20

You may make the following additional assumptions: • Sound is incident on and radiates from only the bottom of the duct – not the sides or top.

436

Noise Control: From Concept to Application • There is a substantial partition in the ceiling space to reduce sound transmission from one office to the next sufficiently so that the sound transmission through this part of the partition can be ignored. • Attenuation of the sound propagating in the duct is negligible. • The ambient noise level in each office due to non-speech noise sources is 35 dBA. (a) For each 1/3-octave band from 100 Hz to 3150 Hz, calculate the difference in sound power level incident on the outside of the air conditioning duct in the source office and the sound power level propagating in one direction inside the duct. (b) Calculate the difference in sound power level propagating in one direction inside the duct and the sound power level radiated into the receiving office. (c) For each 1/3-octave band from 100 Hz to 3150 Hz, add twice the ceiling TL to the sum of items 1 and 2 to obtain the TL for air conditioning duct transmission. (d) Combine the TL for the wall, duct and air conditioning outlets (one in each room), taking into account the relative areas associated with each, to find the overall TL in each frequency band. (e) Find the overall TL averaged over all frequency bands from 100 Hz to 3150 Hz. Tabulate all results. (f) Add the average overall TL to the ambient noise level and use this result to enter Table 2.10 to find the speech privacy condition. (g) Look at the relative values in the tabulated results to determine what would increase the speech privacy rating.

3. For a machine enclosure to be ventilated with a cooling fan, design a lined duct muffler (for the 500 Hz 1/3-octave band only) that will be mounted on the enclosure to supply air to the cooling fan, such that its length is as short as possible. The maximum allowed external cross-sectional dimensions of the duct are 400 mm × 400 mm and the maximum allowed air flow speed is 7 m/s. The required volume of airflow is 0.263 m3 /s. You may use acoustic material with no plastic liner and no internal solid partitions but with a 30% open area perforated steel facing. 4. Calculate the reduction in sound pressure level in the 1000 Hz 1/3-octave band at an observer location 1.5 m above the ground at a horizontal distance of 35 m from an industrial fan exhaust located 2 m above the ground, if an unlined 20 m long vertical exhaust stack of 0.4 m diameter is added to the exhaust and the exhaust is re-orientated from pointing directly at the observer to pointing up vertically. Ignore any excess attenuation effects except for the ground effect. However, you may assume that the ground between the exhaust and the measurement location is concrete and there is incoherent addition of the direct and ground-reflected waves at the observer. The temperature of the air exiting the exhaust stack is 20◦ C. Ignore sound radiation from the exhaust stack walls. Use the theoretical exhaust stack directivity curves calculated using the Davy model. 5. The tube shown in Figure 8.54 is terminated at the right end by a perforated plate of thickness 1.6 mm, open area 10% and hole diameter 2.0 mm. A 300 Hz, sound introduced into the left end of the tube produces a standing wave in the tube which has a standing wave ratio of 15 dB when the temperature in the tube is 20◦ C. (a) What is the normal incidence sound absorption coefficient of the perforated sheet at 300 Hz for sound incident from the left hand side?

437

Muffling Devices

2m

0.4 m

M

FIGURE 8.54 Arrangement for Problem 5.

(b) What is the effective length of the holes in the perforated sheet if the Mach number, M , of the cross flow over the perforated panel is 0.15? (c) Calculate the acoustic impedance that would be expected to be measured by measuring the distance from the inside edge of the perforated sheet, of the first sound pressure level minimum in the tube if the Mach number, M , of the cross flow over the perforated panel is 0.15 and if the acoustic resistance of each hole in the perforated panel consists only of the contribution from this flow. You may ignore the contribution to the impedance of the mass of the steel part of the perforated panel. 6. Consider the muffler system shown in Figure 8.55. 4 3 1

7 2

5

6

FIGURE 8.55 Arrangement for Problem 6.

(a) Assuming that the load impedance is included with item 7 and that the impedances of each element may be treated as lumped, draw an equivalent electrical circuit for the system illustrated. (b) Derive an expression (in terms of the impedances, Zi , i = 1, ....7, of each of the above 7 elements) for the Insertion Loss if the shaded source on the left is a constant volume-velocity source. 7. An acoustic enclosure requires a ventilation system. To exhaust the air, a dissipative muffler in the form of a duct lined on all four sides is needed. An attenuation of 9 dB is needed at 125 Hz, 15 dB at 1000 Hz and 15 dB at 2000 Hz. Calculate the length of lined square section duct needed if the maximum allowed outer cross-sectional area is 1 m2 and the minimum allowed internal cross-sectional area is 0.25 m2 . 8. A plenum chamber shaped like a rectangular parallelepiped is to be included in an air-conditioning duct that has a noise problem in the 630 Hz 1/3-octave band. The allowable space limits the cross-sectional size of the plenum chamber to 3 m× 3 m. The length is 4 m. The inlet and outlet ducts have cross-sectional dimensions of 0.5 m × 0.5 m and the inlet duct enters the chamber at one end at the centre top, while the outlet duct is attached at the other end at the centre bottom. (a) Sketch the arrangement.

438

Noise Control: From Concept to Application (b) If the mean Sabine absorption coefficient of the chamber walls is 0.2 in the 630 Hz 1/3-octave band, what would be the reduction in sound power level transmitted to the exit duct as a result of inserting the chamber, assuming that the plenum chamber does not affect the sound power output of the source and that the downstream duct is sufficiently long that the amplitudes of the waves reflected back upstream from the duct exit are negligible at the exit of the plenum chamber (use both the Wells and ASHRAE methods)? (c) What would be the increase in noise reduction if the Sabine absorption coefficient of the chamber were increased to 0.5 (use both the Wells and ASHRAE methods)? (d) How much additional noise reduction (for the configuration in part (a) above) would be obtained if a partial partition were placed in the centre of the chamber so that the line of sight between the inlet and outlet openings in the plenum chamber no longer existed (use only the Wells method)? The partition extends the full width of the chamber and only has an opening, 0.2 m high, at the bottom. (e) If the sound pressure level in the upstream duct were 90 dB in the 630 Hz octave band, what would be the sound pressure level in the downstream duct for the configuration on part (b)? Use both the Wells and ASHRAE methods.

A Properties of Materials

The properties of materials can vary considerably, especially for wood and plastic. The values listed in the table in this appendix have been obtained from a variety of sources, including Simonds and Ellis (1943); Eldridge (1974); Levy (2001); Lyman (1961); Green et al. (1999). The data vary significantly between different sources for plastics and wood and sometimes even for metals; however, the values listed in this appendix reflect those most commonly found. Where values of Poisson’s ratio were unavailable, they were calculated from data for the speed of sound in a 3-D solid using the equation at the end of this table. These data were unavailable for some plastics so for those cases, values for similar materials were used. For wood products, the value for Poisson’s ratio has been left blank where no data were available. Poisson’s ratio is difficult to report for wood as there are six different ones, depending on the direction of stress and the direction of deformation. Here, only the value corresponding to strain in the longitudinal fibre direction coupled with deformation in the radial direction is listed. The speed of sound values in column 4 of the table were calculated from the values in columns 2 and 3. Where a range of values occurred in either or both of columns 2 and 3, a median value of the speed of sound was recorded in column 4. The values in the table should be used with caution and should be considered as approximate only, as actual values may vary between samples. The values for the in situ loss factor refer to the likely value of loss factor for a panel installed in a building and represents a combination of the material internal loss factor, the support loss factor and the sound radiation loss factor. TABLE A.1

Material Air (20◦ C) Fresh water (20◦ C) Sea water (13◦ C) METALS Aluminium sheet Brass Brass (70%Zn 30%Cu) Carbon brick Carbon nanotubes

Properties of materials

Young’s modulus, E (109 N/m2 )

Density ρ (kg/m3 )

— — — 70 95 101 8.2 1000

E/ρ (m/s)

Internal–in situ a loss factor, η

Poisson’s ratio, ν

1.206 998 1025

343 1497 1530

— — —

— 0.5 0.5

2700 8500 8600 1630 1330–1400

5150 3340 3480 2240 27000

0.0001–0.01 0.001–0.01 0.001–0.01 0.001–0.01 0.001–0.01

0.35 0.35 0.35 0.07 0.06

p

Cont. on next page

439

440

Material Graphite mouldings Chromium Copper (annealed) Copper (rolled) Gold Iron Iron(white) Iron (nodular) Iron (wrought) Iron (grey (1)) Iron (grey (2)) Iron (malleable) Lead (annealed) Lead (rolled) Lead sheet Magnesium Molybdenum Monel metal Neodymium Nickel Nickel–iron alloy (Invar) Platinum Silver Steel (mild) Steel (1% carbon) Stainless steel (302) Stainless steel (316) Stainless steel (347) Stainless steel (430) Tin Titanium Tungsten (drawn) Tungsten (annealed) Tungsten carbide Zinc sheet BUILDING MATERIALS Brick Concrete (normal) Concrete (aerated) Concrete (high strength) Masonry block Cork Fibre board Cont. on next page

Noise Control: From Concept to Application Properties of materials (cont.) p a Young’s Density E/ρ Internal–in situ Poisson’s modulus, E loss ratio, ν ρ (kg/m3 ) (m/s) factor, η (109 N/m2 ) 9.0 279 128 126 79 200 180 150 195 83 117 180 16.0 16.7 13.8 44.7 280 180 390 205 143 168 82.7 207 210 200 200 198 230 54 116 360 412 534 96.5

1700 7200 8900 8930 19300 7600 7700 7600 7900 7000 7200 7200 11400 11400 11340 1740 10100 8850 7000 8900 8000 21400 10500 7850 7840 7910 7950 7900 7710 7300 4500 19300 19300 13800 7140

2300 6240 3790 3760 2020 5130 4830 4440 4970 3440 4030 5000 1180 1210 1100 5030 5260 4510 7460 4800 4230 2880 2790 5130 5170 5030 5020 5010 5460 2720 5080 4320 4620 6220 3680

0.001–0.01 0.001–0.01 0.002–0.01 0.001–0.01 0.001–0.01 0.0005–0.01 0.0005–0.01 0.0005–0.01 0.0005–0.01 0.0005–0.02 0.0005–0.03 0.0005–0.04 0.015–0.03 0.015–0.04 0.015–0.05 0.0001–0.01 0.0001–0.01 0.0001–0.02 0.0001–0.03 0.001–0.01 0.001–0.01 0.001–0.02 0.001–0.03 0.0001–0.01 0.0001–0.02 0.0001–0.01 0.0001–0.01 0.0001–0.02 0.0001–0.03 0.0001–0.01 0.0001–0.02 0.0001–0.03 0.0001–0.04 0.0001–0.05 0.0003–0.01

0.07 0.21 0.34 0.34 0.44 0.30 0.30 0.30 0.30 0.30 0.30 0.30 0.43 0.44 0.44 0.29 0.32 0.33 0.31 0.31 0.33 0.27 0.36 0.30 0.29 0.30 0.30 0.30 0.30 0.33 0.32 0.34 0.28 0.22 0.33

24 18–30 1.5–2 30 4.8 0.1 3.5–7

2000 2300 300–600 2400 900 250 480–880

3650 2800 2000 3530 2310 500 2750

0.01–0.05 0.005–0.05 0.05 0.005–0.05 0.005–0.05 0.005–0.05 0.005–0.05

0.12 0.20 0.20 0.20 0.12 0.15 0.15

441

Properties of Materials

Material

Properties of materials (cont.) p a Young’s Density E/ρ Internal–in situ Poisson’s modulus, E loss ratio, ν ρ (kg/m3 ) (m/s) factor, η (109 N/m2 )

BUILDING MATERIALS (Cont.) Gypsum board Glass Glass (Pyrex) WOOD Ash (black) Ash (white) Aspen (quaking) Balsa wood Baltic whitewood Baltic redwood Beech Birch (yellow) Cedar (white–nthn) Cedar (red–western) Compressed Hardboard composite Douglas fir Douglas fir (coastal) Douglas fir (interior) Mahogany (African) Mahogany (Honduras) Maple MDF Meranti (light red) Meranti (dark red) Oak Pine (radiata) Pine (other) Plywood (fir) Poplar Redwood (old) Redwood (second growth) Scots pine Spruce (Sitka) Spruce (Engelmann) Teak Walnut (black) Wood chipboard (floor) Wood chipboard (std) Cont. on next page

2.1 68 62

760 2500 2320

1660 5290 5170

0.006–0.05 0.0006–0.02 0.0006–0.02

0.24 0.23 0.23

11.0 12.0 8.1 3.4 10.0 10.1 11.9 13.9 5.5 7.6

450 600 380 160 400 480 640 620 320 320

4940 4470 4620 4610 5000 4590 4310 4740 4150 4870

0.04–0.05 0.04–0.05 0.04–0.05 0.001–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05

0.37 — 0.49 0.23 — — — 0.43 0.34 0.38

4.0 9.7–13.2 10.8 8.0 9.7 10.3 12.0 3.7 10.5 11.5 12.0 10.2 8.2–13.7 8.3 10.0 9.6 6.6 10.1 9.6 8.9 14.6 11.6 2.8 2.1

1000 500 450 430 420 450 600 770 340 460 630 420 350–590 600 350–500 390 340 500 400 350 550 550 700 625

2000 4800 4900 4310 4810 4780 4470 2190 5560 5000 4360 4930 4830 4540 4900 4960 4410 4490 4900 5040 5150 4590 1980 1830

0.005–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.005–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.06 0.01–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.04–0.05 0.02–0.05 0.04–0.05 0.005–0.05 0.005–0.05

— 0.29 0.29 0.29 0.30 0.31 0.43 — — — 0.35 — — — — 0.36 0.36 — 0.37 0.42 — 0.49 — —

442

Material

Noise Control: From Concept to Application Properties of materials (cont.) p a Young’s Density E/ρ Internal–in situ Poisson’s modulus, E loss ratio, ν ρ (kg/m3 ) (m/s) factor, η (109 N/m2 )

PLASTICS and OTHER Lucite 4.0 1200 1830 Plexiglass (acrylic) 3.5 1190 1710 Polycarbonate 2.3 1200 1380 Polyester (thermo) 2.3 1310 1320 Polyethylene (high density) 0.7–1.4 940–960 1030 (low density) 0.2–0.5 910–925 600 Polypropylene 1.4–2.1 905 1380 Polystyrene (moulded) 3.2 1050 1750 (expanded foam) 0.0012–0.0035 16–32 300 Polyurethane 1.6 900 1330 PVC 2.8 1400 1410 PVDF 1.5 1760 920 Nylon 6 2.4 1200 1410 Nylon 66 2.7–3 1120–1150 1590 Nylon 12 1.2–1.6 1010 1170 Rubber–neoprene 0.01–0.05 1,100–1,400 95–190 Kevlar 49 cloth 31 1330 4830 Aluminium honeycomb Cell Foil size thickness (mm) (mm) 6.4 0.05 6.4 0.08 9.5 0.05 9.5 0.13

1.31 2.24 0.76 1.86

72 96 48 101

— — — —

0.002–0.02 0.002–0.02 0.003–0.1 0.003–0.1

0.35 0.35 0.35 0.40

0.003–0.1 0.003–0.1 0.003–0.1

0.44 0.44 0.40

0.003–0.1 0.0001–0.02 0.003–0.1 0.003–0.1 0.003–0.1 0.003–0.1 0.003–0.1 0.003–0.1 0.05–0.1 0.008

0.34 0.30 0.35 0.40 0.35 0.35 0.35 0.35 0.495 —

0.0001–0.01 0.0001–0.01 0.0001–0.01 0.0001–0.01

— — — —

Loss factors of materials shown characterised by a very large range are very sensitive to specimen mounting conditions. Use the upper limit for panels used in building construction and the lower limit for panels welded together in an enclosure. a

Speed of sound for a 1-D solid, = E/ρ; for a 2-D solid (plate), cL = E/[ρ(1 − ν 2 )]; and for p a 3-D solid, cL = E(1 − ν)/[ρ(1 + ν)(1 − 2ν)]. For gases, replace E with γP , where γ is the ratio of specific heats (=1.40 for air) and P is the absolute pressure. For liquids, replace E with V (∂V /∂p)−1 , where V is the unit volume and ∂V /∂p is the compressibility. Note that Poisson’s ratio, ν, may be defined in terms of Young’s modulus, E, and the material shear modulus, G, as ν = E/(2G) − 1 and it is effectively zero for liquids and gases.

p

p

References Abramowitz, M. and Stegun, I. A. (1965). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Dover, New York, NY, USA. Andersson, P. (1981). Guidelines for film faced absorbers. Noise and Vibration Control Worldwide, 12(January–February):16–17. ANSI S1.11 (2014). Electroacoustics: Octave-band and fractional-octave-band filters: Part 1: Specifications. American National Standards Institute. ANSI S1.6 (2016). Preferred frequencies and filter band center frequencies for acoustical measurements. American National Standards Institute. ANSI/ASA S12.12 (2017). Engineering method for the determination of sound power levels of noise sources using sound intensity. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.18 (2019). Procedures for outdoor measurement of sound pressure level. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.19 (2016). Measurement of occupational noise exposure. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.2 (2019). Criteria for evaluating room noise. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.2-1995 (1995). Criteria for evaluating room noise. American National Standards Institute / Acoustical Society of America. ANSI/ASA S1.25 (2017). Specification for personal noise dosimeters. American National Standards Institute / Acoustical Society of America. ANSI/ASA S1.26 (2014). Methods for calculation of the absorption of sound by the atmosphere. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.65 (2006). Rating noise with respect to speech interference. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.9-1 (2018). Quantities and procedures for description and measurement of environmental sound – Part 1: Basic quantities and definitions. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.9-2 (2018). Quantities and procedures for description and measurement of environmental sound – Part 2: Measurement of long-term, wide-area sound. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.9-3 (2018). Quantities and procedures for description and measurement of environmental sound – Part 3: Short-term measurements with an observer present. American National Standards Institute / Acoustical Society of America. ANSI/ASA S12.9-4 (2015). Quantities and procedures for description and measurement of environmental sound – Part 4: Noise assessment and prediction of long-term community response. American National Standards Institute / Acoustical Society of America.

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Index 1/3-octave band, 44, 45, 50 centre frequency, 45 A-weighted Equivalent Continuous Level, 70 Sound Exposure, 71 Sound Exposure Level, 72 A-weighting, 68, 69 Absorption coefficient, 219, 244, 247 area-weighted average, 247 average, 247 bulk reacting, 221 combined, 247 impedance tube measurement, 216 measurement, 217, 243 measurement in room, 241 measurement, moving microphone, 220 measurement, moving microphone method, 216 measurement, two-microphone method, 233 normal incidence, 25 normal incidence measurement, 218, 219, 233 perforated sheet, 236 porous material with backing cavity, 234 porous material with limp impervious cover, 235 porous material with limp impervious liner, 235 porous material with limp material and perforated sheet cover, 241 porous material with partitioned backing cavity, 235 reverberation room, 244 rigidly backed porous material, 235 Sabine, 216 statistical, 216 two microphone method, 217 two-microphone method, 233, 234 Acoustic camera, 115 Acoustic impedance, see Impedance, acoustic, 56, 57, 348 orifice, 348

Acoustic intensity, see Sound intensity Acoustic liner perforated sheet, 236 Acoustic particle velocity, see Particle velocity Acoustic potential, 11 Acoustic potential function, see also Potential function, 11 plane wave, 12 Acoustic pressure, see Sound pressure, 11 Acoustic resistance, 351, 353 Air absorption, 263 Allowed exposure time, 79 Ambient noise specification, 88 Anechoic duct termination, 344 Anechoic room, 152 Annoyance threshold low-frequency noise, 63 Atmospheric stability, 180 Atmospheric temperature profile, 180 Attenuation coefficient tube, 218 Baffle, 137 Balanced Noise Criteria (NCB) curves, 93 Barrier, 317, 320–322, 341 diffraction, 317 directivity factors, 318 finite width, 320 Fresnel number, 317 image source, 319 indoor, 335, 341 Insertion Loss, ISO, 334 noise reduction, 54 outdoor, 318, 335 terrain, 334 thick, 322 Transmission Loss, 317 Beamforming, 115 depth of field, 115 disadvantages, 116 dynamic range, 115 frequency range, 115 spatial resolution, 115 spherical array, 115 457

458

Noise Control: From Concept to Application

spiral array, 115 Bellmouth, 416 Bending stiffness isotropic panel, 280 orthotropic panel, 294 Bending wave, see also Wave, bending, 279 speed, isotropic panel, 280 speed, orthotropic panel, 282 wavelength, 281 Break-in noise, 426, 426 Break-out noise, 346, 424, 426 Building sound radiation, 142 Bulk reacting, 221 Bulk-reacting liner, 395

A-weighted environmental, 100 community noise, 100 environmental noise, 100 speech interference, 85 speech privacy, 99 Critical frequency, 280, 281 orthotropic panel, 283 Cross-over frequency, 260, 261 Cut-on frequency circular ducts, 403 ducts, 15 rectangular ducts, 403 Cylindrical wave, 17 propagation, 17, 17

C-weighted Sound Exposure Level, 72 C-weighting, 68, 69 Characteristic impedance, see also Impedance, characteristic, 12, 57, 214, 220 Circular mufflers self-noise, 423 Close fitting enclosures, 303 CNEL, 73 Coincidence, 281 Coincidence frequency, 281 Combining level reductions, 54 Community complaints, see Noise control strategies, community complaints Community noise criteria, 100 Community Noise Equivalent Level, 73 Complex compressibility, 214 Complex density, 214 Complex propagation coefficient, 214 Compressibility complex, porous material, 214 CONCAWE propagation model, 171, 185 accuracy, 208 atmospheric absorption, 186 barrier attenuation, 188 excess attenuation, 186 geometrical spreading, 186 ground effects, 186 in-plant screening, 189 limitations, 190 meteorological effects, 179 uncertainty, 206, 208 vegetation screening, 190 Convection velocity, 11 Criteria, 86

Damping, 261 resonator, 364 Day-Night Average Sound Level, 73 Decibel addition coherent sounds, 45, 49 incoherent sounds, 50, 53 Decibel averaging, 71 Decibel subtraction incoherent sounds, 52 Decibels, 37 Density complex, porous material, 214 Diffuser, 346 Dipole source, 123, 124, 127 tuning fork, 124 Directivity, 144, 145 effect of reflecting plane surface, 145 exhaust stack, 431, 434 factor, 145 index, 170 Dissipative mufflers, 395, 412 DNL, 73 Dosimeters, 111 Double wall Transmission Loss, 296, 297 design chart, 299 limiting frequency, 297 line support, 297 mass-air-mass resonance frequency, 296 point support, 297 staggered studs, 298 Downwind refraction, 177 Duct anechoic termination, 344 break-in noise, 426, 426 breakout noise, 424, 426

Index cut-on frequency, 15, 402 end reflections, 408 higher order modes, 403 Duct bend Insertion Loss, 407 Duct noise break-in, 426, 426 break-out, 346 whistling, 419 Dynamic viscosity air, 414 EA,T , 71 Ear hearing aid, 75 Effective length, 349 Enclosure, see also Room, 303, 309, 316 close fitting, 303 Insertion Loss, 303 leakage effects, 309, 310, 313 ventilation, 314 vibration isolation, 315 Enclosures, 303 inside noise, 303 End correction, 349, 351, 363, 363 circular orifice, 349 non-circular orifice, 349 orifice, 350 perforated plate, 350, 351 unflanged tube, 350 Energy density, 27, 27 instantaneous, 27 kinetic, 27 potential, 27 time-averaged, 27 Energy reflection coefficient, 24 Environmental noise noise impact, 108 Noise Impact Index, 108 Total Weighted Population, 108 Environmental noise criteria, 100 A-weighted, 100 Environmental noise surveys, 104, 105 duration, 105 measurement locations, 105 measurement parameters, 105 Equal energy trading rule, 79 Equivalent Continuous Level, 70 A-weighted, 70 A-weighted, 8 hour, 71

459 Excess attenuation, see also Sound propagation, 170 atmospheric absorption, 171 atmospheric turbulence, 177 barrier effects, 171 geometric divergence, 170 geometric spreading, 170 ground effect, 171 meteorological effects, 171, 177 reflection from vertical surfaces, 171 source height effects, 171 vegetation effects, 171 Exhaust stack noise reduction, 433 pin noise, 423 Exhaust stack directivity, 431, 434 definition, 433 effect of sound-absorbing material, 431 model studies, 431 temperature gradient effect, 431 Expanded uncertainty, 208 Expansion chamber muffler 1-D wave analysis, 378 constant acoustic-pressure source, 376 constant volume-velocity source, 375 Insertion Loss, 375, 376, 376 lumped element analysis, 375 Transmission Loss, 377, 378 Far field, 143 FFT analysis, see also Frequency analysis Flat room diffusely reflecting boundaries, 273 Flat rooms, 252 Flow noise air conditioning system elements, 424 circular mufflers, 423 mitred bend, 420 mufflers, 419 unlined duct, 419 Flow resistance, see also Flow resistivity, 214 Flow resistivity, see also Flow resistance, 214 ground, 178, 192 Flow-generated noise, 360 Frequency, 13 cut-on, 402 Frequency analysis, 43, 50 overlap processing, 113 Frequency bands, 44

460

Noise Control: From Concept to Application octave, 44

G-weighting, 69 Geometric spreading factor, 171 line source, 171 plane source, 171 point source, 171 Ground reflection, 147, 174, 321 Hearing aid, 75 Hearing conservation program, 84 Hearing damage speech recognition, 75 Hearing damage risk, 75–77, 79 impact noise, 83 impulse noise, 83 ISO1999, 76, 77 quantification, 77 trading rules, 78, 80 Hearing level, 74 Hearing limits, 8 Hearing loss, 75 aging, 75 noise-induced, 75 Hearing threshold, 76 low-frequency, 63 Hearing threshold level, 76 Helmholtz resonator, 361 end correction, 363, 363 flow effect on performance, 366 grazing flow, 363 impedance, 361 Insertion Loss (IL), 365 Insertion Loss, constant acoustic-pressure source, 366 Insertion Loss, constant volume-velocity source, 365 optimum location, 366 quality factor, 364, 366 Transmission Loss, 367, 368 Impedance, 56, 57, 218, 219, 234, 235, 241 acoustic, 56, 57, 348, 351, 353 acoustic, measurement, 220 characteristic, 12, 29, 57, 220, 351 characteristic, porous material, 214 measurement, 220 measurement, in situ, 216 mechanical, 56 orifice with flow, 349 perforated plate, 350

porous acoustic material, 241 radiation, 56 resistance, 351, 353 slits, resistive, 353 specific acoustic, 26, 57, 355 specific acoustic normal, 220 spherical wave, characteristic, 18 volume, 353, 353 Impedance tube, 216–219 propagation loss, 217, 218 Impulse sound pressure level, 74 Incoherent plane source, 140, 141, 143 Indoor barrier, see Barrier, indoor Insertion Loss, 344 definition, 344 duct bends, 407 Helmholtz resonator, 365 muffler measurement, 344 quarter-wavelength tube, 365 side branch resonator, 365, 367 unlined duct, 407 Intensity, see Sound intensity ISO9613-2 propagation model, 171, 190 accuracy, 208 barrier attenuation, 195 double edge diffraction, 197 excess attenuation, 191 industrial site propagation, 191 limitations, 200 miscellaneous effects, 191 reflections from vertical surfaces, 199 uncertainty, 206, 208 vegetation screening, 198 Kirchhoff laws, 360 L10 , 74 L90 , 74 L10 , 111 L50 , 111 L90 , 111 LAFmax , 74 LASmax , 74 LCFmax , 74 LCSmax , 74 Lpeak , 74 LAeq,8h , 71 LAeq , 70 LAE , 72, 73 Lden , 73 Ldn , 73

Index Lep0 d , 71 Leq , 70 LEX,8h , 71 Lex , 71 LImp , 74 Level reductions combining, 54 Line source, 134, 135, 136 coherent, 134 finite length, 135 incoherent, 134 sound pressure, 135 Lined bends Insertion Loss, 407 Lined ducts, 395 Locally reactive, 176 Locally-reacting liner, 395 Logarithmic averaging, 71 Logarithmic decrement, 261 Long rooms, 252 Longitudinal wave, see Wave, longitudinal, 12 speed, plate, 280 speed, rod, 280 Loss factor, 261, 439 Loudness, 62, 63 4 times as loud, 62 band of noise, 63, 68 equal loudness curves, infrasound, 63 low-frequency, 63 perceptible change, 62 phon, 64 sone, 63, 64, 68 twice as loud, 62 Low-pass filter, 386, 395 constant acoustic-pressure source, 388 constant volume-velocity source, 388 high-frequency pass bands, 389 Insertion Loss, 386 Mach number, 236 Material properties, see Properties of materials Maximum sound pressure level A-weighted, 74 C-weighted, 74 fast, 74 slow, 74 Mean square quantities, 26, 27 Mechanical impedance, see Impedance, mechanical, 56

461 loudspeaker, 225 Meteorological attenuation ISO9613, 194 Meteorological effects, 177 Microphone condenser, 108 electret, 108 sensitivity, 109 Minimum audible level, 8 Mitred bend self-noise, 420 Modal damping, 261 Modal overlap, 260, 261 Modulus of elasticity, see Young’s modulus Modulus of rigidity, 288 Monopole source, 120, 123 Mufflers, 343, 404, 423 acoustic performance, 344 bulk-reacting liner, 395 classification, 347 cut-on frequency, 403 design charts, dissipative, 397–402 design requirements, 345 diffuser, 346 dissipative, 345, 395, 412 dissipative, circular section, 401 dissipative, expansion effect, 403 dissipative, high frequency performance, 402 dissipative, IL, 397–402 dissipative, lined 1 side, 401 dissipative, lined 4 sides, 401 dissipative, self-noise, 421 dissipative, splitter, 404 dissipative, temperature effects, 401 electrical analogies, 348, 360, 361 expansion chamber, 375, 386 expansion chamber, Insertion Loss, 375, 376 expansion chamber, Transmission Loss, 377, 378 expansion effects, 403 flow noise, 419 inlet attenuation, 403 Insertion Loss, 344 locally-reacting liner, 395 low-pass filter, 386, 395 noise reduction, 344, 345 performance, 345 physical principle, 345 plenum chamber, 426, 431 practical requirements, 346

462

Noise Control: From Concept to Application

pressure loss, 413, 419 pressure loss, circular section, 418 pressure loss, dynamic, 414, 416 pressure loss, friction, 413 pressure loss, splitters, 414, 418 pressure loss, staggered splitters, 419 protective facing, liner, 396 quarter-wavelength tube, 361 reactive, 345, 360 self-noise, 346, 419, 421–424 side branch resonator, Insertion Loss, 365, 367 side branch resonator, Transmission Loss, 367, 368 splitter, 404 splitter, self-noise, 421 Transmission Loss, 344 Multiplication complex solutions, 14

new facilities, 2, 5, 6 Noise-induced permanent threshold shift, 76 Normal impedance, 220 porous material with limp material and perforated sheet cover, 241 porous material with limp material cover, 235 porous material with non-partitioned backing cavity, 235 porous material with partitioned backing cavity, 235 Normal incidence absorption coefficient measurement, 219 Normal incidence impedance, 220 measurement, 219 NR curves, 90 NRC, 246

NC curves, 92 NCB curves, 93 Near field geometric, 143 hydrodynamic, 143 Neper, 218 Neutral atmospheric conditions, 180 Neutral axis location, 282 NIPTS, 76 Noise control, see Noise control strategies Noise Criteria (NC) curves, 92 Noise effects, 87 behavioural, 87 physiological, 87 psychological, 87 work efficiency, 87 Noise Exposure Level, 71 Noise impact, 108 Noise Impact Index, 108 Noise Rating (NR) curves, 90 Noise reduction, 344 definition, 345 Noise Reduction Coefficient, 246 Noise reductions combining, 54 Noise weighting curves, 88 comparison, 98 Noise-control strategies, 2, 4, 6, 6 existing facilities, 2, 4, 5 modification of the receiver, 4

Octave band, 44, 45, 50 centre frequency, 45 One-third octave band, see 1/3-octave band Orifice impedance, 348 Orthotropic panel, 282, 283, 287 bending stiffness, 282, 283 critical frequency, 283 Outdoor sound propagation, see also Sound propagation, 170, 174, 180 atmospheric absorption, 172, 174 barrier effects, see also Outdoor sound propagation, shielding effects, 197 barrier effects, CONCAWE, 188 CONCAWE, 185 geometric spreading, 171 ground effects, 174 ground effects, CONCAWE, 186 ground effects, ISO9613 model, 191, 194 ground effects, plane wave model, 175, 176 ground effects, simple, 175 ISO9613, 190 meteorological effects, 177, 180 meteorological effects, CONCAWE, 185 meteorological effects, shadow zone, 182 meteorological effects, sonic gradient, 179

463

Index meteorological effects, surface roughness, 180 shadow zone, 182 shielding effects, 197 source height effects, 188 vegetation effects, 198 Panel absorbers, 244, 245 absorption coefficient, 244 empirical absorption prediction, 245 Particle velocity, 7 measurement, 114 Microflown sensor, 114 Peak sound pressure level, 74 C-weighted, 74 Perforated plate impedance, 351 Period, 13 Personal exposure meter, 111 Pin noise exhaust stack, 423 Pink noise, 44 Piston source, 138 circular, 136, 140 directivity, 138 sound intensity, 137 sound power, 139 Plane source coherent, 136, 140 incoherent, 140, 143 incoherent sound power, 143 multiple walls, 142 single wall, 140, 142 Plane source, incoherent, 141 Plane wave acoustic pressure, 14 characteristic impedance, 12 harmonic, 12, 14 harmonic solution, 14 particle velocity, 14 period, 13 propagation, 12, 16 Plane wave propagation ducts, 15 tube, see Plane wave propagation, duct Plenum chamber Insertion Loss, 426, 431 TL with ASHRAE method, 428, 428 TL with Wells model, 427, 428 Point source, 120, 123

Poisson’s ratio, 439 Porous material, 215 absorption coefficient, 216, 219, 221 absorption coefficient measurement, 216 backing cavity, 234 bulk reacting, 221 characteristic impedance, 214 complex compressibility, 214 complex density, 214 limp impervious liner, 235 normal impedance, rigidly backed, 235 normal incidence absorption coefficient, 218 partitioned backing cavity, 235 perforated sheet cover, 236 rigidly backed, 235 Potential function, 11 Power, see also Sound power, see Sound power, 157, 161, 163 Power dissipation, 365 Pressure amplitude reflection coefficient, 25 complex, 25 Pressure loss, 413, 419 Pressure reflection coefficient, 355 Probe tube, 218 Propagation, see also Sound propagation Propagation coefficient, 214 complex, 214 Propagation model CONCAWE, 185 ISO9613, 190 uncertainty, 206 Properties of materials, 439 density, 439 Young’s modulus, 439 Psychological effects of noise, 87 Pulsating sphere, 120, 123 Quadrupole source, 127, 133 lateral quadrupole, 128 longitudinal quadrupole, 129 Quality factor, 261, 364, 366 mean flow effect, 367 Quarter-wavelength tube, 361, 364 flow effect on performance, 366 impedance, 361 Insertion Loss (IL), 365 Insertion Loss, constant acoustic-pressure source, 366

464

Noise Control: From Concept to Application Insertion Loss, constant volume-velocity source, 365 optimum location, 366 quality factor, 364, 366 Transmission Loss, 367, 368

Radiation field, 143 Radiation impedance, see Impedance, radiation, 56 Radiation load, 56 Radiation ratio, see Radiation efficiency Radius of curvature, 179 sound ray, 181 Ray path circular, 180 length, 180 RC curves, 93, 94 Reactive mufflers, 360, 395 Helmholtz resonator, 361 side branch resonator, 361, 374 Receiver control, see Noise control strategies Rectangular enclosure, see Room, rectangular Rectangular room, see also Room Reference sound intensity, see Sound intensity level, reference Reference sound power, see Sound power level, reference Reference sound pressure, see Sound pressure level, reference, 38 Reflection coefficient normal incidence, complex, 234 Reflection effects, 145 directivity, 145 receiver and source near reflecting plane, 147 receiver near reflecting plane, 147 sound power, 145 Refraction atmospheric, 179 Resistance, see Acoustic resistance Resonance frequency, 287 Resonator optimum location, 366 power dissipation, 365 Reverberant field, 264 reduction, 247 Reverberation control, 247 Reverberation time, 269 Fitzroy equation, 269

non-Sabine room, 270 RMS quantities, 26, 27 Room axial mode, 255 cross-over frequency, 261 direct field, 264 flat, 273 high-frequency range, 262, 274 low-frequency range, 252, 260 modal damping, 261 modal density, 260 modal overlap, 260, 261 modal response, 254 oblique mode, 256 rectangular, 254 reverberant field, 264 reverberation time, 269 Sabine absorption coefficient, 269 sound power vs sound pressure, 263 statistical absorption coefficient, 269 tangential mode, 255 Room constant, 161, 241–243, 264 measurement, 241 reference sound source measurement, 241 reverberation time measurement, 242 Room Criteria (RC) curves, 93 Room Criteria Mark II (RC) curves, 94 Rotating vector representation, 14, 21 Sabine absorption coefficient, 269 measurement, 243, 244 Sabine Room, 252, 254 modal response, 254 SEL, 72 Self-noise generation, 346, 419 air conditioning system elements, 424 circular duct, 423 commercial dissipative mufflers, 421 mitred bend, 420 splitter mufflers, 421, 422 unlined duct, 419 Shadow zone, 182 Side branch resonator, 361, 374 flow effect on performance, 366 Insertion Loss, 365, 367 optimum location, 366 Transmission Loss, 367, 368 SIL, 85 Silencers, see Mufflers Simple source, 120, 123

Index Simply supported panel, 287 Single wall Transmission Loss, see Transmission Loss, single leaf wall Software for acoustics, 7 Sonic gradient, 179 wind, 181 Sound Exposure A-weighted, 71 Sound Exposure Level A-weighted, 72 C-weighted, 72 Sound intensity, 27, 28, 31, 38, 151 active, 29, 36 definition, 28 far field, 29 harmonic, 31 instantaneous, 28, 29, 37 instantaneous, spherical wave, 30 level, 38 measurement, 114 measurement, p–p method, 114 measurement, p–u method, 114 meter, 114 plane wave, 29 reactive, 29, 30 reference, 39 single frequency, 28 sound power measurement, 151 time-averaged, 28 units, 32 Sound intensity level, 39 reference, 39 Sound level meter, 109 calibration, 110 dynamic range, 109 fast response, 109 frequency response, 110 impulse response, 110 measurement accuracy, 110 meter response, 109 slow response, 109 Sound power, 31, 37, 38, 151, 156, 157, 161, 163 diffuse field measurement, 155, 158 free field measurement, 152, 155 measurement, see also Sound power measurement, 151, 164 reflection effects, 145 Sound power level, 38 reference, 38

465 Sound power measurement absolute method, 157 diffuse field, absolute method, 157 diffuse field, substitution method, 157 field measurement, 158 field measurement, dual test surface method, 160 field measurement, reference sound source substitution method, 160 field measurement, reference sound source to obtain absorption coefficient, 159 near field measurement, 162 substitution method, 157 Sound pressure, 7 units, 37 Sound pressure level, 37, 54 reference, 38 Sound pressure level addition coherent sounds, 45, 49 incoherent sounds, 50, 51, 53 Sound propagation, see also Outdoor sound propagation atmospheric absorption, 171 barrier effects, 171 CONCAWE, 185 directivity index, 170 excess attenuation, 170, 197 geometric spreading, 171 ground effect, 171 ISO9613, 190 meteorological effects, 171, 177 reflection effects, 171 source height effects, 171 uncertainty, 209 vegetation effects, 171 Sound source localisation beamforming, 115 Sound speed gradient, 177 Sound speed profile, 180 Sound transmission class, 284 Source constant acoustic-pressure, 348 constant volume velocity, 348 volume-velocity, 365 Source control, see Noise control strategies Specific acoustic impedance, see also Impedance, specific acoustic, 57 measurement, 220 Specification, ambient noise, 88 Spectral analysis, see Frequency analysis

466 Spectral density, 43, 44 Spectrum analyser, 113 external clock input, 113 waterfall display, 113 zoom, 113 Spectrum analysis, see Frequency analysis Spectrum level, 45 Speech interference, 85 telephone communication, 86 Speech Interference Level, 85 Speech privacy criteria, 99, 100 Speed of sound, 8 gradient, 179 temperature, 180 vertical gradient, 180 Spherical divergence, see Geometrical spreading Spherical wave, 17 acoustic pressure, 17 characteristic impedance, 18 particle velocity, 17 propagation, 17, 21 Splitter mufflers, 404 entrance losses, 405 exit losses, 405 Insertion Loss, 405 self-noise, 421 Staggered stud construction, 295 Staggered studs, 298 Standards for acoustics, 6 Standing wave, see Wave, standing, 23, 26 amplitude, see Wave, standing amplitude Statistical absorption coefficient, 269 bulk reacting materials, 221 calculation from measurements, 219, 220 effect of air gap, 221 locally reacting material, 221 optimum, 221 Statistical analysers, 111 Statistical noise descriptors, 74 STC, 284 Surface mass, see Surface density Surface roughness, 180 T60, see Reverberation time Temperature gradient, 177 Temporary threshold shift, 75 Threshold of hearing, 63 Total Weighted Population, 108

Noise Control: From Concept to Application Trading rules, 78 Transmission coefficient, 283 composite, 302 crack, 309, 310, 313 slit, 309, 310, 313 Transmission Loss, 283, 288, 290, 298, 300, 301, 302, 344 R, 286 Rw , 286 combined, 302 damping effects, 298 definition, 344 double leaf wall, see also Double wall Transmission Loss, 295, 298 double leaf wall, design chart, 299 double leaf wall, resilient mounting, 295 double leaf wall, Sharp model, 296 double leaf wall, structural resonance, 296 flanking, 284 measured data, 302 measurement, 283, 284 muffler measurement, 344 porous material, 301 single leaf wall, 286 single leaf wall, design chart, 289 single leaf wall, isotropic, 286, 288 single leaf wall, orthotropic, 288 Sound Reduction Index, 286 sound transmission class, 284 STC, 284 STC prediction, 296 steel studs, 296 triple leaf wall, 301, 301 Weighted Sound Reduction Index, 286 Transmission path control, see Noise control strategies Two-microphone method, 233 Uncertainty, 206 combining, 207 expanded, 208 ISO9613 propagation, 208 standard, type A, 206 standard, type B, 207 Units sound intensity, 32 sound power, 32 sound pressure, 37, 38 sound pressure level, 37

467

Index Universal gas constant, 8 Unlined duct Insertion Loss, 407 Vector product, 29 Vertical sound speed gradient, 180 Vertical temperature gradient, 180 Volume velocity, 348 Wave, 280 addition, 21, 22 longitudinal, 12 plane standing, 23 standing, 23 standing amplitude, 23 Wave equation, 11, 11 complex solution, 14 cylindrical coordinates, 17 plane wave solution, 12 spherical coordinates, 17 Wave summation, see Wave, addition Wavelength, 12, 13

trace, 281 Wavenumber, 12, 13 complex, 214 Weighting curves A-weighting curve, 69, 70 C-weighting curve, 69, 70 comparison, 98 G-weighting curve, 69, 70 NC curves, 92 NCB curves, 93 RC curves, 93 Z-weighting curve, 69, 70 Weighting networks, 68, 69 Wells model, 427, 428 White noise, 44 Wind shear, 177 Wind shear coefficient, 180, 181 Wind speed profile, 177 Young’s modulus, 288 Z-weighting, 68, 69